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Page 1: Fluid mech

UNIT 1 FLUID STATICS - --

Structure 1.1 Introduction

0 bjectives

Properties o f ' Fluids 1.2.1 Density 1.2.2 Pressure 1.2.3 Viscosity

1.2.4 Kinematic Viscosity

1.2.5 Bulk Modulus of Elasticity 1.2.6 Vapour Pressure 1.2.7 Surface Tension and Capillarity

1.3 Pressure at a point 1.3.1 Variation of Pressure within a Static Fluid 1.3.2 Absolute and Gauge Pressures

1.4 Force on Plane Areas Immersed in a Fluid

1.5 Force on Curved Surfaces

1.6 Measurement of Pressure 1.6.1 Manometers

1.7 Summary

1.8 Answers to SAQs

1.1 INTRODUCTION

This is your first exposure to engineering fluid mechanics. In your earlier physics courses, you might perhaps have studied a few bits of what we are going to cover here but the emphasis, rigour and approach of coverage will be quite different in this course than in the physics courses. We shall, in fhis course, go through the theory of fluid mechanics only to the extent required to solve practical engineering problems. You will thus find the course quite interesting, especially because you will be yourself solving numerical problems that come straight from the fascinating field of engineering.

This unit being the first one of the course, it is very important that you learn the material contained in this unit thoroughly. It should form a solid foundation for uiiderstandiiig all subsequelit units of the fluid mechanics course. The unit begins with some basic properties of fluids. As you must be already familiar with some of the prelimiiiary properties of fluids we shall not devote much time and space for these. Rather, we will quickly recapitulate the definitions and proceed to the more important ('more important' because you would be studying them for the first time) properties like viscosity and compressibility. Where necessary we shall solve numerical problems so that the definitions will remain firmly fixed in your mind. In the subsequent sections we will discuss the interesting problems of determining the force due to fluid pressure on bodies immersed in fluids. In those sections, we will be dealing with only stationary fluids, and hence if you have gone through a course on statics, you should find it very easy to follow. We shall then go on to study some simple pressure measuring devices, called the manometers. A brief summary of what we have studied in this unit will be given at the end of the unit.

For best results, you are advised to start the unit right from the beginning even if you seem to know most of the material contained in the initial sections. This way you will get tuned to the style of coverage and then understandiiig the subsequent sections becomes easier. You will realise that the material contained in this unit is not only easy to understand but is also very interesting. So, start now, and keep going.

Objectives After studying this unit, you should be able to

* define various fluid properties

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and + determine the pressure at a point in a static fluid, + determine force due to fluid pressure on areas immersed in a static fluid, and

+ carry out manometry computations for determining pressures and pressure differences

1.2 PROPERTIES OF FLUIDS

If you are asked to spell out in one sentence what is the essential difference between the fluids and the solids, perhaps you would say Fluids flow while solids don't. Yes indeed ! That is infact the major difference between the fluids and the solids. Fluids flow under the action of shear (tangential) forces. That is, they do not resist the shear stresses as solids do. To understand this better, imagine a small layer of a liquid on a horizontal flat plate. As long as the plate is horizontal, there is no tangential ,force acting on the liquid and therefore it remains stationary. If you slightly tilt the plate to an inclined position, the liquid starts flowing. The tangential force responsible for the flow in this case is the component of its own weight. If the same experiment is camed out with a solid block instead of the liquid, you may observe that after sufficiently increasing the inclination of the plate the solid block moves as a whole. However, it does not flow - that is, there is no deformity of the block itself.

This infact leads s s to the definition of a fluid : That which deforms continuously under the action of a shear stress is a fluid. Both liquids and gases therefore come under the category of a fluid.

You would perhaps be well conversant with atleast some of the fluid properties such as specific gravity, specific volume etc. We will therefore quickly run through the definitions of these and discuss at some length the viscosity, surface tension and compressibility which are perhaps new to you.

1.2.1 Density The density p (pronounced 'rho') of a fluid is defined as its mass per unit volume.

m that is, p - ; where m is the mass of the fluid when it occupies the volume v. The S I unit of density is kg/m3. Water has a density of 1000 kglm3 at 4OC. The specific volume v, is simply the reciprocal of the density. That is, the specific volume of a fluid is defined as the volume occupied by a unit mass of the fluid.

The specific weight y (pronounced as 'gamma'), also called the unit gravity force is the force due to gravity on (that is, the weight of) a unit volume of fluid.

Since it depends on the acceleration due to gravity, the specific weight chan es from 8 place to place. The specific weight of water at 5OC at sea level is 9806 Nlm . The relative density S, (also called the specific gravity), of a fluid is defined as the ratio of its density to that of water. Water, thus has a specific gravity of 1. If the specific gravity of a fluid is known, say, Sf, its specific weight yfcan be readily calculated as, yf= Sf y,,,, where y, is the specific weight of water. For example, mercury, with a specific gravity of 13.6 has a specific weight of

1.2.2 Pressure A fluid exerts a force normal to a solid boundary or any plane drawn through the fluid. Since many problems may involve bodies of fluid of indefinite extent and, in many cases, the magnitude of the force exerted on a small area of the boundary or plane may vary from place to place, it is convenient to work in terms of the pressurep of the fluid, defined as the force exerted per unit area. If the force exerted on each unit area of a boundary is the same, the pressure is said to be uniform.

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:That is,

Force exerted Pessure - F

or p P - Area of boundary A

If, as is more-s~mmonly the case, the pressure changes from point to point, we consider th~e lemen t of force 6F normal to a small area surrounding the point under consideration. Then we wrlte,

In the limit, as 6.4 -0

pressure at a point,

In SI units, the unit for pressure is ~ / m ' , and is named Pascal (denoted as Pa).

1.2.3 Viscosity Suppose you pour two fluids, water and engine oil on an inclined surface. What do you observe? Water flows faster (or, more easily) than the oil. There seems to be some reluctance for the oil to flow; in other words, the oil offers more resistance to flow than water. The property by virtue of which a fluid offers resistance to flow is called the viscosity. It is the most important property of fluids. Molasses and tar are examples of highly viscous fluids whereas water and air have very small viscosities. We shall now get an expression for the viscosity. Suppose you have a layer of fluid of thickness d on a solid horizontal surface of cross section area A (see figure 1.1). Since there is no tangential force, the fluid is not flowing.

Q F V 0

U Free surface - i

d

I -

5 I I /

I I / v I - ----Cr

x Figure 1.1

Now let us apply a tangential (or shear) force F. Because fluids deform continuously under the action of shear forces, the fluid starts flowing. Experiments show that the fluid immediately in contact with the solid boundary has the same velocity as the boundary. That is, a small film of fluid remains sticking to the boundary. We say there is no slip at the boundary. The velocity keeps increasing towards the surface, where it will be maximum. In figure 1.1, U is the velocity at the surface. Experiments also show that the tangential force F required to produce the velocity U is directly proportional to the cross section area A and the velocity U and is inversely proportional to the thickness of fluid d.

A U That is, F - d (1.1)

Introducing a constant of proportionality and denoting it by 1 (pronounced as 'myu'), we get

A U F - 1 - d (1.2)

Let us see what the constant of proportionality 1 implies. For the same force F, area

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Fluid Statics and Kinematics

of cross section A and thickness of fluid d, a higher p will imply a lower velocity U. That is, the resistance to flow increases as p increases. The constant p is therefore used as a measure of resistance to flow and is variously called as dynamic viscosity, absolute viscosity, coefficient of viscosity or simply viscosity. Rearranging equation (1.2),

The left band side being tangential force per unit area denotes the shear stress, t. The term U/d, being the change in velocity per unit depth of the fluid, is more generally expressed as duldy. Thus, equation. (1.3) is written as

Equation (1.4) is the Newton's law of viscosity. In many theoretical developments, we make an important assumption that the viscosity is zero. When this is so, there will be no shear stresses, regardless of the motion of the fluid.

An interesting behaviour of viscosity in gases and liquids (both of which, you must not forget, are fluids), is that the viscosity of a gas increases with temperature whereas that of a liquid decreases with temperature. It is however independent of pressure, for ordinary pressures. For very great pressures, both gases and liquids show an erratic variation of viscosity with pressure.

To get the units of viscosity, look at equation (1.4). The viscosity cl can be expressed as

Inserting dimensions F, L, T for force, length and time,

we get the dimensions of p as F L - ~ T, The SI unit of viscosity is the Pascal-second (denoted as Pa-s).

Example 1 :

A flat plate of area 1.5 m2 i s w l e d with a velocity of 0.4 mfsec, as shown in figure 1.2, relative to another plate located at a distance of 0.15 mm from it. Find the force required to maintain this velocity, if the viscosity of the fluid separating the plates is 0.1 Pa-s.

Oil of viscosity 0.1 P a s

u = 0.4 m/sec 0.15 nun

Q

F ---------------------------,----- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - I L L - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

I T Figure 1.2

Solution :

The upper plate is moving whereas the lower one is stationary. A velocity gradient is therefore set up in the fluid between the two plates with the velocity being zero immediately adjacent to the lower plate and equal to 0.4 mfsec at the upper plate.

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From Newton's law of viscosity,

where du is the change in velocity over the distance dy.

This is the shear stress acting over the cross section area of 1.5 m2 of the fluid in contact with the upper plate.

Force required to maintain the velocity = z x Area

1.2.4 Kinematic Viscosity The ratio of absolute viscosity p to the density is called the kinematic viscosity, and is denoted by v (pronounced as 'nyu')

You may verify that the dimensions of v are L?' . In fact, it is called the kinematic viscosity because the force term F is absent in it.

1.2.5 Bulk Modulus of Elasticity The bulk modulus of elasticity is a measure of the compressibility of a fluid. It describes how well a fluid can be compressed. Suppose you have a test-tube fitted with an air- tight piston. When you push the piston down, you apply a pressure on the air within; The volume of air decreases as a result. You could achieve this with a reasonable ease. If, however, you carry out the experiment with water, you will see that even to achieve a very small decrease in volume, considerable effort ('pressure') will be involved. That is to say, air can be more easily compressed than water. This is in general true for the gases and liquids. The gases are compressible whereas for most purposes the liquids may be treated as incompressible. An important implication of this is that the density of a gas changes with pressure whereas for most liquids the density remains unchanged with pressure. You must remember however that this is true only for gradual and small changes of pressure. If the change in pressure is sudden or very large then the compressibility of the liquid also becomes significant.

The compressibility of a perfect gas is expressed by the perfect-gas law which you must have studied in one of your chemistry courses. For the liquids, the compressibility is expressed by the bulk modulus of elasticity, K. If the pressure of a volume v of a liquid is increased by dp, its volume will decrease by dv. The bulk modulus of elasticity is defined as

The negative sign indicates that as the pressure increases the volume decreases. Since the denominator, dvlv is dimensionless the bulk modulus of elasticity K has the dimension of pressure. The bulk modulus of elasticity of water at 20°C is 2.2 GPa. (G stands for 'gega', 102. This means that if we apply a pressure of one atmosphere (about 0.1 MPa, M = 10 ) to a volume of 1 cubic meter of water, the change in volume is given by

WINDOWS
Comment on Text
dec. 2007
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Fluid Statics m d Kinematics

which is about 45 cm3. The negative sign only indicates that there is a decrease in volume.

Example 2 : What is the bulk modulus of elasticit of a liquid which is compressed in a 2 cylinder from a volume of 0.0125 m at 80 N/cm2 to a volume of 0.0124 m3 at 150 N/cm2 ?

Solution : Initial volume , V = 0.0125 m3

Final volume = 0.0124 in3

Decrease in volume, - dv = 0.0001 m3

Increase in pressure, dp = 150 - 80 = 70 N/cm2

Bulk Modulus, K - - dv/V

1.2.6 Vapour Pressure At a liquid-air interface, a continuous exchange of molecules takes place. The liquids evaporate because the liquid molecules escape from the surface into a gaseous form, called the vapour. These vapour molecules exert a partial pressure in the space,known as the vapour pressure. When the pressure above a liquid equals the vapour pressure, boiling occurs. It may be interesting for you to know that by reducing the pressure above a water surface to around 2.5 kPa (this being the vapour pressure of water), you may bring about boiling of wa tx even at room temperature.

This property of liquids, in fact, may cause an undesirable effect called the cavitation. In many liquid flow situations, it is possible that very low pressures are produced at certain locations. For example, in figure 1.3 the pressure at point A may drop to a very low value and may equal to the vapour pressure of water. When this occurs,

Low Pressure Cavities

High Pressure

water starts boiling at point A. Thus a rapidly expanding vapour pocket (cavity) is formed. Because of the flow, this pocket of vapour is usually swept away from the

' point A where it originated to a point B of high pressure. Because of the high pressure at B, the cavity collapses (or, the bubbles burst). This continuous growth and decay of the bubbles affect the operating performance of pumps and turbines and may result in erosion of metal parts in the region of cavitation.

1.2.7' Surface Tension and Capillarity The attraction between molecules forms an imaginary film capable of resisting tension at the. interface-of a liquid and a gas or of two immiscible fluids. If you place a small

i-

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F i 1.4

needle on a quiet water surface, you will observe that it is supported there by the film. It is as if this film is held stretched all over the surface. The stretching force required to form this film is called the surface tension. The surface tension is expressed as force per unit length. It is usually denoted by a ('sigma'). The S1 unit of surface tension, thus, is Nlm.

Surface tension effects are generally negligible in most engineering applications. However, an important effect of surface tension, called the capillarity, may need due consideration in many practical problems. If you insert a tube of a small cross section in water you will see that the water level rises in the tube. What you may perhaps not know is that if you insert the tube in mercury, the level goes down in the tube. This phenomenon of rise or fall in the level of a liquid in a tube is called the capillarity. It is effected by surface tension. figure 1.4 shows the capillary rise in a tube. The surface tension is acting all over the surface. From equilibrium considerations, the upward force in the liquid column in the tube must be equal to the downward force at point 0. The downward force is due to gravity, given by z? h y, whereas the upward force is due to the vertical component of the surfac C tens on given by (25c ra ) cos 0. 0 is called the wetting angle and it depends on both the liquid and the material of the tube. For glass and water, 0 is nearly zero and for glass and mercury it is 130". Thus,

2 a cos 8 4 a cos 0 h = 5

y'r Y d

where d is the diameter of the tube.

Example 3 : Calculate the capillary effect in mm. in a glass tube of 4 mm diameter when immersed in a container of (i) water and (ii) mercury. Surface tension of water and mercury are 0.0735 Nlm and 0.510 Nlm respectively. The wetting angle (angle of contact) for water is 0" and that for mercury is 130". Specific gravity of mercury = 13.6 , specific weight of water = 9806 ~ / m ~ .

Solution :

(i) Water :

4 a cos 0 h -

y d

Fluid Ststia

- - 4 x 0.0735 x cos 0"

9806 x 4 x 10-

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Fluid Statics and Kioernatlcs

(ii) Mercury :

The negative sign indicates that the level in the tube is lower than that in the container.

SAQ 1 In the Example 1 (figure 1.2 ) with a different fluid placed between the plates requires a force of 600 Newtons to produce the same velocity (0.4 mls), determine the viscosity of the fluid.

SAQ 2 Determine the increase in pressure required to reduce the volume of a fluid by 1 percent if its bulk modulus of elasticity is 2.1 x lo9 ~ l m ' .

SAQ 3 A glass tube of 2 mm diameter when immersed in a water shows a capillary rise of 13 mm. Find the surface tension.

1.3 PRESSURE AT A POINT

We will now see that the pressure at a point in a static fluid is infact a scalar quantity. That is, the pressure at any point in a fluid at rest is the same in all directions.

This can be readily proved with reference to figure 1.5. Consider a very small, wedge shaped element of fluid at rest whose thickness perpendidular to the plane of the paper is constant and equal to dy. Let p be the average pressure in any direction in the plane of the paper and p, and p, be the average pressures in the horizontal and vertical directions.

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, Gravity Force 1

W - y x y x d r d y d r J I P ~ ~ ~ Y

Figure 1.5

The forces acting on the element of fluid, with the exception of those in the y direction on the two faces parallel to the plane of the paper are shown in the figure. For our purpose, forces in the y direction need not be considered. Since the fluid is at rest, there are no tangential forces. As this is a condition of equilibrium, the sum of the force components in any direction must be equal to zero. Writing such an equation for the components in the x direction,

p dl dy cos a -pxdy dz = 0.

Since dz = dl cos a , it follows that p=pX Similarly, summing up forces in the z 1

direction gives p, dr dy - p dl dy sin a - - y ak dy dz = 0. Since the element is very 2

small, the third term is of a higher order compared to the other and therefore may be neglected. From this, it follows that p = p,, as dr = dl sin a. Can you now show that p = p, ? As you observed, the results are independent of a. Hence the pressure at any point in a fluid at rest is the same in all directions. This is known as Pascal's Law.

13.1 Variation of Pressure Within a Static Fluid Consider the differential element of a static fluid as shown in figure 1.6. Since the element is very small, we can assume that the density of the fluid within the element is constant. Assume the pressure at the center of the element isp and that the dimensions of the element are €a, 6y and 62. What are the forces acting on the element in the vertical direction ? They are, the body force, the action of gravity on the mass within the element, and the surface forces transmitted from the surrounding

Fluid Statics

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Fldd Static8 and Kinematics

fluid and acting at right angles (why?) against the top, bottom and sides of the element. Since the fluid is at rest, the element is in equilibrium and the summation of forces acting on the element in any direction must be zero. To satisfy 2 F, = 0 and 2 F,, = 0 the pressure on the opposite vertical faces (ABCD etc.) must be equal.

F, = 0 leads to

from which * - 0 ax

Similarly for the y direction we get

That is, the pressure remains the same on any horizontal plane.

Summing up forces in the vertical direction and setting equal to zero,

This results in ap / az = - y, which, since p is independent of x and y can be written as

This is the general expression that relates variation of pressure in a static fluid with vertical position. The minus sign indicates that as z gets larger (increasing elevation), the pressure gets smaller.

To evaluate pressure variation in a fluid at rest one must integrate equation (1.8), between appropriately chosen limits. For incompressible fluids ( p = constant), equation (1.8) can be integrated directly. For compressible fluids, however, p must be expressed algebraically as a function of r or p.

For the cases of liquids at rest, it is convenient to measure distances vertically downwards from the free liquid surface. If h is the distance below the free liquid surface and if the pressure of air and vapour on the surface is arbitrarily taken to be zero, we can write, from equation (1.8).

For example the pressure at a distance 2m below the free surface of water is 9806 x 2 = 19,612 ~ / m ' .

As there must always be some pressure on the surface of the liquid, the total pressure at any depth h is given by equation (1.9) plus the pressure on the surface. From equation (1.9), you must note that all points in a continuous body of fluid (of constant density) at rest are under the same pressure if they are at the same depth below the liquid surface.

From equation (1.9), the pressure p may be expressed in the height of a column of the fluid by the relation,

In SI units, p may be expressed in kilo Newtons per square meter in which case if y is expressed in kilo Newtons per cubic meter, h will be in meters. When pressure is expressed in this fashion, it is commonly referred to as pressure head.

1.3.2 Absolute and Gauge Pressures If pressure is measured relative to absolute zero, it is called absolute pressure; when measured relative to atmospheric pressure, it is called gauge pressure. If the pressure

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is below that of the atmosphere, it is designated as a vacuum and its gauge value is the amount by which it is below that of the atmosphere. For example, what is called a high vacuum is really a low absolute pressure. A perfect vacuum would correspond to absolute zero pressure. All values of absolute pressure are positive. The atmospheric pressure is also called the barometric pressure and varies with elevation above sea level.

1.4 FORCE ON PLANE AREAS IMMERSED IN A FLUID

Suppose a rectangular plane area ABCD is immersed horizontally in a fluid, (such as water) at a depth h below the free surface. In figure 1.7(a), this situation is shown in section. You are able to see only the edge AB as all other edges (see figure 1.7(b)) AD, DC and CB are hidden behind AB, because the area is lying horizontally in the fluid. Now, suppose the pressure at A is p. What will be the pressure at B? It will still be p because the point B also lies at the same depth (viz., h) below the free surface. In fact, not only A and B, but all points of the rectangular area ABCD lie at the same depth h below the free surface and therefore the pressure at any point on the

v Free Surface D

5 1 A B

'1 Figure 1.7(a) Figure 1.7(b)

rectangular area ABCD is p. Because the pressure is uniform, the total force on the top side of the area will simply be , F = p x A , where A is the area of the rectangle ABCD. To get its point of action, we proceed from the first principles. Consider an elemental strip of width du at a distance x from the edge AD. The force on this eleinentary area will be dF = phdu, with hdw being the area of the elemental strip. The moment of this force about the point A (you may choose any convenient point for taking moments) is, (dF)x. The total moment produced by forces on all such elementary areas about the point A should be equal to the moment produced by the total (or, the resultant) force about the same point, A. Assuming that this resultant force F acts at a distance x, from the point A, we obtain,

The right hand side of equation (1.1 1) is the sum of moments due to forces on elementary areas. Thus,

b so p h x h xp =

p x b x h

With a similar procedure you will see that y, = hl2, where yp is the distance from point A along AD of the point of action of F. That is, the point of action of the resultant force F due to fluid pressure on a horizontal surface coincides with the centroid of the area. This is true of any shape, as long as it is a plane area and is placed horizontally in the fluid.

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Fluid Statics and Kinematics

Suppose now, that the plane ABCD is placed inclined to the horizontal, as shown in figure 1.8(a). Focus your attention only on the plane area ABCD, the trace AB of which is seen in figure 1.8(a). The plane ABCD is placed normal to the plane of the paper so that the edges BC and AD are hidden behind the edge AB. The plane itself is inclined at an angle 0 to the free surface of the fluid as shown. We shall take the origin of the co-ordinate system to be at the point 0, which is the point of intersection of the line BA (extended) and the free surface. The y axis is parallel to the trace AB (You must note that they axis need not coincide with the edge AB). The x-axis is perpendicular to the y-axis (see figure. 1.8(b)) and runs into the paper in figure 1.8(a), so that in this particular case the edges AD and BC are parallel to the x-axis. Remember that we will adhere to this co-ordinate system irrespective of the shape of the surface being considered. So, in many cases, you may not have any edge (or side) of the surface parallel to any of the axes.

With the axes thus fixed, we will define the following distances. Let G be the centroid of the area with co-ordinates (5 7) and let h be the depth of the centroid below the free surface.

Our interest now is to determine the resultant fluid force on the inclined plane surface ABCD and its point of action. You must have noticed that there is a major difference in the pressure distribution on the rectangular area when it is inclined compared to when the surface is held horizontally. Yes, the difference is that the pressure is no longer uniform on the surface because every point along the edge AB is at a different depth below the free surface and therefore the pressure changes continuously as you move from point to point along AB. However, the pressure is the same along any line parallel to the edge AD or BC because the depth below the free surface of all points on such a line will be the same. We will use this fact in determining the total force and its point of action.

Consider an elemental strip parallel to the x-axis. Let the width of this strip be dy, assumed to be infinitesimally small. This strip (actually, the centre of the strip) is located at a distance y from the origin, along the y-axis. The centre of the trace of this strip is denoted as M in figure. 1.8 (a). The point M (and therefore any point in the elementary strip PQ) is located at a depth h below the free surface. The pressure at any point on this strip will be yh, where y is the specific weight of the fluid. Since the pressure is uniformly distributed on the strip PQ, the force on the strip would be y h d ~ , where d.4 is the area of the strip.

Thus,

Although for this particular case we will be able to express dA in tenns of dy and the width of the rectangle (AD or BC), let us not do it (for the purpose of obtaining a general expression). The force dF will be normal to the line AB (Recall that pressure forces are always normal to'the surface). The system formed by all such forces on elemental strips will thus constitute a system of parallel forces, and the resultant force would be simply the addition of all individual forces. That is,

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From figure. 1.8(a), you will readily see that h = y sin 6

F = $ y y s i n e a A

From the definition of centroid, you know that (Consult your solid geometry text book)

A T = $ y d A A

Equation (1 .l2) may thus be written as

F = y s in0 ( A y )

where h = ys in 0 is the depth of the centroid G below the free surface. y h is the pressure intensity at the centroid G. Denoting this by p,, we can rewrite equation. 1.12(a) as,

I That is, the magnitude of the resultant force due to fluid pressure on a plane surface is given simply by the product of the pressure at the centroid and the area of the surface. Now, that is interesting, is n't it? Also this result is valid irrespective of the inclination, and the shape of the surface. We never used the fact that the surface is rectangular, did we? Equation (1.13) implies that we may rotate the surface about its centroid without changing the total fluid force on it. Look at figure 1.9

Three positions of a plane surface are shown with the centroid fixed at C. That is, we have rotated the plane surface about its centroid. For each of the position, the magnitude of the resultant force is the same, because the depth of the centroid G below the free surface has not changed. A cautionary remark is necessary here.

Equation (1.13) only says that the magnitude of the resultant force is given by the product of the pressure intensity at the centroid and the area. It should not be mistaken that the resultant force itself acts at the centroid. No. The point of action of the resultant force F is, in general, quite different from the centroid. Tha must be clearly understood and remembered.

Having determined the magnitude of the resultant force due to fluid pressure, the next obvious question we need to answer is : where does this resultant force act ? The

Fluid Statics

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Fluid Statics and point of application of this resultant fluid force is called the centre of pressure. We Kinematics shall now derive general expressions for the x and y co- ordinates of the centre of

pressure.

In figure l.lO(a), a body of arbitrary shape is shown immersed in a fluid. The trace AB of the body is seen in the figure. We are interested in getting the co-ordinates x p , yp of the centre of pressure cp. To do this, we shall consider an elementary area dA on the surface, as shown in figure l.lO(b).

Let the fluid force on this elementary area be dF, and the total (resultant) fluid force on the surface be F. By definition, the moment produced by F about any point or axis must be equal to the sum of moments produced by individual forces (forces on individual elementary areas, such as dA). If p is the pressure at the centre of the

v Free Surface

Figure 1.10 (a)

Free Surface / Figure 1.10 (b)

Figure 1.10 (c) Figure 1.10 (d)

elementary area d4, then, taking moments about the x axis, we get

The area element dA is located at a depth h below the free surface.

Also, we know that

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Substituting these in equation (1.14), we get.

yp $A y y sine (U = $ y f s ine (U

Can you identify the integral $ y2d,4 ?. Pmm your solid geometry; you may know that A

this is the moment of inertia about the x- axis, denoted as I,.

That is,

Also, recall that .I t b :. Equation (1.15) may be written as

1 It is usually more convenient to express yp in terms of the moment of inertia about an axis passing through the centcoid and parallel to the x-axis. This may be done by using the parallel axes theorem of the moments of inertia which may be expressed as,

in which I, is the moment of inertia about an axis passing through the centroid of the area and parallel to the x-axis.

I,, + yf A Ix c

YP - Y C A

= Y e + - Yc A

Now, look at that expression. I,,, being a moment of inertia is always positive. And y, and A are also positive. Therefore yp is always greater than ( or atleast equal to) yc. The equation therefore implies that the centre of pressure is always below the centroid. You can very well think of the reason for this. If the pressure distribution is uniform, the centre of pressure coincides with the centroid. Now, because the fluid pressures increase linearly with the depth, the resultant force will be shifted downwards.

To get the x co-ordinate, x,, of the centre of pressure, we shall take moments about the y-axis.

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Fluid Statics sad Kinematics

y s i n e J x y d*

"P * y sin yc A

You may recall that the integral$ x y dA is called the product of inertia and is denoted A

by I,. Therefore,

Again, it is convenient to express x, in terms of the product of inertia about the centroidal axes. (That is, axes parallel to x-axis and y-axis and passing through the centroid). From the parallel axes theorem for the product of inertia, we have,

I x y = Xc Yc A + I x y c

where, I,, is the product of inertia about the centroidal axes. Substituting

In most of the- applications, the area is symmetrical about atleast one of the centroidal axes, in which case I,, is zero. the x-coordinate of the centre of pressure would then be the same as that of the centroid.

We shall now see how to apply the expressions derived above for solving the problems.

Consider the following example : PQR is a triangular plate (with PR = QR ) immersed in water, making an angle 60' with the horizontal, as shown in figure l.lO(c). All required distances are given. We need to find out the magnitude and the poipt of application of the force due to water pressure on the triangular plate. The base PQ of the plate is parallel to the water surface.

First, let us understand how the plate is placed; Look at figure l.lO(d). This gives the dimensions of the triangular~plate. In figure 1.10 (c), the front view of the plate fixed in water is shown. The edge P Q is parallel to the water surface (not to the line MN, ofcourse) and extends into the paper. The point Q is therefore hidden behind the point A. The points G ', Cpl, and R ' of figure (1.10 (c)) are the traces of points G, C, and R respectively. G is the centroid and Cp is the centre of pressure figure (1.10 (c)).

We will now get the magnitude of the force on the plate. We know that the magnitude of the pressure force on a plane surface is given simply by the product of the pressure intensity at the centroid and the area of the plane surface. To get the pressure intensity at the centroid, all we need to know is the depth of the centroid below the free surface. That is, in figure. (l.lO(c)), we must get h, which is quite simple.

- h - 2.5 + P G ' sin 60"

PQ', being the distance of the centroid from the base PQ, is as you know, for this triangle, 1 m. ( = RS/3 = 3/3)

- . h = 2.5 + 1 x sin 60"

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The resultant force due to water pressure on the triangular plate is given by Fluid Statics

F - ' y (Area of triangle)

- 99021 Newtons - 99.021 kN

We have taken the specific weight y of water to be 9806 ~ l m ~ .

To determine the coordinates of the point of action C1, we must choose our coordinate system consistent with those we used to derive equations (1.16) & (1.17). In figure (10(a)), these coordinate axes are shown. Verify that they are infact consistent with what we have used earlier. We will now use the equations (1.16) & (1.17) straightaway. For simplicity we denote I, by IG and y, by

For the Y coordinate of P, denoted yp we have

IG is the moment of inertia of the triangle about the centroidal axis parallel to the x axis. That is, (see figure. l.lO(d) about the axis GI-G2. In this case, therefore,

(where b is the base width and h is the height of the triangle. )

y is the y coordinate of the centroid, which, from figure. l.lO(c), is the distance OG'.

and

A = area of the triai~gle = (112) x 2 x 3 ='3 m 2

= 4.015 m

and

hp = yp sin 60" - 3.477 m.

The x coordinate xp of the centre of pressure Ts given by -

Because the triangle is symmetric about one of the centroidal axes ( about the axis RS which is parallel to the y axis), I, = 0, and therefore,

Xp ' " That is, x, = 1 m.

We therefore got the following answers

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Fluid Shaa and Kinematics

Magnitude of the force, F = 99.021 kilonewtons (kN)

Co-ordinates of the centre of pressure,

1.5 FORCE ON CURVED SURFACE

We are now able to find the magnitude, direction and the point of action of the resultant fluid force on a plane surface. Now, let us recall what we did to obtain the resultant force on a plane surface. We integrated (that is summed) the forces on individual elemental areas to obtain the magnitude of the resultant force. We could do this because the system of forces f i , f2... on individual elemental areas hnstituted a system of parallel forces (see figure l.ll(a)). When the surface is curved, such as shown in figure 1.11@), however, the system of forces on elemental areas will not constitute a system of parallel forces and therefore we will not be able to obtain the resultant force by a single inteption.

To get the resultant force on the curved surface we shall proceed as follows :

Look at figure 1.12. f,, f , ... are the forces on individual elemental areas and they are

v Free Surfaa V ~ r e e surface

?

F i p m 1.11 (a) F w 1.11 (b)

normal to the respective elemental areas. Let F be the resultant force on the curved surface. FH and Fv are respectively the horizontal and vertical components of F. We shall determine FH and Fv instead of attempting to find F directly. Consider the fluid enclosed between the curved surface AB and its projection on a vertical plane A' A' figure. (1.13). In the x'-direction, there are only two forces acting on this volume of

v Free Surface 7

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fluid, viz., F ', the force due to fluid pressure on the projection A' B' and Fk) the force exerted by the curved surface AB on the fluid. Note that FHf is equal and opposite to FPH, the horizontal component of the force exerted by the fluid on the surface AB. Since the fluid within AA' B' B is in equilibrium, the (vector) sum of the forces in any direction must be zero.

From this we draw the important inference that the horizontal component of the fluid force on a curved surface is equal to the fluid force on the projection of the surface in a vertical plane. Also, the line of action of FH1 (and hence of FH) must be the same as that of F. If, for example, the curved surface AB is a cylinder (see figure. (1.14)), with dia d and length 1, its projection on the vertical plane will be a rectangle of dimensions d x I. If the fluid surface is just touching the top of the cylinder, then the

horizontal component of the fluid force on the cylinder is given by *, this being 2

the force exerted by the fluid on the rectangle held vertically. And, this horizontal 2d

component acts at a distance - from the top edge of the cylinder. 3

We shall now determine the vertical component of the resultant fluid force on a curved surface adopting the same reasoning. Consider a section of unit width (normal to the plane of the paper) of the curved surface AB, as shown in figure 1.15. The

volume of fluid A B B'A' is enclosed between the fluid surface and the curved surface AB. Since we are interested in obtaining only the vertical component, let us not worry about the horizontal forces HI and H,. Since the body of fluid withinA B B'A' is in equilibrium the sum of forces in any direction must be zero. The only vertical forces

v B' A ' Free Surface

-.-. F r e e S ~ r f a e ~ T

acting on this body of fluid are its own weight, Wand F,' , the reaction of the surface AB to the vertical component F, of the fluid force F; is, of course, equal and opposite to F, Remember that we are talking about the fluid force on only one side of the surface. Because the fluid and the surface AB are in equilibrium

HI - dia d 1 = length of

the cylinder

'or

That is, the vertical component (F,) of the fluid force on a curved surface is equal to the weight of the fluid enclosed between the curved surface and the free surface of the fluid. Also, because there are only two vertical forces F', and W acting on the body of fluid A A'B'B, the line of action of F,' (and hence of F, itself) must be the same as that of W. That is, the line of action of F, passes through the centre of gravity of the

normal to the A

Paper

B f z FH- (1R)yd I

B

1" * Hz

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Fluid Statics and Kinematics

fluid enclosed between the free surface of the fluid and the boundary of the curved surface.

For example, consider the semicylindrical body of radius r and of unit length (normal to the plane of the paper), shown in figure 1.16.(a).

Figwe 1.16 (a) F l w 1.16 (b)

The vertical component of the fluid force on the upper portion of the body is given by the weight of the fluid enclosed within A C B B'A'. Thus,

The term within the brackets, as you will readily see is the volume of the fluid enclosed within A C B B'A'. Note that the vertical component of force on the lower portion of ACB is also equal in magnitude to the weight of the fluid withinA C B B'A' but it acts in the opposite direction (viz., vertically upwards).

Once we get the horizontal and vertical components of the resultant force, we can determine the magnitude and the point of action of the force from first principles.

Example 3 : Calculate the magnitude and direction of the resultant force due to fluid pressure on the semicircular gate ACB shown in figure 1.16@). The: width of the gate (nonal to the paper) is 3m.

Solution :

We will determine the horizontal and vertical components of the resultant force separately.

The horizontal component FH is given by the force on the projection on a vertical plane of the curved surface. This projection is A'B' and has a width of 3m normal to the paper.

Thus,

where h i s the distance below the free surface of the centroid of the projected area.

and, A = Area of the projection

= Area of the rectangle of width 3m and depth 0.5 m

= 3 x 0.5 = 1.5 m2.

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The vertical component of the force consists of two parts : A downward force, FIV on the portion AC and an upward force FzV on the portion CB of the gate.

We know that

F,, = Weight of water contained in ADECA, and

F,, = Weight of water contained in ECBADE

The resultant vertical force F, is simply the vector sum of F1, and F2, We may thus write,

Fv = F2v - FIV

- y [AreaECBADE -AreaADECA]x 3

= y [Area ACBA ] x 3

= Weight of the volume of fluid equal to the volume of the gate.

The resultant of FH and FV may be got as,

and the angle it makes with the horizontal is given by,

Thus, the resultant fluid force on the gate is 14.99 kN and it acts at an angle 10.9" to the horizontal, normal to the gate (being the force due to fluid pressure).

4 The section of a dam is shown in the figure 1.16 (c). Determine the magnitude of horizontal and vertical component of water pressure on the upstream face per meter length of dam.

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1 I

1.6 MEASUREMENT OF PRESSWRE 1 In this section, we shall study some devices used for measuring pressure. For measuring purposes, pressure is frequently expressed as the height of a liquid column.

I I

In such a case we refer to it as the pressure head. For example a pressure head of 760 rnm of mercury corresponds to the pressure exerted at the base of a column of mercury 760 mm high. Remember that irrespective of the cross-section area of the column, the pressure still remains the same, being force per unit area.

When the pressure is expressed as force per unit area (e.g. NI~') , it is referred to as the pressure intensity. A pressure intensity may be expressed as an equivalent pressure head of any liquid. The pressure intensity, as we have seen earlier, is given by yh; knowing the pressure head in terms of one liquid, we can express it in terms of any other liquid with known specific weight, because

where yl is the speciric weight of liquid 1.

y2 = specific weight of the liquid in terms of which we are interested in expressing the pressure head;

h, and h2 are the pressure heads corresponding to the two liquids.

The atmospheric pressure is normally taken as 760 mm of mercury. To express this pressure head in terms of any other liquid, say, water for example, we use the above expression.

That is, a pressure head of 760 mm of mercury is equivalent to a pressure head of 10.336 m of water. We have used the information that the specific gravity of mercury

is 13.6 and the specific weight of water is 9806 ~ / m ~ . Frequently we will be interested in expressing the pressure head of a liquid as an equivalent height of water column, as demonstrated. In such a case,

where S1 is the s'pecific gravity of the liquid and hl is the pressure head of the liquid.

In many pressure measuring devices we use this concept of converting pressure heads

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into an equivalent height of a convenient liquid (most often, water).

1.6.1 Manometers Manometers are simple devices used to measure the pressure. The simplest of the manometers, called a piemmeter, is simply a glass tube fitted to the pressure source as shown in figure 1.17. The pressure source may be a flow in a pipe. As the glass tube (the piezometer) is open to the atmosphere, the fluid will rise to a certain height h inside the tube to balance the pressure at A.

The height h itself will give the pressure head at point A (above which the height is measured) in tenns of the fluid flowing. Piemmeters are obviously unsuitable for measuring negative pressures, since air would then be sucked into the pressure source.

Also, if very high positive pressures are to be measured, the glass tube has to be very large and it becomes impracticable to use piemmeters. In such cases, the U-tube manometers are used. A simple U-tube manometer is shown in figure 1.18. The liquid in the pressure source will enter the U-tube and attain a level B when equilibrium is reached. If the pressure at A is negative, the point B will be below A and if it is +ve the point B will be above A. To get the pressure at point A, we start at point B where the pressure is known to be atmospheric. In most of the problems, we will be interested in finding the gauge pressure (pressures relative to atmospheric pressure), and therefore we take the atmospheric pressure to be zero. Because the point C is at the same level as B (figure 1.18 ), the pressure at C is equal to that at B. The point C itself is at a depth h below the point A, and there is a continuity of the fluid between A and C. Therefore the pressure at A plus the pressure due to tbe liquid column between A and C must be equal to the pressure at C.

That is,

Figure 1.19 Figwe 1.20

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Knowing the specific weight y of the fluid we will thus be able to determine the pressure at A. When the pressures are large, however, the tube has to be very large, which becomes a limitation of such manometers. To overcome this difficulty, a second fluid, with a relatively high specific weight is rued. The advantage is that the rise in the level of a fluid of higher density will be much smaller than that of a fluid of lower density, for the same pressure difference. Figure 1.19 shows such a manometer. Normally mercury is used as the fluid of the manometer since it has a relatively high specific gravity (S = 13.6) and it is immiscible with most fluids. An advantage of such manometers which use another fluid to measure the pressure of a fluid is that they can be used even when the pressure source is a gas.

In figure 1.19, to determine the pressure p,, at the point A, we proceed as follows. Point B is an interface of fluid 1 and fluid 2; we know that two points along the same horizontal line will have equal pressures (of course, as long as there is a continuity in the fluid between the two points). Thus the pressure at C will be equal to the pressure at B. We also know that the pressure at point D which is open to the atmosphere is zero (since we are talking about gauge pressures; that is, pressures relative to the atmospheric pressure). The obvious procedure to adopt therefore would be to start with pressure at D, add to it the pressure due to the column D C of fluid 2, to get the pressure at C. This is also equal to the pressure at B. From the point B, to reach point A, we need to move up the column of fluid 1 and as we move up the column of the fluid pressure goes on reducing. Thus , we may write,

Pressure at D + y, h, = Pressure at C

= Pressure at B

Pressure at A = Pressure at B - y, h,.

That is,

In many practical problems, we will be interested in the pressure difference between two sources and not in the actual pressures. In such a case we use what is called the differential manometer. The differential manometer is very much similar to the U-tube manometer shown in figure 1.19, except that at both ends of the manometer a pressure source is connected. The differential manometer shown in figure 1.20, can be used to measure the pressure difference between sources A and B. To get the pressure difference PB -PA, we proceed exactly as we did before. That is we reach point E first by adding to p,, (the unknown pressure at A), a pressure yl h,; since E and D are along the same horizontal line with the same fluid (fluid 2) being continuous between them, the pressure at D is equal to that at E. So, we have reached point D. Our aim is to reach point B. But because there are two different fluids (fluid 2 and fluid 3) between points D and B, we cannot reach point B in a single step; First we reach point C by deducting (since we are moving up a fluid column) from the pressure at D a value y2 h2, to get p, and from this we deduct (again we are moving up) y d 3 to get pB , the unknown pressure at B. And this can be written as

PA+ Y I ~ I - Y ~ ~ z - Y ~ ~ ) = PB .

That is,

Since we can measure h,, ht and h, and we know the specific weights of the three fluids, yl, y2 and y3 the pressure difference is got straightaway.

You must have now realised that the procedure involved in determining pressures or the pressure difference between two sources using the manometers is. quite simple and general. We shall recall the stepwise procedure we adopted in the previous discussions. In all manometer problems we use the same procedure.

1. Start at one end (or any meniscus if the circuit is continuous) and write the

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pressure there in appropriate units (say, pascals) or in an appropriate symbol if it is unknown.

2. Add to this, the change in pressure, in the same unit, from one meniscus to the next (plus, if the next meniscus is lower, minus if it is higher).

Continue this procedure till the desired point is reached. n

I You must also have realised that we may have more than one manometer fluids. The I procedure still remains the same.

In some applications, we want to measure small pressures (or pressure differences) accurately. In such cases, we may use the inclined manometer shown in figure 1.21. As the name indicates the limbs of the manometer are inclined to the horizontal. As a result, a large reading (a large difference in the levels of manometer fluid) is produced even for a very small pressure. The accuracy of pressure measurement thus is increased. To better understand this, look at figures 1.22(a) and 1.22(b). In figure 1.22(a), the pressure source A is connected to a vertical manometer. For a very small change in the pressure at A, the level difference y in the manometer limb will also be

Figure 1.22 (a) Figure 1.22 (b)

very small and therefore inaccuracies may arise in reading the levels. In figure 1.22(b), instead of a vertical tube an inclined tube is connected to the same pressure source A. For the same pressure difference that produced a small change in level of y in a vertical manometer, the corresponding difference in reading (along the inclined tube) will be y / sin 0. Since sin 8 may be made as small as desirable by making 0 very small we can get a very large reading corresponding to even very slight variation in pressures.

Fluid Static..

For example, consider the inclined manometer shown in figure 1.21, for measuring the pressure difference between source A and B. The inclined manometer itself is in a plane M (that is, both the limbs of the manometer are in the same plane and there is no elevation difference between the two limbs). To get the pressure difference p, - p,, we use the steps listed above. In almost all problems we neglect the pressures due to air columns, so that, in this case the pressure at C is equal to pressure at B and that at D is equal to the pressure at A. In moving from point C to D, we must deduct the

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mnid statics md pressure due to the 20 mm of alcohol which is inclined at 30°. This pressure is equal Kinematics to the specific weight of alcohol multiplied by the vertical distance between C' and D,

which is 20 sin 30" mm or 0.02sin 30" m.

We write, therefore,

p, - 0.02 y sin 30" - p,

P, - PA - 0.020 x (0.8 x 9806) sin 30"

SAQ 5 In figure 1.20 fluid 1 is water (y - 9.806kN/m3), fluid 2 is mercury ( S = 13.6), and fluid 3 is oil (S = 0.8). If hl = 15 cms, hp = 30 cms and h3 = 15 cms. Determine the pressure difference between A and B.

1.7 SUMMARY

This unit introduces you to some basic concepts of fluid mechanics. We have mostly dealt with fluids at rest, and thus, the unit is named as Fluid Statics. This unit paves way for understanding the subsequent units. We have discussed the basic fluid properties with some very rudimentary numerical examples which only make sure that you understand the definitions right. The variation of pressure in a static fluid and force on plane and curved surfaces are dealt with in a greater detail. The numerical examples solved, although very few in number, are aimed at equipping you with procedures to solve any problem of similar natures. Simple pressure measuring techniques using manometers are discussed in the last section.

It is emphasised that, to assimilate completely the material contained in this unit, a large number of numerical problems must be solved. You are therefore urged to solve numerical examples from the textbooks recommended. It makes the process of learning enjoyable.

1.8 ANSWERS TO SAQs

SAQ 1 For the configuration shown in figure 1.2

du From Newton's Law of Viscosity r - p -

dy

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SAQ 2

Reduction of 1 % of volume implies that

Increase in pressure - dp

SAQ 3 for water

u = hyd 4 cos 8

SAQ 4 Horizontal component

FH = y (Projected area x j;)

Vertical component

Fv I y (Volume TNMS )

= 9 . 8 0 6 ~ 45 x 3

= 1323.81 kN

SAQ 5 Given data

y , - 9.806 w/m3 hl = 15 x 10- m

y2 = 13.6 x 9.806 w/m3 & = 30 x 10- m

~3 = 0.8 x 9.806 kN/m3 & = 15 x 10- m

for the configuration shown in figure 1.20.

+ yl hl = Y2h2 + Y3h3 + PB

PA - PB = 13.6 x 9.806 x 30 x + 0.8 x 9.806 x 15 x - 9.806 x 15 x

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UNIT 2 BUOYANCY pp - - - - - - -

Structure 2.1 Introduction

Objectives

2.2 Buoyant Force

2.3 Stability of Bodies Immersed in a Fluid 2.3.1 Stability of Submerged Bodies

2.3.2 Stability of Floating Bodies.

2.4 Liquids in a Container Subjected to Linear Acceleration

2.5 Liquids in a Container Subjected to Rotation

2.6 Summary

2.7 Answers to SAQs

2.1 INTRODUCTION

This unit forms the concluding part of Fluid statics. For most part of this unit, we will still be dealing with fluids at rest, as we did in Unit 1. Determination of buoyant force, its point of action, metacentric height are some of the important exercises you will be carrying out in this unit. However, towards the end of the unit we shall deal with some special cases of fluids under motion where the methods of fluid statics still apply.

Objectives After studying this unit, you should be able to

* determine the magnitude, direction and point of action of the buoyant force,

* conclude whether a floating body is stable against angular displacements, and

* determine the inclination/shape assumed by the fluid free surface when the fluid con- tainer is subjected to: a) linear acceleration and b) rotation.

2.2 BUOYANT FORCE

A body immersed completely or partially in a fluid is acted upon by an upward buoyant force equal to the weight of the fluid displaced by the body. This force acts through the centre of gravity of the displaced fluid. Buoyant force FB may therefore be written as,

FB' Y" (2.1)

where V is the volume of the fluid displaced (equal to the volume of the submerged portion of the body) and y is the specific weight of the fluid. For example, consider a rectangular block immersed in water as shown in figure 2.1 The volume of the portion

1 m Figure 21

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of block immersed in the fluid is equal to 1 x 1 x 0.5 - 0.5 m3. The specific weight of water being equal to 1000 ~ / m ~ , the weight of water displaced ahd hence the magnitude of the buoyant force is,

FB - yV - 1000 x 05 = 500 N

This force acts vertically upwards through the centre of gravity of the water in the portion ABCDEFGH. You can readily see that for the block to be in equilibrium the buoyant force must be equal to the weight of the block, since there are no other forces acting in the vertical direction. As the weight of the block increases, a greater buoyant force will be necessary to keep it in equilibrium and therefore more and more portion of the block will be submerged in the fluid. The point of action of the buoyant force, which coincides with the centre of gravity of the fluid displaced is called the Centre of Buoyancy.

Example 1 :

What fraction of an iceber would be above the free surface in the ocean, if the B density of ice = 920 kg/m and density of sea water = 1030 kg/m3.

Solution :

In figure 2.2, Vl is the volume of the iceberg above the free surface and V2 is the volume below it.

Total weight of the ice berg

Buoyant Force FB - Weight of the fluid displaced

- p w x g x V 2

= 1030 x 9.81 x V2 Newtons

Since the iceberg is in equilibrium, the two forces must be equal.

That is, one part of 9.36 parts is above the free surface.

Example 2 :

Find the density of a metallic body which floats at the interface of mercury (specific gravity, 13.6) and water such that 40% of its volume is submerged in mercury and 60% in water.

Solution :

Let the volume of the metallic body be V.

Then volume of the body immersed in mercury

Similarly volume of the body immersed in water 0.6 V m3

Now, for equilibrium, weight of the body = Total buoyant force on the body

= buoyant force due to water + buoyant force due to mercury.

= weight of water displaced + weight of mercu j displaced

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Weight of the body = p g V

- 6040 kg/m3

SAQ 1 A piece of irregularly shaped metal weighs 300 N in air. When the metal is completely submerged in water it weighs 232.5 N. Find the volume of the metal.

SAQ 2 A hollow cube 1 m on each side weighs 2.4 kN. The cube is tied to a solid concrete block weighing 10 kN. Will these two objects tied together float or sink in water ? Sp. gravity of concrete is 2.4.

23 STABILITY OF BODIES IMMERSED IN A FLUID

A wooden bar with a heavy lead ball attached at its lower end is shown floating in a fluid in figure 2.4(a). If you disturb this position slightly by tilting the bar and releasing it, you will observe that the bar quickly gets back to its original position of equilibrium. This original position of equilibrium AB is then called the position of stable equilibrium. In figure 2.4@), the same bar is shown with the lead ball attached to the upper end. The bar is still in equilibrium. However, if you slightly tilt the bar and release it, the bar will completely turn over and will not regain its original position of equilibrium. Such a position is said to be one of unstable equilibrium.

Bodies floating in a fluid are in stable equilibrium against vertical linear displacement because a vertical displacement causes a change in buoyant force on the body and returns the body to its original position of equilibrium. For example, if the spherical ball shown in figure 2.5(a) is pushed vertically downward (producing a vertical linear displacement), and released, it quickly bounces back to its original position of

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Figure 2 4 (a) 2 4 (b)

equilibrium. This is because, as more portion of the body is submerged, the buoyant . force increases (see figure 2,5(b)). As this increased buoyant force is more than that

required to keep the body in equilibrium the body is pushed up, till the buoyant force becomes equal to the weight of the body, which occurs in its original position.

An interesting engineering problem is the stability of floating bodies against angular displacements. Ships, boats and Bubmarines often have to experience sudden angular

displacements. It is important that they are stable against such small angular displacements. Let us now see under what conditions the bodies are stable when they are immersed in a fluid.

L

2.3.1 Stability of Submerged Bodies In case of completely submerged bodies, the position of centre of gravity (CG) and

' centre of buoyancy (CB) get fixed with respect to the body and hence the stability qf the body is completely determine4 by the relative positions of the CG and CB. Look at figure 2.6(a). The centre of buoyancy CB is above the centre of gravity CG. The

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Fluid Staria and bematics

buoyant force FB acts vertically upwards at CB and the weight of the body, W acts downwards at CG. If you slightly tilt the body as shown in figure 2.6(b), the f o m FB and W, will form a couple that tends to restore the body to its original position. If, however, the centre of buoyancy CB is below the centre of gravity CG as shown in figure 2.7(a), then the couple formed by FB and W in the new position (figure 2.7@)) will only increase the angular displacement and therefore the body will be unstable. Thus, for a completely submerged body to be stable against small angular displacements, the centre of buoyancy must lie above the centre of gravity.

E%cl- 2 7 (a) Fk- 2 7 (b)

2.3.2 Stability of Floating Bodies Any floating body with its centre of gravity below the centre of buoyancy is also in stable equilibrium for the same reasons as explained earlier in the case of completely submerged bodies. However, in practical situations, the CG of floating bodies like sbips and vessels may often lie above the CB and still they may be in stable equilibrium. We shall now see under what conditions the floating bodies with their CG above the CB will be stable against angular displacements.

W

I -3 n \ f k t / FB = W

Figure 2.8 (a) Figure 2.8 (b)

In the case of floating bodies, although the CG is fixed with respect to the body, the CB keeps varying with rotation, since the submerged volume varies. If the point CB shifts to the right of CG for a clockwise rotation as shown in figure 2.8(a) then a restoring couple is set up (figure 2.8(b)) and the body will be stable. Remember, however, that this is true only for small angular displacements. If, for a clockwise angular displacement the CB shifts to the left of CG, figure 2.8(c), then a clockwise moment is set up and hence the body will be unstable. Whether CB shifts to the left or right of the CG depends on the shape of the body.

WINDOWS
Comment on Text
dec.2007
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Figure 2.8 (c) 6

To make this more clear, let us consider the simple rectangular floating body as shown in figure 2.9(a).

b

For a small angular displacement, B' is the new position of the centre of buoyancy, figure 2.9@). The point of intersection of the vertical through B ' (which is the line of ation of Fg in the new position) and the line BOG (extended) is called the Meta Centre. The position of meta centre in a floating body determines whether the body will be stable against angular displacement or not. If M is above G, as shown in figure 2.9@), a restoring moment will be set up and therefore the body will be stable.

If, on the other hand, the point M is below G it will be unstable. The position of metacentre M with respect to G, thus determines the stability of a body against angular displacements.

I The distance MG is called the Metacentric height. It is positive if the point M is above G and negative if it is below G. It can be shown that

where % is the metacentric height, I is the moment of inertia of the cross section at the level of water surface about its centmidal axis, V is the volume of water displaced and is the distance GB. The negative sign is taken in equation (2.2) if the point G is above B and positive sign is taken if it is below B. For stability % must be positive.

Example 3 :

A rectangular pontoon is 5m long, 3m wide and 1.2m high. The depth of immetsion is 0.80m in sea water. If the CG is 0.6 m above the bottom of the pontoon, dertermine the metacentric height. Density of sea water = 1025 kg/m3.

Solution :

Figure 2.1qa) shows how the body is floating in water. In figure 2.10@) the cross section of the b o d y a t h e water level is shown; Since G is above B, the negative sign is used for BG in equation (2.2).

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5 m

Rprc 210 (b)

- I - i.e. GM=--BG v I = MI about axis Y-Y in figure 2.10@).

b d I-

12

5 x (3)3 45 4 -= -m 12 4

V = Volume of the pontoonsubmerged

= 3x0.8x5=12m3

BG-AG- AB= 0.6-0.41 0.2m

45 1 :. GM=-X--0.2 4 12

= 0.9375 - 0.2 = 0.7375

Since GM is positive, the pontoon is stable against small angular displacements.

Example 4 : A cube of side a and relative density S floats in water. Determine the conditions for its stability against angular tilt.

Ia --j

V v .G

'T X B

1 7- XI2

_t mn=Z11(4

Solution :

For equilibrium weight of the cube = Buoyant force on the cube.

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3 2 :. y a S = ya x , where y is the specific weight of water

CG of cube is at 'a and CB at (from bottom) 2 2

For stability,

V = Submerged volume

2 = a x

X Since -= S , ( from (A))

a

or 6 s 2 - 6 S + 1 > 0

From solution of 6 S 2 - 6S + 1 - 0, we get S - 0.789 or 0.211.

:. 0.211 z S r 0.789

This is the condition to be satisified for stability.

SAQ 3

Would the wooden cylinder (sp. gr. 0.61) be stable if placed vertically in oil (sp. gr. 0.85) as shown in figure 2.11 (b) ?

dia

. Buoyancy

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I Fluid Statics and Kiacmatks

SAQ 4 A cube of side length L and sp. gravity 0.8 floats in water. Is the cube stable ?

2.4 LI UIDS IN A CONTAINER SUBJECTED TO 8 E LI AR ACCELERATION

Under certain conditions there may be no relative motion between the particles of a fluid mass, yet the fluid mass as a whole may be in motion. For example a liquid being transported in a tanker moves as a single body without any relative motion

between particles. If the liquid is being transported at a uniform velocity, the conditions are those considered in fluid statics. But if the liquid mass is subjected to acceleration, then other forces need to be considered. You know that for a fluid at absolute rest, the hydro-static pressure distribution applies. Let us now see what happens to a fluid which is in an accelerated motion.

Consider a volume element of a fluid whose container is subjected to accelerations a, a, and a, in the three directions as shown in figure 2.12. With pressure at the centre of the element 0 being p, pressures of the centre o f the sides are as indicated in the figure. (For an explanation of the convention used in amving at these pressures, you may refer to your unit 1 on fluid statics, where the hydrostatic pressure equation was derived with this convention). Since the element as a whole is in motion we may apply the Newton's second law of motion in the three directions to get.

Where m is the mass of the element. Notice that the weight of the fluid mg is accounted for in the z-direction. Since the mass m of the element is equal to the density times volume,

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m = p (Ar Ay Az),. the above set of equations (2.3) to (2.5) result in

when the acceleration in the three directions is zero, axE a,, = a,- 0, the above equations result in,

which is just the hydrostatic pressure distribution you have studied in your first unit. When the container is in an accelerated motion, the liquid level in the container assumes an inclined position instead of the horizontal level it has in purely static conditions. This is because of the pressure distribution given by equations (2.6), (2.7) and (2.8). With a little treatment of these general pressure distribution equations, it may be shown that the inclination 8, of the free surface in a container, (see figure 2.13) in the x - z plane is given by

Figure 2.13

For a purely horizontal acceleration, a, = 0 in equation (2.13), in which case,

Example 5 :

A thin walled, open topped tank in the form of a cube of 500 mm side is initially full of oil of sp. gravity 0.88. It is accelerated uniformly at 5 m/sec2 up a long straight slope, tan- '(114) to the horizontal.

Buoyancy*

Calculate a) the volume of oil left in the tank when no more spill occurs b) the pressure at the lowest corner of the tank

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Fluid Stnties and Kinematics

"V."e.V.I.

See figure 2.14 (a) for tbe definition of various terms.

Pigum 2.14 (a)

We have, -x t an8 =-

a, + g

At the point when no more spill occurs, the liquid surface will be as shown in figure 2.14. To get the volume of water left in tbe tank, we need to determine the distance C.

We shall proceed as follows:

Similarly, 1

a,- a s i n $ - 5 x m

Now,

tanO+tan@ tan (0 + I#) =

1 -tan 0tan I#

From figure, C

t an@+$)=- 0.5

:. C = 0.775 x 0.5 - 0.3875 m

:. Volume left = Volume of the tank - Volume of portion BEF

= 76.5 litres.

Further,

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-'- P - - pa, x - p (g + a,) z + constant

(We have substituted for * and * and then integrated) ax az

Withtheorig inatB,p=Oatx=Oandz=O

:. constant = 0

:.PO-pa,x-p(a,+g)z

We want the pressure at A.

Co-ordinates of A - (0.5 sin 4 , - 0.5 cos 4)

-'- PA = - p(a cos +) - (0.5 sin+) - p(a sin + + g) (- 0.5 cos +)

- pg (0.5 cos 4)

SAQ 5 A tank shown in figure 2.141(b)contains oil of sp. gravity 0.80. If it is given an acceleration of 5.0 m/sec2 along a 30" inclined plane in the upward direction, determine the slope of free surface and pressure at B.

2.5 LI UIDS IN A CONTAINER SUBJECTED TO s. RO ATION

When a vessel containing a liquid is rotated about an axis, the fluid elements experience both centrifugal and gravitational body forces. Consider a cylindrical container with liquid rotated about a vertical axis at a constant angular speed o. Under the action of centrifugal and gravitational forces the liquid line assumes a curved free surface, as shown in figure 2.15.

Buoyancy

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Fluid Statics and Kinematics

For the fluid element A in figure 2.15, the acceleration a, in the r-direction is given by 2 a , - o r

The fluid element will be in equilibrium, if

- * + pu2r = 0, and ar

dz The slope of the free surface, - is therefore obtained as,

dr

Integration of equation (2.11) yields,

02r2 z=- + C

2g

With z = 0 at r = 0, c = 0

The maximum height to which the liquid level can rise is thus, - w 2 ~ 2 where R is the 2g

radius of the container.

Example 6 :

A cylindrical vertical container 50 cm internal dia, is rotated about its axis. The container has a height of 1 m and was initially filled to 60 cm. Calculate the speed of rotation at which the water shall begin to spill over the container, and the pressure at a point 20 cm radial position and 5 cm above the base. (See figure 2.16)

Solution : At the condition of spilling over

1 021t2 - = 40 cm (half the maximum height) 2 2g

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. 0 2 - 0.4 x 4 x 9.81 - 251.136 (0.25)~

& - 15.84 radlsec

The point whose vertical position is 5 cm above the base corresponds to z

2 1OOOx251.136x(0.2) +1000x9B1x0.15 0

2

= 6.494 x lo3 N / m2

If the cylinder in the above example was closed with a lid and was that the point at the centre of the base was just clear of water, what been the angular speed ?

rotated so would have

2.6 SUMMARY

Now.let us summarise what you have learnt in this unit.

a) The buoyancy force on a wholly or partially submerged surface is equal to the weight of the fluid displaced and it acts vertically upwards through the centre of buoyancy.

b) The point of intersection of the buoyancy force from its new position after a slight angular tilt and the line joining the centre of buoyancy in the undisturbed position & the centre of gravity of the body is called the Meta Centre. For stability the metacentre should be above the centre of gravity.

c) The free surface of a liquid in a container under acceleration inclines upward in a direction against that of the acceleration. Under the condition of rotation of the container at a constant speed the free surface assumes a paraboloid shape.

2.7 ANSWERS TO SAQs

SAQ 1

SAQ 2 W = Weight of hollow cube plus solid concrete block.

Fb, = buoyant force on hollow cube

Fb2 = buoyant force on solid concrete block

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Fluid Statics l ad Kinematics

Since [ W- 12.41 < + Fb, - 13.96 the two objects will float in water I SAQ 3

The first step is to determine the submerged depth of the cylinder when placed in oil.

Fl)= w

The centre of buoyancy is located - 0'9333 - 0.466in from the bottom of the 2

cylinder.

That is, the metacentre is located 0.030 m above the centre of buoyancy. This 1.30

places the metacentre - - 0.466 - 0.030 or 0.154 m below the centre of 2

gravity. Since the metacentre is below the centre of gravity, the cylinder is not stable.

SAQ 4 The cube's centre of gravity is at 0.5 L above its bottom. Since the cube's sp.gr. is 0.8, it will float at a submerged depth of 0.8 L, and its centre of buoyancy will be at 0.4 L above its bottom.

I MB- - = V Lx L x 0.8L - 0.1042L That is, the metacentre is located 0.1042 L above the centre of buoyancy and 0.1042 L + 0.4 L - 0.5 L or 0.00$2 L and therefore the cube is stable.

SAQ 5 'Ptre acceleration a, is resolved as

g - 5 sin 30" - 25~m18ec2

Slope of warer'surface is given by

or 0 = 19.38"

Depth at B - 2 - 1.5 tan 0 - 1.472m 5 = (1 + :) x hb (See the solved example in text for hint).

Y

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:. Pb = 1.847 x 9.81 x 1000 x 0.87

= 1576359 ~ / m ~

= 15.763 kN/m2

Since the forces continue to be the same two body forces even when the container is closed, the equipressure surface will still be paraboloid. However, as the water is not allowed to spill out and it in turn begins to readjust the volume there is no true free surface.

With this hint, you should be able to work out the problem. You will get an angular speed of 19.81 redlsec as the answer.

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UNIT 3 FUNDAMENTALS OF FLUID MOTION

Structure Introduction

Objectives

Types of Fluid Flow 3.2.1 Laminar and Turbulent Flows 3.2.2 Steady and Unsteady Flows 3.2:3 Uniform Flow and Non-Uniform Flow 3.2.4 isothermal and Isentropic Flow

Methods of Fluid Flow Analysis

Description of Flow Field-Lagrange's and Euler's Approaches

Streamlines

Differential Equation of a Stream Line 3.6.1 Stream lbbe

Path Line and Streak Line 3.7.1 Path Iine 3.7.2 Streak Line

Translation, Deformation and Rotation of a Fluid Element in Motion 3.8.1 Translation 3.8.2 . Rotation 3.8.3 Angular Deformation 3.8.4 Volume Deformation

Acceleration : Total Acceleration - Local and Convective Components of Acceleration

Equation of Continuity

Summary

Answers to SAQs

In the previous units you have studied the physical properties of the fluid, pressure distribution on submerged plane and curved surfaces when the fluid is at rest. You also know the magnitude and direction of pressure as well as its measurement. The concept of buoyancy, metacentre was also introduced. Previous unit has also covered relative equillibrium of a liquid subjected to linear acceleration and rotation. In this unit, we shall be dealing with a few fundamentals of fluid motion and its related aspects. Fluid kinematics deals with space-time relationship without considering forces involved in the flow. When fluid is in motion, the relative position of each particle is not fixed from time to time. Depending upon the type of flow, type of fluid each particle will have its own velocity and acceleration at any instant of time. The terms velocity and acceleration apply at a point in a fluid. In rectanglar coordinate system, the velocity components along the x, y and z directions are denoted by y v and w respectively. The velocity vector at a point is denoted by

Objectives After reading this unit, you should be able to

* identify different types of flow, * differentiate between streamlines, pathlines and streaklines, * determine the acceleration in a fluid flow, and

derive the continuity equation for a fluid flow.

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Ruid statics and Kinematics 3.2 TYPES OF FLUID FLOW

There are 96 combinations in which fluid flow can be classified. However, the following combinations are usually encountered in our day to day life which are required to be known in detail.

FLUID m w

m Uniiorm Non-dorm

Figure 3.1 : CI.ssiLlution d Fluid Flow

I

3.2.1 Laminar and Turbulent Flows

kmlnmx Turbulent

If fluid particles move in smooth paths in layen or laminae with one layer sliding over an adjacent layer, the flow is said to be laminar. Paths of individual particles do not cross or intersect. Contrary to this, flow is said to be turbulent if the fluid particles move in irregular paths, there being momentum exchange from one portion of fluid to another.

Figure 2 2 (a) I h m h u Flow Figure 3.2 (b) I Turbullmt Flow

r Bot.Uonml 1rrot.tiod

3.2.2 Steady and Unsteady Flows A flow is considered to be steady if the dependent fluid flow variables at any point in the flow do not change with time. Mathematically, this can be stated as

I 1 &.dJ Umbad J

I I Cempradble Inoemprrmdble

a - (dependent fluid variables) m 0 at (3.1)

1 h t h d battOpia

Thus, for steady flow

Generally velocity vector is represented as

where y v, w are the velocity components of the fluid,p is the pressure and p is the specific mass of the fluid.

For example, if the volume rate of fluid at a given cross section remains constant with time, the flow is said to be steady flow.

Whereas in unsteady flow or transient flow, fluid flow exhibits variations at a fixed point in a space with respect to time.

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F i 3.3 : Pipclioc of constant crow section

Q is the volume rate of flow. The same can also be expressed as weight rate of flow

W = y Q where y = specific weight of fluid in ~ / r n ~

M = Mass rate of flow

M - p x Q where p = specific mass of fluid in kg/m3

At cross section 1-1 in figure 3.3

and at time tl, Q - Ql.

Fm&mentab of Pluld Motion

t 1 i At time t,, Q = Q, I

I I Hence, - I aQ = 0. at

For turbulent flows, due to fluctuations in flow at any section (figure 3.4) if temporal I mean velocity Ti does not vary with time, flow is called steady flow. If temporal mean

velocity varies with time, the flow can be treated as unsteady flow

Unsteady Steady - A

Time t - Figure 3.4 : Turbulent flow - Temporal Mean Velocity

3.2.3 Uniform Flow and Non-Uniform Flow If at any given instant velocity vector of fluid remains same in magnitude and direction at all points, the flow is uniform. This stipulates the velocity components to be the same at different points in the flow. If the velocity varies from point to point at any instant, the flow is called non-uniform.

For uniform flow,

a au av aw -(dependent fluid variables) I 0 i.e. - = - - - , s o as ax ay az (3.5)

In case of ideal fluid flow, velocity at a fixed boundary is not zero while in the case of real fluids, the velocity at fixed boundary is zero. This is usually referred to as No-slip condition. In ideal fluids, at any section, all fluid particles move with the same vetocity. .

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~ e i d rt.tia .nd 3.2.4 Isothermal and Isentropic Flow Khca8tkr

If gas flows at constant temperature, flow is said to be isothermal. If during the gas flow,

Fipm 3,s (a) Vdodty in M ideal Figure I 3.5 (b) Vdocitg h 1.mtnu F i 3.5 (c) Vclodty lo h i d (dip condition at the n o w turb&nt flow

k d u y ) (Rul Hdd) (Rul

no heat enters or leaves the boundaries of the fluid, the fluid flow is adiabatic. Frictionless flow is called isentropic flow. Detailed description of these are beyond the scope of this course.

For compressible and incompressible fluid flows reader may refer any book on these topics. Rotational and imtational flows are seperately treated in Unit 4.

3.3 METHODS OF FLUID FLOW ANALYSIS

All fluid flows must belong to any one of the three types of flow viz, one-dimensional flow, two-dimensional flow and three-dimensional flow.

In one-dimensional flow, fluid property and parameters of flow are constant at any cross section normal to the flow. In otherwards, if the dependent variables are functions of only one space coordinate, say x it is ldimensional flow (figure 3.6(a)).

If the dependent variables are functions of two space coordinates x and y then, the flow is said to be Zdimensional (figure 3.6@)).

If the fluid flow parameters are functions of all the three space coordinates x, y and z, then the flow is said to be 3dimensional (figure 3.6(c)).

Numerous example for 3-dimensional flow are available in nature. For example, flow around trees, flow in a river, flow within fluid machines are examples of three-dimensional flows.

A flow is said to be axisymmetric if the velocity profile is symmetrical about the axis.

(8) Study id& tlaid &W tbFoUgb 8 (b) S h d y BOW t h w 8 pee (C) Study I ~ w 8 conveq#hg scctioa axed w d

ONEDIMENSIONAL FU)W TWO-DIMENSIONAL FLOW THREE-DIMENSIONAL FLOW

This means that the dependent variables cannot vary in the cimmferential direction. An axisymmetric flow is essentially a twodimensional flow.

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, 1 SAQ 1 Identify the type of flow for problems listed in column A dhoosing the type from column B.

A B (a) flow in a pipe of constant (i) non-uniform

cross section

(b) u - a + b t (ii) unsteady and non-uniform

(c) flow in a river during a (iii) unsteady, non- uniform and turbulent rainstorm

(d) flow in a pipe of varying (iv) uniform cross section

SAQ 2 Select the correct answer from the multiple choices given

(a) One-dimensional flow is

(i) steady uniform flow

(ii) uniform flow

(iii) flow which neglects changes in a transverse direction

(iv) restricted to flow in a straight line

(v) none of the above.

(b) The correct practical example of steady uniform flow is

(i) motion of water around a boat in a lake.

(ii) motion of river around bridge piers.

(iii) pipeline carrying varying flow.

(iv) constant flow through a reducing pipe section.

(v) constant flow through a long, straight pipe.

3.4 DESCRIPTION OF FLOW FIELD

The value of velocity of a fluid particle can be treated as a continuous position of space and time. There are two approaches usually followed to describe the flow field. They are Lagrangian approach and Eulerian approach.

In the Lagrangian approach, the observer observes the movement of a single fluid particle instant to instant. In otherwords, the observer here traces the movement of individual fluid particles. As the fluid flow consists of movement of large number of individual

FundPmentrls of Fluid Motioo

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Fluid statics md Kinematics

fluid particles it is not only impractical to watch the movement of individual fluid particles but also cumbersome to keep track of their movement. The flow quantity say, velocity is defined as a function of time and function of position of the fluid particle at initial instant t,. The initial position <of the particle at instant to is given by

4 - At a later time t, the particle is at position Fwhich is r (ro , t). The velocity vector

2

Or ( ) ( Y- yo and W - ( $ ) Z - z0 (3.8)

Thus, one can very well realise how cumbersome this approach can be. This approach finds its place in standard works on Classical Mechanics of Fluid which is very much in the realm of Mathematicians.

Whereas in Eulerian approach, the observer sits at one place and watches the movement of fluid particles as they flow past by h& side. Here, the observer notes the movement of many fluid particles as they flow past the observer. Thus, the velocity therefore is a function of space and time. This approach is much more practical and is adopted by Engineers in their study of Fluid Mechanics.

+ + - - 4 v = velocity vector = V (x,y, z, t) = i u + j v + k w (3m9)

where u-u(xa,z,t); vPv(x~,z, t) and w.=w(x,y,z,t)

This approach is much easier to follow and is widely used in Fluid Mechanics.

3.5 STREAMLINES

Stmamline : Streamline is a line tangent to which at every point indicates the direction of flow. We can say that streamline is an instantaneous picture of entire flow field.

Since, the component of velocity normal to a streamline is zero, there can be no flow across a streamline. Further, the instantaneous velocity at a point in a fluid flow field

F i 3.7 (a) : Shr.mlincs Figure 3.7 (b) : Expanded view of portion circled at F i i 3.7 (a)

must be unique both in magnitude and direction. This clearly means that the same point cannot belong to more than one streamline.

If we consider steady flow, the velocity components do not change with time. Hence, the velocity vector also does not change with time. This essentially means that the same streamline pattern holds good at all times. Whereas in unsteady flow the streamline pattern changes from instant to instant.

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3.6 DIFFERENTIAL EQUATION OF A STREAMLINE 4

Let us consider an elementary displacement element 6s along a streamline where the --+

velocity V is given by - b y + +

V = u l + v j + w k and g= tixZ6yj<&k4 (3.10) --+

From the definition of streamline it is clear that V must be directed along s. Hence,

I + --+ or (v6z - w6y) (u6z - wbr) j + (u6y - vbr) k = 0

which gives

aX 6.2 - 3 =- u v w

1

This is called the differential equation of a streamline in three-dimensional flow.

For a 2-D flow in the x - y plane

which in the limit & + 0 becomes

We can clearly say that the slope of a plane streamline is the ratio of velocity components.

3.6.1 Stream Tube A bunch of closely spaced streamlines bound by a imaginary tubular stream surface is called a streamtube (figure 3.8).

F i ' 3.8 : StrePm tube

As there cannot be any flow across a streamline it is quite obvious that the flow enters the stream tube at one end and leaves at the other end.

Example 1 : 4 4 4

The velocity vector V is given by V = i x - j- y. Determine the equation of streamlines. Plot them.

Solution : 4 --+

Wehave ~ x 6 s = O - ? ?

where V = r x - j y

and g = r& + 7 6 y + +

1.e. ( i x - j y )x (F 'kx+ j*dy )=~

FunbmentaL of Fluid Motion

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Fluid statics and Kinematics

i.e. (xdy + ydr) r- 0

Integrating we get xy = C where C is an arbitrary constant.

. The equation of stream lines is xy = C.

On plotting it, we see that it represents stream lines of steady Zdimensional flow at a 90' Y

F i 3.9 : Streamline pattern

33 3

comer (figure 3.9). We can also see that velocity vector V - i x - j y is everywhere tangent to the stream line xy - C.

3.7 PATH LINE AND STREAK LINE

3.7.1 Path Line Path line is the path taken by a single fluid particle at a given time. Path line thus shows the direction of the velocity of same particle at successive instants of time (figure 3.18).

Figure 3.10 : Path Line dz u = g v = * and w = -

dt' dt dt

o rx - Sudt, y =Jvdt and I - l w d t (3.15)

This constitutes the equation of the path line.

3.7.2 Streak Line A streak line is the locus of locations at an instant of time of all the fluid particles that has passed through a fixed point in the flow field.

SAQ 3 A streamline is a line

(i) connecting the mid points of flow cross sections

(ii) defined for uniform flow only

(iii) drawn normal to the velocity vector at every point

(iv) tangent to which at every point indicates the direction of flow

(v) showing the path of the particle.

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SAQ 4 Velocity components for a three-dimensional flow is given by u = - x, v = 2y and w - 3 - z. Find the equation of streamline passing through (1,1,2).

3.8 TRANSLATION, DEFORMATION AND ROTATION OF A FLUID ELEMENT IN MOTION

As a fluid element moves, it can be translated, deformed or rotated as shown in figure 3.11. Of the four possible modes of changes the fluid element undergoes the translation and rotation does not impose any stress on the system and hence not of much use in fluid dynamics. The angular deformation is shear strain which gives rise to tangential stresses and thus very important to be considered.

The linear deformation results in change in volume. Let us derive expressions for all these four possible changes the fluid element undergoes during its motion.

3.8.1 Translation Let the velocity at the left hand lowennost comer of the fluid element have components u along x-axis and v along y-axis. In time 6t, the element will occupy new position such that along x-axis it has moved a distance u6t and along y-axis a distance v6t. This amply describes the translation of the fluid element [figure 3.11(a)].

a~ - -by bf d u -by bf

u br

" Lpar--- -'

U

(a) Translatron (b) Rotation (c) Angular deformation

av - by br

C

(d) Volume d~stortion

Figure 3.11 : Changes the fluid element undergoes while in motion

3.8.2 Rotation Let us consider anti-clockwise rotation as positive [figure 3.11 (b)]. The rotation depends

au av av on the rate - and -. For rotation as shown in figure 3.11 (b) -is positive since v

ay ax ax au

increases with increase in x. Whereas - is negative because, u is in negative direction ay

with increase in y.

Fundnmeotds of Fluid Motion

WINDOWS
Comment on Text
dec.2007
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Fluid statics and Kinematics

Similarly

Therefore, the average rate of rotation in the positive direction will be

Similarly the rotation about axes parallel to x and y can be obtained

as 1 aw 1 au wx 5 ( - ] and wy - (% - 2) (3.18)

This rotation is related to vorticity which will be further explained in Unit 4.

The resultant rotation will be + 7' + 4

w = t o x + j o y + k o ,

1 1 -+ 2- x v4) = 1- curl v

It will be shown in Unit 4 that, for irrotational flow these components of rotation are zero.

3.8.3 Angular Deformation av au

In figure 3.11 (c) -and - are positive. da is positive while df3 is negative. This is ax ay

because the face is rotated clockwise. Thus, the rate of angular deformation is

parallel to z-axis

Similarly ex = ($ + g) parallel to x-axis

and parallel to y-axis

As stated earlier, these rates of angular deformations cause tangential stresses.

3.8.4 Volume Deformation The rates of linear deformation of each of the three sides of a cube [figure 3.11 (d)] of

au av aw sides 6x, 6y and 62 are +, - 4 y and --dz respectively. The increment in volume by

ax a y az

each of these deformation are % ~ 6 ~ 6 z , %6~& and %by6z. Therefore, the rate of ax ay a~

increase of entire volume per unit volume will be

This equation is called dilatation or div v 3 r v . V-1t will be proved later in this unit that for incompressible fluid flow, div ?= 0. This relation is also the elementary continuity equation.

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3.9 ACCELERATION : TOTAL ACCELERATION- LOCAL AND CONVECTIVE COMPONENTS OF ACCELERATION

To obtain equations of motion for a fluid flow we have to make use of Newton's second law of motion. This needs proper definition of acceleration in terms of time t and coordinates x, y, z of space.

Our knowledge of elementary mechanics tells us that velocity is the rate of change of displacement with time. Thus, one writes the velocity component say u as a function of x, y, z and t. This is in close agreement with Eulerian approach which is used in Fluid Mechanics. It is also a well known fact that any property of the fluid element which is a function of space and time will change as it moves from point to point due to (i) passage of time and (ii) variation in its position.

Let us suppose that the fluid element is at a position (x, y, z) at any time t. F(x, y, z, t) describes completely its position at that instant. After an increment of time 6t, the fluid element has reached a new position (x + u6t, y + v&, z + w6t). Thus, the new value of the function is the old value of the function plus the change due to time and space variations.

This can be rewritten as

This simplifies to

D In the above expression -stands for total derivative or substantial derivative.

Dt

Vectorically this equation can be expressed as

The first term on the right hand side of the equation gives rate of change of a quantity at a fixed point whereas the second term gives the rate of change of quantity at a fixed time t. We will come to this point a little later.

+ D v ' If the above function F represents velocity vector V (x, y, z, t) then -will indicate

D t acceleration that the fluid element undergoes as it passes a given point. Thus, the accceleration vector ;is

where

The three components o f acceleration vector are

DU au au au au a,, acceleration alongx-axis is - = - + u - + v - + w -

Dt at ax ay az

Fundamentals of Fluid Motion

DV a v av av av a,, acceleration alon y-axis is - = - + u - + v - + w -

Dt at ax ay az

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Fluid etatics and Kinematics aw aw aw

a, acceleration along z-axis is 5 = aW + u - + v - + w - ~t at ax ay az

au av aw In the above, the terms -, -, -represent "local accelerations" at a point where as the

at at at - + + + remaining terms V . v V or V grad V represent convective acceleration.

au - is the local acceleration i.e. the time derivative of u at a certain placeA(x?y,z) in the at

. field of flow. In steady flow, it is zero. The difference between substantial acceleration and local acceleration is "Convective Acceleration" so called because of the change in the velocity component due to convection movement of the fluid particle. This is generally not zero even in steady flow.

Further if we consider two-dimensional steady flow, the acceleration components a, and ay becomes

and

3.10 EQUATION OF CONTINUITY

Let us consider a rectangular element of fluid having lengths 6n, 6y and & parallel to the coordinate axes x,y,z. Let the velocity components at the centroid P(x, y, z) be u, v, w. Let p be the specific mass of fluid.

The rate at which the fluid is entering the surface ABCD is

is based on the assumption that the rate of change of mass

Similarly, the rate at which the fluid is leaving the surface EFGH is

Figure 3.12 : Mass flow in and out of the rectangular parallellopiped

Therefore, the net rate at which the fluid is gained between the two faces in the x direction is the difference between the rate at which the fluid is entering the surface ABCD and the rate at which the ,fluid is leaving the surface EFGH.

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Thus, the rate of gain of mass in x direction is Fundnmentnk of Fluid Motion

Similarly, the rates at which the fluid is gained between the faces along y and z axes will be

a - q p v ] & x 6y & and - --[pw]&x $ 6 ~ respectively. ay az

Thus, the rate at which the matter is gained in the element is the sum of the three gains, that is

a (pv) + - (pw) &x 6y 62

az a I The rate at which the mass is increasing is

a ( p &x 6y 6z) at

p&x6y6z represents the mass contained in the rectangular parallellopiped.

So, the rate at which the mass is increasing in the element must equal to the rate at which the matterr is gained within the fluid element.

sic av aw or * + u * + a + a + p - + - + - -0

at ax ay az [ax ay a z ]

These two equations (3.34 & 3.35) are known as Equation of Continuity.

Equation (3.34) is the general form of continuity equation while the equation (3.35) is the vectorial form of the continuity equation.

For incompressible fluid flow specific mass remains constant and the equation of continuity becomes

+ --+ V' V-0 ordiv V-0 (3.36)

Thus, for incompressible fluid flow the continuity equation is given by equation (3.36) or (3.37).

For any possible fluid motion, steady or unsteady, uniform or non-uniform, the equation of continuity must be satisfied.

Example 2 :

A fluid flow field is given by x 4 i? y2zJc (2ryz + y~)kT'~rove that it is a case of possible steady incompressible flow. Calculate the velocity and acceleration at the point (2, 1, 3).

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Fluid statics and Kinematics

Solution :

In the given flow field

u = &, v - yZz and w - - ( w z + y Z )

(i) For a case of possible steady incompressible flow, the elementary continuity equation must be satisfied.

Hence, the given flow field is a possible case of steady incompressible flow since it satisfies the continuity equation.

(ii) Velocity at (2, 1, 3)

v"-& G y 2 z j 5 ( - w z - y t )

-2'. 1 G l 2 - 3 - j < ( - 2 . 2 . 1 . 3 - 1 .3') k- 4 4 - 4i + 3 j - 21k-

Resultant velocity - 44' + 3' + (-21)' units - 416 + 9 + 441 units - units - 21.59 units

(iii) Acceleration at (2, 1, 3)

= 123 units -7' . . 2- 28i - 3j + 1 2 ~ -

Resultant acceleration - d(28)' + (- 3)' + (123)' units - 126.18 units

Example 3 :

Given below are two components of velokity of a 3-Dimensional steady incompressible fluid flow. Determine the third velocity component.

u - 1 + J + 2 , ~ - r y 2 - ~ 2 + x y

Solution :

Since, they are the velocity components of a 3-dimensional steady, incompressible fluid flow they must satisfy the continuity equation.

viz

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Integrating we get,

2 w = (- 3x2 - 2xyz + -) + constant of integration

3

Here, the constant of integration can not be a function of z. But, it can be a function of x and y.

is the missing velocity component.

SAQ 5 The following velocity components for steady, incompressible flow are

0 ) u-&-3y, v-x-2yandw-0 I (ii) u - P - x y + t , ~ = ~ - 4 r y + ~ ~ a n d w-2ry-yz+y2

:-I I

(iii) u - & + ? a n d v - - & y

Is the equation of continuity satisfied in these cases?

In this unit, we have learnt that

* There are so many ways the fluid flow can be classified. * In one-dimensional flow the dependent fluid variables are functions of only one

space coordinate. * In 2-dimensional and 3-dimensional flows they are functions of 2 and 3 space

coordinates respectively. * Eulerian approach to describe the flow field is more practical and hence used

extensively in fluid mechanics. * Streamline is a line tangent to which at every point gives the direction of flow. * Translation and Rotation of fluid element do not impose any stress on the system. * Angular deformation gives rise to tangential stresses. * Acceleration consists of local and convective components. * For any possible fluid motion, steady or unsteady, uniform or non~uniform, the

equation of continuity must be satisfied.

3.12 ANSWERS TO SAQs

SAQ 1

(a) - (iv)

(b) 4 (ii)

(c) 4 (iii)

( 4 - (9

Fundamentab of Ruid Motion

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Fluid statics .nd I

Kinenlmtia SAQ 2

(a) - (iii) because in one-dimensional flow only one velocity component in the direction of flow exists.

SAQ 3 (iv) by the definition of streamline.

SAQ 4 The differential equation of a streamline is

d x d z = I-

U V W

In this case it is - x 2y 3 - Z

1 Integrating - logx - -logy + A

2

Since, streamline is passing thtoughx - 1, y = 1

i.e. 0 - 0 +A i.e.A= 0

Considering the second differential equation

Integrating - logx - - log (3 - z) + B

Since streamline is passing through x = 1, z = 2, B = 0

x = 3 - z

is the equation of streamline passing through (1,1,2).

SAQ 5 -

(i) Equation of continuity for a 3-D steady, incompressible fluid flow is

i.e. 2 - 2 + 0 - 0. Hence, this satisfies the continuity equation.

(ii) 4x-y-4x+2y-y-0. Hence this satisfies the continuity equation.

(iii) This is a 2-D flow

au av This too satisfies the continuity equation. - + - = & - 4 ~ = 0 . ax ay

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UNIT 4 IDEAL FLUID FLOW

Structure

4.1 Introduction Objectives

4.2 Stream Function (2-Dimensional)

4.3 Velocity Potential

4.4 Properties of Stream Function and Velocity Potential

4.5 Rotational and Irrotational Flows

4.6 Circulation of Flow - Circulation and Vorticity

4.7 Laplace Equation * "

4.8 Vortex Flow - Forced Vortex and Free Vortex

4.8.1 Vortex Motion

4.9 Flow Net

4.1 0 Plotting of Flow Net-Methods

4.1 1 Uses of a Flow Net

4.1 2 Applications

(i) Flow Past a Circular Cylinder (ii) Flow Through a Sluice Outlet

4.13 Summary

4.14 Answers to SAQs

4.1 INTRODUCTION

In the previous unit, Fundamentals of fluid motion, you have studied different types of nuid flow, the concept of streamline acceleration and its components and derived the equation of continuity. In this unit we shall be dealing with the Ideal fluid flow and its related aspects

Ideal fluid flow is an important branch of fluid mechanics. Ideal fluid may be defind as that in which friction is absent, that is, its viscosity is zero. Thus, in ideal fluid tangential stresses or shear stresses are absent when fluid is in motion. Ideal fluid is an imaginary fluid conceived by mathematicians to simplify the problems of fluid motion. Ideal fluid flow has uniform flow velocity at a cross-section. It exhibits 'slip' condition at the fixed boundary whereas in real fluid flow, there is 'no slip' condition at the fixed boundary. That is, in real fluid flow if the boundary is at rest, the fluid particles are also at rest whereas in ideal fluid flow, fluid particles at the fixed solid boundary move. Hence shear stresses are absent in ideal fluid flow. Ideal fluid flow is also known as potential flow. Theory developed for ideal fluid flow can be applied to real fluid flow with certain degree of approximation.

Objectives

After studying this unit, you should be able to

* define stream function and velocity potential,

* ldentify whether a flow is rotational or irrotational,

* define and compute circulation and vorticity,

* derive Laplace equation from fundamentals,

* describe vortex flow and also forced vortex & free vortex, and

* define flownet and enumerate its uses and applications.

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Fluid Statics and Kinematics

4.2 STREAM FUNCTION (2-DIMENSIONAL)

The velocity components u and v are related to a scalar field function called stream function Q(x,y). This stream function is generally used for two-dimensional motions. It has the dimensions of volume per unit length per unit time.

Let P (xo ,yo) be a fixed point in the x - y plane which has unit thickness in z-direction and R(x, y) an arbitrary point in the same plane. Let there be two

I

arbitrary paths PQR and PSR between points P and R. I

Because of law of conservation of mass, no fluid is created or destroyed within the region bounded by these two paths.

Hence the flow rate entering across path PQR will be same as that of flow rate across PSR. The flux across PQR will be same as that across PSR. Since point P is fixed, flux is a function of position of R and time t. This flux is denoted by Q known as stream function and written as Q = ~ ( x , y , t)

Above fact is represented here to relate velocity components with stream function.

F i r e 4.1 Relationship between velocity components and strenm function

Let Q, and q2 be the neighbouring streamlines represented by ARB and CPD (figure 4.1). points R and P are points on streamlines having stream functions of q1 and and flux across PQR is q2 - ql = d q . Along streamline ARB or CPD, or any streamline the stream function remains constant and hence chl, = 0, because there cannot be any flow across it.

By law of conservation of mass,

flux across PQR = flux across RS + flux across PS

chl, udy - V ~ X

By rules of partial differentiation, we get

d q = $ . d x + $ . d y and I#= r $ . d r + r $ ~ - . d y , ~ (4.2)

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Idem1 Fluid Flow Co~par ing above equations (4.1) and (4.2), velocity components are defined as under:

u s w and v - - 2.9 ay ax

This is valid for both rotational and irrotational flow. The rate of flow 6Q between any two streamlines is defined as difference between the two stream functions

w = ' 4 2 - '41

The continuity equation for two-dimensional flow is given by

Substituting values of u and v from.equation (4.3), we obtain

This proves that existence of I# means a case of possible fluid flow. The flow may be rotational or irrotational. It shows that '4 is a continuous function of x and y and hence the order of differentiation can be interchanged.

Example 1 : If for a two-dimensional flow, the stream function is given by '4 = y ( 3 2 - y2), obtain the velocity components. What is the velocity at point (2, 1) ?

Solution :

and

w By definitions of velocity components in I#, I# = 32y - y3 , u = - = 3x2 - 3y2 ay

u atpoint ( 5 1 ) = (3 x 4) - (3 x 1) - 12 - 3 = 9

vat point(& 1 ) - - 6 x 2 x 1= - 12

Hence IVI = 6 7 = d ( 9 ) ' + ( - 1 2 ) ~

-m = 15

We now give you some exercises based on the concepts discussed so far.

SAQ 1 The stream function exists for

a) Two dimensional flow only.

b) Three dimensional flow.

c) Irrotational flow only.

d) Rotational flow only.

e) All types of flow.

f) Two-dimensional, rotational and irrotational flow. Select the correct answer from the above.

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Fluid Statics and Kinematics

SAQ 2

The x and y components of velocity in two-dimensional incompressible fluid flow are given by u = 2ry and v = a2 + x 2 - 3. Determine the expression for stream function.

4.3 VELOCITY POTENTIAL

Similar to stream function, there is another scalar field function $(x,y) called velocity potential. It is used to define velocity components in irrotational flow. The definition of velocity potential is applicable to ideal fluid flow only.

The velocity vector is expressed as -by 4 -b v 8 u + j v + k w ...( For 3-dimensional flow)

and 7 , 7, V - r u + J V ...( For Zdimensional flow)

It is also written as

<= grad 4 - $ 4 = (I-2 + j-2 + k - a ax ay az) +

Therefore u = 3 v = @ and W E

ax, ay ' 3 az

Hence velocity component is defined as a gradient of velocity potential in that direction. It is to be noted that for a velocity potential to exist, the flow has to be irrotational. This is proved in the following discussion vide equation (4.17).

If we substitute this definitions of velocity components in equation of continuity, we get

Therefore,

which is Laplace's equation in 4 . This is to say that if the flow is irrotational, it must satisfy Laplace Equation.

Example 2 :

Find the velocity components and resultant velocity for the velocity potential given by 4 = x2 - y2. Determine stream function also.

Solution :

4 - 2 - y 2

and

By definition

Therefore Q = 2xy + constant , and Q - 2yx + constant

Hence 1) - 2xy (constant is zero as at x = 0 and v = 0. u - 0. v = 0).

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Example 3 : Find the velocity potential, resultant velocity and its direction at point (2, 4) for the stream function represented by q -2 +#

Solution :

Given q = 2 + yZ

Therefore u a 3 = 2 y and v = - 3 s - 2 x ay ax

IVI = ~ = ~ - ~

At point (2, 4), velocity components are u = 2 x 4 = 8 units, and

i v = - 2 x 2 = - 4 units.

Resultant velocity =m= m= 2 x 4.472

and

that is tan (180 - 0) or tan (360 - 8) = - tan0

Therefore, resultant velocity makes an angle of (180 - 26" 34') - 153" 26' with x-axis,

j or (360 - 26" 34') = 333" 26' with x-axis.

Streamlines are concentric circles as q = 2 +y2.

Example 4 :

The velocity potential is given by $ -2 -#. Does this represent a possible flow field ? If it so, prove that the flow is irrotational.

Solution : To have a possible flow, it must satisfy continuity equation

Given : $ - 1 - y 2

Equation of continuity in Zdimensional is

a a Hence

- ( a ) + - ( - 2 y ) = 2 - 2 - 0 . ax JY

This proves that this is a possible flow field.

Flow to be irrotational, it should satisfy Laplace's equation

Given $ - 2 - y 2

*= 2r, and & = 2 " ax a 2

Similarly, *=- 2y and ay

Ideal Fluid Flow

This proves that flow is irrotational.

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Fluid Statics md Kinematics . 4.4 PROPERTIES OF STREAM FUNCTION AND

VELOCITY POTENTIAL

From the above discussion, we-may summarize the properties of stream function and potential function.

The properties of stream function (9) are:

(1) If stream function (9) exists, it is a possible case of fluid flow which may be rotational or irrotational. It exists for 2-Dimensional flows only.

(2) If stream function (9) satisfies the Laplace equation, it is a possible case of irrotational flow.

(3) If stream function (9) satisfies equation of continuity but does not satisfy Laplace equation, it is a case of rotational flow.

The properties of potential function (4) are :

(1) If velocity potential (4) exists, the flow should be irrotational. This means the velocity potential is applicable only for ideal fluid flow. It exists for 2D or 3D flows.

(2) If velocity potential (4) satisfies the Laplace equation, it is a case of possible steady incompressible irrotational flow.

(3) If Laplace equation is not satisfied by ,velocity potential, the flow is either rotational or non-existing.

All real fluid flows are rotational while ideal fluid flows are irrotational.

Example 5 :

For a two-dimensional flow, stream function is given by I# a 2xy, prove that the flow is irrotational.

Solution :

us*-2x and v--*=-2y. a~ ax

For x - y plane, vorticity is given by

Hence, flow is irrotational.

4.5 ROTATIONAL AND IRROTATIONAL FLOWS

The ideal fluid flow consists of two types differing from each other both physically and mathematically. They are rotational and irrotational flow. If the fluid particles within a flow have a rotation about any axis, the flow is said to be rotational. If fluid particles do not suffer rotation, the flow is an irrotational one. The non-uniform velocity distribution of real fluids close to the boundary causes particles to deform with a small degree of rotation whereas, the flow is considered irrotational if the velocity distribution is uniform across a section in a flow field.

Path of fluid particles

x

X

F~gwe 4.2 Irrotational Motion

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The Prandtl hypothesis (which provides an important link between ideal fluid flow and real fluid flow) states that for fluids of low viscosity (e.g. air, water), the effects of viscosity are appreciable only in a narrow region surrounding the fluid boundary. For incompressible flow situations, results of ideal fluid flow can be applied to real fluid flow to a satisfactory degree of approximation.

To understand rotational and irrotational flow, one has to understand the concepts of circulation and vorticity.

F i e 4.3 Rotational motion

Now would you like to try your hand at these exercises?

i SAQ 3 Tick (fi the correct answer or most appropriate response from among the alternatives given:

Irrotational flow is such that

a) Circulation is zero.

b) Stream function exists.

c) Vorticity is not zero.

d) Velocity potential satisfies Laplace equation.

SAQ 4 Following are the velocity components

a) u = 3x + 5y + 62, v - 5x - 3y, w = 5x + 5y + 5

Do these two cases represents irrotational flow? Determine velocity potentials.

4.6 C I R C U L A T ~ N OF FLOW: CIRCULATION AND VORTICITY

Circulation r (gamma) is defined as the line integral of a velocity around a closed curve in a fluid flow.

For a three-dimensional flow

-4 7. ZL = 4 (uuk + vdy + wdz) (4.6)

Ideal Fluid Flow

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Fluid Statica and Kinematics For a two-dimensional flow

/ Fim 4.4 I Circalation uoand a c a ~ e

Consider a rectangular element having sides 6u and 8y parallel to x and y axis. We may obtain circulation around any closed curve with the help of above equation (4.8). Convention is that anticlockwise direction of circulation is taken as positive.

ABCD represents rectangular fluid element. At centre 0, velocity components u and v are represented. At mid-points of the sides of the rectangle, corresponding velocities are written at points 1,2,3 and 4.

Hence circulation around an elementary rectangle is given by

I -- X

Figure 45 I CioLtiw in x - J p h e

av ti^ - u+aU.& ax- v - - . - ( ay 2 ) ( ax 2)t (4.8a)

On simplifying, the above equation reduced to av au

or lim - 6r I&-&- 6 r 6y-mO 6 x 8 ~ ax a~ t; (zeta) (4.9)

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which is the circulation in x -y plane and along an axis parallel to z-axis. The limiting value of circulation per unit area is known as Vorticity. Similarly, vorticity along axes parallel to x and y are

aw av au aw 5 (xi) --- - and q (eta) - - - - ay az az ax

The above equations are 3-vorticity components of vorticity parallel to x, y and z axis.

For a rotational flow, vorticity 5 (xi), q (eta) and c (zeta) are not equal to zero, whereas in irrotational flow these components are equal to zero.

1 Thus for irrotational flow these components

4

Mathematically, vorticity can be defined as curl of velocity vector V ; ---C ---C < - v x 7- curl v

where ---C---C I <miE+j';+fi-S

Also sometime vorticity is expressed by Greek letter SZ and

absolute value of vorticity is I L1 I - I F I - Example 6 :

Find the vorticity for the fluid motion having velocity V ~ S ---C---C

V.- i ( ~ A X Z ) + Y ~ A ( Z + ~ - I ) ]

where c is numerical constant. Solution :

For a given velocity, velocity components are u - 2A xz, v - 0 and w - A ( c 2 + g - 2 ) Now the components of vorticity are

Hence the flow is irrotational

Example 7 : The velocity components for a flow are given as u - 6y, v - 0, w - 0. Calculate vorticity.

Solution :

This flow is one-dimensional

The vorticity will be 5 - 0, q - 0 and 5 - - 6

Ideal Fluid Flow

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Fluid Statics and Kinematics

4.7 LAPLACE EQUATION

If you have followed the ideas introduced in this section, then you should be able to solve this exercise.

SAQ 5

Find the circulation around the square enclosed by the lines x - 1, y - 1 for a two-dimensional flow given by u - x + y, v - 2 - y at centre 0.

Thus for an incompressible, irrotational flow :- + +

By equation of continuity div V or V . V = 0

which gives the equation of continuity for three-dimensional fluid flow as

1 Y

As defined earlier substituting the values of u, v and w in equation (4.12), we get

( -1 , l )a

0

Above equation (4.13) is known as Laplace's equaiton. Hence, to obtain solution for any incompressible, irrotational flow, one has to solve Laplace's equation with the prescribed boundary conditions.

. b ( l , 1 )

X

In case of two-dimensional flow the equation (4.13) will be

( - 1 , -1) d c (1, - 1 )

Components of rotation as derived in earlier chapter are given by

aw about y-axis WY-.(z-z)

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about z-axis

For the flow to be irrotational, each component of rotational or vorticity should be zero.

For a two-dimensional flow in x - y plane,

o, = 0 and substituting the values of u and v in 9

Hence we get

Equation (4.16) is known as Laplace's equation for a two-dimensional flow.

Similarly by substituting the values of u and v in 4, in o, we get

Hence if a flow is imtational, a velocity potential must exist and it should satisfy Laplace's equation also. Further it proves that 4 is a continuous function of x and y. Therefore order of differentiation can be interchanged.

Here are some exercises for you.

SAQ 6

State whether the following statements are true or false. If false then correct it.

a) Potential function satisfies Laplace equation whereas stream function for rotational flow only satisfies the Laplace equation.

b) Laplace equation holds good only for imtational flow.

aV =* can be used for irrotational and rotational flow. ax ay

d) In case of real fluid, the flow imediately near the boundary is rotational one.

4.8 V O * X ~ O W - FORCED VORTEX ANDFREE VORTEX

4.8.1 Vortex Motion Flow having circular streamlines or flow along curved path of a rotating mass of a fluid is known as vortex motion. The vortex flow is of two types namely

(a) Forced vortex or Rotational flow

(b) Free vortex or Irrotational flow

The flow in free vortex is known as Rankine combined v0rtex.a~ it is a combination of above two types.

(a) Forced Vortex

In this type of vortex flow, application of external torque is required. When a container filled with liquid is rotated about its vertical axis, the liquid surface no longer remains horizontal and pressure variations occur due to centrifugal effects.

The vorticity of flow everywhere within the liquid is non-zero and finite. Hence the flow is called rotational . From elementary mechanics, it is known that the tangential velocity at radius r is

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Fluid Statics and Kinematics angular velocity (w) times radius r. If V is the tangential velocity and o (omega) is the angular velocity, then

V = m (4.18)

that is V a r Therefore velocity increases as we move away from the centre and streamlines are concentric circles.

Figure 4.6 illustrates the distinguishing characteristics of a forced-vortex or rotational vortex. As the streamlines are circular, the velocity vector being every where tangential has no radial component.

An expression for its stream function may be obtained as below for radial co-ordinates (r, 8)

Velocity ! distribution

i (a) Free Surface in a rotating Tank (b) Stnuallee pattern and velacity dislribution in a

horizontal plme

Flgarr 4.6 I Chumderistia, of a Forced-vortex

V - r o but

and radial component V, = *-0 r ae

Substituting these values and integrating,

we get

30 9 - -- 2

+ constant

Constant is zero as at r = 0, 9 - 0.

Rotational aspects can be visualized by dropping a small match stick denoted by AB. Numbers 1, 2, 3, 4 and 5 represents the various positions of match sticks as it moves along with the rotating liquid. The direction of AB changes its orientation from place to place which clearly indicates the rotation of the object about its mass centre. This illustrates that vorticity exists at every point indicating aspects of rotational flow.

(b) Free Vortex

The flow in a free vortex is a combination of rotational and irrotational flow and known as Rankine combined vortex. The flow within the core is rotational while outside the core, the flow is irrotational. Figure 4.7 shows the characteristics of a free-vortex or irrotational vortex flow. In free-vortex, no external torque is necessary to produce this motion. Examples of this are a cyclone, vortex in a river during flood at a deep pit, emptying of a sink etc.

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Idtd. Fluld Flow Torque T can be expressed as,

--C

where F i s position vector and M is momentum which can be written as 4 -D

M - m V --C

In above, v is velocity vector and m is the mass of fluid. Thus equation (4.23) becomes

Writing equation (4.24) in scaler form, we get

Free Surface at r = w

lnootional Flow

(b)Vdodty &hibution Figure 4.7 : Chmracttriaties of m Fret Vortex

If external torque is zero, equation (4.25) gives,

that is rV- C = constant (4.27)

This means that V increases as r decreases while in forced vortex V increases as r increases (V = no)

The circulation around a circular streamline is given by

- constant (4-28)

The circulation is constant for all the streamlines as it is independent of radial distance. The stream function for free vortex may be obtained as under.

and r (Gamma) = Vx Z m (circulation)

therefore

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Fluid Statics and Kinematics and

substituting these values,

--- I' log r + constant B 2.z

Let at r - ro,\C,-0,

r hence constant B - - 2.z logro

r '0 Now, \C, - - log- 2.z r

The circulation r which is constant for a vortex is knwon as vortex strength.

Irrotational aspects can be visualized by dropping small match stick AE (figure 4.6). Its positions is shown at various points as the liquid rotates. Its orientation remains the same with respect to its mass centre. There is no rotation. This indicates irrotational aspects of the flow.

4.9 FLOW NET

As stated earlier, along streamline the value of stream function is constant. Similarly if we draw lines having equal values of velocity potential, we get equipotential lines along which velocity potential remains constant that is d@ = 0.

A grid obtained by drawings series of equipotential lines and streamlines is known as Flow net. Flow net is one of the methods for solving and analyzing two-dimensional irrotational flow problems.

The following derivation proves that equipotential lines are normal to the stream lines.

Along a stream line, & - 0 and along an equipotential line, d$ - 0.

y~

Pi- 4.8 I Flow net for 2dimcndon.l

As \C, and 4 are functions of two independent variable x and y, i.e \C, = \C, (x, y)

and 4 - 44x9 Y) By rules of partial differentiation,

which gives

Similarly

Hence

4 = - for comtant *. a

dx w a y

- - for constant 4 way

WINDOWS
Comment on Text
dec-2007
WINDOWS
Note
Unmarked set by WINDOWS
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By definition,

and

Substituting these in -equations (4.30) and (4.31)

We obtain

and

* = 1 for constant U) dx u

* - - Y for constant 4 dx v

Now, * represents slopes of streamline and equipotential lines. dx

Therefore - (2 for U) - constant) x (2 for 4 = ...( 4.32)

which proves the fact that streamlines and equipotential lines are normal to each other. In other words, they intersect at right angles.

Flow net is a graphical solution for two-dimensional, irrotational flow. Combination of equipotential lines and streamlines give rise to small, tiny squares. There can be only one flow net for a given flow. With the help of flow net, pressure and velocity can be found. Spacing between streamlines and equipotential lines can be varied. To get accurate results, one must draw finer net work.

4.10 PLOTTING OF FLOW NET - METHODS

The plane ideal fluid flow pattern can be plotted with the help of followings techniques :

(1) Graphical ~ e t h o d

(2) Analytical Method

(3) Analogue Method

(4) Numerical Method

(5) Direct Method

(6) Hydraulic Models

(1) Graphical Method :- It is the simplest but it requires experience and skilled drawing. In graphical plotting, the properties of flow net should be kept in mind. (i) Equipotential lines should meet streamlines at right angles (ii) Rigid boundary is considered as one of the streamlines (iii)Equipotential line should meet rigid boundary orthogonally. (iv) Spacing of streamlines and equipotential lines in uniform flow should be

equal and should result in perfect squares. (v) Spacing of streamlines and equipotential line in non-uniform flow i.e in

100 V Conductors

--

Zem volt

(a) Streamlines

Ideal nuid Flow

Figarc 4.9 1 Ele&iul mdogy method

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Comment on Text
dec-2007
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converging or diverging boundaries should be done skillfully so that the resulting pattern will have near-squares

(vi) Diagonal lines drawn through the squares in uniform flow will result in perfect squares and near-squares in non-uniform flow.

(2) Anelytlcal Method :- It is also known as mathematical method. It requires the equations of stream function and velocity potential in terms of x and y. By assigning different constant values of 9 and $ and plotting them will give rise to flownet.

(3) Analogue Method :- If the characteristics of two o r more systems which are apparently different, can be expressed in identical mathematical fqrms, they are said to be analogous. The technique of electrical analogy is based on the similarity of flow of electric current through conductors in an electrical system on one hand and flow of water in a pipe in hydraulic system on other hand.

Heat flow is also analogous to water flow in idealized circumstances. Electrical potential V is identical to $ and field strength vector E stands for velocity field u. Discharge Q cooresponds to electric current i. Both the systems use Laplace equation for their solution.

(4) Numerical Method :- It requires to express Laplace's equation in finitedifference form and solving for the distribution of $ numerically by trial and error.

(5) Direct Method :- It involves velocity measurements and plotting velocity profile by actual measurements at number of axial stations. From velocity pyofiles, points of equal discharges along the length are obtained and they are joined to get a streamline.

(6) Hydraulic Models :- Hele-Shaw apparatus is used in which by injecting a dye streamlines can be traced. The flow net is completed by drawing equipotential lines on the streamlines obtained as above.

4.11 USES OF A FLOW NET

Some of the important uses of flow net analysis are as under :

(1) For given boundaries, the velocity and pressure can be obtained if velocity and pressure at any reference section are known.

Figure 4.10 : Flow net for a now past a cylinder @~t.tion.l flow)

(2) Seepage loss in earth dam and canals can be estimated.

(3) Uplift pressure below the weir floor can be estimated.

(4) Streamlining of the outlet's shape can be done.

4.12 APPLICATIONS

(i) Flow Past a Circular Cylinder :- On the basis of ideal fluid flow theory, when a cylinder is placed in a free stream or uniform flow, it should experience no drag force. According to D'Alembert, be discovered that if a cylinder or any other symmetrical body is placed in a real fluid, it experiences a drag force. This paradox is called D9Alembert Paradox.

WINDOWS
Comment on Text
dec.2007
WINDOWS
Comment on Text
dec.2007
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lded Fluid Flow In actual flow, the cylinder experience viscous effects of the fluid particularly in the region close to the surface of the body. The flow pattern in the vicinity of the cylinder boundary deviates appreciably from the pattern of ideal fluid flow analysis.

Figure 4.11 : Actual flow pattern for now overs cylinder (Red fluid flow)

The pressure distribution on the surface is also not the same that is developed by ideal fluid flow theory. Figures (4.10) and (4.11) show this difference. The flow separates at points P and Q and vortex rings are formed on the downstream. The pressure distribution mainly deviates in the downstream

I region. (ii) Flow through a sluice outlet

Figure 4.12 shows a flow underneath a sluice gate. Here channel bottom, sluice gate and free surface represent boundary conditions in drawing flow net

Figure 4.12 : Flow through sluice outlet

4.13 SUMMARY

To sum up what has been done in this unit

* Stream function and velocity potential are important scalar field functions to describe fluid flow

* Velocity potential exists for irrotational flow only * Solution of Laplace equation will give $-lines and v-lines (equipotential lines

and streamlines) * Flow net is a graphical solution of Laplace equation and can be drawn for

irrotational flow * Rotational and Irrotational aspects are the important phenomena of ideal fluid

flow.

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Fluid Stmtics m d Kincmmtics

4.14 A N S W W S TO SAQs

SAQ 1

(4

SAQ 2

u - 9 - b y (given) ay

Integrating equation (i) with respect to y.

And

Integrating equation (iii) with respect to x,

2 9- -a%--+x?+cz+f (y) 3

Comparing equations (ii) and (iv),

(iii)

Hence 2

f (x) = - a2x - - 3

f b ) = O

C 1 = C z = ~ a s u - O w h e n x = O a n d y = O b u t v = a 2 w h e n x - O a n d y - 0

Hence 3

v=-a2x- -+$+c 3

SAQ 3

(a) and ( 4

SAQ 4

(a) For condition of irrotationality, curl V-must be zero

i.e.

7 1- k- 4 - -

curl V - V x V - a - a - a - ax ay az

1

curl F- f [ L ( S x + h + 5 ) - $ h - 3 y ) a~ a I

= 5 r+ J7

Hence the flow is rotational and velocity potentials do not exist for rotational -

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(b) For irrotational flow, curl V-should be zero

which is not zero. Therefore the flow is rotational. Hence velacity potentials not exist.

t SAQ 5

1 . r - circulation- f V< df

Putting the values of x and y at proper places and using equation (4.8a) for integration.

1 - 1 - 1 1

We get r = I ( x + i ) d r + l ( l -y)dy + s ( x - 1)dr + j ( l -y)dy - 1 1 1 - 1

SAQ 6

(a) False. Potential function satisfies Laplace equation whereas stream function for irrotational flow only satisfies in the Laplace equation.

(b) True

au (') 2- - can be used only for irrotational flow. ax ay

(d) True

Ideal Fluid Flow

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Notes

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UNIT 5 DIMENSIONAL ANALYSIS AND SIMILITUDE

Structure 5.1 Introduction

Ohjectlves

5.2 Fundamental Dimensions

5.3 Physical Quantity and Dimensions

5.4 Dimensional Homogeneity

5.5 Non-Homogeneous Equations

5.6 Dimensional Homogeneity in Functions of Variables 5.6.1 Summation 5.6.2 Product

5.7 Non-dimensional Parameters

5.8 Complete Set of Dimensionless Products

5.9 n-theorem 5.9.1 Significance of the x-theorem

5.10 Dimensional Analysis: Choice of Variables 5.10.1 Criticism of Dimensional Analysis

5.1 1 Determination of Dimensionless Parameters : Rayleigh Method

5.12 Method of Repeating Variables

5.13 Examples in Dimensional Analysis 5.13.1 Rise in Capillary Tuhr 5.13.2 Head Characteristics of Pump 5.13.3 Drag on a Ship 5.13.4 Fall Velocity of a Sphere 5.13.5 Discharge over a Sharpedged Weir 5.13.6 Velocity in an Open Channel 5.13.7 Pipe Orifice

5.13.8 (l'elerity of a Gravity Wave

5.14 Model Similitude

5.15 Scale Models in Submerged Flows

5.1 6 Free S urlace Problems

5.17 ~uriace Tension

5.18 Summary

5.19 AnswerstoSAQs

5.1 INTRODUCTION

The gelieral equations of motion together with the equations of state and continuity presented in the earlier and ensuing units form the basis for investigation of all tlow phenomena. However, application of these basic principles to solve specific problems have not always been successful due to inatlicmatical difficulties encountered in finding solutions to nonlinear differential equations qatisfying complex boundary conditions. Only the simplest of problems have been amenable to rigorous analytical solutions. A few others have found approximate solutions following simplifying (but justifiable in a certain range of variables) assumptions to some of the terms in the equations of motion. Yet another class of practical solutions have arisen based on experimental information and hypothesis and intuitive understanding of the physical nature of the flow phenomena. In recent years, with the advent of large computers, numerical solutions have been possible to a large number of complex problems using finite difference, finite elementhoundary element techniques.

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IDIpuk.-I In thi conce exha1 suffic amon vari01 exper analy

Objc After

In Ne expre

wherc itself mass of for founc dime] Force dime] them heat 1 funda probl electr entiti1 dime] chose usefu when

The fc c o r n sys ter (or H

s unit, we shall lay emphasis on the class of solutions that arise out of physical :pts guided by past experience and supported by experimental evidence. An rstive experimental programme guided by dimensional analysis often provides :ient insight into many complex flow problems to establish functional relationships g the variables governing the flow phenomena. Empirical constants linking the us significant dimensionless parameters can be determined by both laboratory iments and field observations. We shall now discuss the principles of dimensional sis and its application to model similitude later.

I ectives studying this unit you should be able to I

* identify fundamental dimensions involved in the definition of physical quantities and to express physical quantities as powers of fundamental dimensions,

* convert the magnitude of a physical quantity from one unit of measurement to another,

* verify the dimensional homogeneity of theoretical and empirical equations and to render non-homogeneous equations into dimensionally homogeneous ones,

* identify variables governing a flow phenomena and establish a reduced number of non-dimensional n-parameters,

* learn the significance of inajor n-parameters such as Euler, Reynolds, Froude, Weber and Mach numbers, and ,

* study laws of similitude and plan experiments and present experimental results in a generalised form.

FUNDAMENTAL DIMENSIONS

wtonian Mechanics, the law relating force to rate of change of momentum is ssed as

: the constant of proportionality has been universally accepted as unity. This in has defined relationships among dimensional magnijudes contained in force, and acceleration. For example, the weight of unit mdss must be equal to g units .ce. As such, in Newtonian Mechanics, the fundamental dimensions generally I convenient for use are mass MI, length [L], and time [m. Then, the lsions of force can be expressed as [ M L T - ~ 1. However, we may have chosen : [F], length [L] and time [TI as fundamental dimensions in whichyase the nsion of mass would have been [FL-' T~ 1. In problems related t 6 iodynamics, another dimension for heat [HI is required and the dimension for

. Alternatively, it is more convenient to use heat [HI as the then the corresponding dimension for heat is [FL]. In

ems relating to flow of electricity through conductors, a fresh dimension of .ic charge may be used. These dimensions mentioned above in defining physical es are not the only ones employed and they may be varied. Basically, the nsions chosen must be independent of each other. Whatever be the dimensions in, they tell us the relative nature of physical quantities. They serve also a very 1 purpose in determining the variation of numerical values of physical quantities basic units of measurement are changed.

ollowing table (Table 5.1) illustrates the dimensions contained in several quantities ionly met with in problems related to fluid mechanics. Though the M - L - T - 8 n is generally employed in flow problems, the table describes the F - L - T - 8 [-L-T-8)systems also.

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Table 5.1 : Dimensions of Physical Quantities Dimensional Analysis

and Similitude

Dimensions in M-L-T-8 Dimensions in F-L-T-81 H-L-T-8

Mass

Mass density, p

Dynamic viscosity, p

Kinematic viscosity, v

Surface tension, a

Bulk modulus, K

Modulus of elasticity,E 1 Force

Pressure intensity, Stress

Rate of angular deformation, Angular velocity o

Velocity

Discharge

Acceleration

Momentum (angular)

Momentum (linear)

Momentum or Torque

Work,Energy,Heat

Power

Enthalpy,Quantity of heat

Specific heat

Thermal conductivity

Entropy

[LT -'I [LT -'I [L3 T-' ] [L3T -I]

[LT [LT

[ML~T -'I [KTI

[MLT -'I [F T]

[ML~T -7 [-I

[ML~T -2] [=I

[MLZT 41 [FLT -'I

[MLZT -21 [%I or CHI

[L? -2 8- ] [L2T -' 8-'1

[MLT" 8- '1 [ lT18 - '1 or [HL-'T-' 8-'1

[ M L ~ T - ~ ~ - 'I [FLe- '1 or [He- '1

5.3 PHYSICAL QUANTITY AND DIMENSIONS

We start with a general statement:

Bvby physical quantity can be expressed as products of powers of fundamental dimensions.

Suppose a physical quantity designated as u has dimensions M, L, T and 8 contained in it, then the above statement would mean

For example, if the physical quantity u is entropy then

a = l , b = 2 , c = - 2 andd=-1 sothat,

u (entropy) = w' L2 T ' 0- '1

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The proof for the validity of this statement can be found in advanced textbooks on dimensional analysis. The application of the rule will be highlighted in the ensuing sections.

One of the important applications js found in the establishment of the general law for transformation of the magnitude of a physical quantity from one unit of measurement to another. The physical quantity u is given by

u = M ~ L ~ T ~ B ~ (5.3)

in any one standard set of units of measurement such as the C.G.S or M.K.S. or F.P.S. system. Let the new unit of measurement of mass be a times the old unit, that of L be P times corresponding previous'unit and so on. The magnitude of IZ in the new system is given by

M L T B ii = (.J(jiJ(;)[xJ

Example 5.1 :

As an illustration, if the dynamic viscosity of a fluid at a given temperature (say 15°C) is given as 0.766 x 10-"b/ft.sec in British units, find the corresponding value in the C.G.S system

Solution :

We have the relationships

The unit of time is the same in both systems and hence y = 1. Dynamic viscosity has the dimensions [ML-'T -'I and therefore we have a = 1, b = -1, c = -1, d = 0. The new value of p in C.G.S. units will be

' - p = - 453'6 x 0.766 x = 1.14 x lo-' gm 1 crn sec. 30.48

SAQ 1

(a) If [F-L-TJ are used as fundamental dimensions, what is the dimension of mass [MI.

(b) What is the dimension of quantity of heat H, in tenns of [M-L-TJ and [F-L-TI systems.

(c) Convert the force expressed in F P S units - 10 1bflsq.ft. into ~ l r n ~ .

(d) Show that 1 poise (1 dyne - sec/cm2) = 2.09 x 10" lbf - sec/ft2.

(e) . Having shown that 1 lbf = 4.448 N, find the surface tension values of mercury in contact with air and water in Nlm, if their corresponding values are given as 0.0324 lbflft and 0.027 lbflft respectively.

, (f) A 20 lb mass (slug) is moving at a velocity of 120 milesihour. Express the kinetic energy in terms Joules (N-m)

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5.4 DIMENSIONAL HOMOGENEITY D h d d AMlyS&I

and Sbnilitade

It has been seen that every physical quantity has dimensions and magnitudes expressed in certain units of measurement. Hence any fundamental relationships between two or more of these physical quantities necessarily must show equivalence in both dimensions and magnitudes. More explicitly, we may state that the two sides of an equation must not only balance numerically but also in dimensions. As a corollary, it follows that if certain numerical constants appear in a dimensionally homogeneous equation, the value of the constant should remain unaltered even if the system of units of measurement are changed.

It is known that the stagnation pressure p, at the nose of a cylinder placed in a uniform flow of velocity Uo is given by the expression

p being the density of fluid. It is seen that this equation is dimensionally homogeneous; the dimensions of the two sides are identical

The numerical constant, that appears on the right hand side of the above equation

remains unaltered of units of measurement employed.

5.5 NON-HOMOGENEOUS EQUATIONS

Very often, scientists and engineers formulate empirical relationships among variables governing a phenomena based on experiments, field observations and intuitive reasoning. Such empirical relationships may not be dimensionally homogeneous. We may modify these empirical non-homogeneous relationships in the light of the requirements of dimensional homogeneity. We may discuss one such formula in detail. The velocity in open channel is given by the Manning's equation in metric units.

where n is a resistance factor dependent on the type of wetted surface (perimeter) of sides and bottom of the channel, R is the hydraulic radius defined as the ratio of area to the perimeter (A / P ) having the dimension of length [L] , S the slope of the channel is dimensionsless. Thus if n is treated as a numerical constant, the equation does not satisfy dimensional homogeneity

The factor n being dependent on height, shape and spacing of roughness elements can at best have the dimension of length. Hence we conclude that the constant 1.0 on the right hand side of the equation must have dimensions of length and time. A critical appraisal of the equation reveals that the flow in the open channel occurs by force of gravity and hence acceleration due to gravity must appear in the equation (5.6). By inspection, it is found that the factor G m u s t appear on the right hand side to balance the dimension of time. We now modify equation (5.6) as

cdg R 2 / 3 1 / 2 v=- S , Cis dimensionless n

. The dimensions of two sides of equation (5.6a) are :

We qdiclude that the factor n must necessarily have the dimension of [L1I6] to make the e@ation dimensionally homogeneous. Similarly many other non-homogeneous equations may be appropriately interpreted or evaluated afresh and rendered dimensionally homogeneous. Further it will help us to evaluate the magnitude of empirical constants, when units of measurements are altered. For example, we may now convert the mannings

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equation (5.6) in metric units to the FPS system. We had found that the dimension of the

or [L' l 3 T' ''1. Since the unit of time remains unaltered in both . ,.,

systems and 1 m = 3.28 ft, the corresponding constant in FPS units is (3.281~' = 1.486 Hence in foot-second units, Manning's equation becomes

The value of n is considered to be having the same values in both metric and British systems and de ads on the characteristics of boundary roughness. Further discussion on 7 the value of n may be found in a later unit on open channels.

SAQ 2

(a) Chezy's empirical formula for avera e velocity V in a steady uniform flow in open channels is given as V = C 5- RS, where R is the hydraulic radius (areatperimeter = A IP) and S is the slope of the channel bottom

(i) Is the equation dimensionally homogeneous ? If not find the dimension of C . Remember that free surface flows are governed by gravitational force.

(ii) If V = 75 in MKS units; find the equivalent expression in FPS units.

(b) Another empirical formula for velocity in open channels is due to Bazin :

Comment on the dimensions of a and m.

(c) Kennedy's equation relating mean velocity V to depth of flow in open channels in alluvial soil in FPS units is V = 0.84 y:M. Determine the

dimension of the coefficient (0.84) and modify it for use in MKS units.

(d) The perimeter P of an alluvial channel in equilibrium carrying a discharge Q is given by the empirical equation P = 2.67 @in FPS units. Find the corresponding expression in SI units.

5.6 DIMENSIONAL HOMOGENETY IN FUNCTIONS OF VARIABLES

Let the functional relationships among certain quantities u,ul,~,u3,. . .,u, be expressed as

Here u is taken as the dependent variable which is a function of the independent variables

By virtue of the general law stated in section 5.3, the quantities may be expressed as powers of the dimensions contained in them.

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5.6.1 Summation If the functional relationship is of the summation type (algebraic addition or subtraction).

By dimensional homogeneity, we observe that (equating like powers of dimensions M, L, T, 0 and etc)

and etc.

Therefore, each quantity must have the same dimension.

5.6.2 Product If the functional relationship (5.7) is in the form of a product

u =.ulml uZm2 . . . unmn (5.12)

- - IM(n1 ml + 9 + ... + a , %) ~ ( b ~ ml + b2 m2 + .... + bn m,) T (cl ml + c2 m2 ... en I,,) O(dl ml + d2 m2 +... + d n m n)]

Equating the powers of like dimensions, we get

M : alm,+a,,m,+ ...+ a,m, = a

L : blm, + b2m, + . . . + b,ni, = b

and etc.

5.7 NON DIMENSIONAL PARAMETERS

A special case of products given in Sectioq 5.6.2 arises when u is a nondimensional product of the variables u1,%, ..., u,. Then the powers of all dimensions contained in u are zero, viz., a = b = c = d = 0 with the regult that equation (5.14) can be written as

I@ : alml +%& ... + a , m , , = ~

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While Ulere are ti u~lknowns in t t ~ , . 1 1 1 ~ ~ ...- t l~ , , i l l equatio~is (5.15) there are only as lnruly equalions ;l$ die number of dimensions contained in the physical clua~itities u. u,. u, .... rr, , . For example, if there are only three dimensions M. L, and T (;IS is thc case in most problems in lluid ~necliaiics), Uiere will he only rlirce equations. Thus if the number of unknowns is more than the nu~ilber of lineiuly independent equations (say r ) , then a unique solutio~i f(;r the unknowns is not possible. Then, the solution to any r ~iunlber of unknowns nlay be expressed in ternis of the otller ( n - r-) unknowns. The number of linearly independent eclualions is equal to the rank of the dimnens~onal matrix formcd by the cocffic~ents 01 m , , I/!,, ..., nil,

( I , (I? (1; . . . (4, h, t)? hl . . . t),, C ,

. ' " Cll

(1, (1, ( I 3 . . . (Il1 (5.16)

The rank of a matrix is tlle Iiigllcst order of the dclermia;ul( (that ciul be for~iicd from the above matrix) whosc value is not zero. It is readily seen thal the liigliest order of Ule deternlinanl can not exceed llic nunitwr of ccluiltions p. The riuik of the matrix ca i be equal t o y or less thiul it (1. I p) . Barring a few cxccptions, the rank 01' Ulc matrix is equal to the number of di~nensions, but it is alwi~ys wise to deter~iline tllc rank of the ~nalrix before atlenlpling to solve 1- ualaiowns In rerlils 01' tlic reniaining ( n - r) unknowns.

We can assign arbitrary but lineilrly independent values to (n - r ) unknowns and obla~n solutions for the renli~ini~lg I . unknowns. The set of di~nensionlcss products so formcd which are independent of each other is crlllcd it complete set. I 5.8 COMPLETE SET OF DIMENSIONLESS PRODUCTS

The procedure of obtaining non-dimcnsional products is best illustrated by al illustrative example. Lel it be required to find a conlplele sel of dimensiorlless products of the following variables.

Force F, length L, velocity V, dc~lsily p, viscosity u , gravitational acceleration g, velocity of sound wave c iuld surl'ace tension o

= F lrrl L~II? v 111, ,,111~ p1115 Il l , , Ill7 I l l X (5.17) 1 I f

Substituting the di~llcrls~o~ls in die pllysical cluantitics I

Should u be dimensionless M () L (' T ". on cqu;l,fing the like powers of the dimensions on the two sides of the preceding ecluatio~l

M ~ ) : m i + )I /4 + I / l 5 + N18 = ( I i L" : t11, + tll? + I l l 3 - 3tt14 - tIl5 + W16 + Yf17 = 0

We have eight unknowns and three ecluatioris. Ille di~nensional mtlrix formed by the coefficients nl is

1 0 0 1 1 0 0 1 1 I 1 - 3 -1 1 I 0

- 2 0 - 1 0 -1 - 2 - 1 - 2 (5.20)

The rank of the n ~ ~ t r i x is found to tw 3, same as the number of dimensioils or number of rows in the matrix. Hence the solutioi~ to any three of the unknowns can be expressed in terms of 8 - 3 = 5 sets of assigned values to the other u Let us arbitrarily choose ~ 1 , . nl3, 1 1 1 ~ as unknowns and solve them in lerms

N L , n15, tt16, nl7 m d

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We now assign (n - r ) = 5 independent sels of values for m,, t115, m,, 111, and nl,.

We assign the following convenient 5 sets of values to rcduce 'uilhmetical work.

3rd set 0 0 1 0 -2 0

Assigned

~a lu ' c s of till n15 ' 4 , ~ 1 , tn,

1st set 1 0 0 0 0

4th set 0 o o I 0 0 - 1 0

Delerinined from equation (5.18)

t1l2 "13 "14

-2 -2 -1

5Lh set 0 0 0 0 1 -1 -2 -1

2nd sel 0 1 0 0 0 -1 - 1 - 1

irom the first set

The second set yiclds F () L- ' V- ' 1 0 . 0 0 L p - ' p g c. 0 = PVL

Thc third set gives L V - ~ ,,u co 0o = &, v-

F' Lo V - ' C The fourth sct yields ,,o cl 0~ = - v

FO L - 1 V - 2 0 The fifth set yields ,,-I co = - p v 7 L

The complete set of non-dimensional parameters resulting in thc preceding assigned values is

If wc had assigned the values of m, = - 1 in the 2nd set: nl, = - 1 in the third set: nt, = - 1 in the fourth set and m, = - 1 in the fifth set and thc othcr assigned values remaining unaltered, the complete set would have been

llc last four parameters are Thesc parameters

are known as the

of powers of the 5 non-dimcrisional parameters already found. The basic complete set for the chosen eight physical quantities (F, L, V , p, p, g, c, c ) are the dimensionless numbers designated as

E, Re, F, M, W. r

SAQ 3

(a) Conlbine the following into dimensionless groups

(ii) p , V , E (modulus of elasticity)

(iii) F , p , g , T '

(iv) frequency n , V , D , p , p

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(vi) Cp (specific heat at constant pressure), F , k (thermal conductivity)

(vii) Shear strees T, specific weighty, hydraulic radius R , slope S.

(b) Whal is the rank of the following matrices.

(1) 1 0 0 (ii) 0 0 1 (iii) 0 1 1 1 1 1 1 1 -2 1 -2 0

-2 0 -1 0 0 -2 0 -2 -2 I

5.9 7C - THEOREM

The discussion presenled in thc previous Sections 5.3 lo 5.8 form the basis for an importar~t conclusion know11 as the n-iheorcnl (soinclimes called Buckingham n-Ll~eorem) in diillei~sio~~al analysis. The gencral statement of the n-theorem is:

If n physical quanlities are rccluired to describe a physical phenomena and if these quantities have y din~ensions contai~~ed in them, tllc relationships can be reduced to one 1

comprising of (n - r) non-dimensional products, r I p being the rank of the n x p matclx.

From the previous exainple in Section 5.5, if a physical phenomena is governed by 8 I

variables sucl~ as F, L, V, p, F g, c, and o. llle relalionship among the variables can be expressed in the functional l'or111:

.f (F, L, V. p. F, R, c, 0) = 0 (5.24)

These variables have 3 dii ensi ions among them and the dimensional matrix is

which is the matcix given in equation (5.20) formed of the coefficients in equation (5.19). The llumber of dimellsions is 3 'and the rank of the above matrix is also 3. Hence the above functional form of dimensionill variables in equation (5.24) can be reduced to one of 5 non-dimensional parameters

9 (nI,n2,n39n4.ns) = 0 (5.25)

or more generally 9 (nl,n2,n3,..., - = 0 (5.25a)

where n1,n2,. . . ,n,, - , are non-dimensional parameters in the cwmplete set for grouping of variables as shown in Section 5.8.711e preceding functional (5.24) can be now written as

The proof of the n-theorem is beyond the scope of the present treatment and can be found in references [I], [2] and [3]. Here, we shall confine ourselves to the application of the IT- theorem.

Once we h u e obtained a set of non-dimensional parameters as in equation (5.25), we can derive combinations of the basic n - paraineterss already derived, provided the number of

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n- parilnielers remains the sanie. For cx:uiiple. anorlier perniissiblc combin;ttion of equation (5.26) is

It may be verified thal the n - parameters still remain indepcndc~~l. Also. we lnay drop o ~ i c or more of the n - parameters, if we know from prior knowletlgc or pliysical understanding of Lhc problem thal these piuamelcrs have negligible influence o n Lhc flow prnble~il under consideration.

5.9.1 Significance of the n-theorem The inlportitncc of the rcquircnlcnls of dinlensional Iioniogencity iuid tlic applicalio~i of tlic x-tlicorcni sliould not be undcresti~natcd

(i t ) Dinlcnsional hooiogeneity liclps 11s lo baliulce the tlinicns~ons of empirlci~l equaliolls ant1 check analyl~cally derlvcd ones for homogeneity.

( b ) It helps us to detcrmlne the dinlensions. iT any. of cnlpiric;~l conslimls.

(c) It helps us to find the value of cnipirical co~istiulls whcli units o f ~ i l ~ i t s u r c ~ n ~ ~ i t itrc a1 tcred.

Tlie role of the n-tlleorcni is more extensive.

(a) When llle nunibcr of variiibles governing tlie phenolne~ia is large, 11 is difficult to correliltc the cflcct of each one of these variables on Uic depclident variable. Hence b e n-theorem is of great Imporlalice in as nlucli as it groups Lliesc variitblcs into non- dimensional paranlcters and reduces tlie nunibcr of vi~i;tblcs by Lhc cxtelit of the rank of the dimensional matrix involvetl (generally equal to tlic nunibcr din~ensions), Correlatio~i among the rcduceti ~iu~iiber oS variables becomes eilslcr and more meanaiglul.

(b) Tlie di~nc~isitriless parameters arc generally ratios of two hrces such as

Reynolds nun1I)er : Ratio of inertial force to viscous force.

Froude number : Ratio of inertial force Lo gravitational forcc.

Weber number : Ratio of inertial i r c c Lo surlice tension force.

Thc llow pattern remains qnchruiged if Uiese non-dilnensional numbers remain the same despite the fact that indk@ual values ol' viuiablcs may cli;mgc. For example, even if viscosity cliiuiges, velocity, daisity, and the linear dimension involved in the

Rey~iolds number fi niay be so cllosen tliat tlie Reynolds nunikr renlai~is tlie CL

sanle. Thus two difkren( lluids lnay liavc the sanie tlow pattern. Alternatively Ulc linear dimensions can be altered as is dolie 111 phys~citl niodclling i111d Ule other variables changed to keep Uie Reynolds, Froude, Webcr and otlicr non- dimensional parameters unaltered. These n-paraluelers are therefore, capitblc of wider and more general interprctatioli than is possible with numerous dinlensional variables.

( c ) Tliese n - paritnieters are non-di~nensional affording us the l'acility of using tlie salllc graphical or aialyt~cal relationsh~ps regardless of units of mcasurenient.

(d) The grouping of these variables into a fewer nun~bcr oinon-di~ilcnsio~ial parameters acts as a guide to my experimental programme, thc ilinl o C which is to eslablisli fundamental relationship among tliese significimt n-pilrametcrs and supplerne~lt it by experimentally determined coefficients.

(e) A distinctive advantage of the n-theorem is found in model studies. Expcrinients in the rnodel are carried out at identical values of Lhe govcnii~ig or sigriificant X-parameters as obtained in the prototype. Since the flow processes are si~iiilar at identical values of n-parameters, results obtained from models can he scalcd up to prototype values. The laws of ~~ iode l sinlilitude will be amplified ill a later section.

5.10 DIMENSIONAL ANALYSIS : CHOICE OF VARIABLES

The first step in dimensioilal analysis is tlie choice of governing vnriahles relev;ult to the problem under investigation. This task is 1101 an easy one and is wrought with pitfalls. The choice of the dependent variables (that is tlie unknown variable Uiat we wish to find out) is not a difficult one-it comes specified with ttie probleni land how it is posed. Having

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~ k l d ~&r"der-~ chosen the dependent variable, we proceed to test out other variables (which are independent of each other) that are likely to have a significant effect or influence on the dependent variable. The following precautions are taken in clioosing the independent variables.

(a) Care should be taken to make a complele list of all the variables. To assist in the choice of variables, we nlay classify the v;uiables into

(i) Kinetic variables such as velocity

(ii) Dynamic variables such as force, i~cceleratioli 1. (iii) Geometric variables such as diameter of a pipe, size and shape of inunersed

bodies, length, height and shape of weir ctc.

(iv) Fluid properties such as density, viscosity, surface tension, bulk modulus, vapour pressure.

(b) We should not onul any ilnportanl v;dable by oversight or wrong interpretation. For example, elastic properties of materials, gravitational acceleration though they appear I

to be numerical conslmts, yet they arc dynanlic variables intluencing the flow phenomena and should not be ignored. Omission of an importanl varaible will exclude an inlporla~il n-parameter.

(c) We should not include unnecessary or unconnected variables n.ot related to the problem which will only confuse the issue. Extraneous values which have nothing to do with the problem sllould be avoided. For example, surface tension has no role to play in a pressure flow in closed conduit tlowing full.

(d) We should not chose two or more variables dependent on each other. for example if discharge Q and area of tlow A are chosen, then velocity should not be chosen as yet another var~ablc. s~nce V is dependent on Q and A: Q = VA . 7'he choice of right vdables comes with experience and inluitive insight into the problem. Dimensional A~ialysis is a powerful tool in the hands of an experienced investigator.

5.10.1 Criticisms of Dimensional Analysis (a) The main criticism against di~llensional luialysis is that, it does not give us any

indication aboul tlie choice o f v6uiables.

(b) It does no1 give an explicit functional relaliol~ship among the n-parameters.

(c) It docs not readily supply tlie numerical constanls that appear in the functional relatio~iships but are to he found experi~nentally.

(d) Dimensioaal analysis ~ i ~ . ~ s l bc supported by experimental investigation to obtain graphical or analytical relalionships among the n-parameters.

(e) Physical understaiding of the problem and experience of the investigator weighs heilvily in the application of dimensional ruialysis and inlerpretation 6f results.

5.11 DETERMINATION OF DIMENSIONLESS PARAMETERS : RAYLEIGH METHOD

Once the variables have been chosen, the grouping of these varaibles into (n - r) dimensio~lless parameters can be done either by the Rayleigh method or by the method of repeating variables. The Rayleigh melhod has already been described in Section 5.8. We shall demonstrate its application by aoolller cxa~nplc.

Let it be required to find the pressure loss in a circular conduit in which the flow is steady 1 and uniform. The pressure loss Ap, over a lengtli of pipe L, is tlie dependent dyna~ilic I variable. The i~idepe~ldcnt variables are: I Kinematic : Velocity (average velocity V is chosen) i Geometric : Dianietcr of pipe I ) , Length of pipc L, roughness of boundary

symbolized by roughness height, k. It represents dynanuc boundary resistance to How.

Fluid ! properties : De~~sity p and dynamic viscosity p of lluid at tlie given constant temperature.

The accel~ration clue to gravity does uot come inlo the picture directly as the flow is

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determination of pressure intensity y at any point. The functional relationship among the DLrko3-* variables can be written as d s i m w d e

A y = f (V , (1, L, k, p, p ) (5.27)

The total number of variables is 7 'and the dimensional nlalrix formed between these variables and the dimensions contained in them is given below:

The rank of the matrix (highest order of the non-vanishing determinant) can be easily verified to be 3, which in this case is same as Ule number of dimensions.

There are (7 - 3) = 4 7~ - parameters. These parameters can be found by the product function

( A y ) "1 ( V ) m2 (d ) "'3 (L)"'4 (k ) '"5 ( p ) "'6 ( p ) "'7 or in terms of their din~ensions

Since t l~e parameters are non-dimensional, wc get the following Ulree equatioi~s ( M O LP T O )

There are seven unknowns with three equations. We solve for three chosen unknowns say m2 , m, and m, in terms of arbitrarily assigned four sets of values for mi, m,, m5 and m,. We select the simplest four independent sets of values :

Assigned values of Determined values of

m1 m4 m~ n17

1st set 1 0 0 0

With the first set

m2 m3 m6 -

-2 0 -1

2nd set 0 1 0 0

3rd set 0 0 1 0

L With the second set : n 2 = -

D

With the third set k

: 7 C 3 = - D

With the fourtll set : n4 = 1 PVD

Q,= m f 4 . A . 4 - I

0 - 1 0

0 -1 0

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The reciprocal of the last parameter is the Reynolds number, Re = @ . Since the CL I

pressure gradient is constant (or the pressure falls linearly with length), we can modify 1 the relation 1 ,

Dividing throughout by g, we get the head loss hf = 4 Y

Comparing the preceding expressio~i with the Darcy-Weisbach equation for head loss for turbulent flow in pipes,

We find that the pipe resistance factor f

f = Re)

The dimensional analysis has given us the equation for head loss in pipes and in addition has thrown an insight into the dependence of pipe resistance factor on relative roughness

& and Reynolds number.

5.12 METHOD OF REPEATING VARIABLES ! (a) In this method, a set of repeating variables is so chosen that these variables are

dimensionally independent of each other and contain among them all the dimensions contained in all the variables.

(b) The number of repeating variables cliosen is equal to the rank of the dimensional matrix.

(c) The dependent variable should not be chosen as a repeating variable.

(d) Very often, the most common repeating variables chosen are density p [ML-3], velocity V [L T - ' I aid a linear dimension [L]. It is seen that these are independent in dimensions. I

(e) These repeating variables (r in number) are then combined successively (one by one) i :

with each of the remaining (n - r) variables to form (n - r) number of n - parameters I :c I I

This method is illustrated using the same example of Section 5.11 employed for demonstration of the Rayleigh method.

As before, pressure loss in the pipe, Ap is dependent on the average velocity V , diameter D, roughness height k, length of pipe L, & fluid properties p and p.

1

The number of variables is 7; rank of the matrix is 3; the n - parameters will be 7 - 3 = 4 , 2

in number. Let us choose the repeating variables as p, V and D and combine them with Ap, L, k and p, one at a time to get 4 x- parameters. The correct exponent to each variable is found by the Rayleigh method already outlined or by inspection.

By inspection, for the combination of p, V, D and Ap , if we place Ap [ML-'T -'I in the numerator, p which contains mass must appear in the denominator.

Next we balance the time T and find v2 must be placed in the denominator and AP incidentally the linear dimension also cancels out, leaving -, as the n - parameter. P v

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1 Thus

AP AP combination of p, V , D and Ap gives = -7 PV D PV

combination of p, V , D and L gives p O v O ~ L - D D '

Note that p alone contains mass and V alone contains time and hence both should be raised to zero power given as L ID.

combination of p, V , D and k gives p O ~ O k k -

D D '

The argument is same as above,

combination of p, V . D and p gives = Reynolds number P

p and p contain mass and hence if p is chosen in the numerator, p should appear in the denominator p/p = To balance the dimension of T, V should be in the numerator pV/p = [L-'1 and we place D in the numerator to give us the dimensionless number PVD

P The non-dimensional parameter relationship is

Alternatively, adopting a procedure similar to the Rayleigh method, we can combine Ap, p, V and D which is dimensionally equal to

( p)"l ( y m 2 (D)m3 ( A P ) ~ ~ = (ML-3)"'1 (LT')"~ (L)"3 (ML-I r 2 ) " 4 - - ~ r n , + m , ~ - 3 n i , + m ~ + 1 n ~ - m , r r n z - 2 m ,

the powers of M, L and T are equated to zero to get a dimensionless parameter

m, + m4 = 0

-3m1 + m, + NZ, - m4 = 0

-m2 - 2m4 = 0

If m4is assigneda value of 1, m,= - 1,m, = - 2, m, = 0

Other parameters can be derived in a similar fashion.

5.13 EXAMPLES IN DIMENSIONAL ANALYSIS

5.13.1 Rise in Capillary Tube The rise h of liquid level in a capillary tube of radius R is to be detemuned. It is a function of radius of the capillary tube, surface tension o, specific weight y of liquid. Note that y contains p and g

h = f (R, y, 0) The dimensional matrix is

h R y o

M O O 1 1 L 1 1 - 2 0 T 0 0 - 2 - 2

It may be verified that all 3 x 3 determinants ofsthe above matrix have zero value, whereas s h e of the 2 x 2 determinants are non zero. The rank r of the matrix is therefore 2 (and not 3. the number of dimensions). As stated earlier. it is imnortant to find

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The dimensionless n-parameters are thus 4 - 2 = 2. Choose the two repeating variables as h and a and combine them each of the remaining variables R and y individually.

I1

! Combination of h, a and R gives x, = -

I R ' I

I a

and h, a arid y gives x2 = - Y R ~ '

The functional relationship between the two n - parameters becomes h R

Compare the result so obtained with the theoretical one h 20 R - y ~ 2

5.13.2 Head Characteristics of a Pump The head H to which a pump may deliver depends on the discharge Q, rotation speed N of the pump, diameter D, acceleration due to gravity g, & fluid properties p and p

H = f ( Q , N , D , g , p , p ) (5.32)

N has the dimension of [T - '1 There are seven variables and the rank of the dimensional matrix is 3 (same as number of dimensions M, L & T ) giving us 7 - 3 = 4 x-parameters. With p, Q and D as repeating variables, the following successive coinbination of :

H p, Q, D and H gives -

D

Q p, Q, D and N gives - ND?

e2 p, Q. D and g gives - 6?D5

p , Q. D and p gives @ CLD

is a form of Reynolds number which may ?ot be important in highly turbulent flows PD in a pump. It is possible to modify the 71: - parameters

(E] [+I = %2

effectively retaining Q in only one 71:- parameter. Thus the relationship reduces to

The left hand side represents a form of Froude number - (4 5.13.3 Drag on a Ship The ship being partly submerged experiences a surface drag along the sides of the ship due to viscous resistance of water due to ship motion and a wave drag due to the free-surface distortion. The drag force F is thus dependent on the speed of the ship V, acceleration due to gravity g, a characteristic lenglh of ship L, & fluid properties p and p

F=f (V, g, L, p, p) (5.35)

With p, V, and L as repeating variables, the 71: - parameters from the combination of C

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p, V, Land g yields ( F-Froude iiumber) n2 = G

p, V, Land p yields n, = - PVL (Re-Reynolds numbeer) CL

Thus, we find that Euler number is a function of both Froude and Rey~iolds number E = $(F, Re) (5.36)

5.13.4 Fall Velocity of a Sphere

I The terminal fall velocity V of a sphere of diameter D and density p,$, falling freely in a I fluid of density p and dynamic viscosity p, niay be expressed in the functional from,

u = S(R. Q P,, P, P) (5.37)

If we choose tlie repeating variables as p, g and D, successive combinations with other variables will yield :

6 p, R. D and p -+ Reynolds number

CL

U = $ -.- [ a ""I

We will now show the versatility of dimelisional analysis by different choice of repeating variables and subsequent interpretation :

Had we chosen p, U and D as repeating variables, we get

the unknown U appearing in two parametem.

We may reduce the number of variables, since we know that the motivating force is the submerged weight of the sphere.

We can replace the variables ps and g by F

U =f (D, F, p, p)

Again with p, D and U as repeating variables

(Euler number)

p, U, D, and p -+ Q!E (Reynolds number) CL

n o2 The Newtonian drag equation for a sphere is given in the familiar forni with A = - 4

F 1

= CD , drag coefficient - p u 2 . A 2

Hence we conclude from equations (5.40) and (5.41)

c, = 4 (Re)

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Fluid Dynamics - I If we had chosen the repeating variables as F, p and D, an alteniative form for the n-parameters would have resulted.

The term on the left hand side is known as the Stokes number. WUD

If the sphere is falling very slowly, then the accleration effects around the sphere may be neglected as a close approximation. Wheiiever there is no acceleration (inertia forces are absent), mass density has no role to play in fluid motion. Hence we can write equation (5.43) as

which is the Stokes law for slow motion of spheres; the value of the constant in (5.44 a) is analytically found to be 3 n.

5.13.5 Discharge over a Sharp-edged Weir.

The discharge 9 = ' per unit width of tllr weir is dependent on the approach channel B

width Bo , width of weir B, height of weir W, head over weir H, acceleration due to gravity g, & fluid properties p and p as shown in Figure 5.1.

with p, g, Has repeating variables

q2 Q2LB2 p ,g ,H and q + -=-

gH" gH3

B p , g , H and B + -

H

p , g , H and p + ' H a form of Reynolds number, P

Bo B B B we can cross multiply the n-parameters - & - to get - and neglect the effects of - H H BO H

(if B >> H ) and the Reynolds number

comparison with the analytical expressioi~ Q = Cd GBH ' I 2 we find C , coefficient of

to include the effects of weir geometry.

5.13.6 Velocity in an Open Channel The average velocity V in an open chamel of constant cross section A and wetted perimeter P and longitudinal uniform slope S has to be deterrmned. In addition to the above variables, surface roughness height k, acceleration due to gravity g and fluid --,.-,.-Ln,, - ,.-A .. '- L- : - - I . ->->

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v = f (A, P , S, k , g, p, CL)

With p, g and A as repeating variables, combination of

p, g, A and V yields v '

"1 = g p, g, A and P yields

A ff2 = -

p2

Slope S is dimensionless n, = S

p, g, A and k yields k

I t " = q

p, g, A and yields x5 = p 6 ~ ~ ' ~

, a form of Reynolds number CL

Neglecting the effect of Reynolds number in turbulent flow in open channels,

We may leave the result in the above form or with some prior knowledge of flows in open channels (V a 6) the above relationship after some manipulation can be reduced to

I

A Where R is the hydraulic radius = p ,

which is in the form of Chezy equation V = C wit11 C detinecl as a function of

[&)or : 5.13.7 Pipe Orifice Discharge Q through an orifice of diameter d located in a smooth pipe of diameter D is a ,

function of the difference in pressure Ap across the orifice, fluid properties p and p

Choosing the repeating variables as p, Ap and D , we get

p,@, ll and Q + I C - A l - D 2

d p, Ap, n and d + n2 = - D

D 4 A , p,Ap,1> and p+$= --& + Reynolds number

CL

If the effect of the Reynolds number is not significant, we write

d . The coefficient of the orifice meter is a function of 5 ratio.

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i I ~ k k l ~ y t b 8 d ~ r - 1 5.13.8 Celerity of a Gravity Wave The celerity c of a gravity wave is a function of wave height H, wave length h, depth of ocean d and g, but nelgecting the effects of viscosity.

c = f (H , A, d. g) (5.52)

There are only two dimensions L and T in the above variables. The rank of the dimensional matrix is also 2, There are 5-2 = 3 n- parameters. Choice of g and h as repeating variables leads to the parameters :

C g.A and c - t a , = p

gh H

g ,h and H + R , = - h

d g,h and a +a,=-

h

C - - H d m-,[...) (5.53)

H If the wave height is very small (small aniplitude wave), the effect of - can be neglected

h

The actual analytical expression for small amplitude wave is

d For deep water waves, the influence of - is also insignificant c = constant h

which may be compared with the theoretical expression

SAQ 4

(a) In choosing the variables governing flow problems, can we choose the following combination of variables.

(i) Q , v , A , F , p , p (ii) p , A , F , V , p , g

(b) Choose the correct statements from the following.

(i) The number of repeating variables choien must be equal to the rank of the dimensional matrix.

(ii) The repeating variables should combine to form a dimensionless parameter.

(iii) Dependent variable should always be chosen as a repeating variable.

(iv) Repeating variables should be dimensionally independent of each other.

(v) It should be possible to derive one repeating variable from a combination of the other repeating variables chosen.

(vi) The repeating variables should contain among them all the dimensions appearing in the variables.

(c) The most common repeating variables generally used are

(i) p , p , g (ii) p , V , L (iii) p , 0 , p (iv) 6 , V , L.

(d) Physical quantities which have constant values such as g, Bulk modulus K , fluid properties p , p and o when they remain constant at constant pressure and temperature. Is the statement correct ?,If not, give reasons for including them in appropriate situations.

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SAQ 5

F v2 P V D Given the basic .rc - parameters as - - , and - p v 2 ~ 2 ' gD P

Which of the following is not a valid combination

(ii) & y & p ~ 4 ' gD ' gCL

SAQ 6

(a) A rectangular tank containing liquid is moving into a uniform acceleration a, along the x-direction. Find an expression for the variation of depth h in the tank as a function of distance x, acceleration a, and g. Do the fluid properties p and p enter the picture ?

(b) If a liquid in a cylindrical container is rotated about its vertical axis at an angular speed o, find the non-dimensional parameters linking pressure p with radial distance r, angular speed o, and density p.

(c) A V-notch weir has a central angle 8. Write down the variables governing the discharge Q with and without velocity of approach Vo. What is the rank of the dimensional matrix ? Determine the dimensional parameters.

SAQ 7

(a) The velocity c of a capillary wave formed due to a gentle breeze over the surface of a lake is dependent on surface tension o , wave length h, and fluid density p. Determine a functional relationship for the wave celerity.

(b) A cylinder of diameter D and length 1 is placed in a steady uniform stream of velocity V. Find the drag force F on the cylinder as a function of the variables V, D , 1, p and p.

(c) The thrust FT due to an aeroplane propeller is dependent on the angular speed o, speed of aircraft Vo, diameter D of the propeller, fluid properties p and- p and the speed of sound c. Find the .rc - parameters. Choose either p , w , c or p , Vo , D as repeating variables.

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muid Dynunler-I

5.14 MODEL SIMILITUDE Physical Models : Model studies of prototypes are frequently undertaken to obtain both qualitative and quantitative information about flow patterns, velocity distributions, discharges, pressures, forces, head losses and efficiency of devices in proposed hydraulic structures and hydraulic machines. The experimental data can be used to calibrate weirs, orifices, and generally obtain numerical constants in the correlation among n-parameters obtained by dimensional analysis.

Similitude in fluid mechanics would mean relation between full-scale flow (prototype) and the flow in a smaller but geometrically similar model. However, many of the hydraulic models are not geometrically similar but even in such cases, similitude laws have been evolved. We restrict our attention mainly to geometrically similar models. If reliable quantitative information is to be obtained from model studies, one should have geometric, kinematic and dynamic similitudes. Kinematic similitude would mean that flow patterns in the inodel and prototype would be geometrically similar with similar velocity distributions. Dynamic similarity would imply that shear stress and pressure ratios at corresponding points must be a constant. By the same token, integrated values of pressure (or sJiear) on geometrically similar areas or forces in the model and prototyppe must be similar in direction and bear a constant ratio.

The criteria for creating geometric, kinematic, and dynamic similarities between prototype and model are fully met if the relevant values of A-parameters are identical between model and prototype. However, simultaneous satisfaction of the requirement of identical values of all x-parameters poses practical difficulties in modelling. For example, let a free surface flow problem be governed by Reyno1ds;Froude and Weber numbers. Should we prescribe that these non-dimensional numbers be identical, the following relations must be simultaneously satisfied between model and prototype.

a) Reynolds criterion : Pm Vm Lm - P p Vp Lp - C1, PP

b) Froude criterion

c) Weber criterion : om - - =P

Prn V; Lm pp V; Lp

Herein the subscript tn refers to the model and subscript p to the prototype.

v m From the preceding conditions, it may be easily derived (eliminating - in the process) VP

that the linear scale ratio, 5 is linked with fluid properties i d gravitational forces in the LP

model and prototype. 1 -

The choice of available fluids for use in model is very limited-generally air and water are commonly used. Force fields other than the earth's gravitational field are difficult to create. If for convenience, we use the same fluid in the model and prototype and operate the model in the same gravitational field, we obtain the trivial solution that linear scale should be 1 :I.

Here we observe that the physical and economic restrictions on available space, limited choice of available fluids, difficulties in creating force fields other than earth's gravitational force field compel the investigator to make certain sacrifices. Whenever i't is not possible to simulate all dimensionless parameters in the model, recourse is made to select only the most significant parameters and disregard the less important ones. Fortunately, in most cases only two forces happen to be of similar magnitude and other forces are of lesser importance. To be sure, nonsatisfaction of all the parameters does introduce errors in model behavGur and these differences between model and prototype are referred to as scale effects. Correct interpretation of model results and itS1pplication 6- &L- ---AA*--- 2 ----.a- -- .L- ---- - - 2 :-* - - :L: - - - C * L - :--.-..A: --.-..

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The use of scale m&el techniques in bifferent situations will be demonstrated by illustrative exadles in the following sections.

5.15 SCALE MODELS IN SUBMERGED FLOWS

In submerged incompressible flow problems, provided boundary geometry is similar, the

pressure distribution expressed by the Euler number (, - is a function of the

Reynolds number - jP3 \~ 1

E = f (Re) (5.55)

In liquids, cavitation may occur at locations where the pressures are in the vicinity or lower than vapour pressure p,, of the liquid; then the pressure distribution is no longer uniquely governed by Reynolds number alone but also by the cavitation number

K = ( - pv). Then the rellfion~hip becomes

1 - p v 2 2

E = f(Re,K)

But in cavitating flows governed by gravity such as siphons, bridge piers etc. E = f (Re,F)

In transonic and supersonic flows, since compressibility of gas plays an important role, we have to consider the influence of Mach number. But in subsonic flow of gases (i.e. at low Mach numbers), where density changes may be ignored without appreciable error, Euler number remains mainly a function of Reynolds number (equation (5.55)). This feature provides us the added facility of using air in the model as the fluid medium to replace liquids in the prototype (provided as stated earlier, cavitation does not take place in the prototype).

A variety of problems belong to this category of submerged flows such as pressure flows in conduits, pipe fittings, orifices, mouthpieces, nozzles, pipe inlets, drag on submerged bodies and etc.

Example 5.2 :

A 1:10 scale torpedo model was tested in a compressed air tunnel (mass density of airb, 2.6 kg/m3 and kinemetic viscosity v of air 0.65 x lo-' m2/sec). The resistance was found t6 be O.8p kN when the air velocity was 35 m/sec. Find the corresponding velocity andbrce on the prototype torpedo moving under dynamically similar conditions in the ocean (mass density of sea water = 1025 kg/m3 andkinematic viscosity of water = 1.08 x m2/sec)

Solution : \ Since the model is a submerged one, Reynolds number of the model and prototype should be identical and E = f (Re)

= 0.582 m / sec

If the Reynolds number is identical, Euler number is also the same for both model and prototype.

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or since force can be expressed as F = p . A.

Example 5.3 :

The performance of a gate valve located in a 3 m diameter water supply pipe is to be tested on a model valve of 15 cm dia. The velocity of flow fn the prototype is 4 m 1 sec. Suggest a suitable model discharge. KinematicviscoSity of water in model and prototype is 1.0 x lo-* cn121sec

Solution :

The Reynolds number of the prototype is:

and the model velocity with (Re), = (Re),

and the discharge in the model is

n Q, = V , A, = 80 x - (0.15)~ = 1.414 m3 I sec 4

The velocity and discharge to be simulated in the model are too high to be obtained in an experimental laboratory. We know from experience that pipe resistance coefficients and drag coefficients attain nearly constant values at high Reynolds nurrfber. Hence it is sufficient to test the model at a lower Reynolds number, making sure\however that the Reynolds number is still sufficiently high to obtain turbulent flow in the model. Choosing a value of (Re), = lo6 (instead of 12 x lo6), we get

R and the model discharge (2, = V, A, = 6.67 x - x (0.15)~

4

Example 5.4 :

Lift on an aerofoil

An aerofoil of surface area 0.1 m2 is tested for drag and lift in an air tunnel. At an angle of attack of 5". the lift force measured in the model is 32.4 N at a wind speed of 30 m I sec. For a prototype area of 10 m2, what is the lift force at a speed of 100 m 1 sec at the same angle of attack.

Solution :

The drag and lift force coefficients CD and CL respectively of an aerofoil are functions of shape of aerofoil, Reynolds number and angle of attack. If the shape o f the aerofoil and the angle of attack are kept the same between the model and prototype, CD and CL are functions of Reynolds number. However the lift coefficient is not sensitive to changes in Reynolds number and we can assume that CL for the model and prototype is the same in ,the turbulent range.

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The lift coefficient in the model

The prototype (CL)p is also equal to 0.6

Example 5.5 :

A venturi type of contraction has to be tested on a model with a lineqr scale of 1: 10. What pressure should be maintained in the model water tuimel to maintain the same cavitation index. Assume that the pressure head at the critical section in the prototype is (h& Assume that the vapour pressure corresponds to a head of 12.5 cms of water.

Solution :

In terms of head of water, the identity in cavitation parameter K may be expressed as:

If Reynolds criterion has also to be satisfied and assuming vm = v,

5.16 FREE SURFACE PROBLEMS

In turbulent free surface flows commonIy met with in a large number of practical problems such as rivers, estuaries, open channels, gravity waves, weirs, hydraulic pumps etc., gravitational force is predominant in relation to viscous and surface tension forces. In such free surface problems, Euler number is mainly a function of Froude number and it is sufficient to ensure that the flow is turbulent in the model, as well as, prototype and the Reynolds number equivalence is not normally needed.

E = f(F) (5.57)

Euler's number will be identical if Froude numbers of model and prototype are equal in free surface turbulent flows.

Example 5.6 :

The discharge over a weir 150 m long and 15 m in height is to be estimated by means of a model built to a scale of 1:25. What are the scales for velocity and discharge. If the pressure head measured at a given point in the model is 0.01 m, to what pressure does this correspond in the prototype.

Solution :

If the Froude number should be the same for model and prototype.

D I l D ~ d Anslyaim and SimiUt.de

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Since gm = gp,

The discharge ratio is

Also

P or in terms of head , h = - ; pn, = Pp ; Ym = Yf Y

Example 5.7 :

A breakwater model is built to a scale of 136. The prototype breakwates has to withstand waves of height 8 m and wave period 9 seconds. What are the corresponding wave height and periods in the model. If an armor stone of weight W m is stable in the model, what relative size of stone should be used in the prototype breakwater. Mass densities of fresh water in model and sea water in prototype can be assumed as 998 kg/m3 and 1025 kg/m3.

Solution :

Following the Froude criteria, we have

and the celerity of the wave c should satisfy

Since velocity can be written in terms of length and time

we have, L c = - T

9 Tm = - = 1.5 seconds 6

The force scale is obtained by equating the Euler numbers

F m / A m - - FP / AP 1 1 - p v 2 , p m v m 2 P P

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5.17 SURFACETENSION J M m d d h l d y i s

d s d i t a d e

Surface tension forces predominate in capillary waves, liquid level rises in capillary tubes, jet sprays and etc. We then seek equivalence of Reynolds and Weber numbers.

E = ARe, W)

Example 5.8 :

A jet of water emerges from a nozzle 1 cm in diameter at a velocity of 15 m I sec. The jet is found to break into a spray at a distance of 15 cms froin the nozzle. The surface tension of the fluid is o1 = 50 dyneslcm. Another fluid with the mass density p2 = 0.9 pl and kinematic viscosity v2 = 1. lv , , '3, = 75 dyneslcm issues from a geometrically similar nozzle. If the two nozzle flows are to be kinematically similar. determine the scale factors for length, velocity, force and time.

Solution :

The Reynolds and Weber numbers must be identical in the two flow systems.

Reynolds criEi5on:

Also Webers criterion : P I ~ 1 ~ ~ 1 ~ 2 ~ 2 ~ ~ 2 - - - 0 1 '32

The first relation gives

' Vl Substituting this value for - in the second relation v2

Diameter of model nozzle = 0.726 cms

v2 - 1 L1 - Velocity scale - - - . - - 1 Vl 0.9091 & 0.9091 x 0.726

= 1.515

V2 = 1.515 x 15 = 22.73 mlsec

F2 - Force scale - - Fl

Time scale T2 &/V2 L2.3- - - - - - - - - - 0'726 - 0.479 Tl L1/Vl L1 V2 1.515

SAQ 8

What dimensionless parameter is the most significant one for the following flow problems.

(i) Submerged flows

(ii) Free surface flows

(iii) Liquid rise in capillary tubes

(iv) Pressure flow in closed conduits

(v) Ship motion on the sea surface

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(vi) Aircraft flying at low speeds

(vii) Aircraft flying at supersonic speeds

(viii) Capillary waves

(ix) Gravity waves

(x) ~ubm&es and torpedoes moving at great depths.

SAQ 9

(a) If both Reynolds and Froudc numbers are significant, determine the scale for kinematic viscosity in tenns of the length scale, if the gravitational force remains the same for both the model and prototype.

(b) If the model of an aircraft is to be tested at identical Reynolds and Mach numbers, express the requirement for density ratio in terms of the scale ratios for length, viscosity and bulk modulus. Discuss how this can be achieved in practice. Assume the temperatures in the model and prototype are same.

(c) Using the results of example in 5.13.2, determine the head and discharge of 1:4 model of a centrifugal pump that supplies 0.8 m3 at 3 m head when running at 240 rpm. The model operates at 1200 rpm.

SAQ 10

(a) A seale model 1:5 has been made of a sonar device beneath a submarine. If laboratory tests on this device reveal that the drag on it is 15 Newtons at a speed of 18 mlsec, find the corresponding velocity and drag force of the prototype. Assume that density and velocity of water in the model and prototype are identical.

(b) The wave resistance of a ship's model built to a scale of 1:25 is found to be 2.0 Newtons. What is the scale for velocity and time in the model. State the governing parameter in this model similitude. What is the wave resistance in the prototype?

5.18 SUMMARY

Now let us surnrnarise what you have learnt in this unit.

Fundamental dimensions involved in the definition of physical quantities (whether they are scalar, kinematic or dynamic) have been explained. Once the dimensions involved\in a physical quantity is known, transformation from one unit of measurement to another is readily done. We can establish the dimensional homogeneity of equations and modify the constants appearing in non-homogeneons equations to conform to the requirements of dimensional homogeneity. Further dimensional analysis helps us to reduce the number of variables and identify the significant non-dimensional parameters governing a phenomena. These n - parameters pave Ule way for planned experimentation to obtain universal relationships among the parameters. In some cases, where theoretical solution of complicated nonlinear equations is not possible, we can establish meaningful relations among the n - parameters and obtain the missing constants through experimentation.

The fact that flow-and force fields are sinlifar at identical values of n - parameters enables us to make small scale inexpeilsive models to study the performance of

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r I prototypes by simulating the significant n - parameters in the model. Laws of similitude D l m m d d A d y h I lead us to scale up experimental model results to prototype values. Such model and Slnllltpdc

experiments usually form the basis for the preliminary desigii of ships, airplanes, rockets, spillways, weirs, gates and other costly structures.

5.19 ANSWERS TO SAQs

SAQ 1

(a) [MJ = [FL-'T~]

(b) [F-II = [MLZT -2] = FL]

(c) 478.78 ~ / m ~

(e) 0.473 N/m and 0.394 N/m

4 ( f ) 419957 N-m

L SAQ 2 I L (a) t (i) C has the dimension of is the dimension of G. We can

! rewrite the equation of V = C1 is a dimensionless constant. t

(ii) In FPS system C = 135.83

(b) I%] = [GI. If rn has the dimension of [11'], a has the dimension 1 A-

[C] (c) The coefficient 0.84 has the dimension --- ; V = 0.548 yy m / s

- -

(d) p = 4.836 @ m. The coefficient has the dimension [L"" T'O]

SAQ 3

(b) (0 3 (ii) 2 (iii) 2

SAQ 4

(a) (i) No : Q = VA , hence only two independent variables out of Q,V, A is permissible.

(ii) No : F = p A . Choose only two of the variables F, p, A . (b) (i), (iv) and (vi)

(c) (ii)

(d) No. They are dynamic variables governing the gravity forces (g), elastic forces (k ), inertia force (p), viscous force (p) and surface tension force (o) which govern the flow.

SAQ 5

(iv) The first parameter can not be obtained as a conlbination'of the basic parameters . (v) The number of n - parameters should remain'as 3 and not reduced.

SAQ 6 , .

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P , , - constant. Here also p does not have any role. (b) - - P'" r

(c) Neglecting velocity of approach . &5/z = fie) GH

Compare this expression with the theoretical expression 8 0

Q = ,GH"~ tan- 2

Including velocity of approach : -9- H~~

SAQ 7 7

(a) c = constant 4:

FT pod c 1 (c) With p , o , c as repeating variables : - = f, (2. - p o 2 ~ 4 P *w,J

FT With p , b, r as repeating variables 7 = f2 (2;

pvo D

SAQ 8

(i), (iv), (vi) and (x) - Reynolds number

(ii) and (ix) - Froude number

(iii) and (viii) - Weber number

(v) Both Reynolds and Froude number

(vii) Reynolds and Mach number

SAQ 9 3

"P ~ P J

(b) Since c = @@is the same for model and prototype, Vm = Vp for Mach Pm L numbers to be equal. From Reynolds criterion - = -ll, since kinematic PP Lm

viscosity is not affected by pressure (F,, = pp). A wind tunnel using compressed air may be used for model testing.

(c) Qm = 62.5 litres/sec Hm = 1.5625 m

SAQ 10 (a) Vp=3 .6m/ s and F p = 1 5 N

- Tm (b) h-1 - Fp = 31250 N Froudee.s law Vp 5 ' Tp 5 '

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UNIT 6 BASIC EQUATIONS

Structure 6.1 Introduction

Objectives

6.2 Basic Principles

6.3 Concepts of System and Control Volume

6.4 Application of Control Volume to Basic Equations 6.4.1 Continuity Equation

b 6.4.2 Eneigy Equation 6.4.3 Linear - Momentum Equation

I 6.5 Contiriuity Equation i 6.6 Euler's Equation in Streamline Coordinates

6.7 Bernoulli's Equation

6.8 Energy Correction Factor

6.9 Momentum Correction Factor

6.10 ~pplication of Bernoulli's Equation 6.10.1 Pitot Tube 6.10.2 Venturimeter 6.10.3 Suction Effect of a Flowing Fluid 6.10.4 Orifice Flow

6.1 1 Applications of Momentum Equation 6.11.1 Jet Force on a Plane Surface 6.1 1.2 Moving Vanes 6.11.3 Force Exerted on iPipe-Bend 6.1 1.4 Flow through a Nozzle 6.11.5 Momentum lheory for Propellers 6.1 1.6 Jet Propulsion 6.1 1.7 Rocket Mechamcs 6.1 1.8 Losses due to Sudden Expansion 6.1 1.9 Hydraulic Jump

6.12 Summary

6.13 Answers to SAQs

6.1 INTRODUCTION

In this unit, we shall be dealing with the basic equations of fluid flow, their derivation and applications. In the study of fluid dynamics, all the forces which affect the fluid flow are considered. The study of dynamics is mainly concerned with the fluid uhsport phenomena which can be grouped into mass transport, heat (energy) transport and momentum transport. All fluid motions mu$t satisfy the principle of conservation of mass. The principle of conservation of heat (energy) is also valid in a flow process. Newton's second law of motion provides the fundamental relation between the acting force on a particle and the time rate of change of it's momentum Application of the above principles to the fluid flow will yield the governing equations such as the continuity equation, Euler's equation, Bernoulli's equation and momentum equations. These provide useful relationship between pressure, velocity and elevation. Many practical problems can be solved by using the above equations.

I Objectives I After studying this unit you should be able to derive I I * continuity equation,

I * energy equation, * momentum equation,

, * Eoler's eouation. and

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You would also be able to define the concept of system and control volumes, energy and momentum correction and apply of the basic equations to the flow problems.

6.2 BASIC PRINCIPLES

We shall begin our study of fluids in motion from a consideration of the basic principlqs and laws involved.

(a) The principle of conservation of mass states that the mass within a sys tem remains constant with time.

Where M is the total mass of the system.

(b) Newton's second law of motion states that the resultant of all external forces acting on the system is equal to the time rate of change of momentum of the system.

d F = - (MV)

dt in which M is the constant mass of the system,V is the velocity of the centre of mass of the system.

(c) The first law of thermodynamics states the principle of conservation of energy for a system. In other words, the heat added to a system minus the work done by the system is equal to the change in the internal energy of the system.

qh - W = E, - Ei (6.3) in which q h is the heat added to the system, W is the work done by the system, Ef and Ei are the final and initial states of the system in terms of the internal energy. The law can also be expressed as

6.3 CONCEPTS OF SYSTEM AND CONTROL VOLUME

A system relates to a definite mass of material as distinct from surroundings. The closed surface of the system forms its boundary which may vary with time, so that it contains the same mass during changes in its condition. A control volume pertains to a region fixed in space, the size and shape of which are arbitrary. The boundary of a control volume is its control surface. The concept of control volume is useful in analysing situations where flow occurs into and out of the spew. This concept is used in the following sections to derive the equations of continuity, momentum and energy. The relationship between the equations for system and control volume can in general be stated in terms of some property (mass, energy or momentum) within the system. If N be total quantum of such a prOpeRy at any time t and q the amount of this property per unit mass, it can be shown that the rate of increase of N within a system is equal to the time rate of increase of the property N within the control volume (fixed relative to xyz ) plus the net rate of efflux (change) of N across the control volume boundary, stated in the form of an equation,

p being de&ity, dW the element of volume, V the velocity vector and dA is a vector representing an area element of the outflow area cv aql cs refer to control volume and control surface respectively.

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j 6.4 *APPLICATION OF CONTROL VOLUME TO BASIC EOUATIONS

I 6.4.1 Continuity Equation

The equation of continuity is developed from the general principle of conservation of mass which states that the mass M within a system remains constant with time (equation (6.1)). In equation (6.5), let N be the mass of the system and q be the mass per

I unit mass (i.e., q = 1)

In other words, the continuity equation for a control volume states that the time rate of b increase of mass within a control volume is equal to the net rate of mass inflow to the

control volume.

6.4.2 Energy Equation The first law of thermodynamics for a system states that the heat q, added to a system minus the work W done by the system depends only upon the initial and final states of the system (equation (6.3)). The difference in states of the system is a property of the system called as the internal energy E. Let the intemal energy per unit mass be e. Applying equation (6.5), N = E and

pe

The intemal energy e of a substance is the sum of potential , kinetic and intrinsic energies.

6.4.3 Linear - Momentum Equation ,Newton's second law for a systems (equation (6.2)) is used as the basis for finding the linear - momentum equation for a control volume,

I Let N be the linear momentum MV of the system and q be the linear momentum per

unit mass @! . Then by use of equation (6.5), P

In word& the resultant force acting on a control volume is equal to the time rate of increase of linear momentum within the control volume plus the net efflux of linear momentum from the control volume.

6.5 CONTINUITY EOUATION

The reader is referred to Unit 3, Section 3.10 for the derivation of continuity equation. The statement of continuity equation in the control volume approach is given by equar'irl(6.6) as

It may be seen that the first term of the above equation is given by equation (3.32), the second term is given by equation (3.31).

6.6 EULER'S E UATION IN STREAMLINE s. COORDINA ES

In Unit 3, Section 3.5, you have learnt the definition of a streamline. In steady flow, the path of a particle is a streamline. Thus in describing steady fluid flow, the distance along the streamline is a convenient choice as a coordinate. For simplicity, consider the flow in y- plane as in Figure 6.1.

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Fluid Dynan~ics-I

Figure 6.1 : Fluid P d d e Movmg along a Streamline

The equation of motion is written in terms of the coordinates. Since the velocity vector + + must be tangent to the streamline, the velocity field is given by V = V (s,t). The pressure at the centre of the element is p. Applying Newton's second law in the streamline direction (s) to the fluid element of volume ds dn dx, (neglecting viscous forces) we get,

ds dndx - p + -- - T I [ $ ':I dn dx - P g sin b dn dx ds = p d i dn dx as (6.9)

where p is the angle between the tangent to streamline and the horizontal and a, is the acceleration along s direction. On simplification, we get

-- ap - pg sin p = p a. as

dz Since sin 0 = - , the equation (6.10) becomes ds

The total acceleration of fluid particle is given by

Dropping the subscripts on V, as velocity is tangent to streamline.

Equation (6.12) is the Euler's equation along the streamline. For steady flow, the equation reduces to

6.7 BERNOULLI'S EQUATION

The Euler's equation for steady flow along a strean] line given by equation (6.13) can be integrated to yield Bernoulli's equation. Integrrrtion of equation (6.13) yields

v I $ + gz + - = constant. along s . 2 (6.14)

To use equation (6.14), we must specify the relation between p and p. For incompressible flow, p = constant . Equation (6.14) reducks to the Bernoulli's equation,

P v - + gz + - = constant along a streamline P' 2

.(6.15)

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The reslrictions on this equation are steady, incompressible and frictionless flow along a streamline, Applying the equation between any two points 1 & 2 on a stream line,

6.8 ENERGY CORRECTION FACTOR

For the solution of problems in open or closed - channel flow, usually one -dimensional form of analysis is used. The flow is visualised as one large stream tube with average velocity V at each cross section. However, the velocity distribution at any cross-section in real fluid flow is non-uniform (Figure 6.2) due to high viscous and boundary

v resistance. The kinetic energy per unit mass of the fluid given by - does not represent

2 v' the true kinetic energy across the section i.e. average of - taken over the cross section 2

( v being the point velocity). In order to compensate for this discrepancy, a coefficient known as energy correction

v2 factor, a is used. The multiplication of a with - yields the kinetic energy actually

2 passing a section. Considering a cross section, the kinetic energy passing per unit time is

A g u ~ 6.2 : Velocity Distribution

9 in which pvdA is the mass per unit time passing the section and 7 is the kinetic energy L

per unit mass. By equating this to the kinetic energy based on average velocity, V

By solving for, a

The energy correction factor, a is a measure of viscous resistance, generated in a given flow, the effect of which is reflected in tile non-unifonn nature of the velocity distribution.

6.9 MOMENTUM CORRECTION FACTOR

The non-uniformity of velocity distribution also affects the accuracy of momentum flux (the momentum per second passing through a section) computed based on the average velocity of flow. A correction hctor P known as momentum correction factor is used such that PpQV gives the actual momentum tlux passing through a section. Considering an element of area clA where the velocity is v, the momenlunl of fluid passii~g Chis area element is p g clA and the momentum flux is

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Equating the above expression with the momentum flux based on the average velocity, V

Solving for P,

For any type of flow, P is always less than energy correction factor a. For laminar flow in 4

a circular tube p = - and for turbulent flow it varies from 1.01 to 1.05. 3

SAQ 1

The velocity distribution in a turbulent flow in a pipe is given approximately by 1 -

with y the distance from the pipe wall and r, the pipe radius. Find

factor and momentum correction factor.

6.10 APPLICATION OF BERNOULLI'S EQUATION

The Bernoulli's equation is applicable to every problem of incompressible fluid flow where energy considerations are involved. Maany measuring devices such as the Pitot tube, venturimeter, orificeineter etc., utilize the principle of Bernoulli's equation. The details of the structure and operation of these devices are discussed in Unit 7.

6.10.1 Pitot Tube The Pitot tube is a device used for velocity measurement. It measures the stagnation pressure at any point in the flow. The static pressure is measured by a piezometer. The velocity at any point may be evaluated by knowing the static and stagnation pressures. In the case of real fluid (rotational) flow the velocity varies normal to the flow, it being minimum at the boundaries and increasing towards the central zone of flow as shown in Figure 6.3 .

Pitot -tube \ I I l f Total head at

4 Velocity distribution

Figure 6 3 :Total Head in Real Eluid (Rotational) Elow

The Pitot tube can be utilized to study the variation of total head across the flow in a rotational field. The piezometric head in a uniform conduit remains constant in a direction normal to the flow and the total head may be written as

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v 2 Totalhead =,+(: ir )

v" - - - + constant 2g (6.23)

The equation (6.23) show'sthat the total head varies only with the velocity head. In the case of real fluid flow (rotational flow), since the velocity varies throughout the normal section, the total head will also vary in a direction norm11 to the flow.

6.60.2 Venturimeter The venturimeter is a device for measuring the rate of flow in a pipeline. It is based on the Bernoulli principle, that is, when the velocity head increases in an accelerated flow, there is corresponding reduction in the piezometric head. The device may be divided into (i) a converging entrance cone (ii) a cylindrical portion of short length known as the throat and (iii) a-diverging con&(~igure 6.4).

line

Figure 6.4 : ~arist ion of Piezometric Hend in Vetttulimeteter

Applying Bernoulli's equation between points (1) and (2) on the central streamline,

For incompressible fluid, the continuity equation is

Where A, and A, are the cross sectional areas at the inlet and throat respectively. Eliminating V, form equation (6.24), the velocity at the inlet may be expressed as

From equation (6.26) and the continuity equation, it can be seen that the difference in piezometric heads at sections (1) and (2) is related to the flow rate. The higher this difference, the higher will be the flow rate. If the inclination of the venturimeter is changed, the difference in the piezometric heads between the inlet and the throat remains unchanged on account of the fixed positions of the total energy line and hydraulic gradient line. The discharge, therefore, does not de end upon the orientation of the

venturimeter. The change in the piezometric head - + z, 1 - [? + :.)] may be

measured by a differential manometer.

6.10.3 Suction Effect of a Flowing Fluid Consider the flow of an incompressible liquid from the supply tank (A) through a pipe having a gradual coi~traction of cross-sectional area A, at C (Figure 6.5). Let A be the flow area at the outlet located at a depth H below the level of liquid in the supply tank. 1 ,.+ +I.,. a -.., I.,> ,.+,.,,A., T~,.,.-:..~ thA .~~-t:-,,i ..:..,. n ,.---..,.*,.A +- &I.,. -:A-..- 4-1 -.-.

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from which

Wgure 6 5 : Pnneiple of a Jet Pump

section C, the Bernoulli's equation between the free liquid surface and the contracted portion at C can be written as

in which V , = velocity in the tank (A), y, = atmospheric pressure over the free surface, h, = depth of the contracted portion C below the liquid surface and V, , p, are the velocity and pressure at C. From the continuity equation.

Substituting these quantities and simplifying, the pressure difference can be written as

If the pressure at C becomes negative, if a hole is drilled at C, no liquid would flow out through the hole. On the contrary the higher pressure of the atmosphere would force air into the pipe through the hole. If a vertical pipe D is fixed at C with its lower end immersed in the tank (B) having the same liquid as in the tank (A), the liquid in (B) rises in the vertical pipe D and even join the main flow if ( pa - p, ) > y h. n i s condition may be expressed as

(pa - P C ) > Y h

Equation (6.30) shows that for the liquid in the tank (B) to be lifted up and join the main flow, the ratio of the pipe area at the outlet to the contracted area must be greater than the square-root of the ratio (h+hc) / H. This effect resulting from the negative pressure created at C due to the flowing liquid forms the basis of the design of jet pumps.

6.10.4 Orifice Flow Consider an opening in the side of a reservoir as shown in Figure 6.6. The reservoir is assumed to be very large as compared to the size of the opening, so that the velocities at all points in the-reservoir are negligibly small. Applying Bemoulli's equation to points (1) on the reservoir surface and (2) on the centre lime of the emerging jet, considering the horizontal plane through the centre of opening as the arbitrary datum,

n A n A h - y 2 + n + n iL * I \

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E'ipre 6.6 : klow from a Reservoir

In other words, the total energy at the reservoir surface which exists in the potential form is converted into kinetic form as the liquid flows through the opening. The velocity of flow in the jet can thus be given as

V = (6.32)

Which is known as Torricelli's theorem. This theorem is recognised as a special case of the Bernoulli's equation. Instead of point (1) on the reservoir surface, if any other'point is chosen for the application of Bernoulli's equation, say point (3) on the streamline, we

find that z, + = h , and since the velocity at (3) is negligibly small as compared to Y

the velocity at (2). The above expression also yields the same result as expressed by equation (6.32). Example 6.1 :

Air flows steadily and at low speed through a horizontal nozzle, discharging to the atmosphere. At the nozzle inlet, the area is 0.1 m2. At the nozzle exit, the area is 0.02 m2. The flow is essentially incompressible and frictional effects are negligible. Determine the gauge pressure required at the nozzle inlet to produce an outlet speed of 50 mlsec.

Solution :

Given : Flow through a nozzle as shown in Figure 6.7.

The air flow is steady, incompressible and frictionless

-------- - -

+-%A -1

Control volume

p2 = Patm

L,,, , --,,, A

Figure 6.7

P1 v; P2 v,' Basic equations : - + - + g zl = - + - + g z2 P 2 P - 2

Assumptions :

1) steady, incompressible, frictionless flow

2) z , = z2

3) uniform flow at sections (1) and (2)

L

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Apply the Bernoulli's equation along a streamline between points (1) and (2)

Apply the continuity equation to determine V,

-pVIAl + p V2A2 = 0 or Vl A, = V2 A,

For air at standard conditions, p = 1.23 kg/m3

p, - p,, = 1.48 kPa

Example 6.2 :

A U tube act as a water siphon. The bend in the tube is 1 m above the water surface; the tube outlet is 7 m below the water surface . If the flow is frictionless as a first approximation, and the fluid issues from the bottom of the siphon as a free jet at atmosphere pressure, determine the velocity of the free jet and the absolute pressure of fluid in the flow at the bend.

Solution :

Given : water flowing through a siphon as shown in Figure 6.8.

Find : (a) velocity of water leaving as a free jet

(b) pressure at point (A) in the flow - P vL

Basic equation : - + - + gz = constant P 2

Assumptions: (1) neglect friction

(2) steady flow

(3) incompressible flow

(4) flow along a streamline

(5) reservoir is large compared to pipe

0

Rgure 6.8

Apply the Bepoulli's equation between point (1) and (2)

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Since (area),,,i, >> (arealpipe , also P I = P2 = /-)ahn .

v,' &?I = y + RZ2 and v~~ = 2g (zl - z2 )

V2 = d2g(z,-z2) = 4 2 ~ 9 . 8 1 x 7 = 11.7 mlsec.

To determine the pressure at location (A), we write Bernoulli's equation between (1) and (A)

Again Vl -- 0 and from conservation of mass VA = V2 . Hence

1 - - x 999 x (11.7)~ 2

pA = 22.8 kPa (abs)

SAQ 2

At a point in the pipeline where the diameter is 22 cm, the velocity of water is 3.8 m I s and the pressure in 3.6 kp/cm2. At a point 16 m downstream the diameter reduces to 12 cm (Figure 6.9). Calculate the pressure at this point, if the pipe is (a) horizontal (b) vertical with flow downward.

Figure 6.9

SAQ 3

The siphon shown in Figure 6.10 is filled with water and is discharging at 130 litreslsec. Find the losses from point 1 to 3 in terms of velocity head. Find the pressure at point 2 if two -thirds of the losses occur between pbints 1 and 2. ..

Figure 6.19

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ElPkl Dynunia-I SAQ 4 Determine the velocity of the jet at the orifice in the wall of the reservoir as shown in the Figure 6.1 1. Also find th_edischarge through the orifice.

Figure 611

SAQ 5

Water is pumped at the rate of 350 litreslsec through a 35 cm pipe to a hilltop. The hilltop is at an elevation of 60 m and the diameter of the pipeline reduces here to 25 cm. If the pump maintains a pressure of 1.2 kg/cm2 at. the hilltop, what is the pressure at the foot hills ? What is the power required to pump the water.

SAQ 6

Water Foves steadily through the turbine as shown iri the Figure 6.12 at the rate of 0.25 m Is. The pressures at (1) and (2) are 2.0 kg/cm2. ant1 - 0.2 kg/cm2 respectively. Neglecting the heat transfer, determine the horsepower delivered to the turbine from water.

d l = 25cm

6.11 APPLIC,ATIONS OF MOMENTUM EQUATION -

Newton's second law, the equation of motion developed into the linear momentum equation is given as (equation 6.8).

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I

ETpre 6.13 : Contml Volome with Uoifonn M o w d Outtlow N o ~ a l to the Control Surface

Tbis vector relation applied in x direction gives,

Consiaering the flow space consisting of a bundle of stream tubes (control volume) with steady flow as shown in Figure 6.13, the force acting on the control volume is given by integrating the right-hand side of equation (6.34) as,

I ! in which A,, V,, Vxl and A2,V2 ,Vfl are the cross sectional area, velocity and x-component

of the velocity at (1) and (2) respectively. In applying equation (6.34), care must be taken to define the control volume. When the velocity varies over a plane cross section of the control surface, the average velocity may be considered by introducing the momentum correction factor p.

6.11.1 Jet Force on a Plane Surface Consider a jet moving at a velocity V, striking a plane surface as shown in Figure 6.14. After striking the surface, it gets split and flows along the surface.

Control volume

Rpre 6.14 : Force on a Flat Surface due to Change in Momentum

Assuming that the jet before and after the impact remains in the horizontal plane, the effects of gravity are eliminated. Consider the control volume as indicated by the dotted lines. Applying momentum principle, the net force acting on the fluid in the x-direction is given by

Fx = (pQ3V3 - P Q ~ V ~ ) - ~ Q l v , c o s 0 (6.36)

This force, being parallel to the surface, is a shear force exerted by the surface on the fluid. The rate at which y-momentum enters the control volume is p Q, Vl sin 8. The momentum in ydirection leaving the control volume is zero. The force component, F,, exerted by the surface on the fluid is

F, = 0 - ( -p Q, Vl sin 0)

= pQ1Vlsin8 (6.37)

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Hold Dylmda-I The force components exerted by tlie fluid on the surface will be equal in magnitude but opposite in direction to F, and F, , care must be taken while writing the sign of the inflow or outflow term. Flow ovcr a curved surface may also be analysed with similar equations as equations (6.36) and (6.37).

6.11.2 Moving Vanes In Figure 6.15, a moving vane is shown with the liquid flowing onto it tangentially. Let u be the vane velocity. The control volume in this case encloses the liquid in contact with the vane, with its control surface normal to the flow at sections 1 and 2 . Let A,, V,, Q, be the area, velocity and discharge at the inlet respectively. The impinging jet is deviated through an angle 0 by the moving vane. The application of equation (6.34) to the control volume yields.

F, = p (V, - u ) ~ A l (1 -cos 0) (6.38)

Figure 6.15 : (a) Moving Vane (b) Vane viewed as Stendy State Pmblent by Superposition of u to the ],eft.

For a series of vanes, the above relations become,

F, = p Q1 (V1 - U) (1 - cos 0)

F, = p Q, (V, - U ) sin 0 (6.41)

The forces resulting from the motion over the moving vanes have turbo machinery applications. Example 6.3 :

Detenniiie the force components for a moving vane as show1 in Figure 6.16 due to the water jet and also calculate the rate of work d o ~ e on the vane.

( a (b )

Figure 6.16 : Jet acting on n Moving Vane

Solution :

F, = p (V, - u ) ~ A, (1 - cos 0)

= 1000 x (100 - 60)' x 0.005 x (1 - cos 160")

= 15.52 kN

F, = p (V, - u ) ~ A, sill 8

= 1000 x ( I 00 - 60)~ x 0.005 x (sin 1600) = 2.74 kN

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The power exerted on the vane is

u F, = 60 x 15.52

SAQ 7 Find the force exerted on a fixed vane, when a jet discharging 50 litres/sec water at 40 m / s is deflected through 120'.

SAQ 8 Determine the power that can be obtained from a series of vanes curved through 155", moving 18 m / s away from a 75 litres/sec water jet having a cross section of 25 cm2. Draw the polar vector diagram and calculate the energy remaining in the jet.

6.11.3 Force Exerted on a Pipe -Bend Consider the pipe bend as shown in Figure 6.17 which changes the direction of flow through an angle 8. For simplicity, it is assumed that the axis of the bend is in the horizontal plane. The control volume selected includes the inlet and outlet sections (1) and (2) of the pipe bend. Let V,, p, , A , and V2, p,, A , be the average velocity, pressure and the area of flow at sections (1) artd (2) respectively; L e t - F X M Y be the components

( 1 1

Figure 6.17 : Momentum Equation Applied to a Reducer Bend

of the force exerted on the fluid by the pipe bend in the x and y directions respectively. The other external forces acting over fluid in the control volume are p, A, at (1) and p2A2 at (2). The momentum equation in the x direction may be written as

p, Al - p2 A2 cos 8 + F, = p Q (V2 cos 8 - V, ) (6.42)

Similarly the momentum equation in the y-direction is

- p2 A, sin 8 + F, = p Q (V, sin 8 - 0 ) (6.43)

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The above two equations together with Bernoulli's and continuity equation enable the determination of Fx and Fy. By knowing Fx and Fy , the to@ force, F can be determined. The force exerted by the fluid on the pipe bend will be equal in magnitude but opposite in the direction to F. Example 6.4 :

Determine the forces exerted by the pipe boundary on the fluid for the following case,p, = 2.2kg/cm2, V, =4.4m/sec,D, =30cm, 8 = 60°, D2= 15cmand p=102msl/m3.

( 2 )

I Figure 6.18

Solution :

Velocity at section (2),

Applying Bernoulli's equation between sections (1) and (2)

P2 - = 7.199 mof water Y

Momentum equation in the x-direction

pl Al - p2 A2 cos 8 + Fx = pQ (V2 cos 8 - Vl )

Substituting the values

2.2 x 1@ x : x (0.30)~ - 0.7199 x 10" x x (0.15)~ cos 60'' + Fx

. . Fx = - 1351.9 kg

Momentum equation in the y-direction

Fy - p2 A2 sin 8 = pQ (V2 sin 8)

Substituting the values

Fy - 0.7199 x lo4 x x (0.15)~ x sin 60' = 102 x x (0.30)~ x 4.4 x 17.6 sin 60' 4 4

SAQ 9

A 45" deflection angle reducing bend lies in a horizontal plane and tapered from 65 cm diameter to 35 cm diameter at the outlet. The pressure at the inlet is 16 kPa and the flow through the bend is 0.55 m3/s of water. Assuming friction loss of 20% of the kinetic energy at the inlet, compute the magnitude and direction of the resultant force exerted by the water on the bend:

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6.11.4 Flow through a Nozzle In the case of flow through nozzles as shown in Figure 6.19, there is change in the magnitude only and not in its direction. The pressure at the exit being atmospheric is equal to zero. - I

(1 (2 1

L -----f--- i, (1 1

Control voluma

Figure 6.19 : Force on the Nozzle

The force exerted by the nozzle in the direction of flow is given by

p i 4 + F, = PQ (V2- V1) (6.44)

The change in the momentum in the y-direction is zero. Example 6.5 :

What is the force on the elbow-nozzle assembly as shown in the Figure 6.20. Water issues as a free-jet from the nozzle. The interior volume of the elbow-nozzle assembly is 0.15 m3. The elbow-nozzle assembly is in a horizontal plane.

3 crn / Control volume

Figure 6.20

Solution :

Let the control volume be as shown in the Figure 6.20. Applying Bernoulli's equation, the pressure at the entrance to elbow may be obtained.

yielding , - - - 42.1 m of water Y

Momentum entering the control volume

= P Q V I

= 1 0 2 ~ ~ ~ 0 . 1 2 ~ ~ 1 . 8 ~ 1 . 8 4

Momentum leaving the control volume

= P Q V ,

The force exerted by the elbow -nozzle assembly on water be F, . Then

Fx + P, A1 = PQ - Vl)

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Fldd Dpda - I SAQ 10

Water flows at the rate of 0.7 m3/s through the pipe transition. If the pressure at 80 cm section is 900 kg/m2, what is lhe pressure in the 50 cm section ? What force will be required to produce the cliruige in momentum of water as it passes through the uansi tion?

6.11.5 Momentum Theory for Propellers The propeller changes the momentum of the fluid within which,it is submerged and develops a thrust that is used for propulsion. Let us consider a propeller moving to the left with a velocity V , through the stationary fluid which is frictionless and incompressible. Figure 6.21 illusuales the flow conditions upstream and downstream of

Figure 6.21 : Propeller i ~ r a Fluid stream

the propeller. The flow is undisturbed at section 1, upstream from the propeller and gets accelerated as it approaches tlie propeller, owing to the reduced pressure on its upsrream side. As the fluid passes through the propeller, energy is added to it by the propeller resulting in increased pressure at section 3. The velocity V does not change across the propeller, from 2 to 3 on account of same area of flow. In passing through the propeller, the fluid pressure increases causing the reduction of cross section at 4. The pressure at 1 and 4 is that of the undisturbed fluid, which is also the pressure alnng the slipstream boundary. Applying the momentum equation (6.34) to the control volume enclosed by the slipstfeam boundary and sectio~ls 1 and 4, the force acting on the fluid is

in which A is the area swept by the propeller blades. The force F is also equal to the force due to the pressure difference between seclions 2 and 3.

i.e., = ( ~ 3 ~ ~ 2 ) ~ = PAV(V4-V1) (6.46)

Applying Bernoulli's equation between the sections 1 and 2 4 1

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and between sections 3 and 4

Since the velocity V, = V, = V and p, = p4 , equations (6.48) and (6.49) yield

Equating equations (6,42) and (6.50)

Which shows that the velocity through the propeller area is the average of the velocities upstream and downstream from it.

The work done by a propell& per unit time (power transferred) is the product of propeller thrust and velocity

Power = F Vl = pQ (V4 - Vl) Vl (Watts) (6.52)

The power input required to increase the velocity of fluid from Vl to V4 is

Power input = PQ W,'- v:)

2

It can also be expressed as power output plus the power loss

Power input = pQ (V4 - Vl) V1 + PQ (V4 - v1)'

2

The theoretical efficiency e is thus given as

e = - 2Vl Vl output - - - - output + loss V, + V4 V

If AV = V4 - Vl , substituting into the above equation,

Which shows that maximum efficiency is obtained by increasing the velocity of AV

slipstream as little as possible, or for which - is minimum. Airplane propellers under Vl

optimum conditions have the efficiencies of about 85 percent which drops rapidly for the speeds above 650 kmlhr owing to compression effects. Ship propeller efficiencies are only around 60 percent owing to restrictions in diameter.

Example 6.6 :

An airplane travelling 450 kmlhr through still air, y = 12 ~ / m ~ , discharges 1100 m3/s through its two 2.25 m diameter propellers. Determine (a) the theoretical efficiency (b) the thrust (c) the pressure difference across the propellers and (d) the theoretical power required.

Solution :

Aiilane velocity.

Velocity of flow, V = 1100 = 138.3 m / sec 2 x x (2.25)'

Vl Theoretical efficience, e = - v

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b) The velocity, V4 = 2 V - Vl

= 151.6 mlsec

The thrust from the propeller, F = p& (V4 - Vl)

= 32.6 kN

c) The pressure difference across the propellers

Y3 - Y2 = PV (V4 - Vl)

d) The theoretical power

6.11.6 Jet Propulsion In jet engines, air at rest is taken into the engine ,and burned with a small amount of fuel. The gases are then ejected with a much higher velocity th,m in a propeller slipstream. The jet diameter is necessarily smaller than the propeller slipstream. If the mass of the fuel burned is neglected, the propelling force F is given as (Figure 6.22)

in which Vabs = AV is lhe absolute velocity of the fluid in the jet and pQ is the mass per unit time being discharged. The theoretical efticiency is the same as given in equation (6.56). For a given speed V,, the resistance force F is determined by the body and fluid in which it moves. Hence for V,, in equation (6.57) to be very small, pQ must be very large.

Figure 6.22 : Steady State Flow through Jet Engines

Considering the jet propulsion of aircraft, let M, be the mass of air per unit time and R the ratio of mass of fuel burned to mass of air. Then the propulsive force F is given as

F = M& (V2 - Vl) + R Ma, V2 (6.58)

The second term on the right hand side is the mass of the fuel per unit time multiplied by its change in velocity. Rearranging the terms in the above equation,

F = M,, [V2 (1 + R) - Vl] (6.59)

The mechanical efficiency, e which is the ratio of useful work to the sum of useful work and kinetic energy remaining is given as

The efticiency is maximum when V, = V,, as the combustion products are brought to rest and no kinetic energy remains in the jet.

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I I 6.1 1.7 Rocket Mechanics I

The rocket motor carries an oxidizing agent to n ix with its fuel so that a thrust is developed which is independent of the medium through which it travels. Coilsidering the control volume as the on surface of the rocket with a plane area normal lo the jet across the nozzle exit as shown in Figure 6.23.

Figure 6.23 : Control Surface for Analysis of Rocket Acceleration

The control volume has a velocity equal to the velocity of the rocket. Lel r be the air resistance, MR the mass of the rocket body, Mfthe mass of fuel, ti1 the rate at which fuel is being burned and v, the exit gas velocity relativc to the rocket. V, is the actual velocity of the rocket (and of the frame of reference), and V is the velocity of the rocket relative to

dV dV, . the frame of reference. V is zero, but - = -is the rocket accelerillion. The basic

dl dl linear-momentum equation for the y-direction gives

Which becomes

Since V is a function o f t only, the. equation c;m be written as a total differential equation as

c/V dv, rn v, - g (MI. + M, 1 - r -- - - d t d t MR + Mf (6.64)

The mass of propellant reduces with time, i'or constant burning rate rn, MI= M f - m t, 0

with Mfi the initial mass of fuel and oxidizer. Gravity is a function of y, and the air resistance r depends on the Reynolds number, Mach number and on the shape and size of the rocket. By combining the mass of the rocket and fuel together as M R + f , the equation (6.84) can be expressed as

The theoretical efficiency is given as inv, 4E

e = (6.66)

k [ E + 2 )

in which m ~ , =is the available energy input per unit time, E is Ule available energy in the propellant per unit mass. When the propellant is ignited, the ilvailable energy is converted into kinetic energy. The jet velocity relative to the rocket is v, = m. Simplifying equation (6.66),

2 v,

Vr The efficiency is maximum when - = 1. The rocket is lifted and gets accelerated when

i Vl the thrust on it m vr should exceed the total weight plus resistance.

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Rdd--1 SAQ 11 A blower having an efficiency of 75% is to supply 15 m3 of air per minute to a 16 cm pipe under a pressure equivalent of 6 cm of water. If the 32 cm intake pipe draws directly from the atmosphere, what should be horse power of the motor? What will be the pressure in the intake pipe? p& = 1.284 kg/m3

Example 6.7 :

An airplane consumes 1.5 kg fuel for each 25 kg air and discharges hot gases from the tailpipe at V, = 1850 m / s. Determine the mechanical efficiency for airplane speed of 300 m / see

Solutlon :

v2 1850-6.17 For 3 0 0 m / s , - - -- v1 - 300

The ratio of mass of fuel burned to mass of air,

The mechanical efficiency, 1

= 0.281 SAQ 12

If a boat requires a force of 2.5 kN to move it through water at 28 krn / hr with a 16 cm diameter jet, determine the power required by the boat.

Example 6.8 :

Determine tHe burning time for a rocket that initially has a gravity force of 4.8 MN, of which 75% is propellant. It consumes fuel at a constant rate and its initial thrust is 15% greater than its gravity force. v, = 3400 m/s.

Solutlan :

From the thrust relationship

m v, = 1.15 W

m x 3400 = 1.15 x 4.8 x lo6 . m = 1623.5 kg / sec.

The available mass of propellant = 4.8 lo6 = 4.89496 x id kg 9.806

Hence, the burning trme

= 226.1 sec.

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6.11.8 Losses due to Sudden Expansion W B I - The losses due to sudden enlargement in a pipeline may be calculated with both energy and momentum equations. For steady, incompressible, turbulent flow through the control volume between sections 1 and 2 as shown in Figure 6.24, assuming uniform velocity over the flow cross sections, application of equation (6.34) produces

in which pl, A,, V, & p2, A2, V2 are the pressure, area and velocity at sections 1 and 2 respectively.

1 t 2

( a ) Figure 6.24 : Sudden Expa~~sion in a Pipe

At section I, the radial acceleration of tluid particles in the eddy along the surface is small, and so generally a hydrostatic pressure variation occurs across the section. The energy equation applied to sections 1 and 2 yields

in which hl is the loss between sections 1 a11d 2

Solving for - P2) in both h e above equations, Y

Since, A, V1 = A2V2

which indicates that the losses in turbulent tlow are proportional to the square of Ule velocity.

6.11.9 Hydraulic Jump The hydraulic jump is the second application of the basic equations to detennine losses due to a turbulent flow situation. Under certain conditions a rapidly flowing liquid in an open channel suddenly changes to a slowly flowing stream with a larger cross sectional area xld a sudden rise in the elevation of the liquid surface. This phenomenon is known as the hydraulic jump. This is an example of steady non-uniform flow. Here, the rapidly flowing jet expands as shown in Figure 6.25, and converts kinetic energy into potential energy and losses.

Figure 6.25 : Hydraulic Jump in a Rechigular Chmmcl

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PkidDynrla-I continuity, mo~nentum and energy equations. By considering unit width of the channel, A , = y , ; A, = v2 ; where A, , y, and A?, v2 are the area and depth of liquid before and after lhe hydraulic jump respectively. The momentum equation becomes

in which V , and V? are the velocity of flow b c l r c ,and ai'tcr the jump respectively. The energy equatiod applied Sor points on thc liquid surface results,

in which I l j represents losses due to the jump. Eliminating V, in equations (6.72) and (6.73) leads to

The depths y, and v, are referred to as coiljugate depths. Solving lhe energy equation for hj and eliminating V , and V, gives,

l11e hydraulic jump is conlnlonly used at the ends of chutes or bottoms of spillways to destroy much of the kinetic energy in Ule flow. It is also XI effective mixing chamber, because of the violent agitation that takes place in the roller. Example 6.9 :

If 15 m'/s of water per metre of width flows down a spi1lw:iy onto a horizontal flodr XI^ the velocity is 75 nl I s. delenn~nc the downstrea~~r depth required to cause a hydraulic jump and the losses by the jump.

Solution :

-A The depth of water before tllc jump, yl - V x h

YI + The depth of w;iler ;dtcr tllc junlp, v2 = - - + -- 2 R

1 0 2 - Y l ) '

The losses due to the jump, h, = YI V2

The power loss = y Q x losses

SAQ 13

A jump occurs ill a 6.2111 widcchannel c;urying 15.5 m'ls water at a Cptll of 320 nlin. Determine y,, V, and the losscs 111 inetre-Newtons per Newton and kilo Watts.

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6.12 SUMMARY

In this unit, the concepts needed for the analysis of fluid motion are introduced. The basic equations enable us to predict the fluid behaviour. These equations are the equations of motion, continuity, momentum and first law of thermodynamics. The control-volume approach is utilized in the derivation of the continuity, energy and momentum equations. One-dimensional flow theory is developed in this unit with applications limited to incompressible cases, where viscous effects do not predominate. For the frictionless flow, along a streamline of an incompressible fluid, the relationship between the pressure, elevation and velocity is established through Euler's equation. Integration of the Euler equation for steady, incompressible fluid flow without friction yields the Bemoulli's equation. In practical applications of Bemoulli's equation, the restriction of frictionless flow is accommodated by introducing a loss of energy term The momentum principle is derived from Newton's second law and is considered under linear momentum equation category. The momentum equations are applicable to a control volume and are vector equations. In the application of the linear momentum equation, the control volume can be assumed arbitrarily. However, it is usual practice to assume a control volume such that the boundaries are normal to the direction of flow at inlets and outlets. It can also be assumed to be inside the flow boundary and has the same alignment as the flow boundary. In one-dimensional flow analysis, the kinetic energy and momentum correction factors are introduced to correct the average velocity and the momentum flux for the variation of the velocity across the area of cross section. The major applications of the momentum equation include impact of jets on blades, momentum theory for propellers, bends, jet propulsion, rocket mechanics, losses due to sudden expansion and the hydraulic jump.

6.13 ANSWERS TO SAOs

SAQ 1

The average velocity, V = % r v dr in which r = ro - y. nro 0

Substituting for rand v,

t The kinetic-energy factor, a = I A

I The momentum correction factor, P = I

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SAQ 2

Velocity at 12 cm diameter section

(a) Pipeline horizontal

Applying Bernoulli's equation at sections (1) and (2)

(b) Pipeline vertical with flow downward Bernoulli's equation applied to sections ( 3 ) and (2) gives

. . f i = 4.44 kg/cm2

SAQ 3

Applying energy equation between points 1 and 3,

v," Pl v," P3 - + - + z1 = - + - + z3 + losses ,- , %? Y 2g Y

K V: in which a e losses from 1 to 3 are expressed as - .

2g

v," 'The losses are K - = 0.269 m N/N 2~

2 v,2 The energy equation applied between 1 and 2 with losses, - K - = 0.179 m, is

3 2g

P2 - = - 3.71 m of water Y

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SAQ. 4

Applying Bernoulli's equation between the points 1 and 2

= 9.08 m / s

The discharge through the orifice is

Q = A,V*

= 0.046 m q s

SAQ 5

Velocity of flow at 35 cm section.

v1 = 0.35

= 3.64 m / s 71: - x 0.35~ 4

Velocity at 25 cm section

Applying Bernoulli's equation between the pipeline sections at the foothills and at the hill top,

Horse power required, H.P = 75

= 280 H.P SAQ 6

Velocity at 25 cm section = 0.25 x 4 = 5.09 m / s

71: x 0.25~

Velocity at 45 cm section = 0.25x4 = 1.57'm/s 71: x 0.45~

Energy head available at (1) with reference to datum at (2)

Energy head at (2) = 0 - 0.2 x lo4 1 .572 +-

lo3 2 x 9.81

= - 1.874 m

Energy head utilized by the turbine = 22.82 - (- 1.874)

= 24.694 m

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:. Horse power delivered to the turbine = la!! 75

SAQ 7

C o n t r o g volume R O - i'

v 1 /,-/' - Fx

4 #?I--/& IL @

A; Y

Figure 6.26 : Free Jet Impinging un a S~nooth, Fixed Vane

Force, Fx = p Q Vl - p Q Vl cos 8

= 1000x50x10-~x40 - 1ooox50x10-3x40cos1200

= 3 k N

Force, F, = p Q Vl sin 8

= 1 0 x 50 x40 x lo9 x sin 120'

SAQ 8

I b l Figure 6.27 : Flow through Moving Vanes

For a series of vanes, Fx = p Q, (V, - u) (1 - cos 8)

= lOOOx75 x (V - 18)(1 - cos 155')

The jet velocity Vl = = 30 m / s 25 x 1 0 - ~

Substituting the values, Fx = 1716 N

The power obtained = 1716 x 18 = 30.89 kW

The components of absolute velocity leaving the vane as shqwn in Figure 6.27 (b).

vz' Energy remaining in the jet= Q y - 2g

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SAQ 9 ~ a s i e Equations

Figure 6.28

Applying Bernoulli's equation, to sections 1 and 2

By momentum equation in the x-direction,

p1 A, - p2 A2 cos 8 + Fx = pQ (V2 cos 0 - Vl)

i.e., 15000 x ?I x 0.65~ - (-251 x 2 x 0.35~ x cos 459 + Fx 4 4

= 1000 x 0.55 x (5.72 cos 45" - 1.66)

:. Fx = -3.683 kN

Momentum equation in the y-direction yields

Fy - p2A2 sin8 = p Q, (V2sin8 - 0)

Fy - (- 251) x f x 0.35~ x sin 45O = 1000 x 0.55 x 5.72 x sin 45O 4

Hence Fy is 2207.5 N in positive y-direction. /

Resultant force F = = J36832 + 2207.9 = 4293.9 N

Inclination of F to ( -x ) direction, 8 = tan-' ' = tan-' - pi) (ZY)

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SAQ 10

u. 1 Velocity in the 80 cm section, Vl = - - - 1.393 m l s 'It - x 0 .8~ 4 . / Control volume

0 ""'--"' 1

Figure 6.30

Velocity in the 50 cm section,

v - 2 - 11:

0'7 = 3.565 rn/s - x 0 . 5 ~ 4

Applying Bernoulli's equation to 80 cm and 50 cm sections, a a

P2 = 0.351 m of water Y

Let Fx be the force exerted by the nozzle on water. The momentum equation may be written as

~ 1 A 1 + Fx - ~ 2 A 2 = PQ (V2-Vi)

Substituting the values,

. . = - 228.4 kg

The above force acts from right to left. SAQ 11

0

32cm Blower

0 Figure 631

Velocity in 16 cm pipe = 15 = 12.43 rnls 7t

6 0 ~ ~ ~ 0 . 1 6 ~

lZA3 - 3.11 m / s Velocity in 32 cm pipe = - - 4

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Energy at section (2) = Energy at section (1) + Energy supplied by the blower

- V' + = - ~2~ + - P2 - Energy supplied by the blower 2g Y 2g Y

2 2

:. Energy supplied by the blower per unit weight of air = v2 - Vl P2 - P1 +- 21: Y

(12.43)~ - (3.11)~ Energy supplied by the blower = + 588.4

2 x 9.81 1.284 x 9.81

= 54.09 Nm/ N of air

Output of the blower = y Q x Energy

15 = 1.284 x 9.81 x - x 54.09 60

.'. Horse power required 15 0.131 x 9.81 = 5 4 . 0 9 ~ ~ ~ 75 x 0.75

(p& = 0.131 m slug / m3)

= 0.309 H.P. SAQ 12

Here, the method of jet propulsion can be considered in which water is taken in at the bow and discharged out the sters by a 100 percent efficient pumping system. To analyse the propulsion system, the problem is converted to steady state by super position of the boat speed. V, on h a t and surroundings (Figure 6.32).

Figure 6.32

Boat Velocity, V, = 28 loo' = 7.78 m / s

60x 60

Velocity of the jet - - A- = A = 49.74, nd (0.16)~' - 4 n : 4

The propelling force, ' F = pQ (V2 - Vl) = p Q Vhs

i.e., 2500 = 1000 Q (49.74 Q - 7.78)

The efficiency, 1 e = 7 = 0.66

F V1 The power required = - e

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SAQ 13

Q The velocity of flow before jump, Vl = - b y1

'Ihe depth of water after the jump,

The velocity of flow after tbe jump, V2 = l 5 = 1.36 m l s b 6 . 2 ~ 1.84

The losses due to the jump, hL = 02 -yb3 - - f 0 . ~ ~ 1 ' = 1.49 rnN/N 4)'1~2 4 ~ 1 . 8 4 ~ 0 . 3 2

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UNIT 7 FLOW MEASUREMENT

Structure 7.1 Introduction

O b j ~ ~ t i ~ e S

7.2 Measuremmt of Pressure 7.2.1 Slatic Pressure

7.2.2. Dynamic Prearmre 7.2.3 Total h x u r e 7.2.4 Piemmetric Head

7.3 Measurement of Velocity 7.3.1 Pitot Tube 7.3.2 Praudtl Ttrbc

Measurement of Discharge 7.4.1 Orifices

7.4.2 Mouth Pieces 7.4.3 Notchas and Weirs 7.4.4 Ventwimeter 7.4.5 Orifice Meter 7.4.6 Flow N o d e 7.4.7 VenhPi Flume

7.5 Summary

7.6 Answers to S AQs

7.1 INTRODUCTION

Flow measurement is one of the most frequently encountered problems in fluid mechanics by field engineers. Efficient and accurate measurements are absolutely essential to arrive at correct estimates with minimum errors. For this purpose, the engineer should be well equipped with the basic knowledge of various methods to measure fluid properties and phenomena. The purpase of this unit is to give the principle of fluid measurements by various dpGices.

Objectives After studying this d, yod should be able to measure

* static pressure,

* dynamic pressure,

'F velocity, and

The measurement of fluid pressure is essentiaI in Qtermining the velocity of the fluid or the rate of flow using energy equation.

7.21 Static Pressure The static pressure of a fluid in motion indicates the pressure measured without disturbing its velocity. When the flow is 'form and stre-es are gafatlel, the pressure variation across the flow will be ? yqlrostatic. Hence, in swh cases if the pressure intensity at the bomdary is measmed, the pressure at my point may be computed. Figure 7.1 shows a morneter connected to a small hole drilled in the boundary of a pipe line. The manometer reading indicates the static pressure at the boundary. For accurate results, the tube must be normal and flush with the boundary and the diameter should be very small. In the case of rough surface at the boundary, the static pressure can be measured at the mid-section using a static tube as shown in Figure 7.2.

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figure 7.1 : Static Plpssurp M e ~ ~ u n m e n t by Manometer

To manometer - n C 7d -J -static pressure

hole

Figure 7.2 : Static Tube

The tube consists of a smooth cylinder with round nose having radial holes at a minimum distance of 3d from the upstream end of static tube. After the disturbance at the upstream end, the streamlines become parallel again hereafter and hence the flow is undistnrbed. Smaller diameter for the tube increases the accuracy of the measurement. Another way of measuring static pressure is based on the flow net analysis using a cylindrical tube.

7.2.2 Dynamic Pressure This pressure is due to the velocity of the fluid and can be expressed as

The sum of static pressure and dynamic pressure constitutes the total pressure or stagnation pressure. By measuring static pressure and total pressure and connecting to opposite ends of a differential manometer yields the dynamic pressure head. The above procedure to measure the dynamic pressure is essential, since practically it is difficult to read the dynamic pressure head from a free surface using a simple pitot tube.

7.2.3 Total Pressure The total pressure is the pressure which includes the pressure due to hydrostatic force and velocity. To measure the total pressure, a tube having a short right angled limb with a rounded and having a hole is inserted facing the flow. This causes stagnation at the entrance of the tube. At this point the velocity is reduced to zero and the velbcity head is converted into pressure head indicating the total pressure. This tube is known as pitot tube or total head tube.

7.2.4 Piezometric Head Energy of a fluid in motion consists of pressure energy, kinetic energy and potential energy. The sum of pressure head and potential head is h o w n as piezometric head. Piezometers are the simplest manometers which are used to measure the pressure of a liquid at rest or in motion. The piezometer is a tube connected vertically to a small hole in the wall of the conduit with a minimum diameter of 0.625 cm to reduce capillary effects (Figure 7.3).

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Ferc 73 : Measurancd of btPl Hud

Example 7.1 :

Water flows through a pipe of 25 cm diameter at a discharge of 69.5 Ftres/sec. At a section which is 6 m above the-datum, the pipe pressure is 160 kNlm . Determine the total pressure (head) of the flow.

Vebcity, V = ii2 A

Total pressure per unit weight of fluid

v 2 P - - - + - + Z 2g 'I

= 22.43 m

SAQ 1

(i) A piezometer opening is used to measure

a) the pressure of a static fluid

b) the velocity in a flowing fluid

c) the total pressure

d) the dynamic pressure

e) the undisturbed fluid pressure.

(ii) A static tube is used to measure

(a) pressure in a static fluid

(b) the velocity of a flowing fluid

(c) the total pressure

(d) the dynamic pressure

(e) the undisturbed fluid pressure.

Select the correctanswer from the above.

7.3 MEASUREMENT OF VELOCITY

73.1 Pitat Tube Velocity measurement is one of the most important measurements in fluid mechanics. It is useful in determining the discharge, velocity profile and shear stress across a section.

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Pitot tube which is named after Henry Pitot (1730) is used to measure the velocity of the fluid. (Figure 7.3). The difference in the readings of pitot tube and piezbmeter (h) indicates the velocity head.

Example 7.2 :

A pitot tube inserted at the centre of 0.22 m diameter pipe is connected to ;b static tube through inverted U-tube. When the pipe carries water, a difference of 8 cm was indicated in the U-tube. Determine the velocity of flow (Figure 7.4).

t I

Figure 7.4 : Pitot Tube

Solution :

Applying Bernoulli's theorem between points 1 and 2 (dagnation point)

7.3.2 Prandtl Tube The combination of static pressure tube and stagnation pressure (total) tube is known as Prandtl pitot tube. When the boundary surface is rough and the static pressure measurements at the boundary are not possible, this device can be used. The front portion of the tube is rounded (Figure 7.5) to avoid separation of the flow and on its shaft, holes are provided at certain distance where stream lines become parallel. The two tubes are enclosed in the same tube and are connected to a differential manometer. The manometer

To differential m a m e t e r t t

Holes for static p r e s s u m t Flgure 7 5 : Prandtl Pltot Tube

reading indicates the velocity head from which velocity can be determined. A coefficient Cd (0.95 to 0.99) is evaluated to eliminate the error in the velocity caused due to slight imperfection in the fabrication of the device. Hence the actual velocity V,,, is given as

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I

SAQ 2

Tick (J) the correct answer or most appropriate from among the alternatives given

(i) The simple pitot tube measures the

(a) static pressure

(b) dynamic pressure

(c) total pressure

(d) velocity at the stagnation point

(e) difference in total and dynamic pressure.

(ii) IIhe pitot static tube measures

(a) static pressure

(b) dynamic pressure

(c) total pressure

(d) difference in static and dynamic pressure

(e) difference in total and dynamic pressure.

7.4 MEASUREMENT OF DISCHARGE

7.4.1 Orifices An orifice is an opening in the wall of a contamer. It can be of any shape. A standard orifice has a sharp edgg and the flow through such an orifice is shown in Figure 7.6.

Area =

( a ) l b l

Figure 7.6 (a) Elow Uamugh M Ondice (b) An Otitice

As the fluid flows, the streamlines converge to form a jet having a cross section smaller than the orifice, known as vena contracta. Since the fluid patticles can not change their direction abruptly, the position of vena contracta is slightly away from the wall. Streamlines are thus curved between the orifice and vena contracta. However, at the vena contracta the streamlines are parallel and the pressure is atmospheric.

Let a constant head H be maintained above the centre of the orifice. Bernoulli's equation applied between point 1 on the free surface and point 2 at the centre of vena confracta gives

v,' O + O + H = - + O + O 2g

v, = (7.4)

The above expression indicates theoretical velocity of the jet which doesn't include the frictional losses on the periphery of the orifice. Hence a coefficient known as velocity coemcient, C, , is introduced in calculating the actual velocity which is the ratio of the actual velocity V, to the theoretical velocity Vl . Hence, V, = C, (7.5)

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~MdDIwrkr-I The actual discharge Qa through the orifice is the product of actual velocity at the vena contracta and the area of the jet. The area of the jet at vena contracta A, is smaller than the area of the orifice A,, and their ratio is known as coeffldent of contraction (C, ). The actual discharge can thus be given as

Qa = C, C, A, (7.6)

The product of the above two coefficients is known as discharge coefficient C,

i.e., Cd = Cc C,

Hence, a = cd 4 ax If the downstream side of the orifice is submerged, the discharge equation gets as

Qa = Ci 43gT' (7.8)

in which H' is the difference between upstream and downstream 1evels.C; is the new discharge coefficient, the value of which is slightly lesser than C, . The value of coefficient of velocity Cv for circular orifices varies from 0.95 to 0.98. For an orifice of given dimensions, Cv slightly increases with increase in H . For circular orifices, with the diameter less than H, C, is equal to 0.61. The values of C, and C, however are to be determined experimentally. The following are some of the metbods used in determining the hydraulic coefficients, Cv , C, and Cd . (1) Trajectory method : By m u r i n g the co-ordinates of a point 3 on the trajectory

with respect to the centre of vena contracta (Figure 7.6 (a)), the actual velocity, Va can be determined.

Let ' r ' be the time required for the duid partide,to reach the point 3 from vena contracta.

Tben x, = V, t

The vertical distance travelled by the particle at the same time, when there is no initial velocity in that direction is

Eliminating t from the above two equations.

Va Va The coefficient of vetocity C, = - = v, XI

(2) Pitot tube method : The actual velocity Va at the vena contracta is measured using a pitot tube. However, care must be taken while inserting the pitot tube so that the jet velocity is not altered. The coefficient C, can thus be determined knowing the value of vc .

(3) Callipers method : The diameter of the jet at vena contracta may be determined using a callipers. Knowing tbe diameter of the orifice, the coefficient of contraction C, may be determined.

(4) Momentum method : A small tardr suspended on knife edges is shown in Figure 7.7. The tank is balanced initially by putting weight, when the M ~ c e is closed. When the orifice is functioning, a force imparts momentum to the jet by creating an equal and opposite force F which acts against the tank. The tank is now levelled by putting additional weight W. Using the momentum equation,

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I I

EPpm 7.7 : Momeakm method

Also, by taking moments about the knife edge

X Fy = W x :. F = W - Y

From equations (7.1 1) and (7.1 2)

Since the actukl discharge Qa is measured, Va is the only unknown in the equation.

(5) Through measurements : The coefficient of discharge, Cd can be obtained by measuring the area A,, the head Hand the discharge Q, by gravimetric or volumetric means.

Example 7 3 :

A large tank resting on the floor maintains water upto a height of 8 rn A circular orifice of 12 mm diameter is located 2 m above the floor level. For C, = 0.97 and C, = 0.61, determine the discharge and horizontal distance from the tank where the jet will strike the ground. Also determine Cd .

Solution :

The acting head, H = 8 - 2 = 6 m

Theoretical velocity, v = W

Actual velocity, Va = Cv V

= 0.97 x 10.85 = 10.52 m / sec.

Area of the jet at vena contracta = Cc a

= 0.61 x 1.13 x lo4

Actual discharge, Qa = Va x Area at Vena contracts

= 7.258 x lo4 m3/sec

The coefficient of discharge, Cd = Cv Cc = 0.97 x 0.61

Vertical distance travelled by the jet, y = 2 m

If time required is t, then y = ~ g t , 1 2 SO

Horizontal distance travelled during this period, x = V, t = 6.72 m

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Losses in Oriflce Flow :,

The loss of energy in flow through orifice can be determined by applying the Bernoulli's equation between point 1 and vena contracta (Figure 7.6).

Where HL is the head loss from point 1 to vena contracta.

The above equation indicates the losses in terms of H and C, or V, and C,

The orifice discussed above (Figure 7.6) is known as standard orifice, where the sharp edge of the orifice provides only a line contact with the fluid. The characteristic of this orifice is that the thickness of the wall or plate is small compared to the size of the opening. When the head of fluid above the orifice is less than 1-5 times the diameter or height of the orifice, it is classified as large orifice. The discharge through a large rectangular orifice is given as

in which B is the width of the orifice, H,, and H, are the heads above the tog and bottom edges of the orifice.

SAQ 3

(i) Explain briefly how the coefficient of velocity of a jet issuing bough an orifice can be determined experimentally.

Find an expression for head loss in an orifice flow in t e r n of coefficient of velocity and jet velocity.

The head loss in flow through a 8 cm diameter orifice under a certain head is 20 cm of water and the velocity of water in the jet is 7m / sec. If the coefficient of discharge is 0.61, determine

(a) head on the orifice

(b) diameter of the jet, and

(c) .cv +

(ii) A 75 mm diameter orifice discharges 907.6 kg of water in 32.6 sec under a head of 4.90 m. The x and y coordinates of a point on the jet are 4.80 m and 1.3 m respectively, Determine C, , C, and Cd and the head loss.

7.4.2 Mouth Pieces A mouth piece is a short tube connected to the orifice whose length is usually not more than two or three times its diameter. The mouth pieces may be cyl~drical, converging or diverging. The moult4 pieces are also used as measurhgdevices and €he different types are showri in Figure 7.8 . -

l a ) Straight ( b l Conwgent ( c l Di vorgent d l Internal of Borda's

Flpre 7.8 r DiCIermt types of Mouth Picecs

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A standard mouth piece is a cylindrical tube having a length of 2 to 3 times diameter with a sharp cornered entrance (Figure 7.8 a). In such tubes the jet contracts first and then expands filling the tube. Since the tube is full at the outlet, C, is usually considered to be equal to unity and mean value of Cv is about 0.82 .

I I ( a I Standard ( b Y Rounded entmce

Figure 7.9 : Flcrw through Cylindrical Mouth Pieces

In the case of standard mouth pieces, if the entrance is rounded, it reduces the losses due to re-expansion to fill the mouth piece as shown in Figure 7.9. In such cases the coefficient of velocity C, is relatively higher i.e. 0.98. In the case of diverging mouth piece; its length should be equal to 9 times its diameter at the inlet and the angle of divergence equal to 5' for better results. Borda's mouth piece is a short cylindrical re-entrant tube which projects inside. Its length is equal to its diameter and has sharp edges to ensure contraction and in this case the jet will not touch the sides of the tube. The average hydraulic coefficient in this case are C, = 0.98, C, = 0.52, C, = 0.51. A relationship can be established between h e hydraulic coefficients using the principle of momentum as

2 2 c d c v = 2 c v c , = 1 (7.16)

Example 7.4 :

A borda mouth piece 8 cm in diameter has a discharge coefficient of 0.51, what is the diameter of the issuing jet?

Solutian :

We have 2 c d c v = 1

Area of the jet = C, A

4 4 ~ Diameter of the jet = -

It

= 5.77 cm SAQ 4

An external mouth piece converges from inlet upto the vena contracta to the shape of the jet and then diverges gxadually as shown in Eigure 7.10. The diameter at the vena contracta is 3 cm and the head over the centre of the mouth piece is 1.6 m. The head loss in the contraction may be taken as 1% and that in the divergent portion 5% of the total energy head before the jet. What is the maximum discharge that can be drawn through the outlet and what should be the corresponding diameter at the

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Assume that the pressure in the system may be permitted to fall uptp 8 m below atmosphereithe liquid conveyed being water

dl = 3cm Figure 7.10

7.4.3 Notches and Weirs The flow in opeh charnel may be measured by a weir or notch whicb is an obstruction in the channel that causes the liquid to rise behind it and flow over it br through it. By measuring the height of upstream liquid surface, the rate of flow is determined. The notches or weirs may be constructed from a sheet of metal, masonry or from other material. BasicalTy, there is no difference between a notch and a weir except that a notch is of smaller size while a weir will be usually of bigger size. Also, a notch is made of metal plate whereas weir is usually made of masonry or concrete. The common shapes of the notches are rectangular, triangular, trapezoidal, composite, parabolic and proportional. Figure 7.1 1 shows a rectangular notch. The bottom edge over which the

, rawd down ,r Hook gauge

Figure 7.1 1 Sharp Emted ~ectanguiar Notch

liquid flows is called the sill or crest of the notch. The sheet of liquid which springs free. from the crest is known as nappe. If the jet of liquid springs free as it leaves the crest, the weir is known as sharp crested weir. In the case of broad crested weirs, there is a support for the falling nappe over the crest in the longitudiml direction. As the liquid approaches the crest, there is a Qrop in the liquid surface in the form of convex curve known as draw down. The draw down at the weir crest is about 0.15 H where H is the head of liquid surface above the crest. Considering an elementary s@p of area B dh at a depth h below the free surface, the theoretical flow through it can be given as

dQ = Bdh

The theoretical flow passing over the notch can be obtained by integrating the above equation from 0 to H

H

Q = l ~ m d h

However, the losses such as fluid f&tion, draw down at the crest, a non-zero velocity at h = 0 and curved streamlines in the plane of the crest reduce the actual discharge to certain extent. Hence a coefficient C, is introduced to determine the actual discharge

In the above section the length of crest of the notch is the same as the width of the channel. Such notches are called suppressed notch or weir. If the crest length is less

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than the width of the channel, such weirs are known as contracted weirs. In such cases, when the sheet of liquid passes over the crest, there will be lateral contraction of the sheet at both the ends known as end contractions. The value of end correction per end is 0.1 H. Hence the effective length of sheet of liquid is equal to B - 0.2 H . Example 7.5 :

If there is an error of 6% in observing the head over a rectangular weir, determine the error in the discharge resulting from it.

Solution :

The head-discharge relationship for rectangular weir can be written as

Q =, m3l2

We have

dH The error in head measurement, - = 6%

H

:. Error in the discharge = @ = 1 . 5 x 6 = 9% Q

Triangular Notch

If snlaller discharges are to be measured accurately to avoid surface tension effects and to obtain appreciable measurable heads, triangular notch is the best. Figure 7.1 2 shows a triangular notch with a vertex angle 20.

The discharge through an elementary strip of area dA = 2 x dh is dQ = 2x dh where h is the depth of the strip below the liquid surface. From triangle OAB, we have

X - - - tan 0 H-tl

The total discharge Q can be obtained by irllcgratir~g Llic above exprevsion over the limits h= 0 to h= H H

Q = j 2 x d h jZplf 0

The actual discharge however is less due to losses as mentioned earlier and coefficient Cd is introduced in determining it.

- - - -

nor M-4

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5 /2 For a given notch, the discharge varies as H and the discharge-head relationship may be expressed as

in which K is constant for a given notch Example 7.6 :

When water flows through a right angled V-notch, show that the discharge is given 5 / 2

by KH in which K is a constant and His the height of water above the bottom of the notch. If H is measured in cm and Q in litreslsec, and the coefficient of discharge is 0.61, what is the value of K ?

Solution :

The discharge through V-notch is given as

8 5 / 2 Q, = - cd tan 0 H

15

For right angled V-notch, 0 = 90'

If Q is expressed in litres/sec, it has to be converted to cm3/sec and the discharge equation becomes

Trapezoidal Notch

A trapezoidal notch is a combination of rectangular and triangular notch (Figure 7.13). The advantage in this type is that the end contractions are either completely eliminated or considerably reduced increasing the discharge.

i----~---q Figure 7.13 : Trapezoidal Notch

The discharge through such a notch is the sum of the discharges through rectangular and triangular portions.

Where the coefficients of discharge Cd for the two portions may be slightly different.

If the coefficients are the same, equation (7.21) reduces to

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Example 7.7 :

A trapezoidal weir has the dimensions as shown in Figure '7.14. Develop the equation for discharge and find the discharge when the head of water acting is 40 cm. Assume that the coefficient of discharge is 0.6.

Solution a

1:igure 7.14 : I rapzuidd Weir

Consider a strip parallel to weir crest at a depth h and d width h and depth dh . From the geometry,

1 ... x = - ( 0 . 4 - h ) 2

and b = 0.2 + 2 x = 2(0.1 + x )

The discharge through the elementary strip of area h dh is

dQ = bdh 42$ If the head over the crest is H, the total discharge through the weir

For H = 40 cm. the discharge. Qa = 0.6 (0.06071 6 ) /

= 0.1614 m"1sec.

Cipolletti Notch

This is a trapezoidal notch with the sides in the ratio of 1 horizontal to 4 vertical. The advantage in this type of notch is that the decrease in the value of discharge due to end contractions in the rectangular notch is eliminated. The loss is made up by the triangular portion and the same formula for discharge as that of rectangular notch holds good here also. The reduction in the discharge due to end contractions can be b' wen as

The discharge through the triangular portion is

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Since these two discharges must balance

1 Solving for 8, tan 8 = - 4

The above fact was the conclusion drawn based on the experiments by Cipolletti and hence named after him Cipolletti gave an equation for discharge through such a notch as

(2 = 1.86 B H ~ / ~ (7.23)

in which B and H are in metres and Q in m3/sec.

Proportional Weir or Sutro Weir

In this weir, the discharge is linearly proportional to the head H i.e. Q = KH where K is a constant.

In many cases, such as in constant velocity sedimentation tanks, it is desirable to employ this type of weir. The shape of the sides of the weir diverge downward in the form of a hyperbola as shown in Figure 7.15.

Figure 7.15 : Proportional Weir

The weir consists of a rectangular portion joined to a curved portion with the profile equation

The actual discharge in this type of weir is given by

Cd for this weir is mainly dependent on a and B and ranges from 0.60 to 0.65. Even though the weir is considered to be ideal, its construction may be costly and hence not popular.

Parabolic Weir

As the name indicates, .the weir has a parabolic shape as shown in Figure 7.16 and here the discharge is proportiona1,to the head by the following relationship

Figure 7.16 : Parabolic Weir

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The discharge through parabolic weir is given by the following equation

Q, = 1.512 H (7.26)

e the co-ordinates of the

curved boundary.

Broad Crested Weir

Weirs in which the sheet of liquid (nappe) touches the surface of thc crest ;re called 2

broad-crested weirs. The length of such a weir should be > - H as show11 in 3

Figure 7.17 : Broad Cmsted Weir

Applying bernoulli's equation between points 1 and 2

v," H + O + O = - + O + y

2g

v, = 42g (H - v)

The theoretical discharge can thus be given by Q = B Y d2g (H - v) (7.27)

in which B is the length of the weir.

From the above equation, for a given value of H , theoretical discharge is found to be 2

maximum when y = - H which leads to 3

Q = 1.705 B H 3 1 2 (7.28)

Experiments for a well rounded upstream edged weir show the following relationship for the discharge

Qo = 1.67 B H 3 1 2 (7.29)

Which is 2 percent within the theoretical value. The flow, therefore, adjusts itself to the maximum discharge.

rectangular weir is given by t = in which A is the

(ii) If a discliarge of 120 litreslsec is to be measured over a triangular notch with

I a vertex angle of 60°, what will be the head over the notch, given Cd = 0.6. If at this level, there is an error of 1 mm in measuring the head, what is the corresponding error in the discharge.

(iii) A trapezoidal notch has a base width of 0.8 m, and a side slope of 1 horizontal to 2 vertical. Fiiid the discharge for a head of 0.6 m. Assume Cd = 0.62

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7.4.4 Venturimeter 1 I

1

The venturimeter is used to measure the rate of flow in a pipe. It is based on the Bernoulli's principle, that is, when the velocity head increases in an accelerated flow, there is a corresponding reduction in the piezometric head. The device consists of the following parts as shown in Figure 7.18.

(a) a converging entrance come of angle of abolit 20'

(b) a cylindrical throat, and

(c) a divergent part known as diffuser-witla cone angle of 5O to 7'.

Figure 7.18 : Venturimeter

The size of the venturimeter is specified by the pipe and throat diameter, e.g. 10 by 6 cm venturimeter.

The accelerated flow is achieved by the converging cone which increases the velocity in the flow direction. The function of the converging cone is therefore, to convert the pressure energy into kinetic energy. A differential manometer is attached at sections 1 and 2 to measure the difference in the piezometric heads. The discharge through a venturimeter can be given as

in which A, and A, are known areas at 1 and 2: h is the manometer reading indicating the difference in piezomeuic heads. The discharge coefficient varies with Reynold's number, the variation is small and if an average value is considered, the discharge equation reduces to

Q = K & (7.31)

in which K is known as venturimeter constant. The value of K can be determined by calibrating the venturimeter.

Example 7.8 :

Water flows through a 30 cm x 15 cm venturimeter as shown in Figure 7.19, at the rate of 42.5 litreslsec and the differential gauge reads 115 cm. If the relative density of the liquid in the differential gauge is 1.25, what is the coefficient of the meter?

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Solution :

The discharge equation is

:. Coefticient of discharge, Cd = 0.98

:. Coefficient of the meter =

SAQ 6

(i) Estimate the discharge of.kerosene (Relative density = 0.8) through the venturimeter as shown in Figure 7.20.

(ii) For the venturimeter of 150 mrn x 75 rnrn dimensions, determine Ah in the mercury manometer if the pipe carries a discharge of 50 litres/sec. of oil of relative density 0.8. Take Cd = 0.97.

Figure 7.20

7.4.5 Orificemeter Orificemeter is extensively used for flow measurements in pipes: It is also one of the oldest devices for measuring or regulating the flow of fluids. The device consists of q circular hole in a thin flat plate which is clamped between the flanges at a joint in a pipe line so that its plane is perpendicular to the axis of the pipe and the hole is concentric with the pi,pe as shown in Figure 7.21.

p + 3 4 , Orifice meter

Figure 7.21 : Flow (hmugh Orifice Meter

The difference of pressure or differential head between two sections on either side of the nrif ir~ at a r l i c t a n r ~ nf n a n d Dl7 i c rnpac i i r~ r l thrn i ioh r l i f f ~ r ~ n t i a l manrlmptpr

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The fluid flow approaching the orifice gets accelerated and the boundary streamlines assume the shape as shown in figure. The radially inward conlponent of fluid acceleration causes reduction in the flow area of the jet forming vena contracts within a distance of one pipe diameter i.e. at 012. The discharge through the orifice is given by

in which h is the differential head across the orifice; C, is the coefficient of discharge given by

The discharge coefficient C, , in general is dependent on the Reynold's number and the diameter ratio d ID .

7.4.6 Flow Nozzle The tlow nozzle is a truncated form of the venturimeter without the diverging cone. It is simply a constriction with well rounded entrance placed in the pipe line as shown in Figure 7.22. The tlow nozzle is simpler than the venturimeter and can be installed between the flanges of a pipeline.

Figure 7.22 : Flow Ncr~zle

It serves the same purpose, though at an increased energy loss due to the sudden expansion of flow downstream of the nozzle. The discharge equation for the flow nozzle is the same as for the venturimeter. The coefficient of contraction C, = 1 in this case leading to a higher value of coefficient of velocity than for the venturimeter. For accurate results, the flow nozzle must be calibrated in place. The discharge through the nozzle is given by

Q, = CA2 m (7.34)

L v in which C is a coefficient =

A, and A are the cross sectional areas\of /the pipe and flow nozzle respectively, C, is the coefficient of velocity.

The head loss through the nozzle is given by

in which V2 is the velocity of the fluid through the nozzle.

7.4.7 Venturi Flume A venturi flume is an artificial constriction in a channel to measure the flow rate. It is usually conslructed in concrete, masonry, steel or timber. The flume consists of a bell-mouthed entrv. ~aral lel throat and downstream diverging or ti on as shown in

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Figure 7.23 : Venturi blume

) The discharge through venturi flume is given by

in which b,, b2 are the channel widths and y, and y, are the depths of water at sections 1 and 2 respectively and Cd is the coefficient of discharge.

In case the flume is designed to produce critical tlow at the throat, the discharge can be determined by the following formula

However, since the exact position in the throat , where the critical condition occurs is difficult to determine, the discharge may be computed using the depth of water at section 1 as

The value of Cd for the venturi flume varies between 0.94 to 0.99. The advantages of the I flume over the broad crested weir is that the loss of energy in the flume is minimum and

there is no silt deposition problem in the flume. The vei~turi tlume is well suited to 1 measure the flow in streams, small rivers and urification plants. The discharge Capacity P I ranges from a few litreslsecond to about 25 m Isec.

Example 7.9 : I

A venturi flume is 2 m wide at entrance and 1.2 m at the throat. Neglecting the hydraulic losses in the flume, calculate the discharge, if the depth at the entrance and throat is 0.9 m and 0.8 m respectively.

Solution :

From continuity equation

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Applying energy equation at 1 and 2

The discharge

9.81 0.1 x 2 x - 2.5156 = 0.883 m / sec.

7.5 SUMMARY

This unit introduces you to various methods and devices used to measure the fluid flow. Bernoulli's principle is used to evaluate the flow parameters such as pressure, piezometric head, velocity and discharge. Application of Bernoulli's equation to closed conduit and free surface flows are illustrated. The structure and working principles of various flow measuring devices are discussed. The numerical examples solved illustrate the usefulness of the devices in practical problems.

7.6 ANSWERS TO SAOs

SAQ 1

(1) (e)

(ii) (e) SAQ 2

( 0 (c)

(11) (b) SAQ 3

(i) Applying Ben~oulli's equation between the reservoir surface and vena conuacta (See Figure 7.6 (a))

= 2.698 m

v Coefficient of velocity, C, =

Coefficient oim~uaction, C,. = Cd = = 0.635 C, 0.96

:. diameter of the jet = 4, = d < = 8

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(ii) Theoretical velocity, V =

= 42 x 9.81 x 4.9 = 9.805 m / sec.

gt2 The vertical distance travelled, y =

:. Time required to travel this distance, t = 7

- - 6~ = 0.515 sec.

The horizontal distance travelled, x = Vat

:. Actual velocity, v = - = - - 4'8 - 9.32 m / sec. a 1 0.515

Va Coefficient of velocity ,- C = - v = 0.95

The actual discharge, 907.6

= 32.6 x lo00

1 = 0.0278 m3 / sec

Coefficient of discharge, Qa

cd = A-

Cd Coefficient of contraction, C, = - c v

2 Head loss = H (1 - Cv ) = 0.478 m.

SAQ 4

Applying Bernoulli's equation between free surface and section 1 (Figure 7.10), we have

v:. 1 1 . 6 = - 8 + - + - ~ 1 . 6 2g 100

:. V, = 13.71 mlsec.

Maximum discharge, It 32 Q = q x - x 13.71 104

= 9.69 x m3/se'c.

Applying Bernoulli's equation between the free surface and section 2, we have

. . V2 = 5.432 mlsec.

Using continuity equation

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A d d D m m i a - I SAQ 5 (i) The discharge over the rectangular weir of length L and head H is

The rate at which the reservoir level is falling is given by

The time required to lower the reservoir level from H , to H2 is given by

(ii) Using equation (7.19)

0.12 = 0.818 H 5 l 2

:. H = 0.46 m The discharge equation may be written as

Q =& K H " ~

Dividing 2nd equation by the I st

In this case, dH = 0.001 m, H = 0.46 m

(iii) Using equation (7.21 (a))

1 In this case, tan 8 = - = 0.5 2

C, = 0.62

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SAQ 6

(i) Applying Bernoulli's equation between sections 1 and 2

From continuity equation

v, = 4v,

By equating pressures at A-A

i.e. V, = 2.888 mlsec.

Hence Q = 4 % = 0.0227 m3/sec.

(ii)

- - 50 x x 43.1228 x lo4 - 1.95175 x 10" 0.97 x 0.017671 x 4.41786 x x 4.4294

.. h = 6.505 m of oil

- - 6.505 (13.55 - 0.8)

of mercury

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UNIT 8 LAMINAR 'FLOW

Structure 8.1 Introduction

Objectives

8.2 Viscosity 8.2.1 Role of Viscosity as a Momentum Transfer Coefficient

8.2.2 Molecular Exchange of Momentum - Kinetic Theory of Ciases

8.3 Navier - Stokes Equations of Motion

Exact Solutions of Navier - Stokes Equations 8.4.1 Hagen - Poiseuille Flow between Parallel Plates

8.4.2 Hagen - Poiseuille Flow from Navier Stokes Equation 8.4.3 Couette Flow 8.4.4 Couette Flow from Navier Stokes Equation 8.4.5 Combined Hagen Poiseuille and Couette Flows

Flow in Lubricating Film of Oil in Slider Bearings

Combined Hagen - Poiseuille Flow and Couette Flow along Inclined Plates

Laminar Flow along an Inclined Channel with Free Surface

Laminar Flow Through Cylindrical Tubes 8.8.1 Hagen - Poiseuille Flow through Cylindrical Tubes 8.8.2 Use of Navier-Stokes Equation for Solution of Hagen-Poiseuille Flow in Cylindrical

Pipes

Flow in the Annular Region between two Concentric Pipes

Couette Flow in Cylindrical Coordinates

Rotating Cylinders

Torque on the Cylinders

Steady Slow Motion around a Sphere : Stokes Law

Setting Velocity or Fall Velocity of a Sphere

Measurement of Viscosity 8.15.1 Capillary-tube Viscosimeter 8.15.2 Odtwald Viscometer 8.15.3 Concentric - Rotating cylinder Viscometer 8.15.4 Fall Velocity Method

Summary

Answers to SAQs

8.1 INTRODUCTION

Though the property of fluid viscosity was defined in Unit 1 for real fluids, Units 4 and 6 dealt with motion of ideal fluids, wherein we obtained certain basic concepts governing fluid motion. However in real fluids, the property of viscosity introduces shear stresses not considered earlier and loss of energy which is an irreversible process. There are many problems where the viscous shear force is of greater irhportance than the inertial forces such as flow in capillary tubes, fluid layers very close to a boundary, porous media, and lubrication films of bearings. The ratio of inertial force to the viscous force is known as the Reynolds number whose value serves to distinguish the flow as laminar and turbulent. In laminar flow, prevalent at low Reynolds numbers (for example, in pipes for values of Re less than 2000). the fluid particles move in distinct layers, one layer sliding over the adjacent layer. In other words, any lateral movement due to lateral or transverse disturbing (inertial) forces are damped out by viscous damping forces. For larger Reynolds numbers, the inertial forces can cause transverse movements of fluid particles from one layer to another resulting chaotic motion known as turbulent flow. In the present unit, we shall deal with laminar motion.

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Objectives

After studying this unit, you should be able to

* define viscosity and its role in fluid flow,

* develop Navier - Stokes equations of motion,

* obtain exact solutions of Navier Stokes equations as applicable to some laminar flow situations, and

* determine viscosity by devices using the above principles.

8.2 VISCOSITY Since viscous shear forces play a dominant role in laminar tlow, we start with recapitulation of the property of viscosity. Physically, the difference between a solid and a fluid lies in the relative immobility or mobility of the ~nolecules in them. Each n~olecule in a solid has a fixed mean position about which the molecule u~ldergoes vibratory or rotary motions. The molecule in a fluid , on the other hand , by virtue of its mobility may translate in addition to the other types of motion. The distinction between a solid and a fluid lies in their respective behavior under the action of external forces. A solid has a finite tensile strength. A fluid conforms to the shape of its container and can resist co~npressive forces only when so bounded and compressive forces are applied to walls of the container. When subjected to a shearing force, a solid within its elastic limits attains a static equilibrium position whereas, a fluid suffers continuous deformation no matter how small the tangential forces may be.

Viscosity is the property of a fluid that enables it to develop resistance to deformation under the action of shear forces. Let a certain fluid fill the space between two closely spaced

AU

x-axis

stationary plate Figure 8.1 : Sbeer Stress

parallel plates separated by a distance Ay as shown in Figure 8.1. Let the top p l a t e ~ ~ / n o v e a constapt velocity AU by the application of a force AF, white the bottom plate CD remains fixed. Under steady conditions tliere is a balance between external force AF-and the internal force developed due to viscosity. Due to non-slip condition of real fluids, the fluid particles in contact with the lower plate remain stationary whereas those particles in contact with the top plate acquire the same velocity AU of the moving plate. The velocity varies linearly across the narrow space between the plates. It was shown by Newton that the shear stress (obtained by dividing the force by the area of contact of the top plate) was

AU proportional to the velocity gradient ---. At any given point the rate of change of velocity is

Av du given by - dy

Hence based on Newton's fmdings , we write

where p is the constant of proportionality know11 as the coeflcient of dynamic viscosiry having the dimension of [ M L- ' T- ' ] . Here the first subscript refers to the direction of the normal to the plane on which the shear stress is acting and the second subscript to the direction of action of shear stress.

We consider the fluid element PQRS initially rectangular in shape at time t which subsequently assumes the shape P'Q'R'S' at time t + At as shown in Figure 8.1 due to the

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differential velocity between the two layers. P and Q' have moved through a distance of AU . At relative to S and R respectively. The angular decormation of the element is A0 . which from the geometry of the figure is seen to be (considering changes in 0 as positive in the anticlockwise direction)

or the rate of angular deformation at any given point becomes

(10 - clu - clt cly

Therefore, Newton's law of viscosity can be stated that the shear stress is proportional to rate of angular deformation. The coefticient ol' viscosity is dependent on the temperature for any given tluid but is found independent of pressure in the ~iormal range of pressure variations. However at large pressures. viscosity has been found to be dependent on pressure also.

Figure 8.2 : Shear s tms in the x-y plane

111 the dane x-y, if the velocity components t~ and v change in x and y directions, the element PQRS at time t take the shape PQ'R'S at timll t + clt as shown in Figure 8.2.

The angular deformation of the ad-jacent sides. 'P and SR are found to be

I u - ( l a = - - clt

3,.

The mean deformation of the element as a whole is given by that of the diagonal SQ

and the shear stress in the x-y plane being proportional to the rate of angular deformation

Similarly in three dimensional tlows the shear stresses in the zy and xz planes can be shown to be (u, v, w are the velocity components in x, y & z direction respectively.)

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It can also be shown by taking moments of the shear forces acting on the faces of the element about the centre that

There are only three independent shear stresses acting on a three dimentional fluid element.

8.2.1 Role of Viscosity as a Momentum Transfer Coefficient

Consider a uniform flow in the x-direction caused by a constant pressure gradient @ dx'

Figure 8.3 shows a typical velocity distributiol~ of u varying with y. The horizontal momentum at B per unit volume is pu.

Figure 8.3 : Momentum Transfer

The horizontal momentum at level A per unit volume is [pu + d (pu)] and the change in momentum per unit volume is d(pu). There is a momentum gradient or momentum potential across the distance dy. This can be caused only by a shear force since pressure gradient in the y-direction is zero. The shear stress acting on unit area must be equal to the rate of change of momentum

v is known as the coefficient of kinematic viscosity having the dimensions [ L ~ T- '1

8.2.2 Molecular Exchange of Momentum - Kinetic Theory of Gases. The.kinetic theory of gases outlines that the molecules of a gas are constantly in motion with velocities, direction. and length of molecular path varying in a random manner. The very nature of random distribution of these molecular travel characteristics ensures that there is no mean motion of the gas as a whole on tllis account. We define the mean molecular velocity as C and the mean free path of molecules as 1. Let us take an imaginary surface AB (Figure 8.4) across which molecules are in motion both from locations above and below the surface. Should the gas possess a mean motion with a velocity gradient as illustrated in Figure 8.4, then the random motion of the molecules in a direction normal to the flow direction causes an exchange of momentum across the surface AB. The molecules in the upper layer possess greater horizont;il momentum than those in lower layer. The transport of momentum from one layer to another tends to equalise the velocities between the upper and lower layers.

Let us assume that out of N molecules present in a unit volume of gas. equal number of molecules i.e., one third of the molecules , N / 3 are directed along the three-axes directions.

Figure 8.4 : Molecular Exchange of Momentum

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Out of these N / 3 molecules moving in the y-direction half of the molecules (N / 6 ) are moving in a direction downwards and the other half (N / 6 ) are moving upwards across the surface AB. If they are moving with a mean velocity C, the number of molecules moving up or down in time dt is (116) NC dt dA and its mass is (116) m NC dt dA where m is the mass of each molecule. At the upper layer, the molecules had a horizontal momentum of

l and at the lower layer, the momentum is

[ The net transport of momentum transferred by tho molecules motion is thus obtained as

Since the rate of change of momentum per unit time is equal to the tangential force per unit area, 'ryx, we get

since the density p ='Nm. . 1

A comparism of equations (8.1) and (8.6) reveals that p = - pC1 3

A more rigorous analysis gives the value of p = 0.499 pC1

The coefficient of viscosity, p is thus seen to be essentially a momentum transfer coefficient.

It has been shown that dynamic viscosity depends on fluid density, molecular speed and I

molecular free path dimensions. We can therefore expect different trends in the variations in I the value of p for liquids and gases. For example, in gases the molecular activity increases

with increase in temperature leading to an increase in the value of viscosity with temperature. In case of liquids where the molecules are closely packed, cohesion between molecules is more important than molecular activity; Increase in temperature causes a reduction in cohesion with consequent decrease in viscosity of liquids with increase in temperature.

-

8.3 NAVIER -STOKES EOUATIONS OF MOTION

Earlier in Unit 6 , equations of motion for ideal fluids have been derived and we shall now include the shear forces also. Let us consider a three dimentional element of sides dx, dy and dz. For the sake of simplicity, we demarcate the normal and shear stresses acting in the x-direction only as shown in Figure 8.X-

Figure 85 : Stresses in the x-Direction of a 3-Dimensional Element

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aa, & , and that acting on the The normal stress acting on face ADHE (area dy dz) is o, - - -.

ax 2 20, dx

opposite face o, + - -; tensile stresses are considered positive. dx 2

The shear stress acting on bottom face CDHG is z,,, - a 2 d

acting on area dx . dy while ay 2 a2

the top face ADFE is subjected to a shear stress of ryr + -YZ &. The shear stress on the a~ 2 a 2, d z

near face ABCD ot area dx dv is 2, + 7 1 while the opposite farther face experiences a

a ~ z r dz shear stress of 2, - - . - (it may be noticed that we have taken all the stresses into the az 2 . second subscript x). We have the equation of motion in the x-direction

mass x acceleration in x-direction = Forces in x-direction

du p d x d , v d z ; i ; = ( p d x ( l y d z ) X +

aondx dvdz+ r,+ dxdy - ( ~ ~ ~ ] d x d ~ - ( . . - ~ ~

( 2 ; ) [ ;] + T,+- . - d x d y - T,--.- dxdy

Where Xis the component of external body force per unit mass in the x-direction.

On simplification , the equation of motion becomes

It can be shown (the proof is quite involved and out of scope of the present treatment) that

where y is the compressive stress acting on the element. It is the average of the three normal stresses.

and

Using Ule expressions for z~~ and z, from equation (8.3) and equation (8.8) for a, we get

on regrouping, we get

The last term on the right hand side is zero by virtue of equation of continuity for ii~compressible tluids. Hence

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If we write V2 as the operator

The term v V'U defines the viscous force per unit mass and is known as the Navier-Stokes viscous term. In ideal fluids, p = 0 and the equation reduces to the Euler's equation described in the earlier Unit 6

I We can derive similar equations of the motion in the other two directions. For steady flow au av alv

with which we are concerned in the present unit - = - = - = 0. Helicc the Navier- at at at

Stokes equations in cartesian coordinates for steady flow of incompressible fluids are

au au au x-direction: u-+ v-+ w-= X- ax a~ a~

aw aw aw aZw a 2 ~ a2w z-direction: u-+ v-+ w-= Z- &+ v ,+ -

ax av az P az I X, Y, Z are the components of external body force for unit nlass in the x , v & z directions respectively.

We can derive Ulese equations in cylindrical or spherical coordinates either from first principles or transformation of coordinates. It is sufficient now to write the final form of Lliese equations in the cylindrical coordinate system for steady incompresible tlow.

Radial direction :

Tangential dircction :

Axial direction :

R, O, Z being the components of external body forces per unit mass in the radial, tangential and axial directions respectively.

Example 8.1 :

Show that the viscous terms become zero for irrotational flow of incompressible fluids.

Solution :

or irrotational flow . we have frocl Unit 4.

from which we obtain 22, 32, -- - a 2 ~ a 2 ~ and -= - 3 3 - a ~ a y aZ2 2 x 3 ~

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Therefore

= 0, by virtue to equation of continuity

Similarly we can show v2v = v2w = 0 for irrotational flows.

8.4 EXACT SOLUTIONS OF NAVIER-STOKES EQUATIONS

The Navier-Stokes equations (8.1 1 or 8.12) are a set of nonlinear equations for which no general solutions exists. However, for some particular cases of flow with simple boundary conditions, exact solutions are available. In some of the flows presented in the following sections, the nonlinear terms on the left-hand side of equations (8.11) or (8.12) disappear on their own because of the simple flow and boundary conditions. We shall discuss the solutions to these linear equations.

8.4.1 Hagen-Poiseuille Flow between Parallel Plates Hagen, a German engineer and Poiseuille, a ph$sician interested in the resistence of flow of blood through tubes were the first to investigate laminar flow through small tubes.

Shear distribution

/

x-axis -.

distribution

Figu1-c 8.6 (a) : Flow between Parallel Plates

We shall begin with a steady uniform flow between parallel horizontal plates under the influence of a pressure gradient. Ln Figure 8.6, the distance between the plates is taken as 2a and the flow is caused by a pressure gradient.

On the element of size dx dy and unit depth pressure and shear forces are in equilibrium since the flow is uniform and there is no acceleration. The free body diagram of the element is shown in Figure 8.6(b). The direction and magnitude of the stresses on the four faces of

Velocity distribution

Figure 8 6 (b) : Freebody D i a g m of Fluid Element

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LrburRar r the element are shown. Fluid above AB is moving faster and exerts a positive shear force on the element, whereas fluid below CD moviug slower exerts ii negative shear stress. Equating the forces

I du

Since the shear stress in one dimension ~ y , = p dy

we have

@ . It is be noted that is a constant pressure gradient in x-direction and does not vary with Y .

dx

Integrating it once we have

The constant c, is found to be zero since at y = 0 along the centre line, thc velocity is a

du maximum and hence - = 0

dy

The shear stress is linearly distributed. A second integration yields

1 dp y2 u=--- + C, , p clx 2 - c, call he found by putting u = 0 at the boundaries y = _+ a

Hence the velocity distribution is parabolic

& . 111e negative sign is attached to slnce the pressure gradient is negative (Pressure is

dx

decreasing in the direction of x). On the other hand if@ is positive, tho direction of flow dx

will be in the negative x-direction. The discharge per unit width betweell ult: plates is found by integrating the velocity across the spacing

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The average velocity, V is given y 1 %

and the maximum velocity along the centrc line (y = 0)

Therefore,

Power input required to overcome the head loss Ahf over a length 1 of the plate is given by

Where y is the specific weight of the fluid

L

The forceonone plateis[ ~ ~ d x = ~ ~ L . = ~ . a L - L = A ~ - a 0

The force on two plates is 2 Ap . a.

Power required may also be found by multiplying the force on plates by the average velocity

Which is the same expression found in (8.21)

Example 8.2 :

Choose the bottom plate as shown in Figure 8.7, as the x axis and derive the expression for velocity distribution.

Velocity distribution

shear stress S To.-- IJ *a dx - x- axis

Figure 8.7

Solution :

We are starting with the governing equation (8.14)

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I

We integrate twice to give

1 The integration constant can be obtained by the boundary conditions

i Therefore,

andlheshears.tssisgivenby r = y ~ = [ - ~ ) ( a - y i d.v

SAQ 1

For steady laminar flow between parallel plates spaccd at 4 mm, p = 8 x kg1ni.s a~ and - = - 14 k ~ / m ~ . Determine the discharge, nlaxiinum shear stress and ax

maximum velocity

SAQ 2

Kerosene oil flows upwards through incliyd parallel plates at the rate of 2.2 litreslsec per metre widh. Tile distance between the parallel plates is 12 nun and the inclination of the plates is 25' with the horizontal. Determine the difference of pressyre between two sections 12 m apart. Take p = 800 ke/nr3 and p = 0.0021 N sec / m2.

8.4.2 Hagen-Poiseuille Flow from Navier Stokes Equation We can readily obtain the solution for the flow between parallel plates from the Navier-Stokes equation (8.1 1 )

a~ The flow is uniform and hence - = 0 ax av au

Hence from equation of continuity - = - - = 0 ay ax

v = constant, but since v = 0 at the boundaries, v should be zero everywhere. There is no change of w with respect to z and hence all the nonlinear terms on the left hand side vanish. The body forces X and Y are not present.

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a,, a ' ~ x-direction : 0 = - - + p - ax a.?

y-direc tion : 0 = - ih? dv

11 can be a function of x only: u is a fundion of v only and a constant pressure gradient exists in the x-direction.

m e Navier-Stokes equation in x direction rcduces to a linear, second order ordinary differential equation.

Which is tlie same governing equation (8.14) derives from first principles. The integration continues as before iuid will not be repeated hcre.

Example 8.3 :

The wails of a water tank arc 0.5 m thick and walcr in the tank is stored to a depth of 1m (refer Figure 8.8). If the wall develops a crack of Imrn thickness and 20 cms wide at the bottom of the tank. estimate the leakage from the tank in one hour assu~ning that tlie water 0.5m

n level in the tank renlains constant.

7 p=.001 Nseclnl ' , p=998kg/n13

Solution :

If q is Ule dischaqe/unit width

and from equation (8.17)

' nl 2 ~ = 1 nmlor(1=-x- 2 1000

Here p = 0.001 N sec / m2

The head of I m is dissipated in a distance of 0.5 m

Hence

= 5.868 111' 1 hour

20 Q = 5.868 x - = 1.1736 m3/hour = 1173.6 litreslhour

loo

Wall

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Example 8.4 : Lammar Flow

The bearing surface of a Telescope is lubricated by means of an oil introduced under pressure at the mid point of a bearing as shown in the Figure 8.9. Assuming two dimensional flow from an injection slot. What rate of feed in litres per hour would be necessary to preserve a clearance of 0.05 rnm between bearing surface suppotfing

0.05mm a load of 10000 N1sq.m if the oil had a viscosity of 2 x N . s/m2. The bearing is 40 cms wide.

Solution :

The maximum pressure occurs at the centre. 'Ihe lotal upward thrust due to the pressure in the lubricating fluid is Figure 8.9

1 40 2 Pmax X - X 1 = 0.2 p,, per metre length of bearing and this should be equal to 1 0 0 the load from above.

0.2 p,,, = 10000 x 0.4 = 4000

p,, = 20000, ~ / m ' I

The pressure gradient from the centre to h e exit end on both sides is

---- - " 2W00 - 100000 Nlm dx 0.2

The average velocity V is found by equation (8.1 8)

with

= 1.04 166 x 10- m3/sec per meter width of bearing

= 0.375 litreshour/meter width

8.4.3 Couette Flow Couette studied the flow caused by the movement of one of the plates at a uniform velocity U (as was the case in Newton's experiments on viscosity), h e other plate remaining stationary. The f1ow.i~ caused by the transmission of shear force from the moving plate to the interior of the fluid and the pressure gradient is zero.

lving with velocity U

Top plate mo o constant

Constant shear stress.

x-axis

/ ' 0

Stationary boundorv distribution

Figure 8.10 : Couette Flow between Parallel Plates

103

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The x axis is chosen to lie on the stationary boundarymd the spac$lg between the plates is 2a as shown in Figure 8.10.

The only forces acting on the fluid element ABCD are thg sheat4orces on ihe opposite faces AB and CD, and hence we have

-- d2u du dr - --- = O. Since r = - d~ ciy2 dv

Successive integration yield?

u = c 1 y + c 2

we have u = 0 at y = 0, with the result ~2 = 0

U andaga inu=Uaty=2u : c1=-

2a

The velocity distribution is linear

and the shear s.tress is constant

U U The average veioc?ty is -and the discharge per unit width-is - . 2 a = Ua 2 2

The force'required td'draw the top plate of length L at a uniform velocity U is

The power input can also be calculated by considering the powers supplied to the top'and bottom sides of the element. At any point the power P i s obtained by the product of stress, area and velocity.

P = 2 . (dx . 1) . u along surface AR

dP d Along CD P + - dy = 2 . dx . ut+ - ( ~ u ) dy cix dy cry

The net power input to the element is thus

U Substituting the expressions for r = @and u = - y

2a 2a

d 2 ciP = - ( ru ) ciy . dx = -

( 1 ~ ( l ~ k&!Y 4a2 dydx= @-dydx ! 1 4a2

Integrating across the width & length L, we get the total Power

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I This value is the same as that obtained for the power to be supplied to keep the plate moving at a constant velocity. Hence it is inferred that the power supplied to the plate is

b entirely consumed by the internal power loss due to viscous shear.

L 8,4.4 Couette Flow from Navier Stokes Equation

I We may also readily obtain the solution to Couette flow from by direct application of Navier stokes equation (8.1 1). Here we know that the flow is steady and uniform in the x direction only and hence there are no'accelerations. The pressure gradient is also absent. Hence we have

au -- - 0 ; v = w = 0 and & = (, ax ax

i Hence all the tenns on the left hand side of the equation in x direction disappear and the - -

a 2 ~ . only term remaining on the right hand si& is p -. Slnce'it is a function of y only, a?

Successive integration carried out as before will yield

8.4.5 Combined Hagen-Poiseuille and' Couette Flows In addition to the pressure gradient as considered in section 8.6, one ofthe plates is also moved while the other plate remains stationary. We choose the x axis to lie on the stationary boundq as shown in Figure 8.1 1.

d P Pressure gradient - d x

u

S t a t i ~ a f y boundary

Figure 8 1 1 : Combined Hagen-Poiseuille lu~d Couette Flows

The governing equation of motion is the same as equation (8.14)

which on successive integration give

We introduce the boundary conditions u = 0 at y = €I and u = U at y = 2a

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Ekir-1 The final velocity distribution is

and the shear stress

We could have obtained the solution readily by superposing the solutions for Hagen-Poiseuille flow (equation 8.22) and couette flow (equation 8.24)

We can express equation (8.27) in the non-dimensional form

The expression f$ (- 2) is dimensionless and may be replaced by the symbol K

U 2a (8.29)

For various val~l.~ of K (or for various values of , the velocity distribution areshown in dr

Figure 8.12.

I I I I I Stationary platr 0.0 0.4 0.8 1.2 - 0.4

Figure 8.12 : Velocity Profiles in Combined flow

We may have the pressure gradient positive (the pressure is increasing in the x direction) in which case the flow due to pressure gradient will oppose the flow due to the movement of plate in the x direction and these can be reversal of flow.

The average velocity for the combined flow becomes

The average velocity will be zero if K = - 3

It is seen from Figure 8.1 2, that there is a reversal of flow for K = - 2 & - 3

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SAQ 3

Laminar flow takes place between the parallel plates 12 rnm apart as shown in Figure 8.13. The plates x e inclined at 45' with the horizontal. For the oil of viscosity 0.85 kg1m.s. and mass density of 11260 kg/m3, the pressures at two points 11.2 m vertically apart are 82 kNlm2 and 260 k ~ / m ~ when upper plate moves at 2.3 m / s velocity relative to the lower plate but in the opposite direction to the flow. Determine

(i) the velocity distribution

(ii) the maximum velocity, and

(iii) the shear stress on the top plate.

Figure 8.13

8.5 FLOW IN LUBRICATING FILM OF OIL IN SLIDER BEARINGS

The combined Hagen-Poiseuille and Couette flows are of interest in sliders bearings. The Couette flow is caused by the movement of one of the bearing surfaces. The Hagen-Poiseuille flow is caused by pressure gradient resulting from the slight inclination of the two bexing surfaces. Due to the inclination, the spacing between the bearing changes gradually with consequent change in velocity (or spatial acceleration in the x-direction, -

du u -. However we consider the acceleration to be negligible and to this extent the solution ax

is not exact but very close to the real situation. With reference to the Figure 8.14 the spacing varies from 2al to 2a2 over the length L, of the bearing and the inclination, a (being very small) is given by

At any distance x, the spacing is 2a = 2al - a x

city due

I L - - - - - - 4 Figure 8.14 : Slider Bearing

- Top surface moves with a velocity U

Velocity distribution & )a-k d %

Bottom surf- is stati&ry

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m d ~ynrmia-I The velocity distribution at this section is due to both pressume gradient and couette flow. Neglecting all acceleration terms, the preceding solution in Section 8.4.5 is valid (equation (8.27))

Discharge per unit width of bearing

Solving for

Integrating and evaluating the constants by the condition p = po at x = 0 and x = L

Integration of pressure intensity over the length L yields the normal force/unit width of bearing

Example 8.5 :

Two plane boundaries are 6 im apart, the space between them is filled with a liquid of viscosity of 1.2 kg/m.s. What force would be required to move edgewise through the liquid, a plate 3 mm thick and 25 cms square at a velocity of 15 cm / sec.

Solution :

The shear stress in the couette flow in the spaces on both sides of the moving plate is " constant

2 5 c m -

Figure 8.15

where, p = 1.2 kg/m.s, U = 15 cdsec , 2a = 1.5 rnm

The shear force on two sides of the plate 25 cm. sqare

F = 2x A = 1 2 0 x 2 ~ 25x25 = 15 Newtons loo x loo

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Example 8.6 :

The space of 10 mm between two parallel pl.ates is filled with a liquid of viscosity p = 2.0 kg/m.s.. The top plate moves in the negative x direction with a velocity of -2 m / sec. What should be the sign and value of pressure gradient so that the net flow is ,zero. What.are the shear stresses at the two boundaries.

Solution :

For the average velocity to be zero the value of pressure gradient should be such that (from Section 8.4.5) with 2u = 10 mm, U = - 2 m / sec, p = 2.0 kg1m.s.

TO is the shear stress at the bottom boundary y = 0

At the top y =. 2a

Example 8.7 :

In the slider bearing problem of Section 8.5 and Figure 8.14 the spacings 2al , and 2a2 are equal to 6 mm and 4 mm respectively and the length L = 100 cms. Find the location and magnitude of maximum pressure on the bearing surface and the total force. Assume the ambient pressure po = 0 and p = 2.0 kg1m.s. and the velocity of travel of the bearing surface U = 2.0 m / sec.

Solution :

From (8.31) we have

The maximum pressure occurs whenh = 0 from which we get clx

on substitution for q from (8.33)

Substituting and solving forx, we find x = 0.6 m, where the spacing 2a = 4.8 nlm

The maximum pressure is found from (8.32) with pg ='O

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The total force the bearing can support is from (8.34) 1

= 32790 Newtonslm . width

SAQ 4

A slipper bearing moves towards left with a velocity of 1.6 m / s. The data for the bearing are L = 35 cm; a, = 0.04 cm; a* = 0.01 5 cm and p = 0.08 kglms. Find the maximum load which can be sustained by the bearing. Also, find the maximum pressure intensity.

8.6 COMBINED HAGEN-POISEUILLE FLOW AND COUETTE FLOW ALONG INCLINED PLATES

If the plates are inclined, the body force component in the direction of flow also comes into play, or in other words, the component of weight of fluid due to gravitation in the flow direction should also be taken into account in addition to the pressure and viscous shear forces. As shown in Figure 8.16 to the bottom plate is stationary and the top plate is moving with a uniform velocity U. The plates are inclined at an angle 0 to the horizontal direction. The component of the weight of fluid in the element is y dx dy sin 0 Considering all forces on the element shown in Figure 8.16, we have

(-

Figum (L16 : Combined Flow along lndiwd Plates 1

k - ~ k i s

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I Since dx sin 0 = - dh and on simplification

The preceding expression is similar to the one derived for horizontal plates except that the pressure p is now replaced by the piezometric pressure O, + y h). Further integration as outlined in section (8.4.5) yields

i and

We can derive the above equations from Navier-Stokes equation if we recognise lhat the body force is due to gravitational action acting in the vertical direction. Its component per unit mass in the chosen x direction along the bottom plate is X = - g sin 0 and in the y direction Y = - g cos 0 and Z = 0. The equation of motion in the x-direction is

1 which is the same equation as (8.35). I

8.7 LAMINAR FLOW ALONG AN INCLINED CHANNEL WITH FREE SURFACE

This problem illustrated in Figure 8.17 can be considered as one-half of the problem of Hagen-Poiseuille flow between two parallel inclined plates, the free surface coinciding into the x-axis in Figure 8.6 (a). The depth of flow is taken as a.

L lnmr shear dir!;I'pution

L 0

Figure 8.17 : Laminar Flow alang an Inclined Channel

The velocity distribution is parabolic

Since the flow is subjected to uniform atmospheric pressure along the flow direction @ = 0 dx

From the Figure 8.17, it is seen

dh - - = sin 0 = S, slope of the channel. ah

The negative sign is due to the decrease in bed elevation above a given datum in the direction of flow. The expression for velocity can be written as

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u,, occurs at the free surface O, = 0)

The discharge per unit width, q is obtained by the integral

and the average velocity 4 ya2s v = -= - a 3 ~

The bottom shear stress du ~ o = p - 1 = - y a s

dy y = a

The shear stress induced by the flow on the boundary is y a S

Example 8.8 :

Two parallel plates are 8 mm apart and have an inclination of 30" with the horizontal. The fluid filling the space between the plates has a viscosity of 1.2 kg/m.s. and a mass density of 900 kg/m3. The pressume at two points 2.0 m vertically apart are 100 k ~ / m ~ and 300 k ~ / m ~ .

Find the maximum velocity, discharge and shear stresses on the top and bottom plates if the top plates moves (a) at a hiform velocity of 1.2 m / sec in the same direction as the flow (b) at a uniform velocity of 1.2 m / sec in the opposite direction to the flow.

Solution :

From the results of section 8.6

du Maximum velocity occurs at a location where - = 0 dv i

Piezometric pressure difference between the two points is,

2 and this drop takes over a distance of 7 = 4 m

sin 30"

and hence.

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= 1.2109 m / sec

= 0.0106 m / sec

Example 8.9 :

In a certain manufacturing process, a liquid of viscosity 1.5 kg/m.s. and mass density 1200 kg/m3 flows down a sheet of glass inclined at 60" to the horizontal. If the maximum velocity must not exceed 25 cdsec, what should be the limiting thickness of the liquid film.

Solution :

From equation (8.39)

In the given problem um, = 0.25 dsec, y = 1200 x 9.8 Newton/m3

S = sin 60" = 0.866

p = 1.5 kg1m.s. I

= 8.577 x 10- m (8.577 mm)

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E l d d ~ ~ SAQ 5

One plate inoves relative to the other as shdwn in Figure 8.18, p = 0.05 kg/m.s and mass density = 850 kg/m3. Determine the velocity distribution, the discharge and the shear stress exerted oil the upper plate.

Figure 8.18

8.8 LAMINAR FLOW-THROUGH CYLINDRICAL TUBES shall develop exact solutions of Navier - Stokes equations for the c$e of

cyliildrical tubes.

8.8.1 Hagen-Poiseuille Flow through Cylindrical Tubes We follow the comentional cylindrical cosrdimte system y, 0, z. The z-axis being the direction of flow @ perpendicular to the; - 8 plane. We designate the veloCity in the axial direction, z as w.

Comider the cylindrical element s y ~ t r i m l l y p l ~ ~ w i t h respect to thez-axis as shown in Figure 8.19. Since the flow is steady and uniform, the pressure, shear y ~ d gravitational forces acting on the element should be in equilibrium. p varies only with z and w varies only

Figure 1119 : Flow through Cylimlrical Tubes

From the figure dz sin 0 = - dh

2 ~ d z - ~ - d p - y d l - - - r

Thus the shear stress distribution is linear;

I Noting that velocity decreases with r, we have

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Integrating

r 2 d w=- -@+yh)+constantC 4p dz

Since

The final expression for velocity distribution is

which is a paraboloid of revolution.

?he discharge Q through the pipe is

Q ?he average velssity in the pipe, V= -

A

v= nu2

Y" -I E t h ='[-$(P+Yh))= 8P 8p2[ dz(y )) (8.46)

+ h is the piezometric head and is Ule head loss gradlent. Therefore, the head Y loss in the pipe in a length L of the pi

hf = 32pVL 128 pQL L=-- - (8.47)

r 2 r d 2 y n c t

d Substituting a = - and y = pg multiplying and dividiiig the rlght hand side by a I'actor 2Vwe

2 get after rearranging

hf = 64 v 2~

since IS the Reynolds number we write, CI

(8.48)

This may be compared to the Darcy-Weisbach expression which expresses the head loss in a length L of pipe of diameter d in terms of a friction factor and average velocity V as

h - , ~ L v ~ 16 ~ L Q ~ -

f - d . 2 g n 2 d 5 , 2 g (8.49)

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UNIT 9 TURBULENT FLOW Structure

9.1 Introduction Objectives

9.2 Turbulent Flow

9.3 Transition from Laminar to Turbulent Flow 9.3.1 Reynolds' Criterion

9.3.2 Rouse's Criterion

9.4 Mean and Fluctuating Components of Velocity

9.5 Quantitative Description of Turbulence

9.6 Eddy Viscosity and Mixing Length Concepts 9.6.1 Eddy Viscosity

9.6.2 Mixing Length

9.7 lllustrative Problems

9.8 Summary

9.9 Key Words

9.10 Answers to SAQs

9.1 INTRODUCTION

In this unit we will learn about turbulent flow, which is an irregular kind of flow as against the well-ordered laminar flow. Under certain conditions, laminar flow is not possible and we get a transition from laminar to turbulent flow. Most of the flows we encounter in nature are turbulent and hence the importance of this kind of flow. Further, turbulent flow also results in transfer of momentum, mass and heat across the flow and, therefore, assumes importance in many natural phenomena as well as in engineering applications. Objectives After studying this unit you should be able to

* identify transition from laminar to turbulent flow,

* describe turbulent flows,

* understand preliminary concepts related to analysis of turbulent flows, and

* analyse effects of turbulence.

TURBULENT FLOW

We now know that laminar flow is a well-ordered flow in which the fluid flows in parallel layers or laminas and transfer of momentum takes place at the molecular level. Turbulent flow occurs at relatively higher velocities. In turbulent flow chunks of fluid particles migrate from one layer to another. This migration results in transfer of mass as well as momentum across different layers at a macroscopic level. Most of the flows occurring in nature as well as in engineering applications are turbulent. Examples of turbulent flows include flows in natural streams, irrigation canals, pipes and sewers, flow of wind over earth's surface, flow of natural gas and oil in pipe lines etc..

Turbulent flow is an irregular condition of fluid flow in which various flow parameters show a random variation with respect to space and time coordinates such that statistically distinct average values of the flow parameters can be obtained.

Such random variations are observed in velocity components, pressure, concentration of suspended matter, fluid forces on structures such as suspension bridges, chimneys etc.. Qualitative variation of x-component of velocity u with respect to time and space is shown in Figures 9.1 (a) and (b).

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Turbulent Flow

f U t AT GIVEN

Y lNSTANT

0 F

AT A G I V E N LOCATION

0 t - u-

( a ) w. r . t . TIME ( b ) w.r.t . SPACE

Figure 9.1 : Variation of u

Turbulence can be generated in two ways: (1) by friction forces at solid walls (or surfaces) and (2) by the flow of fluid layers with different velocities past the another. These two types of turbulence are, respectively, known as wall turbulence and free turbulence.

Turbulent flow, like all other real viscous fluid flow, is dissipative in nature and requires continuous external source of energy for the continuous generation of the turbulent motion. The source of energy is in the form of mean shear for both types of turbulence and hence turbulent flows are also known as turbulent shear Plows. Viscosity of the fluid also makes the turbulence more homogeneous and less dependent on direction. If the structure of the turbulence is the same everywhere in the flow field, the turbulence is said to be homogeneous. Otherwise, it is non-homogeneous. 'Ihe turbulence is isotropic if it is independent of the direction. Otherwise, it is non-isotropic (or anisotropic). S A Q 1

(i) Differentiate between laminar and turbulent flows.

(ii) List the characteristics of turbulent flows.

(iii) ~ e f h e wall turbulence and free turbulence.

9.3 TRANSITION FROM LAMINAR TO TURBULENT FLOW

93.1 Reynolds' Criterion Prior to 1880, the difference between laminar and turbulent flows was not easily understood. Reynolds' experiment in 1880 clearly demonstrated the nature of these two flows. 'Ihe experimental set-up, Figure 9.2, consisted of a straight length of circular glass tube with a smoothly rounded and flared inlet placed in a large glass-walled tank full of water. Other end of the tube, having a suitable valve to control the flow of water through the

v / / f / / / / / / rn ' / ' / / rn

( a P

Figure 9.2 : Reynold's E~rperimeetd Set-up Ngun 9.3 : Bthaviour of Colound Filament

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tube, was outside the tank. An arrangement was made so that a small reservoir of a liquid dye discharged a coloured filament into the inlet of the glass tube. By observing the behaviour of the coloured filament, Figure 9.3, at different discharges (hence, velocity) of water through the glass tube enabled Reynolds to study the way in which Cater was flowing in the tube. The flow pattern of Figure 9.3(a) shows the condition corresponding to low discharge and Figure 9.3(c) corresponds to much higher discharge. The pattern of the coloured filament indicates that at low discharge, Figure 9.3(a), the water flows in parallel laminas and the filament remains a distinct streak. At higher discharge, Figure 9.3(c), there occurs intermixing of fluid layers and hence the coloured filament gets mixed with water across the entire section of the tube. This flow condition, obviously, corresponds to the turbulent motion. Figure 9.3(b) shows the intermediate stage when the turbulent motion starts. Through these experiments, Reynolds observed that the transition of flow from laminar to turbulent depends on the velocity of flow in the tube, the diameter of the tube and the kinematic viscosity of the liquid. Using the basic concepts and the experimental observations, he concluded that flow in a circular pipe remains laminar if the dirnensionlesss number U D / v , known as the Reynolds number, Re is less than 2100. Under normal conditions, the flow becomes turbulent if the Reynolds number is greater than 3000. When the Reynolds number is between 2100 and 3000, the flow is said to be in transition and is sometimes laminar and at other times turbulent. Under extremely controlled laboratory conditions the flow has been maintained laminar at Reynolds number as high as 50000. We have seen that the flow patterns in laminar and turbulent flows are considerably different. In addition, the relationship between the energy gradient and the velocity of flow is different for the two types of flow, Figure 9.4. While the energy loss is proportional to the first power of the velocity of flow in case of laminar flow, energy loss is proportional to un in which n generally varies from 1.7 to 2.

Fluid Dynamics - 11

TURBULENT l='&l[ V E L O C I T Y , U ( L O G S C A L E )

Figure 9.4 : Energy Loss in Laminar and Turbulent Flow

The Reynolds number, Re (= - uD or -) indicates the importance of the viscous force v U relative to the inertial force. Larger ~eynblds number signifies lesser influence of viscosity upon the flow pattern. An infinite value of Re corresponds to a flow in which the viscous resistance to deformation plays no role in comparison with the intertial resistance to acceleration. On the other hand, a small value of Re indicates more important role of viscosity in the flow. 9.3.2 Rouse's Criterion Turbulence can also be seen as breakdown of laminar flow in some regions of flow due to local disturbances. Under certain circumstances, local fluctuations (i.e.disturbances) in velocity will gradually be dampened by the viscous stresses and the flow eventually becoming laminar. Under other conditions, the viscous stresses may not be sufficient enough to dampen the local disturbances before these have spread throughout the flow region. This flow is termed turbulent flow. It is obvious that same minor disturbances would always be present in a moving fluid. These distrubances would dempen or spread depending upon whether the flow is stable or unstable. The stability of the flow at a given location would be influenced by the velocity gradient duldy, mass density p, viscosity p. and

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Turbulent Flow the wall distance y at the location. Rouse combined these quantities to form a dimensionless I parameter, lmown as stability parameter x , having the following form: I

Obviously, smaller the viscosity p or greater the numerator or this parameter, greater will be the local instability. On the other hand, a larger value of viscosity (1.e. smaller X ) would result in greater stabilizing effect. Also the value of x would be zero at the wall where y is zero and far away from the wall where boundary effects are negligible and hence du.dy = 0. At some intermediate value of y, the stability parameter shall attain a maximum value as

Figure 9.5 : Variation of Stability Permeter

shown in Figure 9.5. Experiments have indicated that the critical value of x is 500. This means that if x is less than 500 at all points in a given flow field the flow is inherently stable. If the value of the stability parameter in a flow exceeds 500, the flow becomes unstable and the flow has a tendencey to become turbulent.

9.4 MEAN AND FLUCTUATING COMPONENTS OF VELOCITY

In a turbulent flow, quantities such as velocities, pressures, forces etc. fluctuate about their mean values. If, for example, one plots the velocity u with time t at a given section, this

I t t VELOCITY ;r

Figore 9.6 : Velocity Variation with Time at a given Seetioo

variation of u with time will be as shown infigure 9.6. Thus, the variation in velocity u could be considered in terms of instantaneous velocity, fluctuating velocity and the time average velocity, i.e. at any instant, the instantaneous velocity u is equal to the sum of the time average velocity Ti and the turbulent fluctuating velocity u'. The component u' can be positive, negative or zero. These fluctuations in velocity result in an extensive intermixing between different fluid layers on a scale much larger than the molecular diffusion in laminar flows.

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Extending the concept of velocity variation to the conventional three co-ordinate directions X, y, z, the corresponding instantaneous velocities u,v and w can be expressed as

The time average velocity u can be found by integrating the instantaneous values of velocity - I

- 1 over a time interval T, thus u = - l u d t

T o

Here the averaging is considered ovet a sufficiently long period of time T. Further, it can be easily shown that the time average of fluctuations u' is zero, i.e.

Therefore, from the above result it can be concluded that the mean value of any fluctuating component in a turbulent flow is zero. This means ?I' = V' = 17 = 0

Statistical analysis of these fluctuations shows that the fluctuations follow a normal or Gaussian distribution. And from the property of the normal distribution, it can be shown that the maximum value of any instantaneous quantity could be as high as three times its time averaged value. Thus if velocity is considered one such quantity, it implies that the maximum value of instantaneous velocity could be as high as three times, the time-averaged mean velocity of flow. SAQ 2

i) List the factors which affect the transition from laminar to turbulent flow.

ii) Why is energy loss in turbulent flow much larger than that in laminar flow?

iii) What factors influence the stability of flow ? Define stability parameter.

iv) Show that ii' = 0

9.5 QUANTITATIVE DESCRIPTION OF TURBULENCE

As pointed out above, turbulent flow results in a complex motion which varies continuously with time. In such a motion, lumps of fluid particles in the form of eddies or vortices of various sizes travel haphazardly due to fluctuations in velocity. Thus one may visualize turbulent flow as a haphazard and ever changing system of eddies superimposed on the mean motion of the fluid. Some of the examples of this disorderly flows are smoke coming out of chimneys, dust patterns behind cars, flow in rivers etc. The diffusion of a column of smoke and its subsequent disappearance into the atmosphere is on account of turbulent mixing.

In' order to get an idea about the vigoumess of turbulent motion, one uses the root mean square value of; name @ . This r.ms. value of u' is a quantitative measure of the intensity of turbulence.

Fluid Dynamics - I1

WINDOWS
Comment on Text
dec.2007
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Turbulent Flow 9.6 EDDY VISCOSITY AND MIXING LENGTH CONCEPTS

9.6.1 Eddy Viscosity It has been already discussed in sections 9.2 to 9.4 that there are rapid fluctuations in velocity components in all the three directions which cause turbulent mixing of layers. This mixing of one layer to the other results in transfer of momentum. Thus, in turbulent flow, there is presence of additional shear which is over and above the viscous shear (discussed in Unit 8 on laminar flow). Therefore, an expression, similar to that of viscous shear for laminar flow, can be expressed as:

Where T and ii represents the time average shear and velocity at any point. The above equation is written in exactly the same way as that proposed by Newton for laminar flow which states that shear at any point is proportional to the velocity gradient at that point. Equation (9.1) can be written in the following simplified form by introducing constant of proportionality,

This equation was proposed by Boussinesq for the first time and the constant of proportionality ,q is called "eddy viscosity". This is analogous to dynamic viscosity p. The difference between them is quite significant. While dynamic viscosity p, is only the fluid property and depends upon temperature of the fluid alone , eddy viscosity q, is dependent upon fluid properties as well as the flow properties. Hence q varies from place to place in any given flow field whereas p remains constant (if the temperature remains constant).

Similar to kinematic viscosity, v eddy kinematic viscosity E, can be expressed as: E = II P

9.6.2 Mixing Length Consider two fluid layers A and B, some distance apart. As shown in Figure 9.7, let the time

I

I w X

Figure 9.7 : Turbulent Mixing

average velocity of these layers be ii, and &respectively. Let us assume that i& > ii,. Hence, it can be easily said that layer A moves with respect to layer B with velocity u,( = ii, - iB). Let small mass of fluid move from layer B to layer A with an average velocity of v,. This mass will now merge with layer A. If this cross motion is assumed to bt uniform over a small area, say a, the mass apv, has moved upwards and started moving with verocity u, and therefore a transfer of momentum equal to pauAvA has taken place. Hence a shear stress now will act on layer A towards left and it will be equal to

The negative sign indicates that the shear stress acts in a direction opposite to that of flow direction. Thus a slow moving layer exerts a retarding force on a fast moving layer and

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similarly the fast moving layer will also exert an equal and opposite force on the slow moving layer. In this manner one can explain that the additional shear stress in turbulent flow is due to the rate of change of momentum of fluid mass due to their motion across imaginary plane parallel to the main flow. The turbi-lent fluctuations u' and v' will naturally be responsible for such a shear. Hence, one can express the time average turbulent shear stress as:

- - 2 = - p u'v'

Now, if one has to compute this turbulent shear stress it is necessary to know the magnitudes of z. Since it is very difficult to measure these quantities, Prandtl proposed a

- mixing length hypothesis which can be used to express ,/. in terms of easily measurable quantities.

According to him, the mixing length 1, can be considered as a transverse distance between two layers such that the lumps of fluid of one layer could reach to the other one and get embedded in it, while retaining their momentum in x direction during their travel through 1.

Prandtl has further assumed that

Further, he assumed that v' is proportional to u' and hence one can write

wherein constant of proportionality is included in 12. From this, Prandtl evaluated the turbulent shear stress 7 as:

Hence, the total shear stress at any point for turbulent flow can be written as.

Wherein the first term on the right side of the equation is viscous shear and the second one is turbulent shear.

SAQ 3 i) What is the role of eddies in a turbulent flow ?

ii) List the similarities and dissimilarities between p and q.

iii) Define mixing length and state the relationship that exists between the turbulent shear stress and the mixing length.

Fluid Dynamics - I1

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Turbulent Flow 9.7 ILLUSTRATIVE PROBLEMS

Example 9.1 :

Below are given the instantaneous values of velocity in m/s measured at a given point in flow at an interval of 1.0 sec.

Determine

Solution :

Total number of values of u = 20 Zu = 108.14

- Zu 108'14 - 5.407 m/s Ans. U = 2 0 = U ) -

As u' = u - Z, values of u' corresponding to u values are:

2.523, -1.717,-1.197, -1.107,4.037, -2.787, -0.557, -0.437,3.133,2.093,2.213, -2.537, 3.343, -1.257,1.023, -0.227, -1.447,4.757,2.153, -2.417

Hence

and - 4 7 - = 0.352 Ans. U

Example 9.2 :

Two open wagons A and B are moving on parallel tracks in the same direction at velocities of 20.0 m/s and 30.0 m/s respectively. If gravel is transported from B to A at the rate of 1000 kg per second, determine the tangential force acting on A.

Solution :

Figure 9.8

As shown in the Figure 9.8, the mass of gravel transferred,from B to A per unit time is 1000 kgls.

Relative velocity of B.w.r.t. A = 10 m/s. Force acting on A = 1000 (10-0) = 10.000 N. Ans.

9.8 SUMMARY

In this unit we have studied the basic concepts of turbulent flow and its characteristics as they differ from laminar flow. Conditions at which laminar flow changes to a turbulent flow were also examined. The role of velocity fluctuations in intense mixing that results in such flows was emphasised. Finally the concept of mixing length was discussed evaluating the turbulent shear stresses in these flows.

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Turbulent Flow : Flow characterised by intense intermixing of fluid particles.

Instantaneous Velocity : Velocity at a given time and is equal to the time average velocity plus the turbulent fluctuation

- .- velocity.

Critical Reynolds Number : Value of Reynolds number at which transition from laminar to turbulent flow is expected to occur.

Isotropic Turbulence : Turbulence in which the mean-square velocity fluctuations in the three co-ordinate directions are equal to each other:

Wall Turbulence : Turbulence, generated due to the presence of solid wall is known as wall turbulence,

Free Turbulence : Turbulence, generated by the flow of fluid layers at different velocities, in the absence of a solid wall is known as free turbulence.

Mixing Length : Average distance perpendicular to the flow which a small fluid mass travels before its momentum is changed by the new environment.

9.10 ANSWERS TO SAOs

Check your answers of all SAQs with respective proceeding text of each SAQ.

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UNIT 10 BOUNDARY LAYER ANALYSIS Structure

10.1 Introduction Objectives

10.2 Description of the Boundary Layer 10.2.1 Boundary Layer 'Ihickuess and its Characteristics 10.2.2 Laminar and Turbulent Boundary Layers aad Laminar Sublayer 10.2.3 Praadtl's Boundary Layer Equations

10.3 Hydrodynamically Smooth and Rough Boundaries 10.3.1 Velocity Dlsmbut~on near Smooth and Rough Boundaries

10.4 Boundary Resistance 10.4.1 Boundary Shear f a Laminar Boundary Layer 10.4.2 Boundary Shear for Turbulent Boundary Layer

10.5 Illustrative Problems

10.6 Summary

10.7 Key Words

10.8 Answers to SAQs

1 0 INTRODUCTION

In this unit we will study the basic concepts of boundary layer as given by Prandtl in 1904. In fact the theory of boundary layer flow constitutes the backbone of the modem fluid dynamics as it has helped in a rapid advances in various branches of engineering especially aeronautical engineering.

It is our common observation that bodies moving through water or air experience lot of frictional resistance during their motion. For example, motion of a ship, aeroplane, automobiles, turbine blades etc. The theory of hydrodynamics which considers the motion of ideal fluids fails to explain this frictional resistance of such moving bodies. Prandtl realised this shortcoming of the hydrodynamics and showed that for the motion of bodies through water or air, the effect of fluid viscosity is limited in a thin layer close to the body. It is this thin layer which is known as boundary layer. Since in several branches of engineering one is concerned with relative motion between solid wall or on immersed body and the real fluid, concepts and analysis of boundary layers assumes prime importance.

Objectives After going through this unit, you will be able to

explain development of boundary layer along a flat plate,

estimate thickness of boundary layer,

identify the hydrodynamic nature of the boundary surface i.e. smooth or rough,

describe flow characteristics inside the boundary layer and their effects on velocity distribution, and

estimate the frictional drag on a flat plate as a result of boundary layer development.

10,2 DESCRIPTION OF THE BOUNDARY LAYER

In order to understand the concept of a boundary layer, let us consider the flow of a real fluid over a thin stationary plate held parallel to the flow in a uniform stream of velocity Uo as shown in Figure 10.1.

Plate held parallel to the flow as in Figure 10.1 is also called plate held at zero incidence and the uniform velocity just upstream of the leading edge of the plate is known as ambient velocity,fieestreamvelocity or potentialvelocity. In the case of real fluids, however small their viscosity may be, the fluid particles adhere to the boundary and hence the condition of no slip prevails. The condition of no slip implies that if the wall or the solid boundary is stationary, the fluid velocity at the wall will be zero whereas if the boundary is moving, the fluid adhering to the boundary will have the same velocity as that of the boundary.

Thus, when the fluid approaches the plate with uniform velocity Uo as shown in Figure 10.1, . . . . . . . . ..

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Further away from the wall, the velocity will change rapidly from zero at the boundary to its uniform velocity U, over a short distance as shown in Figure 10.1. As such there is a velocity

I/ VELOCITY DISTRIBUTION

LEADING EDGE OF THE PIPE PLATE

FFpure 10.1 :Mnilinn Sketch

gradient causing shear or tangential stress on the surface of the plate in the direction of

motion. This shear stress on the surface of the plate is given by the foliowing equation :

The force caused by this stress in the direction of motion is known as surface drag or simply drag whereas the corresponding equal and opposite force exerted by the wall on the fluid is known as shear resistance to the flow. As the fluid passes over the plate, this viscous shear retards more and more fluid consequently the thickness of the boundary layer increases in the downstream direction.

It can be seen from the equation (10.1) that even if the fluid has very small viscosity, the presence of high velocity gradient near the boundary is responsible for high shear stresses. Following important observations could be made on the basis of above description of boundary layer :

1. Viscous effects are confined to a very thin layer, called the boundary layer, near the solid boundary.

2. The thickness of the boundary layer will increase along the downstream direction accordingly velocity gradient will reduce in the downstream direction.

3. Reduction in velocity gradient will mean reduction in z , as per equation (10.1) in the downstream direction.

4. The flow outside the boundary layer can be considered as a potential flow of an ideal fluid i.e. the viscous effects could be ignored.

SAQ 1

(i) What is 'no slip' condition?

(ii) Define free stream velocity.

(iii) What is surface drag?

10.2.1 Boundary Layer Thickness and its Characteristics It was seen above that in case of flow over a flat plate, the thickness of the boundary layer increases in the direction of flow. In many boundary layer problems, the knowledge of this thickness is required. It was also shown that inside the boundary layer, the change in the velocity - zero at the boundary to the free stream velocity U,, takes place asymptotically. Hence there is some difficulty in defining the exact thickness of the boundary layer in which this change in the velocity takes place. Some of the commonly used definitions of boundary laver thickness are as below :

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Boundary Layer Analysis Nominal Thickness 6

It is defined as that distance from the boundary (measured in the y direction) where the velocity differs by 1 percent from the free stream velocity. Thus we have the following condition for the boundary layer thickness or nominal thickness 6.

y = 6 for u = 0.99 Uo

Displacement Thickness 6':

The boundary layer formation and the resulting velocity distribution indicate a reduction in the flow rate as compared to the one that occurs in the absence of the boundary layer. To account for this, a thickness known as displacement thickness 6. is defined. To clarify this concept, consider Figure 10.2 which shows a graphical representation of this displacement thickness.

Fi y r e 10.2: Displacement Thickmas

Let u be the wlocity at a distance y from the boundary, the discharge per unit width through an element of thickness 6y is u.6y. In the absence of any boundary layer, this discharge would have been U,, 6y. Thus, the total reduction in discharge caused by the boundary layer is given by

If we now define 6' such that

w

U06* = ( U U ) dy (10.2) 0

The value of 6. so defined is known as the displacement thickness. In other words, to reduce the total discharge of a frictionless fluid by the same amount as caused by the boundary layer, the solid surface would have to be displaced outwards by a distance 6..

Thus, the concept of displacement thickness allows us to consider the main flow as that of a frictionless fluid past a displaced surface instead of the actual flow past 'the actual surface.

Since for y greater than 6, the local velocity is equal to U, equation (10.2) can be simplified as

Momentum Thickness 0

On lines similar to displacement thickness, we now define momentum thickness denoted as 8. Consider Figure 10.2 again, the fluid passing through an element of the boundary layer carries momentum at a rate (pudy)u per unit width whereas in frictionless flow, the same amount of fluid would have its momentum (pudy )Uo. Thus total reduction in momentum is given by

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I If we now define 0 such that Fluid Dynamics - I1

I the value of 8 so defined is known as momentum thickness. Equation (10.4) can be simplified as,

From above expressions of boundary layer thicknesses, it can be seen that the displacement thickness is smaller than nominal boundary layer thickness while momentum thickness is

I smaller than the displacement thickness.

sb 10.2.2 Laminar and Turbulent Boundary Layers and Laminar Sublayer In the previous unit, we had defined laminar and turbulent flows and pointed out that such flows are identified on the basis of Reynolds number. For example, flow in a pipe is

UD . characterised as laminar if the value of the Reynolds number defined as is less than

2100. Here U is the mean velocity of flow in the pipe, D is the diameter of the pipe and v is the kinematic viscosity of the fluid. Likewise the flow in a bouildary layer may he either laminar or turbulent. Figure 10.3 may be referred to visualise the development of laminar and turbulent boundary layers on a flat plate.

figure 10.3 : Development of Boundary Layer over a Flat Plate

For boundary layer flow along a flat plate, the characteristic Reynolds number is defined as

LAMINAR BOUNDARY

- LAYER

I

uox R, = - where x is the distance along the flat plate measured from the leading edge. It v

has been found that when the fluid of small viscosity flows over a flat plate, the flow in the boundary layer changes from laminar to turbulent if the Reynolds number Rex exceeds a certain value. From the leading edge uplo a certain length, the flow in the boundary layer has all the characteristics of laminar flow, this is irrespective of whether the ambient flow is laminar or turbulent. With increasing thickness of the boundary layer, the laminar boundary layer becomes unstable and the flow within it starts changing into turbulent flow. The region within which this change in flow takes place is known as transition region. Downstream of the transition region, the boundary layer is known as turbulent boundary layer and its thickness increases further. It is worthwhile to mention at this stage that for the case of flow over the flat plate as shown in Figure 10.3, the pressure is uniform as such there is no pressure gradient and secondly the thickness of the boundary layer increases continuously provided there is no separation.(The phenomenon of flow separation is discussed later on). Further, the scale in the y direction as shown in Figure 10.3 is greatlyenlarged and the thickness of boundary layer 6 at any distance x i s very small compared to x.

-_ TF!ANSlTlON - -

The value of Reynolds number Rex at which a laminar boundary layer becomes unstable depends on number of factors such as roughness of the surface, intensity of turbulence in the main stream, pressure gradient etc. In general, the critical Reynolds number Rex defined above varies between 3 x lo5 to 6 x 10'. However, for practical purposes, the value of 1 critical Re= is taken as 5 x 10'.

T U R B U L E N T - BOUNDARY L A Y E R

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Boundary Layer A d y s i s Laminar Sublayer

Turbulent flow is characterised by the presence of random components of velocity in all directions. Since fluid particles can not pass through the solid boundary over which the flow is taking place, these random components of velocity must die out very close to the solid boundary. It therefore, follows that the turbulent flow can not exist immediately in contact with the solid boundary.

Thus even when the main flow possesses considerable turbulence, and even when a greater part of the boundary layer is also turbulent, there is still an extremely thin layer, adjacent to the solid surface, in which the flow has negligible fluctuatiox of velocity. This layer which is very thin and generally denoted by 6' is known as trrarinar sublayer or viscous sublayer, see Figure 10.3. This layer is not to be confused with the laminar boundary layer because the velocity at the outer edge of the laminar boundary layer is equal to the velocity of flow Uo of the main sveam whereas the velocity at the outer edge of the laminar sublayer is different than U, see Figure 10.3.

18.2.3 Prandtl's Boundary Layer Equations As discussed above, the effects of viscosity for flow of real fluids are confined within the boundary layer, with some simplification, Navier-Stokes equations could be simplified to yield boundary layer equations.

The flow p,ast a flat plate of infinite extent is essentially a two dimensional flow and is divided into following two regions :

au (a) The region within the boundary layer where the xlocity gradient, -, is x r y large givjng ay

large shear stress z = p k. It is this region where Navier-Stokes equations apply. JY

(b) The region outside the boundary layer, here velocities are of the order of free stream velocity U,. The flow in this region is governed by the potential flow theory; the viscous effects are negligible and the velocity gradients are also very small. Thus the boundary conditions for analysing the flow in the boundary layer can be written as :

(i) no slip condition at the solid boundary,

(ii) outside the boundary layer, the velocity u tends to become equal to free stream velocity Uo

i.e. u -, Uo for y > 6

6 6 (iii) - < 1.0 i.e. the dimensionless boundary layer thickness - is very small as X X

compared to unity.

Neglecting the body force, the Navier-Stokes equation along the flow direction for flow .outside the boundary layer is reduced to

For steady flow. the pressure p is a function of x only, the partial derivative * can therefore ax

be replaced by the total derivative dl.! and the equation (10.6) thus reduces to ak

1 p + - p Uo = constant 2

The simplified Navier-Stokes equation is thus writ@n as

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(i) in the x direction

for steady flow and substituting the value of @ equation (10.9) is simplified to ah'

(ii) in the y direction

Taking order of magnitude of various quantities, the Navier-Stokes equation in the y direction results in

and the continuity equation becomes

SAQ 2 (i) Define nominal thickness of boundary layer.

(ii) Write down the definitions of Reynolds number for flow in a pipe as well in the boundary layer. Explain the terms involved in these definitions.

(iii) Define laminar sublayer

10.3 HYDRODYNAMICALLY SMOOTH AND ROUGH BOUNDARIES

Consider the flow of a fluid over a solid boundary. If we examine the boundary closely, we will notice that its surface has some irregularities or projections. The height of these irregularities will depend on the nature of the surface. Let us assume that the average height of these projections is k, . One would consider this surface as rough for a large value of k, whereas it will be considered as smooth for smaller values of k,. However, in problems related to flow of fluids, the absolute size of the roughness elements is not a measure of boundary roughness, it is related to fluid and flow properties.

Consider the case of a boundary of roughness height k, and for this case also consider that the thickness of laminar sublayer 6' is much greater than k, , see Figure 10.4 (a). Since all the rough elements are well within the laminar sublayer, these elements do not affect the flow in the turbulent boundary layer. In this case even though the surface is rough, it is defined as

hydrodynamically smooth boundary. Experiments have shown that when ks is less than 0.25, F the boundary is treated as hydrodynamically smooth boundary.

Fluid Dynamies - I1

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Boundary Layer Analysis

1 T U R B U L E N T B O U N D A R Y L R V E R

L H E I G H ~ OF ROUGHNESS ELEMENT

(a) HYDRODYNAMICALLY SMOOTH BOUNDARY

6

HEIGHT OF ROUGHNESS ELEMENT T U R B U L E N T B O U N D A R Y LAYE{

kk , 0.25 6'

L A M I N A R S U B L A Y E R

k/, > 6.0 6'

u - s' ?-LA:~~NAR SUBLAYER

( b) HYDRODYNAMICALLY ROUGH BOUNDARY I

figure 10.4 : Definition Sketch for Smooth a d Rough Boundaries

On the other hand if the thickness of the laminar sublayer 6' is smaller than k, , see Figure 10.4 (b), the roughness elements project outside the laminar sublayer as a result this layer gets completely destroyed. Wakes are formed behind each roughness element and there is considerable energy loss. The boundary in this case is known as hydrodynamically rough

ks boundary and for this " is greater than 6.0. For 0.25 < < 6.0, the boundary is F 6 classified as boundary in transition.

10.3.1 Velocity Distribution near Smooth and Rough Boundaries Let us now study the variation in velocity with distance y from the boundary. We have already seen that for laminar flow over a plane boundary, the velocity distribution is obtained by integrating the following equation.

Here u is the velocity at a distance y from the boundary, and p i s the dynamic viscosity of the fluid. Likewise for turbulent flow we have

Here bars denote the temporal mean values of velocity and the shear stress, and q is eddy viscosity. Therefore to obtain velocity distribution for turbulent flow, one needs to integrate equation (10.12). One difficulty arises in this integration because the eddy viscosity q varies with y. Fortunately, experiments indicate that in turbulent flow the intensity of shear and the eddy viscosity vary insuch a manner that their ratio is, as a first approximation, directly

proportional to (a parameter known as shear wlocity U.) and inversely proportional P

to the distance y from the boundary. For simplicity let us omit bars from equation (10.12), this equation can be rewritten as :

du - 2.5 - - - u* (10.13)

dy Y in which the factor 2.5 is the constant of proportionality determined from experiments. This factor is also expressed as 1/K where K is Karman's constant having its value equal to 0.40.

Integration of equation (10.13) leads to

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Fluid Dynamics - Il

FFpre 105 : Logarithmic Velodty Distribution in Turbulent Flow

The constant of integration C can be evaluated if we write u = 0 at y = y'. This condition yields following :

- - - 5.75 log,, 1 u* Y'

In obtaining equation (10.14) no assumption was made as to whether the boundary is hydrodynamically rough or smooth. Hence equation (10.14) is applicable for both of these boundaries. Further the condition u = 0 at y = y' assumed in obtaining this equation at a first glance would appear to be in disagreement with physical fact. In fact it is not so and could be better understood if we recall that close to the boundary there is a laminar sublayer where effects of viscosity are more pronounced. In this layer, the velocity distribution may be assumed to be linear (which implies constant shear stress in this layer) and one can use equation (10.11) to obtain the velocity distribution in this layer i.e.

Dlviding this by p and simplifying taking U. = $, one get

Thus, a close examination of equations (10.14) and (10.15) will show that the zone of laminar sublayer must extend well beyond the distance y' if a smooth transition is to exist between the velocity distributions for the laminar and turbulent flows see Figure 10.6.

I A I

TRANSiTION (BUFFER ' 1 ZONE

/ / T H I C K N E S S O F LAMINAR

Figure 10.6 : Volwlty Distribution in Different Zones

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Boundary Layer Analysis If one arbitrarily selects the interseqtion of curves given by equations (10.14) and (10.15) as the thickness of laminar sublayer 6 one gets

Nikuradse's experimental data indicate that for hydrodynamically smooth boundaries

substituting this value in equation (10.14) one gets

U - = 5.75 log,, - u*

U* + 5.5 v

This is known as the Karman-Prandtl's equation for velocity distribution for turbulent flow near hydrodynamically smooth boundaries. As can be see from Figure 10.6, equation (10.17)

is not valid close to the boundary. data indicate that equation (10.1) is valid for 9 illl 30 v

while for values of % < 5.0. equation (10.15) is valid. For the transition region i.e. v

5 c - '* < 30, following equation has been suggested v

In the case of hydrodynamically rough boundaries, it is hardly to be presumfl that a laminar sublayer will exist at the boundary if the roughness height k, is greater than 6 . Nikuradse's experimental data indicate that for rough boundaries

and introducing this value of y' in equation (10.14) one gets

- - - 5.5 log,, 1 + 8.5 u* k,

Equation (10.19) is known as Karman-Prandtl's equation for the velocitydistribution in turbulent flow near rough boundaries.

Inspite of inherent limitations of equations (10.17) and (10.19), these are used for all practical purposes to determine the velocity distribution in a vertical section of a given boundary.

SAQ 3 (i) How are boundaries classified as smooth and rough?

(ii) Write down Karman-Prandtl's equations of velocity distributions in turbulent flow riear smooth and rough boundaries.

(iii) Explain why logarithmic law of velocity distribution is not valid at the boundary.

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10.4 BOUNDARY RESISTANCE Fluid Dyeamiw - I1

From the above discussions on boundary layer, it is now clear that the viscous effects are limited in a narrow region close to the boundary. Let us now discuss the magnitude of the resulting forces exerted by the fluid on the boundary. This aspect is proposed to be discussed separately for laminar and turbulent boundary layers.

10.4.1 Boundary Shear for Laminar Boundary Layer Consider again the flow over a flat plate held longitudinally in a moving fluid. The boundary layer would develop on either side of this plate, giving rise to boundary shear stresses. The magnitude of these stresses depends on Re, and the boundary layer thickness 6. As the Reynolds number is a ratio of inertial force to viscous force, a smaller value of it will signify larger viscous forces whereas a large value of Reynolds number would correspond to small viscous forces. Obviously, 6Ix should be a function of R,. Indeed it has been shown both analytically and experimentally that for laminar boundary layer,

Equations (10.20) or (10.21) is known as Blasius equation for the boundary layer thickness.It can be seen from this equation that the boundary layer thickness increases with x and v and decreases with U,.

With increase in the boundary layer thickness, the velocity gradient at the plate will decrease du in magnitude with a consequent reduction in boundary shear stress z since z = p -. dy

Expression for the intensity of shear z, at the surface of the plate can be obtained as :

Substituting the value of 6 from equation (10.21) in equation (10.22). we get,

To = Const. X

after introducing the experimental value of the proportionality constant obtained by Blasius. In equation (10.23), cf is known as local drag coefficient.

The total drag force exerted by the fluid on one side of a plate of width B and length L can now be evaluated as

Evaluation of this integral will show that

In which C, is the mean drag coefficient and its value is given by

UoL where Re, = - v . Equation (10.25) is valid for Re, less than 5 x lo5.

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~ouobvy Layer Analysis 10.4.2 Boundary Shear for Turbulent Boundary Layer As we have studied in earlier units, the turbulent flow is characterised by intense mixing of different fluid layers, as a result velocity distribution in a turbulent boundary layer is more uniform as compared to that in laminar boundary layer. This at the same time produces a very rapid change in velocity near the wall. The corresponding expressions for S, c,. C, for the turbulent boundary layer are as follows :

and

It can be seen from equations (10.21) and (10.26) that the boundary layer thickness increases as 2'' for laminar boundary layer while it increases as f B O for turbulent boundary layer. Two important characteristics of the turbulent boundary layers may be noted. Firstly, the thickness of the turbulent boundary layer increases at a faster rate than the laminar boundary layer thickness. Secondly, the velocity gradients are much larger for turbulent boundary layer than for laminar boundary layer giving large shear stresses.

Equation (10.28) is applicable within the range 5 x Id to 2 x lo7 of Ft,,, provided the boundary layer is turbulent from the leading edge, see Figure 10.7. This means that some special measures must be taken to make the boundary layer turbulent right fiom the leading edge, or one assumes the length of the laminar boundary layer small as compared to the total length of the plate. Beyond %, = 2 x lo7, following equation may be used for C,

In case where the plate is covered with laminar and turbulent boundary layers, following equation as given by Prandtl may be used for C,

+ wIESELSBERGER

- LAMINAR

\

figure 10.7 : The Mean Drag Coefficient for a Plane B d q as a Function 01 the Reynolds Nombet

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The value of constant 1700 in equation (10.30) depends on the value of critical Reynolds number R, at which the laminar boundary layer becomes turbulent and is given in Table 10.1.

TABLE 10.1 : Variation of Constant in equation ( 1 0 3 ) with tbx

1 Equation (10.30) is valid for QL values between 5 x 1@ and lo7 while equation (10.31) is to

1 t

be used for QL > lo7.

R, Constant

Equation (10.31) is applicable for Re, values 5 x 1030 lo9, and the value of A is same as the value of constant given in Table 10.1.

SAQ 4

3 x los

1050

(i) Assuming boundary layer to be laminar, write down the equation that gives the boundary layer thickness at a given section.

(ii) Define local drag coefficient.

5~ lo5

1700

(iii) Is the statement that turbulent boundary layer thickness varies as correct?

10.5 ILLUSTRATIVE PROBLEMS

10"

3300

Example 10.1 5

3~ lo6

8700

Air at 20" C (p, = 1.208 kg/m3, p = 1.85 x lo-' kglms) flows over a 2.0 m wide plate at 10.0 d s velocity. Determine

(i) zo and 6 at place where the boundary layer ceases to be laminar.

(ii) Drag force on one side i f the plate in the laminar region.

Solution :

Let us first calculate the value of x upto which the boundary layer remains laminar

Thickness of the laminar boundary layer at x = 0.765 m

or 6 = 5.41 mm

Local drag coefficient

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Boundary Layer Annlysis

2 ld b = C, pf !$ = 9.33 x 10- x 1.208 x - = 0.056 N/m2 2

Drag force on one side of the plate in the laminar region is given by taking x = L

where

Example 10.2 :

A rectangular plate of sides a and b is towed through water once in a direction of a at the velocity U, and then in the direction of b at the velocity U,,. If in both the cases boundary layer is laminar over the entire length of the plate, determine the ratios of U, and U, which will give equal force on the plate.

Solution :

Case 1 Case 2

u a - a

Case 1 :

Equating FD, and FD2 and simplifying the expression one gets

V3

Answer

10.6 SUMMARY In this unit we have shldied some of the basic concepts of boundary layer alongwith the variois characteristics of laminar and turbulent boundary layers on a flat plate. The importance of laminar sublayer in classifying the nature of boundary roughness was also studied. The nature of velocity distributions over smooth and rough boundaries was then examined. Finally, the various equations for evaluating the boundary layer thicknesses and the drag force on the boundary surface were discussed.

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10.7 KEY WORDS Fluid Dynamics - I1

Boundary Layer

Boundary Layer Thickness

Viscous Layer

Laminar Subfayer

Velocity Distribution

Rough & Smooth Boundaries

Drag Coefficient

Reynolds Number

The layer adjacent to the boundary where fluid is retarted is known as boundary layer.

This is also known as nominal thickness of boundary layer. It is defined as the distance from the boundary where the velocity differs by one percent £rom the ambient velocity Uo

Layer in which viscous forces are more pronounced.

Thickness of turbulent boundary layer in which the flow is laminar is known as laminar sublayer.

Force exerted by the fluid on the body in the direction of flow.

Variation in velocity with distance from the boundary of a given section.

ks Boundary is rough if -gr i 6.0. Boundary is

ks smooth if < 0.25

Designated as CD and is interpreted as drag force per unit area divided by the dynamic head.

It is the raito of inertial force per unit volume to viscous

force per unit volume. It is defined as % where L is

some characteristic length dimension.

10.8 ANSWERS TO SAQs

Check your answers of all SAQs with respective preceding text of each SAQs.

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UNIT 11 PIPE FLOW Structure

11.1 Introduction Objectives

11.2 Energy and Hydraulic Gradient Lines

11.3 Darcy-Weisbach Equation k 11.3.1 Variation o f f with Re and 2 D

11.3.2 Relation Relating f , U and U*

11.4 Resistance of Commercial Pipes

11.5 Illustrative Problems

11.6 Summary

11.7 Key Words

11.8 Answers to SAQs

1 1 . INTRODUCTION

In our earlier units we have already studied the characteristics of laminar and turbulent flows through pipes. Whenever there is flow in a pipe, there is always some energy loss. The amount of this loss depends on the nature of the flow, fluid characteristics and the characteristics of the pipe. In this unit we will study the methods of calculation of this energy loss. Such an estimation is very important in the design of pipe lines of various sizes required to carry the necessary discharge. It is known that this head loss varies directly with the length of the pipe. Since the lengths involved in the design of pipe lines are quite large, one would like to make as precise an estimate of this loss as possible.

Objectives After studying this unit, you should be able to

* understand the meaning of terms like energy gradient line, hydraulic gradient line, head loss due to friction and pipe resistance, and

* calculate head loss due to friction in pipes and obtain hydraulic gradient and energy lines

11.2 ENERGY AND HYDR4ULIC GRADIENT LINES 1)- Figure 11.1 shows flow through a pipe. Two piezometer tubes are installed at sections 1 and 2 of the pipe. The water levels in the tubes indicate the pressures at these sections of the pipe.

-- ENERGY GRADIENT LlNE

I - r HYDRAULIC GRADIENT L I N E .

DATUM LlNE

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10.7 KEY WORDS Fluid Dynamics - I1

Boundary Layer : The layer adjacent to the boundary where fluid is retarted is known as boundary layer.

Boundary Layer Thickness : This is also known as nominal thickness of boundary layer. It is defined as the distance from the boundary where the velocity differs by one percent from the ambient velocity Uo

Viscous Layer : Layer in which viscous forces are more pronounced.

Laminar Subiayer : Thickness of turbulent boundary layer in which the flow is laminar is known as laminar sublayer.

Drag : Force exerted by the fluid on the body in the direction of flow.

Velocity Distribution : Variation in velocity with distance from the boundary of a given section.

Rough & Smooth Boondaries ks t Boundary is rough i f v >' 6.0. Boundary is

Drag Coefficient

Reynds Number

ks smooth if c 0.25

: Designated as CD and is interpreted as drag force per unit area divided by the dynamic head.

: It is the raito of inertial force per unit volume to viscous

force per unit volume. It is defined as where L is

some characteristic length dimension.

10.8 ANSWERS TO SAQs

Check your answers of all SAQs with respective preceding text of each SAQs.

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Fluid Dynamics - II

P I PZ Thus - and - are the pressure heads at sections 1 and 2 respectively. The line joining Y Y these two water levels is known as hydraulic gradient line. The total energy, expressed in terms of head of flowing liquid, in the flow at section 1 with reference to the datum line is epresented as

where Z is the elevation of the centre line of the pipe, p is the pressure in the pipe and U is the mean velocity of flow in the pipe. Suffix denotes the section number. Likewise the total energy at section 2 is given as

The line joining these two total energy levels with respect to the chosen datum is known as energy gradient line or simply the energy line. The loss of head that results when water flows from section 1 to section 2 is represented by hf. Thus, writing the energy equation between section 1 and 2 one gets

Equation (11.3) forms the basic energy equation for solving problems concerning-loss of head. Subsequent discussions in this unit pertain to the estimation of hf.

113 DARCY-WEISBACH EQUATION

Experimental data on flow of water in a long, straight circular pipe indicate that the head loss h, varies in proportion with velocity head and pipe length L and inversely with pipe diameter D. Thus,

where f is a coefficient of proportionality known as Darcy- Weisbach's friction factor. Equation (11.4) is commonly known as Darcy- Weisbach's resistance equation.

The friction factor f depends on the characteristics of flow, fluid and the pipe boundary. The functional relationship for f can be expressed as

Here, v is the kinematic viscosityof the&wing liquid and k, is the equivalent sand grain roughness.

k, . D . The term - u known as relative roughness while - 1s called relative smoothness and is D k, v

the Reynolds number Re.

The physical significance of equation (11.5) may be noted. It signifies that the friction factors for different pipe flows will be the same if the flow Reynolds numbers, and relative roughnesses are the same.

WINDOWS
Comment on Text
dec-2007
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Pipe Flow

11.3.1 Variation off with Re and D

The relationship that relates f with Re and 5 has been obtained on the basis of data D

collected by Nikuradse on flow through artificially roughened pipes. In these experiments the pipes were roughened by fixing a coating of uniform sand grains to the pipe wall.

ks Figure 11.2 shows the variation off with Re and -. D

0.1

0.08

0-0 6

0.04

0.02

0.01

figure 11.2 : StPnt(w1'3 Diagr~m

These data were first plotted by Stanton with logarithmic scales for f and Re as shown in Figure 11.2. As such this diagram is also known as Stanton's diagram. Figure 11.2 shows the following important characteristics :

64 1. Upto R e = 2100, f is given byf = - for all surface roughnesses. This is the limit for

Re laminar flow. Thus the head loss in laminar flow is independent of surface roughness.

ks 2. In a turbulent flow f is a function of Re and -. This variation is such that for large D k

values of Re, f becomes independent of R e and depends only on 2. Equation (11.4) D

2 would then show tha for fully rough turbulent flow in a pipe, hf a U .

3. Although the lowest curve in Figure 11.2 was obtained from tests on hydraulically smooth pipes, many of Nikuradse's rough pipe data coincide with it for 5000 < Re < 50,000. Here the roughness k, is fully submerged within the laminar sublayer and has no effect on friction factor. Blasius has shown that this curve (for 3000 < Re < lo5) may be closely approximated by a line whose equation is

0.316 f = @ITS

4. The data for rough pipes deviate frorn the curve for smooth pipe with increasing Reynolds number. In other words, pipes which are smooth at low values of Re become rough at higher values of Re.

I

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Unfortunately, the excellent results of directly to engineering problems since entirely different, much more variable

Nukuradse as discussed above cannot be applied F l i d Dyuemiu - I1

the roughness patterns of commercial pipes are and much less definable than the artificial roughness

used by Nikuradse.

In order to evaluate the roughness of commercial pipes, let us study the following relationships.

11.3.2 Relation Relating f, U and U* We have studied that for a steady uniform flow in a horizontal pipe, the boundary shear 2,is given by

T o = - 9 0 ax 4

7, = * D Since - & ! = ' P I - PZ - yhf 4 L ax L L

2

Dividing both sides of the above equation by and solving it alongwith equation (1 1.4) 8

one gets

Here U. = 4 is defined as shear velocity.

Relationship involving$ U and U. can also be obtained if we use the equations for mean velocity of flow U for turbulent flow in smooth and rough pipes.

, For turbulent flow in smooth pipe of radius R, mean velocity is given by

u. u* + 1.75 = 5.75 log,, - v

while for rough pipe we have R l- = 5.75 log,, - + 4.75

u. k s

Expressing ? in terms of f as per equation (1 1.8) in equations (1 1.9) and (1 1.10) and u.

simplyfying the resulting equations, the following friction factor equations are obtained:

= 2.03 log, Re @- 0.91 for smooth pipes vf (11.11)

and R = 2.03 loglo -+ 1.68 for rough pipes * (11.12) k s

With a slight modification of numerical constants in equations (10.21) and (10.22) to conform to the experimental data, the final equations which fit data for friction factor in turbulent flow are :

= 2.0 log,, Re @- 0.80 for smooth pipes + (11.13)

R 1 = 2.0 log,, -+ 1.74 for rough pipes' @ (11.14)

k s

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Pipe Flow Equation (10.13) is plotted in Figure 11.2 and is valid for Re values between 4 x lo3 to 4 x lo7. Since equation (1 1.13) is implicit in f, the value off can only be obtained using trial and error procedure. This can be avoided by using following empirical equations:

or alternatively - - - 1.8 log,, Re - 1.5146 G (11.16)

Above equations show that for smooth pipes, f is a function of Re only while for rough pipes, R

it depends only on -. Obviously an intermediate zone must exist in which the friction factor f ks

R depends both on Re and - and this zone is known as transition zone. In order to examine ks

the variation in f in tht: transition zone, let us rearrange equations (11.13) and (11.14) in the form

e T w e e [ = 2 log, - O,80 for smooth pipe and 1.74 for rough p i p . Since R / k s R / k s

1.

- has been shown to be a parameter for classifying hydrodynamically smooth and rough 6'

ks boundaries, this should also classify the transition zone. The value of - can be expressed in

6' ~e terms of - as shown below : R / k s

Since

We have /

On simplification this gives

1 R Thus, equation (11.17) is expected to show a unique variation between - - 2 logl,- and 47 ks

R e @ - for the entire range of roughness or in other words covering all the three zones i.e. R / k s

smooth, transition and rough. Figure 11.3 shows such a variation for Nikuradse's data.

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Fluid Dynamics - I1

I SMOOTH EO- 11.13 NIKURADSE'S DATA

It can be seen from Figure 11.3 that the data start deviating from the curve representing ~e d- smooth pipe for f = 17.0 while they come closer to the curve showing rough pipe for R / k ,

e d - e J = 400. Thus. transition zone starts from = 117 and ends for !%@ = 400. R/ks R /ks R /ks

ks - These two values conespond to 5 = 0.25 and - 6.0 respectively as per 6 6

ks equation (11.18). It should be noted that incidentally, these are the values of, which 6

represent the limits of hydrodynamically smooth and rough bounda~ies.

SAQ 1 Sketch few examples of arbitrary pipe lines comprising of different diameters and various fittings. Sketch HGL and T.E.L. for each of them.

i SAQ 2 What is the meaning of "equivalent sand grain" roughness.

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Pipe Elow SAQ 3 Tick the correct answers.

For rough turbulent flow, does the friction factor depend on

(a) relative roughness

(b) Reynolds number

(c) both relative roughness and the Reynolds number.

11.4 RESISTANCE OF COMMERCIAL PIPE

As pointed out above, the roughness of commercial pipes normally used is quite different than the sand grain roughness adopted by Nikuradse in his experimental data on artificially roughened pipes. Therefore, for commercial pipes it is difficult to express its roughness in terms of diameter of sand grains . Nevertheless, the equivalent sand roughness of these pipes could be evaluated in terms of grain roughness k, which would yield the same limiting value off (for rough condition) as that of the same diameter commercial pipe. Thus, to estimate equivalent sand grain roughness for any commercial pipe, a limiting value off will have to be determined by performing experiments and then using this value in equation (1 1.14). the value of k, could be obtained. Table 11.1 gives the values of k, in mm for some of the commonly used pipe materials. This concept of equivalent sand grain roughness has been used in determining the resistance of commercial pipes.

Table 11.1 : Values of ks

Pipes materials Roughness k, in mm

Riveted steel 1-10

Concrete 0.3 - 3.0

Wood stave 0.2- 1.0

Cast iron 0.25

Galvanised steel 0.15

Asphalted cast iron 0.12

Commercial steel or wrought iron 0.045

Drawn tubing 0.0015

Plotting his test results on commercial pipes, Colebrook and White found that the results of commercial pipes and artificidly pipes are quite different in the transition region. For

R commercial pipes, they showed that the following equation relates f. R and - in the ks

transition region as shown in Figure 11.3.

- R - 2 log,, - = 1.74 - 2 loglo (11.19) G ks

This equation, is not convenient for use because of the trial and m o r procedure involved in the determination of friction factor. This equation has been plotted by Moody, the form of Figure 11.4 which is known as Moody's diagram. Essentially Stanton's diagram shown in Figure 11.2 and Moody's diagram shown in Figure 11.4 a? the same except for the transition region.

An explicit equation for f for commercial pipes covering both smooth and rough pipes has also been proposed as given below : ,- 1

Equation (11.20) will thus enable direct determination off for known values of 5 and Re. R

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Pipe mow SAQ 4

In laminar flow, head loss due to friction is known to be proportional to the , first power of velocity of flow. However, Darcy-weisbach's equation shows that

2 I hf = U . Resolve this contradiction with suitable explanation.

SAQ 5 Distinguish between the actual roughness and the effective roughness of a conduit boundary.

11.5 ILLUSTRATIVE PROBLEMS

Example 11.1 :

A 600 mm diameter rough pipe carries 600 lit./sec of water over a distance of 1 km. Determine hf if ks = 3.0 mm.

Solution :

ks Let us first calculate Re and - values to ascertain that the flow is not in transition 2 R

zone.

and

Moody's diagram shows that equation (11.14) can be used to calculatef.

f = 0.0304

This gives hf = 11.628 m

Example 11.2 :

Determine the type of surface (i.e. rough, smooth or transition) for the following data pertaining to flow in a pipe :

(i) Diameter = 300 mm, Length = 50 m, Drop in Pressure = 4.2 kN/rn2, 3 ks=0.02mm,p = 998 k g l m and v = m 2 / s .

(ii ) 20 = 638.78 ~ / r n ~ , p = 998 k g / m 3 , v.= 10- 6 m 2 I s , and ks = 2.0 mm.

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Solution :

Since T~ = - * ax 4

ks Since r is less than 0.25, the pipe surface is smooth.

(ii)

Since k, ji- 6, the pipe surface is rough.

11.6 SUMMARY

In this unit we studied the basic equations for calculating the head loss in pipes for various ks flow conditions. It was also pointed out that the friction factor f depends on Re and - D

values. A given pipe may behave as a smooth or rough pipe depending on the Raynolds Number.

11.7 KEY WORDS

Friction Factor (f ) It is a coefficient of proportionality in Darcy-Weisbach's resistance equation.

Equivalent Sand Grain Roughness : The roughness of a uniform sand grain coated pipe of the same size as the given commercial pipe which gives the same value off in the completely rough turbulent regime is termed as the equivalent sand grain roughness (k, )

Relative Roughness

Relative Smoothness

ks . The term - is known as relative roughness. D

D . The term - is called relative smoothness.

ks

Fluid Dynamics - I1

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Pipe H o w Head Loss In practical situations where we are transporting fluid, any increase in internal energy is of little use, since it is usually lost in subsequent storage, and the contribution to heating up surroundings, particularly the atmosphere, is certainly usually not economically desirable. We therefore group these terms together, calling the combination the head loss (hf ).

11.8 ANSWERS TO SAOs

SAQ 2

See the text (Art. 11.4)

SAQ 3

(a) is the correct answer

SAQ 4

Compare Hagen - Poiseuillie equation for pipe flow with Darcy-Weisbach 64

equation and obtain the relation f = - and thus show rhat Darcy-Weisbach Re

64 equation also yields hf = U when f is replaced by-.

Re

SAQ 5

It is impossible to describe the geometry of actual roughness of a surface in terms of single length representative of the mean height of protrusions because of infinite number of shapes and sizes of these protrusions.

The effective roughness of a given actual roughness is conveniently expressed in terms of the grain diameter of a uniform sand roughness which would produce the same limiting resistance to flow as does the actual roughness.

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UNIT 12 PIPE FLOW PROBLEMS

Structure 12.1 Introduction

Objectives

12.2 Various Losses in Pipeline 12.2.1 Head Loss Due to Sudden Enlargement 122.2 Head Loss Due to Sudden Contxoction 12.2.3 Head Loss in G R d d Transitions 12.2.4 Head Loss in Pipe Fitlings

12.3 Equivalent Pipe Systems

12.4 Flow through Pipes in Series

12.5 Flow through Pipes in Parallel

12.6 Pipe Network

12.7 Three Reservoir Roblems

12.8 Turbulent Flow in Non-Circular Conduits

12.9 Power Requirements of a Pipeline

12.10 Power Delivered by a Pipeline

12.1 1 Nozzles

12.12 Illustrative Roblems

12.13 Summary

12.14 Key Words

12.15 Answers to SAQs

12.1 INTRODUCTION

In the previous unit on pipe flow we studied basic concepts of fluid mechanics associated with the frictional loss of energy in pipe flow. Now we shall use those concepts to obtain solutions to different kinds of pipe flow problems.

Pipelines are used to convey fluids from one point to another. While conveying the fluids, the pipelines also convey the energy (or power) of the flowing fluid. It is obvious that part of the fluid power would be utilized in overcoming the frictional and form resistances in pipelines. The form resistance is on account of the changes in the shape and or size of the pipeline, the changes in the direction of flow and different types of fittings in the pipeline. The loss(es) of head in overcoming the form resistances is termed the minor losges).

Pipe flow problems can be grouped into two categories depending upon whether the solution is obtainable by direct computations or it requires a trial-anderror method.

Objectives By the end of this unit, you should be able to

* compute frictional and minor losses in a pipeline of given size, characteristics and for known discharge rate,

* compute the flow rate for known pipeline characteristics and the head loss, * compute the diameter of the pipeline for known flow rate and the head loss, * obtain the distribution of flow in a pipe network,

* solve three-reservoir problems, and * solve problems related to transmission of fluid power.

Only incompressible flow has been considered for the subject matter of this unit.

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Pipe Flow Problems

12.2 'VARIOUS LOSSES IN PIPELINES In addition to the loss of head caused by friction in a pipeline, there are'losses due to change in the cross-section and presence of bends, valves and different kinds of fittings. In a long pipeline these additional losses (usually termed minor or secondary losses) may be a small fraction of the ordinary friction loss and hence, are considered negligible in long pipelines. The minor losses may, however, exceed the frictional losses in a shorter pipeline and should, therefore, be accounted for in such situations. These minor losses are generally caused due to sudden changes in the magnitude andtor direction of the velocity of flow. These changes, in turn, generate large-scale turbulence in which the energy is dissipated as heat. The turbulence so geneated affects the flow for a considerable distance downstream of the section where the change in velocity occurred. For the purpose of analysing this complicated flow, it is assumed that the effects of friction and the additional large-scale turbulence can be separated and the additional loss (i.e. the minor loss) is assumed to occur at the device causing it. The total head loss in a pipeline is, obviously, the sum of the friction loss for the pipeline and the minor loss due to various fittings in the pipeline.

The minor losses, in general, cannot be determined theoretically (one exception is sudden enlargement which can be analysed theoretically). Since the losses are proportional to the square of the average velocity of flow (U) in turbulent flow, the minor losses (H,) are generally expressed in the following form :

Here, K is an empirical coefficient which would depend on the type of fitting causing minor loss and the Reynolds number representing the flow condition. Obviously, at high Reynolds number, the value of K remains constant for a given type of fitting.

12.2.1 Head Loss Due to Sudden Enlargement Consider the sudden enlargement of cross-section of pipeline as shown in Figure 12.1. Due to sudden change in the boundary, the fluid emerging from the smaller pipe is unable to follow the boundary. As a result, the flow separates and pockets of eddies are formed in the comers. These eddies result in dissipation of energy as heat.

HYDRAULIC

I I

Assuming that (i) the velocity distribution is uniform at sections 1 and 2, (ii) the mean pressure of the eddying fluid is the same as the pressure at section 1 and (iii) the pipe axis is horizontal, one can write the momentum equation for the control volume between sections 1 and 2 as follows :

Here, p represents the mean pressure, A the cross- sectional area, p the mass density, Q the discharge rate and U is the mean velocity of flow. The subscripts indicate the section at which the values are being considered.

WINDOWS
Comment on Text
dec. 2007
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From the energy equation, one obtains

P1 u: P2 u: - + - + z = - + - + Z + HL Pi? 2 8 P 8 2 8

2 2

P1 - Pz + Ul - u2 .'. HL = - P 8 2 8

2 2 u2 HL = - ( U2 - Ul) + u1 - u2 8 2 8

2

.'. HL = ( u1 - (12)

2 8

(12.2) 2 8

This equation is known as Borda - Carnot equation.

The factor K, known as loss coefficient, depends only on the sizes of the two pipes. When a pipeline discharges into a large reservoir, Figure 12.2, A, is very large compared to A , and, therefore, K is unitv. Thus, exit loss is given as

TOTAL ENERGY LINE HYDRAULIC MADE LINE

U -.- 0 - .-. -* I , /

-c-"Ai a

12.2.2 Head Loss Due to Sudden Contraction In the case of sudden contraction, Figure 12.3, there forms vena contracta immediately downstream of the contraction and thereafter the jet expands. Upstream of the vena contracta, the flow is accelerating and the energy loss is, therefore, relatively small compared to the loss that occurs downstream of the vena contacta in which region the flow is decelerating.

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Rpe hlow Problems

The total loss of head is, obviously, the sum of these two losses, i.e.

Hence,

The loss coefficnet K for sudden contraction depends on the coefficient of contraction Cc which, in turn, depends on the ratio of the areas A, /A, . The observed values of Cc and K are given in Table 12.1. 1

When a pipeline is connected to a reservoir for receiving water from the reservoir, the ratio A, /A, is almost zero and the loss coefficient is 0.5. Thus. for the sharp-edged entrance, the

Table 12.1 : The Coefficient of Contraction and Loss Coefficient for Sharp-edged Sudden Contraction

L

entrance loss is taken as 0.5 2. Foi a rounded or bell-mouthed entry, the loss coefficient 22 -

can be as low as 0.02.

K

0.50

0.46

0.41

0.36

0.30

0.24

0.18

0.12

0.06

0.02

0.00

A2"4,

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 .o

12.2.3 Head Loss in Gradual Transitions Two pipes of different cross-sections are usually connected through a fitting which accomplishes the change gradually. This fiiting is called transition which may be either convergent or divergent. Convergent transition results in accelerating flow which converts the potential energy into kinetic energy and is, therefore, inherently stable and free from separation. The energy loss in gradual contraction is, therefore, very small and is given as

cc 0.617

0.624

0.632

0.643

0.659

0.681

0.712

0.755

0.813

0.892

1 .ooo

in which U, is the average velocity of flow in the contracted section.

In the diverging transitions, however, the flow is retarding and the kinetic energy is converted into potential energy. Further, if the angle of expansion is larger than about 10". the flow separate$ from the boundary and thus results in the formation of eddies. Hence, the

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energy loss in the expanding transitions is relatively more. The head loss due to gradual expansion can be expressed as

I in which the loss coefficient K depends upon the flow condition and the geometry of the expansion. For relatively large Reynolds number, the loss coefficient depends only on the

I geometry of the expansion. The variation of the loss coefficient K with the angle of expansion for a conical diffuser is shown in Figure 12.4.

L

It may be noted that when 8 = 180" (i.e. sudden expansion ), K = 1.0.

SAQ 1 Why head loss in diverging transition is more than that in converging transition?

12.2.4 Head Loss in Pipe Fittings In any pipeline, it is usual to use different kinds of fittings such as Tee joints, valves, bends etc.. The head loss on account of these fittings can be expressed as

The value of K has to be determined experimentally for any type of fitting as it depends on the Reynolds number as well as the geometry of the fitting. Some typical values of K for some pipe fittings are given in Table 12.2.

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Pipe Flow Problems

Table 12.2 : Typical Values of Loss Coefficient K for Pipe Fitings

12.3 EQUIVALENT PIPE SYSTEMS

Pipe Fitting

Standard Tee joint

Standard 90" elbow

Standard 45' elbow

Return bend

Gate Valve fully open

Gate Valve 314 fully open

Gate Valve 112 fully open

Gate Valve 114 fully open

Two pipe systems are said to be equivalent when the same amount of discharge flowing through the two pipe systems causes the same head loss. This concept can be advantageously used to express the form as well as friction effects in terms of an equivalent length of straight pipe of uniform diameter which would result in the same head loss at the same rate of discharge. The total head loss h, in a pipeline can always be expressed as

u2 h , = K- 2 g

K

1.80

1.90

0.42

2.20

0.19

1.15

5.60

24 .O 1

in which K is a loss coefficient which includes the effects of all non-uniformities as well as friction. Another pipeline of straight length 1, and uniform diameter D would be equivalent to the given pipeline having total head loss h, only if the following relation is satisfied.

from which one obtains

Here, 1, is termed the equivalent length of pipe and f is the friction factor of the selected equivalent pipeline.

Likewise two straight pipes of diameters D, and D,, length 1, and I, and friction factorsf, and f, would be equivalent if

Similarly, pipes in series and pipes in parallel can be simplified to equivalent single pipe of specified diameter having an equivalent pipe length 1,.

The concept of equivalent pipe length enables simplification of a compound pipeline into a single pipeline of suitable length i.e. the equivalent pipe length.

SAQ 2 What is meant by "equivalent pipe length"?

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SAQ 3 When are the minor losses really minor ?

When is the terms "minor loss" a misnomer?

Hoid Dynamics - I1

12.4 FLOW THROUGH PIPES IN SERIES When pipes of different cross-sections are connected end to end to form a pipeline, so that ,the fluid flows through each in turn, the pipes are said to be in series. The total loss of head for the entire pipeline would, obviously, be the sum of the friction and minor losses for each pipe together with the losses that might occur at their junctions. The discharge would be the same in all these pipes which have been connected in series.

12.5 FLOW THROUGH PIPES IN PARALLEL When two or more pipes are connected so as to first divide the flow and subsequently bring it together as shown in Figure 12.5, the pipes are said to be in parallel.

Flpre 125

Referring to the Figure, it is obvious that,

Q = Q, + Q, + Q3

and

Here, HL is the headloss between points A and Band HL , H4 and H,, are head losses in

the three pipes.

For known pipe and fluid characteristics, one can easily determine the total discharge Q for known head loss HL by determining Q,, Q, and Q, as HLI, Hq and Hq are all equal to H,.

Other type of problems related to pipes in parallel concerns the distribution of the given amount of total discharge Q and the determination of the head loss H,. Such problems can be solved by the following procedure.

(i) Assume a suitable discharge Q; for pipe 1 and compute the head loss H; with the

assumed discharge Q;.

(ii) For known = HL = ~ i ) obtain Q; and Q;and determine

I: Q' = Q; + Q; + Q;.

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Pipe Flow Problems

(iii) Distribute the total given discharge Q in the following manner

(iv) Check the correctness of the discharges Q,, Q, and Q, by computing the loss in each pipe.

In such problems, minor losses may be either completely neglected or could be considered by adding their equivalent pipe length to the actual length of the relevant pipe.

12.6 PIPE NETWORK For distribution of municipal water, a network of pipelines is used. In addition, the friction equation must also be satisfied for each pipe. It should be noted that Darcy's formula for friction loss takes no account of the direction of flow. When the direction of flow is in doubt, one has to assume it and see if the assumption yields physically possible solution. These pipe networks often are difficult problems to analyse, The fundamental principles of continuity and uniqueness of the head at any junction form the basis of solution of pipe networks. According to the principle of continuity, the total incoming discharge at any junction is equal to the total outgoing discharge at that junction. The uniqueness of the head at a given junction requires that the net head loss (or the algebraic sum of the head loss) round any closed loop in the network must be zero. For the purpose of obtaining the net headloss, the headloss in a pipe of a given loop may be considered as positive if the flow in the pipe is clockwise. One has to obtain the distribution of discharge (magnitude as well as direction and flow) in different pipes of the network for known pipe characteristics. Since the flow direction in many of the pipes of a pipe network may be unknown, it is often very time-consuming if one sdves the resulting equations by ordinary trial and error method, Hardy Cross has developed a systematic method of computation which gives results of acceptable accuracy with relatively small number of trials. The method consists of the following steps :

1. By careful inspection, assume the most reasonable distribution of flows which satisfies the continuity principle at every junction.

2. Obtain head loss h, (usually due to friction i.e. h,) in each pipe by writing

where r is a constant for each pipe. n is usually taken as 2.0. While minor losses within any circuit may be included, the minor losses at the junction points are neglected.

3. Compute the algebraic sum of the head losses around each elementary circuit i.e.

C h f = C r Q n

considering losses from clockwise flows as positive and those from counter clockwise flows as negative.

4. Adjust the flow in each circuit by a correction A Q to balance the head loss in the circuit so that

A Q is determined in the following manner :

For any pipe, one may write

where Q is the correct discharge and"Q, is the assumed discharge. Then, for each pipe,

h, = rQn = r( Q,+ A Q ) " = r ( Q;+ n Q;- ' A Q + . . .)

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I fAQ is small compared with Q,, one may neglect the terms of the series after the second term and

Now for a circuit, with AQ the same for all pipes,

It should be noted that the numerator of this equation is to be summed algebraically with due regard to sign and the denominator is summed arithnatically. The direction of AQ is, obviously, clockwise when it is positive and counter clockwise when negative.

5. After each circuit is given a first correction, the losses will still not balance because of the interaction of one circuit upon another due to which pipes, common to two circuits, receive two independent and different corrections - one for each circuit. The procedure is, therefore, repeated until the corrections become negligible small.

12.7 THREE RESERVOIR PROBLEMS

Pipe system, involving flow among three reservoirs, consists of three pipes meeting at a junction (Figure 12.6). The flow in such a pipe system can also be analysed by satisfying the continuity principle at any junction, uniqueness of head at any point and the Darcy's equation for each pipe. Because of uncertainty in the flow direction in one or more pipes of the pipe system, the solution of three reservoir problem requires trial.

1 DATUM LEVEL t Ffgure 12.6

In Figure 12.6, three reservoirs A, B and C are connected to common junction J by pipes 1, 2 and 3 in which the head losses are hfl , hf2 and hf3 respectively. The characteristics of pipes 1, 2 and 3 as well as surface levels in reservoirs A, B and C are known. The head at junction J is unknown.

For such pipe system, following method is used for the solution :

Since the head at A is the highest and that at C the lowest the direction of flow in pipes 1 and 3 is as indicated by the arrows. The direction of flow in pipe 2, however, is not immediately evident. If hj , the head at J, is intermediate between the heads at A and B then flow occurs from J to B and for steady conditions the following equations apply :

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Pipe Flow Problems 1 I

Since h, is a function of Q, these four equations involve the four unknowns hj , Q,, Q, and Q,. Even when f is assumed constant and minor losses are neglected so that

albegraic solution is tedious (and for more than four pipes impossible). Trial values of h, substituted in the first three equations, however, yield values of Q,, Q, and Q, to be checked I

in the fourth equation. If the calculated value of Q, exceeds Q, + Q,, for example, the flow I rate towards J is too great and a larger trial value of hj is required. Values of Q, - (Q, + Q,) may be plotted against h, as in Figure 12.7 znd the value of h, for which Q, - (Q, + Q,) = 0 readily found. If, however, the direction of flow in pipe 2 was incorrectly assumed no solution is obtainable.

Figure 12.7

For the, opposite direction of flow in pipe 2 the equations are :

It will be noticed that the two sets of equations (12.13) and (12.14), become identical when z, = hj and Q, = 0. A preliminary rrial with h, = z, may, therefore, be used to determine the direction of flow in pipe 2. If the trial value of Q, is greater than that of Q,, that is, if the flow rate towards J exceeds that leaving J, then a greater value of h, is required to restore the balance. On the other hand, if Q, < Q, when hj is set equal to z, , then hj is actually less than 2,

12.8 TURBULENT FLOW IN NON-CIRCULAR CONDUITS

Normally the conduits for carrying a fluid are circular in shape and one can use Darcy- Weisbach equation alongwith Moody's diagram for solving flow problems. Experiments have indicated that the relations developed for circular pipes yield reasonable results if the diameter is replaced by another term known as equivalentdiameter such that the hydraulic mean readius R (i.e. the ratio of area of ckoss- section A and the wetted perimeter P ) remains the same. For a circular section

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i This means that the equivalentdiameter of a non-circular pipe is four times its hydraulic mean radius. Thus, Darcy-Weisbach equation, for non-circular conduits, is written as

and the Reynolds number would be calculated as

t It should, however, be noted that this kind of simplification yields only approximate results depending upon the deviation of non-circular shape from the circular shape. This is on

I account of the assumption involved in the concept of equivalentdiameter that the mean

I shear stress at the boundary is the same as for a circular section. Further, this concept is not applicable to laminar flows.

12.9 POWER REQUIREMENm OF A PIPELINE Problems of flow through pipes involve the estimation of power required to maintain a

! certain flow through the pipe. Since power P is the rate of doing work, it is equivalent to the I

I product of force (i.e. - *. A1 ) and the mean velocity U, i.e. ax

I or P = Q ( P ~ ~ 2 )

For inclined pipes, ( p, - p 2 is replaced by P g (h, - h2) I :. P = p g Q ( h , - h , )

I Here, power P is in watts and Q in m h p in kglm3 and h is in metres.

12.10 POWER DELIVERED BY A PIPELINE Pipelines are often used to transmit water from higher elevation to lower elevation. In the process the water loses its potential energy, part of which is lost in overcoming friction in the pipe and the remaining in the turbine. Thus, if z is the elevation difference and hf is the head loss due to friction in a pipe line carrying discharge Q , then the power P delivered to the turbine is

dP - For maximum power delivery, - - 0. dQ

This means that the power delivered by a given pipe is maximum when the flow is such that one-third of the static head is consumed in friction. However, such a wastage would not be desirable and the pipelines are of such a size that these could deliver water with a loss of only a few percent.

Fluid Dynamics - I1

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Pipe Flow Problems

12.11 NOZZLES

A nozzle is a converging tube which is usually fitted at the end of a pipeline to obtain a high velocity jet at the end of the pipeline. The pressure at the exit end of the nozzle is atmospheric. A practical example of nozzle is the flow from a reservoir to an impulse turbine through a penstock that ends in a nozzle.

12.12 ILLUSTRATIVE PROBLEMS

Example 12.1 :

A 300 mm diameter pipe with friction factor of 0.02 has a pipe fitting with loss coefficient of 1.9 and 200 mm diameter pipe of 50 m length with friction factor of 0.022. Determine their equivalent lengths in terms of 300 mm diameter pipe.

Solution :

Since

Equivalent length for the pipe fiiting = 0.3 = 28.5 m 0.02

Also, since

:. Equivalent length of 50 m pipe in terms of 300 mm diameter pipe

:.Equivalent length for the fitting and 200 mm diameter pipe in terms of 300 mm diameter pipe = 28.5 + 417.66

= 446.16m.

Example 12.2 :

Determine the size of galvanised steel pipe needed to carry water for a distance of 180 m at 85 lit./s with a head loss of 9.0. Take k, = 0.15 mm.

Solution :

Substituting the values,

Taking v = 1.0~ 1u6 m2/s

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Fluid Dynamics - 11

Assume

k From Moody's diagram for Re = 5.48 x lo5 and ' = 0.000812, the value off is 0.019. D Assumption is O.K.

Hence D = 0.1846 m

Example 12.3 :

Determine the head loss due to the flow of 100 lit./s of water through 100 metre length of 15 cm diameter pipe having relative roughness of 0.01.

Solution :

Using Moody's diagram for Re = 8.49 x ld and k ID = 0.01, one obtains f = 0.038

= 41.35 m of water

Example 124 :

Determine the distribution of flow in the pipe network as shown in Figure 12.8 : Assume n = 2

Figure 12.8

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Pipe Flow Problems

Solution :

Let the distribution be as follows :

65, 4

I

,20

d

20 FSgum 129 (a)

For C i d t A For C i d t B

Q', 2rQo ef 2r Qo

2 x 4d= 3200 2 x 2 x 40= 160 4 x 4 d = 6 4 0 0 2 x 4 x 4 0 = 3 2 0

2 x 2 d = 800 Z X 2 x 2O= 80 - 3 ~ 2 d = -1200 2 x 3 x 20= 120

- 4 ~ 4d= -6400 2~ 4 x 40= 320 - 1 ~ P= -25 2~ l x 5 = 10

x re:= -2400 x 2rQ,= 560 x re:= 5175 21 Qo= 450

-m AQ= -[?I = 4.3 AQ= -[=I 5175 = -11.5

Applying these corrections to the assumed distribution, the new distribution IS as follows : -

Fipre 129 (b)

For Circuit A For Circuit B

Q: 2r Qo re; 2r Qo

2 x 44.32 = 3925 2 x 2 x 44.3 = 177.2 - 3 x 31.5~ = - 2976.75 2 x 3 x 31.5 = 189

2 x 2 ~ . 3 ~ = 1181 2 x 2 x 24.3 = 97.2 - 1 x 16.5~ = - 272.25 2 x 1 x 16.5 = 33

- 4 x 24.22 = - 2342.56 2x 4x 24.2= 193.6 + 4 x 24.22 = 2342.56 2 x 4 x 2A.2 = 193.6

x r ~ : = 2 7 6 3 . 4 4 x 2 r Q O = 4 6 8 r&= - 906.44 I: 2rQ,= 415.6

2763.44 AQ= = -5.9 - 906.44 = + 2.2

= -[ 415.6 ] Applying these corrections to the urevious distribution, the revised distribution is as follows:

Rgure 129 (c)

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On further calculation in the same manner one would finally obtain the following disrribution with acceptable level of error.

-re 12s (d)

Examples 12.5 :

For the problem shown in Figure 12.10, find elevation of level of the reservoir C and distribution of discharge. Assume f = 0.03 for all pipes.

FSpn 1210

Solution :

3 Velocity in pipe &om reservoir A = - - X - 3.82 rn/s

(1.0)'

Therefore, total head at P = 100 - 33.47 = 66.53 m

Hence the flow would be from reservoir B to P with head loss h, equal to 80-66.53 = 13.47 m 2

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Pipe Flow Roblems

= 22.47 m

:. Elevation of reservoir level C = 66.53 - 22.47

= 44.06 m.

12.13 SUMMARY

In this Unit we studied the concepts related to hydraulic losses on account of different types of pipe fittings. The methods to solve different kinds of pipe flow problems including pipe network and three reservoir problems were illustrated. There is no alternative to solving many problems on pipe flow to comprehend the methods illustrated in this Unit.

12.14 KEY WORDS

Pipe Fittings : Bends, elbows, joints, valves etc., in a pipe line systems are known as pipe fittings.

Equivalent Pipe : A hypothetical pipe line of uniform diameter resulting in the same head loss as ilr the actual pipe line.

Pipe Netwrk : Pipes of different length and diameters connected in different ways.

Three Reservoir Problem : Problems involving flow among three reservoirs, consists of three pipes meeting at a junction.

Minor Loss : Hydraulic loss in pipe lines on account of different forms and fittings in pipe line and other than those due to friction.

12.15 ANSWERS TO SAOs

SAQ 1

See text (Sec. 12.2.3) SAQ 2

See text (Sec. 12.3) SAQ 3

For long pipelines the combined loss due to all pipe fittings, termed "minor loss" is small compared to the loss due to friction. But, for short pipe lengths, the reverse may be true and the minor loss may be larger than the loss due to friction.

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! UNIT 13 DRAG

* I Structure 13.1 Introduction

Objectives

13.2 Separation 13.2.1 Methods of Controlling Separatioo

13.3 Drag on Immersed Bodies

13.4 Deformation, Friction and Form Drags

13.5 Drag on Sphere

1 3.6 Drag on Cylinder 13.6.1 Karman Vortex Trail

13.7 Drag on Flat Plate

13.8 Drag on an Airfoil

13.9 Illustrative Examples

13.10 Summary

13.11 Key Words

13.12 Answers to SAQs

13.1 INTRODUCTION Several engineering problems such as motion of aeroplanes, submarines and torpedoes; design of fans, turbines, tall buildings, factory shades and bridges etc. require study of flow pattern around these bodies to compute the resistance offered by these bodies to flow. Consider a body placed in an infinite fluid flowing at a steady and uniform velocity. As long as the fluid is real, it will exert a force on the body in the direction of the motion which is known as drag force. The body will in turn, exert a force on the fluid equal in magnitude but opposite in direction. This is known as the resistance. On the other hand, suppose we consider fluid to be stationary and the body in steady and uniform motion, even then drag force will be,exerted on the body. Hence it can be concluded that drag or resitance remains the same whether the body is moving and the fluid is at rest or the fluid is moving and the body is at rest.

The force exerted by the fluid on the body need not necessarily be in the direction of motion. It can make a certain angle with the direction of motion. The component of this force in the direction of motion is known as drag, FD and the component which is perpendicular to the direction of motion is called lift, FL . It is worthwhile to mention here that lift force is present when fluid flow is at an angle with respect to the axis of the body i.e. body is asymmetrical in nature.

Whenever flow takes place around the body, there is always a condition of no slip at the surface of contact as well as there is a thin layer of fluid adjacent to the boundary and hence fluid is retarded near the boundary. Therefore, there exists a velocity gradient and hence shear force is caused. 'Iherefore, it can be said that because of presence of viscosity, however small it may be, the flow pattern gets modified by the development of the boundary layer. If such a surface curves away from the flow, there is tendency for the flow to move away from the wall and hence separation of flow takes place. 'Ihis separation drastically alters the flow pattern and hence the pressure distribution. 'Ihis affects the drag appreciably.

Objectives In this Unit, we are going to study basically the forces on bodies immersed in a fluid. For example, a tall building exposed to a strong flow of wind, a sphere falling in a column of liquid. Thus after going through this unit you should be able to

* identify flow pattern around these submerged bodies,

* appreciate the total force exerted by the fluid on the body as well as the components of this force in the direction of flow and in the direction perpendicular to the flow, \

* identify the basic .geometric and flow parameters i'hfluencing these forces,

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~ r ~ g . LitY and otber * clearly understand the phenomenon of separation of flow and its effect on drag I

Hydrautic Prob~em~ force on the body, and I * estimate the drag force for different shapes of the body. I

1

13.2 SEPARATION

Consider flow through a two dimensional expanding channel as shown in Figure 13.1. As the fluid flows from left to right, the velocity of flow decreases because the cross-sectional area is increasing. This reduction in velocity results in increase in the pressure and hence positive pressure gradient is set up. Positive pressure gradient is one in which pressure increases in the direction of flow. The velocity profiles of various sections along the curved

ap surface are shown in Figure 13.1. Initially, as long as =& is either negative or zero, the fluid

velocity either increases or remains constant. In the region of positive pressure gradient, because of decelerating effects, the boundary layer thickness is increased further and does not remain small. Because of thickening of boundary layer, the velocity gradient near the wall decreases in the direction of flow. It can even become zero as shown in the Figure. A further retardation of flow can start back-flow. The flow separates from the boundary from

ap some distance downstream of the point where - = 0. If all the points below which a reverse ax flow occurs are joined by a smooth curve AB, it represents the separating stream line. However, location of such a stream line is very difficult, because the point where separation starts, depends on roughness, form of the roughness and Reynolds number.

Point of separation, 7 Edge of boundary layer

Separating stream line

FPgerc 13.1 : Scp~rrtioo of BomodPry Layer m an Eqmnliq Flow

Further, such a stream line oscillates and never remains stationary. Since the separation is affected by retardation of the fluid in the boundary layer, it can been seen that the laminar boundary layer is more susceptible to earlier separation than the turbulent boundary layer.

Due to the reversal of flow, eddies are formed in the region close to the wall. This formation of eddies is accompanied by continuous loss of energy since they derive energy from the main flow for their existence. This energxis ultimately lost as heat. As there is intermittent sheddidg of these eddies, the body is also subjected to lateral vibrations which are undesirable. Since the separation occurs near the point of minimum pressure and the pressure in the separation zone is essentially constant, such a pressure difference gives rise to longitudinal and some times lateral boundary forces. The net force in the direction of motion caused by this pressure difference is known as pressure-drag or form drag.

For efficient design of such body shapes, the primary aim is to reduce the tendency for separation and thereby reduce the resistance appreciably. Such problem are encountered in the design of channels and conduit transitions, design of airplanes, ships etc.

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13.2.1 Methods of Controlling Separation The fact that separation originates eddies and these in turn consume energy, calls for control of s'eparation. One way of avoiding separation is to keep the pressure gradient sufficiently small, this is achieved by making the angle of divergence of solid boundaries sufficiently small. This is done by designing diffusers, expanding transitions in open channels with small angle of divergence. The same principle applies to flow past streamlined bodies such as airplane wings.

Separation and backflow takes place when pressure force acts in the opposite direction on fluid particles already slowed down within the boundary layer. Therefore, separation can be delayed if the fluid particles in the boundary layer are accelerated. This can be done by changing laminar boundary layer into turbulent boundary layer by providing artificial roughness. It has been shown in the past that size of the wake behind a sphere can be greatly reduced by providing a thin wire ring as shown in Figure 13.2 (a). This reduces the pressure drag of the sphere appreciably. The same object can be achieved by injecting high velocity fluid in the boundary layer, thereby accelerating the retarded fluid particles as can be seen in Figure 13.2 (c).

WIRE-RING

-+ L

(a F L O W PAST S P H E R E AT

Re =10 5

( b ) 5

FLOW PAST SPHERE AT Re =10 , WAKE S I Z E IS REDUCED BECAUSE OF WIRE-RING

ACCELERATION OF FLUID WITHIN SECTION OF RETARTED FLUID 6 OuNDARY LAYER WITHIN BOUNDARY LAYER

Figure 13.2 : Methods of Boundary Layer Control

The separation can also be controlled by continuously sucking the slow moving fluid from the boundary layer [see Figure 13.2 (d)]. The sucking can be done at one section or along the boundary length. The separation in diverging flows as in bends can be controlled by providing guide vanes as shown in Figure 13.3.

STAGNATION 7 GUIDE V A N E S 7

Figure 1 3 3 : Beads with and without Vanea

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Drag, Lift and other Hydraulic Problems

SAQ 1 (i) Define drag and lift.

(ii) What is positive pressure gradient?

(iii) Define point of separation.

(iv) List the methods of controlling separation.

13.3 DRAG ON IMMERSED BODIES

Let us consider a body held in a flow of uniform velocity Uo as shown in Figure 13.4. Let F be the resultant force acting on the body. This total force F can be resolved in two components; the one in the direction of motion known as drag force, FD and the other perpendicular to the flow known as lift force, F,. Since the velocities at various points near the surface of the body are different, application of Bernoulli's equation will reveal that pressure at these points also varies. At any point on the surface of the body, two types of forces are acting as shown in the figure. One of them is the shear stress, zO, acting in a tangential direction and the other one is pressure force acting perpendicular to the surface. Therefore, this drag force acting in the direction of motion is given by the summation of the components of shear force and pressure force acting over the entire surface of the body.

Figure 13.4 : Dcdinition of Drag and L i i

Hence,

wherein represents summation over the entire surface area. The relative contributionof

each of these to the total drag depends on the shape of the immersed body and on the flow and the fluid characteristics. Thus for the case of flat platecheld parallel to the direction of flow, the second term in equation (13.1) will be zero (i.e. J p cos 8 &I = 0) and hence total

A

drag is only due to viscous shear. On the other hand, if plate is held perpendicular to the flow. the contribution of the first term will be zero (i.e. sin 8 &t = 0) and hence the

total drag is only due to variation in pressure. In between these two extremes, there are numereous body shapes for which contribution of the viscous shear to the total drag varies within wide limits depending upon shape of the body and flow conditions.

The total drag is the sum of deformation drag and form drag. The deformation drag exists at very small velocities i.e. when Reynolds number is very small, preferably less than 0.20. At such small values of Reynolds number, the contribution of inertia force can be neglected as compared to that of viscous force. Under such conditions viscosity influences the flow

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pattern considerably. In the words of Prandtl and Tietjens, The body pushes itselfthrough thefluid which is deformed by it. The resistance caused by this is due primarily to the forces necessary for the deformation of variousfluid panicles. Smaller the Reynolds number greater is the distance from the body upto which deformation takes place.

The deformation drag consists of friction drag or surface drag at the surfadboundary and the pressure drag due to the variation in pressure caused by wide spread deformation. The proportion of these two for a given body depends very much on the Reynolds number. It has been shown in the past that for the case of a sphere in the Stokes' range, one third of the total drag is due to the pressure difference and two third is due to the surface drag (boundary shear). It has been observed that for the case of disc and sphere of same diameters, if Re is less than 0.1, the total drag of the sphere is slightly higher than that of disc. Considering the fact that the shapes of these two objects are different, the above observation regarding the total drag at low Re reveals that the deformation drag depends very much on the projected area and not much on the shape.

With increase in the Re value, the extent of deformation decreases and is limited to a very thin boundary layer. Because of the reduction in the deformation area, the contribution of the deformation drag almost becomes negligible and the contribution due to the friction drag increases. Thus for higher Re, the direct action of the viscous deformation is the friction drag. At higher values of Re, if the shape of the body is such that separation occurs, low pressure area is created at the rear portion of,the body and this produces form drag. Thus the form drag is the result of pressure diffepmce created between front and the rear of the body. The proportions of friction drag and form drag depends very much on the shape of the body and the value of Re. If the body is sharp edged such as flat plate or disc held perpendicular to the flow, the point of separation is almost fixed and contribution of friction drag is negligibly small as compared to that of form drag. On the other hand for the case of flow past cylinder or sphere, the location of point of separation depends on the Reynolds number.

Out of these three types of drags i.e. deformation drag, form drag and friction drag, some can be measured and others can be computed. Since the relative contribution of these to the total drag depends on Reynolds number and body shape, problem of predicting the total drag on a body is not amenable to a complete analytical solution.

SAQ 2 (i) Define deformation drag.

(ii) Differentiate between form drag and friction drag.

(iii) Drag characteristics are greatly affected by Reynolds number and the body shape. Comment.

13.5 DRAG ON SPHERE

To understand the flow past sphere, it is desirable to know the analysis of steady flow of an ideal fluid around a sphere. Solution of the Laplace's equation with the necessary boundary condition will yield an expression for velocity distribution around sphere. Application of Bernoulli's equation will then yield pressure distribution. Since the flow pattern is symmetrical, the pressure distribution will also be symmetrical on the front and rear of the sphere as shown in Figure 13.5 (a). Since there is no viscosity, there is neither form drag nor friction drag indicating that there is no total drag. This inability of the theory of ideal fluid to yield results approximating those observed in experiments is known as D'Alembert's Paradox.

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Drag, Lift and other Hydraulic Problems

( 0 ) IRROTATIONAL FLOW

( b ) Re: 157,200 j CD = 0.471

LAMINAR BOUNDARY LAYER

- + I 0 -1

t J ( c ) Re = 424.500 ; CD =0.143

9u: 12 TURBULENT BOUNDARY LAYER

Rpre 135 : Pressure Distribution around a Sphere

On the other hand when fluid possesses some viscosity, however small it may be, the flow pattern and the resulting pressure distribution are entirely different. If the Reynolds number

uo D P defined as -

P is very small-say less than 0.2, the viscous forces are much more

important than the inertial forces and therefore, Navier Stokes' equation can be integrated to yield the following expression,

wherein C, is the coefficient of drag.

Equation (13.3) reveals that drag coefficient is inversely proportional to the Reynolds number. As mentioned earlier, for this case it can be proved that out of total drag 3zD pUO , the contribution due to form drag (i.e. deformation drag ) is z D pUo and the contribution of the surface drag (friction drag) is 216) pUo . Experimental observations show that upto a value of Re equal to 0.2, Stokes' law can be used to predict drag coefficient with a reasonable confidence provided the sphere is in motion in infinite fluid. Presence of the boundary of the container increases the resistance. For such a case, the drag coefficient can be computed by using the following equation :

7 -3

wherein D' is the diameter of container and D is the diameter of sphere.

Oseen (1927) pointed out that it is only in the vicinity of the sphere that the acceleration terms omitted by Stokes are very small compared to the viscous term. Far away from the

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sphere the case is exactly reverse. Therefore, he improved the Stokes analysis by including certain inertial terms and proposed the equation :

It can be observed from Figure 13.6 that Oseen's solution does not follow experimental trend and the upper limit of applicability of Oseen's solution is Re equal to 2.0.

I

DISC

lo0 - -

SPHERE - lo1 I I I I I I I ' I

1c2 16' lo0 lo1 13 lo3 1 o4 lo6 10'

Figure 13.6 t Variation of CD with Re

As the Re is further increased, the region in which the viscous deformation takes place is reduced considerably and is restricted to the laminar boundary layer. Furthermore, the boundary layer which is laminar increases in thickness and separates somewhat upstream of the point of maximum cross- section. The separation produces a wake behind the body which is laminar for low value of Re. The flow in the wake becomes unstable and continuous eddies ark shed behind the body.

Figure 13.5 (b) shows typical pressure distribution around a sphere when boundary layer is laminar. It can be observed from this figure that the pressure distribution in the zone of accelerating flow is almost the same as that in the potential flow. Since the laminar boundary layer can stand only small adverse pressure gradients, the boundary layer separates at an angle of 80". For Re between 0.50 and lo4, the following empirical equation can be used

It can be observed from Figure 13.6 that CD value changes from 0.4 to 0.5 over a range of Re from lo3 to ld. This is so becausdin the this range of Re contribution of friction drag to the total drag is much small compared to that of fonn drag. This small increase in the value of C, is due to the development of boundary layer and change in the character of the wake.

With further increase in the Reynolds number, the separating boundary layer can no longer remain laminar and it changes into turbulent boundary layer. Simultaneously at the same time, the point of separation shifts further downstream. The separation point is now located at 110 degrees approximately. Due to the change in point of separation, there is reduction in the size of the wake as well as in the value of CD , which reduces from 0.5 to 0.2. Typical pressure distribution at such Re is as shown in Figure 13.5 (e). The Reynods number at which the separating boundary layer changes from laminar to turbulent is known as critical - . . . .. . . . . . . . . . . ..- 4 - 5 . . . n 5 . . "#

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Drag, Lift and other Hydraulic Problems

sphere. The actual value of this critical Reynolds number depends upon the roughness characteristics of the sphere and the level of turbulence present in the flow.

SAQ 3 (i) Define Stoke's law.

(ii) Sketch the variation of CD with Re for a sphere.

(iii) At a certain value of Re, there is a sudden drop in the value of CD in case of a sphere. What is this value of Re? Also explain the reason for such a drop inCD.

13.6 DRAG ON CYLINDER

Analytical solution for the drag experienced by an infinite cylinder placed in uniform stream has been put forward by Lamb. His analytical solution is valid upto a value of Reynolds number equal to 0.2. It has been observed that experimental points start deviating from the results obtained on the basis of theoretical analysis at Reynolds number greater than 0.20. Drag coefficient for cylinder also decreases with increase in Reynolds number in a manner similar to the case of sphere. Oseen's approximate method can also be extended for drag experienced by cylinder for low values of Reynolds number (say Re = 2 or 3). With increase in Reynolds number, the flow pattern becomes asymmetrical about the axis perpendicular to the direction of flow. It has been observed that at Re equal to 4.0, very weak vortices are formed behind the cylinder. This is shown in Figure 13.7 (b). Alongwith these vortices, wake is also formed and flow keeps on recirculating in this region. If the Reynolds number is further increased upto a value of 20, the wake behind the cylinder becomes more predominant and so also the vortex pair as shown in Figure 13.7 (c). If the Reynolds number is again increased, these vortices get stretched and become unsymmetrical and they leave the cylinder with relatively smaller velocity. The various flow patterns around a cylinder for various Reynolds number are shown in Figure 13.7.

IRROTATIONAL FLOW Re = 4.0 (a 1 (h1

Figore 13.7 : Flow Proand pn Laanite CyUder

13.6.1 Karman Vortex Trail As mentioned above when Reynolds number is increased beyond 30, the two vortices behind the cylinder get stretched and become unstable. As a consequence of this, they are transported down and another two vortices are created in place of these transported vortices. This process continues. Under such circumstances, thewake behind the cylinder consists of a series of vortex pairs moving in the downstream direction with relatively 'smaller velocity.

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Kannan was the fust who gave explaination of this phenomenon and hence this is widely known as Karman Vortex Trail. According to him, there are two possible combinations of these vortex pairs as shown in Figures 13.8 (a) and 13.8 (b). Analysis of stability of these vortices revealed that configuration (a) is unstable and configuration (b) is also unstable

a except when - = cosh -' fi = 0.281. b

(a) SYMMETRICAL CONFIGURATION OF VORTEX PAIR

( b ) STAGGERED CONFIGURATION OF VORTEX PAIR

Figure 13.8 : Kannan Vortex T d l i Taylor has given the following relationship for the frequency with which these vortices are

shed from one side of the cylinder

wherein U, is the velocity of flow and D is the diameter of the cylinder. The dimensionless D

quantity f - is commonly known as Strouhl number. It has been found that over the range v,

D of Re from 120 to 20,000, equation (13.7) gives f - = 0.17 to 0.20. Recent experimental

UO D observations have revealed that upto Re equal to lo5, the value off - for cylinder is equal v,

Due to alternate shedding of the vortices, a periodic force is experienced by the cylinder in the direction perpendicular to the flow. If the frequency of this lateral force is equal to the natural frequency of the cylinder, resonance will. take place and lateral forces of large magnitudes will be generated. Such considerations are much more important in aerodynamic study of suspension bridges. Figure 13.9 shows the variation of Strouhl number, S, with drag coefficient C,.

0 0.5 1.0 1.5 2.0

CD

Figure 13.9 : Viuiation of S with CD for a Cylinder

-

Drag

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Ijrag. 1,ift and other Hydraulic Problems

The variation shown in Figure 13.9 could be expressed by the equation :

S A Q 4 (i) Giving sketches, explain Karrnan vortex trail.

(ii) Define Strouhl number and give its significance.

13.7 DRAG ON FLAT PLATE

Flow pattern of irrotational flow for the case of flow past a plate held perpendicular to the direction of flow is depicted in Figure 13.10 (a). Since the flow considered is irrotational, stream line pattern is symmetrical on both sides of the plate.

Figure 13.10 : Mow Pattern around a Mat.Plate

For such a case no separation takes place and due to the symmetrical flow pattern, the pressure distribution will also be symmetrical and hence no pressure drag will be experienced by the plate. But on the other hand, if flow of real fluid is considered, the flow pattern will be totally different as shown in Figure 13.10 (b). The flow will separate at both the ends of the plate and because of this, the pressure behind the plate will be smaller as compared to that on the front side. For such a case CD is a function of Reynolds number for low and moderate values of Re. It has been observed that once Re exceeds about lo3, CD attains a constant values of about. 1.9 if the plate is infinitely long. Flow will not be truely two dimensional if the plate is not sufficiently long and hence the drag coefficient will be smaller. Variation of CD with BIL for high values of Re is shown in Figure 13.11.

2 .o I I I I

a WIESELSBERGER

Figure 13.11 : Vdat ion of CD with B IL for a Rednngular Plate held Perpendicular to the Mow

For the case of disc, CD varies with Re in a similar manner as that for a plate held perpendicular to the flow. But due to the three dimensional character of the flow, the limiting value of drag coefficient is 1.10 which is smaller than the drag coefficient for flat plate (see Figure 13.6).

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13.8 DRAG ON AN AIRFOIL

Consider a case of flow around an airfoil as shown in ~igure 13.12. Airfoil body is a well stream line body and hence flow can only separate at the rear end of the airfoil.

1 . 0 1 , , , , , , I , , , b-------- L ---------------4

MEASURED VALUES 1] OP 0 .~1 ----- THEORETICAL VALUES

Efgure 13-12 : Pressure Distribution around an Airfoil

On account of the nature of streamlining that airfoil body has at relatively high values of Reynolds number, the major contribution to the total drag is even of the friction drag and contribution of form drag (pressure drag) is comparatively very small. Measured pressure distribution and the predicted pressure distribution using theory of irrotational flow are shown in Figure 13.12. Close observation of this figure reveals that both the pressure distributions agree very well except at the rear end. It has been observed that for the case of airfoil, drag coefficient mainly depends on Reynolds number and shape, but a sudden drop in the value of C, as in the case of sphere or cylinder is not found. The main reason for this is that change from laminar to turbulent boundary layer does not alter the wake size significantly to affect the contribution of form drag.

Table 13.1 gives C, values for several geometrical shapes of bodies and Table 13.2 gives a list of drag coefficients of practical body shapes.

TABLE 13.1 : Drag Coefficient for VarlousBody Shapes

Body Profile

Circular Cylinder (Infinite)

Circular Cylinder (Infinite)

Elliptical Cylinders

2 : l

4 : 1

8 : 1

Triangular Cylinders 120" 1200 60" 60" 30" 30"

Hemispheres (hollow)

Semitubular Cylinders

Square Cylinders

Reynolds Number Re

lo4 to2 x lo5

5 x id

5 x lo4 1 6

2.5 x 104 to 105

2.5 x lo4 2 x 1 6

+ lo4 + 104 +# lo4 +< lo4 + lo4 4 lo5

+ m lo4 + a lo3 to 1 6 + 4 x lo4

-,c 4x104 +- 3.5 x 104

-+ 4 l o4x lo5

Drag Coefficient CD 1.20

0.35

0.46 0.60

0.32

0.29 0.20

2.00 1.72 2.20 1.39 1.80 1.00

1.33 0.34 2.30

1.12 2.00

1.60

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Dmg, Lift and otber HydrPulic Robluns

TABLE 13.2 : Drag Coefficient of Some Practical Bodies

13.9 ILLUSTRATIVE EXAMPLES Example 13.1 :

Body Profde

1. Empire State Building, New York, about 380 m high

2. "Trylon" slender pyramidal building 206 m high with 20 m wide base

3. Wind resistance of several ships

4. A man falling down vertically

5. Volkswagen passenger car

6. Minibus

7. Non Streamlined engine with five bogies

Alluminium ball of diameter 2.0 mm having relative density 2.80 is falling freely in a tank containing fluid which has mass density of 910.2 kg/rn3. If the velocity of ball is 0.35 cmls, determine the dynamic viscosity of oil.

Drag Coefficient

1.30 to 1.5 (depending on wind direction)

0.80 to 1.43 (depending on wind direction)

0.20 to 0.70

1 .O to 1.30

0.50

0.73

1.90

Solution :

Initially let us assume flow to be within Stoke's range.

According to Stokes' law, drag force is given by

Submerged weight of the ball is given by

wherein y, and yare the specific weights of the solid and fluid respectively.

Equating the two forces one gets

Now it should be checked whether for this viscosity, Reynolds number is within Stokes' range

Since Re is_ less than 0.2, assumption is justified.

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I

I Example 13.2 : 1 A sphere of diameter 2.5 cm having relative density of 2.65 is freely falling in a tank of I 4 2 oil having mass density 898 kg/m3 and kinematic viscosity of 1.58 x 10- m Is.

Compute the fall velocity of the sphere and the drag force. I

Solution :

~ r a i Force

Also,

Thus, the drag force can be calculated straight way as

= 0.1406 N

I However a trial and error method is required to compute fall velocity because it needs CD which itself depends on fall velocity.

Steps are as follows,

(i) Assume Re

(ii) Compute CD by using equation (13.6) which is

(iii) Use this value of CD in drag equation to calculate U.

(iv) Compute the value of Re using the value of U as obtained in step (iii)

(v) If Re computed in step (iv) is same as in step (i) o.k., otherwise assume a new value of Re and repeat the steps.

Assume RI 100

24 C D = - + - + 0.34 = 0.88 loo ,mi

and

Assume Re = 1 15

This gives CD = 0.828

and Re = 122 0.k.

Hence fall velocity of the sphere is 0.77 m/s.

Example 13.3 :

A truck having a projected area of 6.5 m2 travelling at 80 k& experiences a total resistance of 2 kN. Out of this, the form drag is 60%. Calculate the coefficient of form drag. Assume specific weight of air as 12 ~ / m ~ .

Drag

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Drag, Lift and other Hydraulic Problems

Solution :

Total resistance = 2 x lo3 N.

Form drag = 0.6 x 2 x lo3 = 1.2 x ld N

For form drag, one can write

FD = CD f& x Projected Area 2

(Here U is the velocity of truck.)

.a. CD = 2 2 FD

p U x Projected Area

One gets CD = 0.61.

Example 13.4 :

Electric transmission towers, 10 m high, are fixed 400 m apart to support 10 cables, each of 20 mrn diameter. Determine the moment acting at the base of each tower when a wind flows with a velocity of 100 km/hr. Assume mass density of air as 1.2 kg/m3 and p (air) as 1.6 x kg/rns. Also calculate the frequency of vortex shedding.

Solution :

For drag force

and

Here

Projected Area A = 400 x 0.02 = 8 rn2

Using equation (1 3.6)

Therefore, drag force on each wire of 400 m long

10 Moment at the base = 1315.83 x - = 6579.15 Nm 2

Using equation (13.7) for the frequency of vortex shedding

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13.10 SUMMARY

In this unit the forces exerted on bodies exposed to flow of fluid have been discussed. The importance of flow separation in affecting these forces has also been emphasised. The methods for avoiding or controlling the separation were discussed. It was seen that the laminar boundary layer is more susceptible to separation than the turblent boundary layer.

Commonly encountered body shapes such as sphere, cylinder, flat plate and airfoil were considered while studying their drag characteristics. Relationships were studied for the variation of CD with Re for different body shapes. Attention was also drawn to the vortices and their characteristics that one has in the wake region of these bodies.

1 13.11 KEY WORDS I

Force exerted by fluid on the body in the direction of fluid flow.

Flow conditions in which velocities near the wall are negative, cause flow to separate from the wall, promoting instability, eddy formation and large energy dissipation. This phenomenon is called separation.

Boundary Layer Fluid layer of retarded flow that is formed next to the boundary is known as boundary layer. .In this layer I marked velocity changes take place.

Surface Drag Force due to friction of the fluid against the body is lmown as surface drag.

Pressure Drag Pressure drag or form drag is due to separation of the boundary layer and the resulting pressure difference between the front and the wake region of the boundary.

Drag Coefficient Drag force per unit area divided by the dynamic head is lmown as drag coefficient and is designated by CD .

Strouhl Number It is a dimensionless number used for studying the vortex shedding from an immersed body.

Pressure Distribution : Variation in pressure around the body is lmown as pressure distribution. It could as well define variation in pressure in a particular region of flow, for example, in the wake behind the body.

13.12 ANSWERS TO SAQs

Check your answers of all SAQs with respective preceeding text of each SAQ.

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UNIT 14 LIF'T

Structure 14.1 Introduction

Objectives-

14.2 Circulation in lrrotational Flow

14.3 Magnus Effect

14.4 Computation of Lift Force

14.5 Lift on Aerofoil

14.6 Illustrative Examples

14.7 Summary

14.8 Key Words

14.9 Answers to SAQs

14.1 INTRODUCTION

In the previous unit on drag forces around bodies immersed in fluid, we had observed that the total force exerted by the fluid on the body can be resolved into two components - one in the direction of fluid motion known as the drag and the other perpendicular to the direction of fluid motion known as the lift. This unit is devoted to the elementary concepts about lift force.

Objectives After studying this unit, you should be able to

* understand the cause of lift in irrotational flow,

* compute the lift force exerted on a body, and

* appreciate the lift on an aerofoil.

14.2 CIRCULATION IN IRROTATIONAL FLOW

Circulation refers to such flow which follows a circuitous course back to the starting point. Generally, the circulatory motion is superposed upon the basic urnlatory motion of the fluid and. hence, it becomes essential to express circulation in terms of an arbitrary curve passing through a system of streamlines as shown in Figure 14.1 .

Figure 14.1 : DeClnitioa OtCiredntion

If vL represents the tangential component of the velocity at any point on the curved line then the circulation r around a closed curve is &fined as the integrail of the quantity vL dL around the closed curve. Thus,

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If we have to consider the case of flow in concentric circles (refer Figure 14.2). according to the condition v = or, then the circulation around any stream line (which is a circle of radius r) is obviously equal to o r (2nr ) i.e. 2 n h . One may similarly find circulation around any other curve such as that composed of the two circular arcs and two segments of radii (along which the tangential velocity is zero) as indicated in the figure. In this case,

Fi gum 142 : Circalation a ~ ~ d Two Concentric Circles

Now consider a cylinder placed in an irrotational flow. The streamline pattern is as shown in Figure 14.3 (a). Let us superimpose the streamline pattern of constant circulation around'a cylinder (Figure 14.3 (b)) so that the resulting flow pattern is as shown in Figure 14.3 (c).

1.a 1 IRROTATIONAL I b) FLOW WITH ClRCULAllON

( C 1 IRROTATIONAL FLOW AND CIRCULATORY FLOW SUPERIMWSED

Flpm 14.3 : Genedon of Lift nmcud a Cytiader

Obviously, this has caused increased velocity (and hence decreased pressure) on one side and the decreased velocity on the other. The streamline pattern is, however, still symmetrical about the vertical axis. Therefore, the circulation has not changed the longitudinal force upon the cylinder from its initial magnitude of zero. But, the asymmetry of the flow pattern about the horizontal axis has resulted in decreased pressure intensity in the upper part and increased pressure intensity in the lower part. Thus it is obvious that the circulation in irrotational flow causes a resultant force in the lateral direction.

14.3 MAGNUS EFFECT

Lift

In a real fluid flow, local circulation can be produced through surface drag by rotating the cylinder itself. The circulation decreases instead of remaining constant as in the irrotational case. Nevertheless, this local circulation does produce considerable lateral thrust - commonly known as lift force. This phenomenon was first observed by a German scientist Magnus and hen@ it is named as Magnus effect. The sudden deviation of a ball, which has been chopped (as in volley ball or table tennis) or sliced (as in lawn tennis) by player, from its normal trajectory is simple,illustration of the Magnus effect.

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Drag, Lift and other ~ydraulic Problems 14.4 COMPUTATION OF LIFT FORCE

If U, is the free stream velocity, the velocity v at any point on the surface of a cylinder of radius r can be obtained from the principles of hydrodynamics. Thus,

v=2U0s in0

Similarly, the velocity u, on the surface of the cylinder due to circulation r is given by

Therefore, the resultant velocity u of the superimposed flow

uc For a specified value of --, one can obtain the value of 8 for u = 0. That is . uo

This gives us rhe location of stagnation points. If p, is the pressure in the ambient stream and one applies Bernoulli's equation between any point in the unaffected ambient flow and any point on the surface of the cylinder, one gets,

The lift force FL acting on the surface of the cylinder is given as 2 r

F Ap s ine . L r d e 0

2 u 0 sin e+- 2 o 27cr ) j S i n e d e

which reduces to a very useful relation

The derivation of this expression does not take into account the viscous effects. The actual magnitude of the lift force (or so called Magnus effect) FL must obviously depend upon the Reynolds number of the flow alongwith other parameters such as the density and velocity of the fluid and the size and peripheral speed of the cylinder. However, at higher Reynolds number, viscous effects upon the lift should approach an asymptotic value, just as in the case of drag. Under such circumstances the lift force FL may be expressed as.

which is similar to the expression for the drag force. Here, C, is the lift coefficient and D is the diameter of the cylinder. Thus, L D represent the projected area of the cylinder in a plane normal to the direction of the lift force.

For irrotational flow, therefore,

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This value of the lift coefficient for irrotational flow may be termed as theoretical value which has been compared with the experimental values of C, and the corresponding coefficient of drag C, in Figure 14.4. It is obvious that the local circulation produced through surface drag and viscous shear is about half as effective as the constant circulation required for truly irrotational flow.

Figure 14.4 : Experimental Variation of CL and CD aa a function of uo

14.5 LIFT ON AEROFOIL

In aeroplanes, wings (which are also mown as lifting vanes or aerofoils) are used to produce lift force as they move through the fluid which is air. These vanes (or aerofoils) can be symmetrical or unsymmetrical and are characterised by the geometric chord C and the angle of attack 8, between the geometric chord and the direction of flow (see Figure 14.5).

SYMMETRICAL

LEADING- EDGE U \ \-

TRAILING -- + EDGE a L eat*--- - - - ---

UNSYMMETRICAL

Figure 145 : Nomenclature for an Aerofoil

When a circulatory flow around an aerofoil is superimposed over an irrotational flow around the same aerofoil, the resulting flow pattern appears similar to the one shown in Figure 14.6.

I a 1 IRROTAI FLOW

'IONAL I I b 1 CIRCULATORY FLOW

I c 1 SUPERPOSITION OF f b ) OVER l a )

Lift

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Dmg, Lift and other Bydraulic ProbIems

It has been shown that to make the streamline at the trailing edge tangential to the trailing edge the circulation required is given as

r = n c u o s i n e 0

:. FL = n C L p sin 0,

FL

The theoretical values of CL obtained from this expression are quite close to the actual values in real fluid flow.

SAQ 1

Why do aeroplanes take off and land against the wind?

14.6 ILLUSTRATIVE EXAMPLES

Example 14.1 :

An aeroplane weighing 22,000 Newtons has a wing area of 22 m2 and span of 12.0 m. What is the lift coefficient if it travels at a speed of 360 km/hr in the horizontal direction. Also compute the theoretical value of circulation and angle of attack.

Solution :

. . CL = 0.166

= 2 n sin $

8, = 1.51'

22 22 r = nCUo sine, and C = - = - = 1.833 rn L 12

WINDOWS
Comment on Text
dec.2007
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Example 14.2 : A circular cylinder 2.0 m in diameter and 12 m long is rotated at 360 rpm with its axis perpendicular to the air stream having a velocity of 37.7 rnls. Assuming p to be 1.236 kglm3. Determine (i) circulation (ii) theoretical lift (iii) position of stagnation points and (iv) actual lift. Assume CL = 1.5.

Solution :

Circulation, T = x D u ,

Theoretical lift, FL = p LU,T

= 1.236 x 12 x 37.7 x 236.76

= 132.4 kN

Points of Stagnation

u uc - = 2sin8 + - and u = O u0 u0

Actual lift, 1 7 .

FL = C L X D X L X - p U , 2

14.7 SUMMARY

In this Unit we discussed the generation of lift force on a body immersed in either irrotational flow or viscous flow. Magnus effect was also explained. Lift force on an aerofoil was also briefly described.

14.8 KEY WORDS Lift Force Component of the total force exerted by the fluid on the

body in the direction perpendicular to the flow is known as Lift Force.

Airfoil Body shape which yields high lift and low drag values. Magnus effect Lift force can be caused by superposition of rectilinear

and circulatory flows around a cylinder was first discovered by Magnus in 1852. It is known as Magnus effect.

14.9 ANSWERS TO SAQs SAQ 1

An aeroplane moving against the wind will have greater speed relative to the air and hence greater lift than, when moving in the direction of the wind.

The opposing wind velocity helps in slowing down the velocity of aeroplane whenit lanrtc

Lilt

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UNIT 15 WATER HAMMER AND SURGE TANKS

Structure 15.1 Introduction

Objectives

15.2 Water Hammer 15.2.1 Expression for Rise in Pressure

15.3 Rapid Acceleration of Flow

15.4 Surge Tanks

15.5 Summary

15.6 Keywords

15.7 Answers to SAQs

15.1 INTRODUCTION

In this unit we will be discussing a class of unsteady flows. As you know, the flow is classified as unsteady or steady depending on whether the flow characteristics such as discharge, velocity, pressure etc. at a point change with time or not. The analysis of unsteady flows is rather complicated except that some unsteady flow problems can be solved without much difficulty under simplifying assumptions. Water hammer and surge tanks are two such situations, wherein the flow unsteadiness involves slow periodic or cyclic fluctuations, and it is these Periodic Flows which we propose to discuss.

Objectives After studying this unit, you should be able to

* understand the phenomenon of water hammer,

* compute the rise in pressure due to water hammer, and

* understand the necessity and working of a surge tank.

15.2 WATER HAMMER

Water hammer is a phenomenon encountered in a pipe line, when the velocity of the flowing liquid is suddenly reduced by closing a valve. Such a closure results in sudden and large rise in pressure due to the change in inertia of the fluid and a hammering sound is produced. In many cases this continues with regular periodicity unless the periodic motion is damped out. This phenomenon can cause problem like bursting of pipe lines - specially if they are long and is very important in case of penstocks which carry water from a storage to a power house.

To get a clear concept of the phenomenon of water hammer, let us consider a long pipeline with a valve at the downstream end. The pipeline is discharging some fluid and the valve is suddenly closed.

Phase I When the valve is open, the hydraulic gradient line for the flow is given by CD, when the valve is closed instantaneously the lamina of fluid in contact with the valve is brought to rest and it will be compressed by the rest of the fluid flowing against it causing a rise in pressure p. At the same time, the walls of the pipe surrounding this lamina will be stretched by the excess pressure produced. In other words the kinetic energy of the fluid is converted into the strain energy of water and pipe. This process continues upstream and a pressure wave travels in the upstream direction (from B to A) at a speed equal to the velocity of sound C. When the pressure wave reaches the end A, all the fluid in the pipe is at rest and the whole pipe is at an excess pressure with a hydraulic gradient shown by line EF. This condition occurs at time T = LIC.

Phase I1 The increased pressure p in the pipe causes the fluid to flow from A to the reservoir, thus relieving the excess pressure. A wave of pressure unloading travels back from A to B with a velocity C, and all the strain energy gets converted into kinetic energy.

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Water Hammer and Surge Tanks

H.G.L. (Transient) I - - - - I r - - - - - - I

Pressure Rise - - 7"

Figure 15.1 : Definition Sketch

Figure 15.2 : Pressure Variation at B

2(L-z)

e-

I - Time t

// Valve closed

Figure 153 : Prcssore Variation at M

15.2.1 Expression for Rise in Pressure [A] Slow Closure

2L If the valve is closed slowly, such that the time of closure is several times larger than - C ' the maximum pressure rise will be less than that in case of a rapid closure. This is so because the wave of pressure unloading will reach the valve before its complete closure and will thus prevent any further rise in pressure.

We can get an expression for the r id in pressure in such a case making use of the momentum equation. I

Thus if the initial velocity of flow in the pipe is Vo and the time of closure of the valve is vo

& kc 2 %), the rate of retardation of flow will be -. tc

The mass of water in the pipe will be pAL where p is the mass density of water and A the area of cross-section of thepipe.

If the pressure rise due to closure of the valve is Ap, then the force will be ApA.

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Equating the force to the product of the mass and deceleration we get

vo ApA = pA L - tc

[B] Rapid Closure

2L If the time of closure of the valve is less than --, it is termed as a rapid closure. In fact if the C

valve is closed instantaneously i.e., t, = 0. EQuation (15.1) will give an M i t e pressure rise. 2L

Practically however, the pressure rise is finite and this is at time T = - , this wave reaches C

B and the pipe is at normal pressure with the fluid flowing back into the reservoir.

Phase 111 The reverse velocity from pipe to the reservoir has to be stopped and therefore the pressure at B drops down to -p, thereby sending a wave of rarefaction from B to A with velocity C. - -

3L This wave reaches A at time T = - when all the liquid in the pipe is stationary and at a C

negative pressure p.

Phase IV 4L

The liquid now flows back into the pipe again and at time T = - , the whole liquid in the C

pipe is moving towards B with the original velocity and at normal pressure.

The cycle of events now repeats. The variation of pressure with time at representative points B and M is shown in Figures 15.2 and 15.3.

The pipe friction and damping are not considered in the above discussion. Due to these two effects, the amplitude of this pressure wave will go on diminishing. So because in the case of an instantaneous closure, the compressibility effects became important.

We can obtain the expression for pressure rise in rapid closure as follows :

Let, the initial velocity of flow in the pipe = Vo

diameter of the pipe = D

thickness of pipe walls = C

bulk modulus of elasticity of the fluid = Ef modulus of elasticity for the pipe material = Em

and specific weight of the fluid = Y Then,

n volume of fluid in the pipe = - 0 2 ~

4

original kinetic energy of the fluid

As already mentioned, this kinetic energy is stored as the strain energy in the fluid and the pipe when the valve is closed.

1 Strain Energy of the fluid = - Ap x change in volume.

2

f 2 and Strain energy of the pipe = - x volume .

2E

AP D Where f is the hoop stress in the pipe and f = - 21

To fluid the change in volume of the pipe, we recall the definition of the Bulk modulus of -1 --A:-:&-. v

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Ef = A (change in volume) Water Hammer and

Swge Tanks ( original volume I or Change in volume = x Original volume

Ef

:. Strain energy of fluid = i ~ p 9 L Ef

and Strain energy of pipe = 2 D 2 1 * fl-LI = - n D L f 2Em 4 I2 2Em

Equating the original kinetic energy and the strain energies, we get

Further, we know that the velocity of propagation of a wave (C ) is given by

Where K is the overall volume modulus of elasticity of the fluid.

When the pipe is rigid, i.e. Em + =, K = Ef , While if Em is finite K is given by

Thus the expression for pressure rise equation (15.2) can also be written as

Ap = p CVo

Where

and

2L In case the closure time tc is slightly greater than -, we can assume that the pressure rise is C

2L 1 reduced in proportion to - -, yielding c tc

Where Ap,, is the pressure rise given by equation (15.3)

Example 15.1 :

A 20 cm dia pipe carries water at a velocity of 3.0 m/s, Find the rise in pressure if a valve at the end of the pipe is suddenly closed. Treat the pipe to be rigid and take E,,, = 2.08 x lo9 N I m2.

Solution :

Vo = 3.0 mls

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Drag, Lift and other

Hydraulic Problems Density of water

. . Rise in pressure Ap = p C Vo

= 4322.33 kN/m2

Example 15.2 :

A 100 cm diameter 50 km long steel pipe with a thickness of 10 rnm carries water at a rate of 2.0 m3/s. What wiil be the increase in pressure if a valve at the downstream end of the pipe is closed in (a) 3 seconds (b) 11 seconds.

Take E for steel and water as 2.08 x 10" N/m2 and 2.08 x lo9 N/m2 respectively.

What will be the pressure rise in (a) above if the pipe is treated as being rigid. Solution :

Here El = 2.08 x lo9 Nlm2 and Em = 2.08 x 10" N/m2

:. Time required for pressure wave to travel from the valve to the inlet and back

2L (a) Since closing time tc = 3 secs. is smaller than - (9.85) C

(b) If closing time is 1 lseconds, it is slightly more than 9.85 and hence Ap = 2 L 1

A P ~ , C tc 2L C

- 9.8 sec, tc = 11 sec. Ap,, = 2594.30 kN/m2, - -

If the pipe is treated as rigid

Thus we see that the pressure rise is overpredicted if the elasticity of the pipe is neglected.

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SAQ 1 Water Hammer and Surge Tanks

When will the rise in pressure due to water hammer be more - if the pipe is rigid or if the pipe is elastic - other conditions remaining the same?

SAQ 2

2L If the time of closure of a valve is greater than -, will the pressure rise be more or

C less than the corresponding value for an instantaneous closure.

15.3 RAPID ACCELERATION OF FLOW

Let us consider the pipe AB connected to a reservoir as shown in Figure 15.1. It is apparent that if friction is neglected, the velocity Vo through the pipe is given by

v, =

Let us, however, consider a slightly different situation i.e., having a valve at B which is closed to start with. If this valve is opened suddenly, the velocity of flow in the pipe does not attain a value Vo instantaneously but increases gradually. It can be shown than assuming the fluid to be incompressible and neglecting friction, the time Ttaken for the velocity to attain a value Vcan be given by

1 + - L T=6K-h13] (15.5)

Two facts are apparent from the above equation viz. that the time required to attain Vequal to Vo is infinity, though V may become almost equal to Vo in a relatively short time and that the time required for V to become nearly equal to Vo will be more for a longer pipe.

If on the other hand, we take the compressibility of flow into account, it can be shown that when the valve at B is opened suddenly, the velocity in the pipe increases in steps of

h magnitude V1 = g - where C is the velocity of propagation of a wave as defined earlier.

C 2L

Each such increase takes place at a time interval of - and the total number of time steps C vo requlred for the flow to establish completely will be -. v1

15.4 SURGE TANKS

Surge tanks are large diameter vertical open tanks provided just upstream of hydroelectric power plants.

Before talking about surge tanks, let us try to discuss the problems we may face in hydroelectric power plants. As you know, in such plants turbines convert the potential energy of water into mechanical energy. The water stored in reservoir upstream of dams is carried to the power plant by means of large diameter pipes called penstocks. These penstocks generally are quite long - at times, a few kilometer in length. Depending on the power demand, the discharge through the turbines may have to be increased or decreased. A turbine may have to be shut down sometimes or started as the demand increases. Whenever a turbine is shut down, the closure of the valve at the turbine end is going to result in water hammer and hence increased pressures throughout the penstock. The penstock must therefore, be designed to withstand these increased pressures and in case of long penstocks, this may prove very costly. Likewise, whenever a turbine is started the time required for the

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Drag, Lift and other

Hydraulic ProMerns flow to establish will be considerable in case of long penstocks. It is desirable, therefore, to reduce the length of pipe in which the effects of water hammer pressures or flow establishment are felt. This purpose is achieved by providing surge tanks.

The working of a surge tank can be understood with reference to Figure 15.4. A surge tank is a tank connected to the penstock quite close to the powerhouse.

, Penstock I I

Draft Tube XJ figure 15.4 : Schematic D i ~ g s m of Surge T a d

Let us consider a situation, when due to load rejection, the valve at the turbine is closed suddenly. The water hammer pressure wave travels upstream from C. However, at B this causes a rise in water level of the tank and relieves the penstock upstream from B of the water hammer pressures. Thus only the length BC has to be designed to withstand water hammer.

In case the valve at the penstock is opened suddenly, the surge tank supplies additional flow, till the water in the whole pipe is accelerated.

The surge tank thus acts as a balancing reservoir and helps in both load rejection and increased demand. There are various types of surge tanks viz. simple, restricted orifice, differential etc. Also, the stability of surge tanks is an important criterion for their design. This, however, is beyond the scope of the present course and hence not discussed further.

15.5 SUMMARY This Unit is primarily devoted to a discussion of Water Hammer - a phenomenon which occurs when a valve at the downstream end of a pipeline carrying a liquid is suddenly closed. This causes a rise of pressure in the pipe which is influenced both by the time of closure of the valve, as well as, the elasticity of the pipe. Expressions for this pressure rise have been obtained. A sudden opening of a valve at the downstream end of a long pipeline gives rise to the problem of establishment of the flow, which has also been discussed briefly. Finally the utility of surge tanks for both the aforesaid situations has been indicated.

15.6 KEY WORDS Minor Loss Hydraulic loss in pipe lines on account of different forms

and fittings in pipe line and other than those due to friction. Equivalent Pipe A hypothetical pipe line of uniform diameter resulting in

the same head loss as in the actual pipe line. Pipe Network Pipes of different length and diameters connected in

different ways.

15.7 ANSWERS TO SAQs

SAQ 1

Pressure rise will be more for a rigid pipe (Ref. Example 15.2)

SAQ 2

2L The pressure rise will be less, if the time of closure is more than -

C (Ref. Examole 15.2)

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UNIT 16 HYDRAULIC MODELS

Structure 16.1 Introduction

Objectives

16.2 Laws of Similitude 16.2.1 Geometric Similarity 16.2.2 'Kinematic Similarity 16.2.3 Dynamic Similarity

16.3 Modelling Techniques 1 6.3.1 Drag of an Air& Model 16.3.2 Drag of a Ship Model

16.4 River Models

16.5 Illustrative Examples

16.6 Summary

16.7 Key Words

16.8 Answers to SAQs

INTRODUCTION

There are many problems related to hydraulic engineering and fluid mechanics which defy a complete analytical solution. Model studies are usually conducted to find solutions to these problems. Even if analytical methods are available, experiments are still necessary to verify these methods, because invariably such methods are based on certain assumptions or have some empirical constants. Therefore use of models have steadily increased. To give few examples, an aeronautical engineer obtains data from model tests in wind tunnels; the naval architect tests ship models in towing basins; the mechanical engineer tests models of turbines and pumps; the civil engineer works with models of hydraulic structures and rivers to obtain more reliable solutions to his design problems. Therefore one would like to h o w the various similarity criteria for the design of these models in a manner that they show complete similarity with prototypes. It is rarely feasible to attain this complete similarity; in fact especially for large river models, the horizontal and vertical scale ratios are kept different, resulting in distorted models. In some cases forces that are negligible in prototype, become quite significant in models, this effect is lcnown as scale effect. Some basic principles of dimensional analysis, significance of dimensionless parameters governing the fluid flow phenomena alongwith the types of various similarities required to be attained by the mode1,have already been studied in Unit 5.

In this unit some aspects regarding laws of similitude in respect of few specific fluid flow problems are proposed to be discussed. Need for distorted models and correcting model results for scale effects will then be discussed.

Objectives After studying this Unit, you should be able to

* appreciate need for hydraulic models,

* - write the similarity criteria for designing the models,

* interpret the model results, and

* appreciate the scale effects in these models.

16.2 LAWS OF SIMILITUDE - - - - pp

In all model studies, similarity in behaviour between the model and the prototype must be ensured. There are three types of similarity to be attained :

16.2.1 Geometric Similarity This is similarity of shape. It implies that model and prototype would be geometrically similar if, linear dimension ratios or scale factors are same. In the model and prototype shown in Figure 16.1, it means

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C> 5

Prototype Model

Figure 16.1 : Geometric Similarity in Model and Prototype

In equation (16.1) suffix p is for prototype, m for model and r for length scale. Quite often perfect geometric similarity is not easy to attain. This problem comes especially in building river models requiring study of sediment motion. Nofmally these river models are quite large and their size is limited by the available floor space. But if the vertical and horizontal scale ratios are kept same, the result may be a stream, so shallow that surface tension has a considerable effect and may be flow in the model is laminar instead of turbulent. Therefore, in river models, horizontal and vertical scale ratios are invariably kept different and this results in distorted models. The characteristics of distorted models would be discussed at a later stage of this unit.

16.2.2 Kinematic Similarity Kinematic similarity is similarity of motion. This is attained if velocity ratios at corresponding points in the model and prototype are same.

16.2.3 Dynamic Similarity Dynamic similarity is similarity of forces. In problems related flow of tluids, forces may be .due to many causes : viscosity, gravity, difference in pressure, surface tension etc. For a perfect dynamic similarity, the ratio of all these forces should be same. However, in practice it is not possible to satisfy all these forces in the model. Fortunately, in many instances, some of the forces do not come into picture, or have negligible effect, and so it becomes possible to concentrate on the similarity of the most significant forces governing lhal particular problem. Therefore in order to identify these significant forces, a detailed knowledge of various forces governing the particular problem is required.

Let us now consider some of the modelling techniques and the relevant force ralios that need to be considered in the design of specific models.

16.3 MODELLING TECHNIQUES

In the design of models, the various force ratios are expressed in terms of dimensionless parameters. For example, Reynolds number which signifies the ratio of inertial forcc lo viscous force, Froude number which is the ratio of inertial forcc to gravitational Sorcc, likewise there are many such parameters. Once in a particular problem, the important governing dimensionless parameters are known, lhe model c G be built holding those parameters constant in the model and prototype. The required model scales can lhcn be

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obtained on the basis of this equality and these scales will then be used to convert the model Hydraulic Models

results to prototype results.

To illustrate the above modelling technique, let us study the testing of an aircraft model in a wind tunnel in order to find the drag force FD on the prototype.

16.3.1 Drag of an Aircraft Model In this problem, froces due to gravity and surface tension do not affect the flow field. If compressibility effects are also ignored, the only forces to be considered are viscous and inertia forces, thus for the dynamic similarity, the relevant dimensionless parameter to be considered is the Reynolds number. Thus dynamic similarity is obtained between the model and the prototype

when

provided model and prototype are geometrically similar and their orientation with respect to oncoming flow is identical. Here 1 is some characteristic length of the aircraft may be span of the wing of the aircraft.

Equation (16.2) places no restriction on the fluids of the model and prototype, the prototype could move through air while the model could be tested in water as long as the Reynolds numbecare kept same in model and prototype. If for practical reasons the same fluid is used in model andprototype, equation ( 1 6.2) suggests that the product (U1) must be the same in both. When a male1 aircraft is being tested, the size of the model is naturally less than that of prototype; this means to keep (U1 ) same, the velocities around the model will be larger than the corresponding ones around the prototype. To give the idea of velocity required for this model study, let us assume that the prototype is intended to fly at 300 km/hr speed. Obviously, the velocity in the wind tunnel must be greater than 300 krn/hr since such velocities can not be attained, one way of obtaining a sufficiently high Reynolds number without using inconveniently high velocities is to test the model in air of higher density. Such wind tunnels which use compressed air are known as variable density wind tunnels. In such tunnels v, < vp , the equality of Reynolds number could be achieved easily. Once equality of Reynolds number is obtained, the drag force could be estimated as follows :

since drag coefficient CD will be same both in model and prototype it gives

Equation (16.3) will then give the drag coefficient for the prototype directly using the corresponding model values and no correction for scale effect is needed as Re values are same in model and prototype.

In case variable density tunnel is not available, complete similarity has to be sacrificed and a compromise solution sought. We have already studied that CD varies with Re. merefore if the Reynolds numbers of model and prototype are different, the values of CD will also differ, as such one can not directly transfer the model results to prototype. In this case some extrapolation of model results will be required. For this extrapolation, CD versus Re relationship would be required to be developed from experiments in which models have been compared with their prototypes.

In problems where more than one force ratio is relevant, it is difficult to achieve complete similarity. Similarity is still achieved but with some departure. It is, nevertheless, essential that these departures are justified. An example to illustrate this procedure is regarding model testing of a ship model.

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Drag, Lib nnd other Hydraulic Problems

16.3.2 Drag of a Ship Model The resistance experienced by a ship moving through water is generally due-ts-three causes

(i) Viscosity

(ii) Eddies formed in the wake

(iii) Surface waves

Due to viscosity of the fluid, viscous forces are set up between the fluid layers close to the surface of the ship and those farther away and these forces cause a frictional resistance to the motion of the ship. This part of the total drag is usually termed as skin friction. In addition, as the ship moves forward, part of the flow towards the rear of the ship breaks away from its surface to form a 'wake' of eddying motion. The eddies give rise to a distribution of pressure round the body different from that to be expected in an ideal fluid, and so provide another contribution to the total drag force. This part of the drag is known as pressure or form drag.

These two types of resistances - the skin friction and form drag - are experienced by any solid body moving through the fluid. A ship, however, is only partly submerged in a liquid, and its motion through liquid gives rise to waves on the surface. The formation of these waves requires energy, and since this energy must be derived from the motion of ship, the ship experiences an increased resistance to its motion and this is called wave resistance. Thus for a ship moving through water the total drag is the sum of friction drag (skin friction), form drag and drag due to surface waves. Since the ships are generally streamlined, the form drag is relatively small and within the usual range of Reynolds number, it can be assumed to remain constant. Moreover, it is usual practice to consider this portion of the total drag along with the drag due to surface waves which are associated with gravitational action. As such the resistance to the motion of a ship is affected by viscosity as well as gravity. Making use of dimensional analysis, the functional relationship for the total resistance FD can be written as

The total resistance thus depends both on the Reynolds number and on the Froude number Therefore, for complete similarity these two numbers must be the same in model and prototype. That is,

and u P= um

,k& Trc Since in practice g , = g,,, the combination of above equations give

"P As both model and prototype must operate in water giving - = I. the above condition

"m

demands lP = I , , i.e.model be as large as prototype. Obviously such a condition can not be satisfied. Thus equality of Re and Fr can not be met simultaneously. Therefore, following approach as suggested by Froude may be adopted to determine the total resistance of a ship.

The total resistance FD experienced by a ship may be assumed to be consisting of two portions viz.

(a) the wave resistance F, due to the action of waves;

(b) the frictional resistance Ff due to the frictional effects on the wetted. surface of the ship.

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That is

The total drag FD for the ship may then be obtained by adding the two i.e. Fw and Ff . SAQ 2

(i) What is the physical significance of Reynolds number and Froude number ?

(ii) Is it possible to maintain both Reynolds number and Froude number same in the model & prototype?

FD = Fw + Ff (16.5)

A similar subdivision can also be made for the total resistance FDm of the model. Thus if Fw, and Ffm represent the wave and frictional resistances respectively for the model, then

The frictional resistances may be estimated by assuming that they have the same values as that for a thin flat plate of given length and wetted surface area moving through water at the same velocity. In earlier unit, we have already studied the methods for estimating the drag on a flat plate, hence there is no difficulty in estimating the frictional resistances.

A geometrically similar model of the ship is then tested by towing it in water taking model 1

velocity according to Froude law i.e. u, = u jhT and totd drag FDm is measured. The 1~

\ - / wave drag of the model is then found as

since Ffm for the model has already been estimated as indicated above. Now for dynamic similarity between the model and the prototype considering only wave resistance, we have

The wave resistance Fw for the prototype can be obtained from equation (16.7) as

F w = 2 2 F,, [;m) [:mJ

But from equality of Froude number we have -

1

This gives

Hydraulic Models

(iii) Define wave resistance.

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Drag, Lift and other Hydraulic Problems 16.4 RIVER MODELS

Rivers carry water and sediment, the bed and banks of rivers are generally erodible. Quite often hydraulic engineers are required to build structures across these rivers. For example a dam, a bridge and a barrage. As a result of this construction many changes take place in the river both upstream and downstream of these structures, as well as, around the structures. Model studies are, therefore, required to study these changes. Since these flows are characterised by free surface, gravitational forces dominant over other forces, therefore, invariably these river models are built on the basis of equality of Froude number. Thus assuming horizontal and vertical scales to be same, equality of Froude number and continuity equation for discharge gives

Continuity equation gives Q, = A, u, where A is the area of flow cross- section. 2

Since A, = 1, we have

Equation (16.9) gives the relation between discharge scale and length scale. Obviously two considerations will govern the model dimensions, the available discharge and the available space to build the model. If discharge is going to govern the model scale, the length scale could be found from equation (16.9). Thus if this scale ratio is used to reduce the vertical dimensions of the prototype and size of the prototype roughness, the resulting depth of flow and the height of roughness element at times become unrealistic. The depth of flow in the model could be so small that forces due to surface tension start affecting model results. Also reduced depth of flow could make flow in the model laminar instead of turbulent flow which prototype invariable has.

In view of above, the horizontal and vertical scales are invariably kept different in river models, this introduces what we call as geometric distortion. Another distortion in such models is force distortion because similarity of only Froude number is maintained whereas Reynolds number is allowed to be kept different in the model and in the prototype. Yet another distortion, that is given to river models is material distortion, this is needed to have the motion of sediment on the bed. This requires use of light weight material which could be easily moved in the model under reduced depths of flow. Thus river models are invariably distorted models and it requires lot of experience and judgement to interpret model results and their transfer to predict prototype behaviour.

The other details of these river models are beyond the scope of this unit.

SAQ 3

(i) Why are river models distorted?

(ii) List the various types of distortions that are given to these models.

(ii) River models are built on the basis of equality of Froude number. Comment.

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16.5 ILLUSTRATIVE EXAMPLES Hydraulic Models

Example 16.1 :

I A geometrically similar model of a spillway discharges 0.10 m3/s discharge per metre length of the spillway at a head of 0.14 m. If the scale ratio is 10.0, determine

I the prototype head and discharge per metre length of the spillway.

Solution :

According to Froude's law

Since

As

Also

I Example 16.2 : A ship model 1.0 m long with negligible frictional resistance is tested in a towing tank at a velocity of 0.60 m/s. To what ship velocity does this comespond to if the ship is 60.0 m long. A force of 5.0 N is required to tow this model. Calculate the force required in the prototype.

Solution :

Since frictional force is negligible, for dynamic similarity, Froude number must be same in model and prototype

.I Force ratio

F, = 1: (From Newton's second law of motion)

F~ = F, 1: = 5.0 603 = 1080 k~

16.6 SUMMARY

In this Unit, you have learnt the basic need for model studies in understanding the behaviour of flow in prototype structures. For building these models, the laws governing the phenomenon must be known, in other words, the dimensionless parameters likely to affect the flow should be known. Models are built on the basis of similarity criteria which at times becomes difficult to satisfy; this leads to what is known as scale effect in the model. The various distortions that are given to a river model were also discussed. Thus on the basis of the principles discussed in this Unit, it should be possible to finalise the various scale ratios of a given model for a given set of conditions.

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Drag, Lift and other Hydraulic Prstilems 16.7 KEY WORDS

Model

Scale Ratio

: Structure -built on the bas of similarity criteria - which represents prototype on a smaller size. Occasionally model could be larger than prototype.

: Defined as the value of a particular quantity in prototype to that in the model.

Froude's Model Law : Model built on equality of Froude number

Scale Effect : Model having some of the dimensionless parameters different from the prototype is supposed to have scale effect.

Distorted Models : Models having horizontal and vertical scales different are called distorted models.

16.8 ANSWERS TO SAQs

Check your answers of all SAQs with respective preceeding text of each SAQs.