Fluid Mechanics

MECHANICAL ENGINEERING

FOR

UPSC Engineering Services Examination, GATE, State Engineering Service Examination & Public Sector Examination. (BHEL, NTPC, NHPC, DRDO, SAIL, HAL, BSNL, BPCL, NPCL, etc.)

FLUID MECHANICS

IES MASTER PUBLICATION

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Copyright © 2017, by IES MASTER Publications. No part of this booklet may be reproduced, or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior permission of IES MASTER, New Delhi. Violates are liable to be legally prosecuted.

First Edition : 2017

Typeset at : IES Master Publication, New Delhi-110016 CONTENTS

1. Properties of Fluid ...... 1 – 42

2. Fluid Pressure and Measurement ...... 43 – 65

3. Hydrostatic Forces ...... 66 – 103

4. Buoyancy and Floatation ...... 104 – 131

5. Liquid in Relative Equilibrium ...... 132 – 167

6. Fluid Kinematics...... 168 – 212

7. Fluid Dynamics ...... 213 – 262

8. Momentum Equation & Applications ...... 263 – 294

9. Orifice/Mouthpiece/Notches/Weirs ...... 295 – 322

10. Laminar Flow...... 323 – 367

11. Turbulent Flow ...... 368 – 396

12. Boundary Layer ...... 397 – 444

13. Drag and Lift ...... 445 – 473

14. Dimensional Analysis...... 474 – 528

15. Pipe Flow ...... 529 – 604

16. Open Channel Flow ...... 605 – 636 1

INTRODUCTION

Vapour Pressure and  A substance in liquid or gaseous phase is referred to as fluid. They Cavitation are capable of deforming continuously under the action of shear stress, however small the shear stress may be. Bulk Modulus &  In solid, stress is proportional to strain, but in fluid stress is Compressibility proportional to strain rate. When a constant shear stress is applied, Isothermal Bulk a solid eventually stops deforming at some fixed strain angle, where Modulus as fluid never stops and approaches a constant rate of strain.

Adiabatic bulk TYPE OF FLUID modulus (K ) A Ideal Fluid

Viscosity  An ideal fluid does not have surface tension, viscosity and are Newtonian and non- incompressible. (i.e. bulk modulus =  ) newtonian Fluids  No such fluid exist in practical situation.  However, fluids like air & water have very low value of viscosity Surface Tension and and can be treated as ideal fluids for all practical purposes. Capillary Effect Real Fluid

Fluids which are not ideal are called real fluids. SOME PROPERTIES OF FLUID

mass dm kg  Density = = =  unit is volume dv  m3   Density of liquid is generally assumed to be constant 1  Density of gases  Pressure and  Temperature  Specific gravity or relative density 2 Chapter-1 : Properties of Fluid

density of substance = density of some standard substance at a specified temp. (usually water at 4°C)

NOTE If specific gravity <1 Fluid is lighter than water

Weight of substance N  Specific weight or weight density =    g unit is   Volume of substance m3 

N kN  water = 9810 = 9.81 ,mercury =13.6 w m3 m3 VAPOUR PRESSURE AND CAVITATION

 At a liquid-air interface, a continuous exchange of molecules takes place. The liquids evaporate because the liquid molecules escape from the surface into a gaseous form, called the vapour. These vapour molecules exert a partial pressure in the space, known as the vapour pressure.  When the absolute pressure above a liquid becomes less than or equal to the vapour pressure, boiling starts.  This property of liquids, in fact, may cause an undesirable effect called the cavitation. In many liquid flow situations, it is possible that very low pressures are produced at certain locations. For example, in figure shown below, the pressure at point A may drop to a very low value and may become equal to the vapour pressure of water. When this occurs, water starts boiling at point A. Low Pressure Cavities

r te a w B High Pressure  Thus a rapidly expanding vapour pocket called cavity (because they lead to cavity formation) in the liquid flow is formed.  Because of the flow, this pocket of vapour is usually swept away from the point A where it originated to a point B of high pressure.  Because of the high pressure at B, the cavity collapses (or, the bubbles burst). This process of formation & collapsing of cavity is called cavitation.  Cavitation generates annoying vibrations and noise and causes damage to equipment [because it causes erosion, surface pitting and fatigue failure].  Vapour pressure increases with temperature. Hence at higher temperature, chances of cavitation is more ie., there is greater chance that pressure in fluid flow may fall below vapour pressure.  Liquid with high vapour pressure evaporates readily and are known as volatile liquid. Mercury

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has very low vapour pressure. Hence it can be used in pressure measuring equipments so that it can measure even low pressures without evaporating. Vapour Pressure of water at various temperatures is given below: Temprature Vapour pressureof water in kPa 10 C 1.23 20 C 2.34 100 C 101.3 1 atm 150 C 475.8 Note that if temp is 100°C, vapour pressure of water becomes 1 atm. i.e., becomes equal to atmospheric pressure, thus boiling starts. BULK MODULUS & COMPRESSIBILITY

If original pressure is P in a liquid mass and its volume is V and increase in pressure dP causes change in volume dV then dP Bulk modulus of elasticity (K) = – dV / V stress P [analogous to = Young’s modulus of elasticity] strain Unit of K will be that of pressure.

dP dP P K    dV  d      V    P Note:    V = mass = constant = dV + Vd = 0 d dV = = –  V 1 1 d  Compressibility  = K  dP d  If density does not change with pressure i.e.,  0 (fluid is incompressible) dP ISOTHERMAL BULK MODULUS

 kPa.m3   For ideal gas P = RT R 0.287   Kg.Kalvin  dP      RT d T Cons tan t dP  K =    RT  P . T d

 for isothermal condition in an ideal gas, KT = P i.e. Isothermal bulkmodulus pressure

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Cp Specific heat at constant pressure where, = adiabatic index = = Cv Specific heat at constant volume  m   P   = constant   P = constant (because mass ‘m’ is constant) 

 P = C dP = C(–1) d dP K =  = C = P A d

 Adiabatic bulk modulus  P  Adiabatic index Pr essure

NOTE In case of liquid, effect of compressibility is neglected. However, in some case like water hammer pressure conditon, compressibility can be taken into account.

Example 1 : If 1 m head of water = 10 kPa and 1 atm = 100 kPa, vapour pressure of water at 20°C = 2.5 kPa, at what negative head of water, vapourisation will start. Solution–1 : We know that Vapourisation will start, when vapour pressure > Abs pressure

 Absolute pressure = Atmospheric pressure + Pgauge  2.5 > 100 + (10 × x) [where x = head of water]  10x  –97.5  x  9.75m head of water  x  9.75m head of water  Vapourisation will start when negative head of water is greater than 9.75 m  Below this vapourisation will not start. Example 2 : The bulk modulus of water is 210×104 kN/m2. What pressure is required to reduce the volume of water in a closed container by 2%; Also show that the increase in mass density because of the reduction in volume by 2% will be 2% only.

dp Solution–2 : From the definition of bulk modulus of elasticity k =  ...(I) dV    For 2% reduction in volume V  dV    = 0.02 ...(II) V 

dP 4 2  210 × 10 kN/m = 0.02  dP = 42,000 kN/m2 dP = 42 MN/m2 thus additional pressure required to reduce the volume in closed container by 2% is 42 MN/m2

d dV   For constant mass in the closed contaner =   = – (– 2%) = 2%  V  thus reduction in volume by 2% will lead to increase in mass density by 2%.

Example 3 : The air in an automobile tyre with a volume of 0.015 m3 is at 30ºC and 140 kPa. Determine the amount of air that must be added to increase the pressure to 210 kPa. Assume atm. pressure to be 100 kPa and temperature and volume to remain constant, assuming Ideal gas equation.

Solution–3 : Since volume & temp. is constant hence mass added = 2  1  V

PP2 1  (PP)V2 1 =   V = RT RT  RT

Initial absolute pressure = P1 = (140 + 100) = 240 kPa

Final absolute pressure = P2 = 210 + 100 = 310 kPa

We know that P1 = 1 RT

P2 = 2 RT

P1 1   [at constant Temperature] P2 2 P  310 2      1    1 P1  240

P1  240 240 3       Kg / m   RT  R (273  30) 303R

P 310  240  310 2     Kg / m3   1     P1 240  303R  303R

310 240  Mass added = ( –  ) V =   V [ R = 0.287] 2 1 303R  a 70 70 0.015 =  0.015 kg = = 0.01207 kg. 303R 303 0.287

Example 4 : A pump is used to transport water to a higher reservoir. If the water temp is 30°C [sat vapour pressure at 30°C = 4.25 kPa]. Determine the lowest pressure that can exist in the pump without cavitation.

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Solution–4 : For no cavitation absolute pressure should be more than vapour pressure in saturation process.  Minimum pressure in pump = 4.25 kPa. Example 5 : Calculate the percentage change in density of gas if it is compressed isothermally from 1000 atm to 1001 atm. Solution–5 : We know that from ideal gas-equation P = RT ...(i)  dP = (RT) ...(ii)

dP  d       =   P     Percentage increase in pressure = percentage increase in density (for isothermal condition) 1001 1000  =   100 = 0.1%  1000  Example 6 : The density of sea water at free surface where pressure in 98 KPa is approx. 1030 kg/ m3. Taking bulk modulus of elasticity of sea water to be 2.34 × 109 N/m2 (assume constant) and dp = g dz, where dP = change in pressure, dz = increase in depth, determine the density and pressure at a depth of 2500 m. Disregard the effect of temperature. Solution–6 : We know that bulk modulus of elasticity dp dp K =  ...(i) z dv d v   We also have dp = g dz ...(ii) From (i) & (ii)

g dz 2 g dz   K =    d  d     

 z d  gdz  2  =    K 0   0 1 1 gz    =     K  0 1 gz 1   =  K  0 1    = 1 gz    0 K  At z = 2500

1   =  1 9.81 2500     1030 2.34 109  = 1041.24 kg/m3 ...(III)

P Z Z 1  dP = gdz = gdz    1 gz  P 0 0 0   0 K  1 gz Put  =  0 K

g K    dz = d  dz = d  K g 

1/0  gz/K gz  Kg d  1/0   P – P = = K  0  K log   g  1/0 1/0

 1 1 gz  P – P0 = K log log     0  0 K   1  P –P = K log 0   gz   1  0   K  At z = 2500 m. 1 P = P + K log 0  gz  1  0  K 

 1  = 98 KPa + 2.34 × 109log 103 KPa e  1030 9.81  2500  1    2.34 109  = (0.098 + 25.398) MPa [P = 25.50 MPa]

7 Example 7 : An empirical pressure-density relation for water in P / Pa  3001  / a  – 3000 where suffix ‘a’ refers to atmospheric conditions. Determine the bulk modulus of elasticity and compressibility of water at 5 atmospheric pressure.

7 P   Solution–7 : = 3001  – 3000 ...(I) Pa a 

 Diffn (I) w.r.t.  gives

dP 3001 7 6 d  dP 3001 7  6  P =   a ...(II) P 7 7 a a d a

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