aero sem ppt

8/10/2019 aero sem ppt

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Department of Aeronautical engineering

School of Mechanical engineering

Vel Tech Dr RR & SR Technical University

Course Material

U4MEA10- Fluid Mechanics


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U4MEA10 FLUID MECHANICS

Unit - I Basic Concepts And Properties 9Fluid definition, distinction between solid and fluid - UNIT Is and dimensions - Properties of fluids

- density, specific weight, specific volume, specific gravity, temperature, viscosity, compressibility,

vapour pressure, capillary and surface tension - Fluid statics: concept of fluid static pressure, absolute

and gauge pressures - pressure measurements by manometers and pressure gauges.

Unit II Bernoullis Equation And Boundary Layer Concepts 9

Fluid Kinematics - Flow visualization - lines of flow - types of flow - continuity equation (one

dimensional differential forms)- fluid dynamics - equations of motion - Eulers equation along a

streamline - Bernoullis equation applications - Venturi meter, Orifice meter, Pitot tube - Boundary

layer flows, boundary layer thickness, boundary layer separation - drag and lift coefficients.

Unit - III Flow Through Pipes 9

Viscous flow - Navier - Stokes equation (Statement only) - Shear stress, pressure gradient

relationship - laminar flow between parallel plates - Laminar flow through circular tubes (Hagen

poiseulles) - Hydraulic and energy gradient - flow through pipes - Darcy -weisbacks equation - pipe

roughness -friction factor-minor losses - flow through pipes in series and in parallel - power

transmission.

Unit - IV Dimensional Analysis And Hydraulic Turbines 9

Dimensional analysis - Buckinghams p theorem- applications - similarity laws and models.Hydro

turbines: definition and classifications - Pelton turbine Francis turbine - Kaplan turbine - working

principles - velocity triangles - work done - specific speed - efficiencies -performance curve forturbines.

Unit - V Pumps 9

Pumps: definition and classifications - Centrifugal pump: classifications, working principles, velocity

triangles, specific speed, efficiency and performance curves - Reciprocating pump: classification,

working principles, indicator diagram, work saved by air vessels and performance curves -

cavitations in pumps -priming- slip- rotary pumps: working principles of gear, jet and vane pump.

Text Books

11.. 1.Streeter, V.L., and Wylie, E.B., Fluid Mechanics, McGraw-Hill, 1983.

22.. Kumar, K.L., Engineering Fluid Mechanics, Eurasia Publishing House (P) Ltd., New Delhi

(7thedition), 1995.

33.. Bansal, R.K., Fluid Mechanics and Hydraulics Machines, (5th edition), Laxmi publications (P)

Ltd., New Delhi, 1995.

Reference Books

11..

White, F.M., Fluid Mechanics, TataMcGraw-Hill, 5thEdition, New Delhi, 2003.

22..

Ramamirtham, S., Fluid Mechanics and Hydraulics and Fluid Machines, Dhanpat Rai and

Sons, Delhi, 1998.

33.. Som, S.K., and Biswas, G., Introduction to fluid mechanics and fluid machines, Tata

McGraw-Hill, 2nd

edition, 2004.


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UNIT I

Fluiddefinition, distinction between solid and fluid

Units and dimensions

Properties of fluids - density, specific weight, specific volume,

specific gravity, temperature, viscosity, compressibility, vapour

pressure, capillary and surface tension

Fluid statics: concept of fluid static pressure, absolute and gauge

pressures

Pressure measurements by manometers and pressure gauges.


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BASIC CONCEPTS AND PROPERTIES

PART A

1. Define the following properties.

Density, weight density, specific volume and specific gravity of fluid

(i)Density (or) specific mass (or) Mass Density:

The mass density of a fluid is the mass which is possesses per unit volume

3Mass of fluidMass density ( )= ( / )Volume of the fluid

gm

K mv

(ii) Weight density (or) specific weight (w)

The weight density or specific weight of a fluid is the weight it posses per unit

volume.

3Weight of fluid mass Acceleration due to gravityWeight density(w) = (N/m )Volume of fluid Volume of fluid

w= g

(iii) Specific volume (v)

Specific volume is the reciprocal of specific density. The specific volume of a

fluid is the volume occupied by the unit mass of the fluid.

1

Volume of fluid 1Mass of fluidSpecific volume (v)=Mass of fluid Volume

2. Differentiate between i) ideal fluid & Real Fluid.

(ii) Specific weight and specific volume of a fluid

Ideal Fluid:

A fluid which is incompressible and is having no viscosity is known as ideal fluid.


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Ideal fluid is only an imaginary fluid as all fluids, have some viscosity.

Real Fluid:

A fluid which possess viscosity is known as real fluid

Specific Weight:

The Specific weight of a fluid is the weight it possesses per unit volume.

Weight density Weight of Fluid (w)

(or)

Specific weight Volume of Fluid (v)

w = g

Specific volume:

Specific Volume is the reciprocal of specific density. The specific volume of a

fluid is the volume occupied by the unit mass of the fluid.

3. State the Newtons law of Viscosity.

It states that the shear stress () on a fluid element layer is directly proportionalto the rate of shear strain. The constant of proportionality is called the coefficient of

viscosity.

Mathematically, it is expressed by equationdu

dy

Fluids which obey the above relation are known as Newtonian fluids

Fluids which do not obey the above relation are called Non-Newtonian Fluid.

4.

Distinguish Between Surface Tension and capillarity.

SURFACE TENSION CAPILLARITY

Surface tension is defined as the tensile

force acting on the surface of a Liquid in

contact with a gas or on the surface

between two immiscible Liquids such

that contact surface behaves like a

membrane under tension.

Capillarity is defined as a

phenomenon of rise or fall of a liquid

surface in a small tube relative to the

adjacent general level of Liquid when

the tube is held vertically in the fluid.

Surface tension is expressed in N/m (or) The rise of Liquid surface is known as


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A real fluid in which is not proportional to the rate of shear strain [(or) velocity

gradient] is known as a Non-Newtonian fluid.

8. Define the term Viscosity.

Viscosity: ()

Viscosity is defined as the property of a fluid which offers resistance to the

movement of one layer of fluid over another adjacent Layer of the fluid.

du

dy

This property is due to cohesion and interaction between molecules of the fluid.

9. State Newtons law of viscosity.

Newtons Law of viscosity:

It states that the shear stress on a fluid element layer is directly proportional to

the rate of shear strain. The constant of proportionality is called the coefficient of

viscosity.

. ;du

dy du

dy

The fluid viscosity is due to cohesion and interaction between molecules of

the fluid.

For example

High viscosity FluidTar and caster op

Low viscosity fluidKerosene, Petrol and water

10. What is the effect of temperature on viscosity of water and that of air?

When the viscosity of the liquid decreases with increase in temperature since themolecules present in the Liquid is less.


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When the viscosity of the air increases with increases in temperature.

11. Explain the importance of compressibility in fluid flow.

Compressibility is the reciprocal of the bulk modulus of Elasticity, K which is

defined as the ratio of compressive stress to volumetric strain.

Increase of PressureBulk Modulus K=

Volumetric strain

-dpV =

dv

Compressibility is given by =1/K

12. Explain the phenomenon of capillarity obtain an expression for capillary rise

of a Liquid.

Capillarity is defined as a phenomenon of rise or fall of of a Liquid surface in a

small tube relative to the adjacent general level of liquid when the tube is held

vertically in the Liquid.

The rise of liquid surface is known as capillary rise. While the fall of liquid

surface is known as capillary depression. It is expressed in terms of cm (or ) mm of

Liquid.

Expression for Capillary rise:

Consider a glass tube of small diameter d opened at both ends and is inserted

in a liquid say water, the Liquid will rise in the tube above the level of liquid.

Let h-height of the Liquid in the tube. Under a state of equilibrium, the weight of

Liquid height h is balanced by the force at the surface of the liquid in the tube. But

the force at the surface of the Liquid in the tube is due to surface tension.

Let = surface tension of liquid, =Angle of contact between liquid glass tube.

The weight of liquid of height h in the tube =(Area of tube x h) x g.


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= 2 (1)4

d h g

Where,= Density of liquidVertical component of surface tensile force

= (x Circumference) x Cos = (x d x cos) 2

Equating the equation (1) & (2) we get,

/4d2h g = x d x cos

2

0

cos/ 4

4 cos

when =

4 cosCapillary Rise of Liquid h=

dhd g

hgd

gd

13. What are the types of the fluids?

Types of the fluids:

1.

Ideal Fluid

2. Real Fluid

3. Newtonian Fluid

4. Non-Newtonian Fluid

5.

Ideal Plastic Fluid.

14. Define Pascals Law.

It states that the pressure or intensity of pressure at a point in a static fluid is

equal is equal in all directions.

Px= Py= P2

15. Define Hydraulic Law.

It states that rate of increase of pressure in a vertical direction is equal toweight density of the fluid at that point.


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P = gZ.

16. Define Manometer.

It is defined as the devices used for measuring the pressure at a point in a

fluid by balancing the column of fluid by the same or another column at the fluid.

17. Define absolute pressure.

It is defined as the pressure which is measured with reference to absolute

vacuum pressure.

18. Define Gauge pressure.

It is defined as the pressure which is measured with the help of a pressure

measuring instrument, in which the atmospheric pressure is taken as datum. The

atmospheric pressure on the scale is marked as zero.

19. Define vacuum pressure.

It is defined as the pressure below the atmospheric pressure.

20. What are the different types of mechanical gauges?

There are different types of mechanical gauges.

Mechanical gauges are best suitable for measuring very high fluid pressure.

Incase of steam boilers where manometer can not be used, a mechanical gauge can

be conveniently used.

1. Bourdon tube pressure gauge

2.

Diaphragm pressure gauge

3. Dead weight pressure gauge.

21. What are units and dimension?

S.No Quantity Unit DIMENSIONS


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generally

adopted

MLT SYSTEM FLT SYSTEM

Geometric1. Length M L L

2. Area M2 L2 L2

3. Volume M3 L3 L3

4. Slope

Kinematic

5 Time Sec T T

6 Velocity

(linear)

M/sec LT-1 LT-1

7 Velocity(angular)

Rad / sec2 T-1 T-1

8 Acceleration

(linear)

M/sec LT-2 LT-2

9 Acceleration

(angular)

Rad /sec2 T2 T2

10 Discharge Cum /sec L3T-1 L3T-1

11 Gravitationa

l

acceleration

M/sec2 LT-2 LT-2

12 Kinematic

velocity

M/sec2 L2T-1 L2T-1

Dynamic

13 Mass Kg M FL-1T2

14 Force Newton MLT-2 F

15 Weight Newton MLT-2 F

16 Mass

density

Kg /cum ML-3 FL-4T2

17 Specificweight

Newton/cum

ML-2T2 FL-3

18 Dynamic

viscosity

Newton

/cum

ML-1T-1 FL-3T

19 Surface

tension

Newton/m MT-2 FL-1

20 Elastic

modulus

Newton/m2 ML-1T2 FL-2

21 Pressure Newton/m2 ML-1T2 FL-2

22 Shearintensity

Newton/m2 ML-1T2 FL-2


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23 Work,

energy

Newton m ML2T2 FL

24 Impulsemomentum

Newton sec MLT-1 FT

25 Torque Newton m ML2T-2 FL

26 Power Newton /sec ML2T3 FLT-1

22. Differentiate between fundamental units and derived units.

The fundamental or primary units are the simplest in their form possessing a

single dimension. When the units of measurements of the primary quantities are

defined, the measurements of all other quantities can be easily obtained.Example: Length (L), Time (T), Mass (M), Temperature ()

The derived secondary quantities possess more than one dimension, and are

expressed by a combination of dimensions.

Example:Velocity (LT-1), linear acceleration (LT-2), force (MLT2) etc.


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(ii) Specific weight (w)

Using equation (1.1) w = xg= 700x9.81 N/m3= 6867 N/m3.

We know that specific weightWeight

Volume

or 68670.001 0.001

W Ww or

6867 0.001 6.867 .W N

3. A flat plate area 1.5x106mm2is pulled with a speed of 0.4 m/s relative to another

plate located at a distance of 0.15 mm from it. Find the force and power required

to maintain this speed, if the fluid separating them is having viscosity as 1 poise.

Solution:

Given:

Area of the plate, A = 1.5 x 106mm2= 1.5m2

Speed of plate relative to another plate, du = 0.4 m/s

Distance between the plates, dy = 0.15 mm = 0.15 x 10-3m

Viscosity = 1 poise2

1.

10sN

m

Using equation (1.2), we have

3 2

1 0.4266.66

10 015 10

du N

dy m

(i) Shear force, F= x area =266.66 x 1.5 = 400 N.

(ii) Power* required to move he plate at the speed 0.4 m/sec

= F x u = 400 x 0.4 = 160 W.

4. Calculate the dynamic viscosity of an oil, which is used for lubrication between

a square plate of size 0.8 m x0.8 m and an inclined plane with angle of inclination

30as shown in Fig. The weight of the square plate is 300 N and it slides down

the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is

1.5 mm.

Solution:

Given:


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Area of plate, A = 0.8 x 0.8 = 0.64 m2

Angle of plane, = 30Weight of plate, W = 300 N

Velocity of plate, u = dy

Thickness of oil film, t = dy

= 1.5 mm = 1.5 x 10-3m

Let viscosity of fluid between plate and inclined plane is . Component ofweight W, along the plane = W cos 60=150 N

Thus the shear force, F, on the bottom surface of the plate = 150 N

And shears stress, 2150

/0.64

FN m

Area

Now using equation (1.2), we havedu

dy

where du = Change of velocity = u-0=u=0.3 m/s

dy = t = 1.5 x 10-3m

3

150 0.3

0.64 1.5 10

3

2150 1.5 10 1.17 / 1.17 10 11.7 .0.64 0.3

Ns m poise

5. The space between two square flat parallel plates is filled with oil. Each side of

the plate is 60 cm. The thickness of the oil film is 12.5 mm. The upper plate,

which moves at 2.5 meter per see requires a force of 98.1 N to maintain the speed.

Determine:

(i) the dynamic viscosity f the oil in poise, and

(ii) the kinematic viscosity of the oil in stokes if the specific gravity of the

oil is 0.95. (AMIE, Winter 1977)

Solution:

Given:

Each side of a square plate = 60 cm = 0.60 m


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Area, A = 0.6 x 0.6 = 0.36 m2

Thickness of oil film, dy = 12.5 mm = 12.5 x10-3m

Velocity of upper plate, u = 2.5 m/sec

Change of velocity between plates, du = 2.5 m/sec

Force required on upper plate, F = 98.1 N

Shear stress, 298.10.36Force F N Area A m

(i) Let = Dynamic viscosity of oil

Using equation (1.2),3

98.1 2.5

0.36 12.5 10

duor

dy

3

2 2

98.1 12.5 10 1Ns1.3635 10

0.36 2.5 m

1.3635 10 13.635 .

sN poisem

poise

(ii) Sp. gr. of oil, S = 0.95

Let =kinematic viscosity of oil

Using equation (1.1 A),

Mass density of oil, = S x1000 = 0.95 x1000 = 950 kg/m3

Using the relation,2

1.3635,

950

sNm

we get

= .001435 m2/sec = .001435 x 104cm2/s

= 14.35 stokes. (cm2/s = stoke)

6. The dynamic viscosity of an oil, used for lubrication between a shaft and sleeve

is 6 poise. The shaft is diameter 0.4 m and rotates at 190 r.p.m. Calculate thepower lost in the bearing for a sleeve length of 90 mm. The thickness of the oil


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film is 1.5 mm.

Solution:

Given:

Viscosity = 6 poise

=2 2

60.6

10s sN N

m m fig

Dia. of shaft, D = 0.4 mSpeed of shaft, N = 190 r.p.m.

Sleeve length, L = 90 mm = 90 x 10-3m

Thickness of oil film, t = 1.5 mm = 1.5 x 10-3m

Tangential velocity of shaft,0.4 190

3.98 /60 60

DNu m s

Using the relationdu

dy

Where du = Change of velocity = u 0 = u = 3.98 m/s

dy = Change of distance = t = 1.5 x 10-3m

2

3

3.9810 1592 /

1.5 10N m

This is shear stress on shaft

Shear force on the shaft, F = Shear stress x Area= 1952 x D x L = 1592 x x .4 x 10-3= 180.05N

Torque on the shaft,2

DT Force

0.4180.05 36.012

Nm

Power lost2 2 190 36.01

716.48 .60 60

NTW

7. A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 N

s/m

2

and specific gravity 0.9. A metallic plate 1.2 m x 0.2 cm is to be lifted up witha constant velocity of 0.15 m/sec, through the gap. If the plate is in the middle of


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the gap, find the force required. The weight of the plate is 40 N.

Solution:

Given:

Width of gap = 2.2 cm, viscosity, = 2.0 N s/m2Sq.gr. of fluid = 0.9

Weight density of fluid = 0.9 x 1000 = 900 kgf/m3= 900 x 9.81 N/m3(1 kgf = 9.81 N)

Volume = 1.2 m x 1.2 m x 0.2 cm

= 1.2 x 1.2 x .002 m3

= .00288 m3

Thickness of plate = 0.2 cm

Velocity of plate = 0.15 m /sec

Weight of plate = 40 N.

When plate is in the middle of the gap, the distance of the plate from vertical

surface, of the gap

2.2 0.21 .01 .

2 2

Width of gap thickness of platecm m

Now the shear force on the left side of the metallic plate,

F1 = Shear stress x Area

21

0.152.0 1.2 1.2 1.2 1.2

.01

43.2 .

duArea N Area m

dy

N

Similarly, the shear force on the right side of the metallic plate,

F2 = Shear stress x Area.15

2.0 1.2 1.2 43.2.01

N

Total shear force = F1 + F2 = 43.2 + 43.2 = 86.4 N.

In this case the weight of plate (which is acting vertically downward) and

upward thrust is also to be taken into account.

The upward thrust = Weight of fluid displaced= (Weight density of fluid) x Volume of fluid displaced


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= 9.81 x 900 x .00288 N

.00288Volume of fluid displaced Volume of plate

= 25.43 N.The net force acting in the downward direction due to weight of the plate and

upward thrust

= Weight of plate - Upward thrust

= 40-25.43 = 14.57 N

Total force required to lift the plate up= Total shear force + 14.57 = 86.4+14.57 = 100.97 N.

8. The surface tension of water in contact with air at 20C is 0.0725 N/m. Thepressure inside a droplet of water is to be 0.02 N/cm2 greater than the outside

pressure. Calculate the diameter of the droplet of water.

Solution:

Given:

Surface tension, = 0.0725 N/m

Pressure intensity, p in excess of outside pressure is2 4

20.02 / 0.02 10

Np N cm

m

Let d = dia. of the droplet

Using equation (1.14), we get 44 4 0.0725

0.02 10p ord d

4

4 0.0725.00145 .00145 1000 1.45 .

0.02 10d m mm

9. Find the surface tension in a soap bubble of 40 mm diameter when the inside

pressure is 2.5 N/m2above atmospheric pressure.

Solution:

Given:

Dia. of bubble, d = 40 mm = 40 x 10-3m

Pressure in excess of outside, p = 2.5 N/m2

For a soap bubble, using equation (1.15), we get

38 8 2.5

40 10p or

d


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32.5 40 10/ 0.0125 / .

8N m N m

10. The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm2

(atmospheric pressure). Calculate the pressure within the droplet if surface

tension is given as 0.0725 N/m of water.

Solution:

Given:

Dia. of droplet, d = 0.04 mm = .04 x 10-3m

Pressure outside the droplet = 10.32 N/cm2

= 10.32 x 104N/m2

Surface tension, = 0.0725 N/m

The pressure inside the droplet, inn excess of outside pressure is given by

equation

or 23

4 4 0.07257250 /

.04 10p N m

d

2

4 2

72500.725 /

10

NN cm

cm

Pressure inside the droplet = p + Pressure outside the droplet

= 0.725 + 10.32 = 11.045 N/cm2.

11. Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed

vertically in (a) water and (b) mercury. Take surface tension = 0.0725 N/m forwater and =0.52 N/m for mercury in contact with air. The specific gravity for

mercury is given as 13.6 and angle of contact = 130.

Solution:

Given:

Dia. of tube, d = 2.5 mm = 2.5 x 10-3m

Surface tension, for water = 0.0725 N/m

for mercury = 0.52 N/mSp. gr. of mercury = 13.6


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Density = 13.6 x 1000 kg/m3.(a)Capillary rise for water (= 0)

Using equation (1.20), we get3

4 4 0.0725

1000 9.81 2.5 10h

g d

= .0118 m = 1.18 cm.

(b)For mercury

Angle of constant between mercury and glass tube, = 130

Using equation (1.21), we get3

4 cos 4 0.52 cos130

13.6 1000 9.81 2.5 10

h

g d

= - .004 m = -0.4 cm.

The negative sign indicates the capillary depression.

12. Calculate the capillary effect in millimeters in a glass tube of 4 mm diameter,

when immersed in (i) water, (ii) mercury. The temperature of the liquid is 20C

and the values of the surface tension of water and mercury at 20C in contact with

air are 0.073575 N/m respectively. The angle of contact for water is zero that for

mercury. 1.30. Take density of water at 20C as equal to 998 kg/m3. (U.P.S.C.

Engg. Exam., 1974)

Solution:

Given:

Dia of tube, d = 4 mm = 4 x 10-3m

The capillary effect (i.e., capillary rise or depression) is given by equation

(1.20) as

4 cosh

p g d

where = surface tension in kgf/m

= angle of contact, and = density(i) Capillary effect for water

= 0.073575 N/m, = 0= 998 kg/m3at 20C


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3

4 0.073575 cos0

13600 9.81 10h

= - 2.45 x 10-3m = - 2.46 mm.

The negative sign indicates the capillary depression.

13. Find out the minimum size of glass tube that can be used to measure water

level if the capillary rise in the tube is to be restricted to 2 mm. Consider surface

tension of water in contact with air as 0.073575 N/m. (Converted to SI Units,

A.M.I.E., Summer 1985)

Solution:

Given:

Capillary rise, h = 2.0 mm = 2.0 x 10-3m

Surface tension, = 0.073575 N/mLet dia. of tube = d

The angle for water = 0The density for water, = 1000 kg/m3Using equation (1.20), we get

34 4 0.073575 2.0 101000 9.81

h org d d

3

4 0.0735750.015 1.5 .

1000 9.81 2 10d m cm

Thus minimum diameter of the tube should be 1.5 cm.

14. A hydraulic press has a ram of 30 cm diameter and a plunger of 4.5 cm

diameter. Find the weight lifted by the hydraulic press when the force applied at

the plunger is 500 N.

Solution:

Given:


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Dia. of ram, D = 30 cm 0.3 m

Dia. of plunger, d = 4.5 cm = 0.045 mForce on plunger, F = 500 N

Find weight lifted = W

Area of ram, 22 20.3 0.07068

4 4A D m

Area of plunger, 22 20.045 .00159

4 4a d m

Pressure intensity due to plunger

2 500 / . .00159

Force on plunger FN m

Area of plunger a

Due to Pascals law, the intensity of pressure will be equally transmitted in all

directions. Hence the pressure intensity at the ram

2500 314465.4 /.00159

N m

But pressure intensity at ram

2/ .07068

Weight W WN m

Area of ram A

314465.4.07068

W

Weight = 314465.4 x .07068 = 22222 N = 22.222 kN.

15. The diameters of a small piston and a large piston of a hydraulic jack at 3 ate 3

cm and 10 cm respectively. A force of 80 N is applied on the small piston. Find

the load lifted by the large piston when:

(a)

The piston is 40 cm above the large piston.

(b)Small piston is 40 cm above the large piston.

The density of the liquid in the jack is given as 1000 kg/cm3.


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Solution:

Given:

Dia. of small piston, d = 3 cm

Area of small piston, 22 23 7.068

4 4a d cm

Dia. of large piston, D = 10 cm

Area of larger piston, 2 210 78.54

4

pA cm

Force on small piston, F = 80 NLet the load lifted = W.

(a)When the piston are at the same level.

Pressure intensity on small piston

280 /7.068

FN cm

a

This is transmitted equally on the large piston.

Pressure intensity on the large piston80

7.068

Force on the large piston = Pressure x Area80

78.54 888.96 .7.068

N N

(b)When the small piston is 40 cm above the large piston.

Pressure intensity on the small piston

2807.068F Na cm

Pressure intensity at section A A

Pr int 40 . F

essure ensity due to height of cm of liquida

But pressure intensity due to 40 cm of liquid

= x g x h = 1000 x 9.81 x 0.4 N/m22 2

4

1000 9.81 .40/ 0.3924 / .

10N cm N cm

Pressure intensity at section

80 0.39247.068

A A


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= 11.32 + 0.3924 = 11.71 N/cm2

Pressure intensity transmitted to the large piston = 11.71 N/cm2

Force on the large piston = Pressure x Area of the large piston= 11.71 x A = 11.71 x 78.54 = 919.7 N.

16. What are the gauge pressure and absolute pressure at a point 3 m below the

free surface of a liquid having a density of 1.53 x 103 kg/m3 if the atmospheric

pressure is equivalent to 750 mm of mercury? The specific gravity of mercury is

13.6 and density of water = 1000 kg/m3. (A.M.I.E., Summer 1986)

Solution:

Depth of liquid, Z1= 3 m

Density of liquid, 1= 1.53 x 103kg/m3Atmospheric pressure head, Z0= 750 mm of Hg

7500.75

1000m of Hg

Atmospheric pressure, patm = 0x g x Z0where 0 = density of Hg = 13.6 x 1000 kg/m3and Z0 = Pressure head in teams of mercury.

patm = (13.6 x 1000) x 9.81 x 0.75 N/m2 ( Z0 = 0.75)

= 100062 N/m2

Pressure at a point, which is at a depth of 3 m from the free surface of the

liquid is given by,

p = 1 x g x Z1= (1.53 x 1000) x 9.81 x 3 = 45028 N/m2

Gauge pressure, p = 45028 N/m2.Now absolute pressure = Gauge pressure + Atmospheric pressure

= 45028 + 100062 = 145090 N/m2.

17. The right limb of a simple U-tube manometer containing mercury is open to

the atmosphere while left limb is connected to a pipe in which a fluid of sp. Gr.

0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the

right limb. Find the pressure of fluid in the pipe if the difference of mercury level

in the two limbs is 20 cm.

Solution:

Given:


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Sp.gr. of fluid, S1= 0.9

Density of fluid, 1= S1x 1000 = 0.9 x 1000 = 900 kg/m3

Sp.gr. of mercury, S2= 13.6Density of mercury, 2= 13.6 x 1000 kg/m3Difference of mercury level h2= 20 cm = 0.2 m

Height of fluid from A-A, h1= 20 12 = 8 cm = 0.08 m

Let p = Pressure of fluid in pipe

Equating the pressure above A-A, we get

p + 1gh1 = 2gh2or p + 900 x 9.81 x 0.08 = 13.6 x 1000 x 9.81 x .2

p = 13.6 x 1000 x 9.81 x .2 9.81 x 0.08

= 26683-706 = 25977 N/m2

= 2.597 N/cm2

.

18. A simple U-tube manometer containing mercury is connected to a pipe in

which a fluid of sp. gr. 0.8 and having vacuum pressure is flowing. The other end

of the manometer is open to atmosphere. Find the vacuum pressure in pipe, if thedifference of mercury level in the two limbs is 40 cm and the height of fluid in the

left from the centre of pipe is 15 cm below.

Solution:

Given:

Sp.gr. of fluid, S1= 0.8,

Sp.gr. of mercury, S2= 13.6

Density of fluid, 1= 800Density of mercury, 2= 13.6 x 1000

Difference of mercury level, h2= 40 cm = 0.4 m. Height of liquid in left limb,

h1= 15 cm = 0.15 m. Left the pressure in pipe = p. Equating pressure above datum

line A-A, we get

2gh2 + 1gh1+ p = 0 p = -[2gh2+ 1gh1]

= - [13.6 x 1000 x 9.81 x 0.4 + 800 x 9.81 x 0.15]


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= - [53366.4 + 1177.2] = - 54543.6 N/m2= - 5.454 N/cm2.

19. A single column manometer is connected to a pipe containing a liquid of sp.gr.0.9 as shown in Fig. Find the pressure in the pipe if the area of the reservoir is 100

times the area of the tube for the manometer reading shown in Fig. The specific

gravity of mercury is 13.6.

Solution:

Given:

Sp. gr. of liquid in pipe, S1 = 0.9 Density 1 = 900 kg/m3Sp. gr. of heavy liquid, S2 = 13.6

Density, 2 = 13.6 x 1000

100 lim

Area of reservoir A

Area of right b a

Height of liquid, h1 = 20 cm = 0.2 m

Rise of mercury in right limb

h2 = 40 cm = 0.4 m

Let PA = Pressure in pipe

Using equation (2.9), we get

2 2 1 2 2 1 1

2 2 2

10.4 13.6 1000 9.81 900 9.81 0.4 13.6 1000 9.81 0.2 900 9.81

100

0.4133416 8829 53366.4 1765.8

100

533.664 53366.4 1765.8 / 52134 / 5.21 / .

A

aP h g g h g h g

A

N m N m N cm

20. A differential manometer is connected at the two points A and B of two pipes

as shown in Fig. The pipe A contains a liquid of sp. gr. = 1.5 while pipe B contains

a liquid of sp.gr. = 0.9. The pressures at A and B are 1 kgf/cm2and 1.80 kgf/cm2


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respectively. Find the difference in mercury level in the differential manometer.

Solution:Given:

Sp. gr. of liquid at A, S1 = 1.5 1 = 1500Sp. gr. of liquid at B, S2 = 0.9 2 = 900Pressure at A, pA = 1 kgf/cm2= 1 x 104kgf/m2

= 1.8 x 9.81 N/m2 (1 kgf = 9.81 N)

Pressure at B, pB = 1.8 kgf/cm2

= 1.8 x 104

kgf/cm2

= 1.8 x 104x 9.81 N/m2 (1 kgf = 9.81 N )

Density of mercury = 13.6 x 1000 kg/m3

Taking X-X as datum line.

Pressure above X-X in the left limb

= 13.6 x 1000 x 9.81 x h + 1500 x 9.81 x (2+3) pA

= 13.6 x 1000 x 9.81 x h + 7500 x 9.81 + 9.81 x 104

Pressure above X-X in the right limb = 900 x 9.81 x (h+2) + pB

= 900 x 9.81 x (h+2) + 1.8 x 104

x 9.81Equating the two pressures, we get

13.6 x 1000 x 9.81h + 7500 x 9.81 + 9.81 x 104

= 900 x 9.81 x (h+2) +1.8 x 104x 9.81

Diving by 1000 x 9.81, we get

13.6h + 7.5 + 10 = (h+2.0) x .9 + 18

or 13.6h + 17.5 = 0.9h + 1.8 + 18 = .9h + 19.8

or (13.6-0.9)h=19.8-17.5 or 12.7h = 2.3

2.3

0.181 18.1 .12.7

h m cm

20. A differential manometer is connected at the two points A and B as shown in

Fig. At B air pressure is 9.81 N/cm2(abs), find the absolute pressure at A.

Solution:

Air pressure at B = 9.81 N/cm2

or pB = 9.81 x 104N/m2

Density of oil = 0.9 x 1000 = 900 kg/m3

Density of mercury = 13.6 x 1000 kg/m3


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Let the pressure at A is pA

Taking datum line at X-X


= 1000 x 9.81 x 0.6 + pB

= 5886 + 98100 = 103986


= 13.6 x 1000 x 9.81 x 0.1 + 900 x 9.81 x 0.2 + pA

= 13341.6 + 1765.8 + pA

Equating the two pressure head

103986 = 13341.6 + 1765.8 + pA

pA = 103986-15107.4 = 88876.8

pA = 88876.8 N/m2=2 2

88876.88.887 .

10000

N N

cm cm

Absolute pressure at A = 8.887 N/cm2.

22. Find out the differential reading h of an inverted U-tube manometer

containing oil of specific gravity 0.7 as the manometric fluid when connected

across pipes A and B as shown in Fig. below, conveying liquids of specific

gravities 1.2 and 1.0 and immiscible with manometric fluid. Pipes A and B are

located at the same level and assume the pressures at A and B to be equal.

(A.M.I.E., Winter 1985)

Solution:

Given:

Fig. shows the arrangement. Taking X-X as datum line.

Let PA = Pressure at A

PA = Pressure at B

Density of liquid in pipe A

= Sp. gr. x 1000

= 1.2 x 1000= 1200 kg/m2


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Density of liquid in pipe B

= 1 x 1000 = 1000 kg/m3Density of oil = 0.7 x 1000 = 700 kg/m3

Now pressure below X-X in the left limb.

= pA 1200 x 9.81 x 0.3 700 x 9.81 x h

Pressure below X-X in the right limb

pA 1200 x 9.81 x 0.3 700 x 9.81 x h = pB 1000 x 9.81 (h+0.3)

But pA = pB (given)

-1200 x 9.81 x 0.3 700 x 9.81 x h = -1000 x 9.81 (h+0.3)

Dividing by 1000 x 9.81,1.2 x 0.3-0.7h =-(h+0.3)

or 0.3 x 1.2 + 0.7h = h+0.3 or 0.36-0.3 = h-0.7h = 0.3h

0.36 0.30 0.06

0.30 0.30h m

1 1100 20 .

5 5m cm


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UNIT II

Fluid Kinematics

Flow visualization

Lines of flow

Types of flow

Continuity equation (one dimensional differential forms)-

Fluid dynamics

Equations of motion

Eulers equation along a streamline

Bernoullis equation-applications

Venturi meter, Orifice meter, Pitot tube - Boundary layer flows,

Boundary layer thickness,

Boundary layer separationDrag and lift coefficients.


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BERNOULLIS EQUATION AND BOUNDARY LAYER CONCEPTS

PART A

1. Define Kinematics of flow.

It is defined as that branch of science which deals with motion of particles

without considering the forces causing the motion.

2. What are the methods of describing fluid flow?

The fluid motion is described by two methods they are(i) Lagrangian method and

(ii) Eulerian method

In the Langrangian Method, a single fluid particle is followed during its motion

and its velocity, acceleration , density etc are described . In case of Eulerian method

the velocity, acceleration, density pressure and density etc. are described at a point

in flow field. The eulerian method is commonly used in fluid mechanics.

3. Distinguish between; steady flow and Un steady flow

Steady flow is defined as that type flow in which the fluid characteristics like

velocity , pressure , density etc at a point do not change with time. Thus for steady

flow, mathematically.

0 0 0

0 0 0, , , ,

p0, 0

t

x y z x y z

v

t

Unsteady flow is that type of flow in which the velocity pressure and densityat a point changes with respect to time

0 0 0

0 0 0, ,, ,

0 , 0 x y zx y z

v p

t t

4. Distinguish between uniform Non uniform flows.

Uniform flow is defined as that type of flow in which the velocity at any given time

does not change with respect to space [i.e. Length of direction of the flow]


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t=constant

=0

v

s

v= Changes of velocitys= length of flow in the direction sNon-Uniform Flow:

Non-uniform flow is that type of flow in which the velocity at any given time

changes with respect to space. Thus, mathematically for non- uniform flow.

t=constant

0v

s

5. Distinguish between Laminar and Turbulent flow

Laminar flows is defined as that type of flow in which the fluid particles

move along well defined path or stream line and all the stream lines are straight

and parallel. Thus, the particles move in laminas or layers gliding smoothly over the

adjacent layer. This type of flow is called stream line flow or viscous flow.

Turbulent flow is that type of flow in which the fluid particles move in a Zig-Zag way. Due to the movement of fluid particles in a Zig-Zag way, the eddies

formation takes place which are responsible for high energy loss.

6. Distinguish between compressible and in compressible flow.

Compressible flow is that type of flow in which the density of the fluid

changes from point to point ie. density is not constant for the fluid. Thus

mathematically, for compressible flow.constant

In compressible flow is that type of flow in which the density is constant for

the fluid flow. Liquids are generally incompressible. Mathematically for

compressible flow.constant

7. Distinguish between rotational and in rotational flow

Rotational flow is that type of flow in which the fluid particles while flowing

along stream lines, also rotate about their own axis . And if the fluid particles while


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flowing along stream Lines, do not rotate about their own axis, that type of flow is

called irrotational flow.

8. Define the equation of continuity obtain an expression for continuity equation

for three dimensional flow.

According to Law of Conservation of mass. Rate of flow at section 1-1 =rate of flow

at section 2.2

1A1V1=2A2V2

The above equation is applicable to compressible as incompressible fluids and

is called continuity equation. If the fluid is incompressible then 1 = 2and continuityequation reduces to.

A1V1 = A2V2

9. Explain the term local Acceleration and convective Acceleration.

Local acceleration: is designed as the rate of increase of velocity with respect

to time at a given point in a flow field.

, , is known as local accelerationu v w

t t t

Convective Acceleration:

It is defined as the rate of change of velocity due to the change of position of fluid

particles in a fluid flow.

10. Type of flow line

Path line

Stream line

Streak line or filament lines

Potential lines or Equi-potential lines

Flow net

11. Explain the terms:

(i) Path Line

(ii) Stream Line


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Path Line:

A path line time is defined as , the path or line traced by a single particle offluid during a period of time. Path line shows the direction of velocity of the same

fluid.

Stream Line:

This is an imaginary curve drawn through a flowing fluid in such a way that

the tangent of which at any point . the pattern of flow of fluid may be represented by

a series of stream Lines obtained by drawing a series of curves into the following

fluid such that the velocity vector at any point is tangential to the curves.

12. Define Equipotential line.

A line along which the velocity potential is constant, is called equipotential line.

13. Define flow net.

A grid obtained by drawing a series of equipotential lines and stream lines is

called Flow net.

14. What is the Eulers equation of motion? How will you obtain Bernoullis

equation from if equation of motion.

According to Newtons second law of motion, the net force Fx acting on a

fluid element in the direction of x is equal to mass m of the fluid element multiplied

by the acceleration ax in the x direction

Fx= M.ax

(i)

FgGravity force

(ii) FpThe pressure force

(iii) FvForce due to viscosity

(iv)

Ftforce due to turbulence

(v) Fcforce due to compressibility

Thus in equation, the net force,


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Fx = (Fg)x+ (Fp)x+ (Fv)x+ (Ft)x+ (Fc)x

(i) If the force due to compressibility, Fc is negligible, the resulting net force F x= (Fg)x

+ (Fp)x + (Fv)x + (Ft)x and equation of motions are called Reynolds equation ofmotion.

(ii) For flow, where Ft is negligible the resulting equations of motion are known as

Navier Stoles equation

(iii) If the flow is assumed to be ideal, viscous force (Fp) is zero and equation of

motions are known as Eulers equation of motion.

15. State the Bernoullis theorem?

It states that in a steady ideal flow of an incompressible fluid, the total energy

at any point of the fluid is constant. The total energy consists of pressure energy,

kinetic energy and potential energy or datum energy. These energies per unit weight

of the fluid are;

Pressure energy = pg

pw

Kinetic energy =

2

2

v

g

Datum energy = z

So,2p V

+Z = constant 2g

16. What is a venturi meter?

A venture meter is a device used for measuring the rate of flow of a fluid flowing

through a pipe. It consists of three parts;(i) A short converging part

(ii)

Throat and

(iii) Diverging part

17. What is a orifice meter?

It is a device used for measuring the rate of flow of fluid through a pipe. It is

a cheaper device as compared to venturi meter. It is also works on the same

principle as that of venture meter. It consists of a flat circular plate which has a

circular sharp edged hole called orifice which is concentric with the pipe.


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18. What is pitot tube?

It is a device used for measuring the velocity of flow at any point in a pipe ora channel. It is based on the principle that, if the velocity of flow at appoint becomes

zero, the pressure there is increased due to the conversion of kinetic energy into

pressure energy. In its simplest form, the pitot tube consists of a glass tube, bent at

a right angles.

19. What is the Eulers equation of motion? How will you obtain Bernoullis

equation from if equation of motion.

According to Newtons second law of motion, the net force Fx acting on a

fluid element in the direction of x is equal to mass m of the fluid element multiplied

by the acceleration ax in the x direction.

Fx= M.ax

(vi) FgGravity force

(vii)

FpThe pressure force

(viii) FvForce due to viscosity

(ix) Ftforce due to turbulence

(x) Fcforce due to compressibility

Thus in equation, the net force,

Fx = (Fg)x+ (Fp)x+ (Fv)x+ (Ft)x+ (Fc)x

1.

If the force due to compressibility, Fc is negligible, the resulting net force Fx=

(Fg)x + (Fp)x + (Fv)x + (Ft)x and equation of motions are called Reynolds

equation of motion.

2.

For flow, where Ft is negligible the resulting equations of motion are known

as Navier Stoles equation

3.

If the flow is assumed to be ideal, viscous force (Fp) is zero and equation ofmotions are known as Eulers equation of motion.

20. Types of flows

Uniform flow, non uniform flow, stream line flow, turbulent flow, steady

flow, unsteady flow, compressible flow, incompressible flow, rotational flow,

irrigational flow, one dimensional flow, two dimensional flow, three dimensional

flow etc

21. Define Drag.


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The component of the total Force (FR) in the direction of motion is called drag.

It I denoted by FD.

22. Define Lift.

The component of the total force (FR) in the direction perpendicular to the

direction of motion is known as lift. It is denoted by FL.

PART - B

1. Derive the Bernoullis equation from Eulers equation.

EULERS EQUATION OF MOTION

This is equation of motion in which the forces due to gravity and pressure are

taken into consideration. This is derived by considering the motion of a fluid

element along a stream-line as:

Consider a stream-line in which flow is taking place in S-direction. Consider

a cylindrical element of cross-section dA and length dS. The forces acting on thecylindrical element are:

Pressure force pdA in the direction of flow.

1. Pressure forcep

p ds dAs

opposite to the direction of flow.

2. Weight of element gdAds.

Let is the angle between the direction of flow and the line of action of theweight of element.

The resultant force on the fluid element in the direction of S must be equal to

the mass of fluid element acceleration in the direction S.

s

ppdA - p + ds dA gdAds cos

s

= pdAds a ..... (1)

where asis the acceleration in the direction of S.


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Now as=dv

,dt

where v is a function of s and t.

v ds v v v v ds v

s dt t s t dt

If the flow is steady,v

0t

s

v v a

s

Substituting the value of asin equation (1) and simplifying the equation, we get

p v vdsdA g dAds cos = dAds

s s

Dividing y dsdA,p v v

g cos =s s

orp v v

g cos + v 0s s

we havedz

cos =ds

1 p dz v v p g 0 or gdz vdv 0

s ds s

por gdz vdv 0 ....(2)

Equation (2) is known as Eulers equation of motion.

Bernoullis equation is obtained by integrating the Eulers equation of motion

(2) as

dpgdz vdv = constant

If flow is incompressible, is constant and


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2p v gz + constant

2

or2p v

z cons tantg 2g

or2p v

z constant .....(3)g 2g

Equation (3) is a Bernoullis equation in which

ppressure energy per unit weight of fluid or pressure Head.

g

V2/2g = Kinetic energy per unit weight or kinetic Head.

z = potential energy per unit weight or potential Head.

2. The water is flowing through a pipe having diameters 20 cm and 10 cm at

sections 1 and 2 respectively. The rate of flow through pipe is 35 litres/s. The

section 1 is 6m above datum and section 2 is 4 m above datum. If the pressure at

section 1 is 39.24 N/cm2, find the intensity of pressure at section 2.

Solution:-Given

At section 1, D1= 20 cm = 0.2 m

2 2

1

2

1

4 2

1

A .2 0.0314m4

p 39.24 N/cm

= 39.24 10 N/ m

z 6.0 m

At section 2, D2= 0.10 m

2 2

2

2

2

A 0.1 .00785 m4

z 4 m

p ?

Rate of flow, 335

Q 35lit / s 0.035 m /s.1000


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Now Q = A1V1= A2V2

11

Q 0.035V 1.114 m/sA .0314

and2

2

Q 0.035V 4.456 m/s

A .00785

Applying Bernoullis equation at sections 1 and 2, we get

2 221 1 2 2

1

p v p vz z

g 2g g 2g

or

2 24

21.114 4.456p39.24 10

6.0 4.01000 9.81 2 9.81 1000 9.81 2 9.81

or 2p

40 0.063 6.0 1.012 4.09810

or 2p

46.063 5.0129810

2

p46.063 5.012 41.051

9810

p2= 41.051 9810 N/m2

2 2

4

41.051 9810N/cm 40.27 N/cm .

10

3. Water is flowing through a pipe having diameter 300 mm and 200 mm at the

bottom and upper end respectively. The intensity of pressure at the bottom end is24.525 N/cm2 and the pressure at the upper end is 9.81 N/cm2. Determine the

difference in datum head if the rate of flow through pipe is 40 lit/s.

Solution:

Given:

Section 1, D1= 300 mm = 0.3 m

p1 = 24.525 N/cm2= 24.525 104N/m2

Section 2, D2= 200 mm = 0.2 mp2= 9.81 N/cm2= 9.81 104N/m2


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Rate of flow = 40 lit/s.

or 340

Q 0.04 m / s.1000

Now A1V1= A2V2= rate of flow = 0.04

1221

1

22 2

22

.04 .04 0.04V 0.5658 m/ s.

AD 0.3

4 4

0.566 m/s..04 .04 0.04

V 1.274 m/sA

D .24 4

Applying Bernoullis equation at (1) and(2), we get

2 221 1 2 2

1

p v p vz z

g 2g g 2g

or

24 4

1 21.27424.525 10 .566 .566 9.81 10z z1000 9.81 2 9.81 1000 9.81 2 9.81

or 25 + .32 + z1= 10 + 1.623 + z2

or 25.32 + z1= 11.623+ z2

z2z1= 25.32 11.623 = 13.697 = 13.70 m

Difference in datum head = z2z1= 13.70 m.

4. Explain the working principle of venturimeter.

Venturimeter. A venturimeter is a device used for measuring the rate of a flow of a

fluid flowing through a pipe. It consists of three parts:

(i) A short converging part, (ii) Throat, and (iii) Diverging part. It is based on

the Principle of Bernoullis equation.

Expression for Rate of Flow through Venturimeter


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Consider a venturimeter fitted in a horizontal pipe through which a fluid is

flowing (say water), as shown in figure.

Let d1= diameter at inlet or at section (1),

p1= pressure at section (1)

v1= velocity of fluid at section (1),

a = area at section (1) = 21d

4

and d2, p2, v2, a2are corresponding values at section (2).

Applying Bernoullis equation at sections (1) and (2), we get

2 2

1 1 2 21 2

p v p vz z

g 2g g 2g

As pipe is horizontal, hence z1= z2

2 2 2 2

1 1 2 2 1 2 2 1p v p v p p v v

org 2g g 2g g 2g 2g

But 1 2p p

g

is the difference of pressure heads at sections 1 and 2 and it is equal to h

or 1 2p p

g

= h

Substituting this value of 1 2p p

g

in the above equation, we get

2 2

2 1v vh ...(1)2g 2g

Now applying continuity equation at sections 1 and 2

2 21 1 2 2 1

1

a va v a v or v

a

Substituting this value of v1in equation (1)

2

2 2

2 2 2 2 2 212 2 2 2 1 2

2 2

1 1

a v

av v a v a ah 1

2g 2g 2g a 2g a

or

2

2 12 2 2

1 2

av 2gh a a


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44

2

1 12 2 2 2 2

1 2

1 2

a av 2gh 2gh

a a a a

Discharge, Q = a2v2

1 1 22

2 2 2 2

1 2 1 2

a a aa 2gh 2gh

a a a a

(2)

Equation (2) gives the discharge under ideal conditions and is called

theoretical discharge. Actual discharge will be less than theoretical discharge.

1 2act d2 2

1 2

a aQ C 2gh ....(3)a a

where Cd= Co-efficient of venturimeter and its value is less than 1.

Value of h given by differential U-tube manometer

Case I. Let the differential manometer contains a liquid which is heavier than the

liquid flowing through the pipe. Let

Sh= sp. Gravity of the heavier liquid

So= sp. Gravity of the liquid flowing through pipex = difference of the heavier liquid column in U-tube

Then h

o

Sh x 1

S

(4)

Case II. Ifthe differential manometer contains a liquid which lighter than the liquid

flowing through the pipe, the value of h is given by

h

o

Sh x 1

S

.(5)

where S1= sp. gr. Of lighter liquid in U-tube

So= sp. Gr. Of fluid flowing through pipe

x = difference of the lighter liquid columns in U-tube.

Case III. Inclined Venturimeter with Differential U-tube manometer.

The above two cases are given for a horizontal venturimeter. This case is

related to included venturimeter having differential U-tube manometer. Let the


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differential manometer contains heavier liquid then h is given as

1 2 h1 2

o

p p Sh z z x 1g g S

.(6)

Case IV

Similarly, for inclined venturimeter in which differential manometer contains

a liquid which is lighter than the liquid flowing through the pipe, the value of h is

given as

1 2 l

1 2

o

p p S

h z z x 1g g S

. (7)

5. A horizontal venturimeter with inlet and throat diameters 30 cm and 15 cm

respectively is used to measure the flow of water. The reading of differential

manometer connected to the inlet and the throat is 20 cm of mercury. Determine

the rate of flow. Take Cd= 0.98.

Solution, Given:

Dia. at inlet, d1= 30 cm

Area at inlet, 22 2

1 1a d 30 706.85 cm

4 4

Dia, at throat, d2= 15 cm

2 22

a 15 176.7 cm4

Cd= 0.98

Reading of differential manometer = x = 20 cm of mercury.

Difference of pressure head is given by

or h

o

Sh x 1

S

where Sh= sp. gravity of mercury = 13.6, So= sp. gravity of water = 1

13.6

20 1 20 12.6 cm 252.0 cm of water1


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The discharge through venturimeter is given by

1 2d

2 2

1 2

2 2

a aQ C 2gh

a a

706.85 176.7 = 0.98 2 981 252

706.85 176.7

3

86067593.36 86067593.36

684.4499636.9 31222.9

125756125756 cm / s lit / s 125.756 lit / s.1000

6. A horizontal venturimeter with inlet diameter 20 cm and throat diameter 10 cm

is used to measure the flow of water. The pressure at inlet is 17.658 N/cm 2and the

vacuum pressure at the throat is 30 cm of mercury. Find the discharge of water

through venturimeter. Take Cd= 0.98.

Solution Given:

Dia. at inlet, d1= 20 cm

2 2

1a 20 314.16 cm

4

Dia. at throat, d2= 10 cm

2 22

a 10 78.74 cm4

2 4 2

1p 17.658 N/cm 17.658 10 N/m

for water

4

1

3

pkg 17.658 101000 and 18 m of water m g 9.81 1000

2p

30 cm of mercuryg

= -0.30 m of mercury = - 0.30 13.6 = -4.08 m of water

Differential head 1 2p p

h 18 4.08g g

= 18 + 4.08 = 22.08 m of water = 2208 cm of water

The discharge Q is given by


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47

1 2d

2 2

1 2

a aQ C 2gh

a a

2 2

3

314.16 78.540.98 2 981 2208

314.16 78.74

50328837.21165555 cm / s 165.555 lit/s.

304

7. Explain the Working Principle of Orifice meter.

Orifice Meter or Orifice Plate

It is a device used for measuring the rate of flow of a fluid through a pipe. It is

a cheaper device as compared to venturimeter. It also works on the same principle as

that of venturimeter. It consists of a flat circular plate which has a circular sharp

edged hole called orifice, which is concentric with the pipe. The orifice diameter is

kept generally 0.5 times the diameter of the pipe, though it may vary from 0.4 to 0.8

times the pipe diameter.

A differential manometer is connected at section (1), which is at a distance of

about 1.5 to 2.0 times the pipe diameter upstream from the orifice plate, and atsection (2), which is at a distance of about half the diameter of the orifice on the

down stream side from the orifice plate.

Let p1 = pressure at section (1),

v1 = velocity at section (1),

a1 = area of pipe at section 1), and

p2, v2, a2 are corresponding values at section (2). Applying Bernoullis

equation at sections (1) and (2) we get

2 2

1 1 2 21 2

p v p vz z

g 2g g 2g

or2 2

1 2 2 11 2

p p v vz z

g g 2g 2g

But 1 21 2p p

z z h Differential head

g g


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48

2 2

2 22 12 1

v vh or 2gh = v v

2g 2g

or 22 1v 2gh v ..(i)

Now section (2) is at the vena contracts and a2represents the area at the vena

contracts. If a0is the area of orifice then, we have

2c

o

aC

a

where Cc= Co-efficient of contraction

2 o ca a C .(ii)

By continuity equation, we have

0 c21 1 2 2 1 2 2

1 1

a Caa v a v or v v v

a a .(iii)

Substituting the value of v1in equation (i), we get

2 2 2

0 c 22 2

1

a C vv 2gh

a

or

2

2 2 2 2 20 02 c 2 2 c

1 1

a av 2gh C v or v 1 C 2hg

a a

22

20c

1

2ghv

a1 C

a

The discharge 2 2 2 0 c 2 0 cQ v a v a C a a C from (ii)

0 c

2

20c

1

a C 2gh ....(iv)

a1 C

a

The above expression is simplified by using


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49

2

0

1d c

2

20c

1

a1

aC C

a1 C

a

2

20c

1

c d2

0

1

a1 C

aC C

a1

a

Substituting this value of Ccin equation (iv), we get

2

20c

1

0 d2

200c

11

d 0 d 0 1

2 2 2

1 00

1

a1 C

a 2ghQ a C

aa 1 C1aa

C a 2gh C a a 2gh = ....(1)

a aa1

a

where Cd= Co-efficient of discharge for orifice meter.

The co-efficient of discharge for orifice meter is much smaller than that for a

venturimeter.

8. An orifice meter with orifice diameter 15 cm is inserted in a pipe of 30 cm

diameter. The pressure difference measured by a mercury oil differential

manometer on the two sides of the orifice meter gives a reading of 50 cm of

mercury. Find the rate of flow of oil of sp. gr. 0.9 when the co-efficient of

discharge of the meter = 0.64.

Sol. Given:

Dia. Of orifice, d0= 15 cm

Area, 2 2

oa 15 176.7 cm4

a

Dia. of pipe, d1= 30 cm


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50

Area, 2 2

1a 30 706.85 cm

4

Sp. gr. of oil S0= 0.9

Reading of Differential manometer, x = 50 cm of mercury

Differential head, g

o

S 13.6h x 1 50 1 cm of oil

S 0.9

50 14.11 705.5 cm of oil

Cd= 0.64

The rate of the flow, Q is given equation

0 1d

2 2

1 0

2 2

3

a aQ C . 2gh

a a

176.7 706.85 = 0.64 2 981 705.5

706.85 176.7

94046317.78 = 137414.25 cm / s 137.414 Litres/s.

684.4

9. A sub-marine moves horizontally in sea and has its axis 15m below the surface

of water. A pitot-tube properly placed just in front of the sub-marine and along its

axis is connected to the two limbs of a U- tube containing mercury. The difference

of mercury level is found to be 170mm. Find the speed of the sub-marine knowing

that the sp. Gr. Of mercury is 13.6 and that of sea water is 1.026 with respect of

fresh water. (A.M.I.E., Winter, 1975)

Solution. Given:

Diff. of mercury level, x=170 mm=0.17m

Sp. Gr. Of mercury, Sg=13.6

Sp. Gr. Of sea water, So=1.026


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51

g

o

S 13.6 h=x 1 0.17 1 2.0834m

S 1.026

V= 2gh 2 9.81 2.0834 6.393m / s

6.393 60 60 = km / hr 23.01km / hr.

1000

10. A pitot-tube is inserted in a pipe of 300mm diameter. The static pressure in

pipe is 100mm of mercury (vacuum). The stagnation pressure at the centre of the

pipe, recorded by the pitot-tube is 0.981 N/cm2. Calculate the rate of flow of water

through pipe, if the mean velocity of flow is 0.85 times the central velocity. Take

Cv=0.98.(Converted to S.I. Units, A.M.I.E., Summer, 1987)

Solution. Given:

Dia. of pipe, d=300mm=0.30m

2 2 2Area, a= d (.3) 0.07068m4 4

Static pressure head =100mm of mercury (vacuum)100

13.6 1.36m of water1000

Stagnation pressure =.981 N/cm2=.981 104N/m2

4 4.981 10 .981 10Stagnation Pressure head = 1m

g 1000 9.81

h=Stagnation pressure head Static pressure head

=1.0-(-1.36)=1.0+1.36=2.36m of water

Velocity at centre =C 2gh

=0.98 2 9.81 2.36 6.668m / s

Mean velocity, V 0.85 6.668 5.6678m / s

Rate of flow of water = V area of pipe

3 3 =5.6678 .07068m / s 0.4006m / s.


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52

11. A 30 cm diameter pipe, conveying water, branches into two pipes of diameters

20cm and 15cm respectively. If the average velocity in the 30cm diameter pipe is2.5 m/s, find the discharge in this pipe. Also determine the velocity in 1.5 cm pipe

if the average velocity in 20cm diameter pipe is 2m/s.

Sol. Given:

D1=30cm=0.30m

2 2 21 1 A = D .3 0.07068m

4 4

V1=2.5m/sD2=20cm=0.20m

2 2

2 A = .2 .4 0.0314m ,4 4

V2=2m/s

D3=15cm=0.15m

2 2

3 A = .15 0.225 0.01767m4 4

Find (i) Discharge is pipe 1 or Q1

(ii) Velocity in pipe of dia. cm or V3

Let Q1,Q2and Q3are discharges in pipe , and respectively

Then according to continuity equation

Q1=Q2+Q3 (1)

(i) The discharge Q1in pipe 1 is given by

(ii)

Value of V3

Q2 = A2V2=.0314 2.0=.0628 m3/s

Substituting the values of Q1and Q2in equation (1)

0.1767-0.0628=0.1139m3/s

Q3=.1767-0.0628=0.1139m3/sBut Q3=A3V3=.01767 V3or .1139=.01767 V3

3

.1139 V 6.44m / s.

.01767

12. Water flows through a pipe AB 1.2m diameter at 3m/s and then passes through


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54

The distance, perpendicular to the boundary, by which the free stream is

displaced due to the formation of boundary layer

Expression for *:

Consider the flow of a fluid having free stream velocity equal to U over a

thin smooth plate as shown in figure. At a distance x from the leading edge. The

velocity of fluid at B is zero and at C, which lies on the boundary layer, is U. Thus

velocity varies from zero at B to U at C, where BC is equal to the thickness of

boundary layer i.e.,

Distance BC =

Let y = distance of elemental strip from the plate,

dy = thickness of the elemental strip,

u = velocity of fluid at the elemental strip

b = width of plate.

Then area of elemental strip, dA = b x dy

Mass of fluid per second flowing through elemental strip

= x Velocity x Area of elemental strip= u x dA = u x b x dv (i)

If there had been no plate, then the fluid would have been flowing with a

constant velocity equal to free stream velocity (U). Then mass of fluid per second

flowing through elemental strip would have been

= x Velocity x Area =x U x b x dy (ii)

As U is more than, hence due to the presence of the plate and consequentlydue to the formation of the boundary layer, there will be a reduction in mass flowing

per second through the elemental strip.

This reduction in mass / sec flowing through elemental strip

= mass / sec given by equation (ii) mass / sec given by equation (i)

= Ubdy - ubdy = b(U-u) dy

Total reduction in mass of fluid /s flowing through BC due to plate


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55

0 0

iii

if fluid is incompressible

b U u dy b U u dy

Let the plate is displaced by a distance * and velocity of flow for the distance* is equal to the free-stream velocity (i.e., U). Loss of the mass of the fluid / secflowing through the distance *

= x Velocity x Area= x U x * x b {Area = * x b}

(iv) Equating equation (iii) and (iv), we get

0

*b U u dy U b

Canceling b to both sides, we have

0

*U u dy U

Or

0 0

0

U is constant and can1*

be taken inside the integral

* 1

U u dyU u dy

U U

udy

U

22. A plate of 600 mm length and 400 mm wide is immersed in a fluid of sp.gr. 0.9

and kinematic viscosity (v=1) 10-4m2/s. The fluid is moving with a velocity of 6

m/s. Determine (i) boundary layer thickness, (ii) shear stress at the end of the

plate, and (iii) drag force on one side of the plate.

Solution:

As no velocity profile is given in the above problem, hence Blasiuss solution

will be used.

Given: Length of plate, L = 600 mm = 0.60 m

Width of plate, b = 400 mm = 0.40 m

Sp.gr.of fluid, S = 0.9

Density, = 0.9 x 1000 = 900 kg / m3

Velocity of fluid, U = 6 m/sKinematic viscosity, v = 10-4m2/s


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56

Reynold number 44

6 0.63.6 10

10eL

U LR

v

As ReL is less than 5 x 105, hence boundary layer is laminar over the entire

length of the plate.

(i) Thickness of boundary layer at the end of the plate from Blasiuss solutions is

4ex

4

4.91, where x =0.6 m and R 3.6 10

4.91 0.60.0155 15.5

3.6 10

ex

x

R

m mm

(ii) Shear stress at the end of the plate is2 2

20 4

0.332 900 60.332 56.6 /

3.6 10eL

UN m

R

(iii) Drag force (FD) on one side of the plate is given by

21

2D DF AU C

Where CDfrom Blasiuss solution is

4

2

2

1.328 1.3280.00699

3.6 10

1

2

1900 0.6 0.4 6 .00699 A=L b=0.6 0.4

2

26.78

D

eL

D D

CR

F AU C

N

23. A flat plate 1.5m x 1.5m moves at 50 km/hour in stationary air of density 1.15

kg/m3. if the co-efficient of drag and life are 0.15 and 0.75 respectively,

determine:

i. The lift force,

ii. The drag force

iii. The resultant force, and

iv. The power required to keep the plate in motion (A.M.I.E, Winter 1997)


57/182

57

Solution : Given

Area of the plate, A = 1.5 x 1.5 = 2.25 m2

Velocity of the plane, U = 50 km/hr =50 1000

/ = 13.89m/s60 60

m s

Density of air = 1.15 kg/m3Co-efficient of drag CD= 0.15

Co-efficient of lift CL= 0.75

i) Lift Force (FL) Using equation.

2 21.15 13.89

0.75 2.25 N=187.20N Ans.2 2L LU

F C A

ii) Drag Force (FD) using equation

2 21.15 13.890.15 2.25 N=37.44N Ans.

2 2D D

UF C A

iii) Resultant Force (FR) Using equation

2 2 2 237.44 187.20 N

= 1400+35025 190.85 N

R D LF F F

iv)

Power Required to keep the Plate in Motion

Force in the direction of motion VelocityP=

1000

37.425 13.890.519 kW. Ans

1000 1000

D

kW

F UkW

24. A man weighting 90 kgf descends to the ground from an aeroplane with the

help of a parachute against the resistance of air. The velocity with which the

parachute, which is hemispherical in shape, comes down is 20 m/s. finds the

diameter of the parachute. Assume CD= 0.5 and density of air =1.25 kg/m3.

Solution, Given:

Weight of man W = 90 kgf = 90 x 9.81 N = 882.9 N (1 kgf = 9.81 N)Velocity of parachute U = 20 m/s

Co-efficient of drag CD= 0.5

Density of air = 1.25 kg/m3Let the dia, of parachute = D

Area2 2

4A D m


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58

When the parachute with the man comes down with a uniform velocity, U=20

m/s, the drag resistance will be equal to the weight of man, neglecting the weight

parachute. And projected are of the hemispherical parachute will be equal to4

d2.

Drag, FD= 90 kgf = 90 x 9.81 = 882.9 N (using equation)

2

24

2 2

2

1.25 20882.9=0.5

4 2

882.9 4 2.0D 8.9946 m

0.5 1.25 20 20

8.9946 2.999 m. Ans

D D

UF C A

D

D

25. A kite 0.8 m x 0.8 m weighing 0.4 kgf (3.924 N) assumes an angle, of 12to the

horizontal. The string attached to the kite makes an angle of 45to the horizontal.

The pull on the string is 2.5 kgf (24.525 N) when the wind is flowing at a speed of

30 km/hour. Find the corresponding co-efficient of drag and lift. Density of air is

given as 1.25 kg/m3

Solution, Given:

Projected area of kite, A = 0.8 x 0.8 = 0.64 m2

weight of kite, W = 0.4 kgf = 0.4 x 9.81 = 3.924 N

Angle made by kite with horizontal, 1= 45Pull on the string P = 2.5 kgf = 2.5 x 9.81 = 24.525 N

Speed of wind, U = 30 km/hr =30 1000

/ = 8.333 m/s60 60

m s

Density of air, = 1.25 kg/m3Drag force, FD= Force exerted by wind in the direction

of motion

(i.e. in the X-X direction)= Component of pull, P along X-X


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59

= P cos 45= 24.525 cos 45= 17.34 N

And Lift Force, FL= force excerted by wind on the kite perpendicular to thedirection of motion (i.e, along r-Y direction)

= Component of P in vertically downward

direction + Weight of kite (W)

= P sin 45+ W = 24.525 sin 45+ 3.924 N= 17.34 + 3.924 = 21.264 N.

i) Drag co-efficient (CD). using equation, we have

2

2 2

2

2 2 17.340.624. Ans.

0.64 1.25 8.333

D D

DD

UF C A

FC

A U

ii) Lift co-efficient (CL). using equation, we have

2

2 2

2

2 2 21.2640.765. Ans.

0.64 1.25 8.333

L L

LL

UF C A

FC

A U

26. A jet plane which weights 29.43 kN and having a wing area of 20m 3flies at a

velocity of 950 km/hour, when the entire delivers 7357.5 kW power. 656% of the

power is used to overcome the drag resistance of the wing. Calculate the co-

efficients of lift and drag for the wing. The density of the atmospheric air is 1.21

kg/m3.

Solution, given :

Weight of plane, W = 29.43 kN = 29.43 x 1000 N = 29430 N

Wing area, A = 20 m3.

Speed of plane U = 950 km/hr =950 1000

263.88 /60 60

m s

Engine power, P = 7357.5 kW

Power used to overcome drag

Resistance = 65% of 7357.5 =65

7357.5 4782.375

100

kW

Density of air, = 1.21 kg/m3


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60

Let CD= Coefficient of drag and CL= coefficient of lift.

Now power used in kW to

over come drag resistance =

D

D

F 263.88or 4782.375=

1000 1000

4782.375 1000F

263.88

DF U

But from equation, we have2

. .2

D D

UF C A

2

D 3

4782.375 1000 263.88 20 1.21

263.88 2

4782.375 1000 2C 0.0215. Ans.

20 1.21 263.88

DC

The lift force should be equal to weight of the plane FL= W = 29430 N

But

2 2

L

2

263.88. . or 29430 = C 20 1.21

2 2

29430 20.0349. Ans.

20 1.21 263.88

L L

L

UF C A

C


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62

PART - A

1. What do you understand by the terms major energy loss and minor energy

losses in pipes?

The loss energy in pipe is classified as major energy loss and minor energy

lossed. Major energy loss is due to friction while minor energy losses are due to

sudden expansion of pipe, sudden contraction of pipe, bend in pipe and an

obstruction in pipe.

2. How will you determine the loss of head due to friction in pipes by using (i)Darcy formula and (ii) chezys formula?

Energy loss due to friction is given by darcy formula, hfhf =24fLV

2gd

The head loss due to friction in pipe can also be calculated by Chezys

formula V = cmiWhere,

C = chezys constant

m = hydraulic mean depth = d/4V = velocity of flow

loss of head per unit lengthf

hi

L

3. Derive an expression for the loss of due to

(i) Sudden enlargement (ii) Sudden contraction of a pipe

(i) Loss of head due to sudden expansion of pipe, hc =2

1 2(V - V )

2g

V1= velocity in small pipe

V2= velocity in large pipe

(ii) Loss of head due to sudden contraction of pipe,

hc=

22

2

c

V1 -1

C 2g

Ce= Coefficient of contraction

= 0.37522V

2g[for Cc= 0.62]


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63

= 0.52

2V

2g[if value of Ccis not given]

4. Define the terms:

(i) Hydraulic gradient line

(ii) Total energy line

(i) Hydraulic gradient line:

The line representing the sum of pressure head and datum head of a flowing

fluid in a pipe with respect to some reference line is called hydraulic gradient line(H.G.L)

(ii) Total Energy line:

The line representing the sum of pressure head, datum head and velocity

head of a following fluid in a pipe with respect to some reference line is known as

total energy line [T.E.L]

5. What is a siphon? On what principle it works.

Siphon is a long bent pipe used to transfer liquid from a reservoir at a higher

level to another reservoir at a lower level, when the two reservoirs are separated by a

high level ground.

6. What is compound pipe? What will be loss of head when pipes are connected in

series?

When pipes of different lengths and different diameter are connected end to

end, (in series) it is called as compound pipe. The rate of flow through each pipe

connected is series is same.

7. Explain the terms (i) pipes in parallel (ii) equivalent pipe (iii) Equivalent size of

the pipe?

(i) Pipes in parallel:

When the pipes are connected in parallel, the loss of head in each pipe is


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64

same. The rate of flow in main pipe is equal to sum of the rate of flow in each pipe,

connected in parallel.

(ii) Equivalent size of the pipe:

The diameter of equivalent pipe is called equivalent size of the pipe.

The equivalent size of the pipe is obtained from

31 2

5 5 5 5

1 2 3

LL LL

d d d d

L = Equivalent length of pipe = L1 + L2+L3 and d1, d2, d3 = are diameters of pipe

connected in series.

Equivalent size of the pipes = d

(iii) Equivalent pipe:

A single pipe of uniform diameter, having same discharge and same loss of

head as compound pipe consisting of several pipes of different lengths anddiameters is known as equivalent pipe

8. Explain the phenomenon of Water hammer?

When a liquid is flowing through a long pipe fitted with a valve at the end of

the pipe and the valve is closed suddenly a pressure wave of high intensity is

produced behind the valve. This pressure wave of high intensity is transmitted

along the pipe with sonic velocity. This pressure wave of high intensity is having

the effect of hammering action on the walls of the pipe. The phenomenon is knownas water hammer

9. Find the expression for the power transmission through pipe. What is the

condition for maximum transmission of power and corresponding efficiency of

transmission?

Let H = total head available at inlet of pipe

hf= Loss of head due to friction


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65

Efficiency of power transmission through pipes = fH - h

H

Condition for maximum transmission of power through pipe hf=3

H and maximum

efficiency = 66.67%


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66

PART B

1. Explain the flow of viscous fluid through circular pipe and derive Hagen

Poiseuilles.

FLOW OF VISCOUS FLUID THROUGH CIRCULAR PIPE

For the flow of viscous fluid through circular pipe, the velocity distribution

across a section, the ratio of maximum velocity to average velocity, the shear stress

distribution and drop of pressure for a given length is to be determined. The flow

through the circular pipe will be viscous or laminar, if the Reynolds number (Re*) is

less than 2000. The expression for Reynold number is given by

e

VDR

where = Density of fluid flowing through pipeV = Average velocity of fluid

D = Diameter of pipe and

= Viscosity of fluid.

Consider a horizontal pipe of radius R. The viscous fluid is flowing from left

to right in the pipe. Consider a fluid element of radius r, sliding in a cylindrical fluid

element of radius (r + dr). Let the length of fluid element be x. If p is the intensityof pressure on the face AB, then the intensity of pressure on face CD will be

pp x

x

. Then the forces acting on the fluid element are:

1. The pressure force, p r2on face AB.

2.

The pressure force, 2pp x rx

on face CD.

3. The shear force, 2rx on the surface of fluid element. As there is noacceleration, hence the summation of all forces in the direction of flow must

be zero i.e.

2 2pp r p x r 2 r x 0x

or 2p

x r 2 r x 0

x


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67

orp

.r 2 0x

p r ....(1)x 2

The shear stress across a section varies with r asp

x

across a section is

constant. Hence shear stress distribution across a section is linear.

(i) Velocity Distribution. To obtain the velocity distribution across a section, the

value of shear stressdu

ydy

is substituted in equation (1).

But in the relation = du

ydy

, is measured from the pipe wall. Hence

y = R r and dy = -dr

du du

dr dr

Substituting this value in (1), we get

du p r du 1 por r

dr x 2 dr 2 x

Integrating this above equation w.r.t., r, we get

21 pu r C4 x

.(2)

where C is the constant of integration and its value is obtained from the boundary

condition that at r = R, u = O.

21 p

O R C4 x

21 p

C R4 x

Substituting this value of C in equation (2), we get


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68

2 2

2 2

1 p 1 pu r R

4 x 4 x

1 p = - R r ....(3)4 x

In equation (3), values of ,p

x

and R are constant, which means the velocity,

u varies with the square of r. Thus equation (3) is a equation of parabola. This shows

that the velocity distribution across the section of a pipe is parabolic. This velocity

distribution is shown in figure (b).

(ii) Ratio of Maximum Velocity to Average Velocity

The velocity is maximum, when r = 0 in equation (3). Thus maximum velocity, Umax

is obtained as

2

max

1 pU R

4 x

.(4)

The average velocity, u, is obtained by dividing the discharge of the fluid

across the section by the area of the pipe (R2). The discharge (Q) across the section isobtained by considering the flow through a circular ring element of radius r and

thickness dr as shown in figure (b). The fluid flowing per second through this

elementary ring

dQ = velocity at a radius r area of ring element= u 2r dr

2 21 p R r 2 r dr 4 x

R R

2 2

0 0

1 pQ dQ R r 2 r dr

4 x


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69

R2 2

0

R2 3

0

R2 2 4 4 4

0

44

1 p2 R r rdr

4 x

1 p2 R r r dr

4 x

1 p R r r 1 p R R2 2

4 x 2 4 4 x 2 4

1 p R p2 R

4 x 4 8 x

Average velocity,

4

2

p RQ 8 x

uArea R

or 2p

u R ...(5)8 x

Dividing equation (4) by equation (5),

2

max

2

1 pR

U 4 x 2.01 pu

R8 x

Ratio of maximum velocity to average velocity = 2.0.

(iii) Drop of pressure for a given Length (L) of a pipe

From equation (5), we have

2

2

1 p - p 8 uu R or

8 x x R

Integrating the above equation w.r.t. x, we get


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70

1 1

22 2

8 udp dx

R

1 2 1 2 1 2 2 12 28 u 8 u - p p x x or p p x xR R

2 12

2

8 uL x x L from figure

R

8 uL D R =

2D/ 2

or 1 2 1 2232 uL

p p , where p p is the drop of pressure.D

Loss of pressure head1 2p p

g

1 2f 2

p p 32 uLh ....(6)

g gD

Equation (6) is called Hagen Poiseuille Formula.

2. A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a

horizontal circular pipe of diameter 100 mm and of length 10 m. Calculate the

difference of pressure at the two ends of the pipe, if 100 kg of the oil is collectedin a tank in 30 seconds.

Sol. Given: 20.97

0.97 poise = 0.097 Ns/m10

Relative Density = 0.9

0, or Density = 30.9 1000 900kg /m

Dia. Of pipe, D = 100 mm = 0.1 m

L = 10 m

Mass of oil collected, M = 100 kg

in time, t = 30 seconds

Calculate difference of pressure or (p1p2).

The difference of pressure (p1p2) for viscous or laminar flow is given by

1 2 2

32 uL Qp p where u averagevelocity

D Area

Now, mass of oil/sec

100

kg/s30


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0 0Q 900 Q 900

100

900 Q30

3100 1

Q 0.0037 m / s.30 900

22

Q .0037 .0037u 0.471 m/s.

AreaD .1

4 4

For laminar or viscous flow, the Reynolds number (Re) is less than 2000. Let us

calculate the Reynolds number for this problem.

Reynolds number, eVD

R

where0 900, V = u 0.471, D 0.1 m, = 0.097

e

.471 0.1R 900 436.91

0.097

As Reynolds number is less than 2000, the flow is laminar.

2

1 2 22

32 uL 32 0.097 .471 10p p N/m

D .1

= 1462.28 N/m2= 1462.28 10-4N/cm2= 0.1462 N/cm2.3. A fluid of viscosity 0.7 Ns/m2 and specific gravity 1.3 is flowing through a

circular pipe of diameter 100 mm. The maximum shear stress at the pipe wall is

given as 196.2 N/m2, find (a) the pressure gradient (b) the average velocity and (c)

Reynold number of the flow.

Sol. Given:2

Ns0.7

m

Sp. gr. = 1.3

Density =3

1.3 1000 1300 kg/m Dia. Of pipe, D = 100 mm = 0.1 m

Shear stress, 0= 196.2 N/m2

Find

(i) Pressure gradient,dp

dx

(ii) Average velocity, u

(iii)

Reynold number, Re


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(i) Pressure gradient,dp

dx

The maximum shear stress (0) is given by

0

p R p D p 0.1 or 196.2 = -

x 2 x 4 x 4

2p 196.2 4

7848 N/m per mx 0.1

Pressure Gradient = -7848N/m2per m.

(ii) Average velocity, u

2 2

max max

1 1 1 p 1 pu U R U R

2 2 4 x 8 x

21 p R8 x

21 D 1

7848 .05 R = .058 0.7 2 2

3.50 m/s.

(iii) Reynold number, Re

e

u D u D u DR

v /

3.50 0.1 = 1300 650.00.

0.7

4. Calculate: (a) the pressure gradient along flow, (b) the average velocity, and (c)

the discharge for an oil of viscosity 0.02 Ns/m2 flowing between two stationaryparallel plates 1 m wide maintained 10 mm apart. The velocity midway between

the plates is 2 m/s. (Delhi University, 1982)

Sol. Given:

Viscosity, 2.02 Ns/m

Width, b = 1 m

Distance between plates, t = 10 mm = .01 m

Velocity midway between the plates, Umax= 2 m/s.


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(i) Pressure gradientdp

dx

22

max

1 dp 1 dpU t or 2.0 = - .01

8 dx 8 .02 dx

2dp 2.0 8 .02

3200 N/m per m.dx .01 .01

(ii) Average velocity (u )

Using equation maxU 3

2u

max2 U 2 2

u 1.33 m/s.3 3

(iii) Discharge (Q) = Area of flow 3u b t u 1 .01 1.33 .0133 m /s.

5. Derive Darcy-Weisbach equation.

Expression for loss of head due to Friction in pipes

Consider a uniform horizontal pipe, having steady flow as shown in figure.

Let 1-1 and 2-2 are two sections of pipe.

Let p1= Pressure intensity at section 1-1,

V1= Velocity of flow at section 1-1,

L = length of the pipe between sections 1-1 and 2-2

d = diameter of pipe,f = frictional resistance per unit wetted area per unit velocity,

hf= loss of head due to friction

and p2, V2= are values of pressure intensity and velocity at section 2-2.

Applying Bernoullis equations between sections 1-1 and 2-2,

Total head at 1-1 = Total head at 2-2 + loss of head due to friction between 1-1

and 2-2


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or2 2

1 1 2 21 2 f

p V p Vz z h

g 2g g 2g

But z1= z2pipe is horizontal

V1= V2as dia. Of pipe is same at 1-1 and 2-2

1 2 1 2f f

p p p ph or h ....(i)

g g g g

But hfis the head lost due to friction and hence intensity of pressure will be

reduced in the direction of flow by frictional resistance.

Now frictional resistance = frictional resistance per unit wetted area per unit

velocity wetted area velocity2

or F1= f dL V2 [ wetted area = d L velocity = V = V1= V2]= f P L V2 [ d = Perimeter = P] .(ii)

The forces acting on the fluid between section 1-1 and 2-2 are:

1.

Pressure force at section 1-1 = p1A

where A = Area of pipe

2.

Pressure force at section 2-2 = p2A

3. Frictional force F1as shown in figure.

Resolving all forces in the horizontal direction, we have

p1A p2A F1 = 0 (1)

or 2 2

1 2 1 1p p A F f ' P L V [ From (ii), F f 'PLV ]

or2

1 2

f ' P L Vp p

A

But from equation (i), p1p2= ghf

Equation the value of (p1p2), we get

2

f

f ' P L Vgh

A


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or 2ff ' P

h L V ...(iii)g A

In equation (iii),2

P Wetted perimeter d 4

A Area dd

4

2

2

f

f ' 4 f ' 4LVh L V ...(iv)

g d g d

Puttingf ' f

,g 2

where f is known as co-efficient of friction.

Equation (iv), becomes as2 2

f4.f LV 4f. L. Vh . ...(2)2g d d 2g

Equation (2) is known as Darcy-Weisbach equation. This equation is commonly used

for finding loss of head due to friction in pipes.

Sometimes

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