ME Vol 2 FM

GATE

MECHANICAL ENGINEERING

NODIA AND COMPANY

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Multiple Choice Questions GATE Mechanical Engineering Vol 2, 1e Copyright © By Publishers ISBN 9-788192-27629-8 Information contained in this book has been obtained by authors, from sources believes to be reliable. However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineering or other professional services.

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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 MATHEMATICS Linear 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 DESIGN Engineering 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 SCIENCES Fluid 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 ENGINEERING Engineering 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 APTITUDE Verbal 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 1 BASIC CONCEPTS AND PROPERTIES OF FLUIDS

Common Data For Q. 1 and 2 In 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- m3 /kgK . FM 1.1

The pressure in the tire at air temperature of 322 K 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) 0.0140 kg (D) 0.3484 kg

FM 1.3

Consider Carbon dioxide at 12 atm and 400cC . What will be the density of Carbon dioxide and c p at this state and the new pressure when the gas is cooled isentropically to 150cC ? (For Carbon dioxide k = . and R = m2 s2 ) (A) ρ = 0.797 kg/m3 , c p = 4 . (B) ρ = 1.3 # 10-4 kg/m3 , c p = (C) ρ = 7.97 kg/m3 , c p =

kg −

kg −

, p2 = kg − , p2 =

kPa

, p2 =

5.5 kPa

5.5 kPa

, p2 = 5.5 Pa kg − A Cane of beverage contains 455 ml of liquid. The mass of cane with liquid is 0.369 kg while an empty cane weighs 0.193 N . 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 (D) ρ = 7.97 kg/m3 , c p =

FM 1.4

FM 1.5

The specific gravity of a gas contained in a tank at the temperature of 25cC 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 : (p1, T1)= (1.31 MPa, 250cC), Rsteam = 4 m2 s2− and k = . ). If the steam expands isentropically to a new pressure of 414 kPa, what will be the density ρ1 and the density ρ2 ? (A) ρ1 = 5.44 kg/m3, ρ2 = 5.04 kg/m3 (B) ρ1 = 2.28 kg/m3, ρ2 = 5.44 kg/m3 (C) ρ1 = 5.44 kg/m3, ρ2 = 2.28 kg/m3 (D) ρ1 = 5.04 kg/m3, ρ2 = 5.44 kg/m3

FM 1.7

A 30 m3 cylinder contains Hydrogen at 25cC and 200 kPa What amount of

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Page 4

Basic Concepts and Properties of Fluids

FM 1

Hydrogen must be bled off to maintain the Hydrogen in cylinder at 20cC and 600 kPa ? (R = 0.2968 kPa.m3 /kg.K) (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 30cC and 1 atm. If Rair = m s− 2 2 , Rwater = 461 m /s −K and vapor pressure of saturated water at 30cC 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 80cC . 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 30cC water ( pv = 4.242 kPa ) 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 150cC and 400 kpa? (k = 1.33, R = 461 m2 /s2−K ) (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 9268 N/m3 and dynamic viscosity of 131.5 N s/m2 . What will be the kinematic viscosity of the liquid in m2/sec ? (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 = 2 pLR2 . The weight of 20cC gas within the blimp 3 for (a) helium at 1.1 atm and (b) air at 1.0 atm, is ( RHe = 2077m2 /s2 − , Rair = 287 m2 /s2−K ) (A) WHe = 60.97kN , Wair = 401.1kN (B) WHe = 401.1kN , Wair = 6.97kN (C) WHe = 6.2kN , Wair = 40.9kN (D) WHe = 40.9kN , Wair = 6.2kN

FM 1.15

The oil having viscosity of 4.56 # 10−2 N − s/m2 , 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 0.129 m2 , what value of F is required to translate

−

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Page 5

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FM 5

the plate with velocity of 1 m/ sec ? (A) 2.32 N (C) 232 N FM 1.16

FM 1.17

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

(B) F = :

(C) F = ; h2 − h E VA m2 m

(D) F = :

m m + 2 D VA h h2

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 m/sec , what will be the magnitude of the shearing stresses on plate 1 and plate 2 respectively, that act on the fixed plates ?

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

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

V V - Vm (A) μA ; p + p h1 h2 E μAVp (C) h1 FM 1.19

m2 m − VA h2 h D

(A) F = ; h + h2 E VA m m2

(A) 10 N/m2 , 15 N/m2 (C) 15 N/m2 , 15 N/m2 FM 1.18

(B) 23.2 N (D) 0.232 N

(B) μA (Vp - Vm) h2 (V - Vm) V (D) μA ; p - p E h1 h2

A Newtonian fluid having the specific gravity of 0.91 and Kinematic viscosity of 4 # 10−4 m2 / sec , flows over a fixed surface. The velocity profile near the surface

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FM 1

is given by the relation: u = sin py a 2d k U What will be the magnitude of the shearing stress developed on the plate in term of U and δ ? (B) 5.71 δ N/m2 (A) 0.571U N/m2 U δ (C) 5.71U N/m2 (D) 0.571 δ N/m2 U δ FM 1.20

A 50 cm # 30 cm # 20 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 ? (g = 10 m/s2)

(A) 55 N (C) 6.42 N

(B) 55.55 N (D) 414.75 N

FM 1.21

A closed rectangular container is half filled with water at 45cC . 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 45cC is9.593 kPa) is (A) P > Psaturation (B) P < Psaturation (C) P = Psaturation (D) Not determined

FM 1.22

Consider two parallel plates as shown in figure below. If the fluid is glycerin (ρ = 1264 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 V = m s and the Reynolds number respectively ?

(A) 100 Pa, 460 (C) 10000 Pa, 4.6 FM 1.23

(B) 10 Pa, 4600 (D) 1000 Pa, 46

The velocity profile in a pipe flow is given by u = u ( − rn Rn), 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

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FM 7

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

(A) πnμu L (C) 2nπμu 0 L

(B) nμu R (D) 2nπμu 0

FM 1.24

Consider air at 20cC with μ = 1.8 # 10-5 Pa - s . Its viscosity at 400°C by (a) The Power-law (n=0.7) (b) the sutherland law (S = 110 K) respectively, are - s, μs = 1. # 10 -s (A) μp = . 1 # 10- s, μs = -s (B) μp = . 1 # 10. # 10 - s, μs = -s (C) μp = . . 1 # 10 # 10 - s, μs = -s (D) μp = 1. # 10. 1 # 10-

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

mgh sin q mA mgA sin q (C) V = mh (A) V =

FM 1.26

mgh cos q mA mgA cos q (D) V = mh (B) V =

A thin layer of glycerin flows down on an inclined plate of unit width with the velocity distribution: u = y −y U h h If the plate is inclined at an angle α with the horizontal, the expression for the surface velocity U will be

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FM 1

(A) U = h sin a gm (C) U =

gh m sin a

(B) U =

gh sin a m

(D)U =

gh m sin a

FM 1.27

A shaft of 8.0 cm diameter and 30 cm length is pulled steadily at V = m s through a sleeve of 8.02 cm diameter. The clearance is filled with oil of ν = 0.003 m2 /s and S.G. = . , the force required to pull the shaft is (ρw = 998 kg/m3) (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

3.

Dynamic viscosity

4.

Capillary depression

5.

Kinematic viscosity

6.

Capillary rise

c.

μ/ρ

d. Mercury in glass

Codes a 1 1 2 2

(A) (B) (C) (D) FM 1.29

b 2 2 1 1

c 4 3 3 5

d 6 4 6 4

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.

3.

Shear forces

4.

Surfaces Tension

Viscosity

d. Specific gravity Codes (A) (B) (C) (D)

a 1 1 4 4

b 4 4 1 1

c 2 3 2 3

d 3 2 3 2

FM 1.30

The hydrogen bubbles have diameter D - . 1 mm . Assume an ‘‘air-water” interface at 30cC and surface tension σ = 0.0712 N/m . 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 # 10−2 N/m . The

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FM 9

excess pressure inside the rain drop is (A) 973.3 Pa (C) 9.73 Pa

(B) 97.33 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 20cC water surface. How much lift force is required if σ = 0.0728 N/m ? (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 20cC and the values of the surface tension of water and mercury at 20cC 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 60cC . What will be the capillary rise if the fluid is (a) water ( σ = 0.0662 N/m , ρ = 983 kg/m2 , θ , 0c) and (b) Mercury ( σ = 0.47 N/m , ρ = 13500 kg/m3 , θ , 130c) ?

(A) hw = 0.0275m , hm =− 0.0456 m (C) hw = 0.0275 m , hm =− 0.00 1 m

(B) hw =− 0.0275 m , hm = 0.00 1m (D) hw = 0.0137 m , hm =− 0.00456 m

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

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FM 1.38

Page 10

Basic Concepts and Properties of Fluids

FM 1

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 θ < 45c . If the gap is very narrow, what will be the expression for the capillary rise h in the annulus gap ?

s cos q rg (ro − ri) s cos q (D) h = rg (r o − r i ) 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 s cos q rg (ro − ri) (C) h = s cos q rg (ro − ri) (A) h =

FM 1.39

Demo Ebook

(B) h =

Common Data For Linked Answer Q. 40 and 41 A Frustum-shaped body is rotating at a uniform angular velocity ω = 200 rad/s 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 20cC .

FM 1.40

The power required at the top surface to maintain this motion is πμω2 D3 πμω 2 D 4 (B) (A) 24h 32h (C)

FM 1.41

πμω 2 D 4 4h

(D)

πμω 2 D 2 16h

The reduction in power required at the top surface when oil viscosity is 0.0078 Pa − s at 80cC , will be (A) 5.29 W (B) 67.824 W (C) 62.533 W (D) No reduction

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FM 11

A fluid of surface tension σ = 0.0728 N/m and contact angle θ = 0c is filled between 0.75 mm apart two parallel plates as shown in figure. If the density of fluid is ρ = 998 kg/m3 , the capillary height h will be

(A) 2 mm (C) 20 mm

(B) 10 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 0.12 cm 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 (C) 0.173 N − s/m2 FM 1.45

(B) 0.0231 N − s/m2 (D) 0.0346 N − s/m2

A layer of water having the viscosity of 1.2 # 10−3 N − s/m2 flows down on inclined fixed surface with the velocity distribution as given by:

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FM 1

u = y −y U h h If the velocity of water U = m sec and h = m , what will be the magnitude of the shearing stress that the water exerts on the fixed surface in N/m2 ?

(A) 7.20 (C) 7.2 # 10−3

(B) 0.720 (D) 0.072

FM 1.46

A 2.5 mm diameter aluminum sphere ( ρ = 2700 kg/m3 ) 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 ω rad/s 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 ?

2T (ri - r0) πωr i L T( 0 - ) (C) μ = 2πω 3 (A) μ =

FM 1.48

T( 0 - ) 2πω 03 T( 0 + ) (D) μ = 2πω 3 (B) μ =

The velocity profile for laminar one-dimensional flow through a circular pipe is given as u (r) = u max ( − r2 R2), where R is the radius of the pipe and r is the radial distance from the centre of the pipe. If an oil at 40cC flows through a 15 m long pipe with R = 0.0 m and maximum velocity of u max = m s , what will be the friction drag force applied by the fluid on inner surface of the pipe when μ = 0.0010 kg/m - s ?

(A) 0.0942 N (C) 0.856 N

(B) 0.942 N (D) 0.916 N

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Basic Concepts and Properties of Fluids

FM 13

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 m/ sec2 is (A) 550 kN (B) 55 N (C) 5.5 N (D) 55 kN

FM 1.50

A cylindrical rod of diameter D , length L and density ρs falls due to gravity inside a tube of diameter Do . The clearance, (Do − D) 2 # 10 , then the flow is turbulent over the entire surface and friction coefficient is 0.074 C f = 0.074 = 0.00252 1/5 = Re (2.2 # 107) 1/5 rV Now Drag force FD = C f A # = 0.00252 # (1 # 3) #

999.1 # (30/3.6) 2 2

= 262.3 N and Power needed to overcome it P = FD # V = 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 And

nx = 5 U

τw = 0.332U 3/2

(1.12 # 10−6) x = 7.48 # 10−3 x m 0.5 rm μ ν= x ρ

= 0.332 # (0.5) 3/2

1000 # 1.12 # 10−3 x

= .

N m x Thus, at the trailing edge ( x = m ), δ = 7.48 # 10−3 6 = 0.0183 m τw = 0.124 = 0.0506 N/m2 6 FM 7.47

Option (D) is correct. Let the flow is turbulent and for turbulent flow δturb = 0.16 = 0.16 1/7 x (Rex ) (rVx/m) 1/7 . 0.16 or = x (998 # 20 # x/0.001) 1/7 0.16 # (0.001) 1/7 . = x (998 # 20) 1/7 # x 1/7 (0.16) 7 # 0.001 . b x l = 998 # 20 # x (0.001) 7 # 998 # 20 or = 7.44 # 10−9 x = (0.16) 7 # 0.003 x = 0.0442 m

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FM 282

External Flow

FM 7

FM 7.48

Option (D) is correct. The flow becomes turbulent where the Reynolds number becomes equal to the critical Reynolds number Re cr = 5 # 105 . Re cr = Vxcr n − # . # xcr = Re cr # n = # = . V The thickness of the boundary layer at xcr = . m is xcr = 4.91 # 0.976 = 0.00678 m δx = 4.91 # 1/2 (5 # 105) 0.5 Re cr = 0.678 cm

FM 7.49

m

Option (C) is correct. For flow remains laminar . δ = . = L Re L (rair UL m) 5.0 δ = or = 0.00315 2.44 1.2 # 15.5 # 2.44 1.8 # 10−5 or and drag Where

Hence

δ = 0.00315 # 2.44 = 0.00768 m = 7.7 mm r FD = CD # # U # A = CD = . ReL

1.328 = 0.00084 1.2 # 15.5 # 2.44 1.8 # 10−5 FD = 0.00084 # 1.2 # (15.5) 2 # (2.44 # 1.22 # 2 sides) 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, rV FD = CD # A # 1.25 # (110/3.6) 2 = 5158 N 2 Now, the total resistance forces on the trunk is = 8.84 #

Ftotal = FD + Fbearing + Frolling And the power required to overcome these forces is to be, P = Ftotal # V = (FD + Fbearing + Frolling) # V 1

= [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 rV FD = CD A # = 7 # 1.25 # V 2 = 4.375V 22 2 2

CD A =

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External Flow

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P = [FD + Fbearing + Frolling] # V2 423000 = [4.375V 22 + 350 + 8339] # V2 = 4.375V 23 + 8689V2 By solving above equation for V Therefore

2

V = 32.07 m/s = 116 km/h FM 7.51

Option (B) is correct.

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 rV Convex side FD = CD # A # = (0.4) # p #

(0.08) 2 1.25 # (15) 2 2 4 #

CD = .

= 0.283 N Concave side

FD = CD # A # = (1.2) #

= 0.848 N Taking the moment about the pivot, M max = FD # .

rV

p # (0.08) 2 1.25 # (15) 2 # 2 4

− FD # .

CD = .

= (FD − FD ) # 0.50 2 2

1

= (0.848 − 0.283) # 0.25 = 0.1412 N− m FM 7.52

Option (A) is correct. For sea-level air, take ρ = 1.225 kg/m3 and μ = 1.78 # 10-5 kg/m-s . Each ball moves at a center line velocity Vb = w # rb = 42 # b 0.56 + 0.0735 l 2 2 = 42 # (0.28 + 0.03675) = 13.3 m/s Then the drag force on each baseball is approximately, r FD, ball = CD # V b # p D = 0.47 # b 1.225 l # (13.3) 2 # p # (0.0735) 2 = 0.215 N 2 4

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External Flow

FM 7

Now the drag force on each rod is calculated. Firstly Vrod = wravg = 42 # b 0.28 l = 5.88 m/s 2 r And FD rod = CD # # V rod # Drod # Lrod = 1.2 # b 1.225 l # (5.88) 2 # 0.007 # 0.28 2 , 0.0498 N Then with two balls and two rods, the total driving power required is P = 2FD, ball Vball + 2FD, rod Vrod = (2 # 0.215 # 13.3) + (2 # 0.0498 # 5.88) = 5.72 + 0.58 = 6.3 Watt FM 7.53

Option (C) is correct.

Subscript a and w denotes the portion of the iceberg in the air and in the water respectively. We have va = 1 v and vw = v where v = volume of the iceberg 7 For steady motion

Where with Thus

FDa = FDw FDa = 1 CDa ra (U − Ub) 2 Aa and FDw = CDw rw U b Aw 2

Ub = speed of the iceberg 1 C ρ (U - U ) 2 A = 1 C r U 2 A b a 2 Da a 2 Dw w b w

(U − Ub) C r A r A = Dw w w = w w Assume CDa = CDw CDa ra Aa ra Aa Ub If D is a characteristic length, then v - D and A - D 1v 3 v 7 a Hence = = D a3 or Da = b l vw 6 v Dw Dw 7 Aa = Da = So that b Dw l b l Aw or

...(i)

Thus from equation (i), (U − Ub) = 1026 # (6) 2/3 = 2754.29 - 2760 1.23 Ub U − Ub = 2760 = 52.5 Ub U = 53.5 & U = 0.0187U or b Ub FM 7.54

Option (B) is correct. For sea-level air, take ρ = 1.225 kg/m3 and μ = 1.78 # 10-5 kg/m-s . Let “a”

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External Flow

FM 285

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 ΣMo = Fa #

c − Wa # . cos c − Fb . sin c + Wb . cos

. sin

c =

Fa − Fb = Wa − Wb ρ ρ or CDa # # U # π D a - CDb # # U # π D b

or

= (S.G.) # rwater # g a p D a3 − p D b3 k 6 6 ρ or π # # U 2 # 6CDa # D a2 - CDb # D b2@ = p # S.G. # rwater # g 6D a3 − D b3@ 2 6 4 or 1 # CDa # π # ρ # U 2 6D a2 - D b2@ = p # S.G. # rwater # g # 6D a3 − D b3@ 8 6 or 1 # 0.47 # π # 1.225 # U 2 # 7]0.02g2 - ]0.01g2A 8 = 7.86 # 999 # 9.8 # p # 7]0.02g 3 − ]0.01g 3A 6 6.78 # 10−5 U 2 = 0.282 U = 4160 U = FM 7.55

4160 , 64.5 m/s

Option (A) is correct. FD =

Drag force

p

# p cos qdA = # p cos q # (br dq) q=

=2

#

p

p cos q (br) dq

...(i)

0

for π # θ # π 2 = 1 rU 2 :1 − 6 qD for 0 # θ # π p 2 2 p = 1 rU 2 if θ = 0 , p =− rU if θ = π 2 2 p =− rU

where and i.e. π

p

Thus

# p cos θdθ =− rU # cos qdq=− rU

and

# p cos θdθ = 12 rU #

π

p

π

2

0

p/2

sin q

p p

= rU

...(ii)

6 :1 − p qD cos qdq

p/2 = 1 rU 2 :sin q − 6 (cos q + q sin q)D p 2 0 p 1 6 6 2 = rU ;1 − a k − b− lE p 2 p 2 = 1 rU 2 : 6 − 2D p 2

...(iii)

Thus from equation (i), (ii) and (iii), we have p

p cos qdq = 2br ;1 rU 2 b 6 − 2 l + rU 2E p 2 0 = 1 rU 2 b 12br l p 2

FD = 2br

So that

CD =

#

2 1 12br FD = FD 2 rU # ^ p h = 2 2 2 1 1 1 2 rU # (2br) 2 rU # (2rb) 2 rU A

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External Flow

FM 7

= 6 = 1.91 p 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 = wr and the strip frontal area is Ddr . The total torque (2 tubes) is

R

R

T = 2 # rdF = 2 # r 9CD 0

0

= 2 # CD #

or Power

FM 7.57

r (wr) 2 Ddr C 2#

R r w 2 # D # r3 dr # 2 0

= CD # r # w 2 # D # R = 1 CD # r # w 2 D # R 4 4 T = 1 # 2.3 # 998 # w 2 # (0.075) # (1.0) 4 = 43.04 w 2 4 P = T#w 20000 = 43.04 w 2 # w ω 3 = 20000 = 464.7 43.04

ω = 7.75 rad/s or ω = 7.75 # 60 , 74 rpm 2p Option (D) is correct.

For equilibrium, moment M = l1 FD + bl1 − D2 l FD 2 2 Where l = 2 m , l2 = 2. m , D = . 2 m and b2 = 2 m . FD = 1 CD ArU 2 = 1 CD rU 2 # l1 D1 2 2 1

2

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...(i)

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FM 7

External Flow

FM 287

and

1 1.2 1.23 (20) 2 (20 0.12) - 708.5 N # # # # 2# FD = 1 CD rU 2 l2 b2 2 = 1 # 0.08 # 1.23 # (20) 2 # (2.5 # 2) = 98.4 N 2

...(ii)

...(iii)

By combining equation (i), (ii) and (iii), we obtain M = 20 # 708.5 + b 20 − 2 l # 98.4 = 7085 + 1869.6 2 2 = 8954.6 N − m - 8955 N − m

FM 7.58

Option (C) is correct. For air at 20cC , take ρ = 1.2 kg/m3 and μ = 1.8 # 10-5 kg/m - s . Since the dimensions are large and the flow is turbulent. The drag when the leading edge is not at x = , x

# tw dA = # aC f r u

F = x

=

x2

k # dA

τw = C f #

x2

# c 0.0272brU

x1

=c

ρ

#u

x

r 2 0.027 # ;(rUx 1/7 E # 2 u # b # dx /m)

x1

=

x

2

Cf =

m 1/7 m # b rU l # x−1/7 dx

m 1/7 0.027brU 2 7 x 6/7 x m # # b l : D 3 6 rU x

. (rUx m)

and dA = b # dx

2

1

0.031brU 2 m 1/7 6/7 x − x 16/7@ =c m # b 2 rU l 6 2 or

−5 1/7 F = 0.031 # 2 # 1.2 # (33) 2 # b 1.8 # 10 l # (56/7 − 26/7) 2 1.2 # 33

= 10.87 N Since the dislodging friction force or or

F = mW 10.87 = m # 90 μ = 10.87 = 0.12 90

Hence, coefficient of solid friction between board and roof is μ = 0.12 FM 7.59

Option (D) is correct.

Let or Where

dT = torque from the drag on element dA of the blade. ...(i) dT = (FD, top + FD, bottom) y = 2 # b 1 rU 2 CD dA l y 2 U = wy and ω = 100 # 1 # 2π = 10.47 rad/s 60 1

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FM 288

External Flow

FM 7

The maximum Re l will occur at point (1) where y = L Re l = Ul = wLl = n n

# # = # − # Thus, at all point on the blade Re x < Re x = 5 # 105 and the flow is laminar. or

1

cr

CD = 1.328 = 1.328 n Re l Ul

So that From equation (i),

dT = rU # .

n Ul

# (l dy) y = .

rU

# nl y dy

But with U = wy dT = 1.328rw 3/2 nl y 5/2 dy

= 1.328 # 1.23 # (10.47) 3/2 61.46 # 10−5 # 0.1@ 1/2 y 5/2 dy = 0.0669y5/2 dy N− m Thus the net torque on the five blades is 0.8 0.8 T = 5 dT = 5 0.0669y5/2 dy = 5 # 0.0669 # 2 6y7/2@0 7 y=0

#

#

= 0.0438 N− m

***********

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FM 8 OPEN CHANNEL FLOW

FM 8.1

Water at 20cC 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 20cC ρ = 998 kg/m3 and μ = 1.002 # 10-3 kg/m -s ) (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 m3/s . If the flow is to be critical, the maximum depth is (A) 2.58 m (B) 4.30 m (C) 3.44 m (D) 6.88 m

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 m3/s (B) 25.5 m3/s (C) 12.75 m3/s (D) 5 m3/s

FM 8.4

The ratio of Froude numbers on either side of a hydraulic jump are related by y y (A) b l (B) b l y y y y (C) b l (D) b l y y

FM 8.5

Water (ρ = 999.7 kg/m3, μ = 1.307 # 10-3 kg/m -s) 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 m3/s 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 m/s, 2.38 m/s (B) 2.37 m/s, 2.38 m/s, 0 (C) 2.38 m/s, 2.37 m/s, 0 (D) 2.37 m/s, 0, 2.38 m/s

GATE Mechanical Engineering in 4 Volume NODIA FM 290

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Open Channel Flow

FM 8

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 m3/s (B) 6.03 m3/s (C) 27.11 m3/s (D) 5.56 m3/s

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 m3/s (C) 14.36 m3/s

(B) 22.8 m3/s (D) 8.45 m3/s

FM 8.10

A circular channel of diameter 75 cm is flowing half-full at an average velocity of 3.4 m/s . 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 m # 5 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% (C) 25%

(B) 75% (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 ?

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(A) 2.96 % (C) 23.7 %

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Open Channel Flow

FM 291

(B) 2.37 % (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 = .32D (B) D = .36D (C) D = . D (D) D = . 9D

FM 8.16

The flow rate in the asphalt-lined (n = 0.016) channel shown in figure below is to be 120 m3/s . What will be the elevation drop of the channel per km ?

(A) 8.52 m (C) 0.852 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 (C) 31 cm FM 8.18

(B) 0.0852 m (D) 85.2 m

(B) 6.2 cm (D) 3.1 cm

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 (C) 2.62 N/m2

(B) 4.36 N/m2 (D) 1.75 N/m2

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FM 8.19

(B) 2c (D) 0.01706c

Consider a trapezoidal aqueduct as shown in figure below, carries a normal flow of 60 m3 /s . For clay tile ^n = . h surfaces, the required elevation drop in m/km will be

(B) 0.0038 (D) 0.38

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 m3/s (C) 275.6 m3/s FM 8.23

(B) 21.62 m3 /s (D) 17.30 m3 /s

Water flows in the symmetrical, unfinished concrete (n = 0.014) trapezoidal channel as shown in figure below at a rate of 25 m3/s and flow depth of 0.69 m. What will be the slope angle (θ) of bottom surface ?

(A) 0.038 (C) 0.00038 FM 8.22

FM 8

The channel shown in figure is built on a slope of 2 m/km and depth is y = m . If the surface are smooth concrete lined (n = . ) except for the diagonal surface, which is gravel with n = . , the flow rate will be

(A) 0c (C) 1c FM 8.21

Page 292

Open Channel Flow

(A) 10.81 m3 /s (C) 12.97 m3 /s FM 8.20

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(B) 43.7 m3/s (D) 36.4 m3/s

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 m3/s to a distance of 2 km. If it is necessary to keep the flow depth below 1 m , the required elevation drop is

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Open Channel Flow

FM 293

(B) 322 m (D) 32.2 m

A channel lined with placed wood is to be carry water at a flow rate of 2 m3/s on a slope of 10 m/800 m . 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 26 Water 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 m3/s (B) 4.26 m3/s (C) 44.2 m3/s (D) 45.56 m3/s

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 m3 /s , what will the normal depth y ?

(A) 20 m (C) 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 m3/s (C) 15.9 m3/s

FM 8.29

(B) 4 m (D) 0.2 m

(B) 0.159 m3/s (D) 0.0159 m3/s

A storm drain constructed of brickwork ^n = 0.015h has the cross section as shown in figure below. If it laid on a slope of 1.5 m/km, the normal discharge

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Open Channel Flow

FM 8

when the water level passes through the center of the circle is

(A) 5.8 m3 /s (C) 46 m3 /s 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 m3 /s (C) 345 m3 /s FM 8.31

(B) 2.90 m3 /s (D) 92 m3 /s

(B) 325 m3 /s (D) 1000 m3 /s

Water flows in a partly full riveted-steel triangular duct ^n = . figure below. If the critical depth is 50 cm, the critical slope is

(A) 0.00205 (C) 0.0410

h as shown in

(B) 0.0205 (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 m3/s upto a distance of 10 km through a 2 m internal diameter circular steel (n = 0.012) drain. For the maximum depth of 1.5 m, the required elevation drop is

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Open Channel Flow

FM 295

(B) 6.37 m (D) 0.0637 m

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 = . (C) y = .

h h

(B) y = . (D) y = .

h h

Common Data For Linked Answer Q. 35 and 36 Water is released from a 5 m deep reservoir into a 1 m wide finished concrete channel (n = 0.012) of bottom slope of 0.23c through a sluice gate with free outflow of depth ratio (y1 /a) = 10 . The water encounters a hydraulic jump. Disregard the bottom slop when analyzing the hydraulic jump.(Cd = . )

FM 8.35

What (A) y (B) y (C) y (D) y

will be the flow depth, velocity and Froude number before the jump ? = . m, V = . m s, Fr = . = . m, V = . m s, Fr = . = . m, V = . m s, Fr = . = . m, V = . m s, Fr = .

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GATE Mechanical Engineering in 4 Volume NODIA FM 296

FM 8.36

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Page 296

Open Channel Flow

The flow depth, velocity (A) y = m, V = (B) y = m, V = (C) y = m, V = (D) y = m, V =

FM 8

and Froude Number after the jump are m s, Fr = m s, Fr = m s, Fr = m s, Fr =

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 (C) 14.3 m/s

FM 8.39

Water flows in a rectangular channel with a velocity of V = m s . A gate at the end of the channel is suddenly closed so that a wave (a moving hydraulic jump) travels upstream with velocity Vw = m s as shown in figure. The depths ahead of and behind the wave respectively, are

(A) 2.61 m , 1.96 m (C) 0.653 m , 1.30 m FM 8.40

(B) 5.72 m/s (D) 2.87 m/s

(B) 0.653 m , 2.61 m (D) 1.96 m, 3.26 m

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 (C) 91 cm

(B) 120 cm (D) 65 cm

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FM 8.41

Page 297

Open Channel Flow

FM 297

Consider a triangular flume as shown in figure, is built to carry the flow rate vo at a depth of 0.90 m . If the flume is to be able to carry up to twice its flow rate vo = vo , what will be the freeboard length l ?

(A) 0.378 m (C) 0.472 m FM 8.42

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(B) 0.283 m (D) 0.236 m

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 (C) 180c

(B) 90c (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 (C) 3.02 m , 25 MW

(B) 1.51 m , 12.5 MW (D) 1.89 m , 15.65 MW

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GATE Mechanical Engineering in 4 Volume NODIA FM 298

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Page 298

Open Channel Flow

FM 8

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 m3/s . 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 49 Water 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 m, 1.97 m (B) 8 m, 4.97 m (C) 1.97 m, 8 m (D) 8 m, 2.44 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 b1 = m , b = m , y1 = 1.9 m and y = 1.5 m , what will be the flow rate ?

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(A) 97 m3 /s (C) 0.97 m3 /s FM 8.51

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Page 299

Open Channel Flow

FM 299

(B) 9.9 m3 /s (D) 4.95 m3 /s

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 m3/s, 1.625 (C) 9 m3/s, 1.625

(B) 1.8 m3/s, 0.575 (D) 1.625 m3/s, 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 m3/s . If the character of flow is to change, the specific energy and the alternate flow depth of water respectively, are (A) 0.874 m, 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) y2 = 0.7 2 m, yc = 1.0 m (B) y2 = 1.0 m, yc = 0.7 2 m (C) y2 = 0.55 m, yc = 0.7 2 m (D) y2 = 1.0 m, yc = 0.55 m

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 (C) Remains same FM 8.55

(B) Yes (D) Not determined

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

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GATE Mechanical Engineering in 4 Volume NODIA FM 300

FM 8.56

Page 300

Open Channel Flow

FM 8

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 Cd = . )

(A) 2.46 m3/s (C) 0.818 m3/s FM 8.57

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(B) 2.26 m3/s (D) 0.145 m3/s

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 (Cd = . ) (A) y = . m, V = . m s (B) y = . m, V = . m s (C) y = . m, V = . m s (D) y = . m, V = . m s

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 Cd = . )

(A) 13 m3/s (C) 10 m3/s

(B) 3.76 m3/s (D) 14.3 m3/s

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FM 8.60

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Page 301

Open Channel Flow

FM 301

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 m3 /s , 0.416 m (C) 1.36 m3 /s , 0.333 m

(B) 2.04 m3 /s , 0.250 m (D) 1.02 m3 /s , 0.167 m

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 m3 /s (B) 2.174 m3 /s (C) 5.33 m3 /s (D) 8.234 m3 /s

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 m3 /s (B) 0.350 m2 /s (C) 0.350 m3 /s (D) 3.50 m2 /s

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 m/300 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 (C) 0.311 m , Impossible

(B) 0.311 m , Possible (D) 0.415 m, Possible

***********

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GATE Mechanical Engineering in 4 Volume NODIA FM 302

Demo Ebook

Page 302

Open Channel Flow

FM 8

SOLUTIONS FM 8.1

Option (D) is correct. The Reynolds number of the flow is rVy # # . = . = Re = # − m . # Which is greater than critical value of 500. Therefore, the flow is turbulent. Now the Froude number is Fr = V = = . gy . # 0. Which is greater than 1, therefore the flow is supercritical.

FM 8.2

Option (C) is correct 0 = 0 y #y 0 y Fr = V = gy 6 . 1 # y@ 1 o V = v = A

We have Also

where y = depth = . y

For critical flow Fr = 1. Thus y = (6.39) 2/3 = 3.44 m FM 8.3

Option (A) is correct. Since flow is critical, then Fr = V = 1 gy V =1 gy (5) y =V = g . 1 y = yc = .55 m Thus flow rate becomes vo = VAc = V # b # y = 5 # 4 # 2.55 = 51 m3/s

FM 8.4

Option (B) is correct. From continuity vo = Vy = constant or

FM 8.5

gy V gy1 Fr2 = V = # Fr1 V1 gy1 V1 # gy o y1 y =V # = v # o1 # y y V1 v

y1 y = b 1l y y

Option (D) is correct. The hydraulic radius is

^pR h R Ac = = perimeter pR = 2 = 1m 2

Rh =

Circular channel is half full

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Open Channel Flow

FM 303

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. pR 2 Ac = 2 = pR = p # 2 = p = 1.570 m yh = 2 4 4 2R Top width Now Froude number . Fr = V = = . gyh . # . Since Fr = 0.637 < 1, therefore the flow is subcritical. FM 8.6

Option (A) is correct.

The flow area and average velocity are 2y bb + b + tan q l Ac = #y 2 4 + 4 + (2 # 0.6/ tan 45c) 2 = # 0.6 = 2.76 m 2 o V = v = m s = . . Ac Ac . . Hydraulic depth y = yh = = 0.5308 m = = y Top width + # . b+ tan c tan q . Then Froude number Fr1 = V = = . gyh . # . Since Fr > 1, therefore the flow is supercritical. FM 8.7

Option (B) is correct.

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GATE Mechanical Engineering in 4 Volume NODIA FM 304

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Page 304

Open Channel Flow

FM 8

Since c = gy It follows that the tank depth is = y =c = m g If the tank accelerates upward with acceleration a , the effective acceleration of gravity is geff = g + a =

+ =

m s

Thus = c = geff y = m s # If the tank accelerates horizontally with acceleration a , the effective acceleration is geff = Thus In orbit FM 8.8

g +a =

+

=

m s

c = 13.87 # 0.408 = 2.38 m/s geff = 0 (weightless). So c =

Option (A) is correct. The flow area, wetted perimeter and hydraulic radius of this channel are Ac = b # y = # = m perimeter = b + y = + # = m and Rh = Ac = = . m p Now from Manning equation, flow rate vo = a Ac R h S n = 1 # 2 # (0.5) 2/3 # (0.0105) 1/2 = 10.76 m3/s 0.012

FM 8.9

Option (B) is correct. The flow area, wetted perimeter and Hydraulic radius of the channel are Flow area Ac = 2h # h = 2 # 2 # 2 = 4 m2 2 2 Wetted perimeter = 2# h = 2# 2 = 4 2 m sin 45c sin 45c Ac Hydraulic Radius = 4 = 1 = 0.7071 m Rh = perimeter 4 2 2 Hence, Discharge rate vo = a Ac (Rh) # S n = 1 # 4 # (0.7071) 2/3 # (0.008727) 1/2 0.013 = 22.8 m3/s

FM 8.10

Option (D) is correct For a half-full channel, A = pR The volume flow rate is

, perimeter = pR , Rh = R and b = R .

vo = VA = . # p R = 3.4 # p # ]0.375g2 = 0.75 m3 /s 2 And

VC =

gh =

gAC = b

.

# 6p # ] . .

g

@

= .

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m s

GATE Mechanical Engineering in 4 Volume NODIA FM 8

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Page 305

Open Channel Flow

FM 305

So, from manning’s formula SC = FM 8.11

]0.016g2 # ]1.7g2 = 0.0069 n2 VC2 = ]1g2 # R h4/3 ]1g2 # ]0.1875g 4/3

Option (C) is correct.

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 2/3 1/2 2/3 perimeter1 2/3 vo2 = (a/n) Ac R h S 2 = Rh 2/3 = Ac /perimeter2 = = G ; ; E perimeter2 E Rh Ac /perimeter1 vo1 (a/n) Ac R h2/3 S 11/2 For condition 1: 2

2

1

1

2

2

1

1

perimeter1 = 6 # 5 = 30 cm For condition 2: perimeter 2 = 4 # 5 = 20 cm Substituting these values, we get vo2 = 30 2/3 = 1.31 (31% increase) vo1 :20 D FM 8.12

Option (C) is correct. The hydraulic radius A = 3 # 1 = 0.6 m perimeter 3 + 1 + 1 For finished concrete, flow rate (From Manning’s formula) vo1 = 1 AR h2 3 S 1 2 = 1 # 3 # ( . ) 2 3 # ( . 1 ) 1 2 = 6.65 m3/s n . 12 Rh =

Also for asphalt vo2 = 1 AR h2 3 S 1 2 = 1 # 3 # ( . ) 2 3 # ( . n . 1

1 ) 1 2 = 5 m3/s

Percentage reduction = 6.65 − 5 # 100 = 25% 6.65 FM 8.13

Option (A) is correct. The flow area, wetted perimeter and hydraulic radius for this flow are Ac = y # b = 3. m2 perimeter = 2y + b y Since = 0.4 & y = 0.4b b From equation (i) and (ii),

...(i) ...(ii)

0.4 # b # b = 3.6 0.4b2 = 3.6 b2 = 3.6 = 9 0.4

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GATE Mechanical Engineering in 4 Volume NODIA FM 306

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Page 306

Open Channel Flow

FM 8

b = 3m y = 0.4 # 3 = 1.2 m and Perimeter = 2 # 1.2 + 3 = 5.4 m Ac And hydraulic radius = 3.6 = 0.6667 m Rh = perimeter 5.4 Thus the flow rate vo = a Ac Rh S n = 1 # 3.6 # (0.6667) 2/3 # (0.002) 1/2 = 6.14 m3/s 0.02 Hence, the critical depth is o yc = = v G = = G = 0.753 m gb #] g Since y > yc, the channel at these flow condition is classified as mild and the flow is subcritical. FM 8.14

Option (C) is correct. From manning’s equation, the flow rate vo = a AR h S n

a = , if S.I. units are used.

Without the center board A = bb b l = b And

and perimeter = b + b2 + b2 = 2b

b2 A = 2 =b Rh = 4 Perimeter 2b

vowithout = a # b # b b l S n With the centerboard vowith = 2vo2 Thus

Where

And Thus

A = b b l and perimeter = b2 + b2 + b2 = 2 b 2 b2l Rh = =b b 3b l 6 2 2 2/3 vowith = 2 a b b l b b l S 01/2 n 2 6 2

...(i) ...(ii) 3b 2

...(iii)

Dividing equation (iii) by (i) to obtain 2 2/3 2b b l # b b l 6 vowith = 2 = 0.763 2 vowithout b b 2/3 b 2 l#b4l 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 = 3vo1 Where and vo = a A R h S vo = a AR h S n n With n = n , S = S , A = pD .

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..(i)

GATE Mechanical Engineering in 4 Volume NODIA FM 8

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Page 307

Open Channel Flow

FM 307

pD2 A = 8 =D Rh = pD 4 Perimeter 2 p D2 1 A 1 and = 8 = D1 Rh = pD 4 Perimeter1 2 1 Thus from equation (i), we get AR h = 3A1 R h21/3 D1 2/3 πD2 D 2/3 = 3 p (D ) 2 # # 1 #b b l 8 8 4 l 4 Hence FM 8.16

D

= 3 (D1) 8/3 or D =

D

Option (A) is correct. The flow area, wetted perimeter and hydraulic radius of channel are Flow area Ac = 1 # (10 + 5) # (2.2) = 16.5 m2 2 p = 5 + 2 # (2.2) 2 + (2.5) 2 = 11.66 m

Wetted perimeter

. = . Rh = Ac = m p . Substituting the given parameters into Manning’s equation vo = a Ac R h S n And hydraulic radius

a = constant = 1 m1/3 /s , S = Slope Thus 120 = 1 # 16.5 # (1.415) 2/3 # S 01/2 0.016 S 01 2 = 120 # 0.016 2/3 16.5 # (1.415) S 0 = 0.008524 Therefore, the elevation drop Δz across a pipe length of L = 1 m must be Δz = S 0 # L = 0.00 2 # 1000 = 8.52 m Where

FM 8.17

Option (A) is correct. The wave angle and depth give Fr = 1 = 2.37 sin 25c V = 2.37 or c V = 2.37 # c = 2.37 # gy

Fr = V c c=

gy

= 2.37 # 9.81 # 0.35 = 4.39 m/s Flow rate per meter width m3 /s vo = V # y = . # 0. = 1.54 m Hence

(1. ) 2 1 o2 1 v yC = ; E = ; g . 1 E

o2 1 yC = e v2 o bg

, 0.62 m = 62 cm FM 8.18

Option (A) is correct.

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GATE Mechanical Engineering in 4 Volume NODIA FM 308

Demo Ebook

Page 308

Open Channel Flow

FM 8

Average shear stress given by the relation τw = gRh S Where

2

For a wide flat plate

FM 8.19

γ = 0.85 gH O = 0.85 # 9800 = 8330 N/m3 A = by and Perimeter = b A = y = 8 # 10−3 m perimeter

So that

Rh =

and Thus

S = sin 3c τw = 8330 # 8 # 10−3 # sin 3c = 3.49 N/m2

Option (D) is correct.

Flow rate Where

a=

Also

vo = vo + vo = a A R h S + a A R h S n n (for S.I. units), S = . , n = . , n = . A = 1 # 1.0 # 3 = 1.50 m2 2

...(i)

Perimeter1 = 7]1.0g2 + ]3.0g2A = 3.16 m A1 Rh = = 1.50 = 0.475 m Perimeter1 3.16 1/2

A = 3 # 1.5 = 4.5 m2 , Perimeter 2 = 0.5 + 3 + 1.5 = 5 m A2 Rh = = 4.5 = 0.90 m 5 Perimeter2

and

Hence from equation (i), we get 1 1.50 # (0.475) 2/3 # (0.002) 1/2 0.025 # + 1 # 4.5 # (0.90) 2/3 # (0.002) 1/2 0.012 vo - 17.3 m3/s vo =

or FM 8.20

Option (C) is correct. From Manning’s equation

Where flow area Hydraulic Radius

vo = a Ac ^Rh h S n 5 + 5 + ^2 # 0.69h 2 Ac = # 0.69 = 3.92 m 2 Ac 3.92 Rh = = = 0.565 m perimeter 5 + 2 # 0.69 sin 45c

a = 1 m1/3 /s n = 0.014 for unfinished surface vo = 25 m3/s Substitute these numerical values in equation (i), Manning coefficient

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...(i)

GATE Mechanical Engineering in 4 Volume NODIA FM 8

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Page 309

Open Channel Flow

FM 309

1 3.92 # (0.565) 2/3 # S 01/2 0.014 # 25 # 0.014 S = 3.92 # (0.565) 2/3 2 S = ; 25 # 0.014 2/3 E = 0.01706 3.92 # (0.565) S = tan q = 0.01706 θ = tan−1 (0.01706) = 0.98 , 1c 25 =

Since FM 8.21

Option (D) is correct. The geometry leads to these values

And So that

A = by + y cot q = 5 # 3.2 + (3.2) 2 cot 40c = 28.2 m2 perimeter = b + y cosec q = 5 + 2 # 3.2 # cosec 40c = 14.96 m A Rh = = 28.2 = 1.885 m perimeter 14.96

Then flow rate

vo = 60 =

or FM 8.22

n

AR h S

1 28.2 # ^1.885h2/3 # S 01/2 0.014 #

S = 0.00038 m/m = 0.38 m/km

Option (D) is correct. 2y y Ac = 1 # bb + b + 2 tan 30c l # 2 = 1 # b4 + 4 + 2 # 2 = 14.93 m2 2 tan 30c l # y Wetted perimeter = b + = + # = m sin c sin c Ac Hydraulic radius = 14.93 = 1.244 m Rh = 12 perimeter Flow area

Bottom slope Hence the flow rate

tan θ = tan (0.057) = 0.001 vo = a Ac R h S n = 1 # 14.93 # (1.244) 2/3 # (0.001) 1/2 0.015 = 36.4 m3/s

FM 8.23

Option (D) is correct.

The flow area, wetted perimeter and hydraulic radius of the channel are b + b + 2y 1.5 + 1.5 + 2 # 1 1 = 2.5 m2 Ac = #y = # 2 2

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Page 310

Open Channel Flow

perimeter = b +

y +y =

FM 8

+ # ] g +] g =

m

Ac = 2.5 = 0.5776 m perimeter 4.328 From Manning’s equation vo = a Ac R h S n 10 = 1 # 2.5 # (0.5776) 2/3 # S 01/2 0.022 Rh =

S 0 = 0.0161 Therefore the elevation drop Δz across a pipe length of L = m is Δz = S 0 L = 0.0161 # 2000 = 32.2 m FM 8.24

Option (D) is correct From manning’s equation, the flow rate is ...(i) vo = a AR h S 01 n Let subscript t and r denotes the triangle cross-section and the rectangular crosssection respectively.

We have vor = vot = 2 m3 /s , S 0r = S 0t = 10 and nr = nt 00 So that equation (i) becomes A where Rh = Ar R hr = At R ht erimeter Thus Ar = 2y r2 , Perimeterr = 4yr So that Also So that

...(ii)

2y r2 1 = yr 2 4yr At = 1 (2yt) yt = y t2 , Perimetert = 2 ( 2 yt) 2 y Rht = t 2 2

Rhr =

Thus from equation (ii), 2/3 2y r2 b 1 yr l = y t c 1 yt m , 2 1.26 y r8/3 = 0.5 y t8/3 yr = 0.707yt The amount of wood is proportional to the wetted perimeter. Since

Perimetert = 2 2 yt = 2 2 yt = 1.00 4yr 4 # 0.707yt Perimeterr

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

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FM 311

to divide the channel into two subsections. For subsection 1. Ac = 3 # 2 = 6 m2 perimeter1 = 3 + 1 + 1 + 1 = 6 m Ac Rh = = 6 = 1m perimeter1 6 1

1

1

Ac = 10 # 1 = 10 m2 perimeter2 = 10 + 1 = 11 m Ac Rh = = 10 = 0.909 m perimeter 2 11 Applying the Manning equation to each subsection, the total flow rate through the channel becomes vototal = vo1 + vo = a # Ac R h S 1 + a # Ac R h S 1 n1 n 2

1

= a #=

1

Ac # R h A R + c # h G # S1 n1 n 1

1

6 (1) 2/3 10 # (0.909) 2/3 = 1 #; # + E # (0.00873) 1/2 0.020 0.050 = 45.56 m3 /s FM 8.26

Option (D) is correct. For entire channel Ac = Ac + Ac = 1 + = 1 m perimeter = perimeter1 + perimeter 2 = 6 + 11 = 17 m Ac Rh = = 16 = 0.941 m perimeter 17 1

Hence, the effective Manning coefficient for entire channel a A R S1 neff = # c # o h # v 1 16 (0.941) 2/3 # (0.00873) 1/2 = # # = 0.0315 45.56 FM 8.27

Option (C) is correct For a V-channel A = y cot q = y cot

c

perimeter = 2y cosec q 2 and Thus or

y y A = cos q = cos 45c 2 2 perimeter 2 vo = 1 AR h S 1 n 1/2 2/3 y 11.35 = 1 # y 2 cot 45c # a cos 45ck # b 1 l 2 0.014 400 Rh =

11.35 = 1.7857 y 8/3 y

= 11.35 = 6.356 1.7857

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GATE Mechanical Engineering in 4 Volume NODIA FM 312

Page 312

Open Channel Flow

FM 8

y = ]6.356g 3/8 , 2.0 m

or FM 8.28

Demo Ebook

Option (B) is correct.

From geometric considerations cos θ =

R−y = . − . . R

= .

θ = cos−1 (0.5) = 60c or 60 # p = p radians 180 3 Ac = R (q − sin q cos q)

Flow area

= (0.5) 2 9 p − sin a p k cos a p kC = 0.1535 m2 3 3 3 R (q − sin q cos q) Rh = Ac = = q − sin q cos q # R p Rq 2q p − sin p cos p 3 3 # 0.5 = 0.1466 m = 3 p 2# 3 Thus the flow rate can be vo = a Ac R h S n = 1 # (0.1535) # (0.1466) 2/3 # (0.002) 1/2 0.012 Hydraulic radius

= 0.159 m3 /s FM 8.29

Option (B) is correct. The section properties are A = p R2 + R2 = R2 a1 + p k 4 4 = 12 # a1 + p k = 1.785 m2 4 perimeter = 1 # 2pR + 1 + 1 4 = 1 # 2p # 1 + 1 + 1 = 3.57 m 4 A So Rh = = 1.785 = 0.5 m and S = perimeter 3.57 Hence

vo = =

n

.

= .

AR h S

1 1.785 # ]0.5g2/3 # ]0.0015g1/2 0.015 #

= 2.90 m3 /s

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m m

GATE Mechanical Engineering in 4 Volume NODIA FM 8

FM 8.30

Demo Ebook

Page 313

Open Channel Flow

FM 313

Option (D) is correct We 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. ...(i) vototal = vo + vo For

So

vo =

A Rh S n #

A = ^y + y h # b = ^ + . h # = m Rh = 132 = 5.077 m ^6 + 20h 1/2 vo = 1 # 132 # ]5.077g2/3 # b 1 l = 344.95 m3 /s 0.022 2640 , 345 m3 /s

For

vo =

A Rh S n #

A = b # y = 150 # 3.6 = 540 m2 Rh = So

Thus

vo = vototal

A = . + y +b

= .

m

1/2 1 540 # ]3.52g2/3 # b 1 l # 2640 0.075

= 324.26 m3 /s , 325 m3 /s = vo + vo = 345 + 2 # 325 = 995 , 1000 m3 /s

FM 8.31

Option (B) is correct. The cross-section properties are yC = 0.5 m h − yC − . = . b = . h And Since or or

h=

] g −] . g = .

= 0.423 m AC = 1 # ^0.423 + 1h # 0.5 = 0.356 m2 2 gAC b o g A g A VC = # C or v = # C b b AC ) # . A g ( . vo = C # = . b vo = 1.046 m3 /s VC =

perimeter = 1 + 0.577 + 0.577 = 2.15 m Rh =

o VC = v AC

0.5 = 0.577 sin 60c

A = 0.356 = 0.165 m 2.15 perimeter

Now the critical slope, from the Manning’s formula is ( . ) # . # . n g A SC = # # C = . #] . g b Rh , 0.0205

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GATE Mechanical Engineering in 4 Volume NODIA FM 314

FM 8.32

Demo Ebook

Page 314

Open Channel Flow

FM 8

Option (A) is correct.

For Rectangular channel Ac = b Rh =

perimeter = 3b

2 Ac = b =b perimeter 3b 3

vorect. = a Ac # R h S n

Flow rate

= a b # bb l n

#S

]b g =aS # n ] g For circular channel flowing half-full 2 Ac = pD , perimeter = pD 8 2 Ac Rh = =D perimeter 4 vocir. = a # Ac R h S n

and flow rate

= a # pD # b D l # S n

= a S # pD n # Setting the flow rates in the flow channels are equal, we get vocir. = vorect. a S n#

FM 8.33

#

πD = a #S # b n ] g ] g # 8/3 8/3 πD = b 2/3 8 # ]4g2/3 ]3g 3/8 b = p # ]3g2/3 = 0.655 = D 8 # ]4g2/3 G b = 0.655D

Option (C) is correct.

From geometrical consideration

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

Demo Ebook

Page 315

Open Channel Flow

FM 315

cos φ =

and Flow area

y−R = R

− =

φ = cos−1 (0.5) = 60c = 60 # p = p radian 180 3 θ = p − f = p − p = 2p = 120c 3 3 Ac = R q − n q o q

= ]1g2 :2p − sin 2p # cos 2p D = 2.527 m2 3 3 3 2p − sin 2p cos 2p A q − q q sin cos 3 3 #1 c = Rh = #R = 3 p 2 perimeter 2q 2# 3 = 0.6034 m Substituting the given values in Manning’s equation, we get vo = a Ac R h S n 12 = 1 # 2.527 # (0.6034) 2/3 # (S 0) 1/2 0.012 2 12 # 0.012 S =; 2/3 E = 0.00637 1 # 2.527 # (0.6034) Therefore, the elevation drop Δz across a pipe length must be Δz = S L = = 63.7 m FM 8.34

#1

Option (A) is correct.

Flow rate

vo = a AR h2 S 1 2 n

From figure

l =

Area

A

or

A

and Thus Therefore

Perimeter Rh vo

2h , b = 2 (h − y) , l = y tan 60c tan c s sin c 1 [h2 − (h − y) 2] = 1 lh − 1 b (h − y) = 2 2 tan 60c 1 [2hy − y2] = tan 60c y = l + 2ls = 2 b h + tan c sin c l 2hy − y2 A = = y Perimeter 2 ah + cos 60c k 2 2hy − y 2 a 1 2 (2hy − y ) = # S1 2 n y tan c >2ah + H cos c k

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GATE Mechanical Engineering in 4 Volume NODIA FM 316

Demo Ebook

Page 316

Open Channel Flow

FM 8

o For the maximum flow rate dv = , which is equivalent to dF = , where dy dy (2hy − y2) 5/3 (y + h cos 60c) 2/3 By differentiation and simplification this gives F y /

5 (y + h cos 60c) (h − y) − (2hy − y 2) = 0 or 4y 2 + (5h cos 60c − 3h) y − 5h2 cos 60c = 0 Which can be written as y 2 y 8a k − a k − 5 = 0 h h 1 ! 1 + 4 (8 # 5) y = =− 0.731 or + 0.856 16 h The negative root has no physical meaning. Thus y = 0.856h So that

FM 8.35

Option (C) is correct. For depth ratio of 10 and the discharge coefficient for underflow is Cd = 0. discharge rate through the slice gate vo = Cd ba gy1 y1 Since = 10 a y a = 1 = 5 = 0.5 10 10 Thus

FM 8.36

, the

vo = 0.58 # 1 # 0.5 # 2 # 9.81 # 5 = 2.872 m3 /s

For wide channels, hydraulic radius is the flow depth Rh = y . Then from Manning’s equation vo = a # Ac R h S 01 n 2.872 = 1 # (b # y2) # (y2) 2/3 # [tan (0.23c)] 1/2 0.012 2.872 = 1 # 1 # y2 # y22/3 # (0.004) 1/2 0.012 = 0.545 y2 = 2.872 # 0.012 1 # (0.004) 1/2 y2 = (0.545) 3/5 = 0.6947 m Now, flow velocity and Froude number before the jump are o 2.872 V2 = v = = .1 m s 7 by2 1 # 0. .1 Fr2 = V2 = = 1. 8 gy2 .81 # 0. 7 Option (D) is correct. For after the jump condition, flow depth y = 0.5y2 (− 1 + 1 + 8Fr22 ) = 0.5 # 0.6947 # (− 1 + 1 + 8 # (1.584) 2 = 1.25 m y 7 Velocity .1 = 2.3 m/s V = 2 # V2 = 0. y 1. 2 # 2. 0 Froude Number Fr 3 = V = = 0. 7 gy .81 # 1.2

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

FM 8.37

Thus

Open Channel Flow

FM 317

3 = 1 + 8Fr 12 − 1 16 = 1 + 8Fr 12 Fr1 = 1.37 V = Fr1 # gy1 = 1.37 # 9.81 # 4 = 8.58 m/s , 8.6 m/s

Option (D) is correct. The bore moves at speed V and induced a velocity ΔV behind it. If viewed in a frame fixed to the wave as above, the downstream V = V − DV . Vy Since V = = . # = . m s y Thus

FM 8.39

Page 317

Option (A) is correct. With the Jump-height ratio y = 1 6 1 + 8Fr 12 − 1@ y 2 6 =1 or 1 + 8Fr 12 − 1@ 2 #6 4

or FM 8.38

Demo Ebook

ΔV = V − V = 8.6 − 5.73 = 2.87 m/s

Option (B) is correct For an observer moving to the left with speed Vw = m s the flow appears as shown below.

Thus treat as a jump with V = m s , V = m s Since AV =A V y or =V = = y V From the depth ratio y = 1 6− 1 + 1 + 8Fr 12 @ = 4 y 2 Fr1 = 3.16 Fr1 = V1 1/2 (gy1) ( ) y = V = = . gFr . #( . ) y = 4y1 = 4 # 0.653 = 2.61 m

However So that and FM 8.40

m

Option (C) is correct. The upstream velocity is V =

And

R S n# h = 1 # ]0.3g2/3 # ]tan 1cg1/2 = 4.23 m/s 0.014 . Fr1 = V = , . gy . # .

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GATE Mechanical Engineering in 4 Volume NODIA FM 318

Demo Ebook

Page 318

Open Channel Flow

FM 8

So from jump-height ratio y = 1 6 1 + 8Fr 12 − 1@ = 1 # 8 1 + 8 (2.465) 2 − 1B , 3.02 y 2 2 or FM 8.41

y = 0.3 # 3.02 , 0.91 m = 91 cm

Option (A) is correct. Using the subscript 0 for the design conditions and 2 for denote conditions. We have vo = 2vo0 Flow rate vo = a A R h S where a = n A = 2 :1 # 0.9 # 0.9D = 0.81 m2 2 and

perimeter 0 = 2 [ (0.9) 2 + (0.9) 2 ] = 2.55 m A0 Rh = = 0.81 = 0.318 m perimeter 0 2.55

Hence

vo =

or

vo

Also

vo

Where

A =y

. #( . n # . S = n = a A Rh S n

)

S ...(i)

and perimeter 2 = 2 2 y2

y 22 A2 = = 0.354y2 perimeter 2 2 2 y2 Hence with n = n and S = S Rh =

vo =

y ( . n # #

. vo =

or

S n

y)

#S ..(ii)

y

From equation (i) and (ii) with vo = vo , we obtain or However

0.500y 28/3 = 2 # 0.377 y = 1.167 m y − . = l sin c l =

So that FM 8.42

y2 − 0.9 1.167 − 0.9 = = 0.378 m sin 45c sin 45c

Option (D) is correct. Flow rate

vo = a AR h S n

or with a , n , S constant

dvo = aS ;R dA + A b l R − dRh E h h n dq dq dθ o Thus for a given flow rate dv = and for the minimum area dA = dθ dθ equation (i) gives dRh = 0 dθ A Also Rh = perimeter

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...(i)

GATE Mechanical Engineering in 4 Volume NODIA FM 8

Since

Page 319

Open Channel Flow

FM 319

A = 1 b # (l cos q) and perimeter = 2l 2

where or

Demo Ebook

b # l cos q 1 = b cos q 4 2l it follows that sin θ = b or l = b 2l in q A = 1 b b b cos q l 2 2 sin q Rh =

1 2

..(ii)

or b = 2 A # tan q Thus equation (ii) becomes Rh = cos q (2 A # tan q ) = 1 A (sin q cos q) 1/2 2 4 dRh = 1 (sin q cos q) 1/2 1 A−1/2 dA So that #2 2 dθ dq

With dA = dθ

+ 1 A # b 1 l (sin q cos q) −1/2 (cos2 q − sin2 q) 2 2 (i.e. minimum area), dRh = when cos2 θ - sin2 θ = 0 dθ

θ = 45c i.e. the best hydraulic cross-section occurs with a right angle triangle. or FM 8.43

Option (A) is correct. The depth- ratio across a hydraulic jump is given by the relation y = 1 6− 1 + 1 + 8Fr 12 @ y 2 1.2 = 1 7− 1 + 1 + 8Fr 2 A 1 0.3 2 (9) 2 = 1 + 8Fr 12 or Fr1 = 3.16 Since Fr1 = V1 1/2 (gy1) V = 3.16 # 69.81 # 0.3@ 1/2 = 5.42 m/s The power dissipated is given by oL P = gvh Where dimension less head loss hL = 1 − y2 + Fr 12 1 − y1 2 b y2 l E y1 2 ; y or and Thus

FM 8.44

2 (3.16) 2 hL = (0.3) =1 − 1.2 + 1 − b 0.3 l 1G = 0.504 m ' 0.3 2 1.2 vo = A V = y bV = . # # . = . m s P = 9.8 # 81.3 # 0.504 = 401 kN − m/s = 401 kW

Option (D) is correct. With the jump height ratio y = 1 6 1 + 8Fr 12 − 1@ 2 y 140 = 1 1 + 8Fr 12 − 1@ 2 #6 40 7 =

1 + 8Fr 12 − 1

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GATE Mechanical Engineering in 4 Volume NODIA FM 320

Demo Ebook

Page 320

Open Channel Flow

or

8 = 1 + 8Fr 12 64 = 1 + 8Fr 12 Fr 12 = 63 = 7.875 8

or

Fr1 , 2.81

Since

FM 8

V = Fr1 gy1 = 2.81 # 9.81 # 0.4 = 5.56 m/s

+ = m E = y +V = g # ^y2 − y1h3 ]1.4 − 0.4g 3 Thus hf = = 0.45 m = 4y1 y2 4 # 0. 4 # 1. 4 Therefore percentage dissipation h = f = , 23% E

And

FM 8.45

Option (B) is correct. From the depth ratio y = y 3.6 = 0.9 or But

Thus And

1 7− 1 + 1 + 8Fr 2 A 1 2 1 7− 1 + 1 + 8Fr 2 A 1 2

Fr1 = 3.16 Fr1 = V gy V = 3.16 59.81 # 0.9? 1/2 = 9.39 m/s vo = A V = by V = # . # . y y hL = y = − + Fr ' − b l 1G y y

= (0.9) =1 − 3.6 + 0.9 The power dissipated is given by oL P = gvh

= 845 m3/s

2 (3.16) 2 1 − b 0.9 l 1G = 1.51 m ' 2 3.6

= 9.80 # 845 # 1.51 = 12500 kN − m/s = 12500 kW = 12.5 MW FM 8.46

Option (D) is correct. The average velocities before and after the jump are o V = v = = m s b#y # . o V = v = = . m s b#y # Head loss

( ) −( . hL = y − y + V − V = ( . ) − ( ) + g # .

)

= 6.33 m The mass flow rate of water is o = rvo = m g s # = Then the dissipated mechanical power becomes o # hL = 70000 # 9.81 # 6.33 Pmechanicl = mg = 4346811 W - 4.35 MW

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

FM 8.47

Demo Ebook

Page 321

Open Channel Flow

FM 321

Option (C) is correct.

Froude number before the hydraulic jump is Fr1 = V = = . gy . # . Fr1 > 1, therefore the flow is supercritical before the jump. The flow depth after the jump is y = 0.5y1 8− 1 + 1 + 8Fr 2 B 1

= 0.5 # 0.35 # 8− 1 + 1 + 8 # (6.476) 2 B = 3.035 m From continuity equation Vy =Vy y V =V = # . = . y . Now from the energy equation, head loss is 2 2 hL = (y1 − y2) + V 1 − V 2 2g (0.35 − 3.035) +

m s

(12) 2 − (1.384) 2 2 # 9.81

hL =− 2.685 + 7.242 = 4.56 m FM 8.48

Option (A) is correct. Flow depth before the sluice gate is y =V #y = = m . # V Now Froude number after the jump Fr 3 = V = = . gy . # Then flow depth y is y = 0.5y 3 [− 1 + 1 + 8Fr 32 ]

= 0.5 # 3 # 8− 1 + 1 + 8 # (0.7373) 2 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 = hL ...(i) Es Now, velocity before the jump y V = #V = = 6.092 m/s y . # Head loss due to jump

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GATE Mechanical Engineering in 4 Volume NODIA FM 322

Demo Ebook

Page 322

Open Channel Flow

FM 8

hL = y − y + V − V = g

−

= 0.0460 m Specific energy before the jump is

]

Es = y + V = + g # Substituting values in equation (i), we get Dissipation ratio = hL = = 0.0119 Es FM 8.50

+

− #

g =

m

Option (B) is correct Applying the Continuity and Energy equations to eliminate V Continuity : vo = V y b = V y b Energy: E = y +V = y +V g g

...(i) ...(ii)

Combining equation (i) and (ii) to eliminate V . V = 2g (y1 − y2) + V 12 = 2g (y1 − y2) + V 22 # c

y2 b2 2 y1 b 1 m

from eq. (i)

yb = 2g (y1 − y2) V = −c y b mG V1 R 2g ^y1 − y2h W 2 S V =S y2 b2 2 W SS )1 − c y1 b1 m 3 WW X T g ^y − y h or vo = V y b = > − − − − H "b y − b y , Substitute values, we get vo = = FM 8.51

2 2 # 9.81 # ]1.9 − 1.5g 3 −2 −2 −2 G , 9.9 m /s −2 − 2 1 . 5 3 1 . 9 ] g #] g ] g #] g 1

Option (C) is correct. When frictional effects are negligible and the flow section is horizontal, the specific energy remains constant. Es = Es y +V = y +V g g vo vo =y + y + g # (b # y ) g # (b # y ) vo2 vo2 = + 1+ 0.5 2 # 9.81 # (5 # 1) 2 2 # 9.81 # (5 # 0.5) 2 0.5 = 0.006114vo2 vo = 9.0 m3/s Hence, the downstream velocity and Froude number are o V = v = = . m s b#y # . . and Fr2 = V = gy . # . = 1.625

o V= v by

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

FM 8.52

Demo Ebook

Page 323

Open Channel Flow

FM 323

Option (A) is correct.

The specific energy of water is to be Es = y + V g where

y = flow depth = 0.25 m V = Velocity of water =

Discharg e 0.7 = 3.5 m/s = 0.25 # 0.8 Ac

(3.5) 2 = 0.874 m 2 # 9.81 Now alternate depth y is to be determined by ( . ) o Es = y + v =y + gb y # . # . #y 0.874 = y + 0.0 y Solving above equation y = 0.815 m There are three roots of this equation. One for subcritical (y = 0.815), one for supercritical (y = 0.25 m) 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 m to 0.815 m. Thus

FM 8.53

Es = 0.25 +

Option (B) is correct. Specific energy of flow is to be Es = y + V g V2 = E −y s 2g V =

2g (Es − y) =

= 3.636 m/s Now, the flow rate becomes vo = V # b # y = . Alternate depth is determined from Es = Es Es = y +

2 # 9.81 # (1.224 − 0.55)

# # 0.

=

m s

vo gb y

( ) # .8 # 2 1.224y 2 = y + 0. 04 y − . 4y + 0. 04 = 0 By solving above equation 1.224 = y +

= y + 0. 04 y #y

y = 1.03 m Now the critical depth of flow

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GATE Mechanical Engineering in 4 Volume NODIA FM 324

Demo Ebook

Open Channel Flow

o yc = e v o gb FM 8.54

Page 324

==

FM 8

] g G #] g

= 0.742 m

Option (B) is correct. The upstream Froude number and critical depth are . Fr1 = V = = . gy . # . (V by ) o y V and yc = = v G = = G =; g E gb gb ]1.2g2 # ]2.5g2 1/3 =; E = 0.972 m 9.81 Since Fr11, the flow is subcritical and the flow depth decreases over the bump. The upstream over the bump and critical specific energies are ( . ) = . m Es = y + V = ( . ) + g # . Es = Es − Dzb = .

− .

= .

m

Ec = 3 # yc = 3 # 0.972 = 1.46 m 2 2 It is show that Es < Ec . 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. FM 8.55

Option (B) is correct.

The upstream Froude number and the critical depth are Fr1 = V = = . gy . # (V by ) o V y ] g #] g yc = = v G = = E G =; g E =; . gb gb = 1.868 m Since Fr1 > 1, the upstream flow is supercritical and flow depth increases over the bump. The upstream, over the bump and critical specific energies are

] g

Es = y + V = + g # .

= .

m

Es = Es − Dzb = .

= .

m

− .

Ec = 3 # yc = 3 # 1.868 = 2.802 m 2 2 The flow depth (y2) over the bump is determined from y − (Es − Dzb) y + V y = 0 g

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

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Page 325

Open Channel Flow

FM 325

y − Es y + V y = 0 g y −

#y +

] g

#

#] g = 0

=0 y − y + 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 + Dzb = 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 = 8 Cd # 2g tan b q l # H 5/2 15 2 Where Thus

H = Heigh οf weir = 1.5 − 0.5 = 1 m vo = 8 # 0.60 # 2 # 9.81 # tan b 60c l # ]1g5/2 15 2 = 0.818 m3 /s

FM 8.57

Option (B) is correct. The discharge rate through the triangular weir is vo = 8 # Cd # tan b q l # 2g # H 5/2 15 2 For constant Cd and H , discharge depends on tan b θ l 2 Therefore

vo

= k # tan b

l

k = constant = 8 Cd # 2g # H 5/2 15

Where and

c

vo

c

= k # tan b

l

50c vo c = tan _ 2 i = 0.4663 = 0.391 vo c tan _ 1002 c i 1.1917 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%

Then

FM 8.58

Option (A) is correct.

The discharge rate per unit width vo = Cd ab gy = 0.57 # 0.4 # 1 # 2 # 9.81 # 2.4 = 1.56 m3 /s When the frictional effects are negligible in horizontal flow, the specific energy

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GATE Mechanical Engineering in 4 Volume NODIA FM 326

Demo Ebook

Page 326

Open Channel Flow

FM 8

remains constant. Es = Es Since and

vo g b#y vo Es = y + g b#y Es = y +

= . +

# .

. #

# .

= 2.421 m

= .

. # . # #y Solving above equation y = 0.238 m for flow depth as the physically meaningful root (positive and less than 2.4 m). o o . And downstream velocity V = v = v = = 6.55 m/s Ac by # . Thus

FM 8.59

2.421 = y +

Option (D) is correct. The depth ratio y a and the contraction coefficient y a for this flow are y y = . = = 5 = 10 and a 0.5 a . The discharge coefficient corresponding to these values of (y1 /a) and (y2 /a) is determined from a experimental graph between Cd and (y1 /a, y2 /a) is Cd = . . Then the discharge rate becomes vo = Cd ba gy = . # # . # # . # = 14.3 m3/s

FM 8.60

Option (C) is correct. For broad- crested weir, the flow rate is given by the relation vo = Cwb b g b l And

H

where Cwb = Broad crested weir coefficient

0.65 = 0.65 1/2 = 0.563 H 1/2 _1 + 10..55 i _1 + H i 3/2 vo = 0.563 # 4 # (9.81) 1/2 # b 2 l # (0.5) 3/2 = 1.36 m3/s 3 y min = yc = H = 2 # 0.5 = 0.333 m 3 Cwb =

w

Thus and FM 8.61

Option (C) is correct.

The flow rate measure by a rectangular weir is given by vo = 2 # Cd # 2g # b # H 3/2 3 And Cd = 0.598 + 0.0897 # H Hw = 0.598 + 0.0897 # 0.60 = 0.6469 1. 1

...(i) Hw = heigh of weir

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GATE Mechanical Engineering in 4 Volume NODIA FM 8

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Page 327

Open Channel Flow

FM 327

The condition H Hw < is satisfied. Since 0.60/1.1 = 0.55, then the water flow rate through the channel vo = 2 # 0.6469 # 2 # 9.81 # 6 # (0.6) 3/2 = 5.33 m3 /s 3 FM 8.62

Option (B) is correct. # g #b l # H H + b Hw l H = 0.5 m and Hw = . o q =v = b

We have Substitute

0.65 2 3/2 (0.5) 3/2 1/2 (9.81) b # # 1/2 3l # b1 + 20..05 l = 0.350 m2/s

q =

Thus

FM 8.63

.

Option (D) is correct. For rectangular sharp crested weir, flow rate is vo = 2 Cwr 2g bH 3/2 where Cwr = rectangular weir coefficient 3 And Cwr = 0.611 + 0.075 b H l Hw Thus

H = 3 − 2.2 = 0.8 m and Hw = . vo = 2 ;0.611 + 0.075 # b 0.8 lE # 62 # 9.81@ 1/2 b (0.8) 3/2 3 2.2

= 1.349b m3/s o o b= . V = v = v = . y by by A For uniform flow (From manning’s formula) where a = vo = a AR h S n S = 2 = 0.00667 300

...(i)

Also for a wide channel A = by

So that

and perimeter1 = 2y1 + b by1 A1 = . y1 if b >> y Rh = perimeter1 (2y1 + b)

Thus with n = . vo = 1.349b =

1 by (y ) 2/3 (0.00667) 1/2 0.014 # 1 # 1 #

or y = 0.415 m Now from equation (i), V = 1.349 = 3.25 m/s 0.415 . So that Fr1 = V = = 1.61 gy 6. # . @ Since Fr1 > 1, it is possible to produce a jump.

***********

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FM 9 TURBO MACHINERY

FM 9.1

Match List-I with List-II and select the correct answer using the codes given below List-I 1.

Dynamic pump

Q. Fans

2.

Positive-Displacement turbine

R. Water-meters

3.

Positive-Displacement Pump

S.

4.

Dynamic turbine

P.

Wind mills

Heart

Codes : P (A) 2 (B) 1 (C) 4 (D) 3 FM 9.2

List-II

Q 3 4 1 2

R 4 3 2 1

S 1 2 3 4

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 (A) (B) (C) (D)

P 3 2 2 3

Q 4 1 4 1

R 2 3 3 2

S 1 4 1 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) Vout = Vin (B) Vout 1Vin (C) Vout 2Vin (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

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FM 329

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 20cC is delivered by a pump with 1500 L/min 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 L/ min. 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 m3/h of water at 20cC 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 20cC with 35 m3/ min 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 (C) 63 m FM 9.9

(B) 58 m (D) 48 m

Match List I with List II and select the correct answer using the codes given below : List-I P.

List-II 1.

Kinetic Energy

Q. Pumps

2.

Momentum Exchange

R. Impulse-Turbine

3.

Diffuser

S.

4.

Pressure-rise

Draft tube

Reaction-Turbine

Codes : P (A) 3

Q 4

R 1

S 2

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(B) (C) (D)

4 2 1

3 1 4

2 4 3

FM 9

1 3 2

FM 9.10

Consider the following statements : A water turbine governor 1. 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 20cC is sprinkled at 14 m3/h by a 36 cm diameter turbine as shown in figure below. If the nozzle exit diameter is 8 mm, the appropriate rotation rate will be

(A) 1070 rpm (C) 1030 rpm

(B) 107 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 statements If 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.

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R. the required shaft power goes up by S. the output shaft power of the pump Which of the above is/are TRUE ? (A) P and R (C) Q and S

a factor of about 4. goes up by a factor of about 2. (B) P only (D) P and S

FM 9.15

A pump delivers water at a rate of 3 m3/s 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 DA = . m and spins at NA = rpm . At its best o efficiency point, vA = 2 m /s , HA = 2. m of water and bhp A = 132 MW . The turbine-B will spin at 240 rpm and its net head will be HB = . m . What will be the diameter DB , voB and bhp B such that it operates most efficiently ? (A) DB = . m , voB = m /s , bhpB = 126 MW (B) DB = 2. m , voB = m /s , bhpB = 1057 MW (C) DB = . m , voB = m /s , bhpB = 14 MW (D) DB = 2. m , voB = 42 m /s , bhpB = 341 MW

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 m3/s 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 m3 /s , 0.8 m (B) 0.0328 m3 /s , 8 m (C) 3.28 m3 /s , 12 m (D) 0.328 m3 /s , 80 m

FM 9.18

Which one of the following relation is true for dimensionless parameters of two dynamically similar pumps ? /4 /4 o /2 o /2 (A) DB = DA # b HA l # c voB m (B) DB = DA # b HB l # c voB m HA HB vA vA /2 /4 o /4 o /2 (C) DB = DA # b HA l # c voB m (D) DB = DA # b HA l # c voA m HB HB vA vB

FM 9.19

Which of the following relation is true for specific speed of turbine and specific speed of the pump ? (B) NST = NSP # hturbine (A) NSP = NST (C) NST = NSP (D) NST = NSP # hturbine hturbine

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 m3/s 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

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(A) 0.5 (C) 2 FM 9.22

Page 332 FM 9

(B) 1 (D) 0.25

Performance data for a very small (D = 8.25 cm) model water turbine, operating with an available head of 15 m are as follows : vo m h

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 m3/s 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 (C) 23.5 MW FM 9.24

(B) 23.5 kW (D) 2.35 MW

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 m3/s Fluid is light oil S.G. = 0.91 Efficiency η = 75% What will be the input power ?

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FM 333

(A) 1525 W (C) 760 W FM 9.25

(B) 858 W (D) 1715 W

A liquid (S.G = 0.9) flows through the pump with the flow rate 7.57 # 10−3 m3 /s . 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 z2 − z1 = 0.5 m , what will be the actual head rise across the pump ?

(A) 5.5 m (C) 11.5 m

(B) 8.6 m (D) 14.4 m

Common Data For Q. 26 and 27. Water at 40cC ( γ = 9.731 # 103 N/m3, pv = 7.376 # 103 N/m2 ) 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) 1.55 kW (C) 1.1 kW (D) 4.14 kW

FM 9.27

What will be the NPSH A at the pump inlet ? (A) 15.75 m (B) 9.45 m (C) 6.3 m (D) 12.6 m

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 ha = 52.0 − 1.01 # 10 vo2 with ha in meters when vo is in m3/s . 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 (KL = 1) ?

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(A) 5.29 m3 /s (C) 0.0529 m3 /s

FM 9

(B) 0.00529 m3 /s (D) 0.529 m3 /s

FM 9.29

Consider a hydraulic turbine which provided with 4.25 m3/s 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 1100 kW , 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 m3/s . 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 30 mm 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

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(A) 23.4 MW (C) 46.8 MW

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FM 335

(B) 17.55 MW (D) 29.25 MW

Common Data For Linked Answer Q. 33 and 34 Water 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

FM 9.34

FM 9.35

What will be the nozzle diameter D that gives the maximum power output of the turbine ? fD D (A) D = (B) D = ^l D h ^ fl D h fD D (C) D = (D) D = ^ fl D h ^ fl D h For the maximum power to be developed by the turbine, the velocity head at the nozzle exit is (B) 2H (A) 1 H 3 (C) 2 H (D) 3 H 3 2 Water at 20cC with flow rate of 3.5 m3/s 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 (C) 477 kW

(B) 47.7 kW (D) 239 kW

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GATE Mechanical Engineering in 4 Volume NODIA FM 336

FM 9.36

FM 9

A centrifugal pump rotates at 1000 rpm. Water enters the impeller normal to the blades (α1 = 0c) and exits at an angle of 35c from radial (α2 = 35c). The inlet radius is r = cm , at which the blade width b = cm . The outlet radius is r = cm , at which the blade width b = cm . The volume flow rate and pump efficiency are 0.0573 m3 /s and 76% respectively. What will be the net head produced by this pump and required brake horsepower, respectively ? ( ρwater = . kg m ) (B) 4.87 m, 20763 W (D) 1.55 m, 8662 W

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 (C) 69.25c FM 9.38

Page 336

Turbo Machinery

(A) 4.87 m, 27320 W (C) 1.55 m, 6583 W FM 9.37

Demo Ebook

(B) 41.55c (D) 55.4c

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 (C) 357 rpm , 112.12 kg/s

(B) 286 rpm , 89.7 kg/s (D) 250 rpm , 76.8 kg/s

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FM 337

Common Data For Linked Answer Q. 39 and 40 A radial flow centrifugal pump delivers gasoline (ρ = 680 kg/m3) at 20cC . The pump has d = cm , d = cm , b = cm , b = . cm , β1 = 25c, β2 = 40c and rotates at 1160 rpm. FM 9.39

FM 9.40

FM 9.41

The flow rate in m3/hour is (A) 17.23 (C) 28.72

(B) 0.2872 (D) 1038

What will be the head ? (A) 48 m (C) 32 m

(B) 16 m (D) 64 m

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 (C) 522 kW

(B) 261 kW (D) 696 kW

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) 3.04 N− m (C) 4.05 N− m (D) 6.1 N− m

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GATE Mechanical Engineering in 4 Volume NODIA FM 338

FM 9.43

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Turbo Machinery

FM 9

If the resisting torque is negligible, what would be the angular velocity of the manifold ? (A) 1.705 rev/s (B) 17.05 rev/s (C) 1.278 rev/s (D) 12.78 rev/s

Common Data For Q. 44 and 45 A Francis radial-flow hydroturbine is being designed with the following dimensions, where location 2 is inlet and 1 is outlet : 1 m , r1 = 1 2 m , b1 = 2 2 m r2 = 2 m , b2 = Where r = radius , b = blade width Runner blade angle at inlet β2 = 66.2c Runner blade angle at outlet β1 = 36.1c Volume flow rate vo = 340 m3/s 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 α2 through which the wicket gates should turn the flow and the swirl angle α1 ? (A) α2 = 30c, α1 = 10c (B) α2 = 60c, α1 = 80c (C) α2 = 80c, α1 = 60c (D) α2 = 10c, α1 = 30c

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 α1 = 60c and an inlet rotor tip speed U1 = m s 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 m2 /s2 (C) 7.8 m2 /s2 FM 9.47

(B) − 15.6 m2 /s2 (D) 15.6 m2 /s2

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 α1 = 55c and β1 = 30c

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FM 339

while at the outlet β2 = 60c. What will be the horse-power ? ( ρair = . (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 β = 165c. 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 m3/min of water at 20cC . If the blade thickness is constant at 20 cm, the theoretical power developed at 100% efficiency will be

(A) 800 kW (C) 250 kW FM 9.50

(B) 400 kW (D) 375 kW

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 m2 /s2 , 68.9% (C) 17.4 m2 /s2 , 84.3%

(B) 13.05 m2 /s2 , 42.15% (D) 34.8 m2 /s2 , 63.25%

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GATE Mechanical Engineering in 4 Volume NODIA FM 340

FM 9.51

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Page 340

Turbo Machinery

FM 9

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 m2 /s2 (C) 505 m2 /s2

(B) 303 m2 /s2 (D) 404 m2 /s2

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 β 2 is 25c as shown in figure. The uniform blade height is 50 mm . When the flow through the pump is 0.16 m3/s , the shaft power required to turn the impeller is (A) 72.3 kW (B) 90.38 kW (C) 45.18 kW (D) 54.22 kW

FM 9.53

Consider a hydraulic turbine runner as shown in figure. Relative to the rotating runner, water enters at section (1) ( radius r 1 = 1. m ) at an angle of 100c from the tangential direction and leaves at section (2) ( radius r 2 = . m ) 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 m3/s . The runner speed is 130 rpm in the direction shown. What will be the shaft power developed ?

(A) − 12.8 MW (C) 9.6 MW

(B) − 9.6 MW (D) 12.8 MW ***********

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SOLUTIONS FM 9.1

Option (C) is correct. P.

4.

Dynamic turbine

Q. Fans

1.

Dynamic pump

R. Water-meters

2.

Positive-Displacement turbine

3.

Positive-Displacement Pump

S. FM 9.2

Wind mills

Heart

Option (B) is correct List-I

FM 9.3

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

Option (C) is correct. Conservation of mass requires that oout o in = m m ρin Vin Ain = rout Vout Aout The cross-sectional area is proportional to the square of diameter r Vout = Vin in : Din D = Vin : Din D rout Dout Dout For Dout < Din , the ratio (Din /Dout > 1). Hence Vout > Vin

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.

FM 9.5

Option (D) is correct. o = voDp Since Pwater = rgvH

Δp = ρgH

1500 270 = 6.75 kW 1000 # 60 # output power η = = 6.75 9.00 input power =

and Efficiency

η = 75% FM 9.6

Option (C) is correct. The ideal power of the pump is

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FM 9

Pideal = rgHvo = 998 # 9.81 # 1.6 #

18 60 # 100

18 L/ min. =

18 m3/s 60 # 1000

= 4.7 W Hence, the actual power (bhp) is bhp = Pideal # hpump = 4.7 # 0.70 = 3.3 W FM 9.7

Option (B) is correct. For water at 20cC , take ρ = 998 kg/m3 . The power relation is o rgvH P = h 998 # 9.81 # b 125 l # H 3600 or (22 # 745.7) = 22 hp = 22 # 745.7 W 0.71 H = 22 # 745.7 # 0.71 # 3600 , 34 m 998 # 9.81 # 125 Pressure rise

FM 9.8

Δp = rgH =

# .

#

= 332873 Pa - 333 kPa

Option (B) is correct. The energy equation for the system. H pump = Dz + f L V d g o V = v = p A

Velocity

, #( . )

. m s

(18.6) 2 H pump = 11 − 4 + 0.0144 # 20 + 12 + 8 # 0.2 2 # 9.81

Thus

, 58 m FM 9.9

Option (A) is correct. P.

Draft tube

3.

Diffuser

Q. Pumps

4.

Pressure rise

R. Impulse-Turbine

1.

Kinetic Energy

S.

2.

Momentum Exchange

Reaction-Turbine

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 20cC take ρ = 998 kg/m3 . Each arm takes 7 m3/h . Vrel

vo = 2 = p Aexit

3 #( .

)2

= 38.7 m/s (At max power)

u = wR = 1 Vrel = 1 .3 m s 2 or

0.18ω = 19.35

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ω = 19.35 = 107 rad/s 0.18 ω = 107 # 60 , 1030 rpm 2p 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. gH gH CH = 2 2 = 2 2 wD N D For full scale turbine A: gHA CH A = NA DA and for Quarter scale Turbine B : (DB = DA /4)

ω = 2πN 60

gHB gHB = N B D B N B # DA b l CH for both turbine must be same, therefore CH, A = CH, B CH B =

gHA gHB # = NADA NB DA N B = N A # HB # HA NB = NA # HB # HA FM 9.14

= 428 #

10 4 = 988 rpm 30 #

Option (D) is correct. Affinity or scaling laws DB voB = wB wA # b DA l voA HB = wB DB net head HA a wA k # b DA l r bhpB net Power = B a wB k # b DB l rA wA DA bhpA True : Rotation rate appears with an exponent of 1 in the affinity law for capacity. Thus, the change is linear. 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. 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. 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.

For capacity For For P. Q. R.

S.

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FM 9.15

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FM 9

Option (D) is correct w vo 6g # ha@ For ω = 60 rad/s , vo = m s , g = m s and ha = Ns =

Specific speed

m

60 # 3 = 1.98 69.81 # 20@3/4 , the pump is a mixed-flow pump. Ns =

For Ns = 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 = DA # HB # b NA l = ( . ) # #b l = 2.073 m HA NB . 3 and voB = voA # b NB lb DB l = #b l # b . l = 342 m /s NA DA 3 5 r and bhpB = bhpA # b B l # b NB l b DB l rA NA DA 5 3 = 132 # b 998 l # b 240 l # b 2.073 l = 340.71 , 341 MW 998 300 1.5

FM 9.17

Option (B) is correct. For similarity the model pump operate at the same flow coefficient, so that o o c v m =c v m ωD m wD p where the subscript (m ) refers to the model and (p) to the prototype. Thus, vom = wm # c Dm m vop wp Dp , and vop = . m s , and with ωm = ωp , Dm D p = vom = 1 # b 1 l # 4.1 = 0.0328 m3/s 5 3

We get Also So that

c

gha gha 2 2m =c 2 2m ω D m w D p g ha = p # b wm l # c Dm m # ha gm wp Dp m

and with g p = gm , ωm = ωp , Dm D p = We get FM 9.18

ha

m

p

, and ha = m 2 = 1 # (1) 2 # b 1 l # 200 = 8.00 m 5 p

Option (A) is correct. Since the two pumps are dynamically similar, dimensionless pump parameter head coefficients CH must be the same for both pumps CH, A = CH, B gHB gHA = ωA D A wB D B ωA = HA DB ...(i) ωB HB # b DA l Similarly, dimensionless pump parameter capacity coefficient Cvo must be same for both pumps Cvo, A = Cvo, B

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voB wB D B voB wA ...(ii) wB # voA get HA DB voB # # HB DA voA o DB HA voB or DB = DA # b HA l # c voB m b DA l = HB # vo H vA B A

voA = ωA D A DB b DA l = From equation (i) and (ii), we DB b DA l =

FM 9.19

Option (B) is correct. o1/2 Pump specific speed : NSP = wv 3/4 (gH ) w (bhp) 1/2 Turbine specific speed: NST = (gH) 5/4 # r1/2 Dividing and multiplying equation (ii) by vo , w (bhp) 1/2 vo1/2 NST = 5/4 1/2 # 1/2 (gH ) # r vo (bhp) 1/2 o1/2 By Rearranging, NST = = wv 3/4 G # 1/2 (gH) r # vo1/2 # (gH ) 1/2 b p NST = NSP # c o m = NSP # hturbine rvgH o1/2 bhp where o = ηturbine and ωv 3/4 = NSP ρvgH (gH)

...(i) ...(ii)

From eq (i)

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 .

Here

voB = voA # wB # b DB l = . wA DA DB = DA , NB = NA and ω = 2πN 60

#

= . #

−

m s

pN B HB = HA # a wB k # b DB l = HA # e o #( ) wA DA pN A = HA # b NB l = . # b l = . m NA Now, the ratio of specific speeds 1/2 (gHB) 3/4 NSA = wA voA voA 1/2 HB 3/4 wA = # # # b c m 3 4 1 2 / / o w HA l NSB vB B (gHA) wB # voB 3/4 o 1/2 = 2pNA # 60 # c voA m # b HB l 60 2p N B HA vB − o . # = NA # c voA m # b HB l = # #b . l c NB HA vB . # − m NSA = 1 NSB and

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FM 9.22

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FM 9

Option (C) is correct. We use best efficiency point (BEP) data to calculate power. At BEP η = 65% , N= rpm , vo = . m = . m s and at BEP o #h Pmodel = rg # vH = 998 # 9.81 # 0.00464 # 15 # 0.65 = 443 W First establish the BEP coefficients from the model turbine data. o . = . Cvo = vm = ND ) #( . .

CH =

gHm = Nm D

CP =

Pm = rN m D

b

# l #^ .

#b

h

=

l #( .

.

)

= .

Now enter new data for geometrically similar turbine o Cvo = 0.198 = vt 3 = 0.19 3 Nt D t Nt # D t or Nt # D t = 0.19 = 0.96 0.198 gHt = . # CH = 12.45 = Nt # Dt Nt # Dt or N t # D t = 9.81 # 46 = 36.25 12.45

...(i)

..(ii)

From equation (i) and (ii) Nt = . Dt or

or

Then

and N t # D t = 36.25

. c D m # D t = 36.25 t ( . ) # D t = 36.25 Dt (0.96) 2 = 0.02542 Dt = 36.25 Dt = 0.399 b 0.40 m and Nt = . = . = rps Dt ( . ) Pt = CP # r # N t # D t = 1.6 # 998 # (15) 3 # (0.40) 5 = 55185 W

or FM 9.23

Pt = 55185 = 74 hp 745.7

Option (C) is correct. For flow from section (1) to section (2), energy equation gives 2 p p2 V 22 + + gz2 = 1 + V 1 + gz1 + wshaft net in − loss 2 2 ρ r Since p2 = p1 = patm , V1 = and wshaft net out =− wshaft net in 2 wshaft net out = g (z1 − z2) − V 2 − loss 2 o to get For power, we multiply equation (i) by the mass flow rate m 2 o (z1 − z2) − m oV 2 − m o loss Pshaft net out = mg 2

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o z − z − rvoV − rvo Pshaft net out = rvg

o = rvo m

= (999 # 30 # 9.81 # 100) − 999 # 30 #

(2) 2 − (999 # 30 # 20 # 9.81) 2

= 23.5 # 106 N − m/s = 23.5 MW FM 9.24

Option (A) is correct. Convert p = p =

= 13332 Pa = 66661 Pa

p = rgH

γoil = 0.91 # 9790 = 8909 N/m3 o V = v = p . = . s A #( . ) o V = v = p . = . s A #( . ) p p H = +V + z − −V − z g g g g

and

So, the head is

(5.79) 2 (1.00) 2 = 66661 + + 0.65 − − 13332 − −0 8909 2 # 9.81 8909 2 # 9.81 = 11.3 m Therefore input power Pinput = FM 9.25

o gvH = h

#( .

)# .

Option (C) is correct. The head rise gained by fluid flowing through a pump is p −p ha = + z − z +V − V g g − o Since = . m s V = v = .p # A ( . ) From continuity equation

.

= 1525 W

...(i)

VA =VA V =Vb

l = .

#( ) = .

m s

h Thus, from equation (i), with p =− (hHg) (gHg) =− ( . )^ # and p = 80 # 103 N/m2 , 80 # 103 + (0.095) (133 # 103) (3.19) 2 − (0.797) 2 + + ha = 0.5 2 # 9.81 0.9 # ^9.80 # 103h ha = 11.5 m FM 9.26

Option (A) is correct.

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

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p V p + + z + hp = + V + z + f l V γ g g g D g where p = p = , V = , V = m s , z = m , and z = Thus,equation (i) becomes 2 z + h p = V 2 b1 + f l l 2g D

FM 9

...(i)

200 (3) 2 hp = 1 + 0.0152 # b 0.05 lF − 3 = 25.3 m 2 # 9.81 < Hence, Power gained by fluid o p P = gvh = (9.731 # 103) # p (0.05) 2 # 3 # 25.3 4

FM 9.27

= 1.45 # 103 = 1.45 kW Power gained by fluid and Power supplied to fluid = Efficiency = 1.45 = 2.07 kW 0.7 Option (D) is correct. p p We have ...(i) NPSH = s + V s − v g g g where ps and Vs refer to the pressure and velocity at the pump inlet, respectively. p p V Also, + + z = s + V s + zs + h L g g γ g So that with p = patm , V = , zs = and hL = patm p ...(ii) + z = s + Vs g g γ and therefore from equation (i) and (ii), the available NPSH is p p ..(iii) NPSH A = atm + z − v g g With z positive (since pump is below reservoir) and ΣhL = . Thus, from equation (ii) with patm. = Pa 3 (7.376 # 103) NPSH A = 101 # 10 3 + 3 − = 12.6 m 9.731 # 10 (9.731 # 103)

FM 9.28

Option (C) is correct.

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From Bernoulli’s equation at section 1 and 2, 2 p p1 V 12 ...(i) + + z1 + h p = 2 + V 2 + z2 + ΣhL γ g g 2g and with p1 = p2 = , V1 = V2 = , and z2 − z1 = m , equation (i) becomes ...(ii) h p = 33 + ΣhL The head loss term can be expressed as 2 ΣhL = bKL + f # l lV D 2g With KL = 1, f = . 2 , l = m , D = .1 m , equation (ii) can be written as 2 h p = 33 + ;1.0 + 0.02 # 30 E # V 0.1 2 # 9.81 o vo V = v = p ( .1) 2 A

and with

h p = 33 + 61.0 + 6.0@ # 826 # vo2 h p = 33 + 5.78 # 103 vo2 Since the pump equation is h p = 52.0 − 1.01 # 103 vo2 equation (iii) becomes or

...(iii)

...(iv) ...(v)

Equation (iv) and (v) can be equated to determine the flow rate. Thus, 33 + 5.78 # 103 vo2 = 52.0 − 1.01 # 103 vo2 and FM 9.29

vo = 0.0529 m3/s

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 p − p2 loss = 1 ...(i) + g (z1 − z2) − wshaft net out r Since V1 = V2 and wshaft net out =− wshaft net in For power loss through the turbine we need to multiply equation (i) by the mass o . Thus flow rate, m o b p1 − p2 l + mg o (z1 − z2) − Pshaft net out ...(ii) Ploss = m r o = rvo = However m g s # .2 = 2 Also p2 =− (0.25 m of Hg) # rHg # g or =− 0.25 # 13.6 # 999 # 9.81 -− 33300 N/m2 From equation (ii), we get (415000 + 33300) Ploss = 4246 # + 4246 # 9.81 # 3 − 1.1 # 106 999 = 930346 N− m/s - 930 kW

FM 9.30

Option (C) is correct. The efficiency of the pump is given by ideal work required actual work required − loss η = = actual work required actual work required − loss w = shaft net in wshaft net in Now find wshaft net in , by using the energy equation 2 p − pin V out − V in2 + loss wshaft net in = out + r 2

...(i)

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From the volume flow rate, we obtain o o Vout = v = v = Aout pD out p#

FM 9

=

m s

Also from principle of mass conservation, Vin = Vout D out = = m s # D in Thus from equation (i), we get (400000 − 120000) [(28.29) 2 − (3.143) 2] + + 170 wshaft net in = 999 2 Then FM 9.31

= 846 N− m/kg η = 846 − 170 = 0.799 846

Option (C) is correct. Output shaft power for Pelton wheel is ...(i) Pshaft = rwrvo(Vj − wr) ( − cos b) Differentiate equation (i) with respect to ω and set the derivative equal to zero. Pmax = d 6rwrvo(Vj − wr) ( − cos b)@ = 0 dw ρrvo( - cos β) d (Vj ω - ω r) = 0 dω d (V ω - ω r) = 0 dω j Vj − wr = 0 where ωr =Turbine bucket speed and Vj = Jet speed Hence Turbine bucket speed = 1 # jet speed 2

FM 9.32

Option (A) is correct.

The power output for a Pelton wheel is given by o (U − V ) ( − cos b) Pshaft = rvU For maximum power β = 180c and U = V Thus,

Pshaft max =− rvoV

From the Bernoulli’s equation at section (0) and (1), p0 V 0 p + + z 0 = 1 + V 1 + z1 + f l V γ g g g D g

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FM 351

where p = p = , z =

,z = and V = z = z +V + f l V g D g π D 2V = p D 2V A V = AV or 4 1 1 4

Hence Also

V = bD l V = b . l V = . . D

That is

...(ii)

V

So that equation (ii) becomes. 975 = 250 +

V 12 1 + 0.032 # b 1020 l (0.0494) 2E 2 # 9.81 ; 0.9

V = 114.3 m/s

or

vo = A V = p ( . ) #

Hence

. = .59 m s

Therefore, from equation (i), we get Pshaft max =− (999) # 3.59 #

(114.3) 2 = 23.4 # 106 N m/s 2 −

= 23400 kW = 23.4 MW FM 9.33

Option (B) is correct. Since the power output of the pelton wheel is o (U − V ) ( − cos b) Pshaft = rvU So the maximum power output occurs with β = 180c and U = V . Thus, = rvoV (In magnitude) P shaft

...(i)

At section (0) and (1), from Bernoulli’s equation p0 V 0 p + + z 0 = 1 + V 1 + z1 + f l V γ g g g D g But p 0 = p1 = 0 , V0 = 0 , and z 0 − z1 = H 2 2 Thus, h = V1 + f l V 2g D 2g Since A1 V1 = AV or π D 12 V1 = π D2 V 4 4 We have From (ii), and equation (i) gives

...(ii)

V1 = c D m V Therefore, D1 2 4 2 h = V 1 ;1 + f l D 14 E or V 1 = 2g 2g DD

h D

V 12 = 2 H 2g 3

or FM 9.35

Option (C) is correct. For water, take ρ = 998 kg/m3 u = wr = b Here

and

Thus, FM 9.36

2 = 3 # V1 2g l lH 2 b2 # f D

1

#

p

l # . = 9.90 m/s

α2 = 30c, α1 = 90c 3. 5 Vn = 3.5 = , 7.96 m/s 2pr2 b2 2p # 0.7 # 0.1 Vt = Vn 2 = 7.96 = 13.8 m/s tan a2 tan 30c Vt = Vn1 = Vn1 = 0 tan a1 tan 90c o Vt = 998 # 3.5 # 9.90 # 13.8 Ptheoretical = rvu = 477213 W , 477 kW

Option (B) is correct. Normal velocity components at Inlet and Outlet are, o 0.573 V ,n = v = = 4.224 m/s 2pb1 r1 2 # p # 0.12 # 0.18 o 0.573 V ,n = v = = 2.715 m/s 2pr2 b2 2 # p # 0.24 # 0.14 and tangential velocity components at Inlet and Outlet are, V ,t = V ,n # tan a = . # tan c = V ,t = V ,n # tan a = . # tan c = . m s Now, the net head produced by the pump is H = w # 6r V , t − r V , t@ = pN # r V , t g #g H = 2p # 1000 # 0.24 # 1.9 = 4.87 m 60 # 9.81

V ,t =

and required brake power at 76% efficiency is o # hpump bhp = rgvH = 998.0 # 9.81 # 0.573 # 4.87 # 0.76 = 20763 W FM 9.37

Option (D) is correct. Since then But Vθ = V sin Hence,

FM 9.38

o (r Vq − r Vq ) = , T =m r Vθ = r Vq c = sin c = . m s and Vθ = V sin q = 1.9 # 13 = 25 sin q # 1.2 or θ = 55.4c

Option (B) is correct. Torque

# sin q

o (r Vq − r Vq ) T =m

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FM 9

Substituting numerical values,

o 6 # # sin c − − 60 = m # # sin c@ o = 89.7 N s/m = 89.7 kg/s m P = wT ω =P =− = 30 # b 1 l # 60 = 286 rpm − 2p T

or Also,

−

So that FM 9.39

Option (D) is correct.

We have

N = 1160 rpm ω = 2p # 1160 = 121.5 rad/s 60 u = wr = . #b . l = .

m s

Vn = u tan b = . # tan c = 5.1 m/s vo = 2pr1 b1 # Vn1 = 2p # 0.09 # 0.10 # 5.1 = 0.2882 m3/s = 0.2882 # 3600 , 1038 m3/hour FM 9.40

Option (C) is correct. o 0.2882 Vn = v = = 3.71 m/s 2pr2 b2 2p # 0.165 # 0.075

and Finally FM 9.41

u = wr = . # . = 20.05 m/s Vt = u − Vn cot c = 20.05 − 3.71 # cot 40c = 15.60 m/s o Vt = Pideal = rvu # . # . # . = 61297 W 61297 Head H = P o = , 32 m (rgv) 680 # 9.81 # 0.2882

Option (A) is correct From the shaft power equation with Vθ = o Vq We have Pshaft = rvo(U Vq − U Vq ) = rvU p# where U =r w= . # = . m s To evaluate Vθ we use the exit velocity triangle shown below.

Thus,

Vθ = U −

with

Vr =

Vr tan c

vo = 0.20 = 7.07 m/s 2pr 2 b2 2p # 0.15 # 0.03

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FM 355

Vθ = 47.1 − 7.07 = 37.0 m/s tan 35c

Hence,

Pshaft = 999 # 0.20 # 47.1 # 37.0 - 3.48 # 105 N m/s = 348 kW

From eq. (i), FM 9.42

Demo Ebook

−

Option (C) is correct. From the moment of momentum equation, the shaft torque o (r Vq − r Vq ) T =m For this continuously distributed outflow with Vθ = , it becomes. T =2

# rV dmo q

o= . where Vθ = m s and dm

FM 9.43

# r # ( ) dr

#

0.4

Thus,

T =2

or

T = :53.9 r D = 4.05 N − m 2 0.1

r # 3 # 0.003 # 999 # (3) dr

r = 0.1

2 0.4

Option (A) is correct.

T =2

With

# rV dmo = 0

o = rWh dr = dm

where

...(i)

q

# #( .

) dr = 8.99 dr and

From the figure Vθ = W − wr = − wr Thus, from equation (i) 0 =2

#

0.4

0.1

r (3 − wr) (8.99 dr) =

#

0.4

0.1

(3r − wr2) dr

= b 3 r2 − 1 wr3 l = 0.225 − 0.021w 2 3 0.1 ω = 10.7 # 1 = 1.705 rev/s 2p 0.4

or FM 9.44

Option (A) is correct. Normal component of velocity at inlet and outlet are o 340 V ,n = v = = 37.0 m/s 2pr2 b2 2 # p # 2 # 0.731 o 340 V ,n = v = = 17.33 m/s 2pr1 b1 2 # p # 1.42 # 2.2 and tangential velocity components at inlet and outlet are V ,n 37 V ,t = wr − = 2p # 180 # 2 − 60 tan b tan (66.2c) and

V ,t

= 21.38 m/s V ,n = wr − = tan b

.

# .

−

ω = 2πN 60

. = 3.0 m/s tan ( . ) c

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=V V tan α2 = V

Since

V

t

n t n

FM 9

# tan a =

α2 = tan−1 (0.58) = 30c V t = V n # tan a V tan α1 = t = V n

and

α1 = tan−1 (0.174) = 9.9c , 10c FM 9.45

Option (B) is correct. The shaft power is estimated from the Euler’s turbo-machine equation Pshaft = rwvo(r V t − r V t) = 998 # 18.85 # 340 # [2 # 21.38 − 1.42 # 3.0] = 2.46 # 108 W b 246 MW . Since the irreversibility neglected. Therefore bhp # H = o= rgv # . #

Assuming ηturbine = Net Head

= 73.9 m FM 9.46

Option (B) is correct The energy transfer per unit mass of fluid is given by the relation wshaft = U Vq − U Vq o = 2pr 1 bVr1 = 2pr 2 bVr 2 m V = r V = # r cos c cos

Also or

= m s

Vθ = V sin a = # sin c = . m s and Vθ = wshaft =− U Vq =− ( ) # . =− 15.6 m2 /s2

Then Thus FM 9.47

c

Option (A) is correct.

The average radius Thus, Also or

Then and Finally

R = R1 + R2 = 0.5 + 0.4 = 0.45 m 2 2 p u = wR = b # l # ( . ) = 56.6 m/s

u = Vn (cot a + cot b ) = Vn (cot a + cot b ) u . Vn = Vn = = (cot c + cot c) cot c + cot

c

Vn = Vn , . m s vo = Vn A = . # p ( . ) − ( . ) = 6.56 m3/s gH = u − uVn (cot a + cot b ) = (56.6) 2 − (56.6) # (23.2) (cot 55c + cot 60c) = 1520 m2/s2 o P = rvgH = 1.205 # 6.56 # 1520 = 12000 W b 16 hp

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FM 357

Option (D) is correct. The volume flow rate of the jet is vo = Vj # p D j =

# .

#

( .

)

= 0.785 m3/s

ω = 2pN = 2 # p # 270 = 28.26 rad/s 60 60

and

The ideal shaft power is Pideal = rwrvo(Vj − wr) ( − cos b) = 998 # 28.26 # 1.8 # 0.785 (100 − 28.26 # 1.8) # (1 − cos 165c) = 3.85 MW and actual shaft power is PActual = Pideal # hturbine = . FM 9.49

# .

= .

Option (A) is correct. The inlet (2) and outlet (1) velocity vector diagrams are shown below. The normal velocities are.

b l o 8 = = 5.31 m/s Vn2 = v = 2p # 1.2 # 0.2 A2 2p # r2 # b2 o o Vn1 = v = v = = 7.96 m/s 2p # . # . 2 2 r p A1 1 b1

From these we can compute the tangential velocities at each section. u2 = wr2 = # b 2p l # 1.2 = 10.1 m/s u1 = wr1 =

Then,

FM 9.50

#

2p

# . = 6.70 m/s

Vt2 = Vn2 cot 2 c = 5.31 cot 25c = 11.4 m/s Vt1 = Vn1 cot c = 7.96 # cot 30c = 2.11 m/s Ptheoretical = rvo(u2 Vt 2 − u1 Vt 1) = 998 # 8 # [(10.1 # 11.4) − (6.70 # 2.11)] = 800000 W , 800 kW

Option (C) is correct. We know Where and Thus,

loss =

p 1− p 2 + wshaft r

p 1 − p 2 = stagnation pressure drop across rotor = Dps since Vθ2 = wshaft = U2 Vq2 − U1 Vq1 =− U1 Vq1 2 2 wshaft =− (9) # (12 cos 30c) =− 93.5 m /s

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Also

FM 9

Δp s = p − p + r V − V = 0.07 + 1 # 1.23 # 6(12) 2 − (6) 2@c 1 3 m 2 10

Thus and FM 9.51

= (0.07 + 0.0664) kPa = 0.1364 kPa 3 loss = 0.1364 # 10 − 93.5 = 17.4 m2/s2 1.23 − wshaft η = = = 0.843 = 84.3% Dp s b l d r n

Option (D) is correct. From shaft power equation ...(i) wshaft = U Vq − U Vq Since the relative velocity at the exit is in the radial direction (see the figure below), Vθ = U = m s

Also, from the inlet conditions, =− Vr tan c, where the minus sign means that Vθ of U . (with ρ1 = ρ2 ) = Vr A = W A Vr = W # A = W # pr b = W # r b =b r pr b A

Vθ is in the opposite direction From conservation of mass We have Vr A Thus, Also,

U = r w and U = r w

or

r =U = = . r U Vr = 16 = 32 m/s 0. 5

Thus So that From eq. (i), FM 9.52

Vθ =− Vr # tan c =− ( ) tan c =− . m s wshaft = 16 # 16 − 8 # (− 18.5) = 404 m2/s2 = 404 N m/kg −

Option (A) is correct. Power Pshaft = Tshaft w = Tshaft # and Tshaft = rvo(r Vq − r Vq ) o Vq With Vθ = Tshaft = rvr

pN

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...(i)

GATE Mechanical Engineering in 4 Volume NODIA FM 9

From figure So that For

Since the flow rate is, or

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FM 359

cot β2 = U − Vq Vr ...(ii) Vθ = U − Vr ot b r = 0.5 = 0.25 m and ω = 2π # 900 = 94.2 rad/s 2 60 U =r w= . # . = . m vo = 2pr 2 b2 Vr 2 o 0.16 Vr = v = = 2.04 m/s 2pr 2 b2 2p # 0.25 # 0.05

Thus, from equation (ii), Vθ = (23.6 − 2.04 cot 25c) m/s = 19.22 m/s From equation (i), we get Tshaft = 999 # 0.16 # 0.25 # 19.22 = 768 N m Pshaft = 768 # 94.2 = 72346 - 72.3 kW −

So FM 9.53

Option (A) is correct Note 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. o (U Vq − U Vq ) ...(i) Pshaft = m o = rvo = where m g s # = and U = wr with ω = 2p # 130 = 13.6 rad/s 60 Hence, and

U = wr = U = wr =

. # . = . m s . # . = . m s

From the figure,

Vθ = U + W cos

c

...(ii)

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and since or Thus, from (ii),

FM 9

vo = 2pr 1 bW1 sin 80c 30 = 2p # 1.5 # (0.45) W1 sin 80c W = 7.18 m/s Vθ = 20.4 + 7.18 cos 80c = 21.6 m/s

vo = 2pr 2 bW2 sin 50c 30 = 2p # 0.85 # (0.45) W2 sin 50c or W = 16.3 m/s Thus, Vθ = U − W cos c = − cos Substitute these values in equation (i), we get Similarly, since

c = 1.08 m/s

Pshaft = (29970) 611.56 # 1.08 − 20.4 # 21.6@ =− 1.28 # 107 N m/s =− 12.8 MW −

***********

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