Fluid Mechanics 1

TABLE OF CONTENTS

Introduction: Flow parameters and useful conversion factors

Chapter 1: Statics

Chapter 2: Hydrostatic forces

Chapter 3: Pipeline flows

Chapter 4: Flow measurement

Chapter 5: Forces in pipe systems

Chapter 6: Pumps and pump selection

Chapter 7: Dimensional analysis

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The study of fluid flow requires a good understanding of the parameters which define the character of the flow as well as some fundamental laws such as the conservation of Mass, Momentum and Energy. Both liquids and gases can be dealt with using the same theory but only if the gases are in the incompressible part of the flow regime. i.e. the density is constant everywhere. It is also assumed, in this course, that all properties or parameters are constant with time (Steady State).

Frequently problems in Fluids are not easily expressed in equation form (analytical solutions) and this stems mainly from the presence of viscosity in the flow as well as the turbulent behaviour of the flow. These concepts will be expanded later on in the degree.

This course will rely entirely on the Metric system and therefore the standard SI units (System Internationale) i.e. m (Metre), kg (Kilogram) and s (Second) will be used throughout. However, a set of tables have been provided at the end of this section to allow the British system of units to be converted to SI units.

Common flow variables are,

Pressure…….Force per unit area. i.e. N/m2 note that 1 atm= 1 bar = approximately 105 N/m2 where 1 Pa is 1 N/m2 . Thus 1 bar =100kPa. It should be noted that differences in pressure from one point in a flow to another, drive the movement of fluid and hence a thorough conceptual understanding of pressure and pressure gradients is vital. There is a further categorisation of pressure into gauge pressure or absolute pressure.

Gauge pressure implies a pressure measured above the surrounding Atmospheric pressure. In the case of a transducer measuring gauge pressure, one side of the diaphragm will be vented to atmosphere whilst the other side will then be exposed to the pressure to be measured. Such gauges are common and relatively inexpensive.

Absolute pressure, on the other hand, is pressure measured with respect to the absolute zero pressure condition. An absolute pressure gauge (they are rare) therefore requires, for example, that one side of a measuring diaphragm is evacuated to a perfect vacuum. The other side will then be exposed to the pressure to be measured.These gauges are expensive due to the need to create the vacuum in the capsule and to ensure it is maintained without leaking.

Static pressure would correspond to the pressure that is measured in a stationary liquid. To ensure that the static pressure in a moving flow is not influenced by a component of the Dynamic pressure, the static pressure tappings are drilled through the container/pipe walls at right angles to the material of the wall.

Dynamic pressure. This takes into account that there is a component of energy (the Kinetic energy) which can be recovered as an increase in Static pressure should one bring the flow to a standstill. The Dynamic pressure is expressed as a product i.e.

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………………where V is the flow velocity in m/s and the density of the flow.

Total pressure is given by the sum of the Static and Dynamic pressures i.e.

Density. The term ρ, (rho is the Greek symbol reserved for density) is commonly used to denote the density of a fluid and is a measurement of the mass/ per unit volume i.e. kg/m3.

i.e. fresh water has ρ= 1000 kg/m3. Air ρ= 1.2 kg/m3. Hydrogen ρ= .083 kg/m3. Mercury ρ= 13600 kg/m3. Steel ρ= 7800 kg/m3.

In fluids the density is nearly always constant i.e. they are incompressible. Gases on the other hand are compressible and when the pressure is suddenly relieved or released, this leads to a massive expansion of the stored gas. This is akin to the release of the energy stored in a spring. For this reason Pressure vessels are often tested to determine their compliance with a pressure rating by pressurising them with water; an incompressible liquid which will not project steel plate in all direction should a rupture occur!!!!.

Specific Gravity or SG this is simply a ratio of the density of the liquid under consideration to the density of water. In other words the SG of water is 1 whilst that of Mercury is 13.6

Specific Weight this is the density of the material multiplied by the gravitational constant g. It is noted that Force is the product of mass and acceleration hence by multiplying the density by gravity, we are, in effect, converting a “mass density” to a “force density”. Specific weight will be used infrequently but it is as well to remember that mass is reserved as a descriptor for the amount of matter in an object or flow whilst weight is reserved for use as a parameter indicating force i.e. mass*acceleration .

The above parameters are primary characteristics of a fluid. There are a number of other parameters which are extremely useful but stem from a combination of the above such as the Reynolds number, Viscosity etc. These are often referred to as secondary variables.

Viscosity A simple qualitative explanation of viscosity would be its “stickiness”. i.e. swimming across a pool of treacle would require an enormous amount of power compared to a swim in water yet the densities might not be too different. The difference, however, lies in the swimmer requiring a lot of energy to shear throughthe treacle i.e. the molecules are tightly attracted to each other. If we were to heat the treacle, the swim would be easier as the viscosity would drop. Hence viscosity is usually strongly dependent on temperature.

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In equation form,

where, τ, is the shear stress and, μ, the coefficient of viscosity and the

derivative is the rate of change of velocity with distance, y, from say a surface.

The following viscosity table is appropriate for conditions at 20 Celsius and 1 atm. only.

Fluid Viscosity, ,kg/(m.s)

Density,, kg/m3.

Air 1.8E-5 1.2Water 0.001 998Alcohol 0.0012 789Mercury 0.0015 13580SAE30 oil 0.29 891Glycerine 1.5 1264

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Useful Conversion Factors Parameter Convert To Multiply by

Acceleration ft/s2 m/s2 .3048

Area ft2 m2 .092903

Density slug/ft3

lbm/ft3kg/m3

kg/m3515.3816.019

Energy BtuCalories

JoulesJoules

1055.14.1868

Force lbf Newtons 4.4482

Length ft m 0.3048

Mass SlugLbm

kgkg

14.45940.45359

Mass Flow Slug/slbm/s

kg/skg/s

14.594.45359

Power ft.lbf/shp

WattWatt

1.3558745.70

Parameter Convert To Multiply byPressure lbf/ft2 Pa 47.880

Viscosity lbf.s/ft2

g/(cm.s)N.s/m2

N.s/m247.8800.1

Density slug/ft3

lbm/ft3kg/m3

kg/m3515.3816.019

Volume flowrate

ft3/sgal/min

m3/sm3/s

.0283176.3090 E-5

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Fluid Dynamics: Chapter 1 Statics

The earliest forms of pressure measurement relied on Manometers or U-tubes and aneroid capsules. Both devices could be used with gases and liquids, though aneroid capsules are best suited to air pressure measurements.

U-TubesThe U-tube comes in various guises i.e. tilted tube, U-tubes with two liquids, U tubes with dissimilar diameter ends etc. and examples are shown below.

In such cases the pressure that is being applied is obtained by referring to the column height of liquid displaced by the pressure difference being measured,

Pressure = gH

Where, = (rho) is the density of the liquid in the column, kg/m3

g = Gravitational acceleration, 9.81 m/s2

H = Column height of displaced liquid, m

In addition, the following laws hold, Pressures measured at the same depth in a liquid of constant density are equal

provided the fluid is at rest. The pressure acting at a point acts equally in all directions.

Aneroid Pressure GaugesAneroid pressure measuring devices rely on an evacuated corrugated box. The vacuum in the box is prevented from collapsing the box by a spring. With changing pressures, the corrugated surface experiences a force sufficient to overcome the opposing spring force and the surface moves a small distance.

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This movement is amplified by a system of levers and gears to drive a needle over a calibrated dial. There are numerous designs of aneroid barometer which employ this basic approach.

Bourdon GaugeThe Bourdon gauge is equally common and uses much the same approach. Here, however, the interior of a circular tube is exposed to the pressure of interest. An increase in pressure sees the tube trying to straighten. This movement, in turn drags a series of linkages which are coupled to gears to produce a needle rotation proportional to pressure.

If the volume surrounding the tube is evacuated to near vacuum conditions then the Bourdon gauge will record the Absolute Pressure.

Venting this same volume to the ambient atmospheric pressure means the pressure is relative to atmospheric, i.e. records Gauge pressure.

Often such Bourdon tubes are submerged in Glycerine. This damps out needle vibration in pulsating flows which may otherwise lead to needle failure from metal fatigue.

A common approach is to have a second indicator needle which is not attached directly to the tube but instead is dragged ahead by the main needle as the pressure increases. Conversely, it is left behind when the needle recedes. This provides a useful indication of the maximum pressures reached. The indicator needle can then be returned to the zero position by hand.

The following sketch indicates the internal workings of a typical Bourdon gauge.

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Examples of U-tube calculations

Example (1)

Consider a U tube that is measuring pressure in a water pipeline. This water comes in contact with Mercury (Hg) in the manometer as shown below. The left hand Hg meniscus is 300mm below the pipe centre line whilst the right hand meniscus is 200mm above the same line.

300mm

200mm

Hg

water

C

BA

Cl

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Determine the pipeline pressure at its centre.

H2O=1000kg/m3 ,Hg=13600kg/m3

(same level in a liquid of constant density (at rest))................................(1)

Consider the left hand limb

(pressure increases with depth)........................................(2)

Consider the right hand limb

(pressure increases with depth ),

note Pc is at atmospheric pressure i.e. Zero gauge pressure)......................................(3)

Relying on Eqtn (1) we may equate Eqtn (2) and (3) but bear in mind that the pressure at “C” = 0 if we are working with a gauge pressure reference.

Thus, ......................................................(4)

Solving yields, kN/m2

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Example 2If the angle of a U-tube manometer limb is 30 degree from the horizontal, the sensitivity of the measurement will double. Prove this to be the case.

From simple trigonometry, the length of the liquid column as measured along the tube length will double for a given vertical height change in the liquid level, i.e.

Thus liquid column length =Y/0.5 =2Y (thus proven).

Example 3The sensitivity of a U-tube manometer can also be increased by enlarging the ends of the tubes and using dissimilar liquids in the same device. Oil SG =0.95 and water SG=1 (Density=1000kg/m3) are used as shown in the sketch below with water, being the denser, gathering at the lowest point in the “U” and oil floating on the surface. Care is taken to keep the liquids in their respective limbs as shown. The diameter of the tube is 5mm whilst that of the ends is, equally 35mm giving a surface area difference of 1225/25=49 times (the ratio of the square of the radii).

Recalling that the pressure at the same level in a liquid of constant density is constant provided that the liquid is at rest, we can equate the pressure at the Oil/water interface in the right limb, B, to the same height in the left limb. Here, i.e at “A”, we may define the pressure, quite simply, as that caused by the column height of liquid above that point plus the unknown pressure applied to the free water /air surface,

If the unknown pressure that is applied to the free water surface is greater than that existing above the free surface of the oil, the water surface will ride down on the LHS whilst on the RHS the oil will rise up in the 35mm diameter tube section. The volume of fluid displaced downwards for, say, a “y” mm change in surface level is given by,

P1 P2

B

ha

Oil

Water

A

y

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This volume reappears in the RHS limb as an identical volume of oil.

Example 4The example above used equal diameters in the two expanded tube ends. If these are made different as shown in the sketch below, it is possible to produce a force amplifying device or Jack as shown below.

A force of 1000N is applied to a piston , “A” to support a mass, “m” at B. For equilibrium to occur both forces must be equal otherwise movement would occur. The force at A is 1000N and leads to a pressure beneath the piston of

Seeing the pressure at the same level in a liquid of constant density is constant, the pressure beneath the RHS piston is identical i.e. P1. However, this is acting over a much larger area and hence supports a much larger load. i.e.

Equating these two equations gives,

Clearly this force is significantly greater than the applied force of 1000N and the difference in piston areas has acted so as to amplify the applied force. In a jack, though, it will be necessary to provide a practically acceptable displacement (stroke) of the load, “mg”. If this were to be a vertical lift of 100mm then we would have to displace,

A

Mass, m

OIL

B

1000N

P1 P1

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as compared to in the LHS chamber.

Clearly then Y is much greater than 0.1m and hence a number of pumping actions are required in the LH chamber._____________________________________________________________________Tutorial Question 1Use diameters of 15mm and 60mm for the LHS and RHS chambers respectively to calculate the load that is counterbalanced by the applied force of 1000N and also determine the number of 10cc strokes needed to lift the load by 0.1metre.

Tutorial Question 2Using the dimensions of Question (1) above, calculate the mass that could be held in equilibrium if the piston supporting the mass was positioned 100mm below the piston that is subjected to the 1000N force.

Example 5

A U-tube manometer is used to measure a small pressure head of gas. The U-tube has one side enlarged. It contains oil of SG=0.87 on the enlarged side, the oil overlays fresh water of SG=1.0. The oil/water interface occurs in the small tube below the enlarged end. The gas pressure is equivalent to 40mm of water head and the surface of separation moves 35mm when the pressure is applied. Determine the diameter of the enlarged end that leads to these changes when the smaller diameter of the tube is 8mm.

When P1= P2 initially i.e. no pressure applied yet, equal pressures apply at xx i.e.

.......................................(1)

therefore, by rearranging Eqtn (1)

P1=40mm Wg

Oil

Diam 8mm

x x35mm

X

y y Y

y

P2

ah

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When pressure P1 above the oil is increased above P2 the oil/water interface drops a distance “y”=35mm in the left tube and the level in the right tube rises by a distance “y”=35mm. The level in the enlarged tube also drops but by a distance “x” such that an identical volume is displaced. This volume is given by

Volume = Area.x and in the small tubes this same volume is given by Volume=area.y

Therefore Area.x = area.y or rearranging, x = y.(area/Area)

Now referring to the sketch above, we have a new position of the oil/water interface at Y-Y and on the LHS,

on the RHS we may in a similar manner (and at the same level in the liquid) write,

But these equations represent the same pressure and may be equated to give,

But “h” and “x” were defined above and hence may be substituted in the above equation to give finally

.............................................................(2)

Now the pressure difference was given as

and y = 35mm.

Thus we may substitute in Eqtn (2) to give

392.4=0.035*9.81(2*1000-870+870*(area/Area))

solving yields D= 65.8mm diameter for a small tube diameter of d= 8mm.

Example 6 Oil of density 800kg/m3 flows in a pipeline as shown in the adjacent sketch. This pipeline isl connected at its centreline height to a U-tube manometer containing a liquid of density 1250kg/m3. Determine the pipeline pressure in metres of oil as well as its equivalent in mWg at the pipe centreline. (Answer -2.03m of liq SG=0.8)

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Example 7The following manometer is both inclined at a small angle as well as being stepped in diameter to improve sensitivity. It is required to measure a pressure of 20mm water gauge. The large tube diameter is 56mm whilst the smaller tube is 7mm in diameter. The manometer fluid has an SG of 0.75.

Calculate the horizontal angle of inclination of the small tube if an accuracy of +/- 2% of the pressure reading is required and therefore, that the linear measurement of the liquid column displacement may be accurately read to +/- 0.5%.

Example 8: In a U tube manometer the addition of water will immediately lead to the meniscus in either limb adopting the same horizontal level. This presumes that equal pressures apply at the water levels. The question that is posed here is, “Will the addition of a volume of oil on one side cause the levels to be different?” This may be solved by considering the mass of liquid in the individual limbs. First assume each limb has 100mm of water depth.

Now add oil such that the RH limb drops by 20mm to 80mm and the LHS must therefore rise to 120mm. The RH limb now has a volume of oil in it which together

A

2.5m

0.3m

SG=0.8

SG=1.25

Diam56mm

20mm Wg

Diam=7mmOil

B

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with the 80mm column of water must exactly balance the LH side column of 120 mm of water. i.e.

Here H is the unknown height of the oil in the RH limbAfter significant cancellation and using 1000kg/m3 and 900kg/m3 for water and oil density respectively we may write,

1000*.12=1000*0.08 +900H or

H = 0.04/0.9=0.0444m

Thus the RH limb is of height 80mm+44.4mm=124.44mm. Thus although the pressures experienced by either meniscus is the same, the heights of the two limbs are different by 4.44 mm.

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