me vol 2 fm

GATEMECHANICAL ENGINEERING

NODIA AND COMPANY

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Multiple Choice QuestionsGATE Mechanical Engineering Vol 2, 1e

Copyright © By PublishersISBN 9-788192-27629-8

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PREFACE

This book doesn’t make promise but provide complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exam as well as fresh set of problems.

The book is categorized into units which are sub-divided into chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts are techniques which are absolutely necessary. Again time is crucial factor both from the point of view of preparation duration and time taken for solving each problem in the book are those which take the least distance to the solution.

But however to make a comment that the book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books.

Authors

SYLLABUS

ENGINEERING MATHEMATICSLinear Algebra: Matrix algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus: Functions of single variable, Limit, continuity and differentiability, Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives, Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.

Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Cauchy’s and Euler’s equations, Initial and boundary value problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace equation.

Complex variables: Analytic functions, Cauchy’s integral theorem, Taylor and Laurent series.

Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Poisson,Normal and Binomial distributions.

Numerical Methods: Numerical solutions of linear and non-linear algebraic equations Integration by trapezoidal and Simpson’s rule, single and multi-step methods for differential equations.

APPLIED MECHANICS AND DESIGNEngineering Mechanics: Free body diagrams and equilibrium; trusses and frames; virtual work; kinematics and dynamics of particles and of rigid bodies in plane motion, including impulse and momentum (linear and angular) and energy formulations; impact.

Strength of Materials: Stress and strain, stress-strain relationship and elastic constants, Mohr’s circle for plane stress and plane strain, thin cylinders; shear force and bending moment diagrams; bending and shear stresses; deflection of beams; torsion of circular shafts; Euler’s theory of columns; strain energy methods; thermal stresses.

Theory of Machines: Displacement, velocity and acceleration analysis of plane mechanisms; dynamic analysis of slider-crank mechanism; gear trains; flywheels.

Vibrations: Free and forced vibration of single degree of freedom systems; effect of damping; vibration isolation; resonance, critical speeds of shafts.

Design: Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints, shafts, spur gears, rolling and sliding contact bearings, brakes and clutches.

FLUID MECHANICS AND THERMAL SCIENCESFluid Mechanics: Fluid properties; fluid statics, manometry, buoyancy; control-volume analysis of mass, momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoulli’s equation; viscous flow of incompressible fluids; boundary layer; elementary turbulent flow; flow through pipes, head losses in pipes, bends etc.

Heat-Transfer: Modes of heat transfer; one dimensional heat conduction, resistance concept, electrical analogy, unsteady heat conduction, fins; dimensionless parameters in free and forced

convective heat transfer, various correlations for heat transfer in flow over flat plates and through pipes; thermal boundary layer; effect of turbulence; radiative heat transfer, black and grey surfaces, shape factors, network analysis; heat exchanger performance, LMTD and NTU methods.

Thermodynamics: Zeroth, First and Second laws of thermodynamics; thermodynamic system and processes; Carnot cycle. irreversibility and availability; behaviour of ideal and real gases, properties of pure substances, calculation of work and heat in ideal processes; analysis of thermodynamic cycles related to energy conversion.

Applications: Power Engineering: Steam Tables, Rankine, Brayton cycles with regeneration and reheat. I.C. Engines: air-standard Otto, Diesel cycles. Refrigeration and air-conditioning: Vapour refrigeration cycle, heat pumps, gas refrigeration, Reverse Brayton cycle; moist air: psychrometric chart, basic psychrometric processes. Turbomachinery: Pelton-wheel, Francis and Kaplan turbines — impulse and reaction principles, velocity diagrams.

MANUFACTURING AND INDUSTRIAL ENGINEERINGEngineering Materials: Structure and properties of engineering materials, heat treatment, stress-strain diagrams for engineering materials.

Metal Casting: Design of patterns, moulds and cores; solidification and cooling; riser and gating design, design considerations.

Forming: Plastic deformation and yield criteria; fundamentals of hot and cold working processes; load estimation for bulk (forging, rolling, extrusion, drawing) and sheet (shearing, deep drawing, bending) metal forming processes; principles of powder metallurgy.

Joining: Physics of welding, brazing and soldering; adhesive bonding; design considerations in welding.

Machining and Machine Tool Operations: Mechanics of machining, single and multi-point cutting tools, tool geometry and materials, tool life and wear; economics of machining; principles of non-traditional machining processes; principles of work holding, principles of design of jigs and fixtures

Metrology and Inspection: Limits, fits and tolerances; linear and angular measurements; comparators; gauge design; interferometry; form and finish measurement; alignment and testing methods; tolerance analysis in manufacturing and assembly.

Computer Integrated Manufacturing: Basic concepts of CAD/CAM and their integration tools.

Production Planning and Control: Forecasting models, aggregate production planning, scheduling, materials requirement planning.

Inventory Control: Deterministic and probabilistic models; safety stock inventory control systems.

Operations Research: Linear programming, simplex and duplex method, transportation, assignment, network flow models, simple queuing models, PERT and CPM.

GENERAL APTITUDEVerbal Ability: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction.

Numerical Ability: Numerical computation, numerical estimation, numerical reasoning and data interpretation.

CONTENTS

FLUID MECHANICS

FM 1 Basic Concepts and Properties of Fluids FM 3

FM 2 Pressure and Fluid Statics FM 33

FM 3 Fluid Kinematics & Bernouli Equation FM 80

FM 4 Flow Analysis Using Control Volumes FM 124

FM 5 Flow Analysis Using Differential Method FM 172

FM 6 Internal Flow FM 211

FM 7 External Flow FM 253

FM 8 Open Channel Flow FM 289

FM 9 Turbo Machinery FM 328

HEAT TRANSFER

HT 1 Basic Concepts & Modes of Heat-Transfer HT 3

HT 2 Fundamentals of Conduction HT 34

HT 3 Steady Heat Conduction HT 63

HT 4 Transient Heat Conduction HT 94

HT 5 Fundamentals of Convection HT 114

HT 6 Free and Force Convection HT 129

HT 7 Radiation Heat Transfer HT 155

HT 8 Heat Exchangers HT 181

THERMODYNAMICS

TD 1 Basic Concepts and Energy Analysis TD 3

TD 2 Properties of Pure Substances TD 28

TD 3 Energy Analysis of Closed System TD 52

TD 4 Mass and Energy Analysis of Control Volume TD 76

TD 5 Second Law of Thermodynamics TD 106

TD 6 Entropy TD 136

TD 7 Gas Power Cycles TD 166

TD 8 Vapor and Combined Power Cycles TD 199

TD 9 Refrigeration and Air Conditioning TD 226

***********

FM 1BASIC CONCEPTS AND PROPERTIES OF FLUIDS

Common Data For Q. 1 and 2In an automobile tire the pressure is 245 kPa and the air temperature is 298 K. The volume of tire is 0.050 m3 and gas constant of air is 0.287 kPa- /m kgK3 .

FM 1.1 The pressure in the tire at air temperature of 32 K2 when volume of tire is constant, will be (A) 336 kPa (B) 26 kPa

(C) 310 kPa (D) 1854.02 kPa

FM 1.2 What amount of air should be come out to obtain pressure to its original value at same temperature ?(A) 0.1812 kg (B) 0.1672 kg

(C) 0.014 kg0 (D) 0.3484 kg

FM 1.3 Consider Carbon dioxide at 1 atm2 and 400 Cc . What will be the density of Carbon dioxide and cp at this state and the new pressure when the gas is cooled isentropically to 150 Cc ? (For Carbon dioxide .k ��= and �� m s �R 2 2= )

(A) 0.797 /kg m3ρ = , 4�.�kg �

�cp−

= , ��kPap2 =

(B) 1.3 10 /kg m4 3ρ #= - , ��kg �

�cp−

= , ��5.5 kPap2 =

(C) 7.97 /kg m3ρ = , ��kg �

�cp−

= , ��5.5 kPap2 =

(D) 7.97 /kg m3ρ = , ��kg �

�cp−

= , ��5.5 �Pap2 =

FM 1.4 A Cane of beverage contains 455 ml of liquid. The mass of cane with liquid is 0.369 kg while an empty cane weighs 0.1 3 N9 . What will be the specific weight, density and specific gravity of liquid respectively ?(A) 0.977 /kN m3, 99.6 /kg m3, .0 0996

(B) 9.77 /kN m3, 996 /kg m3, .0 996

(C) 9.77 /N m3, 996 /kg m3, .9 96

(D) 97.7 /kN m3, 996 /kg m3, .0 996

FM 1.5 The specific gravity of a gas contained in a tank at the temperature of 25 Cc is 2 10 3#

− . If the atmospheric pressure is 10.1 kPa , the gage pressure is(A) 70 kPa (B) 7 kPa

(C) 0.7 kPa (D) 70 kPa

FM 1.6 Consider steam at state near the saturation line : ( , )p T1 1 (1.31 , 250 )MPa Cc= , 4�� .��)m s �andR ksteam

2 2−= = . If the steam expands isentropically to a new

pressure of 414 kPa, what will be the density 1ρ and the density 2ρ ?(A) 5.44 / , 5.04 /kg m kg m1

32

3ρ ρ= = (B) 2.28 / , 5.44 /kg m kg m13

23ρ ρ= =

(C) 5.44 / , 2.28 /kg m kg m13

23ρ ρ= = (D) 5.04 / , 5.44 /kg m kg m1

32

3ρ ρ= =

FM 1.7 A 30 m3 cylinder contains Hydrogen at 25 Cc and 200 kPa What amount of


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Hydrogen must be bled off to maintain the Hydrogen in cylinder at 20 Cc and 600 kPa ? ( 0.2968 . / . )kPa m kg KR 3=(A) 271.35 kg (B) 206.99 kg

(C) 478.34 kg (D) 64.36 kg

FM 1.8 Wet air with 100% relative humidity, is at 30 Cc and 1 atm. If �� m s �Rair� �

−=, Rwater 461 /m s K2 2

−= and vapor pressure of saturated water at 30 Cc is 4242 Pa, what will be the density of this wet air using Dalton’s law of partial Pressures ?(A) 1.12 /kg m3 (B) 1.09 /kg m3

(C) 0.03 /kg m3 (D) 1.147 /kg m3

FM 1.9 In a formula one race, at the start of the race the absolute pressure of a car tire is 362.5 kPa and at the end of the race the absolute pressure of car tire is measured to be 387.5 kPa . If the volume of the tire remains constant at 0.022 m3 then percentage increase in the absolute temperature of the air in the tire is(A) 6.9% (B) 69%

(C) 0.69% (D) Not increased

FM 1.10 A compressed air tank contains 24 kg of air at a temperature of 80 Cc . If the reading of gage mounted on the tank is 300 kPa, what will be the volume of tank in m3 ?(A) 404 (B) 4.04

(C) 0.404 (D) 40.4

FM 1.11 A small submersible moves in 30 Cc water ( 4.242 kPapv = ) at 2-m depth, where ambient pressure is 133 kPa. Its critical cavitation number is .Ca 0 2�. . At what velocity will cavitation bubbles form ?(A) 22.72 m/s (B) 32.66 m/s

(C) Zero (D) 32.13 m/s

FM 1.12 What will be the speed of sound of steam at 150 Cc and 400 kpa? (k = 1.33, R = 461 /m s K2 2

− )(A) 50.9 m/s (B) 509 m/s

(C) 30.3 m/s (D) 303 m/s

FM 1.13 A liquid has a weight density of 9 /N m268 3 and dynamic viscosity of . /N s m131 5 2−

. What will be the kinematic viscosity of the liquid in /secm2 ?(A) 0.0139 (B) 1.39

(C) 0.139 (D) 13.9

FM 1.14 A 72 m long and 30 m diameter blimp is approximated by a prolate spheroid

whose volume is given by v LR32 2p= . The weight of 20 Cc gas within the blimp

for (a) helium at 1.1 atm and (b) air at 1.0 atm, is ( 2077 / �m sRHe

2 2−= , Rair 287 /m s K2 2

−= )(A) 60.97kNWHe = , 401.1kNWair = (B) 401.1kNWHe = , 6.97kNWair =(C) 6.2kNWHe = , 40.9kNWair = (D) 40.9kNWHe = , 6.2kNWair =

FM 1.15 The oil having viscosity of 4.56 10 /N s m2 2# −

− , is contained between two parallel plates. The bottom plate is fixed and upper plate moves when a force F is applied. If the distance between the stationary and moving plates is 2.54 mm and the area of the upper plate is . m0 129 2, what value of F is required to translate


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the plate with velocity of 1 / secm ?(A) . N2 32 (B) 23.2 N

(C) 232 N (D) 0.232 N

FM 1.16 A thin moving plate is separated from two fixed plates by two fluids of different viscosity as shown in figure below. If the contact area is A, the force required for the flow to be steady laminar viscous flow, is

(A) F h h VA�

�

2

2m m= +; E (B) F h h VA

2

2

�

�m m= −: D

(C) F h h VA2

2

�

�m m= −; E (D) F h h VA

�

�

2

2m m= +: D

FM 1.17 A large movable plate is located between two large fixed plates. Two fluids having the different viscosities are contained between the plates. If the moving plate has a velocity of 6 /secm , what will be the magnitude of the shearing stresses on plate 1 and plate 2 respectively, that act on the fixed plates ?

(A) 10 /N m2, 15 /N m2 (B) 20 /N m2, 15 /N m2

(C) 15 /N m2, 15 /N m2 (D) 15 /N m2, 20 /N m2

FM 1.18 A thin flat plate of area A is moved horizontally between two plates, one stationary and one moving with a constant velocity Vm as shown in figure below. If velocity of flat plate is Vp and dynamic viscosity of oil is μ, the force must be applied on the plate to manage this motion is

(A) A hV

hV Vp p m

1 2μ + -

; E (B) ( )�A V V hp m 2μ -

(C) hAVp

1

μ (D)

( )A h

Vh

V Vp p m

1 2μ - -

; E

FM 1.19 A Newtonian fluid having the specific gravity of 0.91 and Kinematic viscosity of 4 10 / secm4 2#

− , flows over a fixed surface. The velocity profile near the surface





is given by the relation:

Uu sin y

2dp= a k

What will be the magnitude of the shearing stress developed on the plate in term of U and δ ?

(A) 0.571 /N mU 2

δ (B) 5.71 /N mU2δ

(C) 5.71 /N mU 2

δ (D) 0.571 /N mU2δ

FM 1.20 A 50 30 20cm cm cm# # block of 15 kg mass is to be moving at a constant velocity of 0.8 /m s on an inclined plane. If a 0.8 mm thick oil film with a dynamic viscosity of 0.006 Pa s− is there between the block and inclined plane, what amount of force is required in x -direction ? ( 10 / )m sg 2=

(A) 55 N (B) 55.55 N

(C) 6.42 N (D) 414.75 N

FM 1.21 A closed rectangular container is half filled with water at 45 Cc . If the air in remaining half section of container is completely escaped. The absolute pressure in the escaped space at same temperature (saturation pressure of water at 45 Cc is9.593 kPa) is(A) P P> saturation (B) P P< saturation

(C) P Psaturation= (D) Not determined

FM 1.22 Consider two parallel plates as shown in figure below. If the fluid is glycerin (ρ1264 /kg m3= , μ 1.5 /N s m2

−= ) and the distance between plates is 9 mm. What will be the shear stress required to move the upper plate at � �m sV = and the Reynolds number respectively ?

(A) 100 , 4Pa 60 (B) 10 , 4Pa 600

(C) 10000 , 4.Pa 6 (D) 1000 , 46Pa

FM 1.23 The velocity profile in a pipe flow is given by ( � )u u r R� n n�= − , where r is the

radial distance from the centre. If the viscosity of the fluid is μ then the drag force applied by the fluid on the pipe wall in the direction of flow across length L





of the pipe is ( )radius of circular pipeR = .

(A) n u L�π μ (B) �n u R�μ(C) n u L2 0πμ (D) n u2 0πμ

FM 1.24 Consider air at 20 Cc with 1.8 10 sPa5μ # -= - . Its viscosity at 400°C by (a) The Power-law (n=0.7) (b) the sutherland law (S = 110 K) respectively, are(A) �.��1 10 � , 1.� 10 �� s �� sp s

� �μ μ# #- -= =- -

(B) �.��1 10 � , �.�� 10 �� s �� sp s� �μ μ# #- -= =- -

(C) �.�� 10 � , �.��1 10 �� s �� sp s� �μ μ# #- -= =- -

(D) 1.� 10 � , �.��1 10 �� s �� sp s� �μ μ# #- -= =- -

FM 1.25 Consider a block of mass m slides down on an inclined plane of a thin oil film as shown in figure below. The film contact area is A and its thickness is h . The terminal velocity V of the block is

(A) sinV Amgh

mq= (B) cosV A

mghm

q=

(C) sinV hmgA

mq= (D) cosV h

mgAm

q=

FM 1.26 A thin layer of glycerin flows down on an inclined plate of unit width with the velocity distribution:

Uu h

yhy�

�

�

= −

If the plate is inclined at an angle α with the horizontal, the expression for the surface velocity U will be





(A) sinU h��

gma= (B) sinU h

�

�

m ag=

(C) sinU h�

�

mg a= (D) sinU h� �

m ag=

FM 1.27 A shaft of 8.0 cm diameter and 30 cm length is pulled steadily at �� m sV = through a sleeve of 8.02 cm diameter. The clearance is filled with oil of 0.003 /m s2ν = and . . .S G ��= , the force required to pull the shaft is ( 998 / )kg mw

3ρ =(A) 793 N (B) 795 N

(C) 79.3 N (D) 7.95 N

FM 1.28 Match List I (Properties of fluids) with List II (Definition/ Result) and select the correct answer using the codes given below :

List-I List-II

a. Ideal fluid 1. Viscosity does not vary with rate of deformation

b. Newtonian fluid 2. Fluid of zero viscosity

c. /μ ρ 3. Dynamic viscosity

d. Mercury in glass 4. Capillary depression

5. Kinematic viscosity

6. Capillary rise

Codes a b c d(A) 1 2 4 6(B) 1 2 3 4(C) 2 1 3 6(D) 2 1 5 4

FM 1.29 Match List I (Fluid properties) with List II (Related terms) and select the correct answer using the codes given below :

List-I List-II

a. Capillarity 1. Cavitation

b. Vapour pressure 2. Density of water

c. Viscosity 3. Shear forces

d. Specific gravity 4. Surfaces Tension

Codes a b c d(A) 1 4 2 3(B) 1 4 3 2(C) 4 1 2 3(D) 4 1 3 2

FM 1.30 The hydrogen bubbles have diameter �.�1 mmD - . Assume an ‘‘air-water” interface at 30 Cc and surface tension 0.0712 /N mσ = . What will be the excess pressure within the bubble ?(A) 1.42 kPa (B) 2.85 kPa

(C) 28.5 kPa (D) 14.2 kPa

FM 1.31 The surface tension in a rain drop of 3 mm diameter is 7.3 10 /N m2#

− . The





excess pressure inside the rain drop is(A) 973.3 Pa (B) 97.33 Pa

(C) 9.73 Pa (D) 97.33 kPa

FM 1.32 A shower head emits a cylindrical water jet of diameter 0.73 mm into air. The pressure inside the jet is approximately 300 Pa greater than the air pressure. What will be the surface tension of water ?(A) 0.0365 N/m (B) 0.73 N/m

(C) 0.365 N/m (D) 0.073 N/m

FM 1.33 A thin wire ring of 6 cm diameter is lifted from a 20 Cc water surface. How much lift force is required if 0.0728 /N mσ = ?(A) 0.274 N (B) 0.0274 N

(C) 0.137 N (D) 0.0137 N

FM 1.34 A 4 mm diameter glass tube is immersed in water and mercury. The temperature of the liquid is 20 Cc and the values of the surface tension of water and mercury at 20 Cc in contact with air are 0.0734 /N m and 0.51 /N m, respectively. The angle of contact for water is zero and that for mercury is 128c. What will be the capillary effect for water and mercury in millimeters, respectively ?(A) 4.60, 3.82 (B) 2.35, 7.48

(C) 3.82, 4.60 (D) 7.48, 2.35

FM 1.35 The system shown in figure below is used to estimate the pressure inside the tank by measuring the height of liquid in the 1 mm diameter tube. The fluid is at 60 Cc . What will be the capillary rise if the fluid is (a) water ( 0.0662 /N mσ =, 983 /kg m2ρ = , 0c,θ ) and (b) Mercury ( 0.47 /N mσ = , 13500 /kg m3ρ = ,

130c,θ ) ?

(A) 0.0275mhw = , 0.0456 mhm =− (B) 0.0275 mhw =− , 0.00�1mhm =(C) 0.0275 mhw = , 0.00�1 mhm =− (D) 0.0137 mhw = , 0.00456 mhm =−

FM 1.36 A glass tube of 4.6 mm diameter is inserted into milk and milk rises upto 3.5 mm in the tube. If the density of milk is 960 /kg m3 and contact angle is 15c, the surface tension of milk is(A) 0.2315 /N m (B) 0.025 /N m

(C) 0.0236 /N m (D) 0.02315 /N m

FM 1.37 A liquid film suspended on a rectangle wire frame of one movable side of 12 cm. What amount of surface tension is required if the movable side of frame is to be moved with 0.018 N ?(A) 0.075 /N m (B) 0.00432 /N m

(C) 0.055 /N m (D) 0.75 /N m





FM 1.38 In figure shown, a vertical concentric annulus with outer radius ro and inner radius ri is lowered into the fluid of surface tension σ and contact angle 45< cθ. If the gap is very narrow, what will be the expression for the capillary rise h in the annulus gap ?

(A) ( )

coshg r r�

o i�r

s q=−

(B) ( )cosh

g r r�

o irs q= −

(C) ( )cosh

g r ro irs q= − (D)

( )cosh

g r r�

o i� �r

s q=−

FM 1.39 A solid cylindrical needle of diameter 1.6 mm and density 7824 /kg m3 may float on a liquid surface. Neglect buoyancy and assume a contact angle of 0c. What will be the surface tension σ ?(A) 0.0772 N/m (B) 0.154 N/m

(C) 0.772 N/m (D) 0.0154 N/m

Common Data For Linked Answer Q. 40 and 41A Frustum-shaped body is rotating at a uniform angular velocity 200 /rad sω = in a container. The gap of 1.2 mm on all sides between body and container is filled with oil of viscosity 0.1 Pa s− at 2 C0c .

FM 1.40 The power required at the top surface to maintain this motion is

(A) hD

24

2 3πμω (B) hD

32

2 4πμω

(C) hD

4

2 4πμω (D) hD

16

2 2πμω

FM 1.41 The reduction in power required at the top surface when oil viscosity is 0.0078 Pa s− at 8 C0c , will be(A) 5.29 W (B) 67.824 W

(C) 62.533 W (D) No reduction





FM 1.42 A fluid of surface tension 0.0728 /N mσ = and contact angle 0cθ = is filled between 0.75 mm apart two parallel plates as shown in figure. If the density of fluid is 998 /kg m3ρ = , the capillary height h will be

(A) 2 mm (B) 10 mm

(C) 20 mm (D) 1 mm

FM 1.43 A 56 kg block slides down on a smooth inclined plate. A gap of 0.1 mm between the block and plate contains oil having viscosity 0.4 /N s m2

− . If the velocity distribution in the gap is linear and the area of the block in contact with the oil is 0.4 m2, the terminal velocity of the block is(A) 0.03125 /m s (B) 0.3125 /m s

(C) 3.125 /m s (D) 0.03125 /mm s

FM 1.44 Two 50 cm long concentric cylinders are mounted on a shaft. The inner cylinder is completely submerged in fluid and is rotating at 200 rpm and the outer cylinder is fixed. The fluid film thickness between two cylinders is 0.1 cm2 and outer diameter of the inner cylinder is 20 cm. If the torque transmitted by the shaft to rotate inner cylinder is 0.8 N, the viscosity of the fluid is

(A) 0.0173 /N s m2− (B) 0.0231 /N s m2

−

(C) 0.173 /N s m2− (D) 0.0346 /N s m2

−

FM 1.45 A layer of water having the viscosity of 1.2 10 /N s m3 2# −

− flows down on inclined fixed surface with the velocity distribution as given by:





Uu h

yhy�

�

�

= −

If the velocity of water � �secmU = and ��mh = , what will be the magnitude of the shearing stress that the water exerts on the fixed surface in /N m2 ?

(A) 7.20 (B) 0.720

(C) .7 2 10 3#

− (D) .0 072

FM 1.46 A 2.5 mm diameter aluminum sphere ( 2700 /kg m3ρ = ) falls into an oil of density 875 /kg m3. If the time to fall 75 cm is 48 s then the oil viscosity is(A) 0.0589 /kg m s−

(B) 0.589 /kg m s−

(C) 0.397 /kg m s−

(D) 0.0397 /kg m s−

FM 1.47 Consider a concentric shaft fixed axially and rotates inside the sleeve. If the shaft of radius ri rotates at /rad sω inside the sleeve of radius r0 and length L and the applied Torque is T, what will be the relation for the viscosity μ of the fluid between shaft and sleeve ?

(A) ( )

r LT r r2

i

i�

0μπω

= - (B)

( )��

T � �2

�

03

0μπω

= -

(C) ( )

��T � �2 �

�3

0μπω

= - (D)

( )��

T � �2 �

�3

0μπω

= +

FM 1.48 The velocity profile for laminar one-dimensional flow through a circular pipe is given as ( ) ( � )u r u r R�max

2 2= − , where R is the radius of the pipe and r is the radial distance from the centre of the pipe. If an oil at 40 Cc flows through a 15 m long pipe with 0.0�mR = and maximum velocity of � �m sumax = , what will be the friction drag force applied by the fluid on inner surface of the pipe when

0.0010 /kg m sμ -= ?

(A) 0.0942 N (B) 0.942 N

(C) 0.856 N (D) 0.916 N





FM 1.49 A 1 m diameter cylindrical tank has a length of 5 m long and weight 125 N. If it is filled with a liquid having a specific weight of 10.9 /kN m3, the vertical force required to give the tank an upward acceleration of 2.75 / secm 2 is(A) 550 kN (B) N55

(C) 5. N5 (D) 55 kN

FM 1.50 A cylindrical rod of diameter D , length L and density sρ falls due to gravity inside a tube of diameter Do . The clearance, ( )D D D<<o − is filled with a film of viscous fluid ( , )ρ μ .The expression for terminal fall velocity would be

(A) ( )

V gD D D

�sor m= −

(B) ( )

VgD D D

�s o

mr= +

(C) ( )

Vg D D�s o

mr= −

(D) ( )

VgD D D

�s o

mr= −

FM 1.51 The belt as shown in figure below moves at steady velocity of 2.5 /m s and skims the top of a tank of oil SAE 30 W ( 0.29 / )kg m sμ -= at 20 Cc with �mL = ,

�0 cmb = and �cmh = . What power P in watts is required to remain belt in motion ?

(A) 11 Watts (B) 44 Watts

(C) 109 Watts (D) 1.1 Watts

FM 1.52 Two balls of Steel and Aluminum can float on water due to surface tension effect. The density of steel and aluminium balls are to be 7800 /kg m3 and 2700 /kg m3, respectively. Which metal ball would have maximum diameter to float on water at 2 C0c and what will be the diameter of that ball when surface tension of water at 20 Cc is 0.073 /N m ?(A) steel, 4. mm1 (B) Aluminium, 2.4 mm

(C) Aluminium, 4.1 mm (D) Steel, 2. mm4

FM 1.53 For a cone-plate viscometer of radius �cmR = , the angle 3cθ = and the gap is filled with liquid as shown in figure. If the viscous torque .��T �= and rotation rate is 94.2 /rad s, the liquid viscosity will be

(A) 0.0116 /kg m s−

(B) 0.116 /kg m s−

(C) 0.193 /kg m s−

(D) 0.0193 /kg m s−

FM 1.54 A solid cone of base r0 and initial angular velocity 0ω is rotating inside a conical seat as shown in figure below. If there is no applied torque and air drag is





neglected, the cone’s angular velocity ω is

(A) exp sinr t

mh5

30

03ω ω

μθ= -; E (B) exp sinmh

r t35

003

ω ω θμ= -; E

(C) exp sinmhr t

35

004

ω ω θμ= -; E (D) exp sinmh

r t35

002

ω ω θπμ= -; E

FM 1.55 The rotating-cylinder viscometer as shown in figure below shears the fluid in a narrow clearance ( )R r RΔ = - with a linear velocity distribution in the gap. If the driving torque measured is T and the bottom friction is included then the expression for μ is

(A) ( �)( )

R L RT r R

��μπω

=+-

(B) ( / )

( )� � ��

2 43μπω

=+

-

(C) ( / )

( )� � ��

2 42μπω

=+

- (D)

( / )( )

� � ��

2 43μπω

=-

-

FM 1.56 For a 300 mm long sliding lubricated bearing, the viscosity of oil is 0.008 /kg m s− during steady operation at 80 Cc . The average oil film thickness between the shaft and journal is 1.2 mm. If shaft of 80 mm diameter is rotated at 750 rpm, the amount of torque needed to overcome bearing friction would be(A) 0.0063 N m− (B) 0.063 N m−

(C) 0.63 N m− (D) 6.3 N m−

FM 1.57 0.063 N m−= A disk of radius �cmR = , rotates at 1200 . .r p m inside an oil container of viscosity 0.29 /kg m sμ -= as shown in figure below. The oil film thickness is �mmh = . If the velocity profile is linear and neglecting shear on the outer disk edges, the viscous torque on the disk is





(A) 0.716 N m− (B) 6.83 N m−

(C) 0.0716 N m− (D) 14.3 N m−

FM 1.58 A soap bubble of diameter D� coalesces with another bubble of diameter D� to form a single bubble D� with the same amount of air. For an isothermal process, D� as a function of ,D D� �, patm and surface tension σ is(A) � ( � ) ( � )p D D p D D p D Da a a�

��

��

��

��

��s s s+ = + + +

(B) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s+ = + − +

(C) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s− = − + −

(D) � ( � ) ( � )p D D p D D p D Da a a��

��

��

��

��

��s s s+ = + + −

FM 1.59 A skater of mass m moving at constant speed Vo , suddenly stands stiff with skates pointed directly forward and allows herself to coast to a stop. If blade length is L, water film thickness h , water viscosity μ and blade width is b then how far will she travel (on two blades) before she stops ?

(A) x LbV mho

m= (B) x V mhLb�

o

m=

(C) x LbV mh�

o

m= (D)x V mhLb

o

m=

FM 1.60 Two thin flat plates are tilted at an angle φ and placed in a tank of surface tension σ and contact angle θ as shown in figure below. At the free surface of the liquid in the tank, the distance between two plates are L and width is b into the paper. What will be the expression for σ in terms of other variables ?

(A) ( )

( )cos

tangbh L h2

σ θ φρ φ= -

- (B)

( )cos

tangh L h2σ φ

ρ φ= -

(C) ( )

( )cos

tangh L h2

σ θ φρ φ= -

- (D)

( )( )cos

tangh L h2

σ θ φρ φ= +

+

***********





SOLUTIONS

FM 1.1 Option (A) is correct.

We have p� 310 ,kPa= �.�� mv��= �.��aR = - /m kgK3

T� 298 K= and ��T�=Treating air as an Ideal gas, the final pressure in the tire from the ideal gas law,

Tp v

�

� � Tp v

�

� �=

p� TT p

�

��#= 298

323 310#= 336 kPa= ( )tancons tv v� �=

FM 1.2 Option (C) is correct.Amount of air needs to be bled off to restore pressure ( 310 )kPap2 = is

m m m� �Δ = -

m RTp v

��

�= .. 0.1812 kg0 287 298

310 0 050#

#= =

and m RTp v

��

�= .. 0.1672 kg0 287 323

310 0 050#

#= =

Hence mΔ 0.1812 0.1672 0.014 kg= − =

FM 1.3 Option (C) is correct.

We have p� 10 1013250atm Pa= = T� 400 400 273 673C Kc= = + =From ideal gas law

ρ ( ) ( )

�.��m�g

RTp

��

�

��

#= = =

cp .. ( )

��g �

�kkR

� �� #

−= − = − =

For gas cooled isentropically to �� C �T� c= = , the formula is

pp

�

� TT �k k

�

��

=−

b l

p� ��ap TT

��

..k k

��

��

��

# #= =− −

b bl l 135.5 kPa=

FM 1.4 Option (B) is correct.

Specific weight γ Volume of fluidWeight of fluid= g v

mgv

Wγ ρ= = =

Volume of fluidTotal weight weight of Cane= −

. . . .mg355 10

0 153355 10

0 369 9 81 0 1536 6

# #

#= − = −− −

. 9.77 /kN m355 10

3 476

3

#= =−

Density ρ . ��.�� g m �g mg �� -

g= = =





Specific gravity �.S G ��

waterrr= = =

FM 1.5 Option (D) is correct.We have . � ��.S G �

#= − , �� (�� ) �� T �c= = + = , �� ap �.atm =Density of gas ρ . . Density of �aterS G #= 2 10 1000 2 /kg m3 3

# #= =−

From gas equation p RTr= 2 287 298 171 kPa# #= = (absolute pressure)

Also pabsolute p patmospheric gage= + pgage 171 101= − 70 kPa=


For ideal gas 1ρ RTp

�

�= ( )

. . 5.44 /kg m461 273 250

1 31 10461 5431 31 106 6

3

##

##= + = =

For isentropic expansion

TT

�

� pp k

k

�

��

=−

b l

T� ��.

T pp

�� .

.

kk

��

��

�

� ��

#

## #= =

− −

b cl m 393 K=

Now 2ρ RTp

�

�= 2.28 /kg m461 393414 103

3

##= =

FM 1.7 Option (D) is correct.We have �� ,mv �= �� ,��ap�= �� C �T� c= = + = �� ,��ap�= �� C �T� c= = + =The initial mass of Hydrogen in cylinder

m RTp v

��

�= . 271.35 kg0 2968 298800 30

##= =

Final mass of Hydrogen in cylinder

m RTp v

��

�= . 206.99 kg0 2968 293600 30

##= =

Thus the amount of Hydrogen that must be bled off is

mΔ . .271 35 206 99= − 64.36 kg=

FM 1.8 Option (D) is correct.Dalton’s law of Partial Pressure is

ptotal p p vm R T v

m R Tair watera

aw

w= + = +

or mtotal m m R Tp v

R Tp v

a wa

a

w

w= + = + For an ideal gas

Since ptotal � ��atm Pap pair water= + = = pair 101325 101325 4242 97083 Papwater= − = − =

Now ρ vm m

R Tp

R Tpa w

a

a

w

w= + = +

( )287 30 273

97083461 303

4242# #

= + + 287 30397083

461 3034242

# #= +

1.147 /kg m3=





FM 1.9 Option (A) is correct.We have ��.� ��ap�= ��.� ��ap�= �. ��mv v v ��

�= = =From ideal gas law,

Tp v

�

� � Tp v

�

� �=

TT

�

� pp

�

�= v v� �=

T pp T�

�

��#= .

. .T T362 5387 5 1 0691 1#= =

Increase in temperature T T� �= − 1.069 (1.069 1)T T T1 1 1= − = − . T0 069 1=or 6.9% of T1

FM 1.10 Option (B) is correctWe have ��gm = , �� (�� ) �� T c= = + = , pgage 300 kPa=

From gas equation ρ RTpabsolute=

( )

RTp p

�� .atm gage

�

#

#= + = + 3.96 /kg m3=

Thus, Volume of tank v mr= . 4.04 m3 96

16 3= =

FM 1.11 Option (D) is correct

By definition Cacritical 0.25( )

Vp p2 a v

2r= = −

.0 25 ( )

V9982 133000 4242

2#

= −

V .( )

998 0 252 133000 4242

#= −

32.13 /m s=

FM 1.12 Option (B) is correctThe ideal gas formula Predicts:

Speed of sound a . ( )kRT �� # #, = +

. 509 /m s1 33 461 423# #= =

FM 1.13 Option (C) is correct.We have 9 /N m268 3γ = , . /Ns m131 5 2μ =Weight density γ gr=

ρ . . /kg m9 819268 944 75 3= =

Kinematic viscosity ν ..

944 75131 5

rm= =

0. . /secN m kg139= 0. / .secm139 2=

FM 1.14 Option (A) is correct.The volume of blimp is

v ( )R L32

32 15 722 2# # #p p= = 33929 m3=

From the ideal gas law, the respective densities of helium and air





(a) Heρ � ( )

�� R Tp

��

He

He �

#

#= = =

(b) airρ �� R Tp

��

air

air �

# #= = =

Then the respective gas weights are

WHe �� gvHer # #= = = Wair �� gvairr # #= = =


We have .5 10 /Ns m4 6 2 2μ #= - , . �y �� #= − , � .sec�V �= , . �A �� =

When force F is applied on the plate, shear force comes in the action.

F A#t= yV A# #m= y

Vτ μ=

4.56 10.

0.1292 54 10

123

## # #= −

− . N2 32=

FM 1.16 Option (D) is correct.Assuming a linear velocity distribution on each side of the plate.

F A A1 2t t= + hV A h

V A11

22

m m= +b bl l

hV

hV A h h VA

�

�

�

�

�

�

�

�m m m m= + = +; :E D

FM 1.17 Option (C) is correct.From Newton’s law of viscosity

τ dydu

yUm m= =

1τ . .0 02 0 0086

#= 15 /N m2=

and 2τ . .0 01 0 0046

#= 15 /N m2=

FM 1.18 Option (A) is correct.The magnitudes of shear forces acting on the upper and lower surfaces of the plate are

F ,shear upper A A dydu

,w uppert m= =

A hV

hAV�p p

� �m m= − =

F ,shear lower A A dydu

,w lowert m= =

( )

A hV Vp m

�m= −

Both shear forces are acting in the opposite direction of motion of the plate,





therefore from force balancing

F F F� �shear upper shear lower= +

� �

hAV

hA V Vp p m

� �

m m= + −

A hV

hV Vp p m

� �m= + −

; E

FM 1.19 Option (A) is correct.We have �.��. .S G = , 4 10 / .secm4 2ν #= -

From Newton’s law of viscosity (at the surface of plate)

( )y 0τ = dydu

y �

m==

c m ...(i)

dydu

y �=c m cosU y

2 2 y 0

pd

pd=

=a k: D U

2pd=

From equation (i) τ U2#nr pd= μ νρ=

( . . 1000)S G U2n p

d# # # #=

( . ) . U4 10 0 91 1000 1 574# # # # # d= −

0.571 /N mU 2

d=

FM 1.20 Option (B) is correct.We have �� ,�gm = V 0.8 / ,m s= 0.006 Pa sμ -= y 0.8 8 10mm m4

#= = −

The force balance from figure gives

��FxΣ = � �� cos sinF F F�shear Nc c− − = ...(i)

��FyΣ = �� cos sinF F WN shearc c− − = ...(ii)

Weight W �� Nm g# #= = =

and Fshear As st=

(�.��) (�.� �.�) .A yV

� ��

s �#

m # # #= = −

Fshear 0.9 N=

Equation (ii) gives FN 20

( 20 )cossinF Wshear

cc= +

. 159.95cossin N

200 9 20 150#

cc= + =

By substituting the value of Fshear and FN in equation (i),we get

F cos sinF F�� shear Nc c= + . .cos sin0 9 20 159 95 20# #c c= + 55.55 N=





FM 1.21 Option (C) is correct.We haveThe saturation pressure of water at 45 9.593C kPac =When air is fully escaped, the space is filled with vapor and the container have a two-phase mixture of saturated water vapor.Then vapor pressure �.��aP P (�� )Cv saturation= =c

FM 1.22 Option (D) is correct.Shear stress is given by

τ .. ��aL

V��

�� #m= = =

and the Reynolds Number is

Re ..VL

�� # # bm

r= =


Velocity profile u ( � )u r R� n n�= −

We know that the wall shear stress in pipe flow

wτ drdu

r Rm=−

=

u drd

Rr� n

n

r R�m=− −

=: D u

Rnr

n

n

r R�

�

m=− − −

=; E R

u n�m=

Then the drag force applied by the fluid on the pipe wall becomes

F Aw wt= ( )Rn u R L n u L� ��

�# #m p pm= =

FM 1.24 Option (B) is correct.(a) From the Power-law for air

pμ TT n

00

m= b l 1.8 10 3.221 10 /kg m s293673 .

50 7

5# # # −= =− −

b l

(b) From the sutherland law

sμ ( / ) ( )

T ST T T S.

00

1 50m= ++

; E

1.8 10( )

( / ) ( )673 110

673 293 293 110.5

1 5#

#= ++−

= G

3.225 10 /kg m s5# −= −

FM 1.25 Option (A) is correct.Assume a linear viscous velocity distribution in the film below the block. Then a force balance in x - direction gives:

FxΣ sinW Aq t= −

�sinW hV A maXq m= − = =: D

or sinW θ hV Am=

V sin sinA

hWA

mghm

qm

q= = W mg=

FM 1.26 Option (C) is correct.Due to the flow of glycerin, shear force acting in the opposite direction to this





flow. The FBD is shown below.

In equilibrium condition

�FxΣ = sinW α l �# #t= b �= sinmg α l#t=

sinvgρ α l#t= m Vr=

sing vgγ α l#t= gρ γ=

�sinl hγ α# # # l#t= τ sinhg a= ...(i)From the Newton’s law of viscosity, shear stress at the plate ( )� 0=

τ dydu

y �

m==

c m hU

hUy� �

y�

�

m= −=

; E hU�m= ...(ii)

From equation (i) and (ii), we get

hU�μ sinhg a=

U sin�2

2

mg a=

FM 1.27 Option (A) is correct.Assuming a linear velocity distribution in the clearance, the force is balanced by resisting shear stress in the oil.

F ( )A RV D Lwall it m pD #= = b l

F R RV D L

i

i

�

m p= − ...(i)

For the given oil

μ ( )�� rn r n# #= =

0.8 998 0.003 2.63 /kg m s7# # −= =Then by substituting in equation (i), we get

F ( . . )

. . . .0 0401 0 0400

2 63 0 4 0 08 0 3# # # #p= − 79 . 793 N2 79 b=


List-I List-II

a. Ideal fluid 2. Fluid of zero viscosity

b. Newtonian fluid 1. Viscosity does not vary with rate of deformation

c. /μ ρ 5. Kinematic viscosity

d. Mercury in glass 4. Capillary depression.

So , correct pairs are a-2, b-1, c-5, d-4.





FM 1.29 Option (D) is correct.

List-I List-II

a. Capillarity 4. Surface tension

b. Vapour pressure 1. Cavitation

c. Viscosity 3. Shear forces

d. Specific gravity 2. Density of water

So, correct pairs are a-4, b-2, c-3, d-2.

FM 1.30 Option (C) is correctFor a droplet or bubble with one spherical surface

pΔ R2s=

.. .

0 005 102 0 0712

5 102 0 0712

3 6#

#

#

#= =− −

28480 28.5Pa kPa-=

FM 1.31 Option (B) is correct.We have 3 ��d = , 7.3 10 /N m2σ #= -

We know that surface tension on liquid droplet is given by the relation,

p d4s= .

3 104 7 3 10

3

2

#

# #= −

−

97.33 Pa=

FM 1.32 Option (D) is correctFor a liquid cylinder, the internal excess pressure is

pΔ Rs=

σ p RD #= ( . )

200 20 00073

#=

.200 0 000365#= 0.073 /N m=

FM 1.33 Option (B) is correctThere are two surface, inside and outside the ring. So the total force measured is

F ( )D D2 2sp ps= = 2 0.0728 (0.06)p# # #= 0.0274 N=

FM 1.34 Option (D) is correct.We have 4 4 ��mm md 3

#= = −

The capillary effect is given by the equation,

h cosg d

4# #rs q=

where σ = Surface tension in /N m

θ = Angle of contactCapillary effect for water

σ 0.0734 /N m= , 0cθ = , ρ 1000 /kg m3=

h .. cos

1000 9 81 4 104 0 0734 0

3# # #

# c= −

7.48 10 7.48m mm3#= =−

Capillary effect for mercury





σ 0.51 /N m= , 128cθ = ρ �� S G #= 13.6 1000 13600 /kg m3

#= =

h .

. cos13600 9 81 4 10

4 0 51 1283

# # #

# # c= −

2.35 10 m3#=− − 2.35 mm=−

Here the negative sign indicates the capillary depression.

In magnitude h 2.35 mm=

FM 1.35 Option (C) is correct.(a) For water, capillary rise

hw . ( . )

.cos cosgD

��

# ## # c

rs q= = 0.0275 m=

(b) For Mercury

hm . ..cos cos

gD�

��

# ## # c

rs q= = 0.00912 m=−

Here negative sign shows the capillary depression.


We have 960 / ,kg m3ρ = �. �.� ��mm mD � �#= = −

R D�= . 1.9 10 m2

3 8 10 33#

#= =−

−

�.�mmh = 0.0025 ,m= 15contact angle cφ =The surface tension of milk

milkσ . . .cos cosgRh

2 2 15960 9 81 1 9 10 2 5 103 3

## # # # #

cfr= =

− −

0.02315 /N m=


We have ��cmb = , 0.12 ,m= �.��F =From the surface tension force relation,

sσ ( . )

.b

F2 2 0 12

0 018#

= = 0.075 /N m=





FM 1.38 Option (B) is correct.From the figure above, the force balance on the annular fluid is

Force in vertical direction Weight of fluid film= ( )cos r r2 2o i#σ θ π π+ ( )g r r ho i

� �r p#= −

h ( )cos

g r r�

o irs q= −


The needle “dents” the surface downward and the surface tension forces are upward as shown in figure. Then a vertical force balance gives:

Vertical forces Weight of needle=

cos L2 #σ θ g D L��#r p=

cos2σ θ g D�

�

r p=

2σ g D�

�

r p= 0 0 1cos"c cθ = =

σ g D� �

�

#r p=

. . ( . )8

7824 9 81 3 14 0 0016 2# # #=

0.0772 /N m=

FM 1.40 Option (B) is correct.The wall shear stress anywhere on the surface of the frustum at a distance r from the axis of rotation is

wτ drdu

hV

hrm m mw= = =

The shear force on the area dA,

dF dA hr dAwt m w= =

Torque dT rdF hr dA

�

mw= =

T h r dAA

�mw= #The shaft power required at top surface is

P ,shaft top T h r dAA

�#w w mw= = #

h r dAA

��mw= # ...(i)

For the top surface dA rdr2p=

Hence P ,shaft top ( )h r r dr��

r

D��

�

�mw p==

#





h r dr hr� ��

� �

r

D

r

D��

�

� � �

�

�pmw pmw#= =

= =: D# h

D32

2 4pmw=

FM 1.41 Option (C) is correct.By putting the value in expression of shaft power at top (20 )Cc ,

P ,shaft top ( . ). ( . ) ( ) ( . )

hD

32 32 0 00123 14 0 1 200 0 122 4 2 4

#

# # #pmw= =

67.824 W=The power is proportional to viscosity. Thus the power required at 80 Cc is

P , , Cshaft top ��c P , ,C

CCshaft top

20

8020m

m#=

c

cc

.. 67.824 5.29 W0 1

0 0078#= =

Therefore, the reduction in the required power input at 80 Cc is

P P, , , ,C Cshaft top shaft top�� −c c . .67 824 5 29= − 62.533 W=

FM 1.42 Option (C) is correctWith b the width of the plates into the paper, the capillary forces on each wall together balance the weight of fluid held above the free surface.

Weight of fluid Surface tension force= ( . )g h b��# # #ρ ( )cosb2# s q=

or h ( . )cos

g ��#rs q=

. ( . ). cos

998 9 81 0 000752 0 0728 0# ## # c= 0.020 20m mm, =

FM 1.43 Option (A) is correct.We have ��gm = , �.�mmy = , �.�mA �= , 30cα = , 0.4 /Ns m2μ =The FBD of the block shown below.

In equilibrium condition

FxΣ 0= sinW ��c At=

sinmg ��c yV A#m= film thicknessy =

V sinA

mgy ��#

cm= W mg=

. ..

0 4 0 410 10 10 0 54

## # #=

− 0.03125 / secm=

FM 1.44 Option (A) is correct.We have �� ,cm mL = = �� ,rpmN = �.�� .��cm mh = = �� ,cmD = �� .� �.�� ,cm mR = = = �.��T =Torque transmitted by the shaft





T h R RL hR L� ��

�

# #mw p pmw= = ...(i)

and ω 20.94 /rad sN60

260

2 200# #p p= = =

From equation (i),we get

μ . . ( . )

. ( . )R L

T h2 2 3 14 20 94 0 075 1

0 8 0 00123 3

# # # #

#

pw#= =

0.0173 /N s m2−=

FM 1.45 Option (D) is correct.We have 1.2 10 /Ns m3 2μ #= - , � �secmU = , �.�mh =From Newton’s law of viscosity

τ dydum= ...(i)

At the fixed surface (at y �= )

dydu h

UhUy� �

y�

�

= −=

; E hU�=

From equation (i) τ hU�

#m= . .1 2 10 0 12 33

# ##= − 0.072 /N m2=

FM 1.46 Option (C) is correct.According to stokes law

μ DLW t3net #p= ...(i)

The net weight of the sphere in the fluid is

Wnet ( ) ( )g v g D6sphere fluid fluid sphere fluid

3

# # #r r r r p= − = −

( ) . ( . )2700 875 9 81 6 0 0025 3# # #

p= − 1.46 10 N4#= −

Then from equation (i),we get

μ ( . ) ( . )

1.46 103 0 0025 0 75

484

# # #

#

p#=

−^ h 0.397 /kg m s−=

FM 1.47 Option (C) is correctAssuming a linear velocity distribution inside the annular clearance, the shear stress is

τ rV

r rr

i

i

�m m wDD= = − ...(i)

This stress causes a force

dF ( )dA r d Lit t q= = ...(ii)The torque of this force about the shaft axis is

dT r dFi= ...(iii)Put equation (i), (ii) and (iii) together

T ( )r dF r r Ld r r rr r Ldi i i i

i

ii

�

�

��

�

# #t q m w q= = = −p p

# ##

r rr L d r r

r L di

i

i

i

�

�

�

�

�

�

�

�mw q mw q= − = −p p

# # r rr L�

i

i

�

�pmw= −

μ ( )

r LT r r2 i

i3

0

pw= −





FM 1.48 Option (B) is correct.The velocity profile is given by

( )u r uRr�max �

�

= −c m

The shear stress at pipe surface is expressed as

sτ drdu u dr

dRr�max

r R r R�

�

#m m=− =− −= =

; E Ru� maxm=

Then the friction drag force

FD ( )A Ru RL� �max

s st m p= = �A RLs p=

FD Lu4 maxpm= ...(i)By substituting the given values in equation (i), we get

FD . ( . ) ( )4 3 14 0 0010 15 5# # # #= 0.942 N=


In the figure WT 125tanWeight of k N= = WL Weight of liquid mg vg vr g= = = =where γ 10.9 /specific weight of liquid kN m3= =

WL . (1) 510 9 10 43 2

#p

# # #= 42.8 kN=

From the Newton’s law of motion in vertical direction

FyΣ may= F W WV T L− − may=

��F ��V − − . 2.759 81125 42800= +b l

F ��V − 12033= FV 54958 55N kN-=

FM 1.50 Option (A) is correct.At terminal velocity, the rod weight should equal the viscous drag.

W Viscous Drag=

g vs #ρ ( )/D D

V DL2o

#m p= −; E

g D L�s�

# #ρ π ( )/D D

V DL2o

m p= −

V ( )gD D D8

s o

mr= −





FM 1.51 Option (C) is correctThe power is the viscous resisting force times the belt velocity.

P Viscous resisting force Velocity#= A Voil belt beltt # #=

hV b L V# # #m= b ^l h V b h

L�m=

By substituting the values, we get

P . ( . ) ( . ) .0 29 2 5 0 9 0 0642

# # #= 108.75 109W W,=


We have �� ,�g msteel�ρ = �� ,�g maluminum

�ρ = �.�� mwaterσ =From surface tension force relation,

F Ds sp s= and ��W mg vg g D�r r p= = =

When the ball floats Fs W= D sπ σ ��g D�r p=

D g�s

rs=

For Steel Dsteel ( ) .( . )

�.� �� mg�

��

steel

s �

#

#rs

#= = = −

2.4 mm=

For Aluminum Daluminum .( . )

�.� �� mg�

��

aluminium

s �

# #

#r

s#= = = −

4.1 mm=Hence Daluminum D> steel

So aluminum ball would be larger in size.

FM 1.53 Option (C) is correctFor any radius r R# , the liquid gap is tanh r q= . Then

dT dA rt= w

.tan cosrr r dr r�m q

w p q= a bk l cosL drq=

T sin sin�� 23

2�2

0

3

qpwm

qpwm= =#

μ sin�

�23

3pwq=





Substituting the numerical values, we get

μ ( . ) ( . )

. sin2 94 2 0 06

3 0 157 33

# #

# # cp

= 0.193 /kg m s−=

FM 1.54 Option (D) is correct.At any radial position r r< � on the cone surface and instantaneous rate ω dT r dAt= w

sinr hr r dr�#m w p q#= 9 :C D sinh r dr� �

#qmw p=

Torque T sinh r dr�r

�

�

�

wmw p= # sinh

r2

04

qpmw= ....(i)

Since for cone T I dtd

�w=− mr dt

d103

02#

w=− For cone I mr103

0 02=

Then from equation (i),

mr dtd

103

02 w− sinh

r2

04

qpm w=

Separating the variables and integrating both the sides,

d�

ωω

ω

ω#

sinmr hr dt

3 210 t

02

04

0#

#

qpm= − #

or ω exp sinmhr t

35

002

w qpm= −; E

FM 1.55 Option (B) is correct.For the fluid in the annular region

TA RdF R dA R RR RLd

�

�

# # #t m w qD= = =p

b l###

RR L� �pmw

D=

Now Tbottom r dA r Rr rdr�

R

�t m w pD= = a k# # R r dr� R

�

�

pwmD= #

RR

42 4pwm

D=

Ttotal RR L

RR�

�� pwm pwm

D D= +

μ ( / )R L R

T R2 43pw

D=+

μ ( / )

( )R L RT r R

2 43pw=

+−

FM 1.56 Option (B) is correct.We have �� . ,mm mL �= = 0.008 / . ,kg m sμ = �.� �.��mm mtfilm = = �� .�� ,mm mD = = ��rpmN =

Torque is given by T Areat Rfilm

�mw# #=

T tR RL t

R L� �film film

� �

#mw p pmw= = �A RLs p=

T ( � )

tN R L� � ��

film

�#pm p= N

602ω π=

T tNR L

604

film

2 3

#

p m=





By substituting numerical values

T .. ( . ) .

60 0 00124 0 008 750 0 04 0 32 3

#

# # # # #p=

FM 1.57 Option (A) is correct.At any r R# , the viscous shear on both sides of the disk

totalτ 2 2 hrt mw

# #= =

and viscous force

dF dAtotal #t= w (� )hr rdr�mw p#=

Then viscous torque

dT .dF r= .hr dr r� �pmw= h

r dr� �pmw=

Integrating both the sides

T .h r dr�r

R�

�

pmw==

# hR

hR�

��

#pmw pmw= =

Substituting the numerical values

T ( . )

. ( . )N0 001 60

0 29 2 0 05 4

#

# # # #p p= N60

2ω π=

( . ). ( . )0 001 60

2 0 29 1200 0 052 4

#

# # # #p= 0.716 N m−=

FM 1.58 Option (A) is correctThe masses remain the same for an isothermal process of an ideal gas

m m� �+ m�= v v1 1 2 2ρ ρ+ v3 3r=

or RTp D RT

p D� ��

��

��π π

# #+9 9 9 9C C C C RTp D�

��p

#= 9 9C C

,RTp v D�

�ρ π= =

� �RT

p rD RT

p rD

��

��

a a��

��s p s p

# #+ + +

; 9 ; 9E C E C �

RTp r

D�

�a �

��s p

#= +; 9E C

P P R�

a�s− =

The temperature cancels out, and we may clean up and rearrange as follows

�p D Da ��

��s+ ( 8 ) ( 8 )p D D p D Da a2

322

13

12s s= + + +

FM 1.59 Option (C) is correct

The skate bottom and the melted ice are like two parallel plates.





τ hVm=

F A hVLbt m= =

Using F ma= to find the stopping distance

FxΣ F hVLb ma m dt

dV�x

m=− =− = = The ‘2’ is for two blades

Separate and integrate once to find the velocity

VdV

V

V

o

# mhLb dt�t

�

m=− #

or log VV

eo

: D mhLb t2m=−

or V V eo mhLb

t2

=m−

Integrate once again to find distance

x Vdt V e dto mhLb

t

�

2

�= =

3 3 m−

# #

LbV mh2

o

m=


Consider the right side of the liquid column, the surface tension acts tangent to the local surface that is along the dashed line at right. This force has magnitude F bs= as shown. Its vertical component is ( )cosF θ φ- as shown. There are two plates, therefore the total vertical force on the liquid column is

Fvertical ( )cosb2s q f= −Then the vertical force holds up the entire weight of liquid column between plates, which is

W ( )tangbh L hr f= −Set W equal to F, we get

( )cos�2σ θ φ- ( )tangbh L hr f= −

or σ ( )

( )cos

tan�� 2 q f

r f= −−

***********

FM 2PRESSURE AND FLUID STATICS

FM 2.1 The barometric reading for a wall is given as 511 mmHg at the top and 5 8.5 mmHg8 at the bottom. For average air density of 1.18 /kg m3, the height of wall is ( 13600 / )kg mHg

3ρ =(A) 205 m (B) 202 m

(C) 210 m (D) 200 m

FM 2.2 A vertical clean glass Piezometer tube has an inside diameter of 4 mm. When a pressure is applied, water at 26°C ( 9790 /N m3γ = , 0.073 /N mσ = , 0cθ = ) rises into the tube to a height of 23.5 cm. After correcting for surface tension the applied pressure will be(A) 147 Pa

(B) 2448 Pa

(C) 2300 Pa

(D) 2154 Pa

FM 2.3 Consider a frictionless piston-cylinder of a gas car as shown in figure. The mass of piston is 4 kg and cross-sectional area is 35 cm2. During the compression stroke of car engine a force of 70 N is exerted on the piston. If the atmospheric pressure is 105 kPa, the pressure inside the cylinder is

(A) 133.5 kPa (B) 13.35 kPa

(C) 60 kPa (D) None of these

FM 2.4 All fluids in figure shown below are at 20 Cc . What will be the pΔ between points A and B ?

Take the specific weights to beBenzene : 8640 /N m3 Mercury : 133100 /N m3

Kerosene : 7885 /N m3 Water : 9790 /N m3





air : 12 /N m3

(A) 16 kPa (B) 13.35 kPa

(C) 29.35 kPa (D) 26.17 kPa

FM 2.5 A one-tone load on the hydraulic lift shown in figure is to be raised by pouring oil into a thin tube. The density of oil is 780 /kg m3 and diameter of hydraulic lift is 1.2 m. The height h , in order to begin to raise the weight should be

(A) 1.34 m

(B) 0.134 m

(C) 1.134 m

(D) 0.1134 m

FM 2.6 A closed cylindrical tank filled with water has a hemispherical dome and is connected to a piping system shown in figure below. The top part of the piping system has a liquid of specific gravity .0 7 and the remaining parts of the system are filled with water. What will be the pressure at point A ?

(A) 33.35 kPa (B) 62.78 kPa

(C) 3.93 kPa (D) 6.278 kPa

FM 2.7 Water flows upward in a pipe inclined at 45c as shown in figure below and the pressure difference between points (1) and (2) in the pipe is 34.4 kPa. What will be the mercury manometer reading h ?





(A) 20 cm (B) 44 cm

(C) 36 cm (D) 12 cm

FM 2.8 The gage pressure of the air in the water tank shown in figure below is 59Kpa. The differential height hHg of the mercury column will be( 13.6). .S G mercury =

(A) 30 cm (B) 36 cm

(C) 13.6 cm (D) 51 cm

FM 2.9 The right leg of the manometer is open to the atmosphere as shown in figure. The gage pressure in the air gap in the tank is 25.68 kPa. What will be the specific weight of the oil in /N m3 ?

(A) 7831 (B) 10815

(C) 11355 (D) 5678





FM 2.10 An inverted U-tube manometer containing oil having specific gravity of 0.95, is located between two reservoirs as shown in the figure. The reservoir on the right, contains water and is open to the atmosphere and the reservoir on the left contains glycerin is closed and pressurized to kPa45 . What will be the depth of water h in the figure ? ( ��.� �� mglycerin

�γ = )

(A) 0.721 m (B) 6.21 m

(C) 7.21 m (D) . m0 0721

FM 2.11 A water tank is divided into two compartments as shown in figure. An oil with density 5��.5 �kg moil

�ρ = is poured into one side and the water level rises a certain amount on the other side to overcome this effect. The oil does not mix with water. What will be the final differential height of water shown in figure ?

(A) 33. 5 cm7 (B) 60 cm

(C) 45 cm (D) 0.3375 cm

FM 2.12 A tank contains water ( 9790 /N m3γ = ) and immiscible oil at 20 Cc as shown in figure below. If the specific weight of oil is 8809 /N m3, what will be the h ?

(A) 26 cm (B) 20 cm

(C) 10 cm (D) 13 cm





FM 2.13 A tank is constructed of a series of cylinders as shown in figure. A mercury manometer is attached to the bottom of the tank. What will be the manometer reading h ?

(A) 37.6 m (B) 3.76 m

(C) 0.0376 m (D) 0.376 m

FM 2.14 The U-tube at right has a 1 cm internal diameter and contains a liquid (S.G. = 1.6) as shown in figure below. If 20 cm3 of water ( 9790 /N m3γ = ) is poured into the right-hand leg, what will be the free surface height in each leg at equilibrium ?

(A) �.�� , ��.��cm cmh hR L= = (B) ��.�6 , ��.�cm cmh hR L= =(C) ��.� , ��.�6cm cmh hR L= = (D) ��.��cmhR = , �.��cmhL =

FM 2.15 Two compartments A and B of the tank are closed and filled with air and a liquid shown in figure below. The liquid having the specific gravity of 0.6. If the pressure gage reads 3.5 kPa and weight of the air is negligible, the manometer reading h will be

(A) 0.0424 m (B) 0.212 m

(C) 2.12 m (D) . m1 912





FM 2.16 In figure given below, both ends of the manometer are open to the atmosphere. The specific gravity of fluid X is

(A) 0.72 (B) 13.6

(C) 6.8 (D) 1.40

FM 2.17 Consider the closed tank as shown in figure below. All fluids are at 20 Cc and air space is pressurized. If the outward net hydrostatic force on the 30 cm by 30 cm panel at the bottom is 3456 N, the pressure in the air space and the reading h on the manometer respectively, are (Take 8720 �N moil

3γ = , 6670 �N mgas3γ = )

(A) 58.6 kPa, 0.44 m (B) 71.8 kPa, 0.54 m

(C) 69.2 kPa, 0.52 m (D) 60.7 kPa, 0.46 m

FM 2.18 A 15 cm diameter piston is located within a cylinder which is connected to 1.3 cm diameter inclined tube manometer as shown in figure below. The fluid in the cylinder and the manometer is SAE 30 oil (specific weight 8.95 /kN m3= ). If a weight W is placed on the cylinder, the fluid level in the manometer tube rises from point (A) to point (B). What will be the weight ?

(A) 1.18 N8 (B) 11.8 N8

(C) 1.18 kN8 (D) 11.8 kN8

FM 2.19 Consider the figure given below. The pressure at A and B are the same as 100 kPa. The water is introduced at A to increase pA upto 230 kPa. If the connecting





tube is uniform 1 cm in diameter and the liquid densities dose not change, what will be the change in the mercury menisci ( hΔ ) ?

(A) 16.2 cm (B) 8.1 cm

(C) 23.4 cm (D) 11.7 cm

FM 2.20 Consider the flow of water through a pipe as given in figure below. For the given values, the pressure difference between the pipe pressure and pressure gage is

(A) 3.6 kPa (B) 4 kPa

(C) 1.3 kPa (D) 7 kPa

FM 2.21 The fuel gage for an automobile tank reads proportional to the bottom gage pressure as shown in figure below. If the tank accidentally contains 3 cm of water plus gasoline (S.G. = 0.68), how many centimeters “h” of air remain when the gage reads “full” in error ?

(A) 1.4 cmh = (B) 0.14 cmh =(C) 1.04 cmh = (D) 14 cmh =

FM 2.22 Consider the system shown in figure below. If a change of 0.7 kPa in the pressure of air causes the glycerin-mercury interface in the right column to drop by 5 mm in the glycerin ( . . 1.26)S G = level in the right column while the pressure in the glycerin pipe remains constant, the ratio of �A A2 1 is





(A) 1.45 (B) 0.290

(C) 0.145 (D) 0.0145

FM 2.23 The sensor A reads 1.5 kPa (gage) as shown in figure below. All fluids are at 20 Cc. What will be the elevations Z in meters of liquid levels in the open piezometer tubes B and C ?

(Take 12 �� mair�γ = , ��20 �� moil

�γ = , 12��0 �� mglycerin�γ = )

(A) 2.��mZB = , 2.1�mZC = (B) 2.1�mZB = , 2.��mZC =(C) 1.�5 mZB = , 2.1�mZC = (D) 2.1�mZB = , 1.�5 mZC =

FM 2.24 A inclined differential manometer shown in figure contains Carbon tetrachloride (specific weight . /kN m15 6 3= ). Initially the pressure difference between pipes A and B is zero. It is measured that for a pressure difference of 0.7 kPa , the manometer gives a differential reading of 30 cm (when measured along the inclined tube). If pipe A and B contains water, what will be the angle of inclination θ in degrees ?





(A) 45 (B) 11.85

(C) 23.70 (D) 47.40

FM 2.25 The containers A and B are cylindrical and are such that p pA B= as shown in figure below. If the oil-water interface on the right moves up a distance h h<Δ, the expression for the difference p pA B− in terms of hΔ , specific weight, d and D will be

(A) ( � ) ( � )h d D d D1 1H O oil2 2 2 2

2γ γΔ + - -6 @

(B) ( � ) ( � )h d D d D1 12 2H O oil

2 22γ γΔ + + -6 @

(C) ( � ) ( � )h d D d D1 12 2H O oil

2 22γ γΔ - + -6 @

(D) ( � ) ( � )h d D d D1 12 2H O oil

2 22γ γΔ - - +6 @

FM 2.26 Water initially fills the funnel and its connecting tube as shown in figure. Oil is poured into the funnel until it reaches a level h > H

2 . What will be the rise in the water level ( )�0 in the tube in terms of l with 0.�mH D= = and 0.03 md = ?

(A) 0.03(1 0.1 )h l �0

1 3= − (B) 0.027(1 0.1 )h l �0

1 3= −(C) 0.3(0.1 )h l �

01 3= − (D) 0.3(1 0.1 )h l �

01 3= −

FM 2.27 Consider a large cubic ice block floating in sea water. The densities of ice and seawater are 920 /kg m3 and 1025 /kg m3, respectively. If the height of ice block below the surface is 87.5 cm then the height of ice block extends above the surface of water is





(A) 0.973 cm (B) 0.0973 cm

(C) 9.73 cm (D) 97.3 cm

FM 2.28 A 5.5 cm diameter circular Pine rod ( . . �.�5S G = ) is 2.2 m long. How much lead ( . . ��.�S G = ) is needed at one end as shown in figure so that the rod will float vertically with 30 cm out of the water ?

(A) 78 N (B) 85 N

(C) 12 N (D) 20 N

FM 2.29 A 61 cm thick block constructed of wood ( . . 0.65)S G = is submerged in water and has a 75 cm thick metal plate ( 26.4 / )specific weight kN m3= attached to the bottom as shown in figure. What amount of force is required to hold the block in its original position ?

(A) N370 (B) N350

(C) 7 N40 (D) 790 N

FM 2.30 A ball of mass 5.0 gm and diameter of 4 cm floats in water at 20 Cc at a depth h . If ��8 �kg mwater

3ρ = and 1.225 �kg mair3ρ = , what will be the h at which the

ball float in water ?(A) 7.3 mm (B) 20 mm

(C) 10 mm (D) 0.73 mm

FM 2.31 The uniform rod ( . . 0.636S G = ) shown in the figure is hinged at B and in static equilibrium when 2 kg of lead ( . . 11.4S G = ) are attached at its end. What is the length of the rod ?

(A) 2.2 m (B) 8 m

(C) 4 m (D) 1.8 m





FM 2.32 Consider the figure shown below. The cylindrical tank has a 35 cm high cylindrical insert in the bottom. The pressure at point B is 160 kPa. If the air pressure outside the tank is neglected, what will be the pressure in the air space and the force on the top of the insert, respectively ?

(A) 166 kPa, 1230 N (B) 166 kPa, 1284 N

(C) 154 kPa, 1284 N (D) 154 kPa, 1230 N

FM 2.33 A uniform block of size L h b# # with b , h L<< is shown in figure below. A uniform heavy ball tied to the left corner causes the block to float exactly on its diagonal. What will be the expression of diameter D of sphere ?

(A) ( . . )

DS G

Lhb1

�1 3p= −; E (B) ( . . )

DS GLhb

11�2

p= −; E

(C) ( . . )

DS GLhb

11�3

p= −; E (D) ( . . )

DS GLhb

11�3

p= +; E

FM 2.34 Consider a cylinder ( . . 0.6S G = ) of 1 m in diameter and 0.8 long. If the cylinder placed to float with its axis vertical in oil ( . . 0.85)S G = as shown in figure, would this cylinder be

(A) Unstable (B) Stable

(C) Not determined (D) None of these





FM 2.35 The tank of water in the figure below accelerates to the right with fluid in rigid body motion. What will be the gage pressure at point A ?

(A) 2938 Pa (B) 1910 Pa

(C) 10010 Pa (D) 3157 Pa

FM 2.36 A 18 cm diameter cane has 18 cm of water, overlaid with 15 cm of oil ( 891 /kg m3ρ =). It is rotated about the centre in rigid-body motion at 150 rpm as shown in figure below. What will be the hΔ and maximum fluid pressure ?

(A) 5.1 cm, 3573 Pa (B) 5.1 cm, 4018 Pa

(C) 10.2 cm, 4072 Pa (D) 10.2 cm, 4964 Pa

FM 2.37 An open tank with diameter D contains water at a depth of meterH when at rest. As the tank is rotated about its vertical axis, the centre of the fluid surface is depressed. If water is spilled from the tank, the relation between initial fluid level and the angular velocity will be

(A) RgH4ω = (B)

RgH2

2ω =

(C) R gH1 2ω = (D) R gH2ω =

FM 2.38 Consider the U-tube fluid in figure below. If the fluid is water at 20 Cc , what will be the uniform rotation rate about axis C for the position shown ?





(A) 195 rpm (B) 107 rpm

(C) 139 rpm (D) 151 rpm

FM 2.39 An inverted hollow cone is pushed into the liquid as shown in figure. If the temperature of air within the cone remains constant, what will be the expression for the distance l that the water rises in the cone as a function of depth d and height H ?

(A) . � ( ) �d H H l10 3 13 3− − −− − (B) 10.3� ( ) 1�d H H l3 3− − −−

(C) 10.3� ( ) 1�d H H l3 3− − −− (D) . � ( ) �d H H l10 3 13 3− − −

FM 2.40 A conical shape plug is situated at the bottom of a pressurized tank as shown in figure. The air pressure is 45 kPa and the liquid in the tank has specific gravity of 2.75. What will be the magnitude of the thrust exerted on the curved surface of the cone within the tank due to the air pressure and the liquid ?

(A) 136.55 kN (B) . kN89 43

(C) 47.12 kN (D) 183.67 kN

FM 2.41 A circular gate ABC is hinged at B as shown in figure below. For arbitrary depth h and gate radius R, the force P just sufficient to keep the gate from opening is





(A) P hR��p g= (B) P R�

�p g=

(C) P R��p g= (D) P R�

�p g=

FM 2.42 Gate AB as shown in figure is 1.2 m long and 0.8 m wide into the paper. If atmospheric pressure effects are neglected, the force F on the gate and its center of pressure position X are

(A) �822� , �.1�1�� mF X= = (B) �822� , 0.01�� mF X= =(C) �822� , 0.�1�� mF X= = (D) �0�80 , 0.�8�� mF X= =

FM 2.43 A water tank with a quarter circle panel at the bottom is shown in figure below. What will be the horizontal and vertical components of the hydrostatic forces on the quarter circle panel ?

(A) �� , 10��k� k�F FH V= =

(B) 2�� , 10��k� k�F FH V= =(C) 10�� , 2��k� k�F FH V= =

(D) 10�� , ��k� k�F FH V= =

FM 2.44 A 122 cm wide gate pivots about the hinge point O as shown in figure. If the water depth h is 1.6 m, the counter weight W is(A) 9 kN (B) 0.9 kN

(C) 9.32 kN (D) 90 kN





FM 2.45 A cylindrical tank is shown in figure. The pressure in the air gap is 8000 Pa gage. If the net hydrostatic force on the annular plane BB is 853 N, the height h and net hydrostatic force on the bottom of the tank are

(A) �� , 8��cm �h Fbottom= = (B) ,cm �h F�0� ��8�bottom= =(C) �� ,cm �h F ��8�bottom= = (D) , 8��cm �h F�0� bottom= =

FM 2.46 A 3 m long cylinder floats in water and rests against a wall as shown figure below. What will be the horizontal force that cylinder exerts on the wall at the point of contact ?

(A) 14.7 kN (B) 58.8 kN

(C) 44.1 kN (D) 28.14 kN

FM 2.47 A water trough of semicircular cross section of radius 0.5 m consists of two symmetric parts hinged to each other at bottom as shown in figure below. The two parts are held together by a cable and turn buckle placed every 4.5 m along the length of the trough. The tension in each cable when the trough is filled to the rim, is

(A) 8663 N (B) 5519 N

(C) 5510 N (D) 10271 N

FM 2.48 A water tank consists of two half cylinders, each weighing 1125 N, bolted together as shown in figure below. If the end caps are neglected and the diameter of water tank is 4 m, the force in each bolt is





(A) 54 kN (B) 46 kN

(C) 23 kN (D) 24 kN

FM 2.49 The tank in figure shown below has a 4 cm diameter plug which will pop out if the hydrostatic force on it reaches 25 N. For 20 Cc fluids, what will be the reading h on the manometer when this happens ?

(A) 69 cm (B) 6.9 cm

(C) 0.69 cm (D) 90.4 cm

FM 2.50 A quarter circle gate BC as shown in figure, is 6 m long and hinged at C . If the weight of the gate and friction at the hinge are negligible then the horizontal force P required to hold the gate stationary, is

(A) 111.60 kN (B) 11.16 kN

(C) 111.60 N (D) 1116 kN

FM 2.51 A semicircular gate AB is shown in figure. It is hinged at B and held by a horizontal force of 366 kN at point A. What will be the height h of water above the gate AB ?





(A) 4 m (B) 8.5 m

(C) 5 m (D) 7.47 m

FM 2.52 Two square cross-sectional gates hinged at one end to a conduit are used to close two openings in it and the conduit is connected to an open tank of water as shown in figure. When the water depth reaches 5 m, it is examine that both gates open at the same time. If the weight of the vertical gate and friction in the hinges is negligible, what will be the weight of the horizontal gate and the horizontal force P , acting on the vertical gate to keep the gates closed until this depth is reached, respectively ?

(A) 497 kN, 471 kN (B) 471 kN, 314 kN

(C) 314 kN, 497 kN (D) 471 kN, 497 kN

FM 2.53 A m5 high, 6 m wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B as shown in figure. What will be the hydrostatic force exerted on the gate by the 5 m high water and the location of the pressure centre ?

(A) 265 kN, 4.25 m (B) 605 kN, 3.5 m

(C) 441 kN, 0.215 m (D) 618 kN, .71 m3

FM 2.54 A vertical lock gate is 5m wide and separates oil (S.G. = 0.82) levels of 5 m and 3 m, respectively as shown in figure. The moment about the bottom required to keep the gate stationary is

(A) 190 kN m− (B) 234 kN m−

(C) 271 kN m− (D) 127 kN m−





FM 2.55 A rectangular plate blocks the end of a freshwater channel as shown in figure below. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from opening by a fixed ridge at point B . If the width of the plate is b , the force exerted on the plate by the ridge is

(A) gh b1130 2ρ (B) gh b30

11 2 2ρ(C) gh b30

11 3ρ (D) gh b3011 2ρ

FM 2.56 A weightless gate AB has length L and width b into the paper, is hinged at B . The liquid level h remains at the top of the gate for any angle θ as shown in figure below. An analytic expression for the force P perpendicular to AB , required to keep the gate in equilibrium is

(A) sinP hb L hL

� ��g q= −b bl l (B) sinP hb L h

L� �

�g q= −b bl l

(C) sinPhb

L hL

� ��g q= +b bl l (D) sinP hb L h

L� �

�g q= +b bl l

FM 2.57 The vertical cross section of a closed 6 m long storage tank is shown in the figure. The tank contains gasoline and the air pressure is 0 kPa3 . The magnitude of the resultant force acting on one end of the tank is

(A) 414 kN (B)818 kN

(C) .5 kN613 (D) 426.72 kN





FM 2.58 A rigid gate OAB is hinged at O and rests against a rigid support at B as shown in figure. If the gate is 3 m into the paper, what minimum horizontal force R is required to hold the gate closed when the weight of the gate and friction at the hinge are neglected ?

(A) 393 N (B) 39.3 kN

(C) 393 kN (D) 3930 kN

FM 2.59 Uniform body A in the figure has width b into the paper and it is in static equilibrium when pivoted about hinge O. What is the specific gravity of this body ?

(A) ��

23 +; E (B) �

�32 +; E

(C) ��

23 1

+−

; E (D) ��

32 1

+−

; E

FM 2.60 A cylindrical Mass m , is connected to a 3 m wide rectangular gate as shown in figure. The height of water level is H . If the friction at the gate hinge and at the pulley are negligible. The expression for cylindrical Mass m will be

(A) ( )�� 4 12γ π- +6 @ (B) ( 1)��4 2

3γ π+ -; E





(C) ( 1)gH H4 2

3γ π+ -: D (D) ( )gH H4 2 1

3γ π+ -; E

FM 2.61 An air container is placed on the Lake floor as shown in figure. A m5 diameter hatch is located on an inclined wall of container and hinged on one edge. The water surface lies m4 above the hinge point. Neglecting the weight of the hatch and friction in the hinge. What will be the minimum air pressure within the container to open the hatch ? ( 10.1 /kN m3γ = )

(A) 28.98 kPa (B) 289.8 kPa

(C) 2.898 kPa (D) 30.6 kPa

FM 2.62 What will be the height H (in terms of R) as shown in figure, for which the hydrostatic force on the rectangular panel is equal to the force on the semicircular panel ?

(A) �.�H R= (B) H R�=(C) .H R��= (D) .H R��=

FM 2.63 A closed tank is filled with liquid ( 10.1 / )kN m3γ = and has a hemispherical dome as shown in the figure. A U-tube manometer is connected to the tank and filled with the fluid having the S.G. of .3 0. If the differential manometer reading is 2.1 m, the vertical force of the water on the dome is

(A) 159.4 kN (B) 150.3 kN

(C) . kN225 45 (D) . kN77 5

***********





SOLUTIONS

FM 2.1 Option (B) is correct.Atmospheric pressures at the top and at bottom of wall are

ptop ( )gh

1000Hg topr= . . 68.175 kPa1000

13600 9 81 0 511# #= =

pbottom ( )gh

1000Hg bottomr= . . 70.51 kPa1000

13600 9 81 0 5285# #= =

From balancing ( )gh airρ p pbottom top= −

. . h1000

1 18 9 81# # . .70 51 68 175= −

h . .. .1 18 9 81

2 335 1000 201 71##= =

202 ,m, which is also the height of wall.

FM 2.2 Option (D) is correctThe capillary rise in the tube

hcap cosR

�gs q= .

. cos9790 0 001

2 0 073 0#

# # c= 0.015 m=

Then the rise due to applied pressure is

hpress . .0 235 0 015= − 0.22 m=The applied pressure is

p hpressg= .9790 0 22#= 2154 Pa=


We have � ��gmpiston = �� cm mA � � �#= = −

��apatm = , ��F .comp =From force balance pA p A W F .atm comp= + +

Thus p p AW F

p Amg F.

..

atmcomp

atmcomp= + + = + +

(105)( )

( . )35 10 1000

4 9 81 604

# #

#= + +− 133.5 kPa=

FM 2.4 Option (B) is correct.Applying hydrostatic relation between point A to B:

�� (�.�) �� (�.��) �� (�.��)pA # # #+ − −





( . ) ( . )9790 0 39 12 0 135# #+ − pB=or p pA B− . . .2592 15972 3784 8 3818 1 1 62=− + + − + pΔ 133 . Pa48 32= 13.35 kPa-


pgage ��

AW

Dmg

Dmg

�� p p= = =

Since pgage ghr=

h gp

D gmg�gage

�#r p r

= =

( . ) .D

m��

� ��

# #

#rp

= = 1.134 m=

FM 2.6 Option (B) is correct.By adding the pressure which are acting on the system (Manometer equation)

( . . ) � �p S G g gA H O H O� �r r# # #+ + pB= pA . . .103000 0 7 1000 9 81 3 1000 9 81 2# # # # #= − − 103000 20601 19620= − − 62779 62.78Pa kPa-=

FM 2.7 Option (D) is correct.The vertical distance between points (1) and (2) is 2.0 45 2tan mc#= =Applying hydrostatic relation at point (1) and (2),

�� 1��1�� 2p h h1 # #+ − − p2= p p1 2− h h133100 9790 9790 2#= − +or 34400 h123310 19580= +

h 12331034400 19580= − 0.12 m= or 12 cm

FM 2.8 Option (B) is correct.For given arrangement, manometer equation is

p gh ghw w Hg Hg1 r r+ − p .atm= p p .atm1 − gh ghHg Hg w wr r= − pgage � � �g h hw Hg w Hg wr r r #= − )p p p. .abs gage atm= +

gp

w

gage

ρ . .S G h hHg Hg w#= − . .S Gw

HgHgρ

ρ =

.45 1000

1000 9 81## . .h13 6 0 3Hg#= − 1�� g mw

�ρ =

hHg 0.36m= or 36 cm

FM 2.9 Option (A) is correctThe two 8 cm legs of air are negligible (only 2 Pa). Begin at the right mercury interface and go to the air gap:

( . . ) ( . . . )133100 0 12 0 09 0 09 0 12 0 08oil# #g+ − + + pair gap=or .27951 0 29 oil# g− 25680=

or oilγ .0 2927951 25680= −

.7831 03= 783 /N m1 3-





FM 2.10 Option (C) is correct.Let pA be the air pressure in the left reservoir.The manometer equation (sum of pressures) can be written as

( . . . . ) .p �� air glycerin oil #g g+ − − − + ( . . )h 0 3 0 3H O2 #g− − − 0=Left hand term takes to zero because water tank is open in atmosphere.

. . .p h�� O � Oair glycerin oil � �# # #g g g g+ + − + 0=

h . . .p �� O

H O

air glycerin oil

�

�# # #g

g g g= + + +

.. . . . . . .

1000 9 8162 10 12 4 10 0 1 0 8 1000 9 81 0 2 1000 9 81 0 63 3

## # # # # # # #= + + +

h 7.21 m=Note - If Reservoir is open in the atmosphere then pressure at the surface of tank takes equal to zero.

FM 2.11 Option (A) is correct.Both fluids are open to the atmosphere. The pressure of water and oil is the same at the contact surface.

pcontact p gh p gh. .atm oil oil atm w wr r= + = + ghoil oilρ ghw wr=

hw . �.�� .��mh ��

w

oiloilr

r# #= = =

33.75 cm=

FM 2.12 Option (C) is correct.Apply the hydrostatic relation from oil surface to the water surface

( �.��) (�.�� .��)p hatm oil waterg g# #+ + − + patm= ( . )h ��oilγ + ( . )��water #g= ( . )h8809 0 10+ .9790 0 18#= h8809 (97 0 0.18) (8809 0.10)9 # #= − h8809 .881 3= h 0.10 10m cm= =


The manometer equation can be written as





�� p h p �� loil Hg� �� g g g g# # # #+ + + + − − =Here both the ends are open in the atmosphere. So p p �� = =

h . . . . .133

9 0 1 9 8 0 1 15 6 0 2# # #= + + 0.0376 m=

FM 2.14 Option (C) is correct.Firstly from figure, the height of water added

��v = (1) h42p#=

h 25.46 cm=Then at equilibrium, the new system must have 25.46 cm of water on the right and a 45 cm of liquid is somewhat displaced, so that “L” is on the right, 0.15 m on the bottom and “0.3 L− ” on the left side. The bottom pressure is constant.

(1.6 ��0)(0.� )p Latm #+ − ��0 (0.2546) (1.6 ��0)p Latm # # #= + + . L4699 2 15664− 2492.534 15664L= + L31328 .2206 67= L 0.0704 7.04m cm= =

Thus Right leg height− − 7.04 25.46 32.5cm ��= + = = Left leg height− − 30.0 7.04 22.96 cm �L= − = =

FM 2.15 Option (D) is correct.Adding the pressures (manometer equation),

� 10p h hH Oair liquid Hg2

2 #g g g# # #− + + − 0= ( )h H Oliquid 2γ γ- p � 10air Hg

2# #g=− − −

h � 3 10

H O ��

�� 2

2

# #g gg= −

+ −

. .. . .

9810 0 6 1000 9 813 5 10 13 6 1000 9 81 3 103 2

# ## # # # #= −

+ −

. . m39243500 4002 48 1 912= + =

FM 2.16 Option (D) is correct.The pressure at the bottom of the manometer must be same regardless of which leg we approach through, left or right:

(��5 0.1) (��0 0.0�) (0.04)� �p left legatm Xg# #+ + +

(��5 0.0�) (��0 0.05) (0.06)� �p right legatm Xg# #= + + +or . .1473 8 0 04 �g+ . .1199 15 0 06 �g= +or .0 02 �γ . . .1473 8 1199 15 274 65= − =

Xγ .. 13732.5 /N m0 02

274 65 3= =

Then S.G. of fluid X is,

. .S G X . .979013732 5 1 40= =

FM 2.17 Option (D) is correct.The force on the panel yields SAE 30 oil (gage) pressure at the centroid of the panel:

F 8450 (0.3 0.4)� � ��G �G# # #= = =





pCG 70416 Pa= (gage)This pressure works 15 cm above the bottom (at centroid of panel).Now from hydrostatic relation from centroid to air space

pair space ( . . ) .70417 8720 0 80 0 15 6670 0 60#= − − − 60747 60.7Pa kPa-=Neglecting the specific weight of air, we move out through the mercury to the atmosphere:

60747 (133100) h#− �( )p gageatm= =

h 0.46 m13310060747= =

FM 2.18 Option (B) is correct.First we have to consider that piston alone forcing the oil.In this condition, the manometer equation can be written as

sinp h ��p oil cg− 0= ...(i)

where pp = pressure on the face of the piston.When weight W placed on the piston head, the pressure pp increases to ppl and

ppl p AW

p= + & pp pp−l AW= ...(ii)

where A = Area of the pistonNow in this condition, manometer equation becomes.

p ( . )sinp h �15 ��oil # cg− +l 0= ...(iii)Subtracting the equation (i) from equation (iii), we get

pp pp−l . sin�15 ��oil # # cg=

AW . .�15 �5oil # #g= From equation (ii)

W . . . ( )8 95 10 0 15 0 5 4 15 103 2 2# # # # # #

p= −

. .671 25 0 0177#= 11.8 N8=

FM 2.19 Option (A) is correctSince the tube diameter is constant, the volume of mercury will displace a distance

hΔ down the left side, equal to the volume increase on the right side: h LΔ Δ=. Apply the hydrostatic relation to the pressure change, beginning at right (air/mercury) interface.

( ) ( )sin sinp L h h LB Hg Wg q g qD D D D+ + − + pA= with L hΔ Δ= 10000 133100( 30 )sinh h0 cD D+ + 9790 ( 30 )sinh h cD D#− + 130000= ( ) ( )sin sinh h133100 1 30 9790 1 30# #c cΔ Δ+ - + 30000= . .h h133100 1 5 9790 1 5# # # #Δ Δ- 30000= h184965 Δ 30000= hΔ 0.162 16.2m cmb =The mercury in the left (vertical) leg will drop 16.2 cm, the mercury in the right (slanted) leg will rise 16.2 cm along the slant and (16.2 30 8.1sin cmc# = ) in vertical elevation.

FM 2.20 Option (A) is correct.For given figure, the pressure equation can be written as





p gh gh ghgage w w gage gage w w� �r r r+ − − ppipe= ppipe � �� p g h S G h hgage w w gage gage w� �r= + − − ...(i)

Since hw� 50 0.50cm m= =

hgage �.�� .�� .��sinL �� q #= = = 0.6667sin 6 68θ = + =

hw� . . .sinL �� #q= = =Substituting the value in equation (i), we get

p ppipe gage− . [ . . . . ]1000 9 81 0 5 2 4 0 04 0 04# #= − − pΔ 3570.8 Pa= 3.6 kPa,

FM 2.21 Option (A) is correct. �.�� mgasoline

�γ #= =When the tank is full, the pressure is

pfull full height of tankgasolineg #= 6657 0.45 2996 Pa#= =Set this pressure equal to 3 cm of water plus ‘X ’ centimeters of gasoline :

pfull . X2996 0 03 6657��# #g= = +

or X ( . )

0.406 40.6m cm66572996 9790 0 03#= − = =

Therefore the air gap

h 45 3 40.6 1.4 cm= − − =

FM 2.22 Option (C) is correct.Starting with air pressure (point A) and moving along the tube by adding (as we go down) or subtracting (as we go up) until we reach the Glycerin (point B ) and setting the result equal before and after the pressure change of air give

p gh gh gh, ,A br br Hg Hg gly gly� � �r r r+ + − p gh gh gh, ,A br br Hg Hg gly gly� �� r r r= + + − p pA A� �− ( ) ( )g h h g h h, , , ,Hg Hg Hg gly gly gly� � � �r r= − − +

gp

wρΔ . . �. .S G h S G hHg Hg gly glyD D= − = ...(i)

. .S Gwρ

ρ =

where hHgΔ and hglyΔ are changes in the differential mercury and glycerin column heights, respectively.The volume of mercury is constant.





A h �Hg left�Δ A h �Hg right�D=

h �Hg leftΔ AA h �Hg right

�

� D#= ...(ii)

We have hglyΔ 0.005 m= and pΔ 0.7 700 /kPa N m2= = hHgΔ h h� �Hg right Hg leftD D= +

h AA hgly gly

�

�#D D= + h A

A�gly�

�D= +c m

By substituting the values in equation (i), we get

.1000 9 81700#

. . .h AA13 56 1 1 26 0 005gly

1

2# #D= + −c m

.1000 9 81700#

. . . .AA13 56 0 005 1 1 26 0 0052

1# #= + −c m= G

AA

�

� .0 145=


Let piezometer tube B be an arbitrary distance H above the oil glycerin interface. Apply the hydrostatic formula from point A to B:

1500 (1.5 )� H�� g g#+ + − ( � )Z Hoil Bg− − − �pB= = 1500 12 2.0 8720(1.5 )H#+ + −

8720( 1 )� HB− − − 0= 1500 24 (8720 1.5) 8720H#+ + − 8720 (8720 1) 8720� HB #− + + 0= �8720 B 23324= ZB 2.67 m=Let Piezometer tube C be an arbitrary distance Y above the bottom. Then

1500 12 2 8720 1.5 12360(1.0 )Y# #+ + + −

12360( )� YC− − �( )p gageC= = 1500 24 13080 (12360 1) 12360Y#+ + + −

12360 12360� YC− + 0= �12360 C 26964= ZC 2.18 m=

FM 2.24 Option (C) is correct.When the pressure differential of 0.7 kPa occurs in the manometer, the pressure p pA B− is changed to p pA B� �− . The left column falls the distance ( )� and the





right column rises by a distance ( )b along the inclined differential manometer.

The manometer equation for final condition can be written as

� � � � � �sin sinp h a a b h bA water CCl water� � ��g g q g q+ + − + − − pB�= sin sinp p a a b bA B water CCl CCl water� � � �g g g q g q− + − − + 0= � � � �sin sinp p a b a bA B water CCl� � �g q g q− + + − + 0= � �� sinp p a bA B water CCl� � �q g g− + + − 0= ...(i)Now from the figure, the differential reading along the manometer is

hΔ sina bq= +

sinh θΔ sina b q= + ...(ii)Now from equation (i) and (ii), we get

( ) ( )sinp p hA B water CCl� � �q g gD #− + − 0=

sinθ ( )( )

hp p

water CCl

A B� �

�g gD=− −−

( . ).

30 10 9810 15 6 100 7 10

2 3

3

# #

#=−−−

( )

. 0.30 5790

0 7 10 4025

##= −

− =

θ (0. ) .sin 402 23 701 c= =−

FM 2.25 Option (A) is correct.From the given condition

pA pB= ( )L hH O�γ + ( )H hoilg= +

or H ( )L h hoil

H O�

gg

#= + −6 @ ...(i)

For a rise h h<Δ , a volume ( /4)d h2π Δ of water leaves reservoir (A), decreasing “L” by ( � )h d D �Δ and an identical volume of oil enters reservoir (B), increasing “H” by the same amount ( � )h d D �Δ .The hydrostatic relation between (A) and (B) becomes.

( � )p L h d D h hA H O�

�g D D+ − + −6 @

� ( � ) �H h d D h hoil�g D D− + + − pB= ...(ii)


� ( � ) �p L h d D h hA H O�

�g D D+ − + −

( ) ( � )L h h h d D h hoiloil

H O ��g gg D D− + − + + −: D pB=





p pA B− L h Dd h hH O

�

�g D D=− − + −b l; E

� �L h h Dd hoil

oil

H O�

�g gg D D+ + + −b l; E

L h Dd h hH O H O H O H O

�

� � � �g g g gD D=− + − +b l

L h h Dd hH O H O oil oil

�

� �g g g gD D+ + + −b l

h h Dd h h D

dH O H O oil oil

� �

� �g g g gD D D D= + − −b bl l; ;E E

h Dd h D

dH O H O oil oil

� �

� �g g g gD D= + − −b bl l; ;E E

or p pA B− hDd

Dd� �

� �

H O oil� ��g gD= + − −c cm m= G


Since the volume of water must be conserved. So,

d l42π D H D h3 2 2 3 2

i2

02

0p p

# #= −b bl l ...(i)

Also from the geometry of the cone

HD

�HD

hD

�i

�

�= =

Di D2= and D H

Dh�

�=

Substitute these values in equation (i),

d l42π D H

HDh h3 16 2 3 2

20

2

0p p# # # #= − b l

d l3 2 D HHD h8

2 2

03

#= − b l ...(ii)

Put �.�mH D= = and �.��md = in equation (ii), we have

( . ) l3 0 03 2#

( . ) ... h8

0 6 0 60 60 62 2

03#

#= − b l

( . ) l3 0 03 2#

( . )h8

0 6 3

03= −

h�� . . l0 027 0 0027= − 0.027(1 0.1 )l= −

h� 0.3(1 0.1 )l /1 3= −


In equilibrium, the weight of ice block floating in a fluid is equal to the buoyant





force acting on it. So

FB Wice= gh Asea submerged cρ # g h Aice total c# #r=

hh

total

submerged sea

ice

rr= .

. .h0 8750 875

1025920 0 9& + = =

h 0.0973 9.73m cm= =

FM 2.28 Option (C) is correct.The weight of Pine (wood) and lead must equal the buoyancy of immersed wood and lead.

W Wpine lead+ B Bpine lead= +or . . . .S G v S G vpine H O pine lead H O lead� �γ γ# # # #+

v v( )H O pine immersed H O lead� �g g# #= +

or . ( . ) . . v0 65 9790 4 0 055 2 2 11 4 9790 lead2

# # # # # #π +9 6C @

( . ) ( . . ) v9790 4 0 055 2 2 0 3 9790 lead2

# # # #p= − +9 C

or . v33 26 111606 lead+ . v44 20 9790 lead= + ( )v111606 9790 lead− . .44 20 33 26= −

vlead . 0.000107 m101816

10 94 3= =

Wlead 11.4 9790 0.000107 12 N# #= =

FM 2.29 Option (C) is correct.The FBD of the block is shown below

where Ww Weight of wood= W �B w Buoyancy force on wood= Wm Weight of metal plate= F �B m Buoyancy force on metal plate= F holding force of block=In equilibrium condition

FVΣ 0= : F F F W W� �B m B w w m+ + − − 0= F W W F F� �w m B w B m= + − − ...(i)

Now Ww . .S G v� O w�g# #=

0.65 9.8 3 1.22 0.611 21

# # # # #= 7.11 kN8=

Wm 26.4 (3 0.15 0.61) 7.25 kN# # #= =

F �B w 9.8 (3 1.22 0.61)v 21

H O w2g # # # # #= = 10.94 kN=





F �B m 9.8 (3 0.15 0.61)vH O m2g # # # #= = 2. kN69=Hence from equation (i), force required to hold the block is

F 7.11 7.25 10.94 2.8 69= + − − 0.7 7 0kN N4 4= =

FM 2.30 Option (A)is correct.For ball to float on water

Wball W Wof water displaced by ball of air displaced by ball= +

m gball ( )g v g R v��

water segment air segment�

# #r r p= + −

or . .0 003 9 81# . ( )h R h998 9 81 3 32

# #p= −

. . ( )R h R h1 225 9 81 34

3 332

# # p p+ − −: D

or .0 02943 998 9.81 (3 0.02 )h h32p

# # # #= −

. . ( . ) ( . )h h1 225 9 81 34 0 02 3 3 0 023 2

# # # #p p+ − −: D

or .0 02943 ( . ) . ( . )h h h h9790 3 0 06 12 3 35 10 3 0 062 5 2# # #

p p= − + − −−9 C

.0 02943 9790 (0.06 ) 12 3.35 10h h32 5p

# # # #= − + −

12 (0.06 )h h32p

#− −

.0 02943 97 (0.06 ) 4.02 10h h90 32 4

#p

#= − + −

.0 02943 614.4 10239.5 4.02 10h h2 3 4#= − + −

or . . .h h10239 5 614 4 0 0290283 2− + 0= gives h 0.0073 7.3m mm, b

FM 2.31 Option (B) is correct.First compute buoyancies,

Brod vwaterg #= 9790 (0.04) 12.3L L42p

# # #= =

Wlead mg= 2 9.81 19.62 N#= = Blead �. .W S Glead= 19.62/11.4 1.72 N= = Wrod . .S G vwater rodg# #=

0.636 9790 (0.04) 7.9L L42p

# # #= =

Sum moments about B:

MBΣ 0= :

cos cosB L B L� �� rod lead# #c c+ �� cos cosW L W L�lead rodc c# #= +

or . .cos cosL L L12 3 2 30 1 72 30# #c c+ . .cos cosL L L19 62 30 7 9 2 30# #c c= +

or 5.3 1.5L L2 + .L L17 3 42 2= + ( . . )L5 3 3 42 2− ( . )L17 1 5= − . L1 88 2 . L15 5= . L1 88 L15=

or L .. 8.2 8 m1 88

15 5 ,= =





FM 2.32 Option (D) is correct.The pressure in the air space can be found by working upwards hydrostatically from point B

160000 (9790) (0.35 0.25) pair#− + = pair .160000 9790 0 60#= − 160000 5874 154126 154Pa kPa,= − =The force on top of the insert is simply the pressure on the insert times the insert area.

Thus ptop 160000 9790 0.35 16000 3426.50#= − = − 156573.5 Pa=

Ftop p Atop top#= . ( . )156573 5 4 0 1 2# #

p= 1230 N=

FM 2.33 Option (C) is correct.The weight of blcok

W Lhbbg=and it acts in the center, at �L � from the left corner, while the buoyancy being

a perfect triangle of displaced water.

B Lhb21 g#=

and acts at �L � from the left corner.Sum moments about the left corner, point C:

MCΣ ( )( / )Lhb L Lhb L0 2 2 3bgg= = − b bl l

or bγ /3g=Then summing vertical forces gives the required string tension T on the left corner.

FZΣ 0 Lhb Lbh T2 bg g= = − −

or T Lhb � bg g= −a k Lbh Lbh

� � �g g g= − =a k �bγ γ=

But T W Bsphere sphere= − ( . .)S G v vg g# #= − . .S G v�g#= −6 @

or Lhb6

γ ( . . 1)S G D63p g# #= −

or D � ( . . )S G

Lhb1p= −

D ( . . )S G

Lhb1

/1 3

p= −; E

FM 2.34 Option (A) is correct.A vertical force balance gives

( . .)S G voil displaced# ( . .)S G vcylinder cylinder#=or . R h0 85 2

# π 0.6 0.8R2p #= h 0.565 m=

The point B at . 0.282m mh2 2

0 565= = above the bottom and the point G at �2. 0.4 m2

0 8= = above the bottom.





Meta center location

MB vI

R h

R�sub

��

�

#p

p= =

( . ) ( . )( . )

.( . )

4 0 5 0 5650 5

4 0 5650 5

2

4 2

# # #

#

#pp= =

0.111 m=Now GB . .0 4 0 282= − 0.118 m=Hence MB MG GB= + MG MB GB= − 0.111 0.118 0.007 m= − =−Since for stability of floating body, MG must be greater than zero (MG �> )

Here �MG � ��< =− , so that cylinder is unstable.

FM 2.35 Option (A) is correct.The slope of the liquid gives us the acceleration

tanθ .ga

�� x= = − =

ax . . .�0 15 0 15 9 81# #= = 1.472 /m s2=Now, we can go straight down on left side using only gravity.

pA g zr D= . .998 9 81 0 3# #= 2938 ( )Pa gage,

or we can start on the right side, go down 15 cm with g and across 100 cm with ax

pA g z a xxr rD D= + . . . .998 9 81 0 15 998 1 472 1 00# # # #= + 1469 1469= + 2938 Pa=


we have ω 150 15.7 /rpm rad s602 150#p= = =

A deflection hΔ up at wall and down in the center:

hΔ .( . ) ( . )

0.051 5.1m cm��

4 4 9 8115 7 0 092 2 2 2

# #

#w= = = =

And the fluid pressure will be highest at point B in the bottom corner.

pB g z g zoil oil water waterr rD D= + . ( . ) ( . ) ( . . )891 9 81 0 15 998 9 81 0 051 0 18# # #= + + .1311 1 2262= + 3573 ( )pa gage=

FM 2.37 Option (D) is correct.The equation for surface of constant pressure is given by

z tanCons t��

22 2w= + ...(i)

For free surface with h 0= and r 0= {from equation (i), constant 0= }

h ��

22 2w= ...(ii)

The initial volume of fluid in the tank is equal to

vi R H�p= ...(iii)And the volume of fluid within rotating tank can be written as (from figure)

dvf ��2p=

vf � ��2 2

� 2 2

0#p w= # from equation (ii)





g r dr gR�

R��

�

� �pw pw#= =# g

R42 4pw=

Since the volume of the fluid in the tank must remain constant. So

vi vf=

R H�π gR

42 4pw=

2ω RgH�

�=

ω RgH

R gH� ��= =

FM 2.38 Option (C) is correct.Let h� be the height of the free surface at the center line.

Then Z h gR

�BB

�

� �

#

w= + , Z h gR

�AA

�

� �w= +

Subtracting the above two equations, we get

Z ZA B− gR h h g

R2 2

A B2 2

0 0

2 2w w= + − −

Z ZA B− g R R2 A B

22 2w= −^ h

. .0 18 0 10− . ( . ) ( . )2 9 81 0 1 0 052

2 2

#

w= −6 @ �.�� , �.�m mR RB A= =

.0 08 ..

19 620 00752

#w=

2ω .. .

0 00750 08 19 62#=

ω . 14.5 /rad s209 28= =

or ω .N60

2 14 5p= =

N . 138.56 139 . .r p m260 14 5

## ,p= =

FM 2.39 Correct option is (C).When cone is pushed into the water, the air get compressed.So, for constant temperature compression of air

p vi i p vf f= ...(i)where subscript andi f denotes the initial and final states.

Now pi patm= pf ( )p d latm g#= + −

vi R H32p=





From similar triangle ABC and CED

EDAB CO

CF=

KR� H l

H= −

K � �HR H l�

#= −

and OD ( )KHR H l2= = −

Thus VCED ( ) ( )HR H l H l3 2

22#

p= − −

vf ( )HH l R H l3

22p

# #= − −b l ( )HR H l3

23

#p= −b l


p R H�atm�π

# ( ) ( )p d l HR H l�atm

��g p= + − −b l6 @

p Hatm # ( ) ( )p d lH

H l�atm �

�g= + − −6 @

patm ( )p d l HH l

atm

�g #= + − −

b l6 @

p H lH patm atm

�

# − −b l ( )d lg= −

d l− pH l

H �atm�

g= − −b l; E

l d pH l

H �atm�

g= − − −b l; E

Substitute patm 101 kPa= and 9.8 /kN m3γ =

l �d H lH��

�= − − −b l; E

� ( )d H H l�� = − − −−6 @

FM 2.40 Option (D) is correct.The FBD of the system is given below.

From figure tan30c d

12=

d 2 30 1.155tan mc= =In equilibrium condition.

FVΣ 0= FC p A Wair= + ...(i)





where W = Weight of the water

FC = force exerted by the cone on the fluid

p Aair = pressure force.

Thus p Aair 45 (1.155) 47.12 kN42p

# #= =

and W d d4 3 3 2 12

2

# #g p p= − b l; E

W . . ( . ) ( . )2 75 1000 9 81 4 1 155 3 3 41 1 1552 2

# # # # # # #p p= −: D

26977.5 (1.555) 3 31

42p

# # #= −b l 136.55 kN=

Hence total thrust exerted on the curved surface of the cone is [from eq (i)]

FC . .47 12 136 55= + 183.67 kN=

FM 2.41 Option (B) is correctThe force F and its line of action are given by

F ( )h A h RCG�

#g g p= =

yCP sin sinh A

Ih R

Rh

R�� CG

XX�

��

#

c c

p

p=− =

−=−

Summing moments about the hinge line Q gives

MQΣ 0=

0 ( )h R hR P R�

��

g p= −^ h

or P �43p g=

FM 2.42 Option (C) is correct.The centroidal depth of the gate is

hCG ( . . )sin4 1 0 0 6 45c= + + 5.1313 m=So that FAB h Awater CG gate# #g= 9790 5.1313 (1.2 0.8) 48226 N# # #= =The line of action of F is slightly below the centroid by the amount

yCP sinh A

ICG

XX # q= . ( . . )

sin5 1313 1 2 0 8

45��12

3

# #

# c=

. . .( . ) ( . ) sin5 1313 1 2 0 80 8 1 2 4512

1 3

# #

# # # c= 0.0165 m=

Thus the position of the center of pressure is at

X 0.6 0.0165 0.617 m,= +

FM 2.43 Option (D) is correctThe horizontal component is

FH h ACG vertg= ( )9790 5 1 9 2# # #= + 1057320 1057N kNb=The vertical component is the weight of the fluid above the quarter circle panel.

FV (� � ) ( )by �uarter circleW Wrectangle= −

( ) ( )9790 2 7 9 9790 4 2 92# # # # #

p= − a k

956874.6 957 kN,=





FM 2.44 Correct option is (A).

The resultant force on the plate is

FR h AH O C2g # #=

where y hC C= . 0.8 mh2 2

1 6= = =

A �.� �.�� .�� h b � �# #= = =

Thus FR 100 9.81 0.8 1.950 2# # #= 15.3 kN-

The location of FR is

yR y AI yC

XCC

:= +

. .( . ) ( . )

.0 8 1 952121 1 22 1 6

0 83

#

# #= + 1.067 m=

In equilibrium condition, taking the moment about point O , �MOΣ = .

( . . )F �� R # − .W ��#=

W .. . 9 kN0 9

15 3 0 533# -=

FM 2.45 Option (C) is correctThe force on annular region BB is

FBB p ABB BB#=

( ) 4 { 0.36 (0.36 (0.10 0.10)) }p h 2 2air waterg p

# # #= + − − +] g< F

(8000 9790 ) 0.36 ( . . )h 4 0 36 0 22 2p# # #= + − −] g7 A

853 ( )( . )h8000 9790 0 08168#= + . . h653 45 799 65= +

h .. 0.25 m799 65

853 653 45= − = 25 cm=

Now the net hydrostatic force on bottom of tank is

Fbottom p Abottom bottom#= pbottom ( �.��)p hair waterg= + +

( . . )8000 9790 0 25 0 12= + + 11622 Pa gage=

Thus Fbottom 11622 (0.36) 1183 N42p

#= =






The FBD of the cylinder is shown below.

From FBD, in equilibrium condition

FP F FR R� �= −

...(i)where FP is force exerts on the wall at the point of contact.

Now FR� h AC� �g= . ( )9 8 22 2 3# # #= 58.8 kN=

and FR� h AC� �g= .9 8 1 21 1 3# # #= +b l 44.1 kN=

From equation (i), we get

FP . .58 8 44 1= − 14.7 kN=


Consider half of the trough whose cross-section is quarter-circle.Horizontal force on vertical surface

FH = p A g R A�.avg r #= = b l

. 20.5 ( . . )1000 9 81 0 5 4 5# # # #= b l

5518.125 N=Vertical force on horizontal surface

F WV = gv g L R�

�

#r r p= = b l

.. . ( . )

1000 9 81 44 5 3 14 0 5 2

# ## #= ; E

866 .45 N3=Then resultant force acting on the surface of the 4.5 m long section of the trough.

FR ( . ) ( . )F F �� H V� � � �= + = +

10271.5 N=and Direction of FR ,

tanθ .. �.��F

F��

H

V= = =

θ .57 5c= downwards from horizontalBy taking moment about A,

MAΣ 0=





T R# �� F RR c= − T �� FR c= − 10271.5 ( . )sin 32 5#= c

5518. N9= 5519 N,

FM 2.48 Option (C) is correct.Consider a 50 cm width of upper cylinder, the water pressure in the bolt plane is

p� hg= 9790 ( ) 9790 4 39160 PaD# #= = =

Then summation of vertical forces on this 50 cm wide free body gives

FZΣ 0 2p A W W Fwater bolt1 1 tank= = − − −

0 ( . ) ( ) . F39160 4 0 5 9790 2 2 0 5 1125 2 bolt2

# # # # #p= − − −

F2 bolt 78320 30741 1125 46454= − − =

F ,one bolt 23227 23N kN246454 ,= =

FM 2.49 Option (B) is correctThe water depth when the plug pops out is

F �5 �h ACGg= =

( . )h9790 4 0 06 25CG2

# # #p= =

or hCG ( . )

0.904 m0 06 979025 4

2# #

#p

= =

But H . �0 0.�0� 0.0� �0sin sinh �0 0�

CG c c= + = +

0.919 m=Thus at mercury-water interface

(��0) ( 0.0�) ��00p H hatm # #+ + − patm=or ( . . )9790 0 919 0 02# + h133100=

h .133100

9790 0 939#=

0.069 6.9m cm= =

FM 2.50 Option (A) is correct.The FBD of the curved surface is shown below.





In equilibrium condition in horizontal direction.

FxΣ 0= F FH�= h AC� �g=

. ( )1000 9 81 22 2 6# # # #= 117.72 kN=

Similarly in vertical direction

FyΣ 0= FV vH O2g #=

( )9810 4 2 62# # #

p= 184.82 kN=

For Hemisphere

y� mR34

34 2

38#

p p p= = =

x� mR3 3

2= =

Now taking the moment about the hinged point C , we get

MCΣ 0= : F x F yV H� �# #+ P �#=

P . .

2184 82 3

2 117 72 38

# # p=+

. 111.60 kN21231 21 100= + =

FM 2.51 Option (C) is correctThe centroid of a semi-circle is at

R34

π 1.273 m34 3# ,p= of the bottom, as shown in the sketch

below.

Thus it is 3.0 1.273 1.727 m− = down from the force.The water force F is

F h ACGg= 9790 ( 1.727)h 2 3 2p# # #= + ] g

138333( 1.727)Nh= +The line of action of F lies below the CG.

yCP sinh A

ICG

XX q=−

( . )( / ) 3

( . ) ( ) sin� 1 727 2

0 10976 3 902

4

#

# # cp

=+ ] g

�.��I RXX�

#=

( . )

.� 1 727

0 629= +−

Then summing moments about B yields the proper support force.

MBΣ 0=





( . ) .( . )

.hh

138333 1 727 1 2731 727

0 629 3 366 103# # # #= + − + −6 ;@ E

0 . ( . ) .h1 273 138333 1 727 138333 0 629 3 366 103# # # # #= + − −

0 .h176098 304121 87011 5 1098000= + − −or h176098 .880890 5= & h 5.00 m=

FM 2.52 Option (D) is correct.The FBD of the horizontal gate is shown below.

For horizontal gate taking moment about hinged point.

MHΣ 0= W �# pA �#= W pA=where p = water pressure on the bottom surface of the gate

A = Area of the gate

W ( ) ( )g � � � �# # #r= − 1000 9.81 3 16 471 kN# # #= =For vertical gate

FR h AH O C2 # #g= 98 0 (5 2) (4 4)1 # # #= + 1098 kN-

The location of FR is yR y AI y

C

XCC

#= +

( )121 4 (4)

7 4 47

3

# #

# #= + 7.190 m=

For equilibrium condition about hinged point

MHΣ 0= �P# ( . )F � � ��R #= + −

P . 496.84 497 kN41098 1 81# -= =






Resultant hydrostatic force on the gate

FR p A gh A�avg cr= =

. . ( )

10001000 9 81 3 5 3 6# # # #= � ��h �

�c = + =

618 kN=Pressure centre is determined as

yP y y AI

CC

XC= + ...(i)

yC 2 3.5 m23= + =

For rectangular section

IXC 3lb12 12

63 3#= = ] g 13.5 m= from X -axis

And A 6 3 18 m2#= =

Therefore, from equation (i),

yP . ..3 5 3 5 18

13 5#

= +

3.71 m= downwards from the free surface of water

FM 2.54 Option (C) is correct.The force diagram of the plate is shown below

F� and F� work at the centre of pressure.

F� �h Aoil C�g=where hC� = vertical distance from the fluid surface to the centroid of the area.

hC� 3 2 5 m= + =Thus F� . . ( )0 9 1000 9 81 5 4 3# # # # #= 5.3 10 N5

#=Force on horizontal part of the gate is

F� h Aoil C� �g=where hC� 4 3 7 m= + =So F� . . ( )0 9 1000 9 81 7 2 3# # # # #= 3.7 10 N5

#=Now location of force F�,

yR� y AI y

C

XCC

� ��

#= +

( )121 (3) (4)

5 4 35

3

# #

#= +

0.27 5= + 5.27 m=





Now taking the moment about the hinged point, �MOΣ = .

��.�� F F R� �# #− + − 0=

R .F F4

2 271 2#= + . . .4

5 3 10 2 27 3 7 105 5# # #= + .

415 731 105

#=

.3 9327 105#= 393 kN-


On the side of the gate where the water measures 3 m, F� acts and has an hCG� of 1.5 m on the opposite side, F� acts with an hCG� of 1 m.

F1 h Aoil CG 11g # #= 0.82 9790 1.5 3 5 180625.5N# # # #= = F� h Aoil CG ��g # #= 0.82 9790 1 2 5 80278 N## # #= =

yCP1 .3 �.5 mh A

L b1�1

1� 1 5 5 31 5

CG 1

13

3

1 #

# #

# # #

# #=−

= − =−] g: D

So F1 acts at 1.5 0.5 1.0 m− = above B.

� �.333 my h AL b1�

1

1� 1 � 51 5

CPCG �

�3

3

�

�#

# #

# # #

# #=−

=− =−] g

So F� acts at 1.0 0.33 0.67 m− = above the B.Taking moments about B, we get

MbottomΣ . .180625 5 1 80278 0 667# #= − 127080 N m−=or Mbottom 127 kN m−=


The average pressure on a surface is the pressure at the centroid of the surface.

p pavg center= gh gh gh� �centerr r r= = =

Then the resultant hydrostatic force on the wall is,

F p AR avg= ( )gh h b gh b2 2

2

# #r r= =





The line of action of the force passes through pressure center

yP h32=

By taking moment about point A,

MAΣ 0= F h y�R P+b l F h h�ridge #= +b l

Fridge y

Fh

hP

R45

4#= +

hgh b

54

2

h h4 3

2 2r#=

+^ h h

gh b gh b54

2 3011

h1211 2

2r r#= =^ h

FM 2.57 Option (A) is correctThe centroid of the gate remains at distance �L � from A and depth �h � below the surface. For any θ , the hydrostatic force is

F = h L b2 # #γb l

The moment of inertia of the gate is

IXX bL12

3

=

yCP sin

b Lh

bL

2

12

3

# #

q= −

and the center of pressure is,

L y2 CP= − from point B

Summing moments about hinge B yields

P L# F L y� CP#= −b l

or P L

F L y� CP

=−b l

sin

L

h Lb L bLh Lb� � ��

�

# # ##

g q

=

−f p

sin sinhb Lh b

bL hb Lh

L

22 12

2

22 6

2 2

## #

##g q g q

=−

=−: :D D

sinsin

hb L hL

hb L hL

221

34 3

2

2#

#

g qg q=

−= −

b l: D

FM 2.58 Option (B) is correct.For resultant fluid force, this cross-section is divided into three parts as shown below.

For Area 1: FR1 Pressure force= �� (� �) 1��k�p Aair 1 # #= = =For Area 2: FR2 p A h Aair C2 2 2g= +





.40 2 4 6 67 24 2 4# # # # #= +

.320 106 72= + 426.72 kN=For Area 3: FR3 p A h Aair C3 3 3# g= +

.40 21 2 4 6 67 3

2 4 21 2 4# # # # # # # #= +

.160 71 15= + 231.15 kN=Total force FR F F FR R R� � 3= + + 160 426.72 231.15= + + 817.87 818kN kN-=

FM 2.59 Option (D) is correct.The water causes a horizontal and a vertical force on the body as shown:

FH R Rb2g= at R3 above O,

FV ��42g p= at �

34

π to the left of O

These must balance the moment of the body weight W about O.

MOΣ 0�� 2 3 4 3

44 3

42

��

2 2 2g gpp

g pp g# #= + − − =b b b bl l l l

. .S G body R b R R b R b R R hb� 3

�� 3

��

s� 3 � �

#gg p

pp

p= = + +b bl l= =G G

s

γγ

��

��

��

1 2 43 4

1 6 43 4

1 23

1 21

2

2

3

2

# #

# #

pp

pp

=+

+=

+

+

;

;

:

:

E

E

D

D

or . .S G

��

�

32

132 1

=+

= +−

: D

FM 2.60 Option (C) is correct.The FBD of the gate and the cylindrical block is shown below.

.

where T = Tension in the string.





W = Wight of cylindrical block.

FB = Buoyancy force

Hydrostatic force on the gate,

FR h A H H H� � ��

C�

# # # #g g g= = = ...(i)

Taking the moment about the hinged point O ,

MOΣ 0= : F H�R # T4=

From equation (i), T H H H23

3 4 82

3

#g g

#= = ...(ii)

Again from the FBD of cylindrical block

FVΣ 0= W T FB= +

mg T FB= + H v8

3g g= + From eq. (ii)

( ) ( )H H8 4 1 13

2#

g g p= + − ( )gH H4 2 1

3g p= + −: D


The forces on the hatch are shown above.

FR h Asea water Cg # #=

where hC sin2 21 2 45# # c= + 2 0.707 2.707 m= + =

Thus FR . . ( )10 1 10 2 707 4 23 2# # # #

p= 85.85 10 N3#=

The location of the FR is

yR y AI y

C

XCC

#= +

where yC 1 3.83sin

m452

c= + =

Thus yR . ( )

( )3.83

3 83 14 1

2

4

# p

p= + I R�XC

�p=

3.89 m=In equilibrium condition, taking the moment about the hinge point �MOΣ =

sin

F y��

R R# c−b l ( )p � ��

# # #p=

. ( . . )85 85 10 3 89 2 833# # − p# p=

p .. .

3 1485 85 10 1 063

# #= 28.98 kPa=

FM 2.62 Option (C) is correct.Force on rectangular panel is





Frect h ACG rect#g= ( /2) ( 2 )H H R H R2g g# # #= =Force on semicircular panel is

Fsemi h ACG semig #= H R R��

��

#g pp= +b l

From the given condition, set them equal.

H R�γ # H R R��

��g p

p#= +b l

RH � R H R2 3

2 32p= +

H � RH R2 3

2 2p= +

or H RH R� �

��p− − 0=

The root of this quadratic equation is

H R R R

22 2 4 1 3

22 2

! # #p p

=

− +b l

R R

22 2 4 3

22#

p p=

+ +a k .R R� � �

� ��

,p p= + +a ak k; E& 0

FM 2.63 Option (B) is correct.The FBD of the dome is shown below.

In the equilibrium condition, the vertical force of the water on the dome is equal to the force which dome exerts on the water.

FVΣ 0= F WD + pA= FD pA W= − ...(i)

where FD = force that dome exerts on the fluid and

pA = pressure force at the base of the dome.Now manometer equation (from the figure) can be written as

(�.� �.�) (�.� �.�)p p� OA gf �g g# #+ + − + − 0= p . . .87 10 3 1000 9 81 2 1 9810 1 23

# # # # #= + − 137 kPa-

and W d d��

��

� ��

��

liquid liquid

��g p g p# #= =b l; :E D

. ( . )10 1 10 12 1 23 3# #

p= 4.57 kN=


FD ( . ) .137 4 1 2 4 572# #

p= − 150.3 kN=

***********

FM 3FLUID KINEMATICS & BERNOULI EQUATION

FM 3.1 Consider steady flow of water through an axisymmetric nozzle as shown in figure. Along the centre line, water speed increases parabolically through nozzle from uin to uout . What will be the expression for center line speed ��u x from x �= to x L= ?

(A) ( )( )

u x uL

u uxin

out in�

�= + − (B) ( )

( )u x u

Lu u

xinin out

��= + −

(C) ( )( )

u x uL

u uxin

out in�

�= − − (D) ( )

( )u x u

Lu u

xinin out

��= − +

FM 3.2 An idealized incompressible flow has the proposed three-dimensional velocity distribution � ( )xy f y zyV i j k� �=− + − . What will be the appropriate form of the function ( )f y which satisfies the continuity relation ?(A) ( ) �f y y constant�= + (B) ( ) �f y y constant= +

(C) ( )f y y constant�= + (D) ( )f y y� constant�

=− +

FM 3.3 A two-dimensional velocity field is given by ( ) (� )x y x xy yV i j� �= − + − + in arbitrary units. At ( , ) ( , )x y 2 4= , the maximum acceleration and its direction respectively, are(A) 142 units, 58.57° (B) 211 units , 58.57°

(C) 211 units, 31.43° (D) 142 units, 31.43°

Common Data For Linked Answer Q. 2 and 3The expression V ( , ) (1 .5 ) ( 0. 1.5 2.5 )u v x y x yi j3 8= = + + + − − − is said to represent the velocity for a steady two dimensional incompressible flow.

FM 3.4 How many stagnation points are there in this flow field ?(A) One (B) Three

(C) Two (D) No stagnation point

FM 3.5 The material acceleration at point ( 2 , 3 )m mx y= = is(A) 14 � , 11.5 �m s m sa ax y

2 2= = (B) 1. � , 1 �m s m sa a1 5 4x y2 2=− =−

(C) 1. � , 1 �m s m sa a1 5 4x y2 2= = (D) 14 � , 11.5 �m s m sa ax y

2 2=− =−





FM 3.6 Polar coordinate of a flow is given by cosV krb�r �q= −b l, sinV k

rb� �θ=- +θ b l.

The flow field will be(A) Unsteady, incompressible (B) Steady, compressible

(C) Unsteady, compressible (D) Steady, incompressible

FM 3.7 The linear couette flow between plates as shown in figure below has velocity

component as , �u hVy v= = . This flow has a stream function but no velocity

potential, because

(A) It satisfy continuity but not satisfy irrotationality

(B) It satisfy continuity and irrotationality

(C) It does not satisfy continuity but satisfy irrotationality

(D) It does not satisfy both continuity and irrotationality

FM 3.8 The velocity potential for a certain inviscid, incompressible flow field is given by the relation

φ x y y2 322 3= − b l

where φ has the units of /m s2 when x and y are in meters.If the pressure at ( , ) (1 ,1 )m mx y = is 200 kPa, elevation changes are neglected and the fluid is water, the pressure at the point ( , ) (2 ,2 )m mx y = is(A) 40.1 kPa (B) 0.1 kPa6

(C) 0.1 kPa10 (D) 0.1 kPa8

FM 3.9 A two-dimensional incompressible flow field is defined by the velocity potential xy x y� �φ = + - . The equation of the streamline is

(A) ( ) 2 Cy x xy21 2 2ψ = - - + (B) ( ) 2 Cy x xy2

1 2 2ψ = - + +

(C) ( ) 2 Cx y xy21 2 2ψ = - + + (D) 2 Cy x xy2

1 2 2ψ = + + +^ h

FM 3.10 Consider the following two-dimensional incompressible flow with � �y xV i j= + , which clearly satisfies continuity. What will be the stream function ( , )rψ θ of this flow using polar coordinates ?

(A) cos sin Cr r23 2 2 2 2ψ θ θ= + + (B) sin cos Cr r2

3 2 2 2 2ψ θ θ= + +

(C) sin cos Cr r23 2 2 2 2ψ θ θ= - + (D) cos sin Cr r2

3 2 2 2 2ψ θ θ= - +

Common Data For Q. 11 and 12The inviscid, incompressible and steady flow of water along the variable area horizontal pipe is shown in figure. The velocity is given by �(� ) � .secmxV i= +





and viscous effects are neglected. ( 1000 /kg mH O3

2ρ = )

FM 3.11 What will be the pressure gradient xp

22 in /N m3 ?

(A) 9000(1 )x− (B) ( )x9000 1− +(C) x9000− (D) ( )x900 1− +

FM 3.12 If the pressure at section (1) is 345 kPa, the pressure at section (2) will be(A) 159 kPa (B) 3315 kPa

(C) 33.15 kPa (D) 331.5 kPa

FM 3.13 The air speed at a given location is 25 /sm and the pressure gradient along the streamline is 125 /N m3. If the effect of gravity is negligible and 1.2� ��g mair

�ρ = , what will be the air speed at a point 0.5 m far along the streamline ?(A) 22 / .secm (B) 2.03 / .secm−(C) 20 / .secm (D) / .secm23

FM 3.14 A frictionless, incompressible, steady-flow field is given by �xy yV i j�= − in arbitrary units. If the density 0ρ = constant and gravity is neglected, an expression for the pressure gradient in the x - direction will be

(A) ( )xp xy y� ��

2

22 r=− −6 @ (B) ( )

xp xy y� ��

2

22 r= −6 @

(C) ( )xp xy y� ��

2

22 r=− +6 @ (D) ( )

xp xy y� ��

2

22 r= +6 @

FM 3.15 An inviscid, incompressible, steady flow fluid has a constant bulk modulus. By integrate “F ma= ” along a streamline, what will be the equivalent Bernoulli equation for this flow ?

(A) E V gz C2V

2

r− + + = (B) E V gz C2V

2

r− + + =

(C) E V gz C2V

2

ρ ρ ρ- + + = (D) E V gz C2V

2

r− + + =

FM 3.16 A constant thickness film of viscous liquid flows in laminar motion down a plate at angle 30cθ = as shown in figure below. The velocity profile is (2 )u Cy h y= −, v w �= = . What will be the volume flux per unit width in terms of the specific weight and viscosity ?





(A) v h�

�

mg=o (B) �v h�

mg=o

(C) �v h�

mg=o (D) �v h�

mg=o

FM 3.17 Consider a steady, two-dimensional, incompressible flow of a Newtonian fluid with the velocity components u xy�=− , v y x� �= − . What will be the pressure field ( , )p x y when the pressure at point ( , )x y0 0= = is equal to pa ?

(A) �(� )p p x y x ya� � � �r= + + + (B) (� )p p x y x y�

�a

� � � �r= + + +

(C) �(� )p p x y x ya� � � �r= − + + (D) (� )p p x y x y�

�a

� � � �r= − + +

FM 3.18 Air blows along an object as shown in figure. It has two entrances a flat door and a mounted back door. The wind blows with velocity V� across the front door. The average velocity across the back door is greater than V� because of the mound and it becomes . V1 08 0. If the wind velocity is 6 /m s, what will be the pressure difference between gate 1 and 2 ?

(A) .4 /N m7 2 (B) 3. 0 /N m7 2

(C) 0.3 /N m7 2 (D) 4.0 /N m7 2

FM 3.19 The height of the water columns in a piezometer and a pitot tube are measured to be cm45 and 50 cm respectively (Both measured from the top surface of the pipe). Both are tapped into a horizontal water pipe. What will be the velocity at the center of pipe ?(A) 17.2 /m s (B) 1.72 /m s

(C) 0.172 /m s (D) 15.6 /m s

FM 3.20 A fire hose nozzle has a diameter of 3 cm. The nozzle must be capable of delivering at least 0.016 /m s3 . If the nozzle is attached to a .5 cm2 diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flow rate ?(A) 500 kPa (B) Pa125

(C) 2 kPa50 (D) 2. kPa50

FM 3.21 A large water reservoir of 10 m diameter is filled with water which is open to the atmosphere. The reservoir has a smooth entrance orifice m2 below the water level where the water leaves the reservoir through a 25 m long horizontal pipe attached to the orifice as shown in figure. If the diameter of orifice is 3 cm, the maximum discharge rate of water through pipe is(A) 0.00443 /m s3 (B) 0.443 /m s2

(C) 0.0443 /m s3 (D) 4.43 /m s3

FM 3.22 A wind blowing past over a home at 18 /m s. It speeds up as it flows up and over the roof. If elevation effects are neglected, what will be the pressure difference between over the roof and inside the home when the speed at the point on the roof is 26.5 /m s ?





(A) 23.26 Pa− (B) 232.6 Pa−(C) 23.26 Pa (D) 232.6 Pa

FM 3.23 The water height in an airtight pressurized tank is 20 m. A hose pointing straight up is connected to the bottom of tank. The gage pressure above the water surface is 04 kPa3 . What will be the maximum height to which the water stream could rise. [The system is at sea level]

(A) 20.8 m (B) 40.8 m

(C) 20.02 m (D) 31.6 m

Common Data For Linked Answer Q. 24 and 25An inviscid, incompressible water flows steadily from a large tank and exits through a vertical, constant diameter pipe as shown in figure. The air in the tank is pressurized to 60 /kN m2.

FM 3.24 What will be the height h to which the water rises up ?(A) 5.12 m (B) 8.1 m2

(C) 4.12 m (D) 6.12 m

FM 3.25 The water velocity in the pipe will be(A) 8 /m s (B) 9 /m s

(C) 10 /m s (D) 15 /m s

FM 3.26 A high pressure liquid jet as shown in figure, can be used to cut various materials. If the viscous effects are negligible, what will be the pressure and flow rate, respectively to produce a 10 m4− diameter water jet with a speed of 700 /m s ?





(A) 2.45 MPa, 55 10 /m s6 3#

− (B) 245 MPa, 5.5 10 /m s6 3#

−

(C) 5.5 10 /m s6 3#

− , . MPa4 9 (D) 24.5 MPa , 5.5 10 /m s6 3#

−

FM 3.27 What will be the velocity at the downstream of the hose as shown in figure when all the viscous effects are neglected ?

(A) 3.5 /m s (B) . /m s0 767

(C) 2.5 /m s (D) Imaginary value

FM 3.28 Water flows through the pipe contraction as shown in figure. If the difference in manometer level is 0.2 m, the flow rate vo will be

(A) 38.9 10 /m s3 3#

− (B) 7.18 10 /m s3 3#

−

(C) 17.5 10 /m s3 3#

− (D) 3.89 10 /m s3 3#

−

FM 3.29 An oil at 20 Cc ( 870 /kg m3ρ = and 0.104 /kg m sμ -= ) flows through a 2 m long straight horizontal pipe. The pressure drop is constant 800 /Pa m= . If Reynolds number ReD of the flow is to be exactly 1000, the flow rate vo is(A) 34 /m h3 (B) 3.4 /m h3

(C) 17 /m h3 (D) 340 /m h3

FM 3.30 A 0.20 m diameter pipe discharges into a 0.15 m diameter pipe. If they are carrying 0.20 /m s3 of water, what will be the difference between their velocity heads ?(A) 6.37 m (B) 4.47 m

(C) 5.47 m (D) 4.87 m

FM 3.31 Two streams of water from two tanks strikes upon each other as shown in figure. The point A is the striking point and all the viscous effects are negligible. What





will be the height h ?

(A) 6.05 m (B) 7.05 m

(C) 8.05 m (D) 5.05 m

FM 3.32 A liquid ( 0.002 /N s m2μ -= , 100 /kg m0 3ρ = ) is forced through the circular tube of mm9 diameter as shown in figure. A differential manometer is connected to the tube to measure the pressure drop along the tube. If the differential reading

hΔ is mm6 and the flow is laminar, what will be the mean velocity in the tube ?

(A) 85.9 10 /m s3#

− (B) 85.9 10 /m s6#

−

(C) 8.59 10 /m s3#

− (D) 8.59 10 /m s4#

−

FM 3.33 Water enters a tank of diameter D , steadily at a mass flow rate of mino , which is open to the atmosphere. An orifice of diameter Do with a smooth entrance (i.e. no losses) is open at the bottom allows water to escape. If the tank is initially empty, the maximum height that the water will reach in the tank is





(A) hg D

m� ��

o

in� �

�

p r= o

= G (B) h g Dm� �

��o

in�

�

rp= o

= G

(C) h g Dm

��

��o

in�

�

rp= o

= G (D) h g Dm

��

��o

in�

rp= o; E

FM 3.34 A liquid ( 789 /kg m3ρ = , 1.19 10 /N s m3 2μ # -= - ) flows through a 10 mm diameter horizontal tube. If the mean velocity is 0.15 /m s, what will be the pressure drop per unit length of the tube and the velocity at a distance of 2 mm from the tube axis, respectively ? (A) 28.55 /N m2, 0.126 /m s (B) 71.32 /N m2, 0.315 /m s

(C) 42.8 /N m2, 0.189 /m s (D) 57.1 /N m2, 0.252 /m s

FM 3.35 Water flows through a variable area pipe with a constant flow rate vo as shown in figure. The viscous effects are negligible and the pressure remains constant throughout the pipe. If D� is the three fourth of the D , what will be the diameter D ?

(A) 0.65 /v z /1 4o (B) . v z0 65 /1 4o

(C) . /z v0 65 /1 4 o (D) . zv0 65o

FM 3.36 310 straight microtubes of 25 cm length each, are bundled together into a ‘honeycomb’ whose total cross-sectional area is 0.0006 m2. The volume flow rate of water at 2 C2c is 1 /m h3 . If the pressure drop from entrance to exit is 1.5 kPa, the appropriate microtube diameter and flow velocity are(A) 1.5� 10 , 0.�63 �m m sD V3

#= =− (B) 1.5� 10 , 0.0�63 �m m sD V�#= =−

(C) 1.5� 10 , �.63 �m m sD V2#= =− (D) 1.5� 10 , �6.3 �m m sD V3

#= =−

FM 3.37 A circular stream of water from a spigot is observed to taper from a diameter of D1 to D2 in a distance of z1. The expression for flow rate is





(A) 1

5.53

DD

A z

1

24

2 1

− b l

(B) .

DD

A z

1

4 43

1

24

2

− b l

(C) .

DD

A z

1

4 43

1

24

2 1

− b l

(D) 1

4.43

DD

A z

1

24

2 1

− b l

FM 3.38 A 1 mm diameter strings are knotted into 2 × 2 cm squares to make a fishnet. If the net plane is normal to flow direction, the horsepower required to tow 28 m2 of this netting at 1.54 m/s in seawater at 20°C ( 1025 / , 0.00107 /kg m kg m s3ρ μ -= =) is(A) 7.5 hp (B) 7.0 hp

(C) 3.75 hp (D) 3.5 hp

FM 3.39 The vent on the tank as shown in figure is closed and the tank is pressurized to increase the flow rate. What will be the pressure p1 needed to produce twice the flow rate of that when the vent is open ?

(A) . kPa132 3 (B) 88.2 kPa

(C) 105.70 kPa (D) 44.1 kPa

Common Data For Linked Answer Q. 40 and 41The water flows through the horizontal branching pipe as shown in figure at a rate of 0.28 /m s3 . The viscous effects are negligible.

FM 3.40 The water speed at section (2) is(A) 6.52 /m s (B) . 2 /m s4 4

(C) 8.83 /m s (D) 7.23 /m s

FM 3.41 What will be the flow rate at section (4) in /m s3 ?





(A) .0 11 (B) 0.055

(C) 0.35 (D) 0.23

FM 3.42 Oil with specific gravity 0.85 flows in the variable area pipe as shown in figure. If the viscous effects are negligible, the flow rate of oil in /m s3 will be

mγ H O2g=(A) .5 5 10 2

#− (B) .4 63 10 3

#−

(C) .4 63 10 2#

− (D) .5 5 10 3#

−

FM 3.43 The water is siphoned from the tank as shown in figure. A water barometer inserted into the tank and it gives a reading of 9.2 m. If the pressure of the vapor in the closed end of the barometer equals to the vapor pressure, what will be the maximum value of h allowed without cavitation occurring ?

(A) 0.075 m (B) 0.095 m

(C) 0.95 m (D) 0.85 m

Common Data For Q. 44 and 45Air flows through the device as shown in figure. When the flow rate is large enough, the pressure within the constriction will be low enough to draw the oil up into the tube. Neglect compressibility and viscous effects. ( �2 � ,N mair

3γ =8.�5 �kN moil

3γ = )





FM 3.44 What will be the flow rate vo in /m s3 ?(A) .0 335 (B) 3.35

(C) 0.00335 (D) 0.0335

FM 3.45 What will be the pressure needed at section (1) to draw the water into section (2) ?(A) 0 (B) 2 kPa

(C) 0.02 kPa (D) 0.2 kPa

FM 3.46 The water flows steadily from a large open tank and discharges into the atmosphere through a 7.6 cm diameter pipe as shown in figure. If the pressure gages at A and B indicates the same pressure then the diameter of the narrowed section of the pipe at A, is

(B) 0.05 3 m8 (B) 0.0 m3165

(C) 0.0633 m (D) 0. m1266

FM 3.47 The oil flows steadily from the two tanks as shown in figure. What will be the water depth hA for tank A ?





(A) 24 m (B) 21 m

(C) 23 m (D) 22 m

FM 3.48 Ethyl alcohol flows through the Venturi meter as shown in figure with a velocity of 4.5 /m s in the .25 cm27 diameter pipe. If the viscous effects are negligible, what will be the elevation h of the alcohol in the open tube, connected to the throat of the Venturi meter ?

(A) 2.4 m (B) 2.096 m

(C) 2.5 m (D) 2.3 m

FM 3.49 The water flows into the sink at a rate of 1.26 10 /m s4 3#

− as shown in figure. When the drain is closed, the water will eventually flow through the overflow drain holes rather than over the edge of the sink. Neglect the viscous effects. What will be the number of drain holes of 1 cm diameter that are needed to ensure that the water does not overflow the sink ? ( 0.61)Cc =

(A) 2 (B) 5

(C) 3 (D) 4

FM 3.50 The water flows into a large tank at a rate of 0.010 /m s3 as shown in figure. The water leaves the tank through 5 holes in the bottom of the tank, each of which produces a stream of 20 mm diameter. For steady state operation, the equilibrium height h will be





(A) 2.50 m (B) 1.63 m

(C) 2.07 m (D) 1.97 m

FM 3.51 Air at standard conditions ( 12.0 / )N m3γ = flows through the cylindrical drying stack as shown in figure. The inclined water filled manometer reading is 20 mm and viscous effects are neglected. What will be the flow rate ?

(A) 10.26 /m s3 (B) . /m s9 36 3

(C) . /m s8 35 3 (D) 1 .26 /m s1 3

FM 3.52 In ideal conditions, the flow rate of seawater through the orifice meter as shown in figure is to be 19 10 /m s4 3

#− with pressure difference ��k�ap p� �− = . If the

contraction coefficient is 0.63, the diameter of the orifice hole d will be

(A) 3 cm (B) 3.6 cm

(C) 1.3 cm (D) 2.6 cm

FM 3.53 A weir of trapezoidal cross section is used to measure the flow rate in a channel as shown in figure. The flow rate is v0o , when H l

�= . What will be the flow rate when H l�

3= ?





(A) ��v v�� =o o (B) ��v v� �� =o o

(C) ��v v�� =o o (D) ��v v�� =o o

FM 3.54 The water flows in a 2 m wide rectangular channel as shown in figure, with up-stream depth of 80 mm. The water surface rises 0 mm2 as it passes over the portion where the channel bottom rises 10 mm. If the viscous effects are negligible, what will be the flow rate through the channel in /m s3 ?

(A) 1.53 (B) 0.205

(C) 0.0064 (D) 0.102

FM 3.55 The water flows down uniformly through the slopping ramp as shown in figure with negligible viscous effects. What will be the depth h2 at the downstream ?

(A) 1.347 m (B) 0.163 m−(C) 0.186 m (D) 0. m093

FM 3.56 Water flows under the inclined sluice gate as shown in figure. If the gate is 2.4 m wide into the paper, the flow rate will be

(A) 3.97 /m s3 (B) 4.96 /m s3

(C) 3.27 /m s3 (D) 2.97 /m s3

FM 3.57 A conical plug regulate the air flow from the pipe of diameter 0.33 m as shown in the figure. The air leaves the edge of the cone with a uniform thickness of 0.02 m. If viscous effect are negligible and the flow rate is 1 /m s3 , the pressure within the





pipe is ( 12 / )N mair2γ =

(A) 580 Pa (B) 490 Pa

(C) 512 Pa (D) 540 Pa

FM 3.58 Water flows from a nozzle of equilateral triangular cross-section as shown in figure. After it has fallen a distance of 82 cm, its cross-section becomes circular with a diameter of 3.3 cm. What will be the flow rate in /m s3 ?

(A) 0.0086 (B) 0.0043

(C) 0.086 (D) 0.043

FM 3.59 The water flows over the unit width spillway as shown in figure. If the velocity is uniform at upstream and downstream and viscous effects are negligible, what will be the velocity at the downstream ?

(A) 7. 4 /m s6 (B) 6.21 /m s

(C) .50 /m s5 (D) 53.95 /m s

***********





SOLUTIONS

FM 3.1 Option (A) is correct.A general equation for parabola in x -direction

u ( )a b x c �= + − ...(i)From boundary conditions

At x �= u uin=At x L= u uout=By setting c �= and from equation (i) at x �= a uin=

and at x L= , b L

u uout in�= −

By substituting these values in equation (i), we get

( )u x ( )

uL

u uxin

out in�

�= + −

FM 3.2 Option (C) is correct.The velocity components are

u xy2 2=− , ( )v f y= , w zy�=−The incompressible continuity equation

xu

yv

zw

22

22

22+ + 0=

or ( � ) �( )� ( )x

xyy

f yz

zy� �

22

22

22− + + − 0=

y dydf y2 2 2− + − 0=

dydf y3 2=

Integrating the above equation,we get

( )f y ( )y dy3 2= # y constant�= +

FM 3.3 Option (B) is correct.The velocity components are

u x y x� �= − + , v (2 )xy y=− +Each component of acceleration :

dtdu u

xu v

yu

22

22= +

ax ( )( ) ( )( )x y x x xy y y2 1 2 22 2= − + + + − − −

dtdv u

xv v

yv

22

22= +

ay ( )( 2 ) ( 2 )( 2 1)x y x y xy y x2 2= − + − + − − − −At ( , ) ( , )x y 2 4= ax ( )( ) ( )( )2 4 2 2 2 1 2 2 4 4 2 42 2

# # # #= − + + + − − − 110=

ay (2 4 2)( 2 4) ( 2 2 4 4)( 2 2 1)2 2# # # #= − + − + − − − − 180=





The maximum acceleration is

amax �� a ax y� � � �= + = +] ]g g 210.95 units=

and its direction

θ tan 1101801= −b l .58 57c=

FM 3.4 Option (A) is correct.For stagnation point, all components of V must equal zero in order for V itself to be zero. So

u 1.25 0x y1= + + = ...(i)

v . . .x y0 5 1 5 2 5 0=− − − = ...(ii)By solving equation (i) and (ii), we get

.x ��=− and .y ��=Since for given velocity field only one value of x and y is obtained. Therefore, only one stagnation point is there.

FM 3.5 Option (C) is correct.At any point ( , )x y in the flow field velocity components are

.u x y� ��= + + , . . .v x y�� =− − −Material acceleration at any point ( , )x y

ax tu u

xu v

yu w

zu

22

22

22

22= + + +

ay tv u

xv v

yv w

zv

22

22

22

22= + + +

For steady two dimensional flow field

tu

22

tv �

22= = and w �=

Then ax uxu v

yu

22

22= +

( . ) ( . ) ( . . . )x y x y1 2 5 2 5 0 5 1 5 2 5 1# #= + + + − − − . x2 4 75= +

ay uxv v

yv

22

22= +

( . ) ( . ) ( . . . ) ( . )x y x y1 2 5 1 5 0 5 1 5 2 5 2 5# #= + + − + − − − − . . y0 25 4 75=− +Material Acceleration at point ( 2 , 3 )m mx y= = is

ax 2 4.75 2 11.5 /m s2#= + =

ay 0.25 4.75 3 14 /m s2#=− + =

FM 3.6 Option (D) is correct.Incompressible continuity relation in polar coordinates :

( )r rrV r

V� �r2

222

q+ q 0=

or cos sinr rr k

rb

r krb� � � �� 2

222θ θ θ- + - +b bl l; ;E E 0=

cos sinr rk r r

br k

rb� � � �2

222θ θ θ- + - +b bl l; ;E E 0=

cos cosr krb

r krb� � � �� #θ θ+ + - +b bl l; ;E E 0=





�� r krb

r krb� � � �� θ θ+ - +b bl l; ;E E 0=

or 0 0=Incompressible continuity equation is satisfied. Hence the flow field will be steady incompressible.

FM 3.7 Option (A) is correct.Check continuity

xu

yv 0

22

22+ = or 0 0 0+ = (satisfied therefore ψ exists)

Now check irrotationality :

xv

yu

hV0 0z 2

222 !ζ = - = - (Not satisfied, flow is rotational, φ does not exist)

FM 3.8 Option (D) is correct.Applying the Bernoulli’s equation at both the points with subscript 1 & 2,

pg

V2

1 12

γ + pg

V2

2 22

g= + ...(i)

With V u v2 2 2= + . For the velocity potential given

ux2

2f= xy4= and vy

x y2 22 2

22f= = −

At point 1, 1 mx = and 1 my = u1 4 1 1 4 /m s# #= = 2 2 0v 1 11

2 2= − =^ ^h h

and V12 16 /m s4 2 2 2= =^ h

At point 2, 2 mx = and 2 my =

So that u2 4 2 2 16 /m s# #= = v 2 2 2 2 022 2= − =^ ^h h

and V22 256 /m s16 2 2 2= =^ h

Thus, from equation (i),

p2 p g V V21 12

22g= + −^ h 200 10 2 9.81

.9 80 1016 2563

3#

##

= + −^^h

h

80.1 kPa=

FM 3.9 Option (B) is correct.First check that 02

4 φ = , which means that incompressible continuity is satisfied,

24 φ

x y2

2

2

2

22

22f f= +

x2

2φ y x2= + , x

22

2

22 φ =

and y2

2φ x y2= − , y

22

2

22 φ =-

Hence x y

2 2 02

2

2

2

22

22φ φ+ = - = , ψ exists

Now use φ to find u and v and then integrate to find ψ

u 2x

y xy2

222f y= = + =

Hence ψ ( )� �� 2 22

= + + ...(i)

v 2 2y

x yx

yxf

22

22

22f y= = − =− =− − from eq. (i)





So ��f x Cx2

2

= − +

The final stream function is thus

ψ ( ) 2 Cy x xy21 2 2= − + +

FM 3.10 Option (C) is correct.In Cartesian coordinates the stream function is

u �y

y22y= = ...(i)

v �x

x22y= − = ...(ii)

By integrating equation (ii), we get

2ψ# x dx2=− #

ψ ( ) ( )x f x x f x22 2

2= − + =− + ...(iii)

differentiate equation (iii) w.r.t. y,

y2

2ψ ( )�y��

0= +

( )dyf x

y3= from eq. (i)

Integrating the above equation,

( )f x# y �y3= # or ( ) Cf x y�� = +

Substitute this in equation (iii),

ψ Cy �23 2 2= − +

But in polar coordinates

y sinr q= and cosx r q=Therefore the desired result is

( , )�ψ θ ( ) ( )sin cos C� �23 2 2q q= − + sin cos C� �2

3 2 2 2 2q q= − +

FM 3.11 Option (B) is correct.Since the flow is steady and inviscid. So equation of motion along the stream line direction is

sin��

22g q− − V

xV

22r=

Here pipe is horizontal i.e. 0cθ = & �(� )�V i= +

Hence xp

22− ( ) [ ( )]�

��3 1 3 1

22r= + + 0 0sin c =

9 (1 ) 9000(1 ) /N m� � 3r=− + =− +

FM 3.12 Option (D) is correct.From previous part of the question

xp

22 ( )�9000 1=− +

p2 ( )� �9000 1 2=− +Substitute limits of p between 345 kPa to p� and for x between 0 to 1 m and





integrate the equation,

p2p

p

��a� ��

�

=

# 9000 ( )x dx1x 0

1=− +

=#

p ��

#− x x9000 22

0

1

=− +: D

p� 345 10 9000 1 213

#= − +b l 345 10 9000 233

# #= −

10 331.5 kPa345 2273

#= − =: D


We have �� sec�V = , �� xp �

22 =

The equation of motion along the stream line direction is

sinxp

22g q− − V

xV

22r=

If neglect the effect of gravity, then

γ �gr= =

So xp

22 V

xV

22r=−

xV

22 . 4.06 sec25 1 23

125 1

#=− =− −

Also Vδ xV x

22 d=

Substitute xV

22 . sec4 06 1=− − and �.�mxδ =

Vδ 4.06 0.5 2.03 /m s#=− =−So the net air speed will be

V Vd+ 2 2.03 .97 /m s5 22 23-= − =

FM 3.14 Option (B) is correct.For this (gravity-free) velocity, the momentum equation is

ux

vy

V V22

22ρ +c m p4=−

(3 ) (3 ) ( ) (3 )xyx

xy y yy

xy yi j i j03 3 3

22

22ρ - + - -; E p4=−

( ) ( ) ( )( )xy y y x yi i j3 3 3 303 2

#ρ + - -6 @ p4=−

xy xy yi i j9 3 302 3 5ρ - +6 @ p4=−

( )xy xy yi j3 9 303 2 5ρ - -6 @ p4=

p4 ( )xy y yi j3 3 302 5r= − −6 @

xp

22 ( )xy y3 30

2r= −6 @

FM 3.15 Option (B) is correctAlong a streamline, the equation of motion is

( )dp d V dz�� r g+ + 0= ...(i)

where gγ ρ= , dp E dV r

r= and mod tanbul� ulus cons tEV = =

Substitute in equation (i),





� �E d d V g dz��

V�

ρρ ρ ρ+ + 0=

E d d V gdz��

V ��

ρρ + +b l 0=

Now integrate this equation between point (1) & (2), we get

E d d V gdz21

Vz

z

V

V

22

1

2

1

2

1

2

ρρ + +

ρ

ρ

b l# ## 0=

( ) ( )E V V g z z1 121

V2 1

22

12

2 1r r− − + − + −: D 0=

It means E V gz2V

2

r− + + = Constant along a streamline


There is atmospheric pressure all along the surface at y = h, hence �xp

22 = . The

x -momentum equation can be

uxu v

yu

22

22ρ +; E x

p g ux2

422 r m=− + +

or uyu� �# #2

2ρ +; E 0 � �x2

4r m= + +

0 0 ( 2 )sin� �r q m= + + −

or C 2 4sin� �m

r qmr= = 30 1/2sin c =

The flow rate per unit width is found by

vo ( )udy Cy h y dy2h h

0 0= = −# # �hy�y �y �y2

h h

0

2

0= −# #

Chy Cy Ch Ch� �

hh

20

�

0

��

= − = −7 ;A E �h32 3=

vo � h32

43

mr

# #= � h61 3

#m

r#= 6

h3

mg= gγ ρ=

FM 3.17 Option (D) is correctThe pressure gradients from the Navier-stokes x and y relations :

uxu v

yu

22

22ρ +c m x

pxu

yu

2

2

2

2

22

22

22m= − + +e o

or ( ) ( )( )xy y y x x2 2 22 2ρ - - + - -6 @ (0 0)xp

22 m= − + +

xp

22 ( )xy x2 2 3r=− +

and similarly for the y-momentum relation

uxv v

yv

22

22ρ +c m y

pxv

yv

zv

2

2

2

2

2

2

22

22

22

22m=− + + += G

or ( ) ( )( )xy x y x y2 2 22 2ρ - - + -6 @ yp 2 2 0

22 m= − + − + +6 @

yp

22 ( )x y y2 2 3r=− + ...(i)

The two gradients xp

22 and

yp

22 may be integrated to find ( , )p x y

p �xp dx tany cons t2

2= =# ( )x y x �y2 2 4

2 2 4

r=− + +c m ...(ii)

then differentiate w.r.t. y,





yp

22 2 ( )x y dy

df2r=− +

or 2 ( )x y y2 3r− + 2 ( )x y dydf2r=− + from eq. (i)

dydf 2 ( ) 2 ( ) 2x y y x y y2 3 2 3r r r=− + + =−

or ( )f y Cy24r=− + ...(iii)

From equation (ii) and (iii),we get

p (2 ) Cx y x y p2 a2 2 4 4r= − + + + =

( , ) (0, 0)At x y = C pa=Finally, the pressure field for this flow is given by

p (� )p x y x y��

a� � � �r= − + +


Applying Bernoulli’s equation at section (1) and (2),

p V z21

1 12

1r g+ + p V z21

2 22

2r g= + +

Neglecting the gravitational effects. So, z z1 2=

Thus p V21

1 12r+ p V2

12 2

2r= +

p p1 2− ( )V V21

22

12r= −

Substitute 1.2� ��g mair�ρ ρ= = , 1.��V V2 �= , V V1 �= and � �m sV�=

p p1 2− 1.23 [(1. 8 6) (6) ]21 0 2 2# # #= −

0.615 (6.48) (6)2 2# −6 @

3.68 3.7 /N m2-=


The application of the Bernoulli’s equation between points (1) and (2) gives

gp

gV z2

1 12

1ρ + + gp

gV z2

2 22

2r= + + At center line z z2 1=

gp p2 1

ρ- g

V V2

12

22

= −





Since point 2 is a stagnation point and thus V �2 =

gV2

12

gp p2 �

r= − � � � �

gg h R g h Rpitot piezo

rr r= + − +

h hpitot piezo= −

V� ( ) . ( . . )g h h2 2 9 81 0 50 0 35pitot piezo # #= − = − .1 715= 1.72 /m s,


We have �.�cmD�= , �cmD2 = , �.�� m sv �=oApplying Bernoulli’s equation at section (1) & (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

For horizontal pipe z z1 2= & at downstream p �2 =

So pg

V2

1 12

γ + gV2

22

=

p1 [ ]g V V2 22

12g= − ...(i)

Now, from continuity equation

vo A V A V1 1 2 2= =

V1 ( . ). �.�2 �m sA

v�� 1�

��1�1 �

2 2#

= = =p −o

V2 (� 1� ). 22.�� m sA

v ��1�2 �

2 2#

= = =p −o

Substitute these values in equation (i), we get

p1 [ ]V V2 22

12r= − [( . ) ( . ) ]2

1000 22 65 3 622 2= − gρ γ=

249959= 2 kPa50-


Consider water level at point (1) and pipe exit at point (2) and applying Bernoulli equation between (1) and (2)

gp

gV z2

1 12

1ρ + + gp

gV z2

2 22

2r= + +

z1 gV2

22

= V gz22 1& = , �, �p p p z Vatm1 2 2 1= = = =





For maximum discharge rate, z� will be maximum i.e. �mz�=

V , max� .gz2 2 9 81 2max # #= = 6.26 /m s=Then maximum discharge rate

vo A V D V�, ,max maxpipeo

�

�

�p

# #= = (0.03)

( . )4 6 262

#p#= 0.00443 /m s3=


Applying the Bernoulli’s equation at section (1) & (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Elevation effects are negligible z z1 2= 0=and V1 18 /m s= , 2�.� �m sV2 = , airγ 1.23 /kg m3=

So p p1 2− ( )g V V2 22

12g= − g

γ ρ=

. [( . ) ( ) ]21 23 26 5 182 2= − 232.6 Pa=

p p2 1− 232.6 Pa=−This negative pressure tends to lift the roof.

FM 3.23 Option (B) is correct.Consider point (1) at the free end of water in tank and point (2) at top of the water trajectory. The Bernoulli equation between these two points.

gp

gV z2

1 12

1ρ + + gp

gV z2

2 22

2r= + + �V V2 1= =

z2 gp p z1 2

1r= − + gp p z g

pz,atm gage1

11

1r r= − + = +

. 20 40.8 m1000 9 81204 1000

##= + =






Applying Bernoulli’s equation at section (1) & (3), we get

pg

V z�1 1

�

1γ + + pg

V z�3 3

�

3g= + +

Here �� mp1�= , V �1 = , �mz1 = , �( )gagep3 = , �V3 = , z h3 =

Hence h p z11g= +

.9 8 1060 10 23

3

#

#= +

6.12 2 8.12 m= + =

FM 3.25 Option (B) is correct.Applying continuity equation at section (2) & (4),

A V2 2 A V4 4= V2 V4= ( section)same crossA A2 4 −=Now applying bernoulli’s equation at section (3) & (4),

pg

V z23 3

2

3γ + + pg

V z24 4

2

4g= + +

Here �p p3 4= = (gage), �V3 = , �.12 mz h3 = = , mz 44 =

Hence gV2

42

z z3 4= −

V42 ( . ) . .8 12 4 2 9 81 80 83# #= − =

V4 9 /m s V2= =


Applying Bernoulli’s equation at section (1) & (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here V �1 - , z z1 2- and p �2 =

So p1 gV V2 2

122

22g r#= = g

γ ρ=

1000 (700) 2.45 10 / 245kN m MPa21 2 5 2# # #= = =

Flow rate vo ( )V A �� 4 1�2 24 2

#p= = − 5.50 10 /m s6 3- #

−






Applying Bernoulli’s equation at section (1) & (2), we get

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p �1 2= = (gage pressure), V �1 = , ��z1 = , ��z2 =−

Hence .0 3 .gV2 0 332

2

= −

V22 . . .0 63 2 9 81 12 36# #= =

V2 3.5 /m s-



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here V �1 = and z z1 2=

So, p1

γ pg

V2

2 22

g= +

V2 ( )g p p2 1 2

g= − ...(i)

Now p1 h1g= & p h2 2g= p p1 2− ( ) .h h 0 21 2g g= − =

Hence V2 . 1.98 /m sg2 0 2

# gg= =

Flow rate vo A V2 2=

(0.05) 1.9842p#= 0.00 89 /m s3 3=

3.89 10 /m s3 3#= −


ReD 1��VDm

r= =

Since Vavg �Lp R2

mD

#=

Thus 1000 � . .( �)

Lp R D

DD�1��

��2

�� 1��

22 2

# ##m

rm

D#= =; ;E E

1000 . .D

0 104870 400 8 4 0 104

3

# ## #

=

1000 . D1 005 106 3#=

or D� .0 00099457= D 0.0998 0.1000 m,=





Now V ��

Lp D ��

mD

#= .

0.1 1.20 /m s2800

4 8 0 104

2

# ##= =] g

Then vo AV= ( . ) .4 0 1 1 202# #

p=

0.009425 /m s3= .0 009425 3600#= 34 /m h3=

FM 3.30 Option (B) is correct.We have �.��mD�= , �.��mD�= , �.�� m sv �=o

Flow rate vo A V A V� � � �= =

V� ( . ).

Av

��

� ��

#= = p

o6.37 /m s=

V� (�.��).

Av ��

� ��

#= = p

o11.32 /m s=

Now the difference between velocity head is

d ( )gV

gV

g V V2 2 212

212

22

12= − = −

. [( . ) ( . ) ] 4.47 m2 9 81 11 32 6 372 2

#= − =


The water jets strikes at point A as shown in figure. We have to take three sections (1), A and (2).Now applying Bernoulli’s equation at section (2) and A, we get

pg

V z22 2

2

2γ + + pg

V z2A A

A

2

g= + +

Here p �2 = (Gage), V �2 = , V �A = , z h1�2 = + , z hA =

Thus z2 p zA

Ag= +

h14 + p hA

g= +

pA 14g= ...(i)Again applying Bernoulli’s equation at section (A) & (1), we have

pg

V z2A A

A

2

γ + + pg

V z21 1

2

1g= + +

Here �VA = , V �1 = , 1�2 k�ap1 = , p 1�A g= , z hA = , 2.�mz1 =

Hence h14 0γγ + + 0 2.5 m172

g= + +





h . .9 8172 2 5 14= + −

17.55 2.5 14 6.05 m= + − =


The flow is laminar flow, so that mean velocity

V Rlp

82

mD=

From manometer equation (see figure)

p h hf gf� g gD D+ − p�=or p p p� � D− = h h ggf f gf fg g r rD D= − = −^ ^ ^h h h

0.007 9.81 68.7 /N m2000 1000 2# #= − =^ h

Thus V 8 0.002 2

. 68.78.59 10 /m s2

0 004 2

3

# #

##= = −

b l

FM 3.33 Option (C) is correct.The Bernoulli equation between water entrance and exit of tank.

gp

gV z�

in inin

�

ρ + + gp

gV z�

out outout

�

r= + +

Since p pin out= patm= , �Vin = and �zout =

Then gV2

out2

zin=

Vout gz2 in=So that the mass flow rate through the orifice

mouto v A V D gz� �out out in��

r r r p #= = =o

zin g Dm

21 4 out

02

2

rp#= o= G

Setting z hmaxin = and m mout in=o o gives the relation for maximum height the water will reach in the tank.

hmax g Dm

21 4 in

02

2

rp= o

= G

FM 3.34 Option (D) is correct.First check Reynolds number to determine if flow is laminar or turbulent.

Re .

�� .�� .�� VD��

<�#m

r # #= = =−

Thus, flow is laminar and mean velocity is given by





V Rlp

82

mD=

So that lpΔ

RV��

m= .

8 . 0.1557.1 /N m per m

20 010

1 19 102

32## #= =

−

b

^

l

h

Since velocity distribution in terms of vmax is

vz v Rr�max

�= − a k: D

and v V�max = , where V is the mean velocity.

It follows that vz 2V Rr1 2

= − a k: D

For mmr �= , vz 2 0.15 0.252 /m s1 52 2

#= − =b l; E

FM 3.35 Option (A) is correct

We have tancons tp = , D D��

�=

Applying Bernoulli’s equation

pg

V z��

γ + + pg

V z��

�

�g= + +

Here p p�= , z ��=

So gV V

212 2− z=

V V�� − gz2= ...(i)

Flow rate vo AV A V� �= =

V �Av

Dv

Dv

�� p p

= = =o o o

V� Av

Dv

Dv

Dv�

��

� � ��

��

#p p= = = =p

o o o o

^ hSubstitute the V & V� in equation (i)

Dv

Dv

964 4

2

2

2

2

π π-o o

c cm m . .D

vD

v��

�

�

�

= −o oc cm m gz2=

D gz��# 3.51v2= o

D� .. 0.179z

vzv

2 9 83 51 2 2

#= =o o

D 0.65 zv /2 1 4

= ob l 0.65

zv/1 4= o





FM 3.36 Option (A) is correct.For water at 20 Cc : 998 /kg m3ρ = and 0.001 /kg m sμ -=Since vo N vtube#= o v310 tube#= o

or 36001 310 310 . .L

D p D128 128 0 001 0 25

15004 4

# ##

mp pD

# #= =

or D� . . .1500 310 3600128 0 001 0 25 6 085 10 12

# # ## #

#p= = −

D 0.00157 m=Also vo V A#=

or V .Av

��= =o

1666.67 /m h=

. /m s36001666 67= 0.463 /m s=



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p �1 2= = (Gage pressure), z �2 =

So, gV z2

12

1+ gV2

22

=

V V22

12− gz2 1= ...(i)

From continuity equation

vo A V A V1 1 2 2= =

V1 Av

1= o

, V Av

22

= o

From equation (i)

Av

Av

22

2

12

2

−o o v

A A1 12

22

12#= −o c m gz2 1=

Av

AA1

22

2

1222

# −oc m gz2 1=

v2o

AA

A gz

1

2

1

22

22

1#=− c m

DD

A gz

1

2

1

24

22

1#=− c m

AA

DD

1

2

1

22

= c m

vo

DD

A gz

1

2

1

24

2 1#=− c m

.

DD

A z

1

4 43

1

24

2 1=− c m





FM 3.38 Option (B) is correct.Considering the string as “Cylinders in cross flow”, the Reynolds Number is

Re .. . ��VD

�� # # -m

r= =

For cylinder .C ��D -

Now drag on one 2 cm strand:

F

�.� (�.��) �.�� .�2 �.�2��NC V DL2 2��2�

D2 2 -

r# # # #= =

Now 1 m2 of net contains 5000 of these 2 cm strands and 28 m2 of net contents 5000 × 28 = 140000 strands total.Now the total force on net

F .140000 0 0243#= 3400 N,

Then the horsepower required to tow the net is

P .FV �400 1 54#= = 5236 W=

P 7.0 hp7465236= =


Applying Bernoulli’s equation at section (1) & (2), when the vent is open.

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p 01 2= = (gage pressure), V 01 = , z 02 = , �mz1 =Thus, V2

2 2 2 9.8 3 58.8gz1 # #= = = V2 7.67 /m s=To have double the flow rate with the vent closed. p 01 =Y and 2 �.�� 15.�4 �m sV2 #= = . From Bernoulli’s equation,

p z11γ + g

V2

22

=

p1

γ .( . )

3 92 9 815 34 2

#= − =

p1 9 9.8 88.2 kPa#= =

FM 3.40 Option (C) is correct.Applying Bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +






V� �� m �A

v��

�

�= = =o

and z� z�= (horizontal pipe)

Thus . .9 869

2 9 832

#+ .

.g

V9 834 5

222

= +

gV2

22

. . . .7 04 0 46 3 52 3 98= + − =

V�� . .3 98 2 9 8 78# #= =

V� 8.83 /m s=

FM 3.41 Option (A) is correct.From continuity equation, the flow rate entering the pipe must be equal to the sum of the flow rates leaving the pipe. So,

v�o v v v� � �= + +o o o

v�o v v v� � �= − −o o o v A V A V� � � � �= − −o

. . . . .0 28 0 0065 8 83 0 0185 6 1# #= − − 0.11 /m s3-



pg

V z2oil

1 12

1γ + + pg

V z2oil

2 22

2g= + +

Here z z1 2= (horizontal pipe), �V1 =

So, gV2

22

p poil

1 2

g= − ...(i)

From the figure, by applying manometer equation,we get

p1 p l p loil oil� �g g= + = + p p� �=and p2 ( )l h h pH Ooil �2g g= + − +Thus p p1 2− ( )h h hH O H Ooil oil2 2g g g g= − = −Substitute this value of p p1 2− in equation (i),

gV2

22

( )h H O

oil

oil2

gg g= −

. .h S G1 1

oil= −b l . . . .S G S G

1H O

H O

2

2

γγ =

V22 . . . .0 1 2 9 81 0 85

1 1 0 346# # #= − =: D

V2 0.59 /m s=

Hence, flow rate vo ( . ) .A V � �1 ��2 22

# #p= = 0.00463 / sm3=

4.63 10 /m s3 3#= −






Applying Bernoulli’s equation at section (2) and (3), we have

pg

V z22 2

2

2γ + + pg

V z23 3

2

3g= + +

Here p �2 = (gage), z �2 = , �.�mz3 = , p pvapor3 = , V �2 =

So, 0 �.�p

gV2

vapor 32

g= + + ...(i)

Also .p �2� #g+ p2=Given p� pvapor= and p 02 = pvapor 9.2g=−

pvapor

γ .9 2=− ...(ii)

From equation (i) and (ii)

. .gV9 2 2 1 83

2

− + + 0=

gV2

32

.7 4=

V32 . . .7 4 2 9 8 145 04# #= =

V3 12.04 /m s=Again applying continuity equation at section (3) and (4),

A V3 3 A V4 4=

( . ) .4 7 6 10 12 042 2# # #

π - ( . ) V4 12 7 10 2 24# # #

p= −

V4 4.32 /m s=Now applying Bernoulli’s equation at section (2) & (4),

pg

V z22 2

2

2γ + + pg

V z24 4

2

4g= + +

Here p 02 = (gage), V 02 = , z h2 = , p 04 = (gage), z 04 =

Hence h gV2

42

=

.( . )

0.95 m2 9 814 32 2

#= =







pg

V z2air

2 22

2γ + + pg

V z2air

3 32

3g= + +

Here p �3 = (gage pressure), z z2 3= (horizontal pipe)

So pair

2

γ gV

gV

2 232

22

= − ...(i)

From continuity equation at section (2) & (3),

A V2 2 A V3 3=

D V4 22

2π D V4 3

23

p=

D V22

2 D V32

3=

V2 DD V V V2�

�� 2

32

3

2

3 3= = =b bl l ...(ii)

From figure p2 hoil #g=− . .8 95 10 0 33

# #=− 2.68 10 Pa3#=− ...(iii)

Substitute the values of V2 & p2 from equation (ii) and (iii) into equation (i),

.12

2 68 103#−

( )g

Vg

VgV

2 24

2153

23

232

= − = −

V32 . . .12 15

2 68 10 2 9 8 291 823

## # #= =

V3 17.08 /m s=

Hence, flow rate vo (�.��) ��.�� .�33� �m sA V �3 32 3p#= = =

Note : Same result is obtained from A V2 2.

FM 3.45 Option (A) is correct.Applying Bernoulli’s equation at section (1) & (3),

pg

V z21 1

2

1γ + + pg

V z23 3

2

3g= + +

Here z z1 3= (horizontal pipe) p �3 = (gage pressure) V V1 3= (because A A1 3= )

Hence p1

γ 0= & p1 0=

FM 3.46 Option (C) is correct.We have �.��mD = , p p p pA B � 3&= =Applying Bernoulli’s equation at section (4) and (2),

pg

V z24 4

2

4γ + + pg

V z22 2

2

2g= + +





Here z ��= & p ��= (Gage pressure), p p� �= (given)

So pg

V z��

�

�γ + + gV2

22

= ...(i)

Again applying Bernoulli’s equation at section (1) & (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p �1 2= = (gage pressure), V �1 = , z �2 = , �.�mz1 =Thus V2

2 . . .gz2 2 9 81 4 8 94 1761 # #= = = V2 9.7 m= ...(ii)

For section (3) & (2), we have

pg

V z23 3

2

3γ + + pg

V z22 2

2

2g= + +

Here V V2 3= (because of A A2 3= ), �p2 = (gage pressure), �z2 =So p3 . ( . . )z �� 2 �3 #g=− =− + 74.48 kPa=− ...(iii)Substituting values from equation (ii) and (iii) into equation (i), we have

.. (4.8 2.4)g

V9 874 48

242− + + − .

( . )2 9 8

9 7 2

#=

gV2

42

. . .4 8 7 6 2 4 10= + − =

V�2 .10 2 9 8 196# #= =

V� 14 /m s=Now from continuity equation at section (2) & (4),

A V4 4 A V2 2=

D V4 42

4π D V4 2

22

p=

D42 . ( . )V

V D 14��

4

222 2

##= = .0 0040=

D4 0.0633 m=

FM 3.47 Option (C) is correct.For steady flow, flow rate remains constant.

v2o v4= o ...(i)Now applying Bernoulli’s equation at section (3) and (4),

pg

V z23 3

2

3γ + + pg

V z24 4

2

4g= + +

Here p p3 4= 0= (Gage pressure), z �4 = , V3 0= , 3 mz hB3 = =

So V4 . 7.67 /m sgz2 2 9 8 33 # #= = =





Flow rate from the section (4) is

v A V4 4 4=o ( . ) .4 0 05 7 672# #

p= 0.0150 /m s3=

Also applying Bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p �1 2= = (Gage pressure), V �1 = , z2 0= , z hA1 =

Thus V2 gh2 A=From equation (i), we get

( . ) gh4 0 03 2 A2

# #π .0 0150=

gh2 A .21 23= gh2 A .450 71= hA 22.995 23 m-=



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here z �1 = , 2�. 1� mz 422

#= − , 4.� �m sV2 =And from continuity equation





A V� � A V� �=

V� AA V D

D V�

��

�

��

�= = b l .. 4.5 10.06 /m s10 2

15 25 2

#= =b l

and p� . .1 83 1 83#g g= =Substitute these values in equation (i),

( � )p

g��

�

γ + . ( . ).g

1 832

4 50 204

2

gg= + +

p�

γ . . . .1 83 1 03 0 204 5 16= + + − .2 096=− ...(ii)

From figure p� hg=−

p�

γ h=− ...(iii)

On comparing equation (ii) & (iii), we get

h 2.096 m=



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + +

Here p p �1 2= = (Gauge pressure), �V1 = , z �2 = and

z1 . 10 3 10 m2 5 21 2 2

# #= + =− −b l

Thus V2 2 9.8 3 10 0.77 /m sgz2 12

# # #= = =−

Now vo nA V nC d V�c2 2 22

2p

#= =

where n = Number of holes required

Cc = Contraction coefficient

vo = Flow rate 1.26 10 /m s4#= −

n C d V

v�c 2

22p

= o

. . ( ) ..

..

3 14 0 61 10 0 774 1 26 10

1 4755 04

2 2

4

# # #

# #= =−

− .3 42=

Hence, 4 holes are needed

FM 3.50 Option (C) is correct.Applying Bernoulli’s equation at section (1) & (2),

pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)





Here p p �� = = (Gage pressure), V ��= , z h�= , z ��=

Now equation (i) becomes

z� gV2

22

=

V�� 2gh= & V� gh2= ...(ii)

It is a steady state operation. So, flow rate remains constant.

v�o v�= o ...(iii)

where v�o 0.01 /m s3=

v�o A V D V5 5 42 2 22

2#p= =

From equation (ii) and (iii),

.0 01 D gh5 4 222

# #p=

gh2 . ( . )

. .5 3 14 0 02

4 0 01 6 372# #

#= =

gh2 ( . ) .6 37 40 572= =

h .. 2.07 m2 9 8

40 57#

= =



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here z �1 = , z h2 = and from continuity equation

A V1 1 A V2 2=

V1 AA V

1

22#= V V

21

42

22

#= =b l





V� V4 1=Substitute these values in equation (i),

pg

V�

� ��

γ + ( )p

gV

h��

�

g= + +

gV

215 1

2

p p h� �

g= − − ...(ii)

The manometer equation can be written as . ( . )sin sinp l h l p�� m� � � �# c cg g g+ + + − − = p p� �− . ( )sin h0 02 30 mc g g g= − + . ( ) h0 01 mg g g= − +

p p� �

γ- 0.01 h1m

gg= − +b l ...(iii)

Substitute value of p p� �

γ- from equation (iii) into equation (ii),

gV15 2

12

0.01 0.0h h1 1 1m m

gg

gg= − + − = −b bl l

V� . g15

0 02 1m

gg= −b l

Substitute �.� �� mm�γ = and 12.0 /N m3γ =

V� . . .0 0131 129 8 10 1 10 68

3

##= − =b l

3. 68 /m s2=

Hence, flow rate vo A V D V��

�#p= = ( ) .4 2 3 2682

# #p=

10.26 /m s3=


We have �� m sv � �#= −o , ��ap p� �− = , .C ��c =


pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + z z1 2=

p p1 2

γ- g

V V2

22

12

= − ...(i)


V1 Av

1= o

( ) . 0.97 /m s

4 5 1019 10

25 3 1419 4

2 2

4

# #

###

p= = =−

−


.9 8 10

16 103

3

#

# .( . )V

2 9 80 972

2 2

#= −

.V ��122 − 16 2 32#= =





V�� . .32 0 941 32 941= + =

V� 5.74 /m s=

Hence v A V� �=o C d V�c�

�# #p=

d � � � �C Vv�

��

c �

�

# # ## #

p= =−o

d � .6 69 10 4#= −

d 2.59 10 m2#= − 2.59 cm= 2.6 cm-


The flow rate is given by

vo AV= ...(i)

where V gH2=From the figure

tanArea f trapez idal sum of parallel sides dis ce between these sides21

#ο ο #=

A ( )tanl l H H21 2 45# c= + + ( )tanH l H ��c= +

From equation (i),

v AV=o ( )tanH l H gH�� #c= +

( 45 )tang H l H2 /3 2 c= + ...(ii)

Now v�o flow rate at H l2= =

and v�o flow rate at H l43= =

From equation (ii), we get

vv

�

�oo

g l l l

g l l l

2 43

43

2 2 2/

/

3 2

3 2

=+

+

b b

b b

l l

l l

vv

�

�oo

0.46l

l

43

47

21

23

6/

/

3 2

3 2

#

#

= =b b

b b

l l

l l

v�o 2.1 v5 0= o







pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here p p �1 2= = (gage pressure), �.��mz1 = , (�.�1 �.1�) �.11 mz2 = + =From continuity equation

A V1 1 A V2 2=

V2 .. .A

A Vh

h V V V�1��

2

1 1

2

1 11 1= = = =


.gV2 0 081

2

+ ( . )

.gV

20 8

0 1112

= +

. [ . ]V2 9 8 1 0 641

2

#− . . .0 11 0 08 0 03= − =

V12 .1 63=

V1 1.28 /m s=Hence, flow rate vo ( . ) .A V �� 2 1 2�1 1 # #= = 0.2048 / 0.205 /m s m s3 3-=



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here p p �1 2= = (Gage pressure), �.�1 �.� �.�1 mz1 = + = , z h2 2= , � �m sV1 =From continuity equation

A V1 1 A V2 2=

V2 . .

AA V h

h V h h��

22

11

2

11

2

#= = = =


0 .3 0.912 9 8

2

#+ +] g 0 .

.h

h2 9 8

0 92

2

2#

= + +b l

.1 37 .h

h��122 2= +

1.�� .��1h h2�

22− + 0=

After solving this equation, we get 1.��mh2 = , 0.186 m, 0.163 m−From these roots h2 .0 163=− is not possible physically

h2 1.347 m= is also not possible because it is greater them �.�mh1 = .So the possible value of h2 is

h2 0.186 m=

FM 3.56 Option (A) is correct.Applying Bernoulli’s equation at section (1) and (2),





pg

V z��

�

�γ + + pg

V z��

�

�g= + + ...(i)

Here p p �� = = (Gage pressure), �.��z�= , �.��z�=And from continuity equation

A V� � A V� �=

V� ..

AA V

hh V V V��

��

� �

�

� �� = = = =


.gV2 1 81

2

+ ( )

.gV2

60 31

2

= +

gV

235 1

2

.1 5=

V�� . . .35

1 5 2 9 8 0 84# #= =

V� 0.92 /m s-

Hence, the flow rate vo ( . . ) .A V �� # #= = 3.97 /m s3-



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here p �2 = (Gage pressure), z �1 = , �.2�mz2 =And from continuity equation

V1 ( . )11.� �m sA

v��1

1 �2

#= = =p

o

V2 . .Av

Rh21

2 �2� ��21

2 # # #p p= = =o 31.85 /m s=






.( . )p

12 2 9 811 7

012

#+ + .

( . ).0 2 9 8

31 850 25

2

#= + +

p12

1 5 . . 6.98 45.021 75 0 25= + − =

p� 45.02 12 540.24 Pa#= = 540 Pa,



pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here p p �1 2= = (Gage pressure), z �2 = , �.�2 mz1 =And from continuity equation

V1 Av

1= o

, V Av

22

= o

A1 = Area of equilateral triangular

a43 2= where side of trianglea =

(0.058) 1.45 10 m43 2 3 2# #= = −

A2 (0.033) 8.55 10 m42 4 2p

# #= = −


Av gz2

1

2

1+oc m A

v2

2

= oc m

vA A1 12

22

12−o

; E gz2 1=

v2o

A A

gz1 12

22

12

1=−

Substitute the numerical values

v2o . .2 9 8 0 82(8.55 10 ) (1.45 10 )

1 14 2 3 2

# #=−

# #− −

. .

..

.136 8 10 47 5 10

16 07289 3 10

16 0724 4 4

# # #=

−=

v2o .0 18 10 4#= −

or vo 0.43 10 0.0043 /m s2 3#= =−







pg

V z21 1

2

1γ + + pg

V z22 2

2

2g= + + ...(i)

Here p p �1 2= = (Gage pressure), �.��z1 = , 1.��z2 =From continuity equation

A V1 1 A V2 2=Considering the unit width.

V1 .. .z

z V V V��1 �� 2��

1

22 2 2# #= = =

Thus equation (i) becomes

( . )

3.8gV

0 20 276 2

2

+ + .gV2 1 052

2

= +

.gV2 1 0 0762

2

−^ h . . .3 8 1 05 2 75= − =

V22 .

. . 5 .0 9242 75 2 9 8 8 33# #= =

V2 7. 4 /m s6=

***********

FM 4FLOW ANALYSIS USING CONTROL VOLUMES

FM 4.1 A 1.5 m3 rigid tank contains air whose density is 1.18 /kg m3. A high-pressure air is allowed to enter the tank until the density in the tank rises to 7.20 /kg m3. The mass of air that entered the tank is(A) 10.8 kg (B) 9.03 kg

(C) 1.77 kg (D) 12.57 kg

FM 4.2 The wind blows through a 2.2 3m m# garage door with a speed of 1.5 /m s as shown in figure. What will be the average speed V of the air through the two 0.91 1.22m m# windows ?

(A) 1.125 /m s (B) 2.2 /m s3

(C) 3.50 /m s (D) 4.50 /m s

FM 4.3 Water is being pumped into a bathtub whose cross-section 3 4m m# . The bathtub has a 5 cm diameter orifice and water is discharged through orifice at a constant velocity of 5 /m s. If the water level in the tub rises at a rate of 2 /mincm, the rate at which water supplied to the pool in /m s3 , is(A) 0.0139 (B) 0.0269

(C) 0.0058 (D) 0.139

FM 4.4 Three pipes steadily delivers water at 20 Cc to a large exit pipe as shown in figure below. The velocity � �m sV�= and the exit flow rate �.�� m sv�

�=o . If increase in v�o by 20% would increase v�o by 10%, the velocities ,V V1 � and V� are

(A) 12 �, �.� �, �.�� m s m s m sV V V�1 � �= = =





(B) �� V V V� � �= = =(C) �� V V V� � �= = =(D) �� V V V� � �= = =

FM 4.5 Water at 20 Cc flows through the piping junction, enters at section 1 with flow rate of 1.26 10 /m s3 3

#− as shown in figure below. A portion of the flow is diverted

through the shower-head, which contains 50 holes of 1 mm diameter at section 3. The average velocity at section 2 is 2.5 m/s. If flow through the shower is uniform , the exit velocity from the shower head jet is

(A) 121 /m s (B) 605 /m s

(C) 12.1 /m s (D) 60.5 /m s

FM 4.6 An oil having a specific gravity of 0.85 is pumped with a water jet pump as shown in figure. The water and oil mixture has an average specific gravity of .0 90 and water flow rate is 0.5 /m s3 . What will be the flow rate in /m s3 at which the pump moves oil ?

(A) 1.5 (B) 0.5

(C) 2 (D) 1

FM 4.7 Water flowing through an 8 cm diameter pipe enters a porous section as shown in figure below, which allows a uniform radial velocity Vw through the wall surfaces for a distance of 1.2 m. If the entrance average velocity is 12 m/s and the exit velocity is 9 /m s, what will be the Vw in cm/s ?

(A) 0.5 (B) 5

(C) 35 (D) 0.05





FM 4.8 Air at the steady rate of 25 /minm3 is drawn into a compressor at standard atmospheric conditions. The compressor pressure ratio, �p pexit inlet is 10 to 1 and through the compressor �p nρ remains constant with .n 1 �= . If the average velocity in the compressor discharge pipe is not to exceed 25 /m s, what will be the minimum discharge pipe diameter ?(A) 6.4 m (B) 0.0064 m

(C) 0.06 m4 (D) 0.64 m

FM 4.9 Consider the river flowing towards a sea at mean velocity of 3 /m s with a rate of 250 /m s3 at a location 90 m above the sea surface. The power generation potential of the entire river at that location is

(A) 230 MW (B) 225 MW

(C) 220 MW (D) 221.75 MW

Common Data For Linked Answer Q. 10 and 11A pump motor unit is used to supply water to a storage tank from a lake at a elevation of 20 m. The pump can supply water at a rate of 0.070 /m s3 with a expanse of 20.4 kW electric power. Any frictional losses in pipes and any changes in kinetic energy is neglected.

FM 4.10 The overall efficiency of the pump-motor unit is(A) %67 (B) %50

(C) %65 (D) %70

FM 4.11 The pressure difference between the inlet and the exit of the pump is(A) 201 kPa (B) 192 kPa

(C) 199 kPa (D) 196 kPa

Common Data For Linked Answer Q. 12 and 13.Air at standard atmospheric condition enters the compressor at a rate of 0.28 /m s3

. It comes out to the tank through a 3 cm diameter pipe with a density of 1.8 /kg m3 and a uniform speed of 214 /m s. (Take 1.�� kg mair

�ρ = )

FM 4.12 What is the mass flow rate in /kg s at which the mass of air in the tank is increasing ?(A) 0.65 (B) 0.073

(C) 0.065 (D) 0.73





FM 4.13 What is the average time rate of change of air density in /kg m s3 within the tank ? (A) 1.56 (B) 1.30

(C) 0.0130 (D) 0.130

FM 4.14 The pipe flow in shown figure below, fills a cylindrical tank. At time t 0= , the water depth in the tank is 65 cm. What will be the time required to fill the remainder of the tank ?

(A) 41.5 s (B) 23 s

(C) 32 s (D) 16 s

FM 4.15 Water is being pumped from a lake to a reservoir of 15 m height by a 7 hp (shaft power) pump. If the mechanical efficiency of the pump is %82 , what will be the maximum volume flow rate of water ?

(A) 35.5 /L s (B) 29 /L s

(C) 3.9 /L s (D) 0.039 /L s

FM 4.16 Water enters the bottom of the cone as shown in figure at a uniformly increasing average velocity V ta= . If d is very small and h 0= at t 0= , the expression for the water surface rise is





(A) ( ) coth t t d��

��

��a q= : D (B) ( ) coth t t d�

� � ��

��

a q= : D

(C) ( ) coth t t d��

��

��a q= : D (D) ( ) coth t t d�

� � ��

��

a q= : D

FM 4.17 The velocity distribution is uniform at the entrance of a 0.9 m wide channel with a velocity V as shown in figure. Further at the downstream the velocity profile is given by u y y� ��= − , where u is in /m s and y is in meter. What will be the value of V in /m s ?

(A) 0.70 (B) 0.35

(C) 0.525 (D) 0.065

FM 4.18 A fluid flowing past incompressibly over a flat plate as shown in figure below, with a uniform inlet profile u Uo= and a exit profile u U 2

3o

3

,η η-

< F, where Yη δ= . What will be the expression for volume flow rate vo across the top surface of the control volume ?

(A) v U bo d=o (B) v U b�5

o d=o

(C) v U b�3

o d=o (D) v U b��3

o d=o

FM 4.19 A syringe plunger is moved forward at the steady rate of 10 /mm s and the vaccine leaks pass the plunger at .0 1 of the volume flow rate, out the needle opening. The inside diameters of the syringe and the needle are 15 mm and 0.525 mm, respectively. What will be the average velocity of the needle exit flow ?

(A) 5.56 /m s (B) 7.42 /m s

(C) 13.42 /m s (D) 6.8 /m s

FM 4.20 Water flows through a horizontal pipe at a rate of 35 /L s. The pipe diameter is reduced from 15 cm to 8 cm by a reducer. The pressure difference at the centre line, between the two sections of pipe is measured to be 30 .kPa If the kinetic energy correction factors to be 1.05, the irreversible head loss in the reducer is(A) 5.44 m (B) 6.75 m

(C) 2.379 m (D) 0.675 m

FM 4.21 Consider an incompressible steady flow between two parallel plates as shown in





figure below. The uniform upstream velocity is, � �m su Uo= = , while downstream velocity profile is � �u az z zo= − , where a is a constant. If �cmzo = and the fluid is gasoline at 20 Cc , what is the value of uma� ?

(A) 3 /cm s (B) 12 /cm s

(C) 9 /cm s (D) 6 /cm s

FM 4.22 A fire hose nozzle is to deliver water that will rise 40 m vertically. What is the stagnation pressure required at the nozzle inlet if (a) no loss is there (b) a loss of 30 /N m kg− is there, respectively ?(A) 392 kPa, 422 kPa (B) 316 kPa, 294 kPa

(C) 294 kPa, 316 kPa (D) 422 kPa , 392 kPa

FM 4.23 The water level in a tank is 16 m above the ground. A hose is connected to the bottom of the tank and the Nozzle at the end of the hose is pointed straight up. The tank is at sea level and water surface is open to the atmosphere. There is a pump in the line leading from the tank to the Nozzle, which increases the pressure of water. If the water jet rises to a height of 30 m from the ground, the minimum pressure rise supplied by the pump to the water line is

(A) 139.4 kPa (B) 135 kPa

(C) 137.34 kPa (D) 140 kPa

FM 4.24 The test section wall in figure shown below contains 12064 holes of 5 mm diameter each. The suction velocity through each hole is 4�0 �minmVr = and the entrance velocity 215 �minmV0 = . For incompressible steady flow of air at 20 Cc , what will be the final velocity Vf ?

(A) 4.63 /m s (B) 4.13 /m s

(C) 35 /m s (D) 31.25 /m s





FM 4.25 The velocity distribution in an open channel flow is characterise by the relation

�� m sU y hV i��=where U = free-surface velocity, y = perpendicular distance from the channel bottom in meter and h = depth of the channel in meter.What is the average velocity of the channel stream as a function of U ?

(A) 0.833 (B) 0.0625

(C) 0.625 (D) 0.0833

FM 4.26 Consider a water jet that is deflected by a stationary cone such as shown in figure below. If the jet velocity and diameter are 30 /m s and 5 cm, respectively and the jet is deflected by 45c, what amount of force is required to hold the cone against the water stream ?

(A) 5�8 , 0�F FRx Ry=− = (B) 5�8 , 0�F FRx Ry= =(C) 0, 5�8 �F FRx Ry= =− (D) 5�8 , 5�8� �F FRx Ry=− =

FM 4.27 The vane turns water jet completely around as shown in figure below. If the water has pressure p and temperature CT c , the maximum jet velocity is

(A) VD

F2 2

2

rp= ; E (B) V

DF

2 2

��2

rp= ; E

(C) VDF�

2

��2

rp= ; E (D) V

DF2

2

��2

rp= ; E

FM 4.28 A jet of water with velocity V is directed in the positive x direction and it is deflected by a flat plate. The plate is moving towards the on coming water jet with velocity 0.5 V. If the jet cross-sectional area is A and a force F is required to maintain the plate stationary then the magnitude of force required to move the plate towards the jet, as shown in figure below, is





(A) AV �ρ (B) . AV0 25 2ρ(C) . AV2 25 2ρ (D) AV2 2ρ

FM 4.29 Air enters in a jet engine at 20 Cc and 1 atm, where ��mA1�= and �� m sV1 =

and leaves at 1 atm, where ��mA��= and 1�� m sV�= as shown in figure

below. If the air-fuel ratio is :30 1, the test stand support reaction Rx will be

(A) 281 kN (B) 356 kN

(C) 205 kN (D) 76 kN

Common Data For Q. 30 and 31The water flows steadily from a tank mounted on a cart as shown in figure. After the water jet leaves the nozzle of the tank, it falls and strikes a vane attached to another cart. Consider the cart’s wheels are frictionless and the fluid is inviscid.

FM 4.30 What is the tension in rope A ?(A) 592 N (B) 320 N

(C) 848 N (D) 490 N

FM 4.31 What is the tension in rope B ?(A) 490 N (B) N636

(C) 536 N (D) 848 N

FM 4.32 Water at 20 Cc flows steadily through a reducing pipe bend as shown in figure below. Known conditions are ��k�ap1 = gage, ��1 cmA1

�= , ��k�ap�= gage,





��cmA��= and �� g sm =o . Neglecting bend and water weight, the total force

which must be resisted by the flange bolts will be

(A) 9.8 kN (B) 14.5 kN

(C) 12 kN (D) 15 kN

FM 4.33 A 300 mm diameter circular plate is held perpendicular to an axisymmetric horizontal jet of air having a velocity of 30 /m s and a diameter of 85 mm as shown in figure. A hole at the centre of the plate is provided which results in a discharge jet of air having a velocity of 40 /m s and a diameter of 35 mm. What will be the horizontal component of force required to hold the plate stationary ?

(A) 9.90 N (B) 8.53 N

(C) . N6 95 (D) . N10 1

FM 4.34 Water flows through a reducing section of pipe as shown in figure below. All fluids are at 20 Cc . If 8 cmD1 = , 5 cmD2 = , 1 atmp2 = , 1� �m sV2 = , and the manometer reading is 58 cmh = . What will be the horizontal force resisted by each bolt when number of bolts is 4 ?( 1�28�� N mHg

�γ = )

(A) 167 N (B) 200 N

(C) 140 N (D) 40 N





FM 4.35 A water tank is drained through a hole of area Ao with fluid velocity V gh�= as shown in figure below, where h is the depth of water above the hole and the cylindrical tank have bottom area Ab . Expression for the time to drain the tank from an initial depth of ho is

(A) t AA

gh�b

o o#= (B) t A

Agh�

o

b o#=

(C) t AA gh�

o

bo#= (D) t A

A gh�b

oo#=

FM 4.36 Water is pumped from a reservoir as shown in figure. The head loss is known to be . /V g1 2 22 , where V is the average velocity in the pipe. The relationship between the pump head and the flow rate is �� h vp

�= − o , where hp is in the meters and vo is in /m s3 . What will be the flow rate vo in /m s3 ?

(A) 0.052 (B) 0.52

(C) 5.2 (D) 0.0052

FM 4.37 Water exists to the standard sea-level atmosphere through the split nozzle as shown in figure below. The weight flow rate at section 2 and 3 is equals to 748 N/s. If ��cmD�= , �0 cmD D2 3= = and �35 k�ap�= (absolute), the force on the flange bolts at section 1 is

(A) 802 N (B) 1768 N

(C) 550 N (D) 3120 N





FM 4.38 Water enters a pump impeller radially and leaves the impeller with a tangential component of absolute velocity of 10 /m s. The impeller exit diameter is 60 mm and the impeller speed is 1800 rpm. If the stagnation pressure rise across the impeller is 45 kPa, the loss of available energy across the impeller and the hydraulic efficiency of the pump respectively, are(A) 8.7 /N m kg− , .0 597 (B) 11.6 /N m kg− , .0 796

(C) 14.5 /N m kg− , .0 995 (D) 5.8 /N m kg− , .0 398

FM 4.39 The water enters a horizontal, circular cross-sectional, sudden contraction nozzle at section (1) with a uniformly distributed velocity of 7.5 /m s and a pressure of 517 kPa as shown in figure. The water exits from the nozzle into the atmosphere at section (2) with the velocity of 30.5 /m s. What will be the axial component of the anchoring force required to hold the contraction in place ?

(A) 15 N21 (B) 759 N

(C) 690 N (D) 1010 N

FM 4.40 Water is flowing through a U-section pipe as shown in figure below. At flange (1), flange (2) and Location (3) the pressures are 100 kPa, 50 kPa and 100 kPa, respectively. If the momentum flux correction factor to be 1.03, the total x and z forces at the two flanges connecting the pipe are

(A) �3.11 , 112�� F FRx Rz=− = (B) �3.11 , 112�� F FRx Rz= =−(C) 112� , �3.11� �F FRx Rz= =− (D) 112� , �3.11� �F FRx Rz=− =

FM 4.41 A nozzle is connected to a vertical pipe and discharges water into the atmosphere at a rate of 0.01 /m s3 as shown in figure. The gage pressure at the flange is 40 kPaand the nozzle has a weight of 200 N. If the volume of water in the nozzle is 0.012 m3, what will be the vertical component of the anchoring force required to hold the nozzle in place ?





(A) 1140 N (B) 1072 N

(C) 12 N81 (D) N954

FM 4.42 Water at 20 Cc flows through a 5 cm diameter pipe as shown in figure, which turns the water flow direction completely around. The pressure at flange 1 is

1�5 ��ap1 = (abs), at flange 2 is 1��ap2 = (abs) and mass flow rate is 23.45 kg/s. What will be the total force which flanges must withstand ?

(A) �53 �Fx = (B) 115��Fx =(C) 24 �Fx = (D) 1�1 �Fx =

Common Data For Linked Answer Q. 43 and 44.The water flows through a horizontal bend and discharges into the atmosphere as shown in figure. When the pressure gage reads 69 kPa, the resultant x-direction anchoring force, FAx , in the horizontal plane required to hold the bend in place is shown in figure and the flow is not frictionless.





FM 4.43 The flow rate through the bend is(A) 0.20 /m s3 (B) 0.02 /m s3

(C) 0.0 2 /m s0 3 (D) 2 /m s3

FM 4.44 What is the anchoring force FAy in y -direction, required to hold the bend in place ?(A) 3251 N (B) 3041 N

(C) 3401 N (D) 341 N

FM 4.45 A liquid jet of velocity Vj and area Aj strikes a single 180c bucket on a turbine wheel rotating at angular velocity ω as shown in figure below. What will be the expression for maximum power in terms of ρ , Aj and Vj ?

(A) P AjV��

max j�r= (B) P A V��

�max j j

�r=

(C) P A V��

max j j�r= (D) P A V��

�max j j

�r=

FM 4.46 Consider a free jet of fluid which strikes a wedge as shown in figure. A portion of the total flow is deflected by 30c and the remainder is not deflected. The horizontal and vertical components of force needed to hold the wedge stationary are FH and FV , respectively. If the effect of gravity is neglecting and the fluid speed remains constant, the force ratio �F FH V is

(A) .2 7 (B) .2 7−(C) .0 27− (D) .0 27

FM 4.47 A liquid jet of velocity Vj and diameter Dj strikes a fixed cone of 60cθ = and deflects back as a conical sheet at the same velocity. What will be the restraining force F ?

(A) �F A Vj j�r= (B) .F A V�� j j

�r=





(C) F A Vj j�r= (D) �F A V�� j j

�r=

FM 4.48 A vertical circular cross section jet of air strikes a conical deflector as shown in figure. A vertical anchoring force of 0.1 N is required to hold the deflector in the place. If the magnitude of velocity of the air remains constant and �.�� kg mair

�ρ =, what will be the mass of the deflector ?

(A) 0.108 kg (B) 1.08 kg

(C) 10.8 kg (D) 0.0108 kg

FM 4.49 The box in figure shown below has three 1.27 cm holes on the right side. The volume flows of 20 Cc water from top and bottom hole is 2��2 �cm sv vtop bottom

�= =o o and from middle is �� cm svmiddle

�=o . What will be the force, which this water flow causes on the box ?

(A) 316 N (B) 126 N

(C) 189 N (D) 379 N

FM 4.50 Water flows through the horizontal tee connection as shown in figure. The flow of water is considered frictionless, incompressible and one-dimensional. If each pipe has an inside diameter of 1 m, the x and y components of the force exerted by the tee on the water respectively, are





(A) 185 kN− , 45.8 kN− (B) 185 kN− , 45.8 kN

(C) 185 kN, 45.8 kN− (D) 185 kN, 45.8 kN

FM 4.51 The tank in figure given below has a mass of 51 kg when it is empty and contains 600 L of water at 20 Cc . Pipes 1 and 2 have area 2.�� 10 mA � 2

#= − and 0.0�� m sv �=o . What should be the scale reading W ?

(A) 6374 N (B) 8800 N

(C) 8300 N (D) 2955 N

FM 4.52 Water is flowing in a 0.1 m diameter uniform pipe as shown in figure. The pipe puts the force on the fluid in the 6 m section. What will be the axial and normal components of the force, respectively ?

(A) 19 N, N327 (B) 27 N, N218

(C) 47 N, N109 (D) 38 N, 436 N

FM 4.53 Water enters in a elbow at diameter 10 cmD1 = , 233 kPap1 = (gage) and exits to atmosphere at 3 cmD2 = as shown in figure below. At a weight flow rate of 150 N/s the force on the flange bolts at section 1 is (Neglect the weight of water and elbow)

(A) 2094 N (B) 264 N

(C) 1566 N (D) 2192 N





Common Data For Linked Answer Q. 54 and 55.Two water jets collide and form one homogeneous jet as shown in figure below. Neglect the effect of gravity.

.

FM 4.54 The speed V at the exit and direction θ of the combined jet respectively, are(A) 5.36 /m s, .21 5c (B) 2.15 /m s, .8 6c

(C) 4.29 /m s, .17 2c (D) 3.22 /m s, .12 9c

FM 4.55 What is the head loss for a fluid particle flowing from (1) to (3) and from (2) to (3) ?(A) 558 /N m s− (B) 419 /N m s−

(C) 837 /N m s− (D) 697 /N m s−

Common Data For Q. 56 and 57.Water enters the rotor at a rate of 0.005 /m s3 along the axis of rotation as shown in figure. The cross section area of each of the three nozzle exits normal to the relative velocity is 18 mm2 and 30cθ = .

FM 4.56 What will be the resisting torque required to hold the rotor stationary ?(A) 225 N m− (B) 150 N m−

(C) 200 N m− (D) 175 N m−

FM 4.57 How fast will be the rotor spin steadily if the resisting torque is reduced to zero ?





(A) 200 /rad s (B) 120 /rad s

(C) 100 /rad s (D) 1 0 /rad s6

FM 4.58 In a pipe flow of water, the distribution of axial direction velocity u is linear from zero at the wall to maximum of uc at the centerline. What will be the average velocity u and the kinetic energy coefficient α, respectively ?

(A) u�c, .5 4 (B) uc , 1.1

(C) u3c , 2.7 (D) u3 c , 8.1

FM 4.59 Consider inward flow radial turbine which involves a nozzle angle 601 cα = and an inlet rotor tip speed � �m sU1 = . The absolute velocity leaving the rotor at section (2) is radial with a magnitude of 12 /m s and the ratio of rotor inlet to outlet diameters is .1 8. If the fluid is water, what will be the energy transfer per unit mass of fluid flowing through this turbine ?

(A) 77.7 /N m kg− (B) 58.3 /N m kg−

(C) 38.8 /N m kg− (D) 97.1 /N m kg−

FM 4.60 The velocity profile in a turbulent pipe flow may be approximated with the expression

uu

Rr1 �

c

n1= −a k

where u = local velocity in the axial direction, uc = centerline velocity in the axial direction, R = pipe inner radius from pipe axis, r = local radius from pipe axis and n = constant.What will be the kinetic energy coefficient α for n �= ?(A) .111 1 (B) .11 1

(C) 0.111 (D) 1.11

FM 4.61 Water flows vertically upward in a circular cross-sectional pipe as shown in figure. The velocity profile over the cross-sectional area at section (1) is uniform and at section (2) the velocity profile is characterise by the relation:

V w Rr k1c= −a k

where V = local velocity vector, wc = centerline velocity in the axial direction, R = pipe radius and r = radius from pipe axis.What will be the expression for the fluid pressure drop between sections (1) and (2) ?





(A) p pRR gh w�

�z� � � �

�

pr r− = − − (B) p p

RR gh w�

�z� � � �

�

pr r− = + +

(C) p pRR gh w�

�z� � � �

�

pr r− = + + (D) p p

RR gh w�

�z� � � �

�

pr r− = − +

Common Data For Q. 62 and 63.A hydraulic jump forms near the downstream end of a river spillway as shown in figure. The velocity of the channel flow is reduced abruptly across the jump. Consider the conservation of mass and linear momentum principles and the width of flow is unity.

FM 4.62 What will be the expression for h�, shown in figure above ?

(A) h h hgV h� �

��

� ��

��

�= − + +b l (B) h h hV g

h� �

��

� ��

��= − + +b l

(C) �h hg

h V h� ��

�

��

�= − +c m (D) �h hg

h V h� ��

�

��

�= − −c m

FM 4.63 If energy conservation is considered, the expression for the loss of available energy across the jump will be

(A) ( )h hg h h4 2 1

1 23− (B) ( )h h

g h h��

� ��−

(C) ( )gh h h h��

� ��− (D) ( )h h

g h h4 2 12 1

3−

***********





SOLUTIONS

FM 4.1 Option (B) is correct.From mass balance

Mass of air entered tan tanFinal mass of k Initial mass of k= − ma m mf i= − v vf ir r= − ( )vf ir r= − tan tancons t for �v = . . .7 30 1 18 1 5#= −] ]g g 9.03 kg=

FM 4.2 Option (B) is correct.Consider steady incompressible flow of wind, then

vgarage dooro v vwindow window= +o o

A Vgarage door garage door �A V A V A Vwindow window window= + =where Vgarage door Normal velocity to the garage door= . sin1 5 30c=Thus . . sin2 2 3 1 5 30# # c . . V2 0 91 1 22# # #=

V .. 2.2 /m s2 22

4 95 3= =

FM 4.3 Option (A) is correct.From mass conservation to the bathtub.

The rate of increase in the amount of water

= The difference between water supply rate and water

discharge rate

mpoolo m msupply discharge= −o o

vpoolo v vsupply discharge= −o o m gvr=o o

vsupplyo v v argpool disch e= +o o ...(i)

Then vdischargeo V Adischarge orifice#=

( . )

0.00982 /m s45 0 05 2

3# #p= =

vpoolo �.�� V Alevel rise pool# # #= = 0.24 / 0.004 /minm m s3 3= =Substituting in equation (i), we get

vsupplyo . .0 004 0 00982= + 0.01382 0.0139 /m s3,=

FM 4.4 Option (A) is correct.For steady flow, we have

v v v� � �+ +o o o v�= o

or V A V A V A� � � � � �+ + V A� �= ...(i)

since v�o V A� �=





or .0 05 �� V � ��

# #p=

V� ( . ). 7.86 /m s

0 090 05 4

2#

#p

= =

and . v0 2 3o . v0 1 4= o

or v�o v2 3= o

v�o . 0.025 /m s20 05 3= =

v�o V A� �=

or V� Av

�

�=o

( )

.( . ).

d4

0 025

4 0 060 025

32 2

# #p p= =

8.84 /m s=from equation (i)

(��) � (��) (��) (��)V � � �� π π π

# # #+ + .0 05=

� � �V �� # + + .0 05=

V� .. . .

0 001250 05 0 00982 0 025= − −

12. 12 /m s14 ,=

FM 4.5 Option (C) is correct.A control volume around sections (1, 2, 3) yields

v1o v v2 3= +o o ...(i)And with 2.� �m sV2 = , flow rate

v2o . ( . )A V 2 � � ��22 22

# #p= = 0.000785 /m s3=

Thus from equation (i),

v3o v v1 2= −o o

. .1 26 10 0 785 103 3# #= −− − 0.475 10 /m s3 3

#= −

Each hole carries v50

3o . 9.5 10 /m s500 000475 6 3

#= = −

or .9 5 10 6#

− ( . ) V4 0 001 jet2

# #p=

Vjet ( . ).0 001

4 9 5 102

6

#

# #p

=−

12.10 /m s=

FM 4.6 Option (D) is correct.For steady flow process

minleto moutlet= o

m m1 2+o o m3= o

v v1 1 2 2ρ ρ+o o v3 3r= o ...(i)

Consider both the water and oil are incompressible. So, 1 2 3ρ ρ ρ= =Thus v v1 2+o o v3= o ...(ii)On combining equation (i) and (ii), we have

v v1 1 2 2ρ ρ+o o ( )v v3 1 2r= +o o

v v11

22r

r+o o ( )v v1

31 2r

r= +o o . . . .S G S GH O H O

H Oliquid

2 2

2

ρρ

ρρ= =

. .v S G v1 2 2+o o . . ( )S G v v3 1 2= +o o





�� v S G S G� � �−o �� v S G� �= −o

v�o ��

��

S G S Gv S G

S G S Gv S G� �

� �

� �

� �

� �= −− = −

−o o

. .. ( . )0 90 0 85

0 5 1 0 90= −−

.. 1 /m s0 05

0 05 3= =

FM 4.7 Option (B) is correctFor a suction velocity of Vw , the cylindrical suction surface area Aw R L2 #p= 2 0.04 1.2 0.3016 m2p# # #= =Since for steady flow v�o v vw �= +o o

or V A� � V A V Aw w � �= +

or ( . )12 4 0 08 2# #

π � ( � )V �� w�

# # #p= +

or Vw ( . ) ( )/ .4 0 08 12 9 0 30162#

p= −

.( . )

0.05 /m s4 0 30163 0 08 2

#

# #p= = 5 /cm s=


We have �� minm m sv ��

�� = =o , �� m sV�= , �.�n = , ��p

pinlet

exit =

It is steady flow process, so mass flow remains constant.

m�o m�= o

A V22 2ρ v1 1r= o

d V42 22

2#ρ π v1 1r= o

d� Vv4

2

1

2

1p rr= o

...(i)

Also pnρ tanCons t=

pp

�

� n

2

1

rr= b l

2

1

ρρ p

p �n

�

� �= b l ...(ii)

Here p� & p� denotes the inlet an exit pressures respectively.Now from equation (i) and (ii), we get

d� pp

Vv4 /n

2

1 1

2

1# #p= o

b l





4101

60 2525/ .1 1 4

# ##p= b l

.4 099 10 3#= − 0.064 m=

FM 4.9 Option (D) is correct.The total mechanical energy of the river water per unit mass becomes

emech . . . .P E K E= +

gh V�

�

= + ( . )( )

9 81 90 23 2

#= +

887.4 /J kg= 0.887 /kJ kg,

The power generation potential of the river water is

Pma� memech= o

m vr=o o 1000 250#= 250000 /kg s=Thus Pma� .250000 0 887#= 221750 kW= 221.75 MW=

FM 4.10 Option (A) is correct.The mass flow rate of water

m vr=o o 1000 0.07 70 /kg s#= =Then the Energy output of the pump

EoutΔ o .m g h ��

# ## #= =o 13.7 kW=

The overall efficiency of pump-motor unit

pump motorη - .. �.��E

E��

in

out

DD= = =

o

67.2% 67%,=

FM 4.11 Option (D) is correct.The change in pressure at inlet to exit must be equal to the useful mechanical energy supplied by the pump

EoutΔ o m p p v p� �

r D= − =o o

pΔ ..

vE

��outD= =o

o 196 kPa=


From the principle of conservation of mass

dtdmsystem m min out= −o o

v A Vin in out out outr r= −o

. . . ( . )1 23 0 28 1 8 4 0 03 2132# # # #

p= −

0.344 0.271= − 0.07 /kg s3=






We know dtdmsystem ( )dt

d Vsystemr= V dtd

systemr=

From previous part of the question

dtdmsystem 0.07 /m s3=

Hence dtdρ .

.. �.�� kg m sV

��

system

�-= =

FM 4.14 Option (B) is correct.For a control volume enclosing the tank and the portion of the pipe below the tank,

dtdv m mout inρ + -o o 0=

or ( ) ( )Ddtdh AV AV4 out in

2

ρ π ρ ρ# + - 0=

Ddtdh

42

#ρ π ( ) ( )AV AVin outr r= −

Ddtdh

42

#ρ π 998 (0.12) 2.5 998 (0.12) (1.9)4 42 2p p

# # # # # #= −

or dtdh 998 (0.12) 2.5 1.9

D442

2p

rp# # #= −] g

998 (0.12) 2.5 1.9( . )4 998 0 75

422

# #

pp# # # #= −] g

( . )

(0.12) 0.60.0154 /m s

0 75 2

2#= =] g

tΔ .( . . )

.. 22.73 s0 0154

1 0 0 650 01540 35= − = = 23 s-


Since pumpη PP

,

,

pump shaft

pump useful=

P ,pump useful P ,pump pump shafth #= 0.82 7 5.74 hp#= =Also Puseful mgz�= vgz�r= o no head loss is there

vo .. .

gzP

�� useful

� # ##

r= = 1 745.7hp W=

0.0291 / 29.1 /m s L s3= = 29 /L s-

FM 4.16 Option (B) is correct.For a control volume around the cone, the mass relation becomes

dtd dv minρ - ob l# 0=

or ( )tandtd h h d t� �

��ρ π θ ρ π α-9 C 0=

Integrate: tanh33 2ρ π θ d t8

2 2r p a#=

( )h t� tan tan

cotd t t d t d83

83

832 2 2 2

2 22 2

2

rp qrp a

qa a q#= = =

( )h t cott d83 1/3

2 2 2a q= : D






By using the control volume indicated in the figure.From the conservation of mass principle

v�o v�= o

V A� � ( )udA y y b dy� �.

A

�

�

��

�

#= = −# #

Where b width of the channel=

( . )V b��# b y y�

��

� .� �

�

��

= −; E V V�=

. V0 23 2 (0.3) (0.3)322 3

#= −

. V0 23 . . .0 18 0 018 0 162= − =

V .. 0.70 /m s0 23

0 162= =

FM 4.18 Option (C) is correct.For the given control volume and incompressible flow,

vino v vtop exit= +o o

or vtopo v vin exit= −o o V A V Ain in exit exit= −

U b dy U b dy��

o o�

�

�

h h= − −d d; E# #

U b U y y bdy��

�o o �

�

�d d d

= − −d

; E#

U b U b y dy y dy��

�o o�

�

�

�d d d

= − −d d

= G# #

��

�U b U b y y�

�

�

� ��

d d d= − −

d

c m

��

�U b U bo od d d= − −b l

vtopo U b83

o d=


The control volume selected for solving this problem is the deforming control volume. From the principle of conservation of mass,

A V v vP leak� �r r r− + +o o 0=





Here tancons tρ = , �v v��leak �=o o

��A V A V A VP� � � � �− + + 0= . A V1 1 2 2 A VP�=

V� . AA V1 1

1P

2

1= c m . dd V1 1

1P

2212

#=

. .1 11

0 52515 10 10

23

# # #= −b l 7.42 /m s=

FM 4.20 Option (C) is correct.Let b be the plate width into the paper. Let the control volume enclose the inlet and outlet. The walls are solid, so no flow through the wall. For incompressible flow vouto vin= o

or ubdzz

�

o

# U bdzoo

zo

= #

( )az z z bdzo

z

�

�

−# U bdzo

z

�

o

= # ( )u az z zo= −

or abz6

o3

U bzo o=

a zU�o

o�=

Now at �z z �o= , u umax=

a z z z az� � �o

oo o

�

= − =9 C

or umax zU z U�

� ��

o

o oo�

�

#= = azU�o

o�=

6 9 /cm s23#= =


The energy equation between sections (1) and (2) gives

gp

gV z2

11

12

1ρ α+ + gp

gV z h2 L

22

22

2r a= + + +

Since z z1 2= and .1 051 2α α= =

Thus hL ( )

gp p

gV V

21 2 1

222

r a= − + − ...(i)

Also V Av

11

= o

( . ).

4 0 150 035

2#

p= 1.98 /m s=

V Av

22

= o

( . ).

4 0 080 035

2#

p= 6.9 /m s6=

By substituting in equation (i), we get

hL . .. [( . ) ( . ) ]

1000 9 8130 1000

2 9 811 07 1 98 6 962 2

##

#= + −

. .3 058 2 382= − 0.675 m=





FM 4.22 Option (A) is correct.To evaluate the stagnation pressure at the nozzle inlet, assume that the stagnation pressure at the nozzle exit is the same as the stagnation pressure at the nozzle inlet and applying the energy equation to the flow from the nozzle exit to the maximum elevation of the water flow.

p� ( )losszg rD= + ...(i)(a) For no loss, equation (i) gives

p� (9.80)(40) 392 kPa= =(b) For loss 30 /N m kg−= , equation (i) gives

p� (9.80 40) (999 ) 422 kPa100030

# #= + =


Applying energy equation between (1) and (2)

gp

gV z h2 pump

11

12

1ρ α+ + + gp

gV z2

22

22

2r a= + +

Since V V �1 2= = and p p patm1 2= =Therefore hpump �� 1� 1�mz z2 1= − = − =A pressure rise corresponds to 1�mhpump = is

pΔ .gh 1�� 1 1�pump # #r= = 137340 / 137.34N m kPa2= =

FM 4.24 Option (B) is correct.The total suction flow leaving is vsuctiono Number of holes vhole#= o

12064 A Vhole r##=

12064 (0.005) 1.895 /m s4 604802 3p

# # #= =

Now v�o v1= o

or A V� � A V1 1=

( . )4 2 5 602152

# #π (0.8) V4

21

p# #=

V1 ( . )( . )

34.99 35 /m s60 0 82 5 215

2

2

#

# .= =

And v2o v vsuction1= −o o

A V2 2 .A V 1 ��1 1= −

or ( . ) V4 0 8 22# #

π ( . ) .4 0 8 35 1 8952# #

p= −

( . ) V4 0 8 22# #

π .15 7=





V� ( . ). 31.25 /m s

0 815 7 4

2#

#p

= =

Since v�o vf= o

or A V� � A Vf f=

( . ) .4 0 8 31 252# #

π ( . ) V4 2 2 f2

# #p=

Vf ( . )( . ) .

2 20 8 31 25

2

2#= 4.13 /m s=


For any flow cross section,

mo Au dAV nA

:r r= = #

Also V n: U hyV i

��

:= = a k

Thus for uniformly distributed density, ρ over area A,

u lh

U hy l dy

h ��

�=a k#

and Uu .h

y d hy

��

��

�

�= = =a ak k#


The mass flow rate of water jet is

mo �� (�.��) �� .� ��g sAV ��r p

# # #= = =

By momentum equation for steady flow

FRx cosmV mVq= −o o

( )cosmV �q= −o

FRy 0= (Due to symmetry about x -axis)

Then FRx 58.9 30 ( 45 1) 518cos Nc# #= − =−





FM 4.27 Option (D) is correct.For a control volume enclosing the vane and the inlet and outlet jets, applying Momentum relation

FxΣ ( ) ( )F m u m u m V m Vout out in in jet jet=− = − = − − +o o o

F− m V2 jet=− o

F 2 2AV V AV 2r r#= =

V AF

DF

2 2 42

#r r p= =

DF�

�rp=

FM 4.28 Option (C) is correct.From the momentum equation for steady one dimensional flow in x direction.

F/ mV mVout in

b b= −o o/ /

F mV AV V AVr r r r�

#r r= = =o

Since jet and plate moves in opposite direction, therefore relative velocity

Vr ( �.� ) �.�V V V= − − =Hence F (�.� ) �.��A V AV� �r r= =

FM 4.29 Option (C) is correct.The density of air

1ρ ( )RT

p��

��#

= = + 1 101325atm Pa=

1.205 /kg m3=For a control volume enclosing the engine

m�o A V1 1 1r= (1.205) (0.25) (500) 150.62 151 /kg s,# #= =and m�o m mf�= +o o

m151 301 151 30

151 151 51#= + = + = +o mm

��

f

�=o

156 /kg s=Now from momentum relation (direction of Rx and flow is same)

FxΣ R m u m u m ux f f� � � �= = − −o o o

( ) ( )156 1800 151 500 0# #= − − and

and

m m m

u ��ssuming

<<f

f

� �

=

o o o

205300 205N kN,=






For a control volume that is fixed and non-deforming, the linear momentum equation is

t

dV dAV V V nCV CS� �

:22 ρ ρ+# # FCV�= /

For steady process t

dVVCV�2

2 ρ# 0=

Thus dAV V nCS�

:ρ# FCV�= /

V V A� �ρ TA= ...(i)Where Tension in ropeT AA = .

V� 2 . . 7 /m sgh 2 9 8 2 51 # #= = =Hence from equation (i),

TA .7 1000 7 0 01# # #= 490 N=

FM 4.31 Option (D) is correctFor control volume CV�, the linear momentum equation can be written as

t

dV dAV V V nCV CS� �

:22 ρ ρ+# # FCV�= /

dAV V nCS�

:ρ# FCV�= / For steady process

V V A� � �ρ TB= ...(i)

Here V� ( ) . ( . )g h h2 2 9 8 2 5 51 2 # #= + = + 12.12 /m s=From principle of conservation of mass

v�o v�= o

V A� � V A� �= ...(ii)

Now from equation (i) and (ii), we get

V V A� � �ρ TB= TB . .12 12 1000 7 0 01# # #= 848 N-

FM 4.32 Option (D) is correct.The inlet and exit velocities are

V� �.�� m sAm

��

��

# #r= = =−o

and V� ��.� �m sAm

��

��

# #r= = =−o

The control volume surrounds the bend and cuts through the flanges.The force balance is

FxΣ F p A p A m u m ubolts gage gage� � � � � � � �# #=− + + = −o o

u V� �=− and u V� �=or Fbolts (250000 491 10 ) (20000 50 10 ) mV mV4 4

2 1# # # #= + =− −− − o o

( )m V V12275 100 2 1= + + +o

12275 100 108(21.6 2.2) 14945 15 kN,= + + + =

FM 4.33 Option (D) is correct.The control volume contains the plate and flowing air as shown in figure above. Applying the linear momentum equation in x -direction.





V V A V V A� � � � � �r r− + FAx=− FAx V V A V V A� � � � � �r r= −

V D V D� ��

��

��

��

# #r p r p= −

[ ]V D V D4 12

12

22

22p r#= −

[(40) (0.085) (30) (0.035) ]42 2 2 2p r# # #= −

1.23[11.56 1.1025]4p#= − �� air

�ρ =

0.965 10.4575#= FAx 10.09 10.1N N-=

FM 4.34 Option (D) is correct.Let the control volume cut through the bolts and through section 2. For the given manometer reading, we may compute the upstream pressure:

p p� 2− ( )hHg waterg g= − (132800 9790) (0.58) 71346 ( )Pa gage#= − =Now apply conservation of mass to find inlet velocity,

v�o v2= o

or D V4 12

1π# # D V4 2

22# #

p=

( . ) V4 0 08 21# #

π ( . )4 0 05 132# #

p=

V� ( . )( . )

5.07 5 /m s0 08

0 05 132

2# ,= =

Finally the balance of horizontal forces gives

FxΣ ( )F p A m V Vbolts gage� � � 2 �=− + = −o

or F bolts� ( )p A m V Vgage� � 2 �= − −o

( . ) ( )A V V V71346 4 0 08 21 1 2 1# # #

p r= − − m A V� �r=o

71346 (0.08) 998 ( . ) ( . )( )4 4 0 08 5 0 13 52 2#

p p# #= − −

158 N=Now force on each bolt

Fbolt 39.5 40N N4158 -= =





FM 4.35 Option (B) is correct.We know that

loss actual total pressure rise across impellerU V �� r= −q

Also U� ( . )( )( )

�.�� m sr ��

�wp= = =

N60

2ω π=

Thus loss 5.66 10 (45 10 ) 99913

# # #= − 11.6 /N m kg−=

and efficiency is given by

η

actual total pressure rise across impeller

U V ��

r=q

. 0.7965 66 10999

45 103

#

#

= =; E

FM 4.36 Option (B) is correctFor a control volume around the tank

dtd dv moutρ + o: D# 0=

A dtdh

bρ m A V A gh�out o or r=− =− =−o

Integrate: h

dhh

�

o

# AA g dt�

b

ot

�=− #

[ ]h h0

0 ��AA g t�

b

o t�#=−

h2 0− AA g t�

b

o#=−

t A gA h

��

o

b o= AA

gh�

o

b o#=


For the control volume shown, the energy equation gives

pg

V z��

�

�γ + + pg

V z h h� s l� �

�

�g= + + + −

Since p p patm� �= = and V ��=

gV z2

22

2+ z h hs l�= + − ...(i)

However hl . gV1 2 2

22

= ...(ii)

and hs h v�� p�= = − o ...(iii)





Since v V A� �=o , from equation (ii) we have

hl .g A

v21 2

2

2

= oc m ...(iv)

and combining equation (i), (iii) and (iv), we get

g Av z2

12

2

2+oc m �� .z v g A

v��

��

�

�

= + − −o oc m ...(v)

or .vgA gA��

��

��

��+ +o c m z z �� = − +

Hence vo .

g d g d

z z

2 4

1

2 4

1 2 2000

20

22 2

22 2

1 2 21

# #p p

=+ +− +

b bl l

R

T

SSSS

V

X

WWWW

( . )( . )

( . )( . )

.

2 9 81 40 07

1

2 9 81 40 07

1 2 20006 20

2 2 2 2

21

p p

=+ +

+R

T

SSSS ; ;

V

X

WWWWE E

. .3441 35 4129 62 200026 /1 2

= + +: D

vo 0.052 /m s3=


The volume flow rate

v�o �.�� specific weightweight flow rate m sv ��

��

�= = = =o

since v�o v v� �= +o o from mass conservation

v�o 0.0764 0.0764 0.1528 /m s3= + = ( )givenv v� �=o o

or v�o ( . ) .A V V� ��

�# #p= = =

V� ( . ). 7.6 /m s

0 160 1528 4

2#

#p

= =

V� ( . )

.V Av

Av

� ��

��

�

�

�

�#

p= = = =o o

V� 9.73 /m s V2= =The control volume incloses the split nozzle and cuts through the flange. The balance of forces is

FxΣ F p A,bolts gage� �=− + ( ��) ( ��) ( )cos cosv V v V v V� � � � � �c cr r r= − + − − +o o o

or Fbolts−





� �� cos cosv V v V v V p A�gage� � � � � � � �c cr r r= − + − − + −o o o

�� cos cosandV V V V� �x x� � � �c c=− =− Fbolts− 2 ( 30 )cosv V v V p A, gage2 2 1 1 1 1cr r=− − −o o andv v V V� � � �= =o o

Fbolts− . ( . ) . .cos2 998 0 0764 9 73 30 998 0 1528 7 6# # # # #c=− −

( ) ( . )135000 101350 4 0 16 2# #

p− − 1285 1159 676 3120 N=− − − =− Fbolts 3120 N=


In this problem we include the nozzle as well as the water at an instant between sections (1) & (2) in the control volume.The atmospheric pressure at section (1) & (2) are same and cancelled out.

For steady process, the linear momentum equation in the horizontal direction (

x -direction) dAV V nCS

:ρ# F= /

V V A V V A1 1 1 2 2 2r r− + p A F p AA A1 1 2= − − ...(i)From the principle of mass conservation

m1o m2= o

V A1 1ρ V A2 2r= ...(ii)From equation (i) and (ii), we have

m V m V1 1 1 2− +o o p A F p AA1 1 2 2= − − ( )m V V1 2 1−o p A F p AA1 1 2 2= − − FA ( )p A p A m V V1 1 2 2 1 2 1= − − −o

FA ( )p D p D V D V V� � �1 12

2 22

1 12

2 1# # #p p r p= − − −

517 10 (7.5 10 ) 043 2 2p

# # #= −−

1000 7.5 (7.5 10 ) ( . . )4 30 5 7 52 2#

p# # # #− −−

228 . 761.7 15 . 15N N2 88 21 18 21-= − =

FM 4.40 Option (D) is correct.We take the entire U-section as the control volume. The x and z components of the anchoring force be FRx and FRz . Then the momentum equations along the x and z axis becomes.

F p A p ARx 1 1 2 2+ + ( )m V m V2 2 1 1b b= − −o o

FRx ( )p A p A m V m V1 1 2 2 2 2 1 1b=− − − +o o ...(i)





FRz 0 0m V3 3b+ = −o

FRz m V� �b= o ...(ii)

Since V� ( ) ( . )A

m

D

m

��

��

�

��

��

# # # #

#r r p p

= = =o o

15.3 /m s=

V� ( ) ( . )A

m

D

m

��

��

�

��

��

# # # #

#r r p p

= = =o o

2.80 /m s=

and V� ( ) ( . )A

m

D

m

��

� ��

�

��

��

# # # #

#r r p p

= = =o o

11.3 /m s=Substituting in equation (i) and (ii),

FRx 100 (0.05) 50 (0.1)4 42 2

#p p# # #=− −

1.03 . .1000

22 2 801000

30 15 3# #− +: D

. . .0 1962 0 3925 0 5362=− − − 1.125 1125kN N,=− −and FRz . .m V �� # #b= =o 93.11 N=


The nozzle and the water in the nozzle are included in the control volume at an instant.From the linear momentum equation in the vertical direction (y -direction),

( )sinm V V�� c −o ��sinp A F W W p Ay nozzle water� � � � c= − − − − p� 0= (Gage pressure)

Fy ( �� )sinp A W W m V Vnozzle water� � � �c= − − − −o

...(i)From the continuity equation

v�o v�= o





V� Av

�

�= o, V A

v�

�

�= o


Fy ( ) sinp A W v v Av

Av��nozzle water� � �

�

�

�

�cg r= − − − −o o oc m

40 10 0.0 200 0.012 9810 1000 0.0153# # # #= − − −

.. . .

.0 020 01 0 707 0 04

0 01# −b l

1600 200 117.72 1. 350= − − − 12 1. 12 N8 45 81-=

FM 4.42 Option (A) is correct.The volume flow rate

vo . �.�� m sm��

��

r= = =o

Let the CV cut through the flanges and surround the pipe bend then the pipe inlet and exit velocities are the same magnitude.

V� ( . )

. �� m sV V Av

� ��

��

#

,p= = = =o

Now subtract patm everywhere, so only p� and p� are non-zero. The horizontal force balance is

FxΣ ( ) ( )F p p A p p A m u m u,x flange atm atm out� � � � � � �=− + − + − = −o o

or (�� ) (�.��)F �,x flange�p

#− + −

( ) ( . )134000 101000 4 0 05 2#

p+ − . ( ) . ( )V V23 45 23 45#= − − + �� )��apatm =

or �� (�.��)F �,x flange�p

# #− +

( . )33000 4 0 05 2# #

p+ . ( )23 45 12 12= − −

��.� ��.��F ,x flange− + + .562 8=− Fx 125.6 64.76 562.8 753 N= + + =


Considering a control volume that contains the bend and the water within the bend between section (1) & (2).Applying the linear momentum equation in x -direction,





��V v V v�� cr r− −o o ��p A F p AAx� � � � c= − + ...(i)

Substitute p ��= (Gage pressure), V Av

��

= o, V A

v�

�= o


��cosAv

Av

�

�

�

�

cr r− −

o o p A FAx� �= −

cosv A A ��

� �c

ρ ρ+o : D F p AAx � �= −

vo cos

A A

F p A��

Ax

� �

� �

cr=

+

−

c m

..

.

. .

1000 0 00930 707

0 01861

6400 69000 0 01861297855116 6#=

+− =: D

0.20 /m s3-

FM 4.44 Option (B) is correctApplying the linear momentum equation in y -direction.

FAy �� sin sinV v Av v�

�c r r= =o o o^ h ��sinA

v�

�

cr= o

From previous part of question

vo 0.20 /m s3=

Hence FAy .( . )

.0 00930 20

0 707 10002

# #= 3041 N-

FM 4.45 Option (B) is correct.Let the control volume enclose the bucket and jet and let it move to the right at bucket velocity V Rw= , so that the jet enters the control volume at relative speed ( )V Rj w− . Then

FxΣ F mu mubucket out in=− = −o o

( )m V R m V Rj jw w= − − − −o o6 7@ A

Fbucket ( ) ( ) ( )m V R A V R V R2 2j j j j#w r w w= − = − −o ( )m A V Rj jr w= −o

( )A V R2 j j2r w= −

and the power is

P ( )RF A R V R�bucket j j�w r w w= = −

Maximum power is found by differentiating this expression and equating to zero (Maxima-Minima principle).

ddP

ω 0=

�� ( ) �dd A R V Rj j

�

ω ρ ω ω- 0=

( )A dd R V R2 j j

2ρ ω ω ω-6 @ 0=

( )dd R V Rj

�

ω ω ω-6 @ 0=

( ) ( ) ( )R V R R V R R� j j�

# # #ω ω ω- - + - 0= ( ) ( )R V R R V R�j jw w w− − + −6 @ 0= R V R2 jw w− + − 0= R Vj3w− + 0= ..(i)





or Rω V3

j=Again from equation (i),

dd P

�

�

ω 3R=− (-ve, maxima)

It means, power will be maximum when RV�jω = . Substituting this in the

expression of power.

Hence Pmax AV

VV

2 3 3jj

jj

2

# #r= −b l 2 AV V3 9

4j

j j2

r # #=

A V278

j j3r=


Applying the horizontal and vertical components of the linear momentum

equation to the contents of the control volume.

t

u dV u dAV nCV CS

:22 ρ ρ+# # FxΣ=

Taking V n: is “ ve+ ” for flow out of CV and “ ve− ” for flow into the CV.

��cosV V A V V A V V A� � � � � � � � �cr r r− + + FH=− ...(i)

and sinV V A�� cr− FV= ...(ii)However V V V V� � �= = = . So equation (i) and (ii) becomes

( )cosV A A A�� cρ + - FH=−

sinV A �� cρ FV=−

and by dividing these two equations, we get

FF

V

H ��sincos

AA A A

��

� cc= + − (iii)

From conservation of mass principle

v�o v v� �= +o o

or A V� � A V A V� � � �= + A� A A� �= + ...(iv)

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

FF

V

H sin

cosA

A A A A��

��

� � � �

cc= + − −

( )

sincosA

A��

��

�

cc= −

.0 27=−

The negative sign indicates that FV is down rather than up as shown in the sketch.

FM 4.47 Option (D) is correct.Let the control volume enclose the cone, jet and the sheet. Then

F FxΣ =− ( )cosm u m u m V mVout out in in j jq= − = − −o o o o

( )cosm V Vj jq= − −o





F− � ��m V V��j jc= − −o

mV

V�j

j= − −ob l mV23

j=− o 60cθ =

F A V V23

j j j#r= m A Vj jr=o

.A V A V23 1 5j j j j

2 2r r= =


Using the stationary, non-deforming control volume shown in the figure above to find the mass of the conical deflector. Applying the linear momentum equation in the vertical direction,

( ��)cosm V V� � c− +o F WA cone=− −or Wcone ( ��)cosm V V FA� � c= − −o

m gcone ( ��)cosA V V V FA� � � � cr= − − ...(i)

However V V� �= and A D��

��p=

Thus equation (i) can be expressed as

mcone ( 30 )cosgD V V V g

F4

A12

1 1 1# cr p= − −

.( )( . )

( . ) ( )( . )

.cos1 23

4 9 810 1 30 30 30 30

9 810 1

2# # cp

#=−

−] g6 @

0.108 kg=

FM 4.49 Option (D) is correct.First we need to compute the velocities through the various holes:

Vtop ( . )

��.�� m sV

� ��

bottom�

�

#

#p= = =

−

Vmiddle ( . )

��.�� m sAv

� ��

middle

middle

�

�

#

#p= = =

−o

Then make a force balance for a control volume enclosing the box:

FxΣ F m u m u�box in in top top= =− +o o u Vin middle=− and u Vtop top=or Fbox ( )v V v V�middle middle top top#r r=− − +o o

. .998 5663 10 44 70 2 998 2832 10 22 356 6# # # # # # #= +− −

252.63 126.34= + 379 N-






We can use the x and y components of the linear momentum equation to evaluate the x and y components of the reaction force exerted by the water on the tee. For the control volume containing water in the tee gives,

Rx p A V v p D V v��

� �#r p r= + = +o o

...(i)

and Ry p D p D V v V v� ��

��

� � � �# #p p r r= − + −o o

...(ii)

The reaction forces in equation (i) and (ii) are actually exerted by the tee on the water in the control volume. The reaction of the water on the tee is equal in magnitude but opposite in direction. From conservation of mass principle,

v v� �+o o v�= o

v v v� � �= −o o o �� (�) �.�� m sv V D� ��

��

� �#

p p#= − = − =o

Also v�o (�) (�) �.�� m sV D� ��

��

� �p p# #= = =

Further V� ( )( . )

6.733 /m sDv

4 41

5 28822

22p p

= = =o

and V� ( )( )

12.73 /m sD

v

4 41

10

32

32p p

= = =o

Because the flow is incompressible and frictionless, we assume that Bernoulli’s equation is valid throughout the control volume. Thus

p� ( )p V V��

��r= + −

200( )

( ) ( . )2999

6 12 73 100012 2

# #= + −6 @ 137 kPa=

Also p� ( )p V V��

��r= + −

200 ( ) ( . )kPa 2999 6 6 733 1000

12 2#= + −b l6 @ 195.3 kPa=


Rx (200000) (1) (6 999 4.712)42p

# # #= +

185000 185N kN= =And the x -direction component of force exerted by the water on the tee is 185 kN−





.From equation (ii), we have

Ry 195300 (1) (137000) (1)4 42 2

#p p

#= −

(6.733 999 5.288) 12.73 999 10# ## #+ − 1.534 10 1.076 10 0.356 10 1.272 105 5 5 5

# # # #= − + − 458 45.8 kN00-− =−and the y -direction component of force exerted by the water on the tee is 45.8 kN+.

FM 4.51 Option (B) is correctLet the control volume surround the tank, cut through the two jets and slip just under the tank bottom. The relevant jet velocities are

V� .. ��.�� .� �m sV A

v��

��

#,= = = =−

o

The scale reads force “P” on the bottom. Then the vertical force balance is

FzΣ �� ( )�P W W m V m V m Vtank water � � � � �= − − = − = − −o o o

P m g v mVtank water water �g# #= − − = o

or P mV m g vwater water� tank # g= + +o

v V m g vwater water� tank# #r g= + +o

(998 0.0834 29.5) (51 9.81) 9790 1000600

# # # #= + +

2455 500 5874 8829 8800 N-= + + =


Using the control volume shown by broken lines and applying the axial and normal components of the linear momentum equation.

FNΣ 0= since there is no momentum flow in the normal direction.

FAΣ 0= since the flow is assumed fully developed and the net amount

of axial direction momentum flow out of the CV is zero.So cosR WN q− 0= or cosR WN q=

Now W v Al d l��

#g g g p= = =

(9.8 10 )( . )

462 N40 1 63

2#p

# #= =

and θ .sin 62 19 51 c= =− from figure

Then RN 462 19.5 436cos N# c= =For the axial direction





sinp A R W p AA� � � �q+ + − 0=

or RA sinp A p A W� � � � q= − − A A� �= ( ) sinp p A W1 2 q= − − ...(i)From the manometer readings

p� h�g= and p h� �g=Thus p p� �− ( )h h1 2g= −and from equation (i), we get

RA ( ) ( . )sinh h A W 19 51 2 cg= − −

(9.8 10 )(3.0 0.5)( . )

(462 19.5 )sin40 13

2

cp

# # #= − −

. .192 4 154 2= − 38 N-


Volume flow rate

vo 0.0153 /specific weightweight flow rate m s9790

150 3= = =

Then the velocities at (1) and (2),

V1 ( . )

. 1.�� m sAv

� �1��1��

1 2#

p= = =o

V2 ( . )

. 21.� �m sAv

� ��1��

2 2#

p= = =o

Mass flow rate mo . 1�.2� �weight flow rate �g sg ��11��= = =

Then the balance of forces in the x-direction gives: FxΣ F p A mu mubolts 1 1 2 1=− + = −o o

Fbolts ( )p A m u u1 1 2 1= − −o ��cosV Vx2 2 c=− cosp A m V V��1 1 2 1c= − − −o^ h

( . ) . ( . . )cos233000 4 0 1 15 29 21 7 45 1 952# # # cp= − − −

18 . 264.29 05 43= + 2094 N-

FM 4.54 Option (C) is correct.For the water flowing through the control volume shown above, the x and y direction components of the linear momentum equation are (Take “ ve+ ” for flow out of CV and “ ve− ” for flow into the CV.)

cosV V A V V A2 2 2 � � �#r q r− + 0= ...(i)

and sinV V A V V A1 1 1 � � �#r q r− + 0= ...(ii)





From the conservation of mass principle, we get

V A V A V A� � � � � �r r r− − + 0= ...(iii)Dividing equation (ii) by (i), we obtain

tanθ ( )

( . )

( )( . )

�.��V AV A

V d

V d

�

�

� ��

� ��

��

�

��

�

��

�

��

�

��

��

#

#

#

#

p

p

p

p= = = =

or θ (0.3086) 17.2tan 1 c= =−

Substitute the values of A V� �ρ from equation (iii) into equation (i), we get

( )cosV A V V A V A��

� � � � � �#r q r r− + + 0=

or V� ( )cos V A V AV A1 1 2 2

22

2

q= + ( )cos V d V d

V d1 1

22 2

222

22

q=

+ A d�

�p=

( . ) [( ) ( . ) ( ) ( . ) ]

( ) ( . )cos17 2 4 0 1 6 0 12

6 0 122 2

2 2

# # #

#

c=

+ 4.29 /m s=

FM 4.55 Option (A) is correct.To evaluate the loss of available energy associated with the flow through this control volume, applying the energy equation,

u V m u V m u V m� � ��

� ��

� ��

�− + − + + +o o ob b bl l l 0= ...(i)

Also the conservation of mass equation gives

m m m� � �− − +o o o 0= ...(ii)Substitute m�o from equation (ii) into (i), we obtain

( )u V m u V m u V m m� � ��

� ��

� ��

� �− + − + + + +o o o ob b bl l l 0=

( ) ( )m u u m u u� � � � � �− + −o o m V V m V V� ��

��

��

��

��

= − + −o ob bl l ...(iii)

The left hand side of equation (iii) represents the rate of available energy loss in this fluid flow. Thus rate of available energy loss is

rate of loss V A V V V A V V� ��

��

��

� ��

��

# #r r= − + −b bl l

or ( ) ( )

d VV V

d VV V

4 2 212

112

32

22

222

32

# #r p= − + −; E

( . )

( . )( ) ( . )

4999 3 14

0 10 4 24 4 292

2 2#

# #= −; E)

(0.12) 6( ) ( . )

26 4 292

2 2

# #+ −; E

784.215 . .0 0481 0 760#= − +6 @

Thus rate of loss 558 /N m s- −






To find the torque required to hold the rotor stationary, we use the moment of momentum torque equation

Tshaft ( )( ) ( )m r V m r Vin in in out out out! != − +q qo o

Tshaft cosmr Vout out q= o ...(i)

We note that mo vr= o ...(ii)

and Vout Av

3 nozzle exit= o

...(iii)

Combining equation (i),(ii) and (iii), we get

Tshaft cos

Av r3 nozzle exit

out2r q=o

...(iv)

For 30cθ = , using equation (iv) to get

Tshaft ( )( . ) . cos3 18 10

999 0 005 0 5 306

2

# #

# # # c= −

200 N m−=

FM 4.57 Option (D) is correct.To determine the rotor angular velocity associated with zero shaft torque, we again use the moment of momentum torque equation to obtain, this time with rotation,

Tshaft ( )cosmr W Uout out outq= −o ...(i)

We note that Uout routw= ...(ii)

and Wout Av

3 nozzle exit= o

...(iii)

Combining equation (i), (ii) and (iii), we get

Tshaft cosvr Av r�out

nozzle exitoutr q w= −o o

c m ...(iv)

From equation (iv), we obtain for T �shaft =

ω .

. 160 /cos rad s3 18 10 0 5

0 005 306

# # #

# c= =−

FM 4.58 Option (C) is correct.For this flow, the velocity distribution is linear and thus





u u Rr�c= −a k

For the average velocity u , we can written as

u �R

u r dru R

rRr d R

r�

�

R

c��

�

�#

p

p= = −a a ak k k

##

u Rr

Rr2 2

131

c2 3

0

1= −a ak k: D u2 2

131

c= −b l

or u u3c= ...(i)

For the kinetic energy coefficient α, we can written as

α R u u

u u r dr

u

u Rr d R

r

�

� � �R

� �

�

��

�

�

�

# #

# #

r p

r p= =

b

a a

l

k k# #

or from the equation (i), put the value of u ,

α uu

Rr d R

r54c

3

0

1

#= a a ak k k#

54 Rr

Rr d R

r1 3

0

1= −a a ak k k#

54 Rr

Rr

Rr

Rr d R

r1 3 33 2

0

1= − − +a a ak k k: D#

54 Rr

Rr

Rr

Rr d R

r3 34 2 3

0

1= − − +a a a ak k k k: D#

Rr

Rr

Rr

Rr54 2

151

33

432 5 3 4

0

1= − − +a a a ak k k k: D

54 21

51 1 4

3= − − +: D 2.72054= =

FM 4.59 Option (A) is correct.Using the moment of momentum energy equation to evaluate the energy transfer per unit mass.

wshaft net out U V ,� �= q ...(i)

The value of V ,�θ can be ascertained with the help of the section (1). Velocity

triangle sketched below.

From the velocity triangle we note that

V ,1θ tanV ��,R 1 c= ...(ii)By principle of conservation of mass, we obtain

V ,R 1 (1�) . �.� �m sV AA V r

r1 �1

, ,R R�1

��

1

�= = = =b l

Thus with equation (ii),

V ,1θ 6.7 60 11.6 /tan m s# c= =and with equation (i), we get

wshaft net out (6.7)(11.6) 77.7 /N m kg−= =





FM 4.60 Option (D) is correct.For flow through surface area A of the control volume, the kinetic energy coefficient

α m V

V dAV n

�

�A�

�

#

:r=

o

#

u R u

u u r dr

u

u Rr d R

r

�

� � �R

� �

�

��

�

�

�

# #

# #

r p

r p= =

a ak k# #

u

u Rr

Rr d R

r� � �c

n

�

� �

�

�

#=

−a a ak k k# ...(i)

For the average velocity, u ,

u �R

u r dru R

r d Rr

�R

��

�

�

#

#

r p

r p= = a ak k

##

2u Rr

Rr d R

r1 /c

n1

0

1= −a a ak k k# ..(ii)

To facilitate the integrations we make the substitution

β Rr1= − ..(iii)

Thus dβ d Rr=− a k ..(iv)

When Rr �= , 1β = or when R

r �= , 0β =

And equation (ii) becomes

u 2 (1 )u dcn1

1

0b b b=− −#

u d2 cn n1 1

1

0b=− b b−

1+_ i#

2u

n n1 1 1 1 1

cn n1 1

1

0

b b=−+ + +

−21+ +

> H

u nn

nn2 1 1 2c=− − + − +a k9 C

( )( )

un nn n n n2

1 2 12

c

2 2

=− − + ++ − −

= G

u ( )( )n n

n u1 2 12

c

2

= + + ..(v)

Combining equation (i), (iii), (iv) and (v), we obtain

α

( )( )

( )

n nn u

u d

1 2 12

2 1

c

cn

2 3

33

1

0

#b b b=

+ +

− −

= G

#

( )( )

( )

n nn

d

1 2 12

2 12 3

1

0n3

b b b=

+ +

− −

= G

#

( )( )

( )( )n n

nn

n n3 3 2

22

1 2 12

2

3

= + ++ +

= ;G E ..(vi)

For n �= , equation (vi) gives





α ( ) [ ( )]

( )( )

( ) [( )( ) ]3 5 3 2 5

2 52 5

5 1 2 5 12

2

3

= + ++ +

) )3 3 .1 11=


For the control volume shown in the figure, the z -component of the momentum equation is

w dAV nCS

:ρ# FzΣ=

or w w A w w rdr�r

R

� � � � ��

#r r p− +=

# p A p A R Wz W� � � �= − − − where dA r dr�p=With A A A� �/= , this equation becomes

p p� �− �AR

AW w A w R

r r dr�z Wc

R

��

�r pr= + − + −a k9 C#

...(i)

But with x Rr/

�w Rr r drc

R� �

�−a k# w R

rRr r dr� �c

r

R�

�

�

�= − +

=c m#

( )w R x x x dx�cx

x� � � �

�

�= − +

=

=

#

w R x x x R w� ��

��

��

c c� �

��

�

�� = − + =: D

Thus A R�p= , equation (i) becomes

p p� �− RR

RW w

RR w�

��z W

c� � ��

��

#p pr

ppr= + − + b l ...(ii)

We can determine wc in terms of w� by using the continuity equation

dAV nCS

:ρ# 0= or since tancons tρ =

A w� � w dA w Rr r dr� �c

r

R

��

# p= = −=

9 C# #

2 ( )w R x x dxcx

2 2

0

1p= −

=#

or R w��π w R2 6

1c

2p= b l





Thus �w wc �= and equation (ii) becomes

p p� �− (� )RR

RW w w�

�z W� � �

��

�#p p

r r= + − + b l

RR

RW w�

�z W� � �

�

p pr= + +

or with W Ah g R hW�g r p= =

p p� �− RR gh w�

�z� �

�

pr r= + +


For unit width of flow, application of the horizontal component of the linear momentum equation to the water in the control volume from section (1) to section (2) gives,

R h h2 2x

12

22g g− + − V h V V h V1 1 1 2 2 2r r=− + ...(i)

where h2

2γ is the hydrostatic force, when width equal to unity.Since the jump occurs over a short distance, we neglect Rx from equation (i). Also from conservation of mass principle

V2 V hh

12

1= ...(ii)

Combining equation (i) and (ii), we get

h hh

2 112

1

22γ - b l; E V h V

Vhh 12

12

11

22

1r= −b l; E

hh1

1

22

− b l; E ghV h

hh

hh2 1

12

12

1

2

12

1

2

rr= −b l; E gγ ρ=

hh1

1

22

− b l ghV

hh

2 1 11

12

1

2= −

b l> H

or hh

hh1 1

1

2

1

2− +b bl l ghV

hh

hh

21

1

12

1

2

1

2

#=−

b

b

l

l

hh

hh

ghV2

1

22

1

2

1

12

+ −b bl l 0= ...(iii)

From equation (iii), we obtain

hh

1

2 1 gh

V

2

1 81

12

!

=− +

or h2 h h

gV h2 2

21 12

12

1=− + +b l

The other quadratic root is not meaningful.

FM 4.63 Option (D) is correct.Application of the energy equation to the flow from point A to point B shown in





the figure above gives

p V gz�B B

B

�

ρ + + �ossp V gz w�A A

A shaft net in

�

r= + + + −

Here �� gage �ressurep pA B= = , w �shaft net in =

Thus jump loss ( )V V g z z2A B

A B

2 2

= − + −

( )V V g h h212

22

1 2= − + − ...(iv)

Combining equation (ii), (iii) and (iv), we obtain (from previous part of question)

Jump loss ( )V V h

h

g h h2

12

12

2

12

1 2=−

+ −b l

( )Vhh g h h2 11

2

2

12

1 2= − + −b l; E

( )ghhh

hh

hh g h h4 11

1

22

1

2

2

12

1 2= + − + −b bl l; ;E E from eq. (iii)

( )ghhh

hh

hh

hh

hh

hh g h h4

11

1

22

1

22

2

12

1

2

1

2

2

2

2# #= − + − + −b b b bl l l l; E

( )ghhh

hh

hh g h h4 12 11

1

2

1

2

21 2= − + − + −b l; E

( )ghh h

h h h h h h g h h41

12

2

23

12

2 1 22

13

1 2= − + − + −; E

( )h hg h h h h h h h h h h4 4

2123

12

2 1 22

13

1 2 1 2= − + − + −6 @

h hg h h h h h h h h h h4 4 41 2

23

12

2 1 22

13

12

2 1 22= − + − + −6 @

h hg h h h h h h4 3 31 2

23

13

12

2 1 22= − + −6 @

Jump loss ( )h hg h h4 1 2

2 13= −

***********

FM 5FLOW ANALYSIS USING DIFFERENTIAL METHOD

FM 5.1 The velocity field of a flow is given by � �� sV x t y t x t xzi j k� � �= + − + +, where ,x y and z are in meters and t is in seconds. For fluid particles on the x-axis, the speed and direction of flow respectively, are(A) 2 /m sx t2 2 , 45c (B) 2 /m sx t2 2 , 30c

(C) /m sx t2 2 , 45c (D) 2 /m sxt2 2 , 45c

FM 5.2 Consider the following steady, three-dimensional velocity field in Cartesian coordinates:

V ( , , ) ( 1.5) ( 3.0) ( 4.5)u v w axy cy dxyi j k2 3= = − + − + −where , anda c d are constants. The flow field is incompressible at(A) a c�=− (B) d a�=−(C) a c3 =− (D) a d2 =−

FM 5.3 The velocity field of a flow is given by ( ) ( )

�m sx y

yx y

xV i j��

�

� � ��

�

=+

−+

,

where x and y are in meters. The fluid speed along the y -axis and the angle between velocity vector and x -axis at points ( , ) ( , )x y 5 5= respectively, are(A) 10 /m s, 45c (B) 10 /m s, 45c−(C) 20 /m s, 45c− (D) 15 /m s, 45c

FM 5.4 Consider the following statements regarding the creeping flow :(a) Unsteady term of Navier stokes equation is ignored.

(b) Inertial term of Navier stokes equation is ignored.

(c) Pressure term of Navier stokes equation is ignored.

(d) Viscous term of Navier stokes equation is ignored.

Which of the statements given above is correct ?(A) 1, 2 and 3 (B) 1, 2 and 4

(C) 1 and 2 (D) 3 and 4

FM 5.5 The velocity field of a flow is given by ( ) ( 1�) �m sx y xyV i k3= − + − , where ,x y are in meters. What will be the location of any stagnation point in the flow field ?(A) ( , )2 1 (B) ( , )1 2

(C) ( , )2 2 (D) ( , )1 1

FM 5.6 The velocity component of a steady, two dimensional, incompressible flow field is 2 4u x xy2= − . What will be the expression for v as a component of x and y ?(A) ( )v xy y f x2 4 2= − + (B) 4 ( )v xy y f x2 2= − +(C) ( )v xy y f x4 2 2=− + + (D) ( )v xy y f x2 4 2=− + +

FM 5.7 For incompressible fluids the volumetric dilatation rate must be zero. For what combination of constants a , b , c and e can the velocity components

u ax by= +





v cx ey= + w 0=be used to describe an incompressible flow field ?(A) a e �+ = (B) �a b e+ + =(C) �b e+ = (D) �a e− =

FM 5.8 A fluid field is given by �m sx y y z yzV i j k� � �= + − , where ,x y and z are in meters. What will be the acceleration at the point ( , , )2 1 3 ?(A) 38 /m s2 (B) 40 /m s2

(C) 20 /m s2 (D) 42 /m s2

FM 5.9 The viscosity of a body lotion as a function of temperature is listed in table below. The specific gravity of the body lotion is about 1.5 and is not a strong function of temperature. The body lotion is squeezed through a small hole of diameter

�mmD = in the lid of an inverted jar. The room and the lotion are at ��CT c=. Assume that Re must be less than 0.1 for the creeping flow approximation, the maximum speed of the lotion through the hole such that the flow can be approximated as creeping flow is

Viscosity of Body lotion at 16 percent moisture content

, CT c ,μ Poise (g/cm-s)

14 600

20 190

30 65

40 20

50 10

70 3

(A) 1.09 /m s (B) 0.109 /m s

(C) 10.9 /m s (D) 0.0109 /m s

FM 5.10 Two velocity components of a steady incompressible flow field are known as:

u ax bxy= + v axz byz2= −where a and b are constants, the velocity component w as a function of ,x y and z is

(A) ( , )w az byz bz f x y33

=− − + + (B) ( , )w az byz bz f x y33

=− + + +

(C) ( , )w az byz bz f x y33

= − + + (D) ( , )w az byz bz f x y33

=− − − +

FM 5.11 Consider a certain region of steady flow with the velocity field

V ( , ) ( ) ( )u v ax b ay cxi j= = + + − +The flow field is(A) incompressible only

(B) inviscid only

(C) incompressible and inviscid

(D) neither incompressible nor inviscid





FM 5.12 The velocity in a certain two-dimensional flow field is given by the equation

V 2 2xt yti j= −where the velocity is in /m s when x , y and t are in meter and seconds, respectively. What will be the magnitude of the velocity and the acceleration respectively at the point �mx y= = at the time t �= ?(A) 8.2 /m s, 0 (B) 0, . /m s5 66

(C) 8.2 /m s, 4.2 /m s (D) 0, 4.2 /m s

FM 5.13 Consider the following steady, two dimensional, incompressible velocity field

V ( , ) (2 3) ( 2 4 )u v x y xi j2= = + + − +The pressure as a function of x and y is(A) cannot be found (B) ( �� )p xr= − −(C) ( � �)p xr= − − (D) ( � �� ) �p x x y�r m= − − − +

FM 5.14 Consider the following statements regarding a laminar boundary layer on a flat plate as shown in figure below

(a) At a given x -location, if the Reynolds number were to increase, the boundary layer thickness would also increase.

(b) As outer flow velocity increase, so does the boundary layer thickness.

(c) As the fluid viscosity increases, so does the boundary layer thickness.

(d) As the fluid density increase, so does the boundary layer thickness

Which one of these is true ?(A) only (a) (B) Only (c)

(C) Both (a) and (c) (D) Both (b) and (d)

FM 5.15 Consider a steady, incompressible, two-dimensional flow of fluid into a converging duct with straight walls as shown in figure below. The volume flow rate is vo and the velocity is in the radial direction only with ur a function of r only. If b be the width into the paper and at the inlet r R= and ( )u u Rr r= , for inviscid flow everywhere, an expression for ur as a function of ,r R and ( )u Rr is

(A) ( )u rR u Rr r= (B) ( )u R

r u Rr r=

C) ( )u R rR ur r= (D) ( )u R R rur r:=





FM 5.16 Air flows parallel to a speed limit sign along the high-way at a speed of �� m sV =. The temperature of the air is 25 Cc ( 1.849 10 / )kg m s5μ # -= - and the width w of the sign parallel to the flow direction is .45 m0 . What will be the boundary layer on the sign ?(A) laminar

(B) Transitional

(C) Turbulent

(D) Laminar for a while and then becomes transitional

FM 5.17 The water flows through the curved hose as shown in figure with an increasing speed of � �m sV t= , where t is in seconds. If sect �= , what will be the resultant acceleration and its direction, respectively ?

(A) 6.71 /m s2, .63 4c (B) 6 /m s2, 2.45 c

(C) 7.61 /m s2, .45 4c (D) 6.71 /m s2, .72 2c

FM 5.18 Air flows into the region between two parallel circular disks from a pipe as shown in figure. The fluid velocity in the gap between the disks is given by �V V R r�= , where R is the radius of the disk, r is the radial coordinate and V� is the fluid velocity at the edge of the disk. If �.� �m sV�= , the acceleration at �.�mr = will be

(A) 7.63 /m s2 (B) 8.62 /m s2

(C) 8.62 /m s2− (D) . /m s7 63 2−

FM 5.19 A hydraulic jump is a rather sudden change in depth of a liquid layer as it flows in an open channel as shown in figure. The liquid depth changes from z� to z� in a relatively short distance with a corresponding changes in velocity from � �m sV�= to � �m sV�= and �.�ml = . What will be the average acceleration of the liquid as it flows across the hydraulic jump ?





(A) 60 /m s2− (B) 120 /m s2−(C) 80 /m s2 (D) 60 /m s2

FM 5.20 A nozzle is used to accelerate the fluid from V� to V� in a linear fashion. If the flow is constant with � �m sV�= at x ��= and �� m sV�= at �mx�= , what will be the local and the convective acceleration, respectively ?(A) 128(2 1)x i+ , 0 (B) 0, (2 1)x i96 −(C) (2 1)x i96 − , 0 (D) 0, 128(2 1)x i+

FM 5.21 Oil flows past a sphere with an upstream velocity of 24 /m s as shown in figure. The speed of the fluid along the front part of the sphere is � sinV V�� 0 q= . If the radius of the sphere is 0.20 m, the streamwise and normal components of acceleration at point A in /m s2 respectively, are

(A) 8863, 7437 (B) 7350, 8950

(C) 7437, 8863 (D) 8950, 7350

FM 5.22 The fluid velocity changes from 7 /m s at point A to 19 /m s at point B along the x -axis as shown in figure. The velocity is a linear function of distance along the streamline. If the flow is steady, what will be the acceleration at point C ?

(A) 1440 /m si 2 (B) 1360 /m si 2

(C) 1560 /m si 2 (D) 1560 /m sj 2

FM 5.23 The temperature of the exhaust in an pipe is given by the relation : (� )�� ( )�cosT T ae c tbx

0 w= + +−

where 80 CT0 c= , a 5= , 0.03 mb �= − , 0.05c = , 125 /rad sω = and the exhaust speed is constant at 1.5 /m s. What will be the time rate of change of temperature of the fluid particles at 4 mx = when t 0= ?(A) 15.8 /C sc− (B) 16.8 /C sc

(C) 16.8 /C sc− (D) 17.5 /C sc

FM 5.24 A fluid flows steadily along the stream line as shown in figure. What is the angle between the acceleration and the stream line at point A ?





(A) 18.5c (B) 0c

(C) 90c (D) 37c

FM 5.25 The velocity field for a steady two-dimensional incompressible flow in the xy-plane is given by V ( , ) 0u v V i jh

y= = + . The flow is(A) Clockwise, Rotational (B) Counter clockwise, Rotational

(C) Irrotational (D) Not determined

FM 5.26 What is the expression for the vorticity of the flow field described by

V x y xyi j� �= −(A) ( )x y k2 2− + (B) ( )x y i2 2− +(C) ( )x y j2 2− (D) ( )x y k2 2+

FM 5.27 The three components of velocity in a flow field are given by

u . . x y3 0 2 0= + − v . .x y2 0 2 0= − w . xy0 5=The vorticity vector as a function of space ( , , )x y z is(A) (0.5 ) (0.5 ) (3.0)y xi j k− + (B) (0.5 ) (0.5 ) (3.0)x yi j k+ −(C) (0.5 ) (0.5 ) (3.0)x yi j k− + (D) (0.5 ) (0.5 ) (3.0)x yi j k+ +

FM 5.28 A velocity field of a flow is given by ( �) ( �)y yV i j= − + − , where V is in /m s and x and y are in meters. What will be the equation of stream line passing through the point ( , )5 3 ?(A) ( )lny x y � �= + − + (B) ( �) �lny x y= + + +(C) ( )lnx y y � �= + − − (D) ( �) �lnx y y= + − +

FM 5.29 The velocity field of a flow is given by �� ( ) �� ( ) �m sy x y x x yV i j� �� =− + + +where x and y are in meters. The equation of streamline will be(A) tancons tx y� �+ = (B) tancons tx y� �+ =(C) tancons tx y� �+ = (D) tancons tx y� �− =

FM 5.30 Consider a uniform flow in the positive x direction combined with a free vortex located at the origin of the coordinate system. If the streamline 0ψ = passes through the point x �= and y �= , what will be the equation of this streamline ?

(A) sin lnU2 4θ πΓ= (B) (4 )sin lnrU r2θ π

Γ=





(C) sin lnrU2 4rθ π

Γ= ^ h (D) sin lnU2 41θ π

Γ= _ i

FM 5.31 A two-dimensional flow field given by the relation

V x y x xy yi j2 2 12 2= + + + +^ ^h hwhere the velocity is in /m s when x and y are in meters. What is the angular rotation of a fluid element located at ( , ) (0.5 ,1.0 )m mx y = ?(A) 0.56 /rad sj (B) 0. /rad sj75

(C) 0.75 /rad sk (D) 0. 5 /rad si7

FM 5.32 The three components of velocity in a flow field are given by

u x y z� � �= + + v xy yz z�= + + w /xz z3 2 42=− − +The volumetric dilatation rate and an expression for the rotation vector respectively, are

(A) 2y z z yi j k2

52− + + −a k , 0 (B) 0, y z z yi j k2 2

52− + + +a k

(C) 0, z y z yi j k25

2 2− + −a k (D) 0, y z z yi j k2 25

2− + + −a k

FM 5.33 For a certain two-dimensional steady flow field it is suggested that the x -component of velocity is given by

u ( )a b x c �= + −where ,a b and c are constants with appropriate dimensions. What will be the expression for v as a function of ,x y and the constants of the given equation such that the flow is incompressible ?(A) �( ) ( )v b x c y f x=− − + (B) �( ) ( )v b x c y f x= − +(C) ( ) ( )v b x c y f x�=− − − (D) ( ) ( )v b x c y f x�= − −

FM 5.34 The radial velocity component in an incompressible, two-dimensional flow field is

vr 2 3 sinr r2 q= + , v 0z =What will be the corresponding tangential velocity component vθ , required to satisfy conservation of mass ?(A) ( )cosv r r f r� ��θ θ= - +θ (B) � � ( )cosv r r f r�θ θ=- - +θ

(C) ( )cosv r r f r� ��θ θ= - +θ (D) � � ( )cosv r r f r�θ θ=- + +θ

FM 5.35 The streamlines in a two dimensional flow field are all concentric circles as shown in figure below. The velocity is given by the equation v rω=θ where ω is the angular velocity of the rotating mass of fluid. What will be the circulation around the path ABCD ?





(A) ( )a b2 21ω θ- (B) ( )b a2 2ω θΔ -

(C) ( )b a2 22ω θ- (D) ( )a b2 2ω θΔ -

FM 5.36 A steady, two-dimensional flow field in the xy -plane has the following stream function

ψ x xy y2 5 32 2= + +The flow field is(A) continuous (B) compressible

(C) incompressible (D) not determined

FM 5.37 The stream function corresponding to the velocity potential �x xy� �φ = - which passes through the origin is(A) x y3 2 3ψ = - (B) 3x y y2 3ψ = -(C) 3x y y2 3ψ = + (D) 3y x y2 3ψ = -

Common Data For Linked Answer Q. 38 and 39Consider a steady two dimensional, incompressible flow field called a uniform stream as shown in figure below. The fluid speed is V everywhere. The Cartesian velocity components are

u V= and v �=

FM 5.38 The expression for the stream function for this flow is(A) Vy Cψ = + (B) Vx Cψ = +(C) Vy Cψ = - (D) Vx Cψ = -

FM 5.39 The 2ψ is a horizontal line at �.�my = and the value of ψ along the x -axis is zero. If �� m sV = , the volume flow rate per unit width between these two streamlines is(A) 5 /m s3 (B) 10 /m s3

(C) 10 /m s2 (D) 5 /m s2

FM 5.40 A uniform stream of speed . �m sV ��= is inclined at angle α from the x -axis as shown in figure below. The cartesian velocity components are

u . cos2 5 a= v . sin2 5 a=If the flow is steady, two dimensional and incompressible, the stream function for this flow is





(A) . ( )sin cosy x C2 5ψ α α= + + (B) 2.5( )cos siny x Cψ α α= + +(C) 2.5( )cos siny x Cψ α α= - + (D) 2.5( )sin cosy x Cψ α α= - +

FM 5.41 The velocity distribution for two dimensional flow of a viscous fluid between wide parallel plates is parabolic and it is given by

u U hy�c

�= − a k: D

The corresponding stream function and velocity potential respectively, are

(A) � ��U y h

y C�

cψ = - +a k: D , Not possible

(B) U y x hy C�

�c

�ψ = - +a k: D , �y xy� �φ = -

(C) Not possible, U y hy C� �

�c

�φ = - +a k: D

(D) y xy�� ψ = - , U y x hy C�

�c

�φ = - +a k: D

FM 5.42 Consider fully developed Couette flow as shown in figure, with top plate moving and bottom plate stationary. The velocity field for above steady, incompressible and two dimensional flow in the xy -plane is given by

V ( , ) 0u v Vhy i j= = +a k

If stream function 0ψ = along the bottom wall of the channel, the value of stream function along the top wall is

(A) zero (B) Vh2ψ =-

(C) Vh2ψ = (D) h

V2ψ =

FM 5.43 The stream function for an incompressible flow field is described by

ψ x y y3 2 3= −where the stream function has the units of /m s2 with x and y in meters.The rate of flow across the straight path AB as shown in figure below is





(A) 1.5 /m s3 (B) /m s2 3

(C) 1 /m s3 (D) .5 /m s0 3

FM 5.44 The velocity components for an incompressible, plane flow are

vr cosAr Br� � q= +− −

vθ sinBr � q= −

where A and B are constants. What is the corresponding stream function ?(A) sinAr B C�ψ θ θ= + +- (B) sinA Br C�ψ θ θ= + +-

(C) sinA Br Cψ θ θ= + + (D) sinAr Br C� �ψ θ= + +- -

Common Data For Linked Answer Q. 45 and 46Consider a steady, two-dimensional, incompressible, irrotational velocity field specified by its velocity potential function as :

φ . ( )x y x y4 5 2 42 2= − + −

FM 5.45 The velocity components u and v respectively, are(A) 9 2, 9 4x y+ − (B) 9 4, 9 2y x− − +(C) 9 4, 9 2y x− + (D) 9 2, 9 4x y+ − −

FM 5.46 The stream function in this region is(A) 9 2 4xy y x Cψ =- - - + (B) 9 2 4xy y x Cψ = + + +(C) 9 2 4xy y x Cψ =- + + + (D) 9 2 4xy y x Cψ = - - +

FM 5.47 Consider the following steady, two-dimensional, incompressible velocity field

V ( , ) (2 1) ( 2 3)u v x yi j= = + + − +What will be the velocity potential function and the flow type ?(A) Rotational, ( )x x f y�φ =- + + (B) Rotational, ( )x x f y�φ = + +(C) Irrotational, ( )x x f y�φ = + + (D) Irrotational, ( )x x f y�φ =- + +

FM 5.48 A reverse flow region may develop by water in the region just downstream of a sluice gate as shown in figure. The velocity profile is to be consist of two uniform regions, one with velocity 3 /m s and the other with velocity 0.9 /m s. If the channel is 6.1 m wide, what will be the net flow rate of water across the portion of the control surface at section (2) ?





(A) 4.50 /m s3 (B) .50 /m s7 3

(C) . /m s3 75 3 (D) . /m s2 95 3

FM 5.49 A fluid having a viscosity of 0.003 /N s m2− flowing with an average velocity of

100 /mm s in a 2 mm diameter tube. What will be the magnitude of the wall shearing stress ( rzτ ) ?(A) 1.5 Pa (B) 1. Pa20

(C) . Pa0 8 (D) 1. Pa0

Common Data For Q. 50 and 51A layer of SAE 30 oil flows down a vertical plate as shown in figure with a velocity profile of

V hV hx x j�

��0

�

= −b l

where V0 and h are constants. The width of the plate is b .

FM 5.50 The shear stress at the edge of the layer ( )x h= is

(A) V h2 0μ (B) hV� 0μ

(C) 0 (D) None of these

FM 5.51 The flow rate across the surface AB is

(A) V bh23

0 (B) . V h1 5 0

(C) V bh32

0 (D) h V b32

0





Common Data For Q. 52 and 53SAE 30 oil at 15.6 Cc ( 0.38 /N s m2μ -= ) flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the channel is 30 /kPa m and the distance between the plates is 4 mm. Consider the flow is laminar.

FM 5.52 The volume rate (per meter of width) of the flow is(A) 42.1 10 /m s4 2

#− (B) 4.21 10 /m s4 2

#−

(C) 4.21 10 /m s2 3#

− (D) 4.21 10 /m s4 3#

−

FM 5.53 The magnitude of the shearing stress acting on the bottom plate is(A) 50 /N m2 (B) 5 /N m7 2

(C) 0 /N m6 2 (D) 5 /N m4 2

FM 5.54 Consider the following statements regarding the D’Alembert’s paradox with the irrotational flow approximation:(a) The pressure drag force on any non-lifting body of any shape immersed in a

uniform stream is zero.

(b) The aerodynamic drag force on any lifting body on any shape immersed in a uniform stream is zero.

(c) The aerodynamic drag force on any non-lifting body of any shape immersed in a uniform stream is zero.

(d) The aerodynamic drag force on any non-lifting body of any shape immersed in a non-uniform stream is zero.

Which of the statement given above is correct.(A) only c (B) b and c

(C) a and d (D) only a

FM 5.55 The two dimensional velocity field for an incompressible, Newtonian fluid is described by the relation V xy x x y yi j12 6 18 42 3 2 3= − + −^ ^h h where the velocity has units of /m s when x and y are in meters. If pressure at point 0.�mx =, �.0 my = is 6 kPa and the fluid is glycerin at 20 Cc ( 1.50 /N s m2μ -= ), the stresses xxσ , yyσ and xyτ at this point respectively, are

(A) 6.02 kPa− , 5.98 kPa− , 45.0 kPa

(B) 45 kPa, 6.02 kPa− , 5.98 kPa

(C) 5.98 kPa, 6.02 kPa− , 45 kPa−

(D) 5.98 kPa− , 6.02 kPa− , 45 Pa

FM 5.56 The stream function for the flow of a non-viscous, incompressible fluid in the vicinity of a corner as shown in figure is given by the equation

ψ sinr2 34/4 3 q=





What will be the expression for the pressure gradient along the boundary /3 4θ π= ?

(A) rp r��

��

22 r= (B)

��

rp

r ��22 r=

(C) ��

rp r ��

22

r= (D) rp

r��

��22 r=−

FM 5.57 A wire of diameter d is stretched along the centerline of a pipe of diameter D .

For a given pressure drop per unit length of pipe and �.�Dd = , by how much the

pressure of the wire does reduce the flow rate ?(A) 31.95% (B) 42.6%

(C) 53.25% (D) 63.9%

FM 5.58 An object having the general shape of a half-body is placed in a stream of fluid and at a great distance upstream the velocity is U as shown in figure below. If the body forces are neglected and the fluid is nonviscous and incompressible, What will be the pressure difference between the stagnation point and point A in terms of U and fluid density ?

(A) . U0 879 2ρ (B) 0.7 U03 2ρ(C) 0. U351 2ρ (D) 0. U527 2ρ

FM 5.59 A viscous fluid is contained between two long concentric cylinders. Assuming that the flow between the cylinders is approximately the same as the laminar flow between two infinite parallel plates. The inner cylinder is fixed. What is the expression for the torque required to rotate the outer cylinder with an angular velocity ω in terms of the geometry of the system, viscosity of the fluid, and angular velocity ?

(A) T r l1

2rr

o4

i

o

p w=−

(B) T r rr l2o i

o3p mw= −

(C) T r rr l2

i o

o3p w= − (D) T r l

12

rr

o

o

i

p w=−





FM 5.60 A bearing is lubricated with an oil having a viscosity of 0.2 /N s m2− through

which a vertical shaft passes as shown in figure below. If the flow characteristics in the gap between the shaft and bearing are the same as those for laminar flow between infinite parallel plates with zero pressure gradient in the direction of flow. What will be the torque required to overcome viscous resistance when the shaft is turning at 80 /minrev ?

(A) 0.355 N m− (B) 0. N m266 −

(C) 0. N m710 − (D) 0. N m444 −

FM 5.61 A drop of water in a rain cloud has a diameter of 45 mμ . The air temperature is 25 Cc and its pressure is standard atmospheric pressure. How fast does the air have to move vertically so that the drop will remain suspended in air ?( �.�� g m sair

�μ # -= - )(A) 0.00596 /m s (B) 0.0596 /m s

(C) 0.596 /m s (D) 5.96 /m s

FM 5.62 A viscous, incompressible fluid flows between the two infinite, vertical, parallel plates as shown in figure. If the flow is laminar, steady and uniform, what will be the expression for the pressure gradient in the direction of flow in terms of mean velocity by using the Navier Stokes equations ?

(A) yp

hV g��2

2 m r= − (B) yp

hV g��2

2 m r= +

(C) yp

hV g��2

2 m r=− + (D) yp

hV g��2

2 m r=− −

FM 5.63 The velocity distribution for free vortex flow in a horizontal, two-dimensional bend through which an ideal fluid flows can be approximated is shown in figure. What will be the discharge (per unit width normal to plane of paper) through the channel ?





(A) q C prD= (B) q C prD=

(C) q C prD= (D) q C p

rD=

***********





SOLUTIONS


We have V 2 [4 ( 1) 2 ] /m sx t y t x t xzi j k2 2 2= + − + +For particle on the x -axis, y z �= =So V 2 2x t x ti j2 2= +Thus on the x -axis the velocity is

V ( ) ( )u v w x t x t� � �� = + + = + + 2 /m sx t2 2=

Direction of flow tanθ �uv

x tx t

��

�

�

= = =

θ 45c=

FM 5.2 Option (A) is correct.From continuity equation, condition for incompressibility

xu

yv

zw

22

22

22+ + 0= ...(i)

xu

22 ( . )

xaxy ay��

22= − =

yv

22 ( . )

ycy cy��

22= − =

zw

22 ( . )

zdxy ��

22= − =

Thus equation (i) becomes

ay cy�� + 0= ay� cy3 2=− a 3c=−Hence at ,a c�=− the given velocity field will be incompressible.


We have V ( ) ( )

/m sx y

yx y

xi j10 10/ /4 4 1 2

2

4 4 1 2

2

=+

−+

Comparing with velocity field, we get

u ( )x y

y10/4 4 1 2

2

=+

, ( )

vx y

x��

�

=−+

The resultant velocity

V u v� �= + �� m sx yy x��

� �

=++ = =

The resultant velocity does not depends on the coordinates.So, velocity remains constant in all directions.

Now tanθ uv

yx�

�

= = −

At ( , )5 5 tanθ 1=− θ 45c=−





FM 5.4 Option (C) is correct.When the creeping flow approximation is made, only pressure and viscous terms remain. The other three terms (i.e. unsteady term, Inertial term and gravity term) are very small compared to these two and can be ignored.


We have V ( ) ( 16)x y xyi k3= − + −On comparing with velocity field, we get

u x y= − , ��v xy�= − ...(i)At the stagnation point, V �= i.e., u v �= =From equation (i) x y− 0= x y=And ��xy�− 0= ��y y�

# − 0= x y=− y� 16= & y 2=Since x y= it follows that

x 2=Hence stagnation point ( , ) ( , )x y 2 2=

FM 5.6 Option (C) is correct.From continuity equation for two-dimensional incompressible flow

xu

yv

22

22+ 0=

xu

22

yv

22=− ...(i)

Since xu

22 ( ) ( )

xx xy x y� � � ��

22=− − = −

Substituting in equation (i),

yv

22 x y4 4=− +

Integrating this equation

v ( )x y y4 4 2= − +# ( )xy y f x4 24 2

=− + +

( )xy y f x4 2 2=− + +

FM 5.7 Option (A) is correct.The volumetric dilation rate for incompressible fluid is

V:Δ 0=

xu

yv

zw

22

22

22+ + 0= ...(i)

With the velocity distribution given

xu a

22 = ,

yv e

22 = ,

zw �

22 =

Thus, from equation (i), we get

a e+ 0=

FM 5.8 Option (B) is correct.For the given fluid flow field

u x y�= , xu xy�

22 = ,

yu x�

22 = ,

zu �

22 = , �

tu

22 =





v y z�= , xv �

22 = ,

yv yz�

22 = ,

zv y�

22 = , �

tv

22 =

w yz�=− , xw �

22 = ,

yw z�

22 =− , �

zw yz

22 =− , �

tw

22 =

The acceleration components ,a ax y and az are

ax tu u

xu v

yu w

zu

22

22

22

22= + + +

2 ( ) 0x y xy y z x yz0 2 2 2 2## #= + + + −

x y x y z2 3 2 2 2= +At ( , , )2 1 3 ax 2 2 1 1 3 23 2 2 2

# # # #= + 16 12 28= + =

ay tv u

xv v

yv w

zv

22

22

22

22= + + +

0 2 ( )x y y z yz yz y0 2 2 2 2# # #= + + + −

2y z y z y z3 2 3 2 3 2= − =At ( , , )2 1 3 ay 1 3 93 2

#= =

az tw u

xw v

yw w

zw

22

22

22

22= + + +

0 ( ) ( )( 2 )x y y z z yz yz0 2 2 2 2# #= + + − + − −

y z y z y z�� =− + =At ( , , )2 1 3 az 1 3 272 3

#= =So, acceleration vector becomes

a a a ai j kx y z= + + 28 9 27i j k= + +Resultant acceleration

a ( ) ( ) ( )28 9 272 2 2= + +

39.92 40 /m s1594 2-= =

FM 5.9 Option (B) is correct.Density of body lotion

lotionρ . .S G lotion waterr#= 1.5 1000 1500 /kg m3#= =

Viscosity of lotion at 30 Cc

lotionμ 65 / 6.5 /g cm s kg m s− −= =Now from Reynolds number definition

Remax V Dma�

lotion

lotion # #m

r=

.0 1 ..V

6 51500 0 004max# #=

Vma� .. . 0.109 /m s1500 0 004

6 5 0 1##= =

FM 5.10 Option (A) is correct.Continuity equation for three-dimensional, incompressible flow field is

xu

yv

zw

22

22

22+ + 0=

zw

22

xu

yv

22

22= − − ...(i)

Since zu

22 ( )

xax bxy a by

22= + = +





yv

22 � �

yaxz byz bz� �

22= − =−


zw

22 a by bz�=− − +

Integrate the above equation, we get

w ( , )az byz bz f x y��

=− − + +

FM 5.11 Option (C) is correct.First check for Incompressibility :From continuity equation

xu

yv

22

22+ 0=

xu

22 ( )

xax b a

22= + =

yv

22 ( )

yay cx a

22= − + =−

Hence xu

yv

22

22+ a a �= − =

Therefore flow is incompressibleSecond check for Inviscid Flow :For flow to be inviscid, the viscous term of Navier stokes equation be zero. Therefore Navier stokes equation, for x -momentum viscous terms

xu a

22 = , �

xu�

�

22 = and

yu �

22 = ,

yu ��

�

22 =

So xu

yu

�

�

�

�

22

22μ += G 0=

Therefore, x -momentum viscous term is zero.

Now for y -momentum viscous terms

,xv c

22 =

xv�

�

22 0= and ,

yv a

22 =− �

yv�

�

22 =

Hence xv

yv

�

�

�

�

22

22μ += G [ ]0 0 0m= + =

Since the viscous terms are identically zero in both components of Navier-stokes equation, this region of flow can indeed be considered inviscid.


From expression for velocity, the components are

u xt�= , v yt�=− and w �=

Since ax tu u

xu v

yu w

zu

22

22

22

22= + + +

ax 2 (2 2 ) ( 2 0)x xt t yt 0# #= + + − + 2 4x xt2= +

Similarly ay tv u

xv v

yv w

zv

22

22

22

22= + + +

ay 2 2 0y xt yt t2 2 0# #=− + + − − +^ ^h h 4y yt2 2=− +At �mx y= = and t �= ,

ax 2 2 4 2 0 4 /m s2# # #= + =

and ay 2 4 0 4 /m s2 2 2# # #=− + =−





Also u 2 2 00# #= = v 2 2 0 0# #=− =So that V 0=and a (4 4 ) /m si j 2= −

a 5.66 /m s4 42 2 2= + − =^ ^h h

FM 5.13 Option (A) is correct.From Navier-stokes equation, x -component of steady-two dimensional flow

udxdu vdy

duρ +; E xp

xu

yu

�

�

�

�

22

22

22m=− + += G ...(i)

Since udxdu (2 3) 2 4 6x x#= + = + , vdy

du ( 2 4 ) 0y x 02#= − + =

xu�

�

22 0= and

yu ��

�

22 =

Hence, the x -momentum is

[ ]x4 6ρ + xp

22=−

xp

22 ( )x4 6r= − −

Now y -component of Navier stokes equation for this flow

udxdv vdy

dvρ +; E yp

xv

yv

�

�

�

�

22

22

22m=− + += G

udxdv ( ) ( )x x x x2 3 8 16 242

#= + = +

vdydv ( ) ( )y x y x2 4 2 4 82 2

#= − + − = −

xv�

�

22 8= and

yv ��

�

22 =

Hence, y -momentum is

[16 24 4 8 ]x x y x2 2ρ + + - �yp

22 m= − + ] g

yp

22 [ ]x x y8 24 4 82r m= − − − +

Now we calculate cross-differentiation x y

p�

222 and

y xp�

222

y x

p�

222 �( � �)� �

y xp

yx

22

22

22 r= = − − =c m

and x y

p�

222 �( � �� ) ��

x yp

xx x y�

22

22

22 r m= = − − − +; E

[ ]x16 24r= − −

Since y x

p�

222

x yp�

222

!

Hence these are not equal, the given velocity distribution is not an exact solution of Navier-stokes equation. Thus we are unable to calculate a steady, incompressible, two-dimensional pressure field with the given velocity field.

FM 5.14 Option (B) is correct.(a) False : If the Reynolds number at a given x location were to increase, all else being equal, viscous forces would decrease in magnitude relative to inertial forces,





rendering the boundary layer thinner.(b) False : Actually as V increases, so does Re and the boundary layer thickness decrease with increasing Reynolds number.(c) True : Since μ appears in the denominator of the Reynolds Number, Re decreases as μ increases, causing the boundary layer thickness to increase.(d) False : Since ρ appears in the numerator of the Reynolds number ( )Re increases as ρ increases, causing the boundary layer thickness to decrease.

FM 5.15 Option (A) is correct.If the flow is inviscid, we could not enforce the no-slip condition at the walls. Thus at any r location, the volume flow rate must be the same.

vo Area velocity rb ur# qD #= =

ur rbvqD= o ...(i)

At radius r R= ( )u Rr Rbv

qD= o ...(ii)

Substituting equation (ii) into equation (i)

ur ( )rR u Rr=

FM 5.16 Option (D) is correct.Since the air flow is parallel to the sign, this flow is that of a flat plate boundary layer and Reynolds number for this flow

Rex VLm

r=

At 25 Cc airρ .. �.�� kg mRT

p��

��atm �

#= = =

Hence Rex .. . .1 849 10

1 184 4 0 45 1 15 1055

#

# ##= =−

For Re 1 105## , the flow is laminar (for smooth flat plate).

1 105# 3 10Re 6# # # , the flow is transitional.

Re 3 106#$ , the flow is turbulent.

Since Rex is greater than 1 105# but just barely so and less than 3 106

# . Thus the boundary layer is laminar for a while and then becomes transition by the trailing edge of fin.

FM 5.17 Option (A) is correct.We have V t�= , sect �= , �mR =The component of the acceleration along the stream line is

as tV V

sV

22

22= +

as 3 0 3 /m s2= + = tV �22 = ,

sV �22 =

And the component of acceleration normal to the stream line is

an ( )

RV t t

��

��

= = =

At sect �= an 6 /m s69 4 2#= =

Now the net acceleration

a � �a as n s ns n= + = +





where s and n are unit vectors.

The Resultant acceleration

a ( ) ( ) 6.71 /m s3 6 452 2 2= + = =Direction of acceleration,

tanθ aa

��

s

n= = =

θ (2)tan 1= − .63 4c=

FM 5.18 Option (C) is correct.The acceleration is given by

a a as ns n= +where s and n are the unit vectors. From the figure it is clearly see that stream

lines are straight i.e. R 3= . Normal acceleration a RV �n

�

= = .

and stream wise acceleration is

as VsV V

rV

22

22= =

as rV R

rV R

rV R�

��

��

#= − =−b l where V rV R�=

Substitute �.�mr = , �.� �m sV�= , �.��mR =

as ( . )( . ) ( . )

8.62 /m s0 6

1 5 0 913

2 22#=− =−

FM 5.19 Option (A) is correct.The flow is only in x -direction and acceleration is given by

a ( )t

V V V:4:22= +

a a uxui ix 2

2= =

Without knowing the actual velocity distribution, the acceleration can be approximated as

ax uxu

22=

V Vl

V V2

1 2 2 1#= + −b bl l .2

4 20 1

2 4#= + −

b bl l

( )3 20#= − 60 /m s2=−

FM 5.20 Option (D) is correct.In the form of linear fashion, the velocity is given by

u ax b= + ...(i)

where a and b are constants

At x ��= , � �m sV�= and from equation (i),

b 8=Also at �mx�= , �� m sV�= and from equation (i),

24 a � �#= + a 24 8 16= − =Now equation (i) becomes

V x16 8= +





and acceleration a � �t

V V V:4:22= +

The local acceleration

t

V22 0=

The velocity is only in x -direction (linear).

a uxua i ix 2

2= =

So, convective acceleration

uxu i

22 (16 8) (16 8)x

xx i

22

#= + +

(16 8) 16x i #= + 128(2 1)x i= +

FM 5.21 Option (C) is correct.We have �� m sV�= , �.��mR =

V 24sin sinV25

25

0 q q#= = 60 /sin m sq= ...(i)

The normal acceleration is given by

an .( )

�� sin

m sRV

��

�c= = =

And the streamwise acceleration is

as sinVsV

sV�� #2

222q= = ...(ii)

Now sV22 V

s#22

22

qq= ...(iii)

From equation (i) V22

θ cos60 q=

Also from the given figure

θ .radiusArc s

0 20= = Let Arcs =

s 0.2 q=

s22

θ .0 2= & s2

2θ .0 21=

From equation (iii) sV22 .

cos0 2

60 q=

Hence from equation (ii), we get

as 60 40 . 8863 /sin cos m s0 260 40 2

#= =

FM 5.22 Option (C) is correct.The acceleration field is given by

a ( )t

V V V:422= +

From the figure, flow is only in x -direction.i.e. ( )u u x= , v �= , w �=

a tu u

xu i

22

22= +b l

uxu i

22=

tu �

22 = ...(i)

Since u is a linear function of x . Let





u C x C� �= + ...(ii)Now applying the boundary conditions on equation (ii) to find out the values of the constants C� and C�.At x �= , � �� su = , which gives

C� 7=At �.��x = , �� su = , which gives

19 .C �1 �1#= +

C1 .0 112 120= =

Now equation (ii) becomes

u (120 7) /m sx= +

xu

22 sec120 1= −

and from equation (i) a (120 7) 120 /m sx i#= +At point C , �.��mx =Hence a (120 0.05 7) 120 1560 /m si i 2

# #= + =

FM 5.23 Option (C) is correct.We have ��CT� c= , a �= , �.��mb 1= − , .c ��= , 125 /rad sω = , 1.� �m su = , v �= , 0ω =

So dtdT ( )

tT T

tT u

xTV:4:

22

22

22= + = + ...(i)

Also T ( )� ( )�cosT ae c t1 1bx� w= + +− ..(ii)

Differentiate equation (ii) w.r.t. t ,

tT22 ( )� ( )�sinT ae c t1 bx

� w w= + −−

Again differentiate equation (i) w.r.t. x ,

xT

22 � ( )�( )cosT t abe1 bx

� w= + − −

Substitute tT22 and

xT

22 in equation (i), we have

dtdT (1 )� ( )� � ( )�( )sin cosT ae c t uT c t abe1bx bx

� �w w w= + − + + −− −

At t �= ,

dtdT ( )( ) ( )uT c abe abuT c e0 1 1bx bx

0 0= + + − =− +− −

Now substitute the values and �mx = , we get

dtdT . . ( . )e5 0 03 1 5 80 1 0 05 .0 03 4

# # #=− + #−

dtdT 18 1.05 0.887 16.8 /C sc-# #=− −


The resultant acceleration is given by

a a an sn s= +where an Normal acceleration= as Streamwise acceleration=

a RV V

sVn s

�

22= +





a ( )

3 3n s33 2

#= + 3 9 /m sn s 2= + ...(i)

Now in terms of unit vector i and j n 30 30sin cosi jc c=− + s 30 30cos sini jc c= +Substitute these values in equation (i),

a 3( 0.5 0.866 ) 9(0.866 0.5 )i j i j= − + + + 6.3 7.1i j= +

Hence tanθ .. .a

a��

x

y= = =

θ ( . ) .tan 1 127 48 51 c-= −

So, angle between the acceleration and stream line is

α 30 48.5 30 18.5c c c cq= − = − =


Velocity field V ( , ) 0u v Vhy i j= = + ....(i)

Since the flow is in xy -plane. Therefore z -component of vorticity

ζz

xv

yu

hV�

22

22= − = −c m h

V=−

Since vorticity is non-zero, this flow is rotational.The vorticity is negative, implying that particles rotate in the clockwise direction.

FM 5.26 Option (A) is correct.The vorticity is twice the rotation vector.

ζ 2 Vw #d= =From expression for velocity, u x y�= , v xy�=− and w �= and with

xω yw

zv

21

22

22= −c m, yω

zu

xw

2122

22= −b l, zω

xv

yu

2122

22= −c m

It follows that

�xω = , �yω = and y x��

z� �ω = - -^ h

Thus ζ i j k2 x y zw w w= + +^ h

y xi j k2 0 0 21 2 2= + + − −^ ^ ^h h h: D x y k� �=− +^ h

Since ζ is not zero everywhere, the flow is not irrotational.





FM 5.27 Option (C) is correct.In Certesian coordinates, the vorticity vector is

ζ yw

zv

zu

xw

xv

yui j k

22

22

22

22

22

22= − + − + −c b cm l m ....(i)

By differentiating u with respect to ,x y and z

xu

22 (�.� �.� ) �.�

xx y

22= + − =

yu

22 1,=− �

zu

22 =

Same for v xv

22 . ,2 0= �.�,

yv

22 =− �

zv

22 =

And for w xw

22 . ,y0 5= . ,

yw x��

22 =

zw �

22 =

Therefore from equation (i), Vorticity vector as a function of space ( , , )x y z is

ζ (0.5 0) (0 0.5 ) {2.0 ( 1)}x yi j k= − + − + − − (0.5 ) (0.5 ) (3.0)x yi j k= − +


We have V ( 1) ( 2)y yi j= − + − ...(i)

From equation (i), the velocity components are

u �y= − , v y �= − , w �=From the equation of stream line

udx v

dywdz= =

ydx

�− ydy

�= −

yy dy y

yy dy�

��

�−− = − − −c m dx=

On integrating, we get

yy dy y dy� �

�− − −# # dx= #

( ) ( ) ( )ln lny y y2 2 2 2− + − − − x C= + ( 2 )&let solvey t− = ( 2) ( 2)lny y− + − x C= +At ( , )5 3 ( )ln1 1+ C5= + C 1 5 4= − =−Hence equation of stream line becomes

( )lny y� �− + − x �= − x ( )lny y � �= + − +


We have V 20 ( ) 20 ( ) /m sy x y x x yi j1/2 /2 4 4 2 4 4 1 2=− + + +

where u ( )y x y20 /2 4 4 1 2=− + and v ( )x x y20 /2 4 4 1 2= +The equation of stream line is

udx v

dy=

( )y x ydx

20 /2 4 4 1 2− +

( )x x ydy

20 /2 4 4 1 2=+





y

dx�−

xdy

�=

x dx� y dy�=−Integrating both the sides, we get

x33

y C3

3

=− +

x y� �+ C=

FM 5.30 Option (C) is correct.The resulting stream function is

ψ Cuniform flow free vortexy y= + +

sin lnUr r C�q pG= − + ...(i)

At ( , ) (4, )x y 0= 0θ = r �= and 0ψ =Thus from equation (i),

0 ln C0 2 4pG= − +

and C ln2 4pG=

So that ψ sin ln lnUr r� �q pG= − −^ h

or ψ sin lnUr r� �q pG= −

The equation for the streamline( 0ψ = ) is therefore

0 sin lnUr r� �q pG= −

or sinθ lnrUr

2 4pG=

FM 5.31 Option (C) is correct.For velocity distribution given,

u 2x y x2= + , v xy y� ��= + + , w 0=

Since zω xv

yu

2122

22= −c m

xv

22 y2 2= and

yu x��

22 =

So that zω y x y x21 2 22 2 2 2= − = −^ h

At 0.�mx = and �.0 my = zω . . . /rad s1 0 0 5 0 752 2= − =^ ^h h

Thus ω 0.75 /rad sk=(Since for a two-dimensional flow field 0x yω ω= = )


Volumetric dilatation rate xu

yv

zw

22

22

22= + +

Thus, for velocity components given

Volumetric dilatation rate x x z x z2 3 0= + + + − − =^ ^h h

This result indicates that there is no change in the volume of a fluid element as





it moves from one location to another.The rotation vectors about the x , y and z axis are given by

xω yw

zv y z y z2

121 0 2 22

222= − = − + =− +c ^ am h k6 @

yω zu

xw z z z

21

21 2 3 2

522

22= − = − − =b ^l h6 @

And zω xv

yu y y y

21

21 2 22

222= − = − =−c ^m h

Thus, ω i j kx y zw w w= + +

ω y z z yi j k2 25

2=− + + −a k

FM 5.33 Option (A) is correct.The x -component of velocity is

u ( )a b x c �= + − ...(i)For incompressible flow, Volumetric strain rate

V DtDV�

xu

yv

zw �

22

22

22= + + =

For two dimensional flow field zw

22 0=

Then xu

yv

22

22+ 0=

yv

&22

xu

22=− ...(ii)

By substituting equation (i) into equation (ii), we get

yv

22 � ( ) � ( )

xa b x c b x c��

22=− + − =− −

Integrate to solve for v v yv dy

22= #

[ ( )] ( )b x c dy f x2= − − +#An arbitrary function of x is added rather than a simple constant of integration. Since this is a partial integration with respect to ,y v is a function of both x and y .

Therefore v ( ) ( )b x c y f x2=− − +

FM 5.34 Option (B) is correct.In cylindrical polar coordinates, the continuity equation for incompressible fluid is

r rrv

rv

zv� �r z

2

2

22

22

q+ +q^ h 0=

Since v �z = , v22

θθ

rrvr

2

2=− ^ h

...(i)

With rvr sinr r2 32 3 q= + , r

rvr

2

2̂ hsinr r4 9 2 q= +

Thus, equation (i) becomes

v22

θθ sinr r4 9 2 q=− +^ h ...(ii)

Equation (ii) can be integrated with respect to θ to obtain

dvθ# sinr r d f r4 9 2 q q=− + +^ ^h h#or vθ 4 cosr r f r9 2q q=− − + ^ h

Where f r^ h is an undetermined function of r .






Circulation Γ dV sABCD

:= #n

v bd v dr v ad v drAB

rBC CD

rDA

q q= + + +q q# # # # ...(i)

Since v �r = and v rω=θ , equation (i) becomes

Γ b d a d� ��

�

�

�

�

w q w q= + + +q

q

q

q

##

b a��

�� w q q w q q= − + −^ ^h h

or Γ b a b a2 12 2 2 2Tw q q w q= − − = −^ ^ ^h h h


We have ψ x xy y2 5 32 2= + +From the definition of stream function, the velocity components are

u ( )y y

x xy y� � ��

22

22y= = + +

u x y5 6= +

and v (� � � )x x

x xy y� �

22

22y=− =− + +

x y4 5=− −For two dimensional flow, the continuity equation is

xu

yv

22

22+ 0=

xu

22 (� �) �

xx y

22= + =

yv

22 ( � �) �

yx y

22= − − =−

Hence xu

yv

22

22+ 5 5 0= − =

Therefore the given flow field is incompressible.


We have u y x

x y� ��

22

22y f= = = −

Integrate with respect to y to obtain

dψ# x y dy3 32 2= −^ h#

or ψ 3 x y y f x32

3

1= − +c ^m h ...(i)





Similarly v x y

xy�22

22y f=− = =−

and integrate with respect to x to obtain

dψ# xydx6= #or ψ x y f y3 2

2= + ^ h ...(ii)

To satisfy both equation (i) and (ii), we get

ψ x y y C3 2 3= − +Where C is an arbitrary constant. Since the streamline 0ψ = passes through

the origin (x �= , y �= ), it follows that C �= and

ψ x y y3 2 3= −

FM 5.38 Option (A) is correct.From the definition of stream function.

u y

V22y= = ...(i)

Integrating with respect to y ,

ψ ( )Vy g x= + ...(ii)

And v � ( )�x x

Vy g x22

22y= − =− + �( )�

xg x

22=−

Here ,v �= for given velocity field.

Thus �( )�x

g x22 0=

Integrate with respect to x , we get

( )g x C=Hence equation (ii) becomes

ψ Vy C= +

FM 5.39 Option (D) is correct.From answer of previous part

ψ Vy C= +At �,y = 0ψ = gives (along x -axis)

C 0= ψ Vy=Then at y �= 0ψ Vy �= =Now at �.�y = 2ψ .10 0 5#= 5 /m s2=

Hence Volume flow rate per unit width between 2ψ and 0ψ is

wvo 5 02 0y y= − = − 5 /m s2=

FM 5.40 Option (C) is correct.From the definition of stream function,

u . cosy

��22y a= =

By integrating with respect to y

ψ . ( )cosy g x2 5 a= + ...(i)Now differentiate with respect to x





x2

2ψ ��x

g x22=

And v . [ ( )]sinx

g x2 522a= =−

Now integrate above equation with respect to x ,

��g x 2.5 sinx Ca=− +By substituting the value of ��g x in equation (i)

ψ 2.5 2.5cos siny x Ca a= − + 2.5[ ]cos siny x Ca a= − +

FM 5.41 Option (A) is correct.From the definition of the stream function

u y

U hy�c

�

22y= = − a k: D

and integrate with respect to y to obtain

dψ# U hy dy�c

�= − a k: D#

or ψ U yhy f x�c �

�

�= − + ^ h; E

Since v x

�22y=− = , ψ is not a function of x

So that ψ U y hy C� �

�c

�= − +a k: D

Where C is an arbitrary constant.

To evaluate the velocity potential, let

u x

U hy�c

�

22f= = − a k: D

and integrate with respect to x to obtain

dφ# U hy dx�c

�= − a k: D#

or φ U x hy x f yc

��= − +a ^k h: D

However v �y h

U xyy

f y� c�

�

22

2

2f= = =− + ^ h

and this relationship cannot be satisfied for all values of x and y . Thus, there is not a velocity potential that describes this flow (The flow is not irrotational).

FM 5.42 Option (C) is correct.From the definition of stream function ( )ψ

u y h

V y22y= = ...(i)

Integrating with respect to y ,

ψ ( )hV y g x�

�

= + ( )hV y g x2

2= + ...(ii)

And v ( )x x h

V y g x��

22

22y=− =− +: D

v �( )�x

g x22=−

Since v 0= for given velocity field. Then





v [ ( )]x

g x022= =−

Integrate with respect to x

��g x C=Substituting in equation (ii), stream function becomes

ψ hV y C2

2= +

Now by limiting conditions on ψ, 0ψ = at y �= (the bottom wall)

0 0 C= + C 0=

Thus ψ hV y2

2=

Now stream function along the top wall, y h= is

topψ hV h2

2#= Vh

2=


We have ψ x y y3 2 3= −Thus vo B Ay y= −At B �x = and �my =So that Bψ 3 1 1 /m s0 12 3 3

#= − =−^ ^h h (per unit width)At A �mx = and y �=

So that Aψ 3 01 0 02 3#= − =^ ^h h

Hence vo � �m s�B�y= − =− (per unit width)

The negative sign indicates that the flow is from right to left as we look from A

to B .

vo 1 /m s3= (In magnitude)

FM 5.44 Option (B) is correct.From the definition of the stream function, for cylindrical polar coordinates,

vr r�22qy= and v

r22ψ=-θ

So that for the velocity distribution given

r dd�

θψ cosAr Br� � q= +− − ...(i)

And r2

2ψ sinBr � q=− − ...(ii)





Integrate equation (i) with respect to θ to obtain

dψ# cosA Br d f r��q q= + +−^ ^h h#

or ψ sinA Br f r��q q= + +− ^ h ...(iii)

Similarly integrate equation (ii) with respect to r to obtain

dψ# sinBr dr f��q q=− +− ^ h#

or ψ sinBr f��q q= +− ^ h ...(iv)

Thus, to satisfy both equation (iii) and (iv), the stream function is

ψ sinA Br C�q q= + +−

Where C is an arbitrary constant.

FM 5.45 Option (D) is correct.From the definitions of potential function

u ��.�( ) � ��x x

x y x y x� ��

22

22f= = − + − = +

v ��.�( ) � �� y y

x y x y y� �

22

22f= = − + − =− −

FM 5.46 Option (B) is correct.By the definition of stream function.

u y2

2y= , vx2

2y=−

u y

x� �22y= = +

Stream function is found by integration of the velocity component.

ψ ( )xy y f x9 2= + +Now differentiate with respect to x

x2

2ψ [ ( )]yx

f x922= +

Since x2

2ψ v=−

v 9 ( )y f x=− − l

y9 4− − 9 ( )y f x=− − l

( )f xl 4=Integrating with respect to x

( )f x 4 tancons tx= +Then stream function ψ 9 2 4 tancons txy y x= + + +

FM 5.47 Option (C) is correct.For the flow to be irrotational, the vorticity must be zero.Since the flow is in xy -plane, the only z -component of vorticity is there.

z -component of vorticity xv

yu

z 22

22ζ = -

xv

22 ( � ) �

xy �

22= − + =

yu

22 ( )

yx� � �

22= + =

Hence zζ 0 0 0= − =





Since voriticity is zero, therefore flow is irrotational.Now from the definition of potential function

u x2

2f= , vy2

2f=

u x

x� �22f= = +

Integrate with respect to x

φ ( )x x f y2 22

#= + + ��x x f y�= + +

FM 5.48 Option (C) is correctBy the principle of control volume.

vneto v va b= −o o V A V Aa a b b= − ( . . ) . ( . . )3 0 37 6 1 0 9 0 55 6 1# # # #= − 3.75 /m s3=


Shearing stress rzτ zv

rvr z

22

22m= +b l

For Poiseuille flow in a tube v �r = .

Therefore rzτ rvz

22m=

Since vz �v Rr �

ma�= − a k: D

And v V�ma�= , where V is the mean velocity.

It follows that rvz

22

RVr�

�=−

Thus, at the wall (r R= ),

rz wallτ^ h R

V�m=−

Hence rz wallτ^ h R

V�m= . /

4 0.003 0.100 1.20 Pa0 002 2

# #= =^ h


We have, velocity component

v hV hx x�

��

�

= −b l ...(i)

From newton’s law of viscosity

τ dxdvm= ...(ii)

Differentiate equation (i) w.r.t. x

dxdv ( )

hV h x��= −

Hence from equation (ii) at x h= ,

τ ( )hV h h� ��

�#m= − =

FM 5.51 Option (C) is correct.The flow rate is given by

vABo vdA v b dxx

h

�# #= =

=# #

hV hx x b dx�

�h

��

�

�#= −b l#

hV b hx x�

�h

��

�

�= −b l#

hV b hx x�

� �

h

��

� �

�= −: D





hV b h h

hV b h�

� ��

��

��

� �

��

�

#= − =b l V bh32

0=

FM 5.52 Option (B) is correct.The volume flow rate per unit width between two parallel plates is given by

q hlp

32 3Tm=

For � �� mh ��

#= = − , 0.38 /N s m2μ -= and �� mlp � �T

#=

q 3 . 4.21 10 /m s0 382 2 10 30 103 3 3

4 2# # # #

##= =

−−^ ^h h

FM 5.53 Option (C) is correct.The shearing stress acting on the bottom plate is

yxτ yu

xv

22

22m= +c m ...(i)

Since velocity profile between two fixed, parallel plates is parabolic in nature

and it is given by

u xp y h2

1 2 2

22

m= −^ h and v �=

It follow that yu

22

xp y2

1 222

m= ^ h, xv �

22 =

and from (i), yxτ xp y

22= ^ h

At the bottom plate y h=− and since xp

lp

22 T=−

Hence yxτ lp h ��

# # #T= = −^ ^ ^h h h

60 /N m2= acting in the direction of flow

FM 5.54 Option (A) is correct.D’Alembert’s paradox states that with the irrotational flow approximation, the aerodynamic drag force on any non-lifting body of any shape immersed in a uniform stream is zero.

FM 5.55 Option (D) is correct.For incompressible, Newtonian fluid the stress in cartesian coordinates are expressed by

xxσ pxu�

22m=− + , yyσ p

yv�

22m=− + , xyτ

yu

xv

22

22m= +c m

For the given velocity distribution

u xy x12 62 3= − and v x y y�� = −with �.�mx = and �.�my = :

xu

22 12 18 12 . 18 . 7.50 sy x 1 0 0 52 2 2 2 1= − = − = −^ ^h h

yu

22 24 24 0.5 1.0 12.0 sxy 1

# #= = = −

xv

22 36 36 0.5 1.0 18.0 sxy 1

##= = = −

yv

22 18 12 18 . 12 . 7.50 sx y 0 5 1 02 2 2 2 1= − = − =− −^ ^h h

Thus for � �� N mp � �#= and 1.50 /N s m2μ -=





xxσ 6 10 2 1.50 7.50 5.98 kPa3# # #=− + =−

yyσ 6 10 2 . . 6.02 kPa1 50 7 503#=− + − =−^ ^h h

xyτ 1.50 . . 45.0 Pa12 0 18 0#= + =^ h

FM 5.56 Option (D) is correct.Along the /3 4θ π= boundary, which is a streamline (i.e. 0ψ = on 3 /4)θ π= .

p V�

�

ρ + = Constant

or rp

22 V

rV22r=− ...(i)

For the stream function given,

vr cosr r��

��

22qy q= =

and along the /3 4θ π= boundary, v �=θ .

So that V v r��

��

r��q p= = =−b l

Since rV22 r9

8 /2 3=− −

It follows from equation (i) that

rp

22 r r3

898/ /1 3 2 3r=− − − −

b bl l r2764

/1 3r=−

FM 5.57 Option (B) is correct.The volume flow rate for axial flow in the annular space between two fixed concentric pipes is given by

vo ( )

lnlp r r

rr

r r8 i

i

i04 4

0

02 2 2

mpD= − − −

a k> H

which can be written as

vo lnl

r prr

rrrr

8 11

o

i

i

i

04

4

0

0

2 2

mp D= − +

−a

a

a

kk

k: D

Z

[

\

]]

]]

_

`

a

bb

bb ...(i)

Since rr

Ddi

�= , equation (i) can also be written as

vo lnl

r pDd

DdDd

8 11

04 4

2 2

mp D= − +

−b

b

b

ll

l; E

Z

[

\

]]

]]

_

`

a

bb

bb

...(ii)

For .Dd ��= , equation (ii) gives

vo ( . )( . )( . )

0.574lnl

r p8 1 0 1

0 11 0 10

44

2 2

mp D= − +

−=6 @

* 4

Thus, for the same pΔ the flow rate is reduced by

% reduction in vo ( . ) . %1 0 574 100 42 6#= − =

FM 5.58 Option (B) is correct.Write Bernoulli equation between stagnation point and point A to obtain

pstag p V��

A A�r= + ...(i)

For half body VA� cosU r

brb� ��

A A�

�

q= + +c m ...(ii)





and r sinb

qp q

=−^ h

At point A, θ 2p=

So that rA sin

bb

2

22p

p pp=

−=

a k or r

bA 2

p= ...(iii)

Substitution of equation (iii) into equation (ii) yields

VA� U � � ��

�p= + +b l 2θ π=

and therefore from equation (i), we get

pstag p U��

A�

�rp

= + +b l .p U��A�r= +

Thus p pstag A− . U0 703 2r=

FM 5.59 Option (B) is correct.The torque which must be applied to outer cylinder to overcome the force due to the shearing stress is (see figure)

dT r dF r r ld r ldo o o o�t q t q= = =^ h dF dAt=

So that T r l d r l�o o� �

�

�t q p t= =

p

# ...(i)

In the gap u Uby=

Since τ dydu

bUm m= = and b r ro i= − , U row= (see figure)

It follows from equation (i) that

T 2 r r rr l r r

r l2o

o i

o

o i

o23

p m w p mw= − = −b l

FM 5.60 Option (A) is correct.The torque due to force dF acting on a differential area dA r ldi q= is (see figure)

dT r dF r ldi i�t q= = Where τ is the shearing stress.

Thus T r l d r l�i i� �

�

�t q p t= =

p

# ...(i)

In the gap u Uby= Where U riw= and b is the gap width.

Also τ dydu

bUm m= =


T r bU l r b

l2 2i i2 3p m p mw= =b l

2 . ..

0.1602

0 075 0 2 602 80

0 25 10

3

3##

##

p p= −b ^^

l hh 0.355 N m−=





FM 5.61 Option (B) is correct.Downward force on the drop is the weight of the drop.

Fdown �o��e of dropgdropr # #=

g D�drop

�

# #r p= ...(i)

The upward force on the drop is the aerodynamic drag force and the buoyancy force.

Fup 3 VD D g6 air

3

pm p r# #= + ...(ii)

For the drop to be remain suspended, equation (i) must be equal to equation (ii).

Then g D�drop

�

# #ρ π VD D g3 6 air

3

#pm p r= +

D g6 drop air

3π ρ ρ-^ h VD3pm=

V [ ]D g18 drop air

2

m r r #= −

where dropρ �� g mwater�r= =

At 25 Cc airρ .. �.�� g mRT

p��

��

#= = =

airμ 1.849 10 /kg m s5# −= −

Hence V ( . )( )

[ . ] .18 1 849 10

45 101000 1 184 9 815

6

#

## #= −−

−

0.0596 /m s=

FM 5.62 Option (D) is correct.With the coordinate system shown u �= , w �= and from the continuity equation

�v y �2 2 = . Thus, from the y -component of the Navier-stokes equations with g gy =− ,

0 yp g

xv�

�

22

22r m=− − + ...(i)

Since the pressure is not a function of x , equation (i) can be written as

dxd v

�

�

pm= l Where p

yp g

22 r= +l and integrated to obtain

dxdv p x C�m= +l ...(ii)

From symmetry dxdv 0= at x 0= . So that C 0�= . Integration of equation (ii)

yields

v p x C��

�m= +l

Since at x h!= , v 0= it follows that C p h��

m=− l^ h and therefore

v p x h22 2

m= −l^ h

The flow rate per unit width in the z -direction can be expressed as

q vdx p x h dx p h� �

�h

h

h

h� �

�

m m= = − =−− −

l l^ h# #

Thus, with V (mean velocity) given by the equation





V hq p h2 3

1 2

m= =− l

It follows that yp

22

hV g��

m r=− −


For free vortex vθ rK= where tancons tK =

So that v Aθ aK= and v b

KB =θ

From the Bernoulli equation at section A and B ,

pg

v�

A A�

γ + θ pg

v�

B B�

g= + q For horizontal pipe z z� �=

or p p pB AT = − g v v Ka b2

121 1 1

A B2 2 2

2 2g r= − = −q q^ bh l ...(i)

Since q lnv dr K rdr K a

ba

b

a

b= = =q ##

or K lna

bq=

Thus from (i), pT lna

bq

a b21 1 1

2

2

2 2r= −b

bl

l

or q lnab

a bp2 1 1 /

2 2

1 2

#Tr= −

−

b l

Therefore q C pTr= Where � �lnC a

ba b

� � �

��

#= −−

b l

***********

FM 6INTERNAL FLOW

FM 6.1 Consider a fully developed laminar pipe flow. If the pipe diameter is reduced by half while the flow rate and pipe length are held constant, the head loss will be(A) Increase by a factor of 2 (B) Increase by a factor of 4

(C) Increase by a factor of 16 (D) Remains same.

FM 6.2 Consider a flow through a 15 m long horizontal pipe at the laminar transition point. The fluid is oil with 890 /kg m3ρ = and 0.07 /kg m sμ -= . If the power delivered to the flow is 1 hp, the flow rate will be(A) 2420 /cm s3 (B) 4840 /cm s3

(C) 3630 /cm s3 (D) 484 /cm s3

FM 6.3 Glycerin at 40 Cc with 1252 /kg m3ρ = and 0.27 /kg m sμ -= is flowing through a 5 cm diameter horizontal smooth pipe with an average velocity of 3.5 /m s. What will be the pressure drop per unit length of the pipe ?(A) 121 kPa (B) 1.21 kPa

(C) 12.1 kPa (D) 0.121 kPa

Common Data For Q. 4 and 5Water at 15 Cc ( 999.1 /kg m3ρ = ) is flowing steadily in a 45 m long and 4 cm diameter horizontal pipe made of stainless steel at a rate of 8 10 /m s3 3

#− . The

friction factor .f ��1��= .

FM 6.4 What will be the head loss ?(A) 36.6 m (B) 3.66 m

(C) 366.0 m (D) 0.366 m

FM 6.5 The pumping power requirement to overcome pressure drop is(A) 1.5 kW (B) 4.5 kW

(C) 3 kW (D) 6.0 kW

Common Data For Q. 6 and 7A light liquid 950 /kg m3ρ =^ h flows through a horizontal smooth tube of diameter 5 cm at an average velocity of 10 /m s. The fluid pressure measured at 2 m intervals along the pipe is as given below:

( )mx 0 2 4 6

( )��ap 304 255 226 200

FM 6.6 The wall shear stress in the fully developed section of the pipe is (A) 163 Pa (B) 325 Pa

(C) 650 Pa (D) 81.5 Pa





FM 6.7 What will be the overall friction factor ?(A) 0.000183 (B) 0.183

(C) 0.00183 (D) 0.0183

FM 6.8 The piston shown in figure below is pushed steadily by a force F , which causes flow rate of �� cm sv �=o through the needle. If fluid has 900 /kg m3ρ = and

0.002 /kg m sμ -= , the force F will be

(A) 2.0 N (B) 3.6 N

(C) 1.35 N (D) 4.0 N

FM 6.9 A compressor that draws in air ( 1.149 / , 1.802 10 / )kg m kg m s3 5ρ μ # -= = - from the outside, through an 12 m long, 20 cm diameter duct. The compressor takes in air at a rate of 0.27 /m s3 . If the friction factor is to be 0.0211, the useful power used by the compressor to overcome the frictional losses in the duct is(Disregarding any minor losses)

(A) 14.5 W (B) 15.4 W

(C) 51.4 W (D) 41.5 W

FM 6.10 In fully developed laminar flow in a circular pipe, the velocity at . �0 5 (midway between the wall surface and the center-line) is(A) �2 max (B) . �0 5 max

(C) . �0 75 max (D) Not changed

(where umax is the maximum velocity)

FM 6.11 The velocity profile in fully developed laminar flow in a circular pipe of inner radius �cmR = in /m s is given by

( )u r 1��4 2

2

= −c m

The maximum velocity in the pipe and the volume flow rate respectively, are(A) 4 / ,m s 0.01005 /m s3 (B) 0.01005 / ,m s 4 /m s3

(C) 0.01005 / ,m s3 4 /m s (D) 4 / ,m s3 0.01005 /m s





FM 6.12 Consider a flow between two smooth parallel horizontal plates of 3 cm apart. If the fluid is 10SAE oil and � �m sV = 870 / 0.104 /kg m and kg ms3ρ μ= =^ h, the head loss per meter is(A) 0.430 /m m (B) 0.487 /m m

(C) 0.325 /m m (D) 0.163 /m m

FM 6.13 Consider laminar flow of a fluid through a rectangular concrete channel with the smooth surfaces of friction factor ( / )Ref 58= . If the average velocity of the fluid is doubled, the change in the head loss of fluid in percentage is (Assume the flow regime remains same)(A) Decrease by 50% (B) Increase by 50%

(C) Increase by 100% (D) Decrease by 100%

FM 6.14 Water at 20 Cc flows from a tank by the pressurized air at a rate of 60 /m h3 as shown in figure below. If coefficient of friction .f 0 01��= , what gage pressure p1 is needed to drive the pipe flow ?

(A) 2.38 MPa (B) 1.2 MPa

(C) 0.238 MPa (D) 0.12 MPa

FM 6.15 A single 6 cm diameter tube consists of seven 2 cm diameter smooth thin tubes packed tightly as shown in figure below. Air at about 20 Cc and 1 atm (

1.2 /kg m3ρ = , 1.8 10 /kg m s5μ # -= - ), flows through this system at 150 /m h3 . What will be the pressure drop per meter length of the pipe ? (Take .f 0 0�50= )

(A) 202.5 Pa (B) 90 Pa

(C) 270 Pa (D) 27.0 Pa

FM 6.16 Oil with a density of 850 /kg m3 and kinematic viscosity of 6 10 /m s4 2#

− flows in a 5 mm diameter and 40 m long horizontal pipe, from a storage tank open to the atmosphere. If the height of the liquid level above the center of the pipe is 3 m and the flow is fully developed laminar, the flow rate of oil through the pipe is(A) 1.8 10 /m s8 6 3

#− (B) 1.8 10 /m s8 4 3

#−

(C) 1.8 10 /m s8 7 3#

− (D) 1.8 10 /m s8 8 3#

−





FM 6.17 A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length l2 . The flow is fully developed laminar and the pressure drop for the first pipe is 1.44 times greater than for the second pipe. If the diameter of the first pipe is D , the diameter D� of the second pipe is(A) . D1 64 (B) 1.37D

(C) . D1 095 (D) . D1 92

FM 6.18 A capillary viscometer measures the time of t �Δ = seconds required for a 8 cm3 of water at 20 Cc to flow through a D diameter glass tube as shown in figure below. If 1�cmL = , �cml = and flow is laminar with no entrance and exit losses, the capillary diameter D will be (Take 0.001 /kg m sμ -= )

(A) . mm1 5 (B) 15 mm

(C) 0.15 mm (D) 0.015 mm

FM 6.19 Oil with 894 /kg m3ρ = and 2.33 /kg m sμ -= , flows at 0.5 /m s through 300 m long and 40 cm diameter cast iron pipe. Neglect minor losses. The pumping power required to overcome the pressure losses, is(A) 0.45 kW (B) 5. kW0

(C) 4 kW5 (D) 4.5 kW

FM 6.20 SAE 30 oil at 20 Cc . /kg m s0 29μ -=^ , /kg m891 3ρ = h flows upward in a 3 cm diameter pipe through a pump from A to B at a rate of 3 /kg s as shown in figure below. At %100 efficiency, what pump power is required ?

(A) 4.8 kW (B) 4 kW

(C) 0.63 kW (D) 3.5 kW

FM 6.21 Oil with 910 /kg m3ρ = and 0.01 /kg m sμ -= flows through a 1.2 m- diameter pipe at a rate of 3 /m s3 . The pressure drop along the pipe and friction factor are





7.6 MPa and .0 0157 respectively. If the pump is %88 efficient, the power required and the length of the pipe respectively, are(A) 26 ,136.5MW km (B) 19.5 , 182MW km

(C) 19.5 , 136.5MW km (D) 26 ,182MW km

FM 6.22 The pump adds 25 kW to the water as shown in figure and causes a flow rate of 0.04 /m s3 . For either case .f ��= and neglect minor losses. What will be the flow rate expected when the pump is removed from the system ?

(A) 0.0289 /m s3 (B) 2.89 /m s3

(C) 0.289 /m s3 (D) 0.00289 /m s3

FM 6.23 Consider the pitot-static pressure arrangement as shown in figure below. Air at 20 Cc is flowing through the pitot tube 1.2 /kg m3ρ =^ , 1.8 10 /kg m s5μ # -= -

h and the manometer fluid is colored water at 20 Cc 998 /kg m3ρ =^ , 0.001 /kg m sμ -= h. If the friction factor of the flow is .f ��= and .V V��avg CL= , the pipe volume flow rate and the wall shear stress respectively, are

(A) 0.109 /m s3 , 1.7 Pa (B) 0.109 /m s3 , 1.233 Pa

(C) 0.128 /m s3 , 1.233 Pa (D) 0.128 /m s3 , 1.7 Pa

FM 6.24 Glycerin at 20 Cc ( 1260 /kg m3ρ = , 1.50 /N s m2μ -= ) flows upward in a vertical 75 mm diameter pipe with a centerline velocity of 1.0 /m s. The head loss and pressure drop in a 10 m length of the pipe respectively, are(A) 8.2 m, 225 kPa (B) 0.11 m, 125 kPa

(C) 6.75 m, 207 kPa (D) 3.43 m, 166 kPa

Common Data For Q. 25 and 26Oil ( 876 /kg m3ρ = and 0.24 /kg m sμ -= ) is flowing through a 1.5 cm diameter pipe that discharges into the atmosphere at 98 kPa. The absolute pressure 15 m before the exit is measured to be 145 kPa.





FM 6.25 If the pipe is horizontal, the flow rate of oil through pipe is(A) 1.62 10 /m s5 3

#− (B) 162 10 /m s5 3

#−

(C) 16.2 10 /m s5 3#

− (D) 162 10 /m s4 3#

−

FM 6.26 The flow rate of oil through the pipe, if the pipe is inclined at 8c upward from the horizontal, is(A) 100 10 /m s5 3

#− (B) 1.00 10 /m s5 3

#−

(C) 0. 10 /m s10 5 3#

− (D) 10.0 10 /m s5 3#

−

FM 6.27 Consider two types of drinking straws, one with a square cross-sectional shape and the other type the typical round shape. The amount of material in each straw and the length of the perimeter of the cross section of each shape are same . Assume the drink is viscous enough to ensure laminar flow and neglect gravity. What is the ratio of the flow rates v

vsquare

round

oo

_ i through the straws for a given pressure drop ? (For square cross section .�ef ��h = and for round shape �ef ��h = ).(A) 0.183 (B) 0.55

(C) 5.5 (D) 1.83

FM 6.28 Water flows from tank A to tank B with the valve closed as shown in figure. If the friction factor is .0 02 for all pipes and all minor losses are neglected, what will be the flow rate into tank B when the valve is opened to allow water to flow into tank C also ?

(A) 0.180 /m s3 (B) 0.00180 /m s3

(C) 0.0180 /m s3 (D) 1.80 /m s3

FM 6.29 Water at 20 Cc flows through a multiple parallel-plate passages heat exchanger as shown in figure below. The available pressure drop is 2 kPa and plate walls are hydraulically smooth. If the desired total flow rate is 0.25 /m s3 , the appropriate number of passages are .f 0 0�8=^ h

(A) 50 passagesN = (B) 30 passagesN =(C) ��passagesN = (D) ��passagesN =





FM 6.30 Oil at 20 Cc ( 888.1 / , 0.8374 / )kg m kg m s3ρ μ -= = is flowing through a vertical glass funnel as shown in figure. The funnel consists of 20 cm high cylindrical reservoir and a 1 cm diameter, 20 cm high pipe. The funnel is always maintained full by the addition of oil from the tank. Neglect entrance losses. What will be the ratio of the actual flow rate through the funnel to the maximum flow rate for the “Frictionless” case ?

(A) 43.91 (B) 0.0232

(C) 2.32 (D) 0.232

FM 6.31 Water at 20 Cc flows upward through an inclined 6 cm diameter pipe at 4 /m s is shown in figure. A mercury manometer has a reading of 13�mmh = . The pipe length between points (1) and (2) is 5 m and point (2) is 3 m higher than point (1). What will be the friction factor of the flow ?

(A) 0.114 (B) 0.07

(C) 0.025 (D) 0.044

FM 6.32 Viscous oil ( . . 0.��S G = , 0.10 Pa sμ -= ) flows from tank A to tank B through the six rectangular slots as shown in figure below. If minor losses are negligible and the total flow rate is 30 /mm s3 , the pressure in tank A will be (Take f 32�0= )





(A) 1.54 kPa (B) 1.15 kPa

(C) 2.31 kPa (D) 1.92 kPa

FM 6.33 A 2 mm diameter and 20 cm long straw delivers the water at 10 Cc with a rate of 3 /cm s3 . If the flow is vertically up, what will be the axial pressure gradient

�p x2 2 ?(Take 1.307 10 /kg m s3μ # -= - , 1000 /kg m3ρ = )(A) 2 /kPa m (B) 10 /kPa m

(C) 4 /kPa m (D) 20 /kPa m

FM 6.34 A tank of water has a 1.5 cm diameter hole at the bottom, where water discharges to the atmosphere. The water level is 3 m above the outlet. Disregarding the effect of the kinetic energy correction factor. If the entrance of the hole is sharp edged, the flow rate of water through the hole is (loss coefficient KL for sharp-edged .0 5= )(A) 1.11 10 /m s3 3

#− (B) 111 10 /m s3 3

#−

(C) 11.1 10 /m s3 3#

− (D) .111 10 /m s0 3 3#

−

FM 6.35 Water at a rate of 0.04 /m s3 , flows in a 0.12 m diameter pipe that contains a sudden contraction to a 0.06 m diameter pipe. If the loss coefficient .K ��L = , the pressure drop across the contraction section is(A) 99.75 kPa (B) 33 kPa

(C) 166.25 kPa (D) 133 kPa

FM 6.36 The water pipe system shown in figure below consists of 1200 m long cast-iron .f ��=^ h pipe of 5 cm diameter, two 45c and four 90c flanged long-radius

elbows, a fully open flanged globe valve and a sharp exit into a reservoir. The minor losses coefficient for the pipe system is as follows

45c long-radius elbow : .K ��, 90c long-radius elbow : �.K �, Open flanged globe valve : .K ��, Sharp exit valve : .K ��,If the elevation at point 1 is 400 m, what gage pressure is required at point 1 to deliver 0.005 /m s3 of water at 20 Cc ( 0.001 / skg mμ -= ) into the reservoir ?(A) .54 MPa4 (B) 3. MPa46

(C) . MPa1 43 (D) .4 MPa6

FM 6.37 Kerosine is to be withdrawn from a 15 cm high kerosine tank by drilling a well rounded 3 cm diameter hole with negligible loss at the bottom surface and attaching a horizontal 90c bend of negligible length. The kinetic energy correction factor is 1.05. What will be the flow rate of water through the bend, respectively





if (a) the bend is a flanged smooth bend and (b) the bend is miter bend without vanes ?

(A) 8.08 / ,L s 4.78 /L s (B) 4.78 / ,L s 6.03 /L s

(C) 6.03 / ,L s 4.78 /L s (D) 8.08 / ,L s 6.03 /L s

FM 6.38 A horizontal pipe has an sudden expansion from ��mD�= to ��mD�= . The water is flowing at 10 /m s and ��ap�= in the small section and the flow is turbulent. If the kinetic energy correction factor to be 1.06 at both inlet and outlet, the downstream pressure is

(A) 300 kPa (B) 278 kPa

(C) 377 kPa (D) 322 kPa

FM 6.39 A 4.5 m diameter tank is initially filled with water 2 m above the centre of a sharp edged 15 cm diameter orifice. The tank water surface is open to the atmosphere and the orifice drains to the atmosphere. Neglecting the effect of the kinetic energy correction factor. The time required to empty the tank is (loss coefficient for sharp edge .K 0 �L = )

(A) 9.6 min1 (B) . min26 4

(C) . min0 264 (D) . min2 64

Common Data For Linked Answer Q. 40 and 41Water at 20 Cc flows through a 10 cm diameter smooth pipe which contains an orifice plate with 5 cm-diameter. The measured orifice pressure drop is 75 kPa. Discharge coefficient 0.605Cd = and non- recoverable head loss coefficient 1.�K = .





FM 6.40 What will be the flow rate in /m hr3 ?(A) 54 (B) 50

(C) 60 (D) 209

FM 6.41 What will be the non recoverable head loss ?(A) 33.4 kPa (B) 6.9 kPa

(C) 52. kPa4 (D) 26.4 kPa

Common Data For Linked Answer Q. 42 and 43Water at 20 Cc ( 9 8 / , 1.002 10 / )kg m kg m s9 3 3ρ μ # -= = - flows through a 50 cm diameter pipe. The flow rate of water is measured with an orifice meter to be 0.25 /m s3 . The diameter ratio β and discharge coefficient Cd are .0 60 and .0 61 respectively.

FM 6.42 The pressure difference indicated by orifice meter is(A) 1.9 kPa (B) 19.0 kPa

(C) 14 kPa6 (D) 14.6 kPa

FM 6.43 What will be the head loss ?(A) 2.207 m (B) 2.0421 m

(C) 0.7734 m (D) 0.940 m

FM 6.44 A 5 cm diameter smooth pipe contains an orifice plate of .0 6β = and it is monitored by a mercury manometer /kg m13550 3ρ =^ h as shown in figure below. What will be the h when the flow rate is 0.334 /minm3 ?(Take .C ��d = )

(A) 75.75 cm (B) 7.52 cm

(C) 57 cm (D) 1.72 cm

FM 6.45 Air at 20 Cc ( 1.204 / )kg m3ρ = flows at high speed through a venturi-meter monitored by a water manometer as shown in figure below. If ��cmh = , what will be the maximum mass flow rate of air that venturi can measure ? (Take discharge coefficient .C ��d = )





(A) 2.73 /kg s (B) 0.273 /kg s

(C) 27.3 /kg s (D) 0.0273 /kg s

FM 6.46 Consider the flow of air at high speed through a venturi monitored by a mercury manometer �� g mHg

�ρ =^ h as shown in figure below. Discharge coefficient Cd and Expansion factor Y for this flow are .0 985 and .0 76 respectively. The upstream conditions are 150 kPa and 353 K. If ��cmh = , the mass flow rate for flow to be compressible is

(A) 0.40 /kg s (B) 3.23 /kg s

(C) 7.27 /kg s (D) 0.90 /kg s

FM 6.47 Ethanol at 20 Cc /kg m789 3ρ =^ , 0.0012 /kg m sμ -= h flows through a 5 cm diameter smooth pipe at a rate of 7 /m hr3 . Three piezometer tubes are installed as shown in figure below. If the pipe contains a thin plate orifice of diameter

�cmd = , the piezometer levels h� and h� will be .Take K 1 5=^ and .f ��= h

(A) �.�mh�= , �.��mh�= (B) �.��mh�= , �.��mh�=(C) �.��mh�= , �.�mh�= (D) �.��mh�= , �.��mh�=

FM 6.48 Consider the parallel-pipe system as shown in figure below. The SAE 10 oil at 20 Cc 870 /kg m3ρ =^ and 0.104 /kg m sμ -= h is flowing laminarly through the pipe system with pressure drop ��ap p� �− = . What will be the total flow rate between 1 and 2 ?

(A) 0.0005 /m s3 (B) 0.0022 /m s3

(C) 0.0027 /m s3 (D) 0.0032 /m s3

FM 6.49 Consider the parallel-pipe system of two identical length and material pipe as shown in figure below. The diameter of pipe A is half of the diameter of pipe B





. If the friction factor to be same in both case and disregarding minor losses, the flow rates in pipes A and B would be

(A) Remains same

(B) Flow rate of A increased by a factor of 0.177.

(C) Flow rate of B increased by a factor of 0.177.

(D) Flow rate of A decreased by a factor of 0.177.

Common Data For Linked Answer Q. 50 and 51Three pipes of same material .f 0 0�7�=^ h are laid in parallel with these dimensions:

Pipe 1 : �00 mL1 = 10 cmd1 =

Pipe 2 : �00 mL2 = 12 cmd2 =

Pipe 3 : �00 mL3 = �cmd3 =The total flow rate is 0.056 /m s3 of water at 20 Cc .

FM 6.50 The flow rate in each pipe is (A) 0.01�� m sv1

3=o , 0.0277 �m sv23=o , 0.011� �m sv3

3=o

(B) 0.01�� m sv13=o , 0.011� �m sv2

3=o , 0.0277 �m sv33=o

(C) 0.0277 �m sv13=o , 0.01�� m sv2

3=o , 0.011� �m sv33=o

(D) 0.011� �m sv13=o , 0.01�� m sv2

3=o , 0.0277 �m sv33=o

FM 6.51 The pressure drop across the system will be(A) 56 kPa (B) 55 kPa

(C) 550 kPa (D) 137.5 kPa

FM 6.52 For the Series -Parallel system of pipes shown in figure below, each pipe is 8 cmdiameter cast iron ( .f 0 0022, ) and the pressure drop 7�0 �Pap p1 2− = . If the minor losses are neglected, what will be the resulting flow rate for water at 20 Cc ?

(A) 101 /m hr3 (B) 23 /m hr3

(C) 62 /m hr3 (D) 39 /m hr3





FM 6.53 Water at 80 Cc ( 3.65 10 /m s7 2ν #= - ) flows with an average velocity of 2 /m s through a 120 mm diameter pipe. If the pipe wall roughness is small enough so that it does not protrude through the laminar sublayer and the pipe is to be considered as smooth ( .f ��= ), what will be the largest roughness allowed to classify this pipe as smooth ?(A) 23.1 mm

(B) 0.0231 mm

(C) 0.00231 mm

(D) 0.231 mm

FM 6.54 The three water-filled tanks are connected by pipes as shown in figure. If minor losses are neglected, the flow rate in /m s3 in each pipe is

(A) .v ��=o , .v ��=o , .v ��=o

(B) .v ��=o , .v ��=o , .v ��=o

(C) .v ��=o , .v ��=o , .v ��=o

(D) .v ��=o , .v ��=o , .v ��=o

FM 6.55 A highly viscous liquid flows under the action of gravity from a large container through a small diameter pipe in laminar flow as shown in figure below. Disregarding entrance effects and velocity heads, the variation of fluid depth in the tank with time, is

(A) 32 lnk hH

b l (B) lnk hH64 b l

(C) lnk hH128 b l (D) lnk h

Hb l

where kgdLD

�

�

n=

FM 6.56 A triangular passages ( 52.9/ )Ref = of heat exchanger with ��cmL = and an isosceles triangle cross section of side length �cma = and included angle 80cβ = is shown in figure below. If the oil 870 / , 0.104 /kg m kg ms3ρ μ= =^ h at 20 Cc flows at 2 /m s, the pressure drop will be





(A) 11.5 kPa (B) 23 kPa

(C) 2.3 kPa (D) 1.15 kPa

FM 6.57 An oil ( �� S G = and 2.2 10 /m s4 2ν #= - ) flows at a rate of 4 10 /m s4 3#

− through a vertical pipe as shown in figure. The manometer reading h will be

(A) 18.5 m− (B) 13.87 m

(C) 13.87 m− (D) 18.5 m

FM 6.58 The water velocity at several locations along a cross section of 5 cm radius pipe is given in table below.

�cmr � �m sV

0 6.4

1 6.1

2 5.2

3 4.4

4 2.0

5 0.0

What will be the flow rate of water ?(A) 0.297 /m s3 (B) 0.0297 /m s3

(C) 2.97 /m s3 (D) 29.7 /m s3

FM 6.59 Oil ( 8900 /N m3γ = , 0.10 /N s m2μ -= ) flows through a 23 mm diameter horizontal tube as shown in figure. A differential U-tube manometer is used to measure the





pressure drop along the tube. What will be the range of h for laminar flow ?

(A) �.��mh # (B) 0.51 0.51m mh# #−(C) 0 0.51 mh# # (D) �.��mh $

FM 6.60 The water at 20 Cc flows from the tank as shown figure below, through the 3 cm long horizontal plastic pipe attached to the bottom of the tank. What time it will take to empty the tank completely, assuming the entrance to the pipe is well-rounded with negligible loss ? (Take the friction factor of the pipe to be 0.022.)

(A) 8.66 hours (B) 82 hours

(C) 86.6 hours (D) 8.2 hours

***********





SOLUTIONS

FM 6.1 Option (C) is correct.In a fully developed laminar pipe flow, head loss is given by

hL ��efD

Lg

VDL

gV

VD DL

gV

��

��

��

n= = =

D DL

gV��

n= D g

LDv��

��

� �n

p= o

> H V

Dv

��p

= o

g D

Lv��p

n= o

where ν = Kinematic viscosity

L = Pipe Length

D = Diameter of pipe

vo = Volume flow rate

If diameter of pipe is reduced by half, then D� D2=

So that hL� ��

g DLv

g DLv

�

��

pn

pn

#= =o o

b l

h16 L=

Hence Reducing the pipe diameter by half increase the head loss by a factor of

16.

FM 6.2 Option (B) is correct.For laminar flow at transition point

Re Vdm

r= 2300=

or Vd890#μ 2300= .Vd ��

�� & #= 0.181 /m s2= ...(i)

Power 1 hp 745.7 W v p minla arD#= = o

.745 7 AV p d VdLV

��

�# #p mD# #= = a ck m

.745 7 8 LV 2p m#= 8 0.07 15 .d

0 181 2

#p# #= b l

or .745 7 .d

��6��= .

.d ��6��& =

d 0.034 m=

From eq. (i) V ... �.�� m sd

�1�1��1�1= = =

Hence vo d V42p

# #= . .4 0 034 5 322p# #= ] ]g g

0.00484 / 4840 /m s cm s3 3= =

FM 6.3 Option (C) is correct.Reynolds number for this flow





Re VDm

r= .. . 811.50 27

1252 3 5 0 05# #= =

Re .811 5 2300<=Hence the flow is laminar and friction factor for this Re is

f . .Re64

811 564 0 07887= = =

Then the pressure drop per unit length ( 1 )mL =

pLΔ fDL V

�

�

#r= 0.0788 .

( . )0 051

21252 3 5

100012

## # #=

12.1 kPa=


Head loss hL fDL

gV�

�

#=

fDL

g Dv

��

�

�

p# #= o; E fD

Lg D

v�

��

�

# #p#= o V

Dv

��p= o

fgDLv

��

� �

�

p#= o

hL . ( . )

. ( )2 9 81 0 04

0 01573 16 45 8 102 5

3 2

# # #

# # # #

p=

−

36.5 m9= 36.6 m,

FM 6.5 Option (C) is correct.Pressure drop is given by

pΔ f DL V

�

�

# #r= �f D

LDv

� �

�rp# # #= o; E

f DL

Dv f

DL v

��

��

� �

�

� �

�

# # # #r

p pr= =o o

pΔ .( . )

. ( )0 01573

2 0 04 100016 45 999 1 8 10

2 5

3 2

## # #

# # # #

p=

−

358.3 kPa= 359 kPa-

Hence the power requirements to overcome this pressure drop is

Ppump � �� v p �D # #= = −o

2.8 kW7= 3 kW,

FM 6.6 Option (A) is correct.The Wall shear stress in the fully-developed region is defined as

Lp

fully developedΔΔ p p

d2 226000 4 w4 6 t= − = =

or wτ . 162.5 Pad4 2

260008

26000 0 05## #= = = 163 Pa,


The overall head loss for �zΔ =

hf . ��.�mgp

��

#rD= = − =

The overall friction factor is defined as

foverall h Ld

Vg�

�f overall �# #= 11.2 . .6

0 0510

2 9 812

##= b bl l

0.0183=





FM 6.8 Option (D) is correctThe velocity of exit from the needle is

V Av

�= o

.. 306 /cm s

4 0 0250 15

2#

p= =] g

The energy equation gives

gp

gV z�

� ��

�ρ + + gp

gV z h h� f f

� ��

� � �r= + + + + ...(i)

With z z� �= , V ��- , �hf �, and hgDLV��

f � �rm= (laminar flow), equation (i)

becomes

gp p� �

ρ- h g

V�f �

��

= + gD

L Vg

V��

��

# # #

rm= +

or gp

ρΔ

. .32 0.002 0.015 3.06

..

900 9 81 0 00025 2 9 813 06

2

2

# # ## # #= +

]]

gg

5.32 0.48 5.8 m= + =

Then F p ApistonD #= �.� .g ��r p# # #= ] g

9.81 900 5.8 . 4.0 N4 0 01 2 ,p# # #= ] g

FM 6.9 Option (A) is correct.The average velocity for this flow

V ( . )

. �.�� m sAv

Dv �

��

c� �

#

#

#p p

= = = =o o

The pressure drop in the duct

pΔ p fDL V

�L

�rD #= =

Substitute f 0.0211, 1.149 / , 12 , 8.594 /kg m m m sL V3r= = = = and D 0.2 m=

So pLΔ . .. ( . )

0 0211 0 2012

21 149 8 594 2

# ##= 53.72 Pa=

Then the required pumping power becomes

Ppump �.�� .��v pD #= =o 14.5 W=

FM 6.10 Option (C) is correct.The velocity profile in fully developed laminar flow in a circular pipe is given by


�

= −; E

At �r R �= ( �)u R � ( �)

uR

Ru�

��

�max max�

�

= − = −; :E D

.u u43 0 75max max= =

FM 6.11 Option (A) is correct.The general velocity profile in fully developed laminar flow is


�

= −; E

We have ( )u r uRr� �

�

= −; E

By comparing these two, we get





u�� 4 /m s=Now the volume flow rate

vo V A V Ravg c avg�p#= =

Since Vavg 2.00 /m su2 2

4max= = =

Hence vo ( . )2 0 04 2# #p= 0.010048 /m s3= 0.01005 /m s3

b

FM 6.12 Option (C) is correct.The half-distance between plates is called “h”, then Hydraulic diameter

Dh 4 4 1.5 6perimeter cmA h4#= = = =

ReDh .. . ��VD

�� h # #

mr= = = (Laminar flow)

For laminar flow between two parallel horizontal plates, the head loss and Pressure drop per meter flow is given by

pΔ .

. .hVL�

��

� �# # #m= =

]]

gg

2773.33 / 2770 /Pa m Pa m-=

Hence hf . �.�� m mgp

�� #r

D= = =

FM 6.13 Option (C) is correct.The friction factor of given channel

f Re58= where Re VDH

mr= Hydraulic diameterDH =

Then Head loss hL� �efDL

gV

DL

gV

��

�H H

� �

# # #= =

VD DL

gV��H H

�#

# #rm=

H29

gDLV

2rm=

If the average velocity is doubled, then

V� 2V=

Hence hL� gDL V

gDLV��

H H� �

#

rm

rm= = = G 2hL1=

Therefore percentage change in Head loss is

(�)hΔ �hh h� ��

L

L L

�

� �#= − = Increase

FM 6.14 Option (A) is correct.For water at 20 Cc , take 998 /kg m3ρ = and 0.001 /kg m sμ -=

V .�

�.�� m sAv

dv

dv

�

��

� ��

# # #

#

p p p= = = = =o o o

]

^

g

h

The energy equation between points (1) (the tank) and (2) (the open jet) :

1�gp

g2�1

2

ρ + + 80g gV

h20

2pipe

f

2

= + + +

or gp1

ρ 80 10gV

h2pipe

f

2

= + + − ...(i)

where hf �.�1�� . ..fD

Lg

V2 ��

�� 2 ��1

��pipe2 2

## # #= =

+ +^ h





170 m,

then from equation (i),

gp�

ρ .. 80 170 102 9 81

8 49 2

#= + + −] g

or p� ��g � ��

��

#r= + + −] g

; E

998 9.81 .8.49

2 9 81 80 170 102

## #= + + −] g

; E

2385659 2.38 MPa,=

FM 6.15 Option (C) is correct.Assume all the flow goes through the seven 2 cm tubes. Thus each tube takes one - seventh of the flow rate :

V .

��.�� m sA

v� ��

��,tubes�

�# #

,p

= =o] g

Re .

. . . ��Vd��

��

#

# #m

r= = =− (Turbulent flow)

Hence pΔ �.�� .. .f d

L V� ��

�� r

# # # #= = ] g

269.32 270 Pa,=

FM 6.16 Option (D) is correct.The pressure at the bottom of the tank is

p ,gage� ghr= .1000

850 9 81 3# #= 25.02 /kN m2=

The pressure loss across the pipe, disregarding minor losses is

pΔ p p p patm� � �= − = − ��.�� k� mp ,gage��= =

Pressure loss for fully developed laminar flow

pΔ DLV

DLv��

� �m

pm= =

o v V D�

�p#=o

D

L v��

# # # #

pr n=

o

Then vo Lp D

128

4

# # #r npD=

128 850 6 10 4025.02 10 ( . )0 005

4

3 4# #p

# # # #

#= −^

^

h

h 1.8 10 /m s8 8 3

#= −

FM 6.17 Correct option is (C)

For laminar flow vo lD p128

4

mp D=

where v v� �=o o and .p p�� Δ Δ=- -

Thus v�o lD p v l

D p128 128

1 2 2 34

334

mp

mpD D= = =− −o D D�=

or D� ( . ) .D pp D D��

��

��

��

��

DD= = =

−

−

c m





FM 6.18 Option (A) is correct.Assume no pressure drop and neglect velocity heads. The energy equation reduces to

gp

gV z�

� ��

�ρ + + gp

gV z h� f

� ��

�r= + + +

or 0 0 L l+ + +^ h 0 0 0 hf= + + + hf L l= +

For laminar flow hf gDLv��

�prm=

o and for uniform draining vo t

vD=

or tvvΔ = o

Lh gD

vgD h

Lv

128

128f f

4 4#

#

mpr pr

m= =

tΔ ( )gD L lLv��

�prm=

+ ( )h L lf = +

or 6 . ( . . )

. .D998 9 81 0 12 0 02

128 0 001 0 12 8 104

6

# # # #

# # # #p

=+

− .

D��

�

��#=

−

D� 4.75 10 12#= −

or D 0.00148 0.0015 1.5m mm- , =

FM 6.19 Option (D) is correct.Reynolds number for this flow

Re .. . .VD

�� # #

mr= = =

Since .Re 76 7 2300<= . Hence the flow is laminar.

The pressure loss for laminar flow is given by

pΔ fDL V

�

�r=

Since f . .Re64

76 764 0 834= = =

Then pΔ . .( . )

0 834 0 4300

2894 0 5

100012

# ##

#= 69.9 kPa=

Therefore pumping power required to overcome this pressure drop

Ppump v p V D p��

# pD D#= =o

. ( . )

.40 5 0 4

69 92

# ##

p= 4.39 4.5 kW,=

FM 6.20 Option (D) is correct.The pipe velocity

V .

�.�� m sAm

��

�# #r p

= = =o] g

Check Red .. . ��VD

�� # #

mr= = = (Laminar flow)

Apply the steady flow energy equation between A and B

gp

gV z�

A AA

�

ρ + + gp

gV z h h�

B BB f p

� �

r= + + + −

or .891 9 81500000#

. 15 h h891 9 81180000

f p#

= + + − V V VA B= = 0and zA =





h hf p− . . 15891 9 81500000

891 9 81180000

# #= − −

or h hf p− 21.61 m=

Where hf � ��

gdLV��

��

� �

� �

# #

# # #

rm= = +

]] ]

gg g 140.4 m=

Now hp ��hf= − 140.4 21.16= − 119. m2=The pump power is then given by

P gvhpr= o mghp= o 3 9.81 119.2# #= 3508 3500 Watts-= or 3.5 kW

FM 6.21 Option (D) is correct.The flow rate gives the velocity and Reynolds number

V .

�.��sm

Av

� ��

�#

p= = =o

] g

Red .. . ��Vd

�� # #

mr= = = (turbulent flow)

Since pΔ f dL V�

�r=

or 7600000 0.0157 . .L1 2 2

910 2 65 2# # #= b ]l g

L 181800 182m km-=

Hence P . �.� ��Wattsv p��

� �� #hD

#= = =o

26 MW=

FM 6.22 Correct option is (A).

Applying the Bernoulli’s equation at section (1) and (2)

pg

V z h212

p1

1γ + + + pg

V z fDl

gV

2 22 2

2

2

2

#g= + + +

Where p p �1 2= = , z h1 = , z �2 = and V �1 =

V2 ( . ). �1.� �m sA

v

��

2 2p= = =o

c m

V ( . ). 1�.1� �m sA

v

� ��

2#

p= = =o

h hp+ .( . )

0.016 . .( . )

133.2 m2 9 8131 8

0 0630

2 9 8114 152 2

## #

#= + =b l

Also hp ( . ) .��. m

vP

�� 1� ��2� 1� ��

�

# #

#g= = =o

Hence h 133.2 63.7 69.5 m= − =





Without the pump h �p = and z gV fD

lg

V� ��

��

#= + ...(i)

Where ��.�mh z�= =

and V� AAV

DD V

� �

�= = b l or .V V V��

��

�= =b l

Thus from equation (i)

.69 5 .

( . ) . .V V

2 9 81

2 25 0 016 0 06302 2

#

#=

+ b l or ��.�� m sV =

So that vo (�.��) ��.�� .�� m sAV �� #

p= = =

FM 6.23 Option (B) is correct.The manometer reads

p p�− ghwater airr r= −^ h . 9.81 . 391 Pa998 1 2 0 040#= − =] ]g g

Therefore, velocity at centre line

VCL . 25.5 /m sp21 2

2 391#rD= = =

Now average velocity

Vavg 0.85VCL= 0.85 25.5 21.675 /m s#= =Thus the flow rate

vo ��.�� . �.�� m sV A � �� p# # #= = =] g

and wall shear stress

wτ f V8 avg2r#=

. 1.2 .80 0175 21 675 2

# #= ] g 1.233 Pa=


For laminar flow in a pipe

Vaverage 1 0.5 /m sV21

21

max #= = =





Thus Re �� VD

�� <# #

mr= = =

The flow is laminar.

So that V ( )sin

lp l D

32

2

mg qD= −

Where 90cθ =

pΔ D

lV l��

m g= + sin90 1=c

( . ). . 9.81 1260 100 075

32 1 50 10 0 52

# # ## #= + gγ ρ=

1.66 10 166Pa kPa5- # =Also, applying Bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1γ + + pg

V z h2 L2 2

2

2g= + + +

With p p p1 2 D= + , V V1 2= , z z l2 1− =

hL .. 1�p l ��1 12��

1 �� 1��

##

gD= − = − 3.43 m=

FM 6.25 Option (A) is correct.Flow rate can be determined from

vo Lp D

128

4

mpD=

Since p p p1 2Δ = - 145 98 47 kPa= − =

Hence vo .( . )

1.62 10 /m s128 0 24 1547000 0 015 4

5 3

# #

# #p#= = −

FM 6.26 Option (B) is correct.For uphill flow with an inclination of 8c,

vuphillo ( )sin

Lp gL D

128

4

mr q pD= −

.[ ( . )] ( . )sin

128 0 24 1547000 876 9 81 15 8 0 015 4

# #

# # # #c p= −

1.00 10 /m s5 3#= −


p1Δ p2D=So hL1 hL2= p hLγΔ =

f Dl

gV2h

11 1

2

1

# f Dl

gV2h

22 2

2

2

#=

where l l1 2= , D Dh1 = and perimeterD Aaa a�

��

h2

22

2 = = =

Thus Df V1 1

2

af V2 2

2

= ...(i)

Since the perimeters are equal

Dπ a4= Therefore a D�p= ...(ii)





and Reh1 V D V Dh� ��

n n= =

Reh2 V D V a V D

�h� � ��

n n np= = = from equation (ii)

Thus, from equation (i)

D

V DV��

��

νb l

.

D

V DV

4

4

56 92

22

pn

p=

b l or �V V�� =

Also v�o A V D V��

�#p= =

and v�o A V a V D V��

�

��

�p= = =

So that vv

�

�oo

1.441 1.83D VD V

VV4 4

162

2

42

1

2

12 p p #= = = =p

p

vroundo . v1 83 square= o



v�o v v� �= +o o

Since D� D D� �= = , it follows that

V� V V� �= + ...(i)

Also for fluid flowing from A to B ,

pg

V z�A A

A

�

γ + + pg

V z f Dl

gV f D

lg

V� � �

B BB

�

��

� ��

��

� ��

# #g= + + + +

Where p p �A B= = , �V VA B= = , ��mzA = and z �B =

Thus zA f Dl

gV f D

lg

V� ��

�

� ��

��

� ��

= + ...(ii)

or 15 . .. [80 40 ]V V0 1 2 9 81

0 0212

22

# #= +

or .18 4 .V V��

��= + , ...(iii)

Similarly, for fluid flowing from A to C ,

pg

V z�A A

A

�

γ + + pg

V z f Dl

gV f D

lg

V� � �

C CC

�

��

� ��

��

� ��

g= + + + +

where �p pA C= = , �V VA C= = , ��mzA = and �zC =

Thus zA f Dl

gV f D

lg

V� ��

�

� ��

��

� ��

= + ...(iv)





Comparing equation (ii) and (iv), we get

f Dl

gV��

�

� ��

f Dl

gV��

�

� ��

=

Since f f� �= and D D� �= l V� �

� l V� ��=

40V22 75V3

2= V� 1.369V3= ...(v)

Solve equation (i), (iii) and (v) for V�, V� and V�. From equation (i) and (v)

V� 1.369 2.369V V V3 3 3= + =from eq (iii) 18.4 ( . ) . ( . )V V2 369 0 5 1 3693

23

2= + V� 1.676 /m s=from eq (v) V� 1.369 1.676 2.29 /m s#= =

Hence v�o (�.1) �.��A V 4� ��

# #p= = 0.0180 /m s3-

FM 6.29 Option (C) is correct.If there are N passages, then cmb ��= for all N and the passage thickness is

�.��H N= . The hydraulic diameter is �D Hh = . The velocity in each passage is related to the Pressure drop.

pΔ f DL V�h

�r# # #= where .f f ��8smooth= =

or 2000 0.028 ..

N

�2 0 5

2 02

998 2

## # #=

Where V . .

. 1 �m sAv

��

c #= = =o

^ h

Thus 2000 0.028 .2.0 N

2 0 5 2998 1 2

## # #= ] g

N . .. 71.57 72 passages0 028 2 0 998

2000 2 0 5 2# #

# # # ,= =


The flow velocity for the frictionless case.

V ,max� . . 2.801 /m s��2 2 9 81 0 401 # #= = =Then maximum volume flow rate





vo ��

V A V D� ��

��

��

# # ##p p= = =

2.20 10 /m s4 3#= −

Now the actual flow rate can be determined by

vactualo ( )sin

Lp gL D

128

4

mr q pD= −

Since the flow is vertically downwards, so 90cθ =- and

p p pinlet outletΔ = - ( )p gh p ghatm cylinder atm cylinderr r= + − =(because at inlet total pressure becomes patm and pressure due to oil in cylinder

ghcylinderρ and at exit atmospheric pressure patm is there)

Therefore vactualo ( )sin

Lgh gL D

128cylinder

4

mr r q p= −

( )

Lg h L D

128cylinder

4

mr p= +

( 90 ) 1sin c− =−

. .. . ( . . ) ( . )

128 0 8374 0 20888 1 9 81 0 20 0 20 0 01 4

# #

# # # #p= +

5.1 10 /m s6 3#= −

So, the ratio of actual flow rate through the funnel to the maximum flow rate is

vv

max

actualoo

..

2 20 105 1 10

4

6

#

#= −

− 0.0232=

FM 6.31 Option (C) is correct.By moving through the manometer, we obtain the pressure change between points (1) and (2).

p gh gh g zw m w1 r r r D+ − − p2=or p p1 2− gh gh g zm w wr r r D= − + gh g zm w wr r r D= − +^ h

9.81 0.135 998 9.81 313568 998# # #= − +^ h

16647 29371= + 46018 46000 Pa-=

Since hf . �gp z �� 1

��w #rD D= − = −

.4 7 3= − 1.7m=

The friction factor f h Ld

Vg2

f 2# #=

1.7 . .5

0 064

2 9 812

## #= 0.0250=


For each slot, flow rate becomes

vo 30 10 5 10 /m s61 9 9 3# # #= =− −

So that V ( )

1.�� 1� �m sAv

� 1 1�� 1�

�

��

# #

##= = =−

−−o

Also ρ . . �.�� 1�� g mS G � O�

2# #= = =r

and Dh ( )( )

1.5perimeter mmA46 2

4 3 1# #= = + =

The energy equation gives





pg

V z�A A

A

�

γ + + pg

V z f Dl

gV

� �B B

Bh

� �

g #= + + +

with z zA B= , p �B = , V V �A B= =

pA f Dl V�

�h

�r= b bl l

3250 .. 850 (1.67 10 )0 0015

0 621 3 2

# # ## #= −

1540 / 1.54N m kPa2- =

FM 6.33 Option (D) is correct.For water at 10 Cc : 1000 /kg m3ρ = 1.307 10 /kg ms3μ #= -

Re �VD DAv

Dv D

�#m

rmr

m pr

#= = =o o V A

vDv�

�p= =o o

Re . .D

v��

� ��

�

# # #

# # #mpr

p= = −

−o 1462= (Laminar flow)

Thus hf . .. .

gDLv��

��

� �

� �

# # #

# # # # #pr

mp

= =− −o

] g

0.204 m,

The axial pressure gradient if the flow is vertically up

LpΔ L

g h zfr D=

+^ h .

. . .0 2

1000 9 81 0 204 0 2# #= +^ h z LΔ =

19816 / 20 /Pa m kPa m,=


The energy equation for a control volume between two points

gp

gV z h� pump

��

��

�ρ α+ + + gp

gV z h h� turbine L

��

��

�r a= + + + +

Since p p� �= ,patm= �,z�= V ��= and h h �pump turbine= =

z� gV h2 L2

22

a= + gV K g

V2 2L2

22

22

a= + h K gV�L L

�

#=

V� Kgz2

L2

1

a= +Since ,12α = then volume flow rate becomes

v V Ac�=o DK

gz4 1

2hole

L

21p

#= +Substituting the numerical values, we get

vo ( . )

..

40 015

1 0 52 9 81 32

##

# #p= + 1.11 10 /m s3 3#= −






Applying the bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1γ + + pg

V z K gV

2 2L2 2

2

222

g= + + + ...(i)

Where V1 ( . ). �.�� sA

v

� �12��

1 2#

p= = =o

V2 ( . ). 1�.1 �� sA

v

� ��

2 2#

p= = =o and z z1 2=

Hence equation (i) becomes,

p p1 2− [ ]K V V V21

L 22

22

12r= + −

999 [0.40(14.1) (14.1) (3.54) ]21 2 2 2# #= + −

or p p1 2− . . 13 kPa499 5 265 80 3# -=

FM 6.36 Option (B) is correct.For water, take 998 /kg m3ρ = and 0.001 /kg m sμ -= . With the flow rate known, we can compute

V .

. 2.�� m sAv

� ��

��2p

= = =o

d ]n g

The energy equation is written from point 1 to the reservoir surface

gp

gV z2

1 12

1ρ + + gp

gV z h K K K K g

V2, ,f valve elbow elbow exit

2 22

2 ��

2

r r= + + + + + + +c c6 @

gp

gV z2

1 12

1ρ + + 0 0 z h K K K K gV2, ,f valve elbow elbow exit2 45 90

2

= + + + + + + +c c6 @

or gp1

ρ z z gV f D

Lg

V2 22 1

12 2

# #= − − + c m

K K K K gV2, ,valve elbow elbow exit��

2

+ + + +c c6 @

500 400 .2.55 . . .

2.552 9 81 0 0315 0 05

12002 9 81

2 2

## #

#= − − +] ]

cg g

m

. . . ..8 5 2 0 2 4 0 3 1 2 9 81

2 55 2

# ##

#+ + + +] ]]

g gg

6 @

gp1

ρ 500 400 0.331 250.56 3.68= − − + + 353.9 354,=

p1 �� .�1 ��gr # # #= = 346 3.46Pa MPa5794 ,=






The energy equation for a control volume between two points

gp

gV z�

��

��

�ρ α+ + gp

gV z h� L

��

��

�r a= + + +

Here �� andV z p p patm� � � �= = = =

So z� gV h2 L2

22

a= + gV K g

V2 2L2

22

22

a= + h K gV�L L

�

=

V� Kgz2

L2

1

a= +

Then the flow rate becomes.

vo A Vpipe �#= DK

gz4

2L

2

2

1pa= +

(a) For Flanged smooth bend ( . )K 0 3L =

vo ( . )

. ..

40 03

1 05 0 32 9 81 52

##

# #p= +

0.006 3 / 6.03 /m s L s0 3= =(b) For Miter bend without vanes ( . )K 1 1L =

vo ( . )

. ..

40 03

1 05 1 12 9 81 52

##

# #p= +

0.00478 /m s3= 4.78 /L s=


From continuity equation for incompressible flow.

A V� � A V� �=

V� ��

AA V

DD

VDD V

��

�

��

��

��

�pp

# # #= = == ;G E

( . )( . )

10 2.5 /m s0 120 06

2

2

#= =

The loss coefficient for sudden expansion

KL ( . )( . )

.AA

DD1 1 1

0 120 06

0 5625argl e

small2

2212 2

2

2 2

= − = − = − =; ; =E E G





and Head loss hL ��

��K gV� � ��

��L

��

## #= = =

The energy equation for the expansion section

gp

gV z�

��

��

�ρ α+ + gp

gV z h� L

��

��

�r a= + + +

Since z z� �= , then

gp

gV�

��

��

ρ α+ gP

gV h� L

��

��

r a= + +

p� p V V gh� L��

��

�

r a a= + − −: D

300 100. ( ) . ( . )

0 21 06 10 1 06 2 52 2

# ##= + −

;

. .9 81 2 87 10001

# #− @

321.53 kPa= 322 kPa,


The average discharge velocity through the orifice at any given time at any

time, in general V� Kgz

12

L= +

We denote the diameter of the orifice by Do and the diameter of tank by D .The amount of water flows through the orifice during a time interval dt is

dv vdt=o o DK

gz dt4 12o

L

2p= + ...(i)

and the decrease in the volume of water in the tank

dv ( )Area dz= − D dz42p= − ...(ii)

From mass conservation, equation (i) must be equal to equation (ii)

DK

gz dt4 12o

L

2π+ D dz4

2p=− ( ve− sign shows decrease in volume)

dt ( )DD

gK z dz�

� �

o

L�

��=− + −

Then the draining time is

dt�t

tf

=# ( )

DD

gK z dz�

� �

o

L

z�

��

�

�=− + −

=#

tf DD

gK z�

� � �

o

L�

��

�

�#=− +

6 @ ( � )DD

gK

��

o

L�

�

# #=− + −

�DD

gK

��

o

L�

�

# #= + DD

gK� �

o

L�

�

#= +





2( . )( . )

..

0 14 5

9 811 0 5

2

2

# #= + 1584 26.4sec min= =

FM 6.40 Option (A) is correctSince flow rate

vo C A p��

d t �r bD

#=−_ i

where ��Dd

��β = = =

0.605 .998 1 0.54 0 05 2 75000

42 #p

# # #=−]]

gg6 @

0.0150 / 54 /m s m h3 3-=


The velocity Vt .

. �.�� m sAv

� ��

t �#

p= = =o

] g

Since head loss due to pressure difference is known as non recoverable head loss. So

plossΔ K V� t�r

# #= 1.8 .2998 7 64 2

# #= ] g

52427 52400 52.4Pa kPa-= =

FM 6.42 Option (D) is correct.For a pressure drop of pΔ across the orifice plate, the flow rate is expressed as

vo ( )

A C p��

o d �r bD=−

...(i)

Since β �.�Dd d

��= = =

d 30 cm=

Then Ao ( . )

0.07069 md4 4

0 32 22# #p p= = =

Therefore from equation (i)

.0 25 (0.07069) (0.61)( ) 1 0.6

p998

24

#

D# #=

− ] g6 @

pΔ 14600 /kg m s2−= 14.6 kPa=

FM 6.43 Option (D) is correct.The head loss between two measurement section is estimated by energy equation.

hL gp p

gV V

�w

� � ��

��

r= − − − gp

gV V

�w

��

��

rD= − − z z� �= ...(i)

Since for constant volume flow rate,

V D��

�

#π V d

��

�p=

V� dD V

�

�#= b l

V� ( . ). �.�� m sA

v��

�� c

�#

#p

= = =o

Substituting in equation (i)

hL gp

gdD V V

gp

gdD V

� �

�

w w

�

�

�

��

�

��

#

r rD D= −

−= −

−b bl l; ;E E





hL . .

.

998 9 8114600

2 9 813050 1 1 27

42

# #

#

= −−b l; E

1.49 0.5521 0.9379= − = 0.940 m,


Since β 0.6 Dd d

5= = = or �.� � �c�d #= =

Now vo . C A p60

0 33412

d 4r bD

# #= =−_ i

C dgh

� �

�d

water

Hg water��

pr br r

# #=−

−

_

^

i

h

where p ghHg wρ ρΔ #= -^ h

or .60

0 334 0.613 .998 1 0.6

. h4 0 03

2 13550 998 9 814

2 # #p# #=

−−

]]

^g

g

h

6 @

.60

0 334 0.00044 16.8 h4#=

0.00557 0.0074 h=

or h .. 0.57 m0 0074

0 00557 2

= =b l 57 cm=

FM 6.45 Option (B) is correct.The pressure drop across the venturi-meter is

pΔ p p gh ghw air� � r r= − = −

( )gh gh1w air airair

wr r r rr= − = −b l

Then flow rate vo ( )

( )( )

A Cp p

A Cgh

��

�

� �o d

airo d

air

air

wair

��

�

#

r b r brr r

=−− =

−

−b l

( )

A Cgh

�

� �o d

air

w

�brr

=−

−b l

where Ao ( . )

0.002827 md4 4

0 062 22#p p= = =

and β �.��Dd

��= = =

Thus vo 0.002827 0.981 0.40

. . .2 9 81 0 4 1 2041000 1

4

# #

# #=−

−

]

b

g

l

6 @

0.2265 /m s3=Then the maximum mass flow rate that venturi-meter can measure is

m vr=o o 1.204 0.2265#= 0.273 /kg s=

FM 6.46 Option (A) is correct.The upstream density is

1ρ �.�� mRTp

��

�

� �

#= = =

and β Dd

��

��= = =

The pressure difference measured by the mercury manometer





p p� �− � �� gh �� Hg airr r #= − = −^ ]h g

pΔ 49200 Pa=Now the compressive venturi formula thus predicts :

mo C YAp p

�

�d t �

� � �

br

#=−

−^ h

0.985 0.76 . .4 0 04

1 32

2 1 48 4920024#

#p# # #=

− b

]

l

g9 C

0.40 /kg s=

FM 6.47 Option (C) is correct.The flow velocity

V .

. �.�� m sAv

D�

��

��

� �#

#p p

= = = =o] g

Also Vt .. �.�� m sA

v��

�� t

�#

# ,p

= =o] g

For �.�Dd

��β = = = and �.�K = , the head loss across the office is

hΔ �.� � �.��.�� .��mh h K g

V�

�t

� �

�

# ##

= − = = =] g; E

Hence h� 0.58 0.58 1 1.58 mh1= + = + =Then the piezometer change between (2) and (3) is due to friction loss

h h hf3 2− = f DL

gV2

2

# #= 0.023 . .. 0.115 m0 05

52 9 81

0 99 2

## #= =] g

or h3 0.115 1.58 0.115h2= + = + 1.695 1.7 m,=


From the pressure drop relation, the flow rates are

paΔ 2� ��dL v�2�a

a a�p

m#= =

o

or vao .. 0.0027 /m s128 0 104 76

21 10 0 083 43

# #

# # #p= =] g

and pbΔ 2��dL v�2�b

b b�p

m= =o

or vbo .. 0.0005 /m s128 0 104 61

21000 0 05 43

# #

# #p= =] g

For parallel pipe system

vo v va b= +o o

0.0027 0.0005= + 0.0032 /m s3=

FM 6.49 Option (D) is correct.For parallel system of pipe, head loss for each pipe must be same.When the minor losses are disregarded, head loss is





hL fDL

gV�

�

= c m f DL

g Dv

��

��

�

# # # p= o

> H

f DL

D gv

��

� �

�

#

#p# #= o

fgD

L v8 2 5

2

##

p= o

Thus vo fLh g D8

.L2

2 5#p#=

When the pipe length, friction factor and head loss is constant, the flow rate

becomes proportional to the D �� in parallel connection.

vo kD .��= k= constant of proportionality

Let diameter of pipe B DB= and diameter of pipe A DA= .

Therefore vBo � �k D .B

��=

vAo � � �. �k D k D��. .A B

�� = = � �gi�enD D�AB=

vAo 0.177 ( ) 0.177k D v.B B

2 5# #= = o

Hence the flow rate of pipe A is decreased by a factor of 0.177.


For parallel pipe system the head losses are the same for each pipe

hf gdfL v

gdfL v

gdfL v8 8 8

215

1 12

225

2 22

215

3 32

p p p= = =

o o o

v v v1 � �+ +o o o 0.056 /m s3= ...(i)

or hf . ..

. ..v v

9 81 0 18 0 0275 900

9 81 0 128 0 0275 800

2 512

2 522

# #

# # #

# #

# # #p p

= =o o

] ]g g

. .

. v9 81 0 08

8 0 0275 6002 5

32

# #

# # #p

=o

] g

hf 204501 73053 416059v v v12

22

32= = =o o o

or v��o 2.8v v73053

20450112

12= =o o or 1.�7�v v� 1=o o

v��o 0.492v v416059

20450112

12= =o o or 0.701�v v� 1=o o

Now from equation (i),

1.�7� 0.701�v v v1 1 1+ +o o o 0.056=

or 3.374v1o .0 056= ] g

v1o ..

3 3740 056= or 0.01�� m sv1

�=o

and v�o =1.673 1.673 0.0186 0.0277 /m sv13

#= =o

v�o 0.7014 0.7014 0.0166 0.0116 /m sv13

#= = =o


Since hf . .. .

g df L v8

9 81 0 18 0 0275 900 0 0166

215

1 1 12

2 5

2

# #

# # #

# #

# # #

p p= =

o

]]g

g

56.35 m=Then pressure drop

p ghfρΔ = 998 9.81 56.35# #=

551 . 550 kPa687 9 ,=





FM 6.52 Option (A) is correct.The head loss is the same through pipes AC and BC (parallel system)

gp

ρΔ h h h hfA fC fB fC= + = +

f dL

gV f d

Lg

V fdL

gV fd

Lg

V� � � �

A

A

B

C

B

B

C

C

� � � �

# # # #= + = +c c c cm m m m

998750000 . . . .

� �0 022 0 08250

2 0 022 0 08150

2� �2 2

# # # #= +b bl l ...(i)

and 998750000 . . . .

� �0 022 0 08100

2 0 022 0 08150

2� �2 2

# # # #= +b bl l ...(ii)

From equation (i) and (ii),

751.50 34.375 20.625� �� 2 2= + ...(iii)

751.50 13.75 20.625� �� 2 2= + ...(iv)

From equation (iii) and (iv),

VA� .

. ..

. . .�34 375

751 50 20 62534 375

751 50 20 625 5 35�2 2

# #= − = − ] g

VA 2.16 /m s=

and VB� .

. . .13 75

751 50 20 625 5 35 2#= − ] g

VB 3.42 /m s= Now for parallel pipe A and B ,

vABo v vA B= +o o

vABo V A V AA A B B= + V V AA B= +^ h D DA B=

. . .2 16 3 42 4 0 08 2p# #= +^ ]h g

0.0280 /m s3=

And for Pipe in series

vABo �.�� m svC�= =o

Hence the total volume flow rate

v v vAB C= =o o o 0.0280 /m s3=

101 /m hr3-

FM 6.53 Correct option is (B)Let roughnessh = . Thus h sd=

Where sδ uV��= ...(i)

And u� lD p

�

� �w

��

#rt

rD= =c dm n where l

D p�wτ Δ=

u� ��8

/2 1 2

= c m since p fDl V�

� �# ρΔ =

Hence u� 80.0125 (2)

0.0791 /m s2 /1 2

#= =; E

and from eq (i), sδ .( . )

2.31 10 0.0231m mm0 07915 3 65 10 7

5##= = =

−−

If the roughness element is smaller than 0.0231 mm it lies within the laminar sublayer.





FM 6.54 Correct option is (D)

Assume that fluid flows from A to B and A to C .

v�o v v� �= +o o

or (0.1) V42

1#π (0.08) (0.08)V V4 4

22

23# #

p p= +

V� 0.64 0.64 . ( )V V V V0 642 3 2 3= + = + ...(i)

For fluid flowing from A to B with p p �A B= = and V V �A B= =

zA z f Dl

gV f D

lg

V� �B �

�

� ��

��

� ��

# #= + +

60 20− 0.015 . . 0.020 . .V V

0 1200

2 9 81 0 08200

2 9 8112

22

# ## # # #= +

or 40 . .V V1 529 2 5512

22= + ...(ii)

Similarly, for fluid flowing from A to C with �p pA C= = and V V �A C= =

zA z f Dl

gV f D

lg

V� �C �

�

� ��

��

� ��

= + +

60 0− 0.015 . ( . )0.020 . ( . )

V V0 1200

2 9 81 0 08400

2 9 8112

32

## # #= +

or 60 . .V V1 529 5 1012

32= + ...(iii)

Solve equation (i), (ii) and (iii) for ,V V� � and V�. From equation (i) and (iii)

60 1.529 (0.64) ( ) 5.10V V V22 3

232

#= + + 60 0. ( ) .V V V626 5 102 3

232+ +

.95 8 ( ) .V V V8 142 32

32= + + ...(iv)

Subtract equation (ii) from equation (iii)

60 40− 5.10 2.55V V32

22= − or .V V� ��

�= − ...(v)

Thus, from equation (iv) and (v)

0 8.14 ( . ) 95.8V V V2 7 8432

32

32= + − + −

This can be simplified to

.V V2 2 7 843 32 − . . V103 6 11 14 3

2= − ...(vi)

Squaring both the sides and rearrange to get

�. ��.�V V�6��

��− + 0= which can be solved by the quadratic formula

to give

V��

. ( . ) .2

19 63 19 63 4 92 52! #= − −

.11 77= or .7 86

Thus V� 3.43 /m s= or �.�0 �m sV�=





Note : The value �� m sV�= is not a solution of the original equations,

equation (i), (ii) and (iii). With this value the right hand side of equation (vi) is

negative (i.e . . . . ( . ) .V103 6 11 14 103 6 11 14 3 43 24 532 2− = − =− ). As seen from the

left hand side of equation (vi), this cannot be. This extra root was introduced

by squaring equation (vi).

Thus v�o (�.��) �.�� .�� m sA V ��

# #p= = =

Also, from equation (iii)

60 1.529 5.10 (2.80)V12 2

#= + or �.6� �m sV�=

or v�o (0.�0) �.6� 0.0�� m sA V ��

# #p= = =

and from equation (i), we get

3.62 0.64 0.64 2.80V2 #= + or 2.�6 �m sV2 =

or v2o (0.0�) 2.�6 0.0�� m sA V � �2 22 3p

# #= = =

FM 6.55 Option (A) is correct.We take point (1) at the free surface of tank and point (2) at the exit of the pipe. Then, the energy equation between these two points.

gp

gV z2

11

12

1ρ α+ + gp

gV z h2 L

22

22

2r a= + + +

Since ,p p patm1 2= = ,V 01 , 0,z2 = and V 02 = (Velocity head negligible)

Above equation becomes

z1 hL= and h hL=where h is the liquid height in the tank at any time t .

Now hL f dL

gV2

2

#=

For fully developed laminar flow

f /Re Vd

64 64n= =

Thus hL Vd d

Lg

V6�2

2

n# #=

dL

gV6�22

n#=

The average velocity

V Av

dv

dv

�

�c

2 2

p p= = =o o o

hL d

Lg d

v6�21 �

2 2n

p# #= o

dL

d gv

g dLv6�

2� 12�

2 2 �#

np p

n#= =o o

h hL = g d

Lv12��p

n= o ...(i)

From mass conservation, above equation must be equal to change of liquid height in the tank.

vo Ddtdh

42p=−


h g d

L Ddtdh12�

��

2

pn p

# #=− gdLD

dtdh32

�

2n=−





dt gdLD

hdh��

�

�n#=−

Integrating from t �= when h H= to t t= when h h=

t �ngdLD

hH��

�

�n= b l


The bottom side of the triangle is

2 2 40 2.57sin cmc,#=

and A 2.57 cos21 2 40c# #= ] g 1.97 cm2=

Perimeter 2 2 2.57 6.57 cm= + + =Then Hydraulic diameter

Dh .. 1.20perimeter cmA4

6 574 1 97#= = =

Re VDD

hh m

r= .. 2010 104

870 2 0 0120# # ,= (Laminar flow)

Then f . . 0.263Re52 9

20152 9= = =

Hence pΔ f DL V�h

�r# # #=

. ..0 263 0 012

0 62

870 2 2# # #= ] b b ]g l l g

23000 23Pa kPa- =

FM 6.57 Correct option is (A)

V ( . )

�.�� m sAv

� ��

�

�

#

#p= = =

−o





So that Re �� VD VD�� <�

#

#m

rn= = = =− ρ

μ ν=

The flow is laminar and flow rate is given by ( 90cθ = )

vo ( )

lp l D128

4

mp gD= −

or p p pD

lv l�� π

μ γΔ = - = +o

...(i)

Thus γ . . �.�� .�� .�� mS G � � ��# #= = =g

and μ ( . . ) 2.2 10 0.87 999 0.191 /N s mS G H O4 2

2nr n r # # # −= = = =−

Equation (i) gives

pΔ ( . )

. 8.53 10 40 020

128 0 191 4 4 104

43

#

# # # ##p #= +

−

. 10 / . /N m kN m1 119 111 95 2 2#= = ...(ii)

From manometer equation,

p� p h h hm� � �g g g= − + −

Where mγ g. . �.� �.�� .�� mS G � �m�

� #= = =and h� h l h�= + − or h h h l� �+ = +Thus p p� �− g( ) ( )p h h h h lm m� � g g g gD= = + − =− − +

...(iii)Combine equation (ii) and (iii), we get

.111 9 (12.74 8.53) 8.53 4h #=− − +or h 18.5 m=−Note: Since h �< , the manometer is displaced in the direction opposite that shown in the original figure.

FM 6.58 Option (B) is correct.The divided cross section of the pipe into 1 cm thick annual regions is shown in table.The flow rate is to be determined by using midpoint velocity values for each section. Therefore

vo � �V dA V r ravg cA

avg out in� �

c

pΣ= = −#

. . (0.01) 0 . . 0.02 0.0126 4 6 1

26 1 5 22 2 2p p# # #= + − + + −b b ] ]l l g g6 6@ @

. . . . . . 0.04 0.0325 2 4 4 0 03 0 02 2

4 4 2 0 2 22 2p p# # #+ + − + + −b ] ] b ] ]l g g l g g6 6@ @

. 0.05 0.04022 0 0 2 2p# #+ + −b ] ]l g g6 @

vo 0.0297 /m s3=

FM 6.59 Correct option is (C)

For laminar flow Re 2100# or VD ��#μρ





Thus the minimum h is h �= (no flow) and the maximum h is for Re 2100= .

2100 .. .V

0 19 818900 0 023# #

=b l

gρ γ=

or V 10.06 /m s=For the flowing fluid, Bernoulli’s equation gives

pg

V z��

�

�γ + + pg

V z fDl

gV

� ��

�

�

�

g= + + + z z� �=

and V V V� �= =

Thus p p� �− f Dl

gV�

�

# # g=

(i)

And for laminar flow

f Re64= or .f ��

�� = =

From equation (i),

pΔ �.�� ..

.( . )

��p p ��

� ��

� �

�

## # #= − =

304 /N m39 2=From manometer equation, we get

( ) . .p H h S G h H� �oil oil� �g g+ + − −g p�= p p p� �Δ = - ( . . )S G hH O2 g= −g

or h ( )

0.5 m7 9800 8900

30439 10#

= − =

Hence 0.5 mh0 10# #


We take point (1) at free surface of the tank and point (2) at the reference level at exit. By applying energy equation for a control volume between these two points

gp

gV z h2 pump

11

12

1ρ α+ + + gp

gV z h h2 turbine L

22

22

2r a= + + + +

Since p p patm1 2= = , �,z2 = h �turbine = , and �V1 ,

z1 gV h2 L2

22

a= + hL fDL

gV2

22

#=

So z1 gV f D

Lg

V2 22

22

22

a #= +

V2 /fL Dgz2

2

1

a=

+ ^ h

/fL Dgz

12=

+ ^ h 12α =





where z is the water height relative to the center of the orifice at that time.

Now, the flow rate of water through the pipe during a time interval dt is

dvo � �

vdt A V dt DfL Dgz dt�

�c

o

o�

�p# #= = =

+o

^ h

where Do is the diameter of orifice.

From conservation of mass

Flow rate of water through the pipe tandecrease in the volume of water k=

1 /

DfL Dgz dt4

2o

o

2π# + ^ h

� �A dz,c tank= − D dz42

#p=−

dt � �

DD

gzfL D

dz�o

o�

�

=−+ ^ h

� �

DD

gfL D

z dz��

o

o�

��

#=−+ −^ h

By integrating above equation from t �= when z z�= to t tf= when z �=

(completely drained tank)

dtt

t

�

f

=# �

DD

gfL D z dz�

� �

o

oz z

�

��

�

�

#=− + −

=^ h #

tf �DD

gfL D z�

�

��

o

o

z

�

� ��

�

#=− + ^ h

�DD

gz fL D� �

o

o�

��#= − ^ h$ .


tf ( . )10

.

. .0 039 81

2 2 1 0 022 0 0325

2

2# ##= +]

bg

l

311965 seconds= 86.6 hours,

***********

FM 7EXTERNAL FLOW

FM 7.1 The resultant force of 00 N4 is acting on a body at angle of 30c with the direction of flow as shown in figure below. What will be the drag and the lift forces acting on the body, respectively ?

(A) 0, 600 N (B) 300 ,N 520 N

(C) 520 ,N 300 N (D) 600 ,N 0

FM 7.2 For un-powered flight (i.e. for which the lift, drag and weight forces are in equilibrium) the glide slope angle θ will be

(A) tan CC

L

D1− c m (B) tan CC

D

L1− c m

(C) tan C CD L1− ^ h (D) tan C

C2L

D1− c m

FM 7.3 A 3 m long and 0.5 m diameter cylindrical tank is installed on top of a vehicle. The average speed of vehicle is to be 26.5 /m s and density of air is 1.028 /kg m3

. What will be the drag force acting on the tank when (a) the frontal area is the front and back of the tank ( .C 0 �D = ) and (b) the frontal area is the sides of the tank ( .C 0 �D = ), respectively ?(A) 64 ,N 0 (B) 64 ,N 433 N

(C) 433 ,N 64 N (B) 0, 433 N

FM 7.4 Water flows at 20 Cc with 25 /cm s, encounters a creep motion. If a smooth ceramic sphere ( . . �.�S G = ) is immersed in the flow of water, what will be the sphere diameter ?(A) 40 mμ (B) . m0 04 μ(C) 4 mμ (D) 0.4 mμ

FM 7.5 A laminar boundary layer formed on one side of a plate of length l and produces a drag FD . For the same upstream velocity, how much must the plate be shortened if the drag on the new plate is to be ��FD ?

(A) l l��= (B) l l��=

(C) l l��

�= (D) .l l0 0��=

FM 7.6 A hydrofoil of 0.5 4m m# cross section moves at 14.5 /m s in sea water. If hydrofoil 1025 /kg m3ρ = and 0.00107 /kg m sμ -= at 20 Cc . What will be its drag for a smooth wall ?(A) 1320 N (B) 3638 N

(C) 251 N (D) 125 N





FM 7.7 The top surface of the passenger bus is 4.0 m wide and 8 m long and the bus is moving at a velocity of 19.5 /m s. If the outdoor air is 1.184 /kg m3 dense and kinematic viscosity of air is 1.562 10 /m s5 2

#− , the drag force acting on the top

surface of the bus is(A) 22 N (B) 20 N

(C) 21 N (D) 42 N

FM 7.8 A 6 /m s stream of SAE 30 oil at 20 Cc ( 891 /kg m3ρ = and 0.29 /kg m sμ -= ) is past over a thin flat plate of 55 110by cm. If the stream is parallel to the long side, the total friction drag is(A) 30 N (b) 362 N

(C) 181 N (D) 60 N

Common Data For Q. 9 and 10A bicyclist is riding down a hill with a slope of 12c into a head wind (measured with respect to the ground). The mass of the cyclist and the bicycle is 95 kg. The air density is to be 1.25 /kg m3. The rolling resistance and friction at the bearings are neglected.

FM 7.9 If the bicyclist has a frontal area of 0.45 m2 and drag coefficient of 1.1 in the upright position, the terminal velocity of the bicyclist is(A) 29 /km h (B) 195 /km h

(C) 62 /km h (D) 90 /km h

FM 7.10 If the bicyclist has a frontal area of 0.4 m2 and a drag coefficient of 0.9 in the racing position, the terminal velocity of the bicyclist is(A) 34 /km h (B) 229 /km h

(C) 73 /km h (D) 106 /km h

FM 7.11 In a laminar boundary flow the net drag on one side of the two plates having the cross section �l l �# parallel to the free stream as shown in figure, is FD . What will be the drag (in terms of FD) on the same two plates when they are connected together as indicated in figure ?

(A) 0.�0�F FD D=l (B) .F F1 �1�D D=l

(C) 0.�3F F0D D=l (D) 0.F F��3D D=l

FM 7.12 For a laminar flow of a fluid over a flat plate if the free-stream velocity of the fluid is doubled, the drag force on the plate is (Assume the flow to remain laminar)(A) �.�3F FD D� 1= (B) F FD D1 �=(C) .F F��3D D1 �= (D) .F F1 1�9D D� 1=





FM 7.13 Wind storms sometimes blow off empty containers of trailer tracks. The dimensions of one type of container are as shown in figure below and the air density is 1.1 /kg m3 .What amount of minimum wind velocity normal to the side of the container would be required to blow off the container ? ( .C ��D = )

(A) 58 /km h (B) 0.715 /km h

(C) 181 /km h (D) 50 /km h

FM 7.14 A 15 m long and 3. m4 wide railway wagon is shown in figure. If the weight of the wagon is 250 kN, the wind velocity necessary to blow over the wagon is (

.C ��D = )

(A) 35 /m s (B) . /m s17 5

(C) . /m s43 75 (D) . /m s26 25

FM 7.15 A 50 mm diameter solid plastic sphere is dropped into water at 20 Cc . If the density of sphere is 1150 /kg m3, the terminal velocity of the sphere in water is (

.C ��D = )(A) 4.43 /m s (B) 0.443 /m s

(C) 0.0443 /m s (D) 44.3 /m s

FM 7.16 A 65 kN weighs tractor-trailer truck is coasting freely with no brakes down an 8c slope at 1000 m standard altitude ( 1.112 / )kg mair

3ρ = . The truck has drag area �.��mC AD

�= . If the rolling resistance is 120 N for every /m s of speed, the terminal coasting velocity will be(A) 33 /m s (B) 66 /m s

(C) 30 /m s (D) 17 /m s

FM 7.17 A dust particle of density 1800 /kg m3 is unsettled during high winds and rise to a height of 350 m. It takes 43 min to fall back to the ground in still air at 1 atm and 15 Cc . If stokes law is applicable, what will be the diameter and velocity of the dust particle, respectively ? ( 1.802 10 / )kg m sair

5μ # -= -

(A) 0.05 ,mm 8.14 /m s (B) 0.05 ,mm 0.136 /m s

(C) 0.5 ,mm 8.14 /m s (D) 0.5 ,mm 0.136 /m s

FM 7.18 A parachutist jumps from a plane, using an 8.5 m diameter parachute in the standard atmosphere. The total mass of parachutist and parachute is 135 kg. If fully open parachute is in quasi steady motion, the time to fall from 2000 m (

1.�� kg mair�ρ = ) to sea level ( 1.�� kg mair

�ρ = ) will be (take .C 1 �D = )





(A) 321 s (B) 168 s

(C) 355 s (D) 337 s

FM 7.19 A heavy steel sphere ( �� S G = ) of 3 cm diameter is attached to a string and should hang at an angle θ when immersed in a stream of velocity �� m sV = of air at standard sea-level ( 1.225 /kg m3ρ = , 1.78 10 /kg m s5μ # -= - ) as shown in figure below. Neglecting the string drag, the angle θ is ( .C ��D = )

(A) 18c (B) 72c

(C) 34c (D) 36c

FM 7.20 A helium-filled balloon at 20 Cc and 1 atm is connected with a string of negligible weight and drag as shown in figure below. The diameter is 50 cm and the balloon material weighs 0.2 N, not including the helium. If the helium pressure is 120 kPa and the tilt angle 87.3cθ = , the airstream velocity U will be (take .C ��D = and

�� kg �RHe −= )

(A) 20 /m s (B) 30 /m s

(C) 15 /m s (D) 40 /m s

FM 7.21 A shortwave radio antenna is constructed from circular tubing as shown in figure. If the wind is blowing with 10 /km hr0 , the wind force on the antenna is(Take

.C ��D = )

(A) 90 N (B) N135

(C) N180 (D) N225





FM 7.22 A fully loaded 400 ton of jet airplane takes off at a speed of 250 /km h. Assume each passenger with luggage is 140 kg and the wing and flap settings air maintained the same. The take off speed when the airplane has 150 empty seats, is(A) 243.5 /km h (B) 24.3 /km h

(C) 257 /km h (D) 25.7 /km h

FM 7.23 Wind is blowing across the 6 mm diameter wire of transmission line at a speed of 40 / .km h The drag force exerted on the wire, when air is at 1 atm and 15 Cc is 55 N. What will be the length of the wire ?( 1.47 10 /m s5 2ν #= - and .C 1 0D = )(A) 1.21 m (B) 121 m

(C) 12.1 m (D) 0.121 m

Common Data For Q. 24 and 25The wing loading on an aircraft is defined as the aircraft weight divided by the wing area. An aircraft has a mass of 2800 kg, a wing area of 30 m2 and a lift coefficient of .0 45 at take off settings.

FM 7.24 The takeoff speed of this aircraft at sea level at standard atmospheric conditions and the wind loading respectively, are (density of standard air 1.225 /kg m3= )(A) 66 / ,km h 94 /N m2 (B) 207 / ,km h 916 /N m2

(C) 916 / ,N m2 207 /km h (D) 94 / ,N m2 66 /km h

FM 7.25 The required power to maintain a constant cruising speed of 300 /km h for a cruising drag coefficient of .0 035, is(A) 186 kW (B) 37.2 kW

(C) 372 kW (D) 18.6 kW

FM 7.26 Two bike racer rides at 30 /km hr through still air. The second racer drafts closely behind the first racer rather than riding alongside. If any forces other than aerodynamic drag is neglected, what percentage will be the power required to overcome aerodynamic drag for the second racer ? (Take 0.��CDND = and

0.50CDD = )(A) . %43 2 (B) . %3 24

(C) . %4 32 (D) . %32 4

FM 7.27 A building is approximately 87.5 m wide and 154 m tall. If the velocity profile against the building is a typical profile for an urban area as shown in figure and the wind speed halfway up the building is 20 /m s, what will be the drag on the building ? (Take .C 1 �D = )

(A) 5.21MN (B) 41.7 kN

(C) 3.13 MN (D) 4.17 MN





FM 7.28 For the triangular two-dimensional object as shown in figure, the lift and drag coefficients based on formal area respectively, are (Neglect the shear forces).

(A) 0, 2.125 (B) 1.70, 0

(C) 0, 1.70 (D) 2.125, 0

Common Data For Linked Answer Q. 29 and 30A 2 m high, 4 m wide signboard is attached to a 4 m wide, 0.15 m high rectangular concrete signpost (density 2300 /kg m3= ) by two 5 cm diameter, 4 m high (exposed part) poles as shown in figure below. The signboard is to withstand 150 /km h winds from any direction. The density of air to be 1.30 /kg m3, the drag coefficient for circular rod is .C 0 �D = , for rectangular signboard .C 2 0D = and the flow is turbulent.

FM 7.29 The drag forces acting on the signboard and 2 poles respectively, are(A) 18.06 ,kN 136 N (B) 180.6 ,kN 13.6 N

(C) 136 ,N 18.06 kN (D) 180.6 ,kN Zero

FM 7.30 What will be the minimum length L of the concrete block for the panel to resist the winds ?(A) 18.5 m (B) 37.1 m

(C) 8.5 m (D) 3.71 m

Common Data For Linked Answer Q. 31 and 32A sign that consists of a 0.45 m high, 0.6 m wide and 0.6 m long rectangular block mounted on top of a cab car. The sign has a frontal area of 0.45 m by 0.6 m from all four sides. The cab car is driven at an average speed of 50 /km h. The density of air is 1.25 /kg m3 and .C 2 2,D rect = .





FM 7.31 The drag force acting on the sign is(A) 928 N (B) 1237 N

(C) 95 N (D) 72 N

FM 7.32 The car is driven 60000 km in a year and the overall efficiency of the engine is 28%. If the density, unit price and heating value of gasoline to be 0.75 /kg L, Rs. 25/L and 42000 /kJ kg, respectively, the increase in the annual fuel cost of car due to this sign is(A) No increment (B) . 2225/Rs year

(C) .12225/Rs year (D) .1 /Rs year225

FM 7.33 A ship is encrusted with tide, the ship requires 7000 hp to overcome friction drag when moving in seawater ( 1025 /kg m3ρ = and 0.00107 /kg m sμ -= ) at 20 Cc. The ship is 150 m long and has a wetted area of 5000 m2 and neglecting the wave drag. How fast would the ship move with the same power if the surface were smooth ?(A) 11 /m s (B) 5.5 /m s

(C) 22 /m s (D) 1.1 /m s

FM 7.34 A 2 m diameter and 40 m high pole is subjected to 22 /m s storm winds at sea level ( 1.225 /kg m3ρ = and 1.78 10 /kg m s5μ # -= - ). What is the estimated wind-induced bending moment about the bottom of the pole ? (take .C ��D = )(A) 760 kN m− (B) 380 kN m−

(C) 95 kN m− (D) 190 kN m−

FM 7.35 A small airplane of 10.2 m2 wing area and 6.22 kN of weight has the cruising speed of 210 /km hr. If the engine delivers 150 kW at this speed and %60 of this power is lost in propeller and to overcome body resistance, what will be the drag coefficient of the wing ?

(A) 4.82 (B) 0.0482

(C) 0.00482 (D) 0.482

Common Data For Q. 36 and 37The resistance to motion of an automobile consists of rolling resistance and aerodynamic drag. The rolling resistance is the product of weight and the coefficient of rolling friction . The total mass of the automobile is 950 kg and it has a frontal area of 1.8 m2. The drag coefficient is 0.32 and the coefficient of rolling friction is 0.04. The maximum power that the engine can deliver to the wheels is 80 kW and air density is 1.20 /kg m3.

FM 7.36 The speed at which the rolling resistance is equal to the aerodynamic drag force, is





(A) 38 /km h (B) 118 /km h

(C) 11.8 /km h (D) 335 /km h

FM 7.37 The maximum speed of this automobile is(A) 200 /km h (B) 98 /km h

(C) 156 /km h (D) 242 /km h

FM 7.38 A 50 cm diameter television-transmitting antenna is on top of a 2.5 m high pipe, which is on top of a tall building and is subjected to normal winds up to /km h150 at 10 Cc and 100 kPa. What will be the drag force acting on the antenna and the bending moment at the bottom of its pipe, respectively ?

(A) 231 ,N 635 N m− (B) 231 ,N 0

(C) 635 ,N 231 N m− (D) 635 ,N 0

FM 7.39 An 0.80 m diameter, 1.2 m high garbage can is found tipped over in the morning due to high winds velocity of 135 /km h during the night. If the air density is to be 1.25 /kg m3 and the drag coefficient of the can to be 0.7, what will be the mass of the can ?(A) 90 kg (B) 443 kg

(C) 180 kg (D) 886 kg

FM 7.40 A tractor-trailer truck has drag area �mC AD�= without the deflector and

�.7 mC AD�= with the deflector added. Its rolling resistance is 112 N for each

/m s of speed. If the truck moves at 24.5 /m s, how much power is reduced due to deflector added ?(A) %17 (B) No Reduction

(C) . %4 25 (D) . %8 5

Common Data For Linked Answer Q. 41 and 42

An airplane weighs 180 kN and has a wing area of 160 m2. The plane is designed

to land at .V V��land stall= and the coefficient of lift and drag for this plane are

�.7�C ,maxL = and .C 0 0�7D = , respectively.

FM 7.41 What is the proper landing speed ?(A) 27 /m s (B) 28 /m s

(C) 32 /m s (D) 39 /m s

FM 7.42 What power is required for take off at the same speed ?(A) 187 kW (B) 50 kW5

(C) 168 kW (D) 280 kW





FM 7.43 A thin flat plate of 50 50cm cm# in size is balanced by a counterweight as shown in figure below. If air at 101 kPa and 25 Cc flows downward over both surfaces of the plate with a free-stream velocity of 36 /km h, the mass of the counterweight that needs to be added in order to balance the plate is ( 1.562 10 /m s5 2ν #= - )

(A) 3.54 gram (B) 17.7 gram

(C) 7.1 gram (D) 71 gram

FM 7.44 Water droplets of 0.06 mμ diameter fall through the air under standard sea-level conditions ( 1.789 10 /N s m5 2μ # -= - ). If the drops do not evaporate and the Reynold’s number is less than unity, what will be the falling velocity of the droplets ?(A) 1.10 10 /m s5

#− (B) 1.10 10 /m s4

#−

(C) 1.10 10 /m s6#

− (D) 1.10 10 /m s7#

−

FM 7.45 A water boat is moving through water ( 999.1 / , 1.138 10 / )kg m kg m s2 3ρ μ # -= = - at 15 Cc with speeds up to 30 /km h. The bottom surface of the boat assume to be a 1 m wide, 3 m long flat surface. The friction drag exerted on the boat by water and power needed to overcome it respectively, are(A) 26.23 ,N 2.186 kW (B) 262.3 ,W 2.186 kN

(C) 262.3 ,N 2.186 kW (D) 26.23 ,kW 2.186 kN

FM 7.46 A smooth flat plate of cross section 6 4m m# is placed in water with an upstream velocity of �.� �m sU = . If the flow is laminar, what will be the boundary layer thickness and the wall shear stress respectively at the trailing edge of the plate ?(

1.12 10 /m s6 2ν #= - )(A) 0.0137 m, 0.0380 /N m2 (B) 0.0183 m, 0.0506 /N m2

(C) 0.183 m, 0.632 /N m2 (D) 0. 29 m02 , 0.0632 /N m2

FM 7.47 Water at 20 Cc and 1 atm ( 998 /kg m3ρ = , 0.001 /kg m sμ -= ) flows past a thin flat plate with 20 /m s. What will be the distance x from the leading edge at which the boundary layer thickness will be 1 mm ?(A) 0.00442 m (B) 0.442 m

(C) 4.42 10 m4#

− (d) 0.0442 m

FM 7.48 Standard air ( 1.562 10 / , 1.184 / )m s kg m5 2 3ν ρ#= =- flows steadily past over a flat plate with a velocity of 8 /m s. At approximately what location will the flow become turbulent and the boundary layer thickness at that location, respectively ?(A) 0.0678 ,cm 0.976 m (B) 0.678 ,cm 0.976 m

(C) 0.0976 ,m 0.678 cm (D) 0.976 ,m 0.678 cm





FM 7.49 A 1.22 2.44 m# sheet of plywood is put on a roof rack of a car as shown in figure below. The car speed is 15.5 /m s and the sheet is perfectly aligned with the airflow. What will be the boundary layer thickness at the end and the drag respectively, if the flow remains laminar ?

(A) 0.77 mm, 7.2 N

(B) 1.54 mm, 1.44 N

(C) 7.7 mm, 0.72 N

(D) 15.4 mm, 14.4 N

FM 7.50 A tractor-trailer truck has a drag area �.��mC AD�= , a rolling resistance of

8339 ,N a bearing friction resistance of 350 N and a maximum speed of 110 /km h on a level road during steady cruising in calm weather with air density of 1.25 /kg m3. Now a aerodynamic deflector is installed to the front of the rig to streamline the flow to the top surface and the drag area is reduced to �mC AD

�=. What will be the maximum speed of the truck with deflector ?(A) 119 /km h

(B) 32.07 /km h

(C) 33.09 /km h

(D) 116 /km h

FM 7.51 The pivot of a wind turbine with two hollow hemispherical cups is stuck as a result of some malfunction as shown in figure below. If the diameter of cups is 8 cm and centre to center distance is 50 cm. The maximum torque applied on the pivot for a wind speed of 15 /m s and air density of 1.25 /kg m3, is ( .C ��,D convex = and .C ��,D concave = )

(A) 0.283 N m− (B) 0.1412 N m−

(C) 0.565 N m− (D) 0.353 N m−

FM 7.52 Two 7.35 cm diameter base balls weighs 145 g each are connected to a rod of 7 mm diameter and 56 cm long as shown in figure below. At sea-level standard air and including the drag of the rod, what power is required to keep the system spinning at 42 /rad s ? ( �.��, �.�C C, ,D ball D rod= = )





(A) 6.3 W (B) 5.70 W

(C) 4.42 W (D) 0.58 W

FM 7.53 An iceberg ( 1026 /kg m3ρ = ) floats with approximately /1 7 of its volume in the air as shown in figure. If the wind velocity is U and the water is stationary, the speed at which the wind forces the iceberg through the water, is

(A) .U U��b =

(B) .U U��b =(C) .U U��b =

(D) .U U��b =

FM 7.54 A thin hinged rod of negligible weight and drag is used to connect two different size steel ( . . �.��S G = ) balls as shown in figure. A stop is used to restrict the counter clockwise rotation. What will be the sea-level air velocity U for which the rod will first begin to rotate clockwise ?( �.��C CDa Db= = )

(A) 50 /m s

(B) 64.5 /m s

(C) 73 /m s

(D) 57 /m s





FM 7.55 The pressure distribution on a cylinder is approximated by the two straight line segments as shown in figure. Neglecting the shear forces. What is the drag coefficient for the cylinder ?

(A) 1.91

(B) 1.43

(C) 2.4

(D) 0.95

FM 7.56 A rotary mixer consists of two 1 m long half-tubes rotating around a central arm as shown in figure below. The fluid is water at 20 Cc and the maximum driving power available is 20 kW. What is the maximum rotation speed ω in rpm ? (

.C 2 3D = )

(A) 148

(B) 206

(C) 74

(D) 103

FM 7.57 A flag of 2 m by 2.5 m is attached to the top of a 2 m0 tall and 0.12 m diameter flag pole. What will be the moment needed at the base of the pole to keep it in place in a 20 /m s wind ?( 0.0�C �D flag = , 1.2C ,D pole = )(A) 11194 N m−

(B) 6716 N m−

(C) 895 N m−

(D) 8955 N m−

FM 7.58 A thin sheet of fiber board weighs 90 N and lies on a rooftop as shown in the figure. Ambient air at 20 Cc and 1 atm at 33 /m s is generates enough friction to dislodge the board. What will be the coefficient of solid friction between board and roof ?





(A) 0.024 (B) 0.24

(C) 0.12 (D) 0.012

FM 7.59 A fan consists of five blades of cross section 0.80 0.10m m# which rotates at 100 rpm. If blades are act as flat plates, the torque needed to overcome the friction on the blades will be ( 1.46 10 /m s5 2ν #= - )(A) 0.438 N m− (B) 43.8 N m−

(C) 4.38 N m− (D) 0.0438 N m−

***********





SOLUTIONS

FM 7.1 Option (C) is correct.The drag and lift forces are determined by decomposing the Resultant force into its components in the flow direction and the normal direction of flow.

Drag force : cosF F ��D R c= 600 30 520cos Nc#= =Lift force : ��sinF FL R c= 600 30 300sin Nc#= =


For steady un-powered flight

FxΣ 0= gives FD sinW q=FyΣ 0= gives FL cosW q=

Thus FF

L

D cossin tanW

Wqq q= = where F

FU ACU AC

CC

L

D

L

D

L

D

��

rr= =

Hence tanθ CC

L

D=

θ tan CC

L

D1= − c m

FM 7.3 Option (B) is correct.(a) The drag force acting on the tank when the frontal area is the front and

back of the tank is

FD C A V�

�

D # #r=

.( . ) . ( . )

0 9 40 5

21 028 26 52 2

##

##p=

63.75 N= 64 N,(b) The drag force acting on the tank when the frontal area is the sides of the

tank is

FD 0.8 (0.5 3). ( . )

21 028 26 5 2

## # #= 433 N=

FM 7.4 Option (C) is correct.For water at 20 Cc take 998 /kg m3ρ = and 0.001 /kg m sμ -= and for creep motion Re �d = .

So Red Vdm

r=





1 .. d

0 001998 0 25# #=

or d .. 4 10 m998 0 25

0 001 1 6

##

#= = − 4 mμ=


Friction drag FF U C A21

Df2r= where . .

�eC

Ul��

Dfl

n

= =

and A bl= where p�ate widthb = .

Thus FF 1.328 0.664UUl

bl U b l21 /2 3 2r

n

r n= =f p

...(i)

Consider two flows with 1 2ρ ρ= , U U� �= , b b� �= , 1 2ν ν= and l l�= .

So from equation (i),

FF

F

F

�

� ll�

�=

So that with FF� F41

F1=

ll� 4= or l l

��=


Reynolds Number for this flow

ReL .. .VL

�� # #

mr= =

6.9 106#= (Turbulent flow)

For smooth wall CD ( )

.Re Re0 031 1440

/L L

1 7= −

( . )

.( . )6 9 10

0 0316 9 10

1440/6 1 7 6

# #= − .0 00306=

Drag force FD (� )sidesC U b L�D�r

# # # # #=

0.00306 (14.5) 4 0.521025 22

## # # #= b l

13 N20=

FM 7.7 Option (B) is correct.The Reynolds number is

ReL .

. �.�� VL��

��

�

#

#n #= = =−

Since Re Re 5 10>L cr5

#= . Thus the flow is combined laminar and turbulent and

friction coefficient Cf .

Re Re0 074 1742

/L L1 5= −

( . )

..

.9 987 10

0 0749 987 10

1742 0 002772/6 1 5 6# #

= − =

Then the drag force acting on the surface becomes

FD . ( ). ( . )

C A V� ��

�� f

� �

# # ##r= =

19.97 20 N,=





FM 7.8 Option (C) is correct.Reynolds Number for this flow is

ReL ..VL

�� # #

mr= = ��cmL =

20278= (Laminar flow)

and drag coefficient

CD ( )

.( )

.Re1 328

202781 328

/ /L

1 2 1 2= = .0 00933,

So, the total drag force due to friction is

FD 2C V bL21

D2r# # # #=

0.00933 891 (6) 2 0.55 1.121 2# # # # # #= 181 N=

FM 7.9 Option (D) is correct.Terminal velocity is determined by setting.

FD sinWtotal # q= �FB = FD sinm g ��total # # c= 95 9.81 12 194sin Nc ,# #=

Since FD C A V�D

�

# #r=

V A CF�

D

D

# ##

r= . . .1 25 0 45 1 12 194# ##=

25. /m s04= 90 /km h,


Since FD sinWtotal q=

C A V�D

�

# #ρ sinm gtotal q=

V sinC Am g�

D

total

# #

# # #r

q=

. . .. sin

0 9 0 4 1 252 95 9 81 12

# ## # # c=

29.3 /m s= 106 /km h,


For case (a):

Drag force FD U C A21

D2r= where . .

ReC

Ul��

Dl

n

= = and A l �=

Thus FD . 0.664UUl

l U l21 1 328 / /2 2 3 2 3 2

# # #r n r n= = ...(i)

For case (b):

FDl U C A21

D2r= Where .C

U l��

D#n

= and (�)A l l l��

#= =

Thus FDl . (0.664 )U

Ull U l2

12

1 3282

1 / /2 2 3 2 3 2# #r n r n= =

By comparing equation (i) and (ii), we see that





FDl 0.707F F2

1D D= =

FM 7.12 Option (A) is correct.For a laminar flow over a flat plate,

FD C A V�f

�

#r= where .

�eC ��

.f ��=

.Re

A V1 3282.0 5

2

# #r=

.VL

A V��.��

�

n

r# #=

b l

Re VLn=

0.664V AL

kV/.

./3 2

0 5

0 53 2n= =

where �.�� tancons tk AL .

.

��

��n= =

Now for Initial condition

FD� kV ��=

If the free stream velocity of fluid is doubled, i.e.

V� V2 1=Then FD� ( )k V �

��= ( ) . ( ) .k V k V F� ��

D��

��

�= = = FD� . F2 83 D1=

FM 7.13 Option (C) is correct.The minimum wind velocity normal to the side of the container to blow the container over will be the terminal velocity.Since terminal velocity is obtained by setting

W F FD B= +Here F �B = , because no buoyancy force is applied by the air.

Hence FD .W m g �� # #= = = 49050 N=Now the velocity of wind is determined by

FD C A V�D

�

# #r=

V . ( ) ( . )C A

F��

� ��D air

D

# ##

# # ##

r= =

50.3 / 50 /m s m s,= 50.3 3.6 181 /km h,#=


If the railway wagon is about to tip around point O , then �MOΣ = or





2.55FD . W0 75=

Hence FD .. ( )

7.3 10 N2 550 75 250 10

53

4# ##= =

Where FD C U A21

D2r=

or U C AF�

D

D

r= . . .( . )

35 /m s1 9 1 23 3 4 152 7 35 10 /4 1 2

# # #

# #= =; E


The terminal velocity of a free falling object is reached when

Weight of solid Drag force Buoyant force= + W F FD B= + FD W FB= −

C AV�

�

Df

# #ρ

gv gvs fr r= −

C D V� �D

f� �π ρ

# # ( )g D6s f

3

r r p= − −

C V�D

�

# gD� �f

s#r

r= −: D

V 1

C

gD

3

4

D

f

s

rr

=−b l

.

. ( . ) 10001150 1

3 0 5

4 9 8 0 05

#

# # #

=−b l

0.443 /m s=

FM 7.16 Option (A) is correct.Summing forces along the roadway gives

sinW θ F FD Rolling= +

C A V C V�D roll�r

# # #= +

or sin65000 8# c 8.64 . 120V V21 112 2

# #= +

.9046 25 4.80 120V V2= + 4.80 120 9046.25V V2 + − = 0

� ��.��V V��+ − = 0

or V 32.63 33 /m s,=





FM 7.17 Option (B) is correct.The velocity of dust particle to travel 350 m in 43 min is to be

V 0.1356 0.136 /tantime

dis ce m s43 60350#

,= = =

This velocity of a free falling particle is reached when

FD W FB= − ...(i)

where FD VD3pm= (stokes law)

W gvsr= and F gvB fr=

v D6

3p=

Substituting these values in equation (i), we get

VD3πμ ( )gv gv g D�s f s f

�

#r r r r p= − = −

D ( )g

V��s fr rm= −

where μ 1.802 10 /kg m s5# −= −

V 0.136 /m s= sρ 1800 /density of dust particle kg m3= = fρ 15density of air at atm and C1 c=

.. �.�� kg �RT

p��

��

#= = =

Hence D . ( . )

. .9 81 1800 1 225

18 1 802 10 0 1365

## # #= −

− 5 10 m5

#= − or 0.05 mm


If acceleration is negligible,

W FD=

or m g# C U D� �D� �r p

# #=

.135 9 81# 1.2 (8.5)U2 42 2r p

# # # #=

or U � .38 89r=

Thus Usea level .. 5.64 /m s1 225

38 89= =

U�� ..

1 006738 89= 6.22 /m s=

Thus the change in velocity is very small, so we can seasonably estimate the

time-to-fall using the average fall velocity.

tfallΔ Vz

avg

D= [( . . )/ ]5 64 6 22 2

2000 0= +− 337 s=






The sphere should hang, so that string tension balances the resultant of drag and net weight.

Drag force FD C V A�D

�r# #= C V D� �D

��r p

# #=

0.5 . (40) (0.03)21 225

42 2p

# # # #= 0.346 N=

Wnet ( )g D6sphere air3

#r r p= −9 C

. . . ( . )7 86 998 1 225 9 81 6 0 03 3# # # #

p= −^ h9 C" ,

1.09 N,

So tan θ ..

FW

��

drag

net= =

θ .. 72tan 0 346

1 091 c,= −b l

FM 7.20 Option (A) is correct.For air at 20 Cc and 1 atm , take �.� �kg mair

�ρ = and 1.8 10 /kg m s5μ # -= - . For Helium �� kg �R −= .

So Heρ RTp= 0.197 /kg m2077 293

120000 3

#= =

The balloon net buoyancy is independent of the flow velocity

Bnet g D�air He�

#r r p= −^ h

(1.2 0.197) 9.81 (0.5) 0. 44 N6 63 ,p# # #= −

Thus, the net upward force is

Fz B Wnet= −^ h

0.644 0.2 0.444 N= − =Now drag FD tanFz q#= 0.444 (87.3 )tan c#= 9.42 N=

Also FD C U A�D

�r# #=

.9 42 0.2 . (0.5)U21 2

42 2p

# # # #=

.9 42 0.0236U 2=

U � ..

0 02369 42 400-=

U 20 /m s=

FM 7.21 Option (C) is correct.The antenna is a composite body consisting of one main pole, one horizontal bar and four vertical rods. Thus.

FD F F F�D D D� � �= + +

[ ]U C A C A C A21 4D D D

21 2 31 2 3r= + +

Where U 100 27.8 /m s185

#= = and .C C C ��D D D� � �= = =

So that FD 1.23 (27.8) 1.4[5 0.04 1.5 0.02 4 1 0.01]21 2# # # # # # #= + +





180 N-

FM 7.22 Option (A) is correctAircraft will take off when lift equals to the total weight.

W F C A V�L L

�

#r= =

V � ( )C AW

C Am g�

L L

#r r= = W mg=

When the airplane is fully loaded, then

Mass m� 400 400000ton kg= = and

Velocity V� 250 /km h=Now, when 150 seats are empty, Mass of airplane

m� ( )m 150 1�01 #= − 400000 21000 379000 kg= − =and Velocity is V�

When the density, lift coefficient and wing area remain constant, the ratio of the velocities of the under-loaded and fully loaded aircraft becomes

VV

1

� ( )/( )/m g C Am g C A

22

L

L

1

2

#

#

rr=

m gm g

mm

1

�

1

�= =

V� V mm

11

�#=


V� 250 243.34 /km h400000379000

#= =

243.5 /km h,

FM 7.23 Option (B) is correct.For given parameter, the Reynolds number

Re .. ( . )

�.5�5 10VD1 ��0 10��0 0 00�

5�

#

#

n #= = =−

b l

The drag coefficient corresponding to this value of .Re 4 535 103#=

CD .1 0=Then, the drag force becomes

FD ( )C A V C L D V� �D D

� �r r# # # # #= =

where A = frontal area for flow past a cylinder LD=

At 1 atm and 15 Cc ρ . ( )

atmRTp

0 �� 151#

= = +





.

101.325 1.225 /kPa kg m0 287 2883

#= =

So 55 1.0 0.006. .L 2

1 225 3 640 2

## # #=

b l

L . . . .1 0 0 006 1 225 3 6

402 55

2

# # #

#=b l

121.22 121 m,=

FM 7.24 Option (B) is correct.An aircraft will takeoff when lift equals the total weight.

W F C A V�L L

�r= =

V . ..

C AW

C Am g� �

��5 ��5 ��

L L

#

# ## #

r r= = =

57.6 /m s= or 207 /km h

And wind loading

Floading . ��5.� �� mA

FAW

�� L �#= = = =

916 /N m2,

FM 7.25 Option (C) is correct.When the aircraft is cruising steadily constant altitude, the net force acting on the aircraft is zero and thus thrust provided by the engine must be equal to the drag force.

FD .. ( �. )

C A V� ��5 ��

��5 ��D

� �

# ##r= =

4466.1 N= or 4.466 kNAnd the power required is to be

P Thrust velocity F VD# #= = . ( / . )4 466 300 3 6#= 372 kW,


PND power when not drafting=

UF U C U A C U A��

��

D D D� �

ND ND ND# r r= = =

Similarly PD power when drafting C U A21

D3

Dr= =

Thus, PP P

ND

ND D− .. .

CC C

�� 5�

D

D D

ND

ND D= − = −

0.432= i.e. a . %43 2 decrease.


For an urban area u Cy .��=Thus with �� m su = at �� my h= =

We obtain C 77

3.5220.0 4= =

] g or .u y�5� .��=

The total drag is FD dF C u dA��

D D�r= =# #





( . ) ( . )C y dy21 3 52 87 5.

Dy

0 4 2

0

154

#r==

#

1.23 1.3 (3.52) 87.5 y dy21 .2 0 8

0

154

# # # #= #

867 . (154)1 81 .1 8

# #=

4.17 10 N6#= 4.17 MN=


FD cos sinp dA dAwq t q= + ## where �wτ =

Thus FD cos cos cosp dA p dA p dA� � �

q q q= + +# # #

� �cos cosp dA p dA��

c c= −# #

21 2 . cosU lb U A2

1 1 20 21 452 2

2# cr r= − −b bl l; E

where A� �

cosl

b��c

= c m, �en�th o�o��ectl =

Hence FD . U lb1 7 21 2r= b l

or CD 1.7

1.70U A

F

U lb

U lb

21

2121

D

2 2

2# #

r r

r= = =

Because of symmetry of the object, �FL =

or CL 0U A

F

21

L

2r= =

FM 7.29 Option (A) is correct.The drag force acting on the sign,

F ,D sign C A V�D

�r# #=

2.0 (4 2). ( / . )

21 30 150 3 6 2

## # #=

18056 N= or 18.06 kNThe drag force acting on the pole,

F ,D pole C A V�D

�r# #=

0.3 (0.05 4). ( / . )

21 30 150 3 6 2

## # #= 68 N=





Therefore, the drag force acting on both poles is

F ,D poles 2 68 136 N#= =

FM 7.30 Option (D) is correct.The resultant drag force on the sign passes through its centre, the drag force on the poles passes through the center of the pole and the weight of the sign passes through the centre of the block.Taking moment about center of the sign post and setting it equal to zero. MΣ 0= (� � �.��) (� �.��) ( ��)F F W L �, ,D sign D poles# # #+ + + + − =where W = the weight of concrete signpost

. ( . )mg gv L�� # # # #r= = = L13538=Substituting the values, we get

. . /L L18060 5 15 136 2 15 13538 2# # #+ − 0= L 3.71 m=


Drag force FD C A V�

�

D # #r=

Where CD .2 2= for a rectangular sign for Normal flow.

A (0.45 0.6)Frontal area m2#= =

ρ 1.25 /kg m3= V 50 / 50/3.6 /km h m= =

Hence FD 2.2 (0.45 0.6). 3.6

50

2

1 252

#

# # #=b l

71.61 N= 72 N,

FM 7.32 Option (C) is correct.The amount of work done to overcome this drag force and required energy input for a distance of 60000 km are

Wdrag F L �� D # #= = 4.3 10 /kJ year6#=

Energy input Einput .. �.�� k�year

W��

�� car

drag�

�#h #= = =

Then the amount of fuel that supplies this much energy is

Heating valuemE

fuel

fuel

fuel

input

r r= =

.

1.54 10489 /L year0 75

420007

#

= =

And the cost of this much fuel is

Cost of fuel ( ) ( )cosAmount of fuel unit t#= 489 25#= .12225/Rs year=

FM 7.33 Option (A) is correct.If the surface were smooth,

P F U#=

C U A U�D�

# #r

#= 9 C





( )

.Re

U A U0 0312/

L1 7

2

# #r

#= = G � �

�CRe��

�DL

��=

( / )

.UL

U A U0 0312/1 7

2

# #r m

r#= = G Re UL

L mr=

7000 hp ( / . )

. 5000U

U U1025 150 0 00107

0 0312

1025/1 7

2

# ## # #= ; E

or 745.7 7000 W# 5428U /3 1 7= −

.5 22 106# 5428U /20 7=

or U �� .5428

5 22 106#= .961 7=

U ( . )961 7 /7 20= 11.0 /m s=


Drag force

FD C U A�D

�r# #= C U D L�D

�r# # #=

.. ( )

0 4 21 225 22 2 402

## # #= 9500 N,

If the flow is uniform, the center of this force should be at approximately mid depth. Therefore the bottom bending moment is

M� �FL �= 9500 240

#=

190000 N m−= 190 kN m−=


We can write Ptotal P P Pbody loss prop wing= + +where P Pbody loss prop+ 0.6Ptotal= 0.6 150 90 kW#= =Thus Pwing 150 90 60 kW= − =

where Pwing F U C U AU��

D D�r= =

or CD . ��

� .

( )�.��

U AP�

��

� �� wing

� �

�

# #

# #

r#

= = =b l

FM 7.36 Option (B) is correct.The rolling resistance of the Automobile

F C WRR RR auto#= C m gRR auto# #= . .0 04 950 9 81# #= 372.8 373N N,=The speed at which the rolling resistance is equal to aerodynamic drag is obtained by setting.

FD FRR=

C A V�D

�

# #ρ 373=

V �� C AD # # r

#= . . .0 32 1 8 1 2373 2# #

#=

32.8 /m s5= 118 /km h,

FM 7.37 Option (A) is correct.The power need to overcome drag and rolling resistance is the product of the sum





of the force drag and the rolling resistance and velocity.

P ( )F F VD RR= +

C AV F V�D RR

�

#r= +c m

C A V F V�D RR

�

# #r= +

80000 . . . V V20 32 1 8 1 2 373

3# # #= +b l

. V V0 3456 3733 + 80000= ...(i)The solution of equation (i) gives

V 55.56 / 200 /m s km h,=


Drag force FD C A V�D

�

# #r=

Where CD 1.1= for thin circular disk

A = Frontal Area d4

2p=

ρ = density of air at 10 Cc and 100 kPa

. �.�� kg mRTp

0 �8� �8��00 �

#= = =

and V 150 / . /km h m s3 6150= =

Thus FD 1.1( . )

1.23 .40 5

3 6150

212 2#p

# # # #= b l

230.72 231N N-=Now the bending moment at the bottom of pipe

Mbottom ( )length of pipe �adius of circular antennaFD #= + ( . . )231 2 5 0 25#= + 635.25 N m−= 635 N m, −


When the garbage can is first tipped, the edge on the wind-loaded side of the can will be off the ground and thus all the reaction forces from the ground will act on the other side. Taking the moment about on axis passing through the contact point.

0McontactΣ = �F H �D # �W D �#=





W DF H

DC A V H

�

�D D#

#

r# # #= =

Where A 1.2 0.8Frontal area #= = ρ 1.25 /kg m3=

V 135 / .135 37.5 /km h m s3 6= = =

D 0.8 m= and H 1.2 m=

Thus W .0.7 ( . . ) . ( . ) .

2 0 81 2 0 8 1 25 37 5 1 22

#

# # # ##=

885.93 886 N,=

Mass of the can m � ��gW

��= = = 90 kg,

FM 7.40 Option (D) is correctFor air at sea-level 1.225 /kg m3ρ = and 1.78 10 /kg m s5μ # -= -

without deflector:

F C A V�D�r

# #= + Rolling Resistance

8 . (24.5) 112 24.521 225 2

# # #= +

2941 2744= + 5685 N=Power required P F V#= 5685 24.5 139282.5 W#= =

.. 187 hp745 7

139282 5 ,=With deflector:

F C A V�D�r

# #= + Rolling Resistance

6.7 . (24.5) 112 24.521 225 2

# # #= +

2463 2744 5207 N= + =Power required P F V#= .5207 24 5#= 127572 171W hp= =Reduction in Power required

PΔ 187 171 16 hp= − =

% Reduction 100 8.5%18716 ,#=

FM 7.41 Option (D) is correct.For air at sea-level, 1.225 /kg m3ρ =

Vstall C AW�,�a�Lr=

. . 32.4 /m s1 225 1 75 1602 180000# ##= =

Thus VLanding 1.2 1.2 32.4Vstall# #= = 38.9 39 /m s,=

FM 7.42 Option (B) is correct.For take off at same speed of 39 /m s, we need a drag estimate





FD C V A�D�r

# # #=

0.087 . (39) 16021 225 2

# # #= 129 0 N6=

Power required P F V#= 129 0 396 #= 50 50W kW5752 5,=


From given values ReL .

( �. ) ( . )�.�� VL

��

��

#

#n #= = =−

Which is less than the critical Reynolds Number ( )Re 5 105#= , Therefore the

flow is laminar. Then

Cf .

( . ). 0.002347

Re1 328

3 201 101 328

. .L0 5 5 0 5

#= = =

and FD �C A V�f

�r# #=

(Because air flows over both surface of the plate so A A�= )

FD . ( . . ). 3.6

36

0 002347 2 0 5 0 5 2

1 1842

# # # #

#

=b l

0.0695 N=The mass whose weight is 0.0695 N,

m .. �.��gg

F��

��D= = = 7.1 gram=


For steady conditions

D FB+ W=

If Re UD 1<n= , FD 3drag DUp m= =

Also W weight v D34

2H O H O

3

2 2g g p#= = = b l

and FB buoyant force v D34

2air air

3

#g g p= = = b l

Since << � Oair �γ γ , we can neglect the buoyant force.

That is FD W=

or DU3π μ D34

2H O

3

2 #g p= b l ...(i)

U D18H O

22

mg=





.

( . ) ( )1.10 10 /m s

18 1 789 109 80 10 6 10

5

3 8 27

# #

# # ##= =−

−−

FM 7.45 Option (C) is correct.The Reynolds number at the end of the bottom surface

Re .

. ( �. )VL��

��

#

# #m

r= = − .2 2 107#=

Since ,Re 2 10>> 5# then the flow is turbulent over the entire surface and

friction coefficient is

Cf .

( . ). .

Re0 074

2 2 100 074 0 00252/ /1 5 7 1 5#

= = =

Now Drag force FD C A V�f

�r#=

0.00252 (1 3). ( / . )

2999 1 30 3 6 2

## # #=

262.3 N=and Power needed to overcome it

P F VD #= . ( / . )262 3 30 3 6#= 2185.9 W= 2.186 kW,

FM 7.46 Option (B) is correct.Laminar flat plate boundary layer thickness is given by the relation

δ 5 Uxn= 5 .

( . )7.48 10 m

xx0 5

1 12 10 63#

#= =−

−

And wτ 0.332U x/3 2 rm= ν ρ

μ=

0.332 (0.5) .x

1000 1 12 10/3 23

# ##=

−

. �N mx

�� =

Thus, at the trailing edge ( �mx = ),

δ 7.48 10 0.0183 m63#= =−

wτ 6

. 0.0506 /N m0 124 2= =

FM 7.47 Option (D) is correct.Let the flow is turbulent and for turbulent flow

xturbδ

( ).

( / ).

Re Vx0 16 0 16

/ /x

1 7 1 7r m= =

or .x

�� ( / . )

.x998 20 0 001

0 16/1 7

# #=

.x

�� ( )

. ( . )x998 20

0 16 0 001/ /

/

1 7 1 7

1 7

# #

#=

.x

��

b l ( . ) .

x998 200 16 0 0017

# #

#=

or x � ( . ) .

( . )7.44 10

0 16 0 0030 001 998 20

7

79

#

# ##= = −

x 0.0442 m=





FM 7.48 Option (D) is correct.The flow becomes turbulent where the Reynolds number becomes equal to the critical Reynolds number Re 5 10cr

5#= .

Recr Vxcr

n=

xcr . �.��Re mV �

� �� cr� �

# # # #n= = =−

The thickness of the boundary layer at �.��mxcr = is

xδ .( ). . 0.00678

Remx4 91

5 104 91 0 976

/ .cr

cr1 2 5 0 5#

#

#= = =

0.678 cm=

FM 7.49 Option (C) is correct.For flow remains laminar

Lδ .

( �).

Re UL��

L airr m= =

or .2 44δ

.. . .

. 0.00315

1 8 101 2 15 5 2 44

5 0

5#

# #= =

−

or δ . .0 00315 2 44#= 0.00768 7.7m mm= =

and drag FD C U A�D�r

# # #=

Where CD .Re

��L

=

.. . .

. 0.00084

1 8 101 2 15 5 2 44

1 328

5#

# #= =

−

Hence FD 0.00084 . (15.5) (2.44 1.22 2 )sides21 2 2

# # # # #=

0.72 N=

FM 7.50 Option (D) is correct.The drag force before deflector at a given velocity of 110 /km h is,

FD� C A V�D

�r# #=

8.84. ( / . )

21 25 110 3 6 2

##= 5158 N=

Now, the total resistance forces on the trunk is

Ftotal F F FD bearing rolling�= + +And the power required to overcome these forces is to be,

P F Vtotal #= ( )F F F VD bearing rolling1 #= + + [ ] ( / . )5158 350 8339 110 3 6#= + + 423100 W= 423 kW,

The maximum velocity that truck can attain at the same power of 423 kW after deflector is installed is determined by setting the sum of bearing resistance, rolling resistance and the drag force in this case equal to 423 kW.The drag force after deflector

FD� C A V�D

��

#r=

. .V V7 21 25 4 3752

2

22

##= = �mC AD

�=





Therefore P [ ]F F F VD bearing rolling 22 #= + + 423000 [ . ]V V4 375 350 83392

22#= + +

. V V4 375 868923

2= +By solving above equation for V�

V� 32.07 /m s= 116 /km h=


The maximum torque occurs when the cups are normal to the wind since the length of the moment arm is maximum.Then the drag force of each cup in this position is

Convex side FD� C A V�D

�

�

r# #=

(0.4)( . ) . ( )

40 08

21 25 152 2

#p# # #= .C ��D�=

0.283 N=

Concave side FD� C A V�D

�

�# #r=

( . )( . ) . ( )

1 2 40 08

21 25 152 2

##

##p= .C ��D�=

0.848 N=Taking the moment about the pivot,

Mmax . .F F�

��

��D D� �# #= − ( ) .F F 2

0 50D D2 1 #= −

( . . ) .0 848 0 283 0 25#= − 0.1412 N m−=

FM 7.52 Option (A) is correct.For sea-level air, take 1.225 /kg m3ρ = and 1.78 10 /kg m s5μ # -= - . Each ball moves at a center line velocity

Vb rbw#= . .42 20 56

20 0735

#= +b l

( . . )42 0 28 0 03675#= + 13.3 /m s=Then the drag force on each baseball is approximately,

F ,D ball C V D� �D b� �r p

# #=

0.47 . (13.3) (0.0735)21 225

42 2p

# # # #= b l 0.215 N=





Now the drag force on each rod is calculated.

Firstly Vrod ravgw=

42 .2

0 28#= b l 5.88 /m s=

And F �D rod C V D L�D rod rod rod�r

# # # #=

1.2 . (5.88) 0.007 0.2821 225 2

# # # #= b l

0.049 N8,

Then with two balls and two rods, the total driving power required is

P 2 2F V F V, ,D ball ball D rod rod= + (2 0.215 13.3) (2 0.049 5.8 )8 8# # # #= + 5.7 0.582= + 6.3 Watt=


Subscript a and w denotes the portion of the iceberg in the air and in the water respectively.

We have va v71= and v v�

�w = where volume of the icebergv =

For steady motion FDa FDw=

Where FDa ( )C U U A21

Da a b a2r= − and F C U A�

�Dw Dw w b w

�r=

with speed of the icebergUb =

Thus ( )C U U A21

Da a b a2ρ - C U A2

1Dw w b w

2r=

or ( )

UU U

b

b�

�− C A

C AAA

Da a

Dw w w

a a

w w

arr

rr= = Assume C CDa Dw= ...(i)

If D is a characteristic length, then v D�- and A D�-

Hence vv

w

a v

v

DD

7671

w

a3

3

= = or DD

��

w

a��

= b l

So that AA

w

a DD

��

w

a� ��

= =b bl l


( )

UU U

b

b�

�− . (6) 27 .1 23

1026 54 29 2760/2 3# -= =

UU U

b

b− 52.52760= =

or UU

b .53 5= & Ub 0.0187U=

FM 7.54 Option (B) is correct.For sea-level air, take 1.225 /kg m3ρ = and 1.78 10 /kg m s5μ # -= - . Let “a”





and “b” denote the large and small balls, respectively. The rod begins to rotate clockwise when the moments of drag and weight are balanced. The moment equation is

MoΣ ��.� �� .� ��sin cosF Wa a #c c#= − �. � �. �sin cosF W�� b bc c− + =or F Fa b− W Wa b= −

or C U D C U D� � � �Da a Db b� � � �ρ π ρ π

# # # # # #-

( . .)S G g D D6 6water a b3 3r p p

# #= −a k

or U C D C D4 2 Da a Db b2 2 2

# #π ρ# # # -6 @ . .S G g D D6 water a b

3 3# # #

p r= −6 @

or C U D D81

Da a b2 2 2π ρ# # # # -6 @ . .S G g D D6 water a b

3 3# # # #

p r= −6 @

or 0.47 1.225 0.02 0.01U81 2 22π# # # # # -] ]g g7 A

7.86 999 9.8 0.02 0.0163 3p

# # # #= −] ]g g7 A

6.78 10 U5 2#

− 0.282= U � 4160=

U 64.5 /m s4160 ,=


Drag force FD ( )cos cosp dA p br d�

�q q q#= =

q

p

=# #

2 ( )cosp br d0

q q=p

# ...(i)

where p U �r=− for 2 # #π θ π

and U21 1 62r p q= −: D for 0 2# #θ π

i.e. p U21 2r= if 0θ = , p U �r=− if 2θ π=

Thus cosp d��

θ θπ

π# cos sinU d U U

�

��

�

� ��r q q r q r=− =− =

p

p

pp# ...(ii)

and cosp d�

�

�

θ θπ

# cosU d21 1 6/

2

0

2

r p q q q= −p

: D#

( )sin cos sinU21 6 /

2

0

2r q p q q q= − +

p

: D

6U21 1 6

22r p

pp= − − −a bk l; E

U21 6 22r p= −: D ...(iii)

Thus from equation (i), (ii) and (iii), we have

FD cosbr p d br U U2 2 21 6 22 2

0q q r p r= = − +

p

b l; E#

U br21 122r p= b l

So that CD (2 ) (2 )U A

FU rb

FU brU

D Dbr

21 2

21 2

21 221 2 12

#

r r rr

# #= = = p^ h





1.916p= =

FM 7.56 Option (C) is correct.Consider a strip of half tube of width dr as shown in figure below. The local velocity is U rw= and the strip frontal area is Ddr . The total torque (2 tubes) is

T 2 2 ( )rdF r C r Ddr2R R

D0 0

2#

r w= = 9 C# #

2 C D r dr2D

R2 3

0

r w# # # #= #

C D R�D

2�

r w# # # #= C D R41

D2 4r w# # #=

or T 2.3 998 (0.075) (1.0)41 2 4w# # # # #= 43.04 2w=

Power P T w#= 20000 43.04 2w w#=

3ω . 464.743 0420000= =

ω 7.75 /rad s=

or ω 7.75 74 rpm260 ,p#=


For equilibrium, moment

M l F l D F2 2D D1

12

1 2= + −b l ...(i)

Where 2�ml�= , 2.�ml2 = , �.�2 mD�= and 2 mb2 = .

FD� C A U C U l D21

21

D D2 2

1 1r r #= =





1.2 1.23 (20) (20 0.12) 708.5 N21 2 -# # # # # ...(ii)

and FD� C U l b21

D2

2 2r=

0.08 1.23 (20) (2.5 2) 98.4 N21 2# # # # #= = ...(iii)

By combining equation (i), (ii) and (iii), we obtain

M . 20 22 .2

20 708 5 98 4# #= + −b l 708 .5 1869 6= +

8954.6 8955N m N m-− −=

FM 7.58 Option (C) is correct.For air at 20 Cc , take 1.2 /kg m3ρ = and 1.8 10 /kg m s5μ # -= - .Since the dimensions are large and the flow is turbulent.The drag when the leading edge is not at x �= ,

F d C u dA�w A

x

x

f

x

x�

�

�

�

�

#t r= = a k# # C u�w f�τ ρ

# #=

( / )

.Ux

u b dx0 0272/

x

x

1 72

1

2

# # #r m

r= ; E# ( �)

.CUx��

�f ��r m=

. b UU x dx2

0 027 / /

x

x 2 1 7 1 7

1

2

# #r

rm= −

c bm l# and dA b dx#=

. b UU x3

0 02767/ /

x

x2 1 7 6 7

1

2rrm

# #= c bm l : D

. b UU x x2

0 031 / / /2 1 7

26 7

16 7r

rm

#= −c bm l 6 @

or F . 2 1.2 (33) .. (5 2 )2

0 0311 2 33

1 8 10 // /2

5 1 76 7 6 7

##

# # # # #= −−

b l

10.87 N=Since the dislodging friction force

F Wm=or .10 87 90#m=

or μ . .9010 87 0 12= =

Hence, coefficient of solid friction between board and roof is

μ .0 12=


Let dT = torque from the drag on element dA of the blade.

or dT ( ) 2F F y U C dA y21

, ,D top D bottom D2

# r= + = b l ...(i)

Where U yw= and 100 10.47 /rad s601

12ω π

# #= =





The maximum Rel will occur at point ( )1 where y L=

or Rel 1 �� Ul Ll��

��

�

#

# #n n

w#= = = =−

Thus, at all point on the blade 5 10Re Re<x x5

cr #= and the flow is laminar.

So that CD . .Re Ul

1 328 1 328l

n= =

From equation (i),

dT . ( ) �.��UUl

l dy y U l y dy�� # # #r n r n= =

But with U yw=

dT 1.328 l y dy/ /3 2 5 2rw n=

1.328 1.23 (10.47) . . y dy1 46 10 0 1/ / /3 2 5 1 2 5 2# ## #= −

6 @

0.0669 N my dy/5 2−=

Thus the net torque on the five blades is

T 5 5 . 5 0.0669dT y dy y0 0669 72/

./ .

y

5 2

0

0 87 2

0

0 8# #= = =

=6 @# #

0.0438 N m−=

***********

FM 8OPEN CHANNEL FLOW

FM 8.1 Water at 20 Cc is flowing uniformly in a wide rectangular channel at an average velocity of 3 /m s. If the water depth is 0.3 ,m the flow is (For water at 20 Cc

998 /kg m3ρ = and 1.002 10 /kg m s3μ # -= - )(A) Laminar and subcritical

(B) turbulent and subcritical

(C) Laminar and supercritical

(D) Turbulent and supercritical

FM 8.2 Water flows in a 3 m wide rectangular channel with a flow rate of 60 /m s3 . If the flow is to be critical, the maximum depth is(A) 2.58 m (B) . m4 30

(C) . m3 44 (D) . m6 88

FM 8.3 Water flows critically through a 4 m wide rectangular channel with an average velocity of 5 /m s. The flow rate of water is(A) 51 /m s3 (B) 25.5 /m s3

(C) 12.75 /m s3 (D) 5 /m s3

FM 8.4 The ratio of Froude numbers on either side of a hydraulic jump are related by

(A) yy �

�

� ��b l (B) y

y �

�

� ��b l

(C) yy �

�

� ��b l (D) y

y�

� �b l

FM 8.5 Water ( 999.7 / ,kg m3ρ = 1.307 10 / )kg m s3μ # -= - in a half-full 4 m diameter circular channel flows at an average velocity of 2.5 /m s. What will be the hydraulic radius and flow regime ?(A) 1 ,m supercritical (B) 1 ,m critical

(C) 2 ,m supercritical (D) 1 ,m subcritical

FM 8.6 Water is discharged at a rate of 27 /m s3 through a trapezoidal channel with a bottom width of 4 m and a side slope of 45c. If the flow depth is 0.6 m, the flow is(A) Supercritical (B) Critical

(C) Subcritical (D) First subcritical, than critical

FM 8.7 Air flows on the surface of a tank at a speed of 2 /m s. How fast would these air waves travel respectively if (a) the tank is in an elevator accelerating upward at a rate of 4 /m s2, (b) the tank accelerates horizontally at a rate of 9.81 /m s2 and (c)the tank is aboard the orbiting Space Shuttle ?(A) 0, 2.37 / , 2.38 /m s m s

(B) 2.37 / , 2.38 / , 0m s m s

(C) 2.38 / , 2.37 / , 0m s m s

(D) 2.37 / , 0, 2.38 /m s m s





FM 8.8 Water flows through a 2 m wide rectangular channel with a manning coefficient of �n ��= . If the water is 1 m deep and the bottom slope of channel is 0.0105, the rate of discharge of the channel in uniform flow is(A) 10.76 /m s3 (B) 6.03 /m s3

(C) 27.11 /m s3 (D) 5.56 /m s3

FM 8.9 Water flows in a V-shaped channel ( . )n 0 013= section as shown in figure below. The bottom slope of channel is .0 008727. For a flow depth of 2 m at the centre, the discharge rate in uniform flow is

(A) 28.8 /m s3 (B) 22.8 /m s3

(C) 14.36 /m s3 (D) 8.45 /m s3

FM 8.10 A circular channel of diameter 75 cm is flowing half-full at an average velocity of 3. /m s4 . If the channel is asphalt lining .n 0 01�=^ h, the critical slope is(A) 0.69 (B) 0.069

(C) 0.00069 (D) 0.0069

FM 8.11 Water flows through two identical channels with square cross sections of 5 5m m#

. Now the two channels are combined to form a single 10 m wide channel and the flow rate is adjusted so that the flow depth remains constant at 5 m. What will be the percent change in flow rate as a result of combining the channels ?(A) %50 increase (B) 31% decrease

(C) 31% increase (D) No change

FM 8.12 Water flows in a finished-concrete ( . )n 0 012= channel of 1 m depth and slope of .0 00114 as shown in figure below. What will be the percentage reduction in flow

when the surface is asphalt ( . )n 0 016= ?

(A) 50% (B) 75%

(C) 25% (D) No Reduction

FM 8.13 Consider a uniform flow in a fine gravel-lined ( . )n 0 02= rectangular channel with a flow area of 3.6 m2 and a bottom slope of 0.002. For a depth-to-width ratio � . ,y b 0 �= the channel should be classified as

(A) Mild (B) Critical

(C) Steep (D) Not determined

FM 8.14 All surfaces of a rectangular channel as shown in figure, are of the same material. By what percent is the flow rate reduced because of the addition of the thin center board ?





(A) 2.96 % (B) 2.37 %

(C) 23.7 % (D) 29.6 %

FM 8.15 Three uniform pipes of diameter D� join to form one pipe of diameter D and each pipe flows half-full. If the Manning coefficient n and the slope are the same for all of the pipes, what will be the diameter D ?(A) .D D�32 �= (B) �.3D D6 �=(C) �.D D�� = (D) �.D D�9 �=

FM 8.16 The flow rate in the asphalt-lined ( . )n 0 016= channel shown in figure below is to be 120 /m s3 . What will be the elevation drop of the channel per km ?

(A) 8.52 m (B) 0.0852 m

(C) 0.852 m (D) 85.2 m

FM 8.17 A steel painted �.��n =^ h rectangular channel flow, creates a 50c full wedge like wave as shown in figure below. If the depth is 35 cm, the critical depth will be

(A) 62 cm (B) 6.2 cm

(C) 31 cm (D) 3.1 cm

FM 8.18 A viscous oil ( . . .S G ��= ) flows down with an average velocity of 50 /mm s through a wide plate at a uniform depth of 8 mm as shown in figure. If the plate is on a 3c hill, the average shear stress between the oil and the plate will be

(A) 3.49 /N m2 (B) . /N m4 36 2

(C) . /N m2 62 2 (D) . /N m1 75 2





FM 8.19 The channel shown in figure is built on a slope of 2 /m km and depth is ��my =. If the surface are smooth concrete lined ( .n ��= ) except for the diagonal surface, which is gravel with .n ��= , the flow rate will be

(A) 10.81 /m s3 (B) . /m s21 62 3

(C) . /m s12 97 3 (D) . /m s17 30 3

FM 8.20 Water flows in the symmetrical, unfinished concrete ( . )n 0 014= trapezoidal channel as shown in figure below at a rate of 25 /m s3 and flow depth of 0.69 .m What will be the slope angle ( )θ of bottom surface ?

(A) 0c (B) 2c

(C) 1c (D) 0.01706c

FM 8.21 Consider a trapezoidal aqueduct as shown in figure below, carries a normal flow of 60 /m s3 . For clay tile .n ��=^ h surfaces, the required elevation drop in

/m km will be

(A) 0.038 (B) 0.0038

(C) 0.00038 (D) 0.38

FM 8.22 A Trapezoidal channel with brick lining ( . )n 0 015= as shown in figure, has a bottom slope of 0.057. What will be the flow rate of water through the channel ?

(A) 11.1 /m s3 (B) 43.7 /m s3

(C) 275.6 /m s3 (D) 36.4 /m s3

FM 8.23 A trapezoidal channel with a bottom width of 1.5 m and sides with a slope of :1 1 is lined with clean earth ( . )n 0 022= and is to drain water at uniform rate of

10 /m s3 to a distance of 2 km. If it is necessary to keep the flow depth below 1 m, the required elevation drop is





(A) 16.1 m (B) 322 m

(C) 3.22 m (D) 32.2 m

FM 8.24 A channel lined with placed wood is to be carry water at a flow rate of 2 /m s3 on a slope of 10 /m m800 . The channel cross section can be either a right angle triangle or a rectangle with a cross section twice as wide as its depth. Which would require less wood ?(A) Triangle (B) Rectangle

(C) Not defined (D) Both requires equal amount

Common Data For Linked Answer Q. 25 and 26Water flows in a channel as shown in figure whose bottom slope is 0.00873.

FM 8.25 What will be the flow rate through the channel ?(A) 10.5 /m s3 (B) 4.26 /m s3

(C) 44.2 /m s3 (D) 45.56 /m s3

FM 8.26 The effective Manning coefficient for the channel is(A) 0.0328 (B) 0.000278

(C) 0.0299 (D) 0.0315

FM 8.27 A clay tile .n 0 01�=^ h channel is laid out on a :1 1400 slope and has a V-shape with an included angle of 90c as shown in figure below. If the flow rate is 11.35 /m s3 , what will the normal depth y ?

(A) 20 m (B) 4 m

(C) 2 m (D) 0.2 m

FM 8.28 Water flows in a partially filled 1 m internal diameter circular channel made of finished concrete ( . )n 0 012= . For a flow depth of 0.25 m at the center with bottom slope of . ,0 002 the flow rate is

(A) 1.59 /m s3 (B) 0.159 /m s3

(C) 15.9 /m s3 (D) 0.0159 /m s3

FM 8.29 A storm drain constructed of brickwork .n 0 015=^ h has the cross section as shown in figure below. If it laid on a slope of 1.5 /m km, the normal discharge





when the water level passes through the center of the circle is

(A) 5.8 /m s3 (B) 2.90 /m s3

(C) 46 /m s3 (D) 92 /m s3

FM 8.30 In flood stage a natural channel often consists of a deep main channel plus two flood plains as shown in figure below. If the channel has the slope :1 2640 everywhere and main channel with clean-earth .n ��=^ h and the sides are heavy brush, what will be the total flow rate ?

(A) 650 /m s3 (B) /m s325 3

(C) /m s345 3 (D) /m s1000 3

FM 8.31 Water flows in a partly full riveted-steel triangular duct .n ��=^ h as shown in figure below. If the critical depth is 50 cm, the critical slope is

(A) 0.00205 (B) 0.0205

(C) 0.0410 (D) 0.205

FM 8.32 Two identical channels, one of rectangular bottom width b and one circular of diameter D with identical flow rates, bottom slopes and surface linings are considered. For the flow height of b in rectangular channel and the half full circular channel, the relation between b and D is(A) .b D��=

(B) .b D��=(C) .b D��=

(D) .b D��=

FM 8.33 Water discharge uniformly at a rate of 12 /m s3 upto a distance of 10 km through a 2 m internal diameter circular steel ( 0. 12)n 0= drain. For the maximum depth of 1.5 ,m the required elevation drop is





(A) 2.82 m (B) 6.37 m

(C) 63.7 m (D) 0.0637 m

FM 8.34 Water flows in a equilateral triangular cross section channel as shown in figure. For a given Manning coefficient n and channel slope, the depth that give the maximum flow rate will be

(A) .y h��= (B) �.y h��=(C) �.y h��= (D) �.y h��=

Common Data For Linked Answer Q. 35 and 36Water is released from a 5 m deep reservoir into a 1 m wide finished concrete channel ( 0.012)n = of bottom slope of .0 23c through a sluice gate with free outflow of depth ratio ( / )y a 101 = . The water encounters a hydraulic jump. Disregard the bottom slop when analyzing the hydraulic jump.( .C ��d = )

FM 8.35 What will be the flow depth, velocity and Froude number before the jump ?(A) �.�� , �.� �, �.��m m s Fry V� � �= = =(B) �.�� , ��.�� , .m m s Fry V �� = = =(C) �.�� , �.�� , .m m s Fry V �� = = =(D) �. , . �, .m m s Fry V�� = = =





FM 8.36 The flow depth, velocity and Froude Number after the jump are(A) �� , �� , ��m m s Fry V� � �= = =(B) �� , �� , ��m m s Fry V� � �= = =(C) �� , �� , �m m s Fry V �� = = =(D) �� , �� , ��m m s Fry V� � �= = =

Common Data For Linked Answer Q. 37 and 38.A bore is a hydraulic jump which propagates upstream into a slower moving fluid as shown in figure below. The slower moving fluid is 4 m deep and the water behind the bore is 6 m deep.

FM 8.37 What will be propagation speed of the bore ?(A) 8.6 /m s (B) 12.15 /m s

(C) 6.5 /m s (D) 17.2 /m s

FM 8.38 The induced water velocity is(A) 8.6 /m s (B) 5.72 /m s

(C) 14.3 /m s (D) 2.8 /m s7

FM 8.39 Water flows in a rectangular channel with a velocity of � �m sV = . A gate at the end of the channel is suddenly closed so that a wave (a moving hydraulic jump) travels upstream with velocity � �m sVw = as shown in figure. The depths ahead of and behind the wave respectively, are

(A) . 1 m2 6 , 1.96 m (B) . m0 653 , .6 m2 1

(C) . m0 653 , 1. m30 (D) 1.96 m, . m3 26

FM 8.40 Consider a uniform flow of water at 30 cm depth down a 1c unfinished-concrete .n ��=] g slope when a hydraulic jump occurs, as shown in figure below. If the

channel is very wide, the water depth downstream of the jump will be

(A) 39 cm (B) 120 cm

(C) 91 cm (D) 65 cm





FM 8.41 Consider a triangular flume as shown in figure, is built to carry the flow rate v�o at a depth of 0.9 m0 . If the flume is to be able to carry up to twice its flow rate v v��=o o , what will be the freeboard length l ?

(A) 0.378 m (B) 0. 8 m2 3

(C) 0. m472 (D) 0. m236

FM 8.42 For the triangular channel as shown in figure, what will be the angle θ for the best hydraulic cross section (i.e. minimum area A for a given flow rate) ?

(A) 0c (B) 90c

(C) 180c (D) 45c

FM 8.43 The water depths upstream and downstream of a hydraulic jump are .0 3 and 1.2 m, respectively. If the channel is 50 m wide, the upstream velocity and the power dissipated, respectively are(A) 5.42 /m s, 401 kW (B) 4.06 /m s, 301 kW

(C) 6.77 /m s, 502kW (D) 2.71 /m s, 202 kW

FM 8.44 A flow through a wide channel undergoes a hydraulic jump from 40 cm to 140 cm, the percent dissipation will be (A) 50% (B) 46%

(C) 25% (D) 23%

FM 8.45 A hydraulic jump occurs at the base of a spillway of a dam as shown in figure. If the spillway is 100 m wide, the head loss and power dissipated by the hydraulic jump respectively, are

(A) 1.13 m, 9.36 MW (B) 1. m51 , . MW12 5

(C) . m3 02 , MW25 (D) . m1 89 , . MW15 65





FM 8.46 During a hydraulic jump in a 10 m wide channel, the flow depth increases from 0.5 m to 4 m. The water flows at a rate of 70 /m s3 . What will be the mechanical power wasted during this jump ?(A) 4.35 kW (B) 43.5 MW

(C) 43.5 kW (D) 4.35 MW

FM 8.47 A hydraulic jump occurs in a wide horizontal channel at a flow depth of 0.35 m and an average velocity of 12 /m s, the head loss associated with hydraulic jump is(A) 1.972 m (B) 9.13 m

(C) 4.56 m (D) 0.271 m

Common Data For Linked Answer Q. 48 and 49Water flowing through a sluice gate undergoes a hydraulic jump as shown in figure below. The velocity of water is 1.5 /m s before reaching the gate and 4 /m s after the jump.

FM 8.48 The flow depths y� and y� respectively, are(A) 8 ,1.97m m (B) 8 , 4.97m m

(C) 1.97 , 8m m (D) 8 , 2.44m m

FM 8.49 What will be the energy dissipation ratio of the jump ?(A) 0.44 (B) 0.0119

(C) 0.59 (D) 0.55

FM 8.50 Consider a channel contraction section as shown in figure below, often called a venturi flume. The losses are neglected and the flow is one-dimensional and subcritical. If �mb1 = , �mb�= , .9 my 11 = and 1.5 my�= , what will be the flow rate ?





(A) 97 /m s3 (B) 9.9 /m s3

(C) 0.97 /m s3 (D) 4.95 /m s3

FM 8.51 A sluice gate is used to control the flow rate of water in a 5 m wide channel. For flow depth of 1.0 m upstream and 0.50 m downstream from the gate, the flow rate of water and downstream Froude number respectively, are

(A) 1.8 / ,m s3 1.625 (B) 1.8 / ,m s3 0.575

(C) 9 / ,m s3 1.625 (D) 1.625 / ,m s3 0.575

FM 8.52 Water flows in a 0.8 m wide rectangular channel at a depth of 0.25 m and discharge at a rate of 0.7 /m s3 . If the character of flow is to change, the specific energy and the alternate flow depth of water respectively, are(A) 0. 74 ,m8 0.815 m (B) 1.5 ,m 0.815 m

(C) 0.0874 ,m 8.15 m (D) 1.5 ,m 8.15 m

FM 8.53 Water flows in a 6 m wide rectangular channel at a depth of 0.55 m with a specific energy of 1.224 m. What will be the alternate depth and critical depth of the flow ?(A) 0.7�2 ,my2 = 1.0�myc = (B) 1.0� ,my2 = 0.7�2 myc =(C) 0.55 ,my2 = 0.7�2 myc = (D) 1.0� ,my2 = 0.55 myc =

FM 8.54 Water is flowing over a 42 cm high bump at a upstream velocity of 2.5 /m s in a wide channel. If the flow depth is 1.2 m, will the flow be chocked over the bump ?

(A) No (B) Yes

(C) Remains same (D) Not determined

FM 8.55 Water flowing in a horizontal open channel with a velocity of 8 /m s and flow depth of 1 ,m encounters a 20 cm high bump. What will be the change in water surface level over the bump ?(A) Remains same (B) Increase of 0.23 m

(C) Decrease of 0.23 m (D) Increase of 0.03 m





FM 8.56 A sharp crested triangular weir with a notch angle of 60c is constructed 0.5 m above the bottom of a 3 m wide channel as shown in figure below. If the flow depth upstream from the weir is 1.5 ,m the flow rate of water through the channel is (Take discharge coefficient .C ��d = )

(A) 2.46 /m s3 (B) 2.26 /m s3

(C) 0.818 /m s3 (D) 0.145 /m s3

FM 8.57 A 100c notch angle sharp crested triangular weir is installed to measure the discharge rate of water in a open-channel. If the notch angle of weir is reduced by half, the percentage change in flow rate is (Assume the head of weir and discharge coefficient remain unchanged)(A) . %39 1 reduction

(B) . %60 9 reduction

(C) . %39 1 increment

(D) No change

FM 8.58 Consider the water flow under a sluice gate with free outflow. The gate is raised to a gap of 40 cm and the upstream flow depth is measured to be 2.4 m. The flow depth and the downstream velocity per unit width are ( .C ��d = )(A) �.�� , �.�� m m sy V� �= =

(B) �.�� , �.�� m m sy V� �= =(C) �.�� , �.�� m m sy V� �= =

(D) �.� , �.� �m m sy V� �= =

FM 8.59 Water is flowing into a channel as shown in figure below under the sluice gate with a 6 m wide and 0.5 m high opening at the bottom. If the flow depth upstream is 5 m and flow depth downstream from the gate is measured to be 2.5 m, the rate of discharge through the gate is (Take discharge coefficient .C ��d = )

(A) 13 /m s3 (B) 3.76 /m s3

(C) 10 /m s3 (D) 14.3 /m s3





FM 8.60 Water flows over a 4 m wide and 1.5 m high broad-crested weir as shown figure below. The free-surface well upstream of the weir is at a height of 0.5 m above the surface of the weir. The flow rate in the channel and the minimum depth of the water above the weir block respectively, are

(A) 1.7 /m s3 , 0.416 m (B) . /m s2 04 3 , 0. m250

(C) 1. /m s36 3 , 0. m333 (D) 1. /m s02 3 , 0. 6 m1 7

FM 8.61 A 1.1 m high sharp crested rectangular weir is used to measured the flow rate of water in a 6 m wide rectangular channel. If the head above the weir crest is 0.60 m upstream from the weir, the flow rate of water is(A) 18.35 /m s3 (B) 2.174 /m s3

(C) 5.33 /m s3 (D) 8.234 /m s3

FM 8.62 What will be the flow rate per unit width q , over a broad-crested weir that is 2.0 m tall and the head H is 0.50 m ?(A) 0.0350 /m s3 (B) 0.350 /m s2

(C) 0.350 /m s3 (D) 3.50 /m s2

FM 8.63 Water flows over the rectangular sharp crested weir in a wide channel, which is lined with unfinished concrete ( .n ��= ) with a bottom slope of 2 /300m m as shown in figure. What will be the downstream depth and will it be possible to produce a hydraulic jump in the channel downstream of the weir ?

(A) 0.415 m, Impossible (B) 0. m311 , Possible

(C) 0. m311 , Impossible (D) 0.415 m, Possible

***********





SOLUTIONS

FM 8.1 Option (D) is correct.The Reynolds number of the flow is

Re .

. �.�� Vy��

��

#

# #m

r#= = =−

Which is greater than critical value of 500. Therefore, the flow is turbulent.Now the Froude number is

Fr . .

.gy

V�� 0 �

� ��#

= = =

Which is greater than 1, therefore the flow is supercritical.


We have V Av

y y��0 �0#

= = =o where depthy =

Also Fr .

�gy

Vy

y��1

�0�1 �#

= =6 @

.y��

��=

For critical flow 1Fr = .

Thus y (6.39) 3.44 m/2 3= =

FM 8.3 Option (A) is correct.Since flow is critical, then

Fr gy

V 1= =

gyV �

1=

y .( )

gV

��15� �

= =

y �.55 myc= =Thus flow rate becomes

vo VA V b yc # #= = 5 4 2.55 51 /m s3# #= =

FM 8.4 Option (B) is correct.From continuity

vo tancons tVy= =

or FrFr

1

2 ��

V gyV gy

V gyV gy

1 1

� �

1 �

� 1

#

#= =

VV

yy

yv

vy

yy

1

�

�

1

�

1

�

1# # #= = o

o yy �

�

1 ��= b l

FM 8.5 Option (D) is correct.The hydraulic radius is

Rh perimeterAc=

��R

R R�

�

pp

= =^ h Circular channel is half full

1 m22= =





When calculating the Froude number, the hydraulic radius should be used.

For non-rectangular channel hydraulic depth is defined as the ratio of the flow

area to top width.

yh Top widthA

R

RR

22

4 42

2c

2

#p

p p p= = = = = 1.570 m=Now Froude number

Fr . .

. .gyV

��

h #= = =

Since 0.637 1Fr <= , therefore the flow is subcritical.


The flow area and average velocity are

Ac

2tanb b y

y2 #q

=+ +b l

( . / )

0.6 2.76tan

m24 4 2 0 6 45 2# c

#= + + =

V . �.�� m sAv

��

c= = =o

Hydraulic depth y .� .

.

tan tanTop widthy A

b y��

��

hc

#cq

= = =+

=+

0.5308 m=

Then Froude number Fr1 . .

. .gyV

��

h #= = =

Since 1Fr > , therefore the flow is supercritical.






Since c gy=It follows that the tank depth is

y ��

��mgc

��

= = =

If the tank accelerates upward with acceleration a , the effective acceleration of gravity is

geff �� m sg a �= + = + =

Thus c � � �� m sg y �� eff #= = =If the tank accelerates horizontally with acceleration a , the effective acceleration

is

geff �� m sg a �� = + = + =

Thus c . . . /m s13 87 0 408 2 38#= =In orbit geff 0= (weightless). So c �=

FM 8.8 Option (A) is correct.The flow area, wetted perimeter and hydraulic radius of this channel are

Ac � � �mb y �# #= = =

perimeter � � � � �mb y #= + = + =

and Rh �.�mpA

��c= = =

Now from Manning equation, flow rate

vo na A R S� �

c h��

��=

. ( . ) ( . )0 0121 2 0 5 0 0105/ /2 3 1 2

# # #= 10.76 /m s3=

FM 8.9 Option (B) is correct.The flow area, wetted perimeter and Hydraulic radius of the channel are

Flow area Ac 4 mh h2

22

2 2 2 2# # #= = =

Wetted perimeter 2 2 4sin sin

mh45 45

2 2c c# #= = =

Hydraulic Radius Rh 0.7071perimeter mA4 2

42

1c= = = =

Hence, Discharge rate vo ( )na A R S� �

c h��

��

#=

. 4 (0.7071) (0.00 72 )0 0131 8 7/ /2 3 1 2

# # #=

22.8 /m s3=


For a half-full channel, ��A R �p= , perimeter Rp= , R R�h = and b R��= .

The volume flow rate is

vo �.�VA R��p

#= =

3.4 .2 0 375 2p# #= ] g 0.75 /m s3=

And VC .. . �

�.�� m sgh bgA

�� C

�

�# #p

= = = =] g6 @





So, from manning’s formula

SC 1 1 .

. .R

n V0 1875

0 016 1 7/ /

h

C2 4 3

2 2

2 4 3

2 2

# #

#= =] ] ]

] ]

g g gg g 0.0069=


We denote the flow conditions for two separate channels by subscript 1 and the conditions for the combined wide channel by subscript 2.The Manning coefficient, channel slope and the flow area Ac remain constant, the flow rate in case 2 can be expressed in terms of flow rate in case 1 as

vv

1

2oo

( / )( / )

//perimeterperimeter

perimeterperimeter

a n A R Sa n A R S

RR

AA

/ /

/ / / / /

c h

c h

h

h

c

c2 3

11 2

2 321 2 2 3

1

22 3

2

12 3

1 1

2 2

1

2

1

2= = = =; = ;E G E

For condition 1:

perimeter1 6 5 30 cm#= =For condition 2:

perimeter2 4 5 20 cm#= =Substituting these values, we get

vv

1

2oo

1.312030 /2 3

= =: D (31% increase)

FM 8.12 Option (C) is correct.The hydraulic radius

Rh 0.6perimeter mA3 1 1

3 1#= = + + =

For finished concrete, flow rate (From Manning’s formula)

v1o . 3 (�.�) (�.��1�)n AR S1��12

1� � � �h2 3

�1 2 2 3 1 2

# # #= = 6. /m s65 3=

Also for asphalt

v2o . 3 (�.�) (�.��1�)n AR S1��1�

1� � � �h2 3

�1 2 2 3 1 2

# # #= = /m s5 3=

Percentage reduction

.. 1006 65

6 65 5#= − 25%=

FM 8.13 Option (A) is correct.The flow area, wetted perimeter and hydraulic radius for this flow are

Ac 3.�my b 2#= = ...(i)

perimeter y b2= +

Since by .0 4= & y . b0 4= ...(ii)

From equation (i) and (ii),

. b b0 4# # .3 6= . b0 4 2 .3 6=

b2 ..

0 43 6= 9=





b 3 m= y 0.4 3 1.2 m#= =and Perimeter 2 1.2 3 5.4 m#= + =

And hydraulic radius Rh .. 0.6667perimeter mA

5 43 6c= = =

Thus the flow rate vo � � � �na A R S� �

c h��

��=

. . ( . ) ( . )0 021 3 6 0 6667 0 002/ /2 3 1 2

# # #= 6.14 /m s3=

Hence, the critical depth is

yc ��

gbv

��

�

� ��

�

� ��

#= =o

] g= =G G 0.753 m=

Since ,y y> c the channel at these flow condition is classified as mild and the flow is subcritical.


From manning’s equation, the flow rate

vo na AR S� �

h��

��= �a = , if S.I. units are used.

Without the center board

A b b b� �

�

= =b l and perimeter b b2b b2 2= + + =

And Rh PerimeterA

b

bb

22

4

2

= = =

Thus vwithouto na b b S� �

� ��

��

# #= b l ...(i)

With the centerboard vwitho 2v2= o ...(ii)

Where A� b2

2= b l and perimeter b b b b

2 2 2 23= + + =

And Rh� b

bb

3 2

26

2

= =b

b

l

l

Thus vwitho 2na b b S2 6

/ /2 2 3

01 2= b bl l ...(iii)

Dividing equation (iii) by (i) to obtain

vvwithout

withoo

0.763b b

b b

2 4

2 2 6/

/

2 2 3

2 2 3

#

#

= =b b

b b

l l

l l

or 100 76.3− 23.7%= reduction

FM 8.15 Option (C) is correct.After joining the three pipes, the net flow rate (vo) becomes

vo v3 1= o ..(i)

Where vo na AR S� �

h��

��= and v n

a A R S� �h�

��

��=o

With n n�= , S S� ��= , A D��p= .





Rh PerimeterA

D

D D

2

84

2

p

p= = =

and Rh� PerimeterA

D

D D

2

841

1

1

12

1p

p= = =

Thus from equation (i), we get

AR �h�� A R3 /

h1 12 3=

D D8 4

/2

2 3

#π

b l 3 ( )D D8 4

/

12 1

2 3

# #p= b l

Hence D �� ( )D3 /1

8 3= or �D D�� =

FM 8.16 Option (A) is correct.The flow area, wetted perimeter and hydraulic radius of channel are

Flow area Ac (10 5) (2.2) 16.5 m21 2# #= + =

Wetted perimeter p 5 2 ( . ) ( . ) 11.66 m2 2 2 52 2#= + + =

And hydraulic radius Rh .. �.��mp

A��c= = =

Substituting the given parameters into Manning’s equation

vo na A R S� �

c h��

��=

Where a 1 /tancons t m s/1 3= = , S� = Slope

Thus 120 . . ( . ) S0 0161 16 5 1 415 / /2 3

01 2

# # #=

S �01 2

. ( . ).

16 5 1 415120 0 016

/2 3#

#=

S0 .0 008524=Therefore, the elevation drop zΔ across a pipe length of 1 �mL = must be

zΔ 0.00��2 1000S L0 # #= = 8.52 m=

FM 8.17 Option (A) is correct.The wave angle and depth give

Fr .sin25

1 2 37c

= =

or cV 2.37= Fr c

V=

V 2.37 2.37c gy# #= = c gy=

2.37 . .9 81 0 35##= 4.3 /m s9=Flow rate per meter width

vo �.� 0.��V y �# #= = 1.5/

mm s

43

=

Hence yC .( . )

gv

��11 �� 2 1 � 2 1 �

= =o; ;E E y

b gv2

2 �

C

1 �

= oe o

0.62 62m cm, =






Average shear stress given by the relation

wτ R Sh �g=

Where γ 0.85 0.85 9800 8330 /N mH O 32 #= = =g

For a wide flat plate A by= and Perimeter b=

So that Rh 8 10perimeter mA y 3#= = = −

and S� sin3c=Thus wτ sin8330 8 10 33

# # # c= − 3.49 /N m2=


Flow rate vo v v na A R S n

a A R S� � � �h h� �

��

��

��

��= + = +o o ...(i)

Where �a = (for S.I. units), .S ��= , �.��n�= , .n ��=

Also A� 1.0 3 1.50 m21 2# #= =

Perimeter1 . . 3.16 m1 0 3 0 /2 2 1 2= + =] ]g g7 A

Rh� .. 0.475Perimeter mA

3 161 50

1

1= = =

and A� 3 1.5 4.5 m2#= = ,

Perimeter2 0.5 3 1.5 5 m= + + =

Rh� . 0.90Perimeter mA5

4 52

2= = =

Hence from equation (i), we get

vo . 1.50 (0.475) (0.002)0 0251 / /2 3 1 2

# ##=

. 4.5 (0.90) (0.002)0 0121 / /2 3 1 2

# # #+or vo 17.3 /m s3-

FM 8.20 Option (C) is correct.From Manning’s equation

vo na A R S� �

c h��

��= ^ h ...(i)

Where flow area Ac 2 0.69

0.69 3.92 m25 5 2#

#= + + =^ h

Hydraulic Radius Rh ..

sinperimeter

A

545

2 0 693 92c

#c

= =+

0.565 m=

Manning coefficient a 1 /m s/1 3= n .0 014= for unfinished surface

vo 25 /m s3=Substitute these numerical values in equation (i),





25 . . ( . ) S0 0141 3 92 0 565 / /2 3

01 2

# # #=

S ��

. ( . ).

3 92 0 56525 0 014

/2 3#

#=

S� . ( . )

. .3 92 0 565

25 0 014 0 01706/2 3

2

#

#= =; E

Since S� 0.01706tanq= = θ ( . )tan 0 017061= − .0 98= 1c,

FM 8.21 Option (D) is correct.The geometry leads to these values

A cotby y� q= + 5 3.2 (3.2) 40cot2 c#= + 28.2 m2=And perimeter � cosecb y q= + 5 2 3.2 40cosec c# #= + 14.96 m=

So that Rh .. 1.88perimeter mA

14 9628 2 5= = =

Then flow rate vo n AR S� � �h��

��=

60 . 28.2 . S0 0141 1 885 / /2 3

01 2

## #= ^ h

or S� 0.00038 /m m= 0.38 /m km=


Flow area Ac tanb b y y2

130

2# #c

= + +b l

2tan2

1 4 4 230

2# # #c= + +b l 14.93 m2=

Wetted perimeter � ��sin sin

mb y��

��

� �#c c

= + = + =

Hydraulic radius Rh . 1.244perimeter mA

1214 93c= = =

Bottom slope tanθ ( . ) .tan 0 057 0 001= =

Hence the flow rate vo na A R S� �

c h��

��=

. . ( . ) ( . )0 0151 14 93 1 244 0 001/ /2 3 1 2

# # #=

36.4 /m s3=


The flow area, wetted perimeter and hydraulic radius of the channel are

Ac . . 2.5 mb b y y2

22

1 5 1 5 2 1 1 2#

##= + + = + + =





perimeter � �� mb y y� � � �#= + + = + + =] ]g g

Rh .. 0.5776perimeter mA

4 3282 5c= = =

From Manning’s equation

vo na A R S� �

c h��

��=

10 . . ( . ) S0 0221 2 5 0 5776 / /2 3

01 2

# # #=

S0 .0 0161=Therefore the elevation drop zΔ across a pipe length of ��mL = is

zΔ S L0= .0 0161 2000#= 32.2 m=

FM 8.24 Option (D) is correctFrom manning’s equation, the flow rate is

vo na AR S� �

h��

01 �= ...(i)

Let subscript t and r denotes the triangle cross-section and the rectangular cross-section respectively.

We have v vr t=o o 2 /m s3= , S S �0010

r t0 0= = and n nr t=So that equation (i) becomes

A R �r hr

�� A R �t ht

��= where �erimeterR Ah = ...(ii)

Thus Ar 2yr2= , 4Perimeter yr r=

So that Rhr yy y4

221

r

rr

2

= =

Also At (2 )y y y21

t t t2= = , 2( )Perimeter y2t t=

So that Rht y

2 2t=

Thus from equation (ii),

y y2 21 /

r r2

2 3

b l y y� �

1 �

t t�

��= c m ,

. y1 26 /r8 3 . y0 5 /

t8 3=

yr 0.707yt=The amount of wood is proportional to the wetted perimeter.

Since PerimeterPerimeter

r

t . 1.00yy

yy

42 2

4 0 7072 2

r

t

t

t

#= = =

The triangle requires the same amount of wood as the rectangle.

FM 8.25 Option (D) is correct.The channel involves two parts with different roughness and thus it is appropriate





to divide the channel into two subsections.For subsection 1.

Ac1 3 2 6 m2#= =

perimeter1 3 1 1 1 6 m= + + + =

Rh1 1perimeter mA66c

1

1= = =

Ac� 10 1 10 m2#= =

perimeter2 10 1 11 m= + =

Rh� 0.909perimeter mA1110c

2

2= = =

Applying the Manning equation to each subsection, the total flow rate through the channel becomes

vtotalo v v1 �= +o o

na A R S n

a A R S� � � �c h c h

1

��1 �

�

��1 �

1 1 � �# #= +

a nA R

nA R

S� �

�c h c h

1

��

�

��

�1 �1 1 � �

## #

#= += G

.( )

.( . )

( . )1 0 0206 1

0 05010 0 909

0 00873/ /

/2 3 2 3

1 2#

# ##= +; E

45.56 /m s3=

FM 8.26 Option (D) is correct.For entire channel

Ac 1� � 1�mA Ac c�

�1= + = + = perimeter 6 11 17perimeter perimeter m1 2= + = + =

Rh 0.941perimeter mA1716c= = =

Hence, the effective Manning coefficient for entire channel

neff va A R S� �

c h��

�1 �

# # #= o

.( . ) ( . )

45 561 16 0 941 0 00873/ /2 3 1 2# # #= .0 0315=

FM 8.27 Option (C) is correctFor a V-channel

A ��cot coty y�� cq= =

perimeter 2 cosecy 2q=

and Rh 45cos cosperimeterA y y

2 2 2 cq= = =

Thus vo n AR S1 � �h��

�1 �=

or 11.35 . 45cot cosy y0 014

12 45 400

1/ /2 2 3 1 2

c c# # #= a bk l

11.35 1.7857 y /8 3=

y �� .. 6.3561 7857

11 35= =





or y .6 356 /3 8= ] g 2.0 m,


From geometric considerations

cosθ .. . .R

R y��

�� = − = − =

θ (0.5) 60 60cos or radians180 31 c p p

#= = =−

Flow area Ac ( )sin cosR � q q q= −

( . ) sin cos0 5 3 3 32 p p p= − a ak k9 C 0.1535 m2=

Hydraulic radius Rh ( )sin cos

pA

RR

�c

�

qq q q= = −

sin cos R2 #qq q q= −

.sin cos

2 3

3 3 3 0 5#

#p

p p p=

− 0.1466 m=

Thus the flow rate can be

vo na A R S� �

c h��

��=

. (0.1535) (0.1466) (0.002)0 0121 / /2 3 1 2

# # #=

0.159 /m s3=

FM 8.29 Option (B) is correct.The section properties are

A R R R4 1 42 2 2p p= + = +a k

1 1 42 p#= +a k 1.785 m2=

perimeter 2 1 1R41 p#= + +

2 1 1 1 3.57 m41 p# #= + + =

So Rh .. 0.5perimeter mA3 571 785= = = and . �.�� m mS ��

��= =

Hence vo n AR S� � �h��

��=

. 1.785 . .0 0151 0 5 0 0015/ /2 3 1 2

# # #= ] ]g g

2.90 /m s3=





FM 8.30 Option (D) is correctWe compute the flow rate in three pieces with the dashed lines in the figure above serving as “water walls” which are not counted as part of the perimeter.

vtotalo �v v� �= +o o ...(i)

For v�o n A R S� � �h

��

��

#=

A� . �� my y b � ��

# #= + = + =^ ^h h

Rh� 5.077 m6 20132=+

=^ h

So v�o . 132 . 344.95 /m s0 0221 5 077 2640

1//

2 31 2

3# # #= =] bg l

345 /m s3,

For v�o n A R S� � �h

��

��

�#=

A� b y� �#= 150 3.6 540 m2#= =

Rh� . �.��my bA

��

� �

�= + = + =

So v�o . 540 .0 0751 3 52 2640

1//

2 31 2

# # #= ] bg l

32 .26 / 325 /m s m s4 3 3,=Thus vtotalo �v v� �= +o o

345 2 325#= +

995 1000 /m s3,=

FM 8.31 Option (B) is correct.The cross-section properties are

yC 0.5 m=

b� .. .

hh y

�� C= − = − � �.� �.��h � �= − =] ]g g

0.423 m=

And AC . 0.521 0 423 1# #= +^ h 0.356 m2=

Since VC bgAC

�=

or VC� b

g AC

�

#= or Av

bg A

C

C�

�

�

#=o V

Av

CC

��

�

= o

or v �o .( . ) .

bA g

�� C

�

� �# #= =

vo 1. /m s046 3=

perimeter 1 0.577 0.577= + + 2.15 m= . .sin60

0 5 0 577c

=

Rh .. 0.165perimeter mA2 150 356= = =

Now the critical slope, from the Manning’s formula is

SC . .

( . ) . .b R

n g A��

��

h

C

��

�

��

�# #

#

# #= =] g

.0 0205,






For Rectangular channel

Ac �b�= 3perimeter b=

Rh perimeterA

bb b3 3

c2

= = =

Flow rate v .recto na A R S n

a b b S��

��

c h��

��

��

��

# # #= = b l

�n

a S b��

�

��

��

��

#=]]

gg

For circular channel flowing half-full

Ac ,D8

2p= perimeter D2p=

Rh perimeterA D

4c= =

and flow rate v .ciro na A R S n

a D D S� ��

��

c h��

��

� ��

��p

# # # #= = b l

na S D

� ��

�

�

��

��

��

##

p=

Setting the flow rates in the flow channels are equal, we get

v .ciro v .rect= o

�n

a S D�

��

�

��

��

��

#

π# #

] g

�na S b�

�

�

��

��

��

# #=] g

4

D8 /

/

2 3

8 3

#

π] g

3b

/

/

2 3

8 3

=] g

Db

43 0.655

8 /

/ /

2 3

2 3 3 8

#

#p= =]

]g

g= G

b . D0 655=


From geometrical consideration





cosφ � �Ry R

�� = − = − =

φ (0.5) 60 60cos radian180 31 c p p

#= = = =−

and θ 3 32 120cp f p p p= − = − = =

Flow area Ac � ��n �o�R� q q q= −

1 2.527sin cos m32

32

322 2

#p p p= − =] g : D

Rh 1sin cos sin cosperimeter

A R2 2 32

32

32

32

c

#q

q q qp

p p p# #= = − =

−

0.6034 m=Substituting the given values in Manning’s equation, we get

vo na A R S� �

c h��

��=

12 . . ( . ) ( )S0 0121 2 527 0 6034 / /2 3

01 2

# # #=

S� . ( . )

.1 2 527 0 6034

12 0 012/2 3

2

# #

#= ; E .0 00637=

Therefore, the elevation drop zΔ across a pipe length must be

zΔ �� 1��S L� #= = 63.7 m=


Flow rate vo na AR S� �

h2 �

�1 2=

From figure l tan

h60

2c

= , ( )tan

bh y

��2

c= −

, sin

l y��s c

=

Area A ( ) [ ( ) ]tan

lh b h y h h y21

21

601 2 2

c= − − = − −

or A [ ]tan

hy y60

1 2 2

c= −

and Perimeter tan sin

l l h y2 2�� s c c

= + = +b l

Thus Rh PerimeterA=

cosh y

hy y

260

2 2

c

=+

−

a k

Therefore vo (2 )tan

cosna hy y

h yhy y S

��1

2��

2 ��2

2 2 �

�1 2

cc

#= −+

−

a k> H





For the maximum flow rate dydv �=o

, which is equivalent to dydF �= , where

��F y ( )

( )cosy h

hy y60

2/

/

2 3

2 5 3

c/

+−

By differentiation and simplification this gives

5( 60 )( ) (2 )cosy h h y hy y 2c+ − − − 0=or 4 (5 60 3 ) 5 60cos cosy h h y h2 2c c+ − − 0=Which can be written as

hy

hy8 5

2− −a ak k 0=

So that hy

( )0.73116

1 1 4 8 5! #= + =− or .0 856+

The negative root has no physical meaning.

Thus y 0.856h=

FM 8.35 Option (C) is correct.For depth ratio of 10 and the discharge coefficient for underflow is .C 0 ��d = , the discharge rate through the slice gate

vo C ba gy�d 1=

Since ay1 10=

a .y10 10

5 0 51= = =

Thus vo . . .0 58 1 0 5 2 9 81 5# # # # #= 2.872 /m s3=

For wide channels, hydraulic radius is the flow depth R yh �= . Then from Manning’s equation

vo na A R S� �

c h��

01 �

#=

2.872 . ( ) ( ) [ ( . )]tanb y y0 0121 0 23/ /

2 22 3 1 2

# # # # c=

2.872 . ( . )y y0 0121 1 0 004/ /

2 22 3 1 2

# # # #=

y �2��

( . ). . .

1 0 0042 872 0 012 0 545/1 2#

#= =

y2 (0.545) 0.6947 m/3 5= =Now, flow velocity and Froude number before the jump are

V2 .. �.1�� m sby

v1 0 ��7

2 8722 #

= = =o

Fr2 . .

. .gyV

�81 0 ��7�1�� 1 �8�

2

2

#= = =

FM 8.36 Option (D) is correct.For after the jump condition, flow depth

y� 0.5 ( 1 )Fry 1 82 22= − + +

0.5 0.6947 ( 1 ( . )1 8 1 584 2## #= − + + 1.25 m=

Velocity V� .. .y

y V 1 2�0 ��7 �1��

�

22# #= = 2.3 /m s=

Froude Number Fr3 �.81 1.2�

. .gyV 2 �0 0 ��7

�

�

#= = =





FM 8.37 Option (A) is correct.With the Jump-height ratio

yy

�

� 1Fr21 1 8 1

2= + −6 @

or 46 1Fr2

1 1 8 12

#= + −6 @

3 1Fr1 8 12= + −

16 1 8Fr12= +

or Fr1 1.37=

Thus V� Fr gy1 1#= 1.37 .9 81 4##= 8.58 / 8.6 /m s m s,=

FM 8.38 Option (D) is correct.The bore moves at speed V� and induced a velocity VΔ behind it. If viewed in a frame fixed to the wave as above, the downstream V V V� � D= − .

Since V� . �.� �m sy

V y�

��

� � #= = =

Thus VΔ V V� �= − 8. 5.76 3= − 2.8 /m s7=

FM 8.39 Option (B) is correctFor an observer moving to the left with speed � �m sVw = the flow appears as shown below.

Thus treat as a jump with � �m sV�= , � �m sV�=Since A V� � A V� �=

or yy

�

� �VV

��

�

�= = =

From the depth ratio

yy

�

� 4Fr21 1 1 8 1

2= − + + =6 @

Fr1 3.16=

However Fr1 ( )gy

V/

11 2

1=

So that y� . ( . )( )

�.��Fr

mgV

��

��

��

�

�

#= = =

and y� 4 4 0.65 2.61 my 31 #= = =

FM 8.40 Option (C) is correct.The upstream velocity is

V� n R S� � �h��

��

#=

. . tan0 0141 0 3 1/ /2 3 1 2c# #= ] ]g g 4.23 /m s=

And Fr1 . .

. �.��gyV

��

�

�

#,= =





So from jump-height ratio

yy

�

� ( . ) 1 3.02Fr21 1 8 1 2

1 1 8 2 46512 2 ,#= + − = + −6 8@ B

or y� 0.3 3.02#= 0.91 1m cm9, =


Using the subscript 0 for the design conditions and 2 for denote conditions.

We have vo 2v0= o

Flow rate v�o na A R S�

h�

� ��

��= where �a =

A� 21 0.9 0.9 0.81 m2 2# #= =: D

perimeter0 2[ ( . ) ( . ) ] 2.55 m0 9 0 92 2= + =

and Rh� .. 0.318perimeter mA

2 550 81

0

0= = =

Hence v�o �.�� (�.��)n S� � �

�

��

# #=

or v�o .n

S��

�

��

= ...(i)

Also v�o na A R S�

h�

� ��

��=

Where A� y��= and 2perimeter y22 2=

Rh� 0.354perimeterA

yy y

2 22

2

2

22

2= = =

Hence with n n� �= and S S�� =

v�o (�.�� )n y y S� � �

��

��

��

# # #=

or v�o .n

S y��

�

��

��= ..(ii)

From equation (i) and (ii) with �v v� �=o o , we obtain

0.500y /28 3 2 0.377#=

or y� 1.167 m=However �.�y�− sinl ��c=

So that l . . . 0.378sin sin

my450 9

451 167 0 92

c c= − = − =


Flow rate vo na AR S� �

h��

��= or with a , n , S� constant

ddv

θo n

aS R ddA A R d

dR��

� �h h

h��

��

q q= + −b l; E ...(i)

Thus for a given flow rate ddv �θ =o

and for the minimum area ddA �θ =

equation (i) gives

ddRh

θ 0=

Also Rh perimeterA=





where A ( )cosb l21 q#= and 2perimeter l=

or Rh cos

coslb l

b2 412

1# q q= = ..(ii)

Since sinθ lb2= or �inl b

� q= it follows that

A sincosb b

21

2 qq= b l

or b 2 tanA q#=Thus equation (ii) becomes

Rh (2 ) ( )cos tan sin cosA A4 21 /1 2

#q q q q= =

So that ddRh

θ ( )sin cos A ddA

21

21/ /1 2 1 2

#q q q= −

( ) ( )sin cos cos sinA21

21 /1 2 2 2

# q q q q+ −−b l

With ddA �θ = (i.e. minimum area), d

dR �h

θ = when cos sin 02 2θ θ- =

or θ 45c=i.e. the best hydraulic cross-section occurs with a right angle triangle.

FM 8.43 Option (A) is correct.The depth- ratio across a hydraulic jump is given by the relation

yy

�

� Fr21 1 1 8 1

2= − + +6 @

..

0 31 2 Fr2

1 1 1 8 12= − + +7 A

( )9 2 1 8Fr12= + or 3.16Fr1 =

Since Fr1 ( )gy

V/

11 2

1=

V� 3.16 . . 5.42 /m s9 81 0 3 /1 2##= =6 @

The power dissipated is given by

P vhLg= o

Where dimension less head loss

yhL

� 1 Fr

yy

yy

2 11

2 12

2

1 2= − + − b l; E

or hL (0.3) 1 .. ( . )

1 1.20.3 0.504 m0 3

1 22

3 16 22

= − + − =b l= G' 1

and vo �.� �� .�� .� �m sA V y bV� � � ��

# #= = = =Thus P 9.8 81.3 0.504 401 /kN m s## −= = 401 kW=

FM 8.44 Option (D) is correct.With the jump height ratio

yy

�

� 1 8Fr21 11

2= + −6 @

40140 1 8Fr2

1 112

#= + −6 @

7 1Fr1 8 12= + −





or 8 1 8Fr12= +

64 1 8Fr12= +

Fr12 7.8758

63= =

or Fr1 2.81,

Since V� 2.81 . .Fr gy 9 81 0 41 1 ##= = 5.56 /m s=

And E� ��

��my gV� � ��

��

��

#= + = + =

Thus hf . .. .

y yy y4 4 0 4 1 4

1 4 0 41 2

2 13 3

# #=

−= −^ ]h g 0.45 m=

Therefore percentage dissipation

��

Eh

��f

�= = %23,

FM 8.45 Option (B) is correct.From the depth ratio

yy

�

� Fr21 1 1 8 1

2= − + +7 A

..

0 93 6 Fr2

1 1 1 8 12= − + +7 A

or Fr1 .3 16=

But Fr1 gyV

�

�=

V� 3.16 . . 9.39 /m s9 81 0 9 /1 2#= =5 ?

Thus vo �� .� �.��A V by V� � � � # #= = = 845 /m s3=

And hL � �Fry yy

yy

� �

� ��

�

� ��

= − + − b l= G' 1

(0.9) 1 .. ( . )

.

. 1.51 m0 93 6

23 16

1 3 60 92 2

= − + − =b l= G' 1

The power dissipated is given by

P vhLg= o

9.80 1.51845# #= 12500 /kN m s−= 12500 12.5kW MW= =

FM 8.46 Option (D) is correct.The average velocities before and after the jump are

V� . �� m sb yv

��

�# #= = =o

V� �.�� m sb yv

��

�# #= = =o

Head loss hL ( . ) ( ) .( ) ( . )

y y gV V

� ��

� ��

��

#= − + − = − + −

6.33 m=The mass flow rate of water is

mo �� g sv �r #= = =o

Then the dissipated mechanical power becomes

Pmechanicl mg hL#= o . .70000 9 81 6 33# #= 4346811 W= 4.35 MW-






Froude number before the hydraulic jump is

Fr1 . .

.gyV

��

�

�

#= = =

1Fr >1 , therefore the flow is supercritical before the jump. The flow depth after the jump is

y� 0.5 Fy 1 1 8 r121= − + +8 B

. . ( . )0 5 0 35 1 1 8 6 476 2# # #= − + +8 B

3.035 m=From continuity equation

V y� � V y� �=

V� �� .. �.�� m sV y

y��

��

�#= = =

Now from the energy equation, head loss is

hL ( )y y gV V

21 212

22

= − + −

( . . ) .( ) ( . )

0 35 3 035 2 9 8112 1 3842 2

#− + −

hL 2.685 7.242 4.56 m=− + =

FM 8.48 Option (A) is correct.Flow depth before the sluice gate is

y� . � �mVV y ��

��

��# #= = =

Now Froude number after the jump

Fr3 .

.gyV

��

�

�

#= = =

Then flow depth y� is

y� . [ 1 8 ]Fry0 5 13 32= − + +

0.5 3 1 1 8 (0.7373)2# # #= − + +8 B

1.97 m=

FM 8.49 Option (B) is correct.Energy dissipation ratio is defined as the ratio of head loss to the energy dissipated.

Dissipation ratio Eh

s

L

�

= ...(i)

Now, velocity before the jump

V� . �yy V ��

��

��# #= = 6.092 /m s=

Head loss due to jump





hL ��

y y gV V

� ��

� ��

��

#= − + − = − + −

0.0460 m=Specific energy before the jump is

Es� �� my g

V� � ��

��

#= + = + =] g

Substituting values in equation (i), we get

Dissipation ratio ��

Eh

��

s

L

�

= = .0 0119=

FM 8.50 Option (B) is correctApplying the Continuity and Energy equations to eliminate V�

Continuity : vo V y b V y b� � � � � �= = ...(i)

Energy: E y gV y g

V� ��

��

��

= + = + ...(ii)

Combining equation (i) and (ii) to eliminate V�.

V�� ( )g y y V2 1 2 1

2= − +

( )g y y V y by b2

1 1

2 21 2 2

22

#= − + c m from eq. (i)

�V y by b

� �

� ��

�� − c m= G ( )g y y2 1 2= −

V�

y by b

g y y

1

2

1 1

2 22

1 2 21

=−

−

c

^

m

hR

T

SSSS

V

X

WWWW) 3

or vo V y bb y b y

g y y��

��

��

��

��

� � ��

= =−

−− − − −

^ h> H" ,

Substitute values, we get

vo . .. . . 9.9 /m s

2 1 5 3 1 92 9 81 1 9 1 52 2 2 2

21

3

# #

# # ,=−

−− − − −] ] ] ]

]g g g g

g= G

FM 8.51 Option (C) is correct.When frictional effects are negligible and the flow section is horizontal, the specific energy remains constant.

Es� Es�=

y gV��

��

+ y gV��

��

= +

( )

yg b y

v��

��

�

# #+ o

( )

yg b y

v��

��

�

# #= + o

V byv= o

1. ( )

v2 9 81 5 1 2

2

# # #+ o

0.5. ( . )

v2 9 81 5 0 5 2

2

# # #= + o

.0 5 0.006114v 2= o

vo 9.0 /m s3=Hence, the downstream velocity and Froude number are

V� . �.� �m sb yv

� ��

�# #= = =o

and Fr2 . .

.gyV

��

�

�

#= =

.1 625=






The specific energy of water is to be

Es y gV�

�

= +

where y 0.25flow depth m= =

V . ..argVelocity of water Disch e

A 0 25 0 80 7

c #= = = 3.5 /m s=

Thus Es 0.25 .( . )

0.874 m2 9 813 5 2

#= + =

Now alternate depth y� is to be determined by

Es ygb yv

��

�

= + o

. .( . )

yy� ��

��

��

�

# # #= +

0.874 .yy

0 0��

��= +

Solving above equation y� 0.815 m=There are three roots of this equation. One for subcritical ( . )y 0 815= , one for supercritical ( 0.25 )my = and third one as a Negative root. Therefore, if the character of flow is changed from supercritical to subcritical while holding specific energy constant, the flow depth will rise from 0.25 0.815m to m.

FM 8.53 Option (B) is correct.Specific energy of flow is to be

Es y gV�

�

= +

gV2

2

E ys= −

V ( ) . ( . . )g E y2 2 9 81 1 224 0 55s # #= − = − 3.636 /m s=Now, the flow rate becomes

vo �.�� 0.�� m sV b y � �# # # #= = =

Alternate depth is determined from E Es s� �=

Es� ygb yv

��

�

= + o

.1 224 .( )

yy� �8� �

��

��

�

# # #= + 0.�04y

y��= +

. y1 224 22 .y 0 �04�

�= + . .y y��4 0 �04�

��− + 0=

By solving above equation

y� 1.03 m=Now the critical depth of flow





yc � �gbv

��

�

� ��

�

� ��

#= =oe ]

]o g

g= G 0.742 m=

FM 8.54 Option (B) is correct.The upstream Froude number and critical depth are

Fr1 . .

. �.��gyV

��

�

�

#= = =

and yc ( )

gbv

gbV by

gy V� � �

�

� ��

��

��

��

= = =o= = ;G G E

.1.2 2.5 0.972 m9 81

/2 2 1 3#= =] ]g g

; E

Since 1Fr1 , the flow is subcritical and the flow depth decreases over the bump. The upstream over the bump and critical specific energies are

Es� (�.�) .( . )

�.��my gV� � ��

��

��

#= + = + =

Es� �.�� .�� .��mE zs b� D= − = − =

Ec 0.972 1.46 my23

23

c# #= = =

It is show that E E<s c� . That is specific energy of the fluid decreases below the level of energy at the critical point, which is minimum energy and this is impossible. Therefore, the flow at specified conditions cannot exist. The flow is choked.


The upstream Froude number and the critical depth are

Fr1 .

�.��gyV

��

�

�

#= = =

yc ( )

.� �

gbv

gbV by

gV y

��

� � � �

�

� ��

��

� ��

��

#= = = =o ] ]g g= = ; ;G G E E

1.868 m=Since 1,Fr >1 the upstream flow is supercritical and flow depth increases over the bump.The upstream, over the bump and critical specific energies are

Es� � .� �.��my g

V� � ��

��

#= + = + =] g

Es� �.�� .�� .��mE zs b� D= − = − =

Ec 1.868 2.802 my23

23

c# #= = =

The flow depth ( )y2 over the bump is determined from

( )y E z y gV y�s b�

��

�

��

� D− − + 0=





y E y gV y�s�

��

�

��

�− + 0=

�� y y � ��

��

��

## #− + ]

]g

g 0=

� �y y��

��− + 0=

By solving above equation, the physically meaningful root of this equation is determined to be y� 1.03 m=Therefore, there is a rise of y y zb� � D− + 1.03 1 0.2 0.23 m= − + =

FM 8.56 Option (C) is correct.The discharge rate of water through the channel is

vo tanC g H158 2 2

/d

5 2q# #= b l

Where H 1.5 0.5 1Heigh f weir mο= = − =

Thus vo 0.60 . tan158 2 9 81 2

60 1 /5 2#

c# # # #= b ]l g

0.818 /m s3=

FM 8.57 Option (B) is correct.The discharge rate through the triangular weir is

vo tanC g H158

2 2 /d

5 2q# # # #= b l

For constant Cd and H , discharge depends on tan 2θ

b l

Therefore v��co tank ��

#= b l

Where k tancons t C g H158 2 /

d5 2

# #= =

and v��co tank ��

#= b l

Then vv��

��

c

c

oo

.. 0.391

tantan

1 19170 4663

0

2100

25

= = =c

c

_

_

i

i

When the notch angle is reduced by half, the discharge rate drops to . %39 1 of original level. There for percent change in the discharge is

Percent Reduction . . . %1 0 391 0 609 60 9= − = =


The discharge rate per unit width

vo C ab gy�d �= 0.57 0.4 1 . . 1.56 /m s2 9 81 2 4 3# ## # #= =

When the frictional effects are negligible in horizontal flow, the specific energy





remains constant.

Es� Es�=

Since Es� � ��.�

. � . ��. �

yg b y

v� � ��

��

��

�

�

�

# # # #= + = +o

2.421 m=

and Es� � ��.��y

g b yv

��

�

�

#= + =o

Thus 2.421 . � ��. �

yy� ��

��

��

�

# # #= +

Solving above equation y� 0.238 m= for flow depth as the physically meaningful

root (positive and less than 2.4 m).

And downstream velocity V� ..

Av

byv

� ��

c � #= = =o o

6.55 /m s=

FM 8.59 Option (D) is correct.The depth ratio �y a� and the contraction coefficient �y a� for this flow are

ay� . 100 5

5= = and ..

ay

�� = =

The discharge coefficient corresponding to these values of ( / )y a1 and ( / )y a2 is determined from a experimental graph between Cd and ( / , / )y a y a1 2 is �.��Cd = .Then the discharge rate becomes

vo �.�� .� .C ba gy� � �� d � # ## # #= = 14.3 /m s3=

FM 8.60 Option (C) is correct.For broad- crested weir, the flow rate is given by the relation

vo C b g H��

�wb

��= b l where Cwb = Broad crested weir coefficient

And Cwb . . 0.563

10 65

10 65

/.. /

HH 1 2

1 50 5 1 2

w

=+

=+

=_ _i i

Thus vo 0.563 4 (9.81) (0.5)32/

//1 2

3 23 2

# # # #= b l 1.36 /m s3=

and ymin y H��

c= = 0.5 0.333 m32#= =


The flow rate measure by a rectangular weir is given by

vo C g b H32 2 /

d3 2

# # # #= ...(i)

And Cd 0.598 0.0897 HH

w#= + heigh of weirHw =

. . .. .0 598 0 0897 1 1

0 60 0 6469#= + =





The condition � �H H <w is satisfied. Since 0.60/1.1 0.55,= then the water flow rate through the channel

vo . . ( . )32 0 6469 2 9 81 6 0 6 /3 2# # # # #= 5.33 /m s3=


We have q .bv

HH

g H�

��

�

��

w

��

��

# # #= =+

o

b

b

l

l

Substitute H 0.5 m= and �.��Hw =

Thus q

.

.. (9.81) (0.5)

1 2 00 5

0 6532

//

//

1 21 2

3 23 2

# # #=+b

bl

l

0.350 /m s2=

FM 8.63 Option (D) is correct.For rectangular sharp crested weir, flow rate is

vo C g bH32 2 /

wr3 2= where Cwr = rectangular weir coefficient

And Cwr 0.6 1 0.075 HH1

w= + b l

H 3 2.2 0.8 m= − = and �.��Hw =

Thus vo . . .. . (0.8)b3

2 0 611 0 075 2 20 8 2 9 81 / /1 2 3 2

# ##= + b l; 6E @

1.349 /m sb 3=

V� . .

Av

byv

byb

y��

� � � �= = = =o o

...(i)

For uniform flow (From manning’s formula)

vo na AR S� �

h��

��= where �a =

S� 0.006673002= =

Also for a wide channel

A� by�= and 2perimeter y b1 1= +

So that Rh� ( )perimeterA

y bby y

21

1

1

11.= = + if b y>> �

Thus with .n ��=

vo 1.349 . ( ) (0.00667)b by y0 0141 / /

1 12 3 1 2

# # #= =

or y� 0.415 m=Now from equation (i),

V� .. 3.25 /m s0 415

1 349= =

So that Fr1 . ..

gyV

��

��

��

#= =

6 @ .1 61=

Since Fr 1>1 , it is possible to produce a jump.

***********

FM 9TURBO MACHINERY

FM 9.1 Match List-I with List-II and select the correct answer using the codes given below

List-I List-II

P. Wind mills 1. Dynamic pump

Q. Fans 2. Positive-Displacement turbine

R. Water-meters 3. Positive-Displacement Pump

S. Heart 4. Dynamic turbine

Codes : P Q R S(A) 2 3 4 1(B) 1 4 3 2(C) 4 1 2 3(D) 3 2 1 4

FM 9.2 Match List I (Machines) with List II (Features) and select the correct answer using the codes given below :

List-I List-II

P. Steam Engine 1. Velocity compounding

Q. Impulse turbine 2. Diagram factor

R. Reaction turbine 3. Continuous pressure drop.

S. Centrifugal compressor 4. Isentropic efficiency

Codes P Q R S(A) 3 4 2 1(B) 2 1 3 4(C) 2 4 3 1(D) 3 1 2 4

FM 9.3 A water pump increases the pressure of the water passing through it. The flow is assumed to be incompressible. If outer diameter ( )Dout is less than inlet diameter ( )Din , how will average water speeds Vout and Vin change across the pump ?(A) V Vout in= (B) V Vout in1

(C) V Vout in2 (D) None of these

FM 9.4 For each statement about centrifugal pump,(1) A centrifugal pump with radial blades has higher efficiency than the same

pump with backward-inclined blades.

(2) At the pump’s shutoff head, the pump efficiency is zero.

(3) A centrifugal pump with forward-inclined blades is a good choice when one needs to provide a large pressure rise over a wide range of volume flow





rates.

(4) At pump’s free delivery, the pump efficiency is zero.

Which of the above is/are TRUE ?(A) 1, 2 and 3 (B) 2 and 3

(C) 1 and 4 (D) 2, 3 and 4

FM 9.5 Water at 20 Cc is delivered by a pump with 500 /minL1 against a pressure rise of 270 /kN m2. The driving motor supplies 9 kW of power. If the change in kinetic and potential energies are negligible, the overall efficiency of the pump is(A) %68 (B) 50%

(C) %71 (D) 75%

FM 9.6 A pump delivers 18.0 / .minL of water at a net head of 1.6 m at its best efficiency point. If the maximum pump efficiency is %70 , the power (bhp) required to run the pump is(A) 403 W (B) 6.72 W

(C) 3.3 W (D) 197 W

FM 9.7 A centrifugal pump delivers 125 /m h3 of water at 20 Cc when the brake horsepower is 22 and the efficiency is %71 . The pressure rise in kPa, is(A) 345 (B) 333

(C) 405 (D) 33.3

FM 9.8 Consider a pump runs at 880 rpm to deliver water at 2 C0c with 35 /minm3 of flow rate through the system as shown in figure below. If the pipe has 20 cm diameter and is made of commercial steel ( . )f 0 0144= , what will be the pump head ?

(A) 55 m (B) 58 m

(C) 63 m (D) 48 m

FM 9.9 Match List I with List II and select the correct answer using the codes given below :

List-I List-II

P. Draft tube 1. Kinetic Energy

Q. Pumps 2. Momentum Exchange

R. Impulse-Turbine 3. Diffuser

S. Reaction-Turbine 4. Pressure-rise

Codes : P Q R S(A) 3 4 1 2





(B) 4 3 2 1(C) 2 1 4 3(D) 1 4 3 2

FM 9.10 Consider the following statements :A water turbine governor1. helps in starting and shutting down the turbo unit.

2. controls the speed of turbine set to match it with the hydroelectric system.

3. sets the amount of load, which a turbine unit has to carry.

Which of these statements are correct ?(A) 1 and 2 (B) 2 and 3

(C) 1, 2 and 3 (D) 1 and 3

FM 9.11 Water at 20 Cc is sprinkled at 14 /m h3 by a cm36 diameter turbine as shown in figure below. If the nozzle exit diameter is 8 mm, the appropriate rotation rate will be

(A) 1070 rpm (B) 107 rpm

(C) 1030 rpm (D) 2054 rpm

FM 9.12 Consider a typical Draft tube for each statement.P. It permits a negative head to be established at the outlet of runner and

thereby increase the net head on the turbine.

Q. Recovers some of the kinetic energy leaving the turbine.

R. Turns the flow horizontally leaving the turbine runner.

S. It permits a negative head to be established at the outlet of runner and thereby decrease the net head on the turbine.

Which of the above is/are FALSE ?(A) P and Q (B) Q and R

(C) R and S (D) S only

FM 9.13 If the full-scale turbine is required to work under a head of 30 m and to run at 428 rpm, then a quarter-scale turbine model tested under a head of 10 m must run at(A) 988 rpm (B) 143 rpm

(C) 341 rpm (D) 428 rpm

FM 9.14 For each statementsIf the rpm of a pump is doubled, all else staying the same,P. the capacity of the pump goes up by a factor of about 2.

Q. the net head of the pump goes up by a factor of about 2.





R. the required shaft power goes up by a factor of about 4.

S. the output shaft power of the pump goes up by a factor of about 2.

Which of the above is/are TRUE ?(A) P and R (B) P only

(C) Q and S (D) P and S

FM 9.15 A pump delivers water at a rate of 3 /m s3 when operating at a speed of 60 /rad s against a head of 20 m. Which type of pump is this ?(A) Radial-flow pump (B) Axial-flow pump

(C) None of these (D) Mixed-flow pump

FM 9.16 Turbine-A has diameter �.��mDA = and spins at ��rpmNA = . At its best efficiency point, ��2 /m svA

�=o , �2.�mHA = of water and 132bhp MWA = . The turbine-B will spin at 240 rpm and its net head will be ��.�mHB = . What will be the diameter DB , vBo and bhpB such that it operates most efficiently ?(A) �.�mDB = , �� /m svB

�=o , 12bhp MW6B =(B) 2.�mDB = , �� /m svB

�=o , 1057bhp MWB =(C) �.��mDB = , �� /m svB

�=o , 14bhp MWB =(D) 2.��mDB = , �42 /m svB

�=o , 341bhp MWB =

FM 9.17 A centrifugal pump having an impeller diameter of 1 m is to be designed so that it will supply a head rise of 200 m at a flow rate of 4.1 /m s3 of water when operating at a speed of 1200 rpm. To study the characteristic of this pump, a /1 5 scale, geometrically similar model operated at the same speed is to be tested

in the laboratory. If both model and prototype operate with the same efficiency (and therefore the same flow coefficient), the required model discharge and head rise respectively, are(A) 0.00328 /m s3 , 0.8 m (B) 0.0328 /m s3 , 8 m

(C) 3.28 /m s3 , m12 (D) 0.328 /m s3 , 8 m0

FM 9.18 Which one of the following relation is true for dimensionless parameters of two dynamically similar pumps ?

(A) D D HH

vv/ /

B AB

A

A

B�4 �2

# #= oo

b cl m (B) D D HH

vv/ /

B AA

B

A

B�4 �2

# #= oo

b cl m

(C) D D HH

vv/ /

B AB

A

A

B�2 �4

# #= oo

b cl m (D) D D HH

vv/ /

B AB

A

B

A�4 �2

# #= oo

b cl m

FM 9.19 Which of the following relation is true for specific speed of turbine and specific speed of the pump ?(A) N NSP ST= (B) N NST SP turbine# h=

(C) N NST

turbine

SP

h= (D) N NST SP turbine# h=

FM 9.20 Which of the following water turbines do not require a draft tube ?(A) Propeller turbine (B) Pelton turbine

(C) Kaplan turbine (D) Francis turbine

FM 9.21 A pump delivers 0.0003 /m s3 of water at a net head of 1.6 m at its best efficiency point. A motor that spins at 1200 rpm is available. If pump is modified by attaching different motor, for which the rpm is half that of the original pump. The ratio of specific speed of both the cases will be





(A) 0.5 (B) 1

(C) 2 (D) 0.25

FM 9.22 Performance data for a very small ( 8.25 )cmD = model water turbine, operating with an available head of 15 m are as follows :

�m hv �o 18.7 18.7 18.3 16.7 11.5

rpm 0 500 1500 2500 3500

η 0 14% 38% 65% 11%

It is desired to use a geometrically similar turbine to serve where the available head and flow rate are 46 m and 0.19 /m s3 respectively. What will be the most efficient horsepower ?(A) 5 hp (B) 37 hp

(C) 74 hp (D) 462 hp

FM 9.23 Consider a hydroelectric power plant operates under the conditions as shown in figure. The head loss associated with flow from the water level upstream of the dam, section (1), to the turbine discharge at atmospheric pressure, section (2), is 20 m. How much power is transferred from the water to the turbine blades ?

(A) 235 MW (B) 23.5 kW

(C) 23.5 MW (D) 2.35 MW

FM 9.24 Consider the test pump as shown in the figure below. The data are :

Intel pressure p1 100 mmHg= (Vacuum)

Outlet pressure p2 500 mmHg= (gage)

Intel diameter D1 12 cm=Outlet diameter D2 5 cm=Flow rate vo 0.01136 /m s3=Fluid is light oil . .S G .0 91=Efficiency η %75=What will be the input power ?





(A) 1525 W (B) 858 W

(C) 760 W (D) 1715 W

FM 9.25 A liquid ( . . )S G 0 9= flows through the pump with the flow rate 7.57 10 /m s3 3#

− . The pressure gage at (1) indicates a vacuum of 95 mm of mercury and the pressure gage at (2) indicates a pressure of 80 kPa as shown in figure. If 0.5 mz z2 1− = , what will be the actual head rise across the pump ?

(A) 5.5 m (B) . m8 6

(C) .5 m11 (D) . m14 4

Common Data For Q. 26 and 27.Water at 40 Cc ( 9.731 10 / , 7.376 10 /N m N mpv

3 3 3 2γ # #= = ) is pumped from an open tank through a 200 m long and 50 mm diameter smooth horizontal pipe ( .f 0 0152= ) as shown in figure and discharge into the atmosphere with a velocity of 3 /m s and at standard atmospheric pressure. Minor losses and the losses in the short section of pipe connecting the pump to the tank are negligible.

FM 9.26 If the efficiency of the pump is 70%, how much power is being supplied to the pump ?(A) 2.07 kW

(B) . kW1 55

(C) . kW1 1

(D) . kW4 14

FM 9.27 What will be the NPSHA at the pump inlet ?(A) 15.75 m (B) . 5 m9 4

(C) . m6 3 (D) . m12 6

FM 9.28 A liquid is pumped from an open reservoir through a 0.1 m diameter vertical pipe into another open reservoir as shown in figure. A valve is located in the pipe and the minor loss coefficient for the valve, as a function of the valve setting is shown in figure by the equation . .h v52 0 1 01 10a

� 2#= − o with ha in meters when vo is in

/m s3 . The fluid levels in the two tanks remain constant. If the friction factor for the pipe is .f 0 02= and all minor losses, except for the valve are negligible, what will be the flow rate when the valve is fully open (K 1L = ) ?





(A) 5.29 /m s3 (B) . 29 /m s0 005 3

(C) . 29 /m s0 05 3 (D) . 29 /m s0 5 3

FM 9.29 Consider a hydraulic turbine which provided with 4.25 /m s3 of water at 415 kPa. A vacuum gage in the turbine discharge 3 m below the turbine inlet center line reads 250 mm Hg vacuum. The supply and discharge pipe inside diameters are identically 800 mm. If the turbine shaft output power is 110 kW0 , the power loss through the turbine is (A) 697.5 kW

(B) 930 kW

(C) 465 kW

(D) 1162.5 kW

FM 9.30 Water moves horizontally through a pump at a rate of 0.02 /m s3 . At the upstream of the pump the pipe diameter and the pressure are 90 mm and 120 kPa, respectively and at the downstream of the pump the pipe diameter and the pressure are 3 mm0 and 400 kPa, respectively. If the loss in energy across the pump due to fluid friction effects is 170 /N m kg− , the hydraulic efficiency of the pump is(A) 0.399 (B) 0.879

(C) 0.799 (D) 0.599

FM 9.31 For a given jet speed, volume flow rate, turning angle and wheel radius, the maximum shaft power produced by a Pelton wheel occurs when(A) the turbine bucket moves at same the jet speed.

(B) the turbine bucket moves at double the jet speed.

(C) the turbine bucket moves at half the jet speed.

(D) the turbine bucket moves at quarter the jet speed.

FM 9.32 Water flows from the head water through the penstock of a Pelton wheel turbine as shown in figure. The effective friction factor for the penstock and control valves is same as .0 032. If the diameter of the jet is 0.20 m, the maximum power output will be





(A) 23.4 MW (B) . MW17 55

(C) . MW46 8 (D) 2 . MW9 25

Common Data For Linked Answer Q. 33 and 34Water supplied from a lake at an elevation H above the turbine to run a Pelton wheel. The penstock that supplies the water to the wheel is of length l and diameter D . The only losses associated with the flow in the penstock are due to pipe friction and the minor losses are negligible.

FM 9.33 What will be the nozzle diameter D� that gives the maximum power output of the turbine ?

(A) �

Dl D

fD�� =

^ h (B)

�D

fl DD

� �� =^ h

(C) �

Dfl D

fD� �� =

^ h (D)

�D

fl DD�

�� =^ h

FM 9.34 For the maximum power to be developed by the turbine, the velocity head at the nozzle exit is

(A) H31 (B) H2

(C) H32 (D) H2

3

FM 9.35 Water at 2 C0c with flow rate of 3.5 /m s3 enters in an idealized radial turbine at 30c and leaves radially inward as shown in figure below. If the flow is absolute and the blade thickness is constant at 10 cm, the theoretical power developed at

%100 efficiency will be

(A) 95.5 kW (B) 47.7 kW

(C) 477 kW (D) 239 kW





FM 9.36 A centrifugal pump rotates at 1000 rpm. Water enters the impeller normal to the blades ( )01 cα = and exits at an angle of 35c from radial ( )352 cα = . The inlet radius is ��cmr�= , at which the blade width ��cmb�= . The outlet radius is ��cmr�= , at which the blade width �cmb ��= . The volume flow rate and pump efficiency are 0.0573 /m s3 and 76% respectively. What will be the net head produced by this pump and required brake horsepower, respectively ? (

��. �kg m�water�ρ = )

(A) 4.87 m, 27320 W (B) 4.87 m, 20763 W

(C) 1.55 m, 6583 W (D) 1.55 m, 8662 W

FM 9.37 Air flows across the rotor as shown in figure below. The magnitude of the absolute velocity increases from 15 /m s to 25 /m s and the absolute velocity at the inlet is in the direction shown. If the fluid puts zero torque on the rotor and air is to be incompressible, the direction of the absolute velocity at the outlet will be

(A) 45c (B) .41 55c

(C) .69 25c (D) .55 4c

FM 9.38 The shaft torque on the turbomachine is measured to be 60 N m−− when the absolute velocities are as indicated in figure. If the magnitude of the shaft power is 1800 /N m s− , the angular velocity and the mass flow rate respectively, are

(A) 215 rpm, 67.3 /kg s (B) rpm286 , . /kg s89 7

(C) rpm357 , . /kg s112 12 (D) rpm250 , 7 . /kg s6 8





Common Data For Linked Answer Q. 39 and 40A radial flow centrifugal pump delivers gasoline ( 680 / )kg m3ρ = at 20 Cc . The pump has ��cmd�= , ��cmd�= , ��cmb�= , �.�cmb�= , 251 cβ = , 402 cβ = and rotates at 1160 rpm.

FM 9.39 The flow rate in /m hour3 is(A) 17.23 (B) 0.2872

(C) 28.72 (D) 1038

FM 9.40 What will be the head ?(A) 48 m (B) 16 m

(C) 32 m (D) 64 m

FM 9.41 The front and side views of a centrifugal pump rotor or impeller are shown in figure below. The flow entering the rotor blade row is essentially radial as viewed from a stationary frame. If the pump delivers 200 /L s of water and the blade exit angle is 35c from the tangential direction, what will be the power required associated with flow leaving at the blade angle ?

(A) 348 kW (B) kW261

(C) kW522 (D) kW696

Common Data For Q. 43 and 44.A 3 mm thickness uniform horizontal sheets of water issue from the slits on the rotating manifold as shown in figure below. The velocity relative to the arm is a constant at 3 /m s along each slit.

FM 9.42 What will be the torque needed to hold the manifold stationary ?(A) 5.06 N m− (B) .0 N m3 4 −

(C) .0 N m4 5 − (D) . N m6 1 −





FM 9.43 If the resisting torque is negligible, what would be the angular velocity of the manifold ?(A) 1.705 /rev s (B) 1 .05 /rev s7

(C) 1. 7 /rev s2 8 (D) 1 .7 /rev s2 8

Common Data For Q. 44 and 45A Francis radial-flow hydroturbine is being designed with the following dimensions, where location 2 is inlet and 1 is outlet :

2��mr2 = , ��1 mb2 = , 1��2 mr1 = , 2�2�mb1 =Where radiusr = , blade widthb =Runner blade angle at inlet 2β .66 2c=Runner blade angle at outlet 1β .36 1c=Volume flow rate vo 340 /m s3=Runner speed N 180 rpm=Gross head Hgross 90.0 m=For the preliminary design, irreversible losses are neglected.

FM 9.44 What will be the angle 2α through which the wicket gates should turn the flow and the swirl angle 1α ?(A) 302 cα = , 101 cα = (B) 602 cα = , 801 cα =(C) ,802 cα = 601 cα = (D) 102 cα = , 301 cα =

FM 9.45 The power output and required net head respectively, are(A) 299.6 MW, 73.9 m (B) 246 MW, 73.9 m

(C) 246 MW, 90.0 m (D) 299.6 MW, 90.0 m

FM 9.46 An inward-flow radial turbine involves a nozzle angle 601 cα = and an inlet rotor tip speed � �m sU1 = as shown figure below. The ratio of rotor inlet to outlet diameters is .2 0. The absolute velocity leaving the rotor at section (2) is radial with a magnitude of 6 /m s. If the fluid is water, the energy transfer per unit mass of fluid flowing through this turbine is

(A) 7.8 /m s2 2− (B) . /m s15 6 2 2−(C) 7.8 /m s2 2 (D) . /m s15 6 2 2

FM 9.47 An axial flow fan has a blade-tip diameter of 1 m and a root diameter of 80 cm and it operates in sea-level air at 1200 rpm. The inlet angles are 551 cα = and 301 cβ =





while at the outlet 602 cβ = . What will be the horse-power ? ( �.�� air�ρ = )

(A) 16 hp (B) 8 hp

(C) 4 hp (D) 32 hp

FM 9.48 The average radius of a pelton wheel is 1.80 m. A jet of velocity 100 /m s is strikes to bucket from a nozzle of 10.0 cm exit diameter. The turning angle of bucket is 165cβ = . If wheel rotates at 270 rpm and the efficiency of the turbine is 82 percent, the output shaft power in MW is(A) 9.67 (B) 0.0547

(C) 5.47 (D) 3.16

FM 9.49 An idealized radial turbine is shown in figure below. The absolute flow enters at 25c with the blade angles as shown. The flow rate is 480 /minm3 of water at 20 Cc. If the blade thickness is constant at 20 cm, the theoretical power developed at

%100 efficiency will be

(A) 800 kW (B) 400 kW

(C) 250 kW (D) 375 kW

FM 9.50 An inward flow radial turbine involves a nozzle angle of 60c and an inlet rotor tip speed of 9 /m s as shown in figure. The radial component of velocity remain constant at 6 /m s through the rotor and the flow leaving the rotor at section (2) is without angular momentum. The ratio of rotor inlet to outlet diameters is .2 0. If the flowing fluid is air and the static pressure drop across the rotor is 0.07 kPa, the loss of available energy across the rotor and the rotor efficiency respectively, are

(A) 21.75 /m s2 2, 68.9% (B) . 5 /m s13 0 2 2, 42.15%

(C) . /m s17 4 2 2, 84.3% (D) . /m s34 8 2 2, 63.25%





FM 9.51 The velocity triangles for water flow through a radial pump rotor are as shown in figure below. What will be the energy added to each unit mass (kg) of water as it flows through the rotor ?

(A) 323 /m s2 2 (B) 3 3 /m s0 2 2

(C) /m s505 2 2 (D) /m s404 2 2

FM 9.52 A centrifugal water pump having an impeller diameter of 0.5 m operates at 900 rpm. The water enters the pump parallel to the pump shaft and the exit blade angle 2β is 25c as shown in figure. The uniform blade height is 5 mm0 . When the flow through the pump is 0.16 /m s3 , the shaft power required to turn the impeller is(A) 72.3 kW (B) .3 kW90 8

(C) . kW45 18 (D) . kW54 22

FM 9.53 Consider a hydraulic turbine runner as shown in figure. Relative to the rotating runner, water enters at section (1) ( radius 1.�mr1 = ) at an angle of 100c from the tangential direction and leaves at section (2) ( radius �.��mr 2 = ) at an angle of 50c from the tangential direction. The blade height at sections (1) and (2) is 0.45 m each and the volume flow rate through the turbine is 30 /m s3 . The runner speed is 130 rpm in the direction shown. What will be the shaft power developed ?

(A) 12.8 MW− (B) . MW9 6−(C) . MW9 6 (D) 12.8 MW

***********





SOLUTIONS


P. Wind mills 4. Dynamic turbine

Q. Fans 1. Dynamic pump

R. Water-meters 2. Positive-Displacement turbine

S. Heart 3. Positive-Displacement Pump

FM 9.2 Option (B) is correct

List-I List-II

P. Steam Engine 2. Diagram factor

Q. Impulse turbine 1. Velocity compounding

R. Reaction turbine 3. Continuous pressure drop.

S. Centrifugal compressor 4. Isentropic efficiency

FM 9.3 Option (C) is correct.Conservation of mass requires that

mino mout= o

V Ain in inρ V Aout out outr=The cross-sectional area is proportional to the square of diameter

Vout V DD V D

Din

out

in

out

inin

out

in� �

rr= =: :D D

For D D<out in , the ratio ( / )D D 1>in out . Hence Vout V> in

FM 9.4 Option (D) is correct.(a) False : Actually, backward-inclined blades yield the highest efficiency.

(b) True : There is no flow rate at shutoff head. Thus pump is not doing any useful work, and the efficiency must be zero.

(c) True : This is the primary reason for choosing forward-inclined blades.

(d) True : There is no head at the pump’s free delivery. Thus, the pump is working against no “resistance” and is therefore not doing any useful work and the efficiency must be zero.


Since Pwater gvH v pr D= =o o p gHρΔ =

2701000 601500#

#= 6.75 kW=

and Efficiency η input poweroutput power= .

.9 006 75=

η %75=

FM 9.6 Option (C) is correct.The ideal power of the pump is





Pideal gHvr= o

. .998 9 81 1 6 60 10018

# # ##

= 18 / . /minL m s60 100018 3

#=

4.7 W=Hence, the actual power (bhp) is

bhp Pideal pumph#= . .4 7 0 70#= 3.3 W=

FM 9.7 Option (B) is correct.For water at 20 Cc , take 998 /kg m3ρ = . The power relation is

P gvHh

r=o

or (22 745.7)# .

. H

0 71

998 9 81 3600125

# # #

=b l

22 22 745.7hp W#=

H .. .

998 9 81 12522 745 7 0 71 3600

# ## # #= 34 m,

Pressure rise pΔ .gH �� # #r= = 332873 Pa= 333 kPa-

FM 9.8 Option (B) is correct.The energy equation for the system.

Hpump z f dL

gV�

�

D= +

Velocity V ( . )

��.� �m sAv

� ��

�#

,p= =o

Thus Hpump 11 4 0.0144 . .( . )

0 220 12 8

2 9 8118 6 2

## #= − + + +

58 m,


P. Draft tube 3. Diffuser

Q. Pumps 4. Pressure rise

R. Impulse-Turbine 1. Kinetic Energy

S. Reaction-Turbine 2. Momentum Exchange

FM 9.10 Option (C) is correct.1. True : Turbine govern helps in starting and shutting down the turbo unit.

2. True : Turbine governor controls the speed of turbine set to match it with the hydroelectric system.

3. True : Governor sets the amount of load, which a turbine unit has to carry.

FM 9.11 Option (C) is correct.For water at 20 Cc take 998 /kg m3ρ = . Each arm takes 7 /m h3 .

Vrel 2( . )A

v

� ��3��

�

exit 2#

p= =o

38.7 /m s= (At max power)

u 1�.3 �m sR V21 �relw= = =

or .0 18ω 19.35=





ω .. 107 /rad s0 18

19 35= =

ω 107 1030 rpm260 ,p#=

FM 9.12 Option (D) is correct.P True : Draft tube increase the net head on the turbine by providing the

negative head at outlet.

Q. True : Draft tube recovers kinetic energy leaving the turbine.

R. True : Draft tube is a diffuser that also turns the flow downstream horizontally of a turbine.

S. False : Draft tube does not decreases the net head on the turbine.

FM 9.13 Option (A) is correct.Since dimensionless turbine parameter head coefficient.

CH D

gHN DgH

2 2 2 2w= = N

602ω π=

For full scale turbine A:

C �H A N DgHA A

A� �=

and for Quarter scale Turbine B : ( / )D D 4B A=

C �H B N DgH

N DgH

�B B

B

BA

B� �

��

#

= =b l

CH for both turbine must be same, therefore

C ,H A C ,H B=

N DgHA A

A� �

N DgH ��

B A

B� �#=

NB� H

N H ��A

A B�# #=

NB �N HH

AA

B# #= 428 4 988 rpm30

10# #= =

FM 9.14 Option (D) is correct.Affinity or scaling laws

For capacity vv

A

Boo

DD

A

B

A

B�

#ww= b l

For net head HH

A

B DD

A

B

A

B� �

#ww= a bk l

For net Power bhpbhp

A

B DD

A

B

A

B

A

B� �

#rr

ww= a bk l

P. True : Rotation rate appears with an exponent of 1 in the affinity law for capacity. Thus, the change is linear.

Q. False : Rotation rate appears with an exponent of 2 in the affinity law for net head. Thus, if r.p.m is doubled, the net head increases by a factor of 4.

R. False : Rotation rate appears with an exponent of 3 in the affinity law for shaft power. Thus, if r.p.m is doubled, the shaft power increase by a factor of 8.

S. True : rotation rate appears with an exponent of 3 in the affinity law for shaft power. Thus if r.p.m is doubled, the shaft power increases by factor of 8.






Specific speed Ns g h

v�

a��

#

w= o

6 @

For 60 /rad sω = , � �m sv �=o , �� m sg �= and ��mha =

Ns .9 81 2060 3

/3 4#

#=6 @

.1 98=

For �N ��s = , the pump is a mixed-flow pump.

FM 9.16 Option (D) is correct.Since the turbine (B) is dynamically similar to the turbine (A).So, at the best efficiency point, from scaling laws.

DB ( . )D HH

NN ��

��

AA

B

B

A# # # #= =b bl l 2.073 m=

and vBo ..v N

NDD ��

��

AA

B

A

B� �

# # #= =o b b b bl l l l 342 /m s3=

and bhpB bhp NN

DD

AA

B

A

B

A

B3 5

# #rr= b b bl l l

..132 998

998300240

1 52 0733 5

# # #= b b bl l l 340.71 341MW,=

FM 9.17 Option (B) is correct.For similarity the model pump operate at the same flow coefficient, so that

Dv

m�ω

oc m

Dv

p�w

= oc m

where the subscript (m ) refers to the model and ( )p to the prototype. Thus,

vmo DD v

p

m

p

mp

�

#ww= oc m

and with m pω ω= , � �D D ��m p = , and �.� �m svp�=o ,

We get vmo 1 4.1 0.0328 /m s51 3

3# #= =b l

Also D

gha

m2 2ω

c m D

gha

p2 2w

= c m

So that ham gg

DD h

m

p

p

m

p

ma

� �

p# # #ww= b cl m

and with g gp m= , m pω ω= , � �D D ��m p = , and ��mhap =

We get ham 1 (1) 200512

2

# # #= b l 8.00 m=

FM 9.18 Option (A) is correct.Since the two pumps are dynamically similar, dimensionless pump parameter head coefficients CH must be the same for both pumps

C ,H A C ,H B=

D

gHA A

A� �ω

D

gHB B

B� �w

=

B

Aωω H

HDD

B

A

A

B#= b l ...(i)

Similarly, dimensionless pump parameter capacity coefficient Cvo must be same for both pumps

C ,v Ao C ,v B= o





D

vA A

A�ω

o

DvB B

B�w

= o

DD

A

B�

b l vv

B

A

A

B#w

w= oo

...(ii)


DD

A

B�

b l HH

DD

vv

B

A

A

B

A

B# #= o

o

DD

A

B�

b l HH

vv

B

A

A

B#= o

o or DB D H

Hvv� �

AB

A

A

B��

# #= oo

b cl m


Pump specific speed : NSP ( )gH

v/

/

3 4

1 2w= o ...(i)

Turbine specific speed: NST ( )

( )gH

bhp/ /

/

5 4 1 2

1 2

# rw= ...(ii)

Dividing and multiplying equation (ii) by v ��o ,

NST ( )

( )bhpgH v

v/ /

/

/

/

5 4 1 2

1 2

1 2

1 2

##

rw=

oo

By Rearranging, NST ( ) ( )

( )bhpgH

vv gH3/4

1/2

/ / /

/

1 2 1 2 1 2

1 2

# #

wr

#= oo= G

NST b�pNvgH

N�

SP SP turbine

��

# #r h= =oc m From eq (i)

where vgHbhp

turbineρ η=o and ( )gH

v N/

/

SP3 4

1 2ω =o

FM 9.20 Option (B) is correct.Since draft tube is used only in reaction turbines. The propeller turbine, Kaplan turbine and Francis turbine are of reaction type, but Pelton turbine is a impulse turbine.

FM 9.21 Option (B) is correct.At homologous points, the affinity laws are used to estimate the operating conditions. Let the original pump be A and modified pump be B .

vBo �. �� .� �� m sv DD � ��

��A

A

B

A

B�

� �

ww

# # # #= = = −o b l

Here D DB A= , N N�BA= and N

602ω π=

and HB ��

(�)H DD H

NN

� ��

AA

B

A

BA

A

B� � ��

ww

pp

# # # #= =a b ek l o

�.� �.��mH NN

��

AA

B� �

# #= = =b bl l

Now, the ratio of specific speeds

NN

SB

SA ( )

( )gH

vv

gHvv

HH

/

/

/

/ / /

A

A A

B B

B

B

A

B

A

A

B3 4

1 2

1 2

3 4 1 2 3 4

##

# #w

w ww= =

o

o oo

c bm l

NN v

vHH

602

260 / /

A

B B

A

A

B1 2 3 4

# # #p

p= oo

c bm l

. .

.NN

vv

HH

��

��

��

B

A

B

A

A

B��

�

� ��

# # ##

##= = −

−

oo

c b c bm l m l

NN

SB

SA 1=





FM 9.22 Option (C) is correct.We use best efficiency point ( )BEP data to calculate power. At BEP %65η = ,

��rpmN = , ��.� � �.�� m � m sv � �= =o and at BEP

Pmodel g vHr h# #= o

. . .998 9 81 0 00464 15 0 65# # # #= 443 W=First establish the BEP coefficients from the model turbine data.

Cvo ( . )

. �.��NDv

��

��m� �

#

= = =o

CH .

. ��.��N DgH

��

�� m

m� � �

�#

#= = =b ^l h

CP ( . )

�.�N DP

��

��m

m� � �

�# #

r= = =

b l

Now enter new data for geometrically similar turbine

Cvo 0.198 .N D

vN D

0 19t t

t

t t3 3

#= = =o

or N Dt t�

# .. 0.960 198

0 19= = ...(i)

CH .12 45= .N D

gHN D��

t t

t

t t� � � �# #

#= =

or N Dt t� �# .

. 3 .2512 459 81 46 6#= = ..(ii)

From equation (i) and (ii)

Nt .D��

t�= and N Dt t

� �# .36 25=

or .D

D��t

t�

��

#c m .36 25=

( . )

DD

��t

t�

��

# .36 25=

Dt� .

( . ).36 25

0 960 02542

2

= =

or Dt 0.399 0.40 mb= and

Nt .

( . ). ��rps

D��

��

t� �= = =

Then Pt C N DP t t� �r# # #=

. ( ) ( . )1 6 998 15 0 403 5# # #= 55185 W=

or Pt . 74 hp745 755185= =

FM 9.23 Option (C) is correct.For flow from section (1) to section (2), energy equation gives

p V gz22 2

2

2ρ + + lossp V gz w2 shaft net in1 1

2

1r= + + + −

Since p p patm2 1= = , V �1 = and w wshaft net out shaft net in=−

wshaft net out ( ) lossg z z V21 222

= − − − ...(i)

For power, we multiply equation (i) by the mass flow rate mo to get

Pshaft net out ( ) lossmg z z mV m21 222

= − − −o o o





Pshaft net out � � ��vg z z vV v��

r r r= − − −o o o m vr=o o

(999 30 9.81 100)# # #=

999 30( )

(999 30 20 9.81)22 2

## # # #− −

23.5 10 /N m s6# −= 23.5 MW=


Convert ��p�= 13332 Pa= p gHr= ��p�= 66661 Pa= oilγ 0.91 9790 8909 /N m3

#= =

and V� ( . ). �.�� sA

v��

��

�#

= = =po

V� (�.��). �.�� sA

v ��

�#

= = =po

So, the head is H pg

V z pg

V z� ��

�

��

�

�g g= + + − − −

.( . )

. 2 9.( . )

890966661

2 9 815 79

0 65 890913332

811 00

02 2

# #= + + − − − −

11.3 m=Therefore input power

Pinput .( . ) .vH

�� # #

hg= =o

152 W5=

FM 9.25 Option (C) is correct.The head rise gained by fluid flowing through a pump is

ha p p z z g

V V�

� ��

��

��

g= − + − + − ...(i)

Since V� ( . ). �.�� m sA

v��

��

�

�#= = =p

−o


V A� � V A� �=

V� �.�� (�) �.� �m sV ��

��

#= = =b l

Thus, from equation (i), with ( )( ) ( . ) �� p h ��Hg Hg��g #=− =− ^ h

and p� 80 10 /N m3 2#= ,

ha . .

( . )( )0.5 .

( . ) ( . )0 9 9 80 10

80 10 0 095 133 102 9 81

3 19 0 7973

3 3 2 2

# #

# #

#= + + + −

^ h

ha 11.5 m=







pg

V z h� p� �

�

�γ + + + pg

V z fDl

gV

� ��

�

��

g= + + + ...(i)

where p p �� = = , V ��= , � �m sV�= , �mz�= , and z ��=Thus,equation (i) becomes

z hp�+ gV f D

l2 12

2

= +b l

hp .( )

. . 3 25.3 m2 9 813

1 0 0152 0 052002

##= + − =b l< F

Hence, Power gained by fluid

P vhpg= o

(9.731 10 ) (0.05) 3 25.343 2

# #p

# #=

1.45 10 1.45 kW3#= =

and Power supplied to fluid EfficiencyPower gained by fluid=

.. 2.07 kW0 7

1 45= =


We have NPSH pg

V p�

s s v�

g g= + − ...(i)

where ps and Vs refer to the pressure and velocity at the pump inlet, respectively.

Also, pg

V z��

�

�γ + + pg

V z h�s s

s L

�

g= + + +

So that with p patm�= , V ��= , z �s = and h �L =

p zatm�γ + p

gV�

s s�

g= + ...(ii)

and therefore from equation (i) and (ii), the available NPSH is

NPSHA p z patm v�g g= + − ..(iii)

With z� positive (since pump is below reservoir) and h �LΣ = .

Thus, from equation (ii) with ��Pap .atm =

NPSHA .

3( . )( . )

9 731 10101 10

9 731 107 376 10

3

3

3

3

#

#

#

#= + − 12.6 m=






From Bernoulli’s equation at section 1 and 2,

pg

V z h2 p1 1

2

1γ + + + p V z hL2 2

2

2g g Σ= + + + ...(i)

and with p p �1 2= = , V V �1 2= = , and ��mz z2 1− = , equation (i) becomes

hp 33 hLΣ= + ...(ii)

The head loss term can be expressed as

hLΣ K f Dl

gV2L

2

#= +b l

With K 1L = , .f ��2= , ��ml = , �.1 mD = , equation (ii) can be written as

hp 33 . . . .V1 0 0 02 0 1

302 9 81

2

##

#= + +; E ...(iii)

and with V (�.1)A

v v

�2p= =o o

equation (iii) becomes hp 33 . . 826 v1 0 6 0 2# #= + + o6 @

or hp 33 5.78 10 v3 2#= + o ...(iv)

Since the pump equation is

hp 52.0 1.01 10 v3 2#= − o ...(v)

Equation (iv) and (v) can be equated to determine the flow rate. Thus,

33 5.78 10 v3 2#+ o 52.0 1.01 10 v3 2

#= − o

and vo 0.0529 /m s3=

FM 9.29 Option (B) is correct.We consider the turbine inlet and discharge to be sections (1) and (2).For flow from sections (1) to (2), energy equation gives

loss ( )p p g z z wshaft net out1 2

1 2r= − + − − ...(i)

Since V V1 2= and w wshaft net out shaft net in=−For power loss through the turbine we need to multiply equation (i) by the mass flow rate, mo . Thus

Ploss ( )m p p mg z z Pshaft net out1 2

1 2r= − + − −o ob l ...(ii)

However mo �� .2� �2�� g svr #= = =o

Also p2 (0.25 )m of Hg gHgr# #=−or 0.25 13.6 999 9.81# # #=− 33300 /N m2-−From equation (ii), we get

Ploss 4246( )

4246 9.81 3 1.1 10999415000 33300 6

# # # #= + + −

930346 / 930N m s kW-−=

FM 9.30 Option (C) is correct.The efficiency of the pump is given by

η actual work requiredideal work required

actual work requiredactual work required loss= = −

loss

ww

shaft net in

shaft net in= −

Now find wshaft net in , by using the energy equation

wshaft net in lossp p V V2

out in out in2 2

r= − + − + ...(i)





From the volume flow rate, we obtain

Vout ��

�� m sAv

Dv

� ��

��out out

� �#p p

= = = =o o

Also from principle of mass conservation,

Vin ��

�� m sVDD

��

outin

out�

�

�

�

#= = =

Thus from equation (i), we get

wshaft net in ( ) [( . ) ( . ) ]

170999400000 120000

228 29 3 1432 2

= − + − +

846 /N m kg−=

Then η 0.799846846 170= − =

FM 9.31 Option (C) is correct.Output shaft power for Pelton wheel is

Pshaft ( )(� )cosrv V rjrw w b= − −o ...(i)Differentiate equation (i) with respect to ω and set the derivative equal to zero.

( )( )cosP dd rv V r �max jw rw w b= − −o6 @ 0=

(� ) ( )cosrv dd V rj

�ρ β ω ω ω- -o 0=

( )dd V rj

�

ω ω ω- 0=

�V rj w− 0=where rω =Turbine bucket speed and �V �j = Jet speed

Hence Turbine bucket speed jet speed21#=


The power output for a Pelton wheel is given by

Pshaft ( )( )cosvU U V ��r b= − −o

For maximum power 180cβ = and U V��=

Thus, P maxshaft vV��

r=− o ...(i)

From the Bernoulli’s equation at section (0) and (1),

pg

V z�0 0

�

0γ + + pg

V z fDl

gV

� �1 1

�

1

�

g= + + +





where p p �� = = , ��z�= , ��z�= and V ��=

Hence z� z gV fD

lg

V� ��

��

= + + ...(ii)

Also A V� � AV= or D V4 12

1π D V4

2p=

That is V .. �.��D

D V V V��

�

�

�

� �= = =b bl l

So that equation (ii) becomes.

975 250 . . . ( . )V2 9 81 1 0 032 0 9

1020 0 049412

2

##= + + b l; E

or V� 114.3 /m s=

Hence vo (�.�) ��.� �.5 �m sA V � 9� �� p#= = =

Therefore, from equation (i), we get

P maxshaft (999) 3.5( . )

23. 10 /N m s9 2114 3

42

6# # # −=− =

23400 23.4kW MW= =


Since the power output of the pelton wheel is

Pshaft ( )( )cosvU U V ��r b= − −o

So the maximum power output occurs with 180cβ = and U V��= .

Thus, Pshaft vV��

r= o (In magnitude) ...(i)

At section (0) and (1), from Bernoulli’s equation

pg

V z�0 0

�

0γ + + pg

V z fDl

gV

� �1 1

�

1

�

g= + + +

But p p 00 1= = , V 00 = , and z z H0 1− =

Thus, h gV f D

lg

V2 2

12 2

= + ...(ii)

Since A V1 1 AV= or D V D V4 412

12π π=

We have V1 DD V

1

�

= c m Therefore,

From (ii), h gV f D

lDD

2 112

414

= +; E or gV

flDD

h2

1

12

514=

+< F

and equation (i) gives





Pshaft fDl D

vgh

fDl D

g D V h

1 1

4/

5 14

5 14

1 2

12

1#r r p=

+=

+

o

c cm m

...(iii)

but from (ii), V� fDl D

gh

1

2/

5 14

1 2=+c m

...(iv)

For this problem f , l , D , and h are constants; D� is variable

Thus, from equation (iii) and (iv)

Pshaft ( )CD

KD1 /

14 3 2

12

=+

where .constK = and .constC fDl

�#= =

Note : P �shaft = as �D�" and as D�"3. To find the D� that gives maximum

power over all, set dDdP

�shaft

�=

dDdPshaft

( ) ( )4 0

CDKD

CD

KDCD

12

123

/ /14 3 21

14 5 2

12

13

#=+

++

−=

b l

or ( ) ( )CD

KDCD

CD1

2 113 0/

14 3 21

14

14

+−

+== G

CD CD1 314

14+ = D C�

��& =

Thus, D� 2f

Dl f D

lD

2

1/ /

5

1 4 1 4

# # # #

= =b bl l


For the flow through the penstock,

pg

V z��

�

�γ + + pg

V z fDl

gV

� ��

�

�

�

g= + + +

where p p �� = = , V ��= , and z z H� �− =

Thus H gV f D

lg

V2 2

12 2

= + ...(i)

But A V� � AV= or D V4 12

1π D V4

2p=

So that V � DD V�

�

��= b l

From previous part, the maximum power occurs if �

DD

f Dl

��

��

#

=b l

or DD�

�b l

f Dl2

1

# #

=





So that V̀ � f D

lV

2

112=

b l

Thus,equation (i) becomes

H gV f D

l

f Dl g

V2 1

2

123

212

12

##

#= + =bb

ll> H

or gV2

12

H32=

FM 9.35 Option (C) is correct.For water, take 9 8 /kg m9 3ρ =

u� �.�r ��

� #w p#= = b l 9.90 /m s=

Here 302 cα = , 901 cα =

Vn� .

. ..

r b23 5

2 0 7 0 13 5

2 2 # #p p= = 7.96 /m s,

and Vt� . 13.8 /tan tan

m sV30

7 96n

2

2

ca= = =

Vt� tan tanV V

900n n

1

1 1

ca= = =

Thus, Ptheoretical vu Vt� �r= o . . .998 3 5 9 90 13 8# # #= 477 477W kW213 ,=

FM 9.36 Option (B) is correct.Normal velocity components at Inlet and Outlet are,

V ,n� . .. 4.224 /m sb r

v2 2 0 12 0 18

0 5731 1 # # #p p= = =o

V ,n� . .. 2.715 /m sr b

v2 2 0 24 0 14

0 5732 2 # # #p p= = =o

and tangential velocity components at Inlet and Outlet are,

V ,t� .tan tanV �� ,n� � # ca#= = = V ,t� �.�� .�� tan tan m sV ,n� � ca# #= = =Now, the net head produced by the pump is

H g r V r V gN r V��

�, , ,t t t� � � � � �#

##

w p= − =6 @ V �, t� =

H . 0.24 1.9 4.8 m60 9 812 1000 7

##p

# #= =

and required brake power at %76 efficiency is

bhp gvH pump#r h= o

. . . . .998 0 9 81 0 573 4 87 0 76# # # #= 20763 W=


Since T ( )m r V r V �� = − =q qo ,

then r V� �θ r V� �= q

But �� .� �sin sin m sV V� � c c= = =θ and V �θ ��sin sinV� #q q= =Hence, 1.9 13# 25 1.2sinq#= or θ .55 4c=


Torque T ( )m r V r V� � � �= −q qo





Substituting numerical values,

60− � �sin sinm �� # # # #c c= −o6 @

or mo 89.7 / 89.7 /N s m kg s−= =Also, P Tw=

So that ω TP

��= = −

− 30 60 286 rpm21

# #p= =b l


We have N 1160 rpm=

ω 121.5 /rad s602 1160#p= =

u� ��. . ��.� �m sr � �� w #= = =b l

Vn� ��.� ��tan tanu �� cb #= = 5. /m s1= vo r b V2 n1 1 1#p= 2 0.09 0.10 5.1p# # #= 0.28 2 /m s8 3= 0.28 2 36008 #= 103 /m hour8 3,


Vn� . ..

r bv

2 2 0 165 0 0750 2882

2 2 # #p p= =o 3. /m s71=

u� ��. �.��r ��w #= = . /m s20 05= Vt� ��cotu Vn� � c= − . 5 3. 40cot20 0 71 c#= − 15.6 /m s0=and Pideal �� .�� . � ��.�vu V � �� t� �r # # #= =o 6 7 W129=

Finally Head H ( ) . .gv

P680 9 81 0 2882

61297# #r= =o 32 m,


From the shaft power equation with V ��=θ

We have Pshaft ( )v U V U V vU V� � � � � �r r= − =q q qo o ...(i)

where U� �.�� .� �m sr �� w #= = =#p

To evaluate V �θ we use the exit velocity triangle shown below.

Thus, V �θ tan

U V��r

��

c= −

with Vr� . .. 7.07 /m sr b

v2 2 0 15 0 03

0 202 2 # #p p= = =o





Hence, V �θ 47.1 . 37.0 /tan

m s35

7 07c

= − =

From eq. (i), Pshaft 999 . 0 47.1 37.00 2# # #= 3.48 10 /N m s5- # −

348 kW=

FM 9.42 Option (C) is correct.From the moment of momentum equation, the shaft torque

T ( )m r V r V� � � �= −q qo

For this continuously distributed outflow with V ��=θ , it becomes.

T rV dm2= q o#

where � �m sV =θ and �.�� (�)drdm r# #=o

Thus, T 2 . ( )drr 3 0 003 999 3.

.

r 0 1

0 4

# # # #==

#

or T 53.9 2 4.05 N mr2

.

.

0 1

0 4−= =: D


With T 2 rV dm 0= =q o# ...(i)

where dmo �� (�.��)dr drWh # #r= = 8.99 dr= and

From the figure Vθ �W r rw w= − = −Thus, from equation (i)

0 2 ( )( . )drr r3 8 99.

.

0 1

0 4w= −# ( )r r dr3

.

.2

0 1

0 4w= −#

r r23

31

.

.2 3

0 1

0 4

w= −b l 0.225 0.021w= −

or ω 10.7 1.705 /rev s21p#= =

FM 9.44 Option (A) is correct.Normal component of velocity at inlet and outlet are

V ,n� . 37.0 /m sr bv

2 2 2 0 731340

2 2 # # #p p= = =o

V ,n� . . 17.33 /m sr bv

2 2 1 42 2 2340

1 1 # # #p p= = =o

and tangential velocity components at inlet and outlet are

V ,t� tanrV ,n

��

�w b= − 2( . )tan60

2 18066 237#

cp

#= − N60

2ω π=

21.3 /m s8=

and V ,t� . .( . ).

tan tanr

V��

��,n

��

�# c

w b= − = − 3.0 /m s=





Since V �t� tanV �n� �# a=

tan 2α �VV

��

�

�

n

t

�

�= =

2α ( . )tan 0 58 301 c= =−

and V �t� tanV �n� �# a=

tan 1α �VV

��

�

�

n

t

�

�= =

1α ( . ) .tan 0 174 9 9 101 c c,= =−

FM 9.45 Option (B) is correct.The shaft power is estimated from the Euler’s turbo-machine equation

Pshaft ( )v r V r V� �t t� � � �rw= −o

998 18.85 340 [2 21.3 1.42 3.0]8# # # # #= − 2.46 10 246W MW8

b#=Assuming ��turbineη = . Since the irreversibility neglected. Therefore

Net Head H .gvbhp

��

# ##

r= =o

73.9 m=

FM 9.46 Option (B) is correctThe energy transfer per unit mass of fluid is given by the relation

wshaft U V U V� � � �= −q q

Also m�o r bV r bV2 2r r1 1 2 2p p= =

or V� � �cos cos

m sr

r V��

��

�

� �

c c#= = =

Then V �θ � �� .�� sin sin m sV� � ca #= = = and V ��=θ

Thus wshaft (�) �.��U V� � #=− =−q 15.6 /m s2 2=−


The average radius R . . 0.45 mR R2 2

0 5 0 41 2= + = + =

Thus, u �� ( . )R �� # #w p= = b l 56.6 /m s=

Also u ( ) ( )cot cot cot cotV Vn n� � � � � �a b a b= + = +

or Vn� ( ).

cot cot cot cotV u

��

n� c c c c= = + = +

Vn� ��.� �m sVn�,=Then vo . ��( . ) ( . ) ��V A �� n

� �# p= = − 6.56 /m s3=

and gH ( )cot cotu uVn�

� �a b= − + ( . ) ( . ) ( . )( )cot cot56 6 56 6 23 2 55 602

# c c= − + 1520 /m s2 2=Finally P vgHr= o . .1 205 6 56 1520# #= 12000 W= 16 hpb





FM 9.48 Option (D) is correct.The volume flow rate of the jet is

vo �� .��( . )

V D� ��

j j�

�p# # #= = 0.785 /m s3=

and ω 28.26 /rad sN60

260

2 270# #p p= = =

The ideal shaft power is

Pideal ( )(� )cosrv V rjrw w b= − −o

998 28.26 1.8 0.785(100 28.26 1.8)# # # #= − (1 165 )cos c# − 3.85 MW=and actual shaft power is

PActual �.�� .�� .��Pideal turbineh# #= = =

FM 9.49 Option (A) is correct.The inlet (2) and outlet (1) velocity vector diagrams are shown below. The normal velocities are.

Vn2 Av

r b2��

2 2 2# #p= =o b l . . 5.31 /m s2 1 2 0 2

8# #p= =

Vn1 . .Av

r bv

2 2 �� 2�

1 1 1 # #p p= = =o o 7.96 /m s=

From these we can compute the tangential velocities at each section.

u2 .r �� 2 1 22 # #w p= = b l 10.1 /m s=

u1 .r �� 2 ��1 # #w p= = 6.70 /m s=

Vt2 cotV 2�n2 c= 5.31 . /cot m s25 11 4c= = Vt1 cotV ��n1 c= 7.96 30 2.11 /cot m sc#= =Then, Ptheoretical ( )v u V u Vt t2 2 1 1r= −o

998 8 [(10.1 11.4) (6.70 2.11)]## # #= − 800000 800W kW,=


We know loss p p wshaft�1 �2

r= − +

Where p p�1 �2− stagnation pressure drop across rotor psD= =and wshaft U V U V U V2 2 1 1 1 1= − =−q q q since V �2 =θ

Thus, wshaft (9) (12 30 ) 93.5 /cos m s2 2c#=− =−





Also psΔ � �p p V V��

� � ��

��r= − + −

0.07 1.23 (12) (6)21

1012 2

3# #= + − c m6 @

(0.07 0.0664) 0.1364kPa kPa= + =

Thus loss .. 93.5 17.4 /m s1 23

0 1364 1032 2#= − =

and η

��

pw

��

s

shaft

rD

= − =d bn l

0.843 . %84 3= =


From shaft power equation

wshaft U V U V� � � �= −q q ...(i)

Since the relative velocity at the exit is in the radial direction (see the figure

below), V �θ �� m sU�= =

Also, from the inlet conditions,

V �θ tanV ��r� c=− , where the minus sign means that V �θ

is in the opposite direction of U�.

From conservation of mass (with 1 2ρ ρ= )

We have V Ar� � V A W Ar� � � �= =

Thus, Vr� W AA W r b

r b W rr

��

��

��

� �

� ��

�

�# # #p

p= = = b b� �=

Also, U� r�w= and U r� �w=

or rr

�

� �.�UU

��

�

�= = =

Thus Vr� . 32 /m s0 516= =

So that V �θ �� (��) �� .� �tan tan m sVr�# c c=− =− =−

From eq. (i), wshaft 6 16 8 ( 18.5) 404 / 404 /m s N m kg1 2 2# # −= − − = =


Power Pshaft T Tshaft shaftN

��w #= = p

and Tshaft ( )v r V r V� � � �r= −q qo

With V ��=θ Tshaft vr V� �r= qo ...(i)





From figure cot 2β VU V

r�

� �= − q

So that V �θ �otU Vr� � �b= − ...(ii)

For r� . 0.2 m2

0 5 5= = and 2 900 94.2 /rad s60#ω π= =

U� �.�� .� ��.� �m �r� #w= = =Since the flow rate is, vo 2 r b Vr2 2 2p=

or Vr� . .. 2.04 /m sr b

v2 2 0 25 0 05

0 162 2 # #p p= = =o

Thus, from equation (ii),

V �θ (23.6 2.04 25 ) /cot m sc= − 19.2 /m s2=From equation (i), we get

Tshaft 999 0.16 0.25 19.2 768 N m2# # # −= =So Pshaft 768 .94 2 72346#= = 72.3 kW-

FM 9.53 Option (A) is correctNote that the shaft power calculated below, Pshaft is less than the power lost by the fluid because some of the power lost by the fluid is due to the fluid and shaft bearing friction while the rest is available at the shaft.

Pshaft ( )m U V U V� � � �= −q qo ...(i)

where mo �� g svr #= = =o

and U rw= with ω 2 13.6 /rad s60130#p= =

Hence, U� ��.� �.� ��.� �m sr� #w= = =and U� ��.� �.�� .�� m sr� #w= = =

From the figure, V �θ ��cosU W� � c= + ...(ii)





and since vo 2 80sinr bW1 1 cp= 30 2 1.5 (0.45) 80sinW1 cp# #=or W� 7.18 /m s=Thus, from (ii), V �θ 20.4 7.18 80 21.6 /cos m sc= + =

Similarly, since vo 2 50sinr bW2 2 cp= 30 2 0.85 (0.45) 50sinW2 cp# #=or W� 16.3 /m s=Thus, V �θ �� cos cosU W� � c c= − = − 1.08 /m s=Substitute these values in equation (i), we get

Pshaft (29970) . . . .11 56 1 08 20 4 21 6# #= −6 @

1.28 10 /N m s7# −=− 12.8 MW=−

***********

me vol 2 fm

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