Fluid Mechanics GATE 1991-2013 Topic Wise Solution

8/14/2019 Fluid Mechanics GATE 1991-2013 Topic Wise Solution

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Mechanical Engineering

INDUSTRIAL ENGINEERING

TOPICWISE GATE SOLUTION

1991-2013

F-108, Katwaria Sarai, Near Mother Dairy Booth, New Delhi-16

Ph. +91-011-64551144, 9810758209

www.drona.org

DRONACHARYA INSTITUTE OF ENGINEERS


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Fluid Mechanics:Fluid properties; fluid statics, manometry, buoyancy; control-volume analysis of mass,

momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoullis

equation; viscous flow of incompressible fluids; boundary layer; elementary turbulent flow; flow through

pipes, head losses in pipes, bends etc. Pelton-wheel, Francis and Kaplan turbines - impulse and reaction

principles, velocity diagrams.

Syl labus of GATE Examination

S.NO. TOPIC PAGE NO.

Fluid Mechanics 69 - 156

1. ............... Properties of Fluids................................................................................ 71- 75

2. ............... Pressure and its Measurement ................................................................. 76 - 80

3. ............... Hydrostatic Forces on Surfaces .............................................................. 81 - 84

4. ............... Buoyancy and Flotation .......................................................................... 85 - 86

5. ............... Fluid Kinematics .................................................................................... 87 - 97

6. ............... Fluid Dynamics...................................................................................... 98 - 111

7. ............... Dimensional and Model Analysis ............................................................. 112 - 115

8. ............... Boundary Layer Theory ......................................................................... 116 - 1249. ............... Laminar and Turbulent Flow ................................................................... 125 - 131

10. ............. Flow Through Pipes............................................................................... 132 - 139

11. .............. Hydraulic Turbines ................................................................................. 140 - 147

12. ............. Centrifugal Pump ................................................................................... 148 - 153

13. ............. Compressible Flow................................................................................ 154 - 155

NOMENCLATURE OF CHAPTERS

Fluid MechanicFluid Mechanic


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Conclusion

S.No. Chapter Name 03 04 05 06 07 08 09 10 11 12 13

1 Properties of Fluids 1 1 2

2Pressure and its

Measurement2 2

3Hydrostatic Forces on

Surfaces2 2 2

4Buoyancy and

Flotation1 1

5 Fluid Kinematics 2 1 1 3 5 2 1

6 Fluid Dynamics 2 4 4 2 2 2 2 2 2 2

7Dimensional and

Model Analysis1 3

8Boundary Layer

Theory2 4 5 2

59Laminar and Turbulent

Flow2 1 2 1 1

10 Flow Through Pipes 4 4 1 2 1

11 Hydraulic Turbines 4 3 4 2 1 1

12 Centrifugal Pump 2 2 2

13. Compressible Flow

Total 13 18 7 17 16 11 8 8 3 5 6

Statistical Analysis

1. Fluid Mechanics has approximate 6 to 8% weightage in GATE.

2. From analysis it is clear that one should focus on Kinematics and Dynamics of Flow, Boundry Layer

Theory, Francis Turbine, Flow through Pipes, Laminar Flow, and Centrifugal Pump.


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Properties of Fluids

1

Year 2008

1. A journal bearing has shaft diameter of 40 mmand a length of 40 mm. The shaft is rotating at 20

rad/s and the viscosity of the lubricant is 20 mPa-

s. The clearance is 0.020 mm. The loss of torque

due to the viscosity of the lubricant is

approximately

(a) 0.040 Nm (b) 0.252 Nm

(c) 0.400 Nm (d) 0.652 Nm

Year 2006

2. For a Newtonian fluid (a) Shear stress is proportional to shear strain

(b) Rate of shear stress is proportional to shear

strain

(c) Shear stress is proportional to rate of shear

strain

(d) Rate of shear stress is proportional to rate of

shear strain

Year 2004

3. An incompressible fluid (kinematic viscosity, 7.4x 107 m2/s, specific gravity, 0.88) is held between

two parallel plates. If the top plate is moved with

a velocity of 0.5 m/s while the bottom one is held

stationary, the fluid attains a linear velocity profile

in the gap of 0.5 mm between these plates; the

shear stress in Pascals on the surface of top

plate is:

(a) 0.651 x 103 (b) 0.651

(c) 6.51 (d) 0.651x103

Year 2001

4. The SI unit of kinematic viscosity (v) is(a) m2/sec (b) kg/(m-sec)

(c) m/sec2 (d) m3/sec2

5. A static fluid can have

(a) non-zero normal and shear stress

(b) negative normal stress and zero shear stress

(c) positive normal stress and zero shear stress

(d) zero normal stress and non-zero shear stress

Year 1999

6. Kinematic viscosity of air at 20oC is given to be

1.6105m2/s. It kinematic viscosity at 70oC will

be vary approximately

(a) 2.2105m2/s (b) 1.6105m2/s

(c) 1.2105m2/s (d) 105m2/s

Year 1996

7. The dimension of surface tension is

(a) ML1 (b) L2 T1

(c) ML1T1 (d) None of these

Year 1995

8. A fluid is said to be Newtonian when the shear

stress is

(a) directly proportional to the velocity gradient

(b) inversely proportional to the velocity gradient

(c) independent of the velocity gradient

(d) none of the above


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73

Chapter-1Answers

1. Ans. (a) 2. Ans. (c) 3. Ans. (b) 4. Ans. (a) 5. Ans. (c)6. Ans. (a) 7. Ans. (d) 8. Ans. (a)

Space for Rough work


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74

Chapter-1Answer & Explanations

Q.1 Ans. (a)

Given: Shaft diameter, d = 40 mm

Shaft length, L = 40 mm

Speed, = 20 rad/s Viscosity, = 20 mPa-s Clearance, y = 0.020 mm

40 mm

40 mm

0.02 mm = 20 Pasm

Shear stress given by Newtons law of viscosity

=du

dy

Here, u = r = 20 0.02 = 0.4 m/s

= 33

0.420 10

0.02 10

= 400 N/m2

Shear force, F = A = 400 d L= 400 0.04 0.04 = 2.0106 N

Torque loss, T = F r = 2.0106 0.02 = 0.0402 NmReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-22, Example-

1.10.

Q.2 Ans. (c)

Exp. Consider a fluid element in a real flow. In a real flow there exist a velocity gradient in the perpendicular

direction of the flow. The change in velocity in two conscutive layer of fluid flow is shown in the figure.

dy

u

du.dtu + du

d

Shear strain, tan d =du dt

dy

If dis small, then tan d ~ d .

Therefore, d =du dt

dy

d

dt

=

du

dt


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75

From Newtons law of viscosity

=du

dy =

d

dt

Hence, for a Newtonian Fluid, the shear stress is directly proportional to rate of shear strain.

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 6, 1.3.3.

or

For a Newtonian fluid, Shear stress, du

dy

du

dy, where

du

dy= velocity gradient

dy

yu

u + du

x

dxdt

dy

dx

dy

dt

, where

dx

dyis shear strain of fluid

Thus

dx

dy

dt

is rate of shear strain

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, 1.8.

Q.3 Ans. (b)

Given, Kinematic viscosity, = 7.4107m2/secSpecific gravity, S = 0.88

Density of fluid, = 0.88 1000 kg/m3

Dynamic viscosity, = = 0.88 103 7.4 107= 0.6512103Pa.s

V = 0.5 m/s

0.5 mm

Now, from Newtons law of viscosity

=3

3

.du 0.6512 10 0.50.6512

dy 0.5 10

N/m2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.

Q.4 Ans. (a)

The SI unit of kinematic viscosity () is m2/s whereas CGS unit is cm2/s which is also known as Stoke.1 m2/s = 104stoke



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Q.5 Ans. (c)

Static fluid has normal stress only. Since fluid starts flowing under the action of shear stress irrespective of

its magnitude. In static fluid, there is no flow. Therefore, there is no shear stress.


Q.6 Ans. (a)

The viscosity of liquid decreases with increase in temperature due to decrease in intermoleculer force of

attraction while the viscosity of gas increases with increase in temperature due to increase in random motion

of the molecules.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.

Q.7 Ans. (d)

Surface tension () is defined as force per unit length. It is also equivalent to surface energy per unit surfacearea. It is mainly due to force of cohesion.

Dimension of =2MLT

L

= MT2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-14, Equation-

1.11.

Q.8 Ans. (a)

A fluid is said to be Newtonian fluid when it obeys the Newtons law of viscosity. For such fluids the viscosity

is independent from the rate of shear strain. For example water, air etc.The other types of fluid is shown in the following figure:

Ideal solid

Bingham

plasticflu

id

Pseudopl

asticfluid

Newtonia

nfluid

Ideal Fluid

Velocity gradient,du

dy

Shearstress,

Dilata

ntflu

id



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77

Pressure and its Measurement

2

Year 2005

1. A U-tube manometer with a small quantity of

mercury is used to measure the static pressure

difference between two locations A and B in a

conical section through which an incompressible

fluid flows. At a particular flow rate, the mercury

column appears as shown in the figure. The

density of mercury is 13 600 kg/m3and g = 9.81

m/s2. Which of the following is correct?

A

B

150 mm

(a) Flow direction is A to B and pA p

B= 20 kPa

(b) Flow direction is B to A and pA p

B=1.4 kPa

(c) Flow direction is A to B and pB p

A=20 kPa

(d) Flow direction is B to A and pB p

A=1.4 kPa

Year 2004

2. The pressure gauges G1and G

2installed on the

system show pressures of PG1

= 5.00 bar and

PG2

= 1.00 bar. The value of unknown pressure

P is

G1

G2

P

Atmospheric pressure

1.01 bar

(a) 1.01 bar (b) 2.01 bar

(c) 5.00 bar (d) 7.01 bar

Year 2000

3. In figure if the pressure of gas in bulb A is 50 cm

Hg vaccum and patm= 76 cm Hg, the height ofcolumn H is equal to

A

HPatm

Hg

(a) 26 cm (b) 50 cm

(c) 76 cm (d) 126 cm

Year 1999

4. If p is the gauge pressure within a spherical

droplet, the gauge pressure within a bubble of the

same fluid and of same size will be

(a)p

4(b)

p

2

(c) p (d) 2p


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Year 1997

5. Refer to figure, the absolute pressure of gas A in

the bulb is

PA

10 cm

AB

C

D

F

E= 13.6 g/ml

2 cm

5 cm

(a) 771.2 mm Hg (b) 752.65 mm Hg

(c) 767.35 mm Hg (d) 748.8 mm Hg

Year 1996

6. A mercury manometer is used to measure the

static pressure at a point in a water pipe as shown

in Fig. The level difference of mercury in the two

limbs is 10 mm. The gauge pressure at the

point A is

10 mmH O2

Hg

WaterA

(a) 1236 Pa (b) 1333 Pa

(c) zero (d) 98 Pa

Year 1994

7. Net force on a control volume due to uniform

normal pressure alone

(a) depends upon the shape of the control volume(b) translation and rotation

(c) translation and deformation

(d) deformation only


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Chapter-2Answers

1. Ans. (a) 2. Ans. (d) 3. Ans. (b) 4. Ans. (d) 5. Ans. (a)6. Ans. (b) 7. Ans. (a)



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80


Q.1 Ans. (a)

AB

150 mm

Writing the pressure balance equation,

pA

= pB+ gh

pA p

B=

150136000 9.81 20.012kPa

1000


Q.2 Ans. (d)

G1

G2

P

Atmospheric pressure1.01 bar

Absolute pressure at 2

Pabs2

= PG2

+P

atm

= 1 + 1.01 = 2.01 bar

Absolute pressure at 1 Pabs1

= PG1

+ Patm

(Atmospheric pressure for G1becomes 2.01 bar)

= 5 + 2.01 = 7.01 bar

Q.3 Ans. (b)

A

HPatm

Hg

Applying pressure balancing equation at free surface

PA+ P

H= P

atm

Patm

= PA P

H

Taking Patm

= 0

Therefore, PH

= 50 cm


Q.4 Ans. (d)

Pressure inside spherical droplet =4

d

Pressure inside soap bubble =8

d

, where is surface tension force and d is diameter..

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-15,

Equation-1.11.


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Q.5 Ans. (a)

PA

17 cm AB

C

D

F

E = 13.6 g/mldatum

5 cm

2 cm

Gauge pressure at A,P

A+

1gh

1=

2gh

2+

1gh

3(Taking the unknown liquid as water)

PA+

171000 9.81

100 =

2 513600 9.81 1000 9.81

100 100

PA

= 2668.32 + 490.5 1667.2 = 1491.12 N/m2

Pab s

= Patm

+ PA

= 1.013105+ 1491.12 = 102791.12 N/m2

Pab s = mghm

m

m

density of mercuryh mercury column height

hm

=102791.12

0.77055m13600 9.81

= 771 mm

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 40, 2.6.2.Q.6 Ans. (b)

10 mm

H O2

Hg

WaterA

Neglecting the depth of water column, gauge pressure is given as

Pguage

= gh =10

13600 9.811000

= 1334.16 N/m2

Q.7 Ans. (a)

Exp. Pressure =Force

Area

Net force = PressureArea.

Therefore, area defined by the shape of the control volume.



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Hydrostatic Forces on Surfaces

3Year 2013

1. A hinged gate of length 5 m,inclined at 30 with

the horizontal and with mass on its left, is shown

in the figure below. Density of water is 1000 kg/

m3. The minimum mass of the gate in kg per unit

width (perpendicular to the plane of paper),

required to keep it closed is

5m

(a) 5000 (b) 6600(c) 7546 (d) 9623

Year 2003

2. A water container is kept on a weighing balance.

Water from a tap is falling vertically into the

container with a volume flow rate of Q; the velocity

of the water when it hits the water surface is U.

At a particularly instant of time the total mass of

the container and water is m. The force registered

by the weighing balance at this instant of time is

(a) mg + QU (b) mg + 2QU(c) mg + QU2/2 (d) QU2/2

Year 2001

3. The horizontal and vertical hydrostatic forces Fx

and Fyon the semi-circular gate, having a width

w into the plane of figure, are

(a) Fx= ghrw and F

y= 0

(b) Fx= 2ghrw and Fy= 0(c) Fx= 2ghrw and F

y= gwr2/2

(d) Fx= 2ghrw and F

y= gwr2/2

Year 1992

4. A 3.6 m square gate provided in an oil tank is

hinged at its top edge (Figure). The tank contains

gasoline (sp. gr. = 0.7) upto a height of 1.8 m

above the top edge of the plate. The space above

the oil is subjected to a negative pressure of 8250

N/m2. Determine the necessary vertical pull to

be applied at the lower edge to open the gate.

45

Negative pressure (8250 N/m )2

Hinge

Gate

Gasoline surface

1.8m

P

Gasoline (S = 0.7)


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Chapter-3Answers

1. Ans. (d) 2. Ans. (a) 3. Ans. (d) 4. Ans. (144.5 kN)



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84


Q.1 Ans. (d)

mg

mg cos 30

30

30

5m

b=1m

G

F

2.5mx

B

h

Depth of centre of gravity from free surface of water,

x = 2.5 sin 30o= 1.25 m

Hydrostatic force, F = gAx = 1000 9.81 5 1 1.25 = 61312.5 NDepth of centre of pressure,

h =2GIx sin

Ax

= 1.25 +

3

2o

11 5

12 sin305 1 1.25

= 1.67 m

For gate to be closed, moment of all forces about the hinge point must be zero. Therefore, taking moment of

all forces about hinge point.

mg cos 30o 2.5 = F h /sin 30o

Therefore, m =0 0

F h

g cos30 2.5 sin30

= 061312.5 1.67

9.81 cos30 1.25

= 9641 kg

Therefore, the nearest possible value is 9623 kg.

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-132, Example-

3.22.

Q.2 Ans. a)

Mass of water strike = AV = QForce on weighing balance due to strike of water = Initial momentum final momentum

= QU Q.0 = QUSince weight of water and container = mg

Total force on weighing balance = mg + QUQ.3 Ans. (d)

Horizontal component of hydrostatic force, Fx= gAx where = density of the liquid, A = surface area, x

= depth of centre of pressure from free surface of liquid

Hence, Fx

= g Ax where projected area, A = w 2rTherefore, F

x= 2 gwrh Projected area (ABCD),


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r + r

2r

r r

w

A

BC

D

Vertical component of hydrostatic force,

Fy

= Weight of water supported by the curved surfaceF

y= g Volume of curved portion

= 2g r w2

where,

2r

2

= Area of semi-circle

=2

g w r

2

where, w is the width of the gate.


Q.4 Ans. ( 144.5 kN)

45

Negative pressure (8250 N/m )2

Hinge

Gate

Gasoline surface

1.8m

P

Gasoline (S = 0.7)

Head of oil equivalent to negative pressure 238 N/m2, h =p 8250

1.2 mw 0.7 9810

This negative pressure will reduce the oil head above the top edge of the gate from 1.8 - 1.2 = 0.6 m of oil.

Calculations for the magnitude and location of the pressure force are thus to be made corresponding to 0.6 mof oil.

Now, x = 0.6 +3.6

sin452

= 1.873 m

Area, A = 3.6 3.6 = 12.96 m2

Pressure, P = wAx = 0.7 9810 12.96 1.873 = 166690 N

Centre of pressure, h =

2

GI sin

xAx

=

3 21 3.6 (3.6) (sin 45 )12 1.873 2.16m

12.96 1.873

Vertical distance of centre of pressure below top edge of the gate = 2.16 0.6 = 1.56 mTaking moments about the hinge.

F sin 45 3.6 = P 1.56

sin45

Hence, vertical force, F =2

P 1.56

3.6 (sin45)

= 2166690 1.56

3.6 (sin45)

= 144465 N = 144.5 kN

Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-101.


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86

Buoyancy and Flotation

4

Year 2010

1. For the stability of a floating body, under theinfluence of gravity alone, which of the following

is TRUE?(a) Metacentre should be below centre of gravity.(b) Metacentre should be above centre of gravity.

(c) Metacentre and centre of gravity must lie on

the same horizontal line.

(d) Metacentre and centre of gravity must lie on

the same vertical line.

Year 2003

2. A cylindrical body of cross-sectional area A,

height H and density s, is immersed to a depth h

in a liquid of density , and tied to the bottom witha string. The tension in the string is

h

(a) ghA (b) (s ) ghA(c) (

s) ghA (d) (h

sH) gA

Year 1994

3. Bodies in flotation to be in stable equilibrium, the

necessary and sufficient condition is that the

centre of gravity is located below the...........


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Q.1 Ans. (b)

Condition of stability in case of Floating bodies is given as:-

1. For stable equilibrium, MG > 0

2. For unstable equilibrium, MG < 03. For neutral equilibrium, MG = 0

M

Meta centre

Centre ofbuoyancy

Centre ofgravityB

G

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-151, 4.3.Q.2 Ans. (d)

h

Free body diagram of the cylindrical body will be

WG

T

FBB

At equilibrium condition

T + weight of body = Buoyancy force

T + Mg = h AgT + (sHA)g = h Ag

T = (h sH) gAQ.3 Ans. metacentre

For floating body the equlibrium conditionds are as follows:-

1. For stable equilibrium, metacentre should be above the centre of gravity.

2. For unstable equilibrium, metacentre should be below the centre of gravity.3. For neutral equilibrium, metacentre should coincide the centre of gravity.

Reference: Fluid Mechanics, R. K. Rajput, Edition 2005, Page-129, 4.3.


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Fluid Kinematics

5

Year 2011

1. A streamline and an equipotential line in a flowfield

(a) are parallel to each other

(b) are perpendicular to each other

(c) intersect at an acute angle

(d) are identical

Year 2009

2. You are asked to evaluate assorted fluid flows

for their suitability in a given labortory application.

The following three flow choices expressed interms of the two dimensional velocity field in the

xy plane are made available

P. u = 2y, v = 3x

Q. u = 3xy, v = 0

R. u = 2x. v = 2y

Which flow (s) should be recommended when

the application requires the flow to be

incompressible and irrota tional?

(a) P and R (b) Q

(c) Q and R (d) R

Year 2008

3. For the continuity equation given by V

to be

valid, where V

is the velocity vector, which one

of the following is a necessary condition?

(a) steady flow

(b) irrotational flow

(c) inviscid flow

(d) incompressible flow

Statement for linked answer questions

4 and 5

The gap between a moving circular plate and astationary surface is being continously reduced,

as the circular plate comes down at a uniform

speed V towards the stationary bottom surface,

as shown in the figure. In the process, the fluid

contained between the two plate flows out

radially. The fluid is assumed to be incompressible

and inviscid.

R

r

Stationarysurface

Movingcircular plate

Vh

4. The radial velocity vrat any radius r, when the

gap width is h, is

(a) vr =Vr

2h(b) vr =

Vr

h

(c) vr =2 V h

r(d) vr =

Vh

r

5. The radial component of the fluid acceleration at

r = R is

(a)2

R

2

3V

4h(b)

2R

2

V

4h

(c)2

R

2

V

2h

(d)

2R

2

V

4h


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89

Year 2006

6. In a two-dimensional velocity field with velocities

u and v along the x and y directions respectively,

the convective acceleration along the x-direction

is given by

(a)u u

u vx y

(b)

u vu v

x y

(c)v u

u vx y

(d)

u uv u

x y

7. A two-dimensional flow field has velocities along

thexandydirections given by u=x2tand v=

2xytrespectively, where tis time. The equation

of streamlines is

(a) x2y = constant

(b) x y2= constant

(c) x y = constant

(d) not possible to determine

Year 2005

8. The velocity components in the x and and y

directions of a two dimensional potential flow are

u and v, respectively. Then ux

is equal to

(a)v

x

(b)v

x

(c)v

y

(d)v

y

Year 2004

9. A fluid flow is represented by the velocity field

V x i y j a a , where a is a constant. The

equation of stream line passing through a point

(1, 2) is:

(a) x - 2y = 0 (b) 2x + y = 0

(c) 2x - y = 0 (d) x + 2y = 0

Year 2003

10. The vector field F xi yj

(where i

andj

are

unit vectors), is:

(a) divergence free, but not irrotational

(b) irrotational, but not divergence free

(c) divergence free and irrotational(d) neither divergence free nor irrotational

Year 2001

11. The 2-D flow with velocity

V x 2y 2 i 4 y j

, is

(a) compressible and irrotational

(b) compressible and not irrotational

(c) incompressible and irrotational

(d) incompressible and not irrotational

Year 1999

12. For the function f = ax2y y3 to represent the

velocity potential of an ideal fluid. D2f should be

equal to zero. In that case, the value of a has to

be:

(a) 1 (b) 1

(c) 3 (d) 3

13. If the velocity vector in 2-D flow field is given by

2 2V = 2xyi + (2y - x )j

, the vorticity vector, curl

V

will be

(a) 22y j

(b) 6yk

(c) zero (d) -4xk

Year 1995

14. The velocity components in the x and y directions

are given by

u xy x y v xy y 3 2 2 43

4,

The value of for a possible flow field involvingan incompressible fluid is

(a) 3

4(b)

4

3

(c)4

3(d) 3


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90

15. The force F needed to support the liquid of density

d and the vessel on top (Fig) is

(a) gd[ha (h H) A] (b) gdHA

(c) gdHa (d) gd (H h) A

Year 1994

16. Stream lines, path lines and streak lines are virtuallyidentical for

(a) Uniform flow

(b) Flow of ideal fluids

(c) Steady flow

(d) Non uniform flow

17. In a flow field, the streamlines and equipotential

lines

(a) are parallel

(b) are orthogonal everywhere in the flow field

(b) cut at any angle

(d) cut orthogonally except at the stagnation points

18. For a fluid element in a two dimensional flow field

(x-y plane), if it will undergo

(a) translation only

(b) translation and rotation

(c) translation and deformation

(d) deformation only

Year 1992

19. Existence of velocity potential implies that(a) Fluid is in continuum

(b) Fluid is irrotational

(c) Fluid is ideal

(d) Fluid is compressible

20. Circulation is defined as line integral of tangential

component of velocity about a..........


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91

Chapter-5Answers

1. Ans. (b) 2. Ans. (d) 3. Ans. (d) 4. Ans. (a) 5. Ans. (c)6. Ans. (a) 7. Ans. (a) 8. Ans. (d) 9. Ans. (c) 10.Ans. (c)11.

Ans. (d) 12.

Ans. (d) 13.

Ans. (d) 14.

Ans. (d) 15.

Ans. (a)

16.Ans. (c) 17.Ans. (b) 18.Ans. (c) 19.Ans. (b) 20.Ans.(closedcontour in afluid flow)



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92


Q.1 Ans. (b)

If and are the stream function and potential function respectively representing the possible flow field.

Slope of stream line represented by is given by

slope (m1) =dy

dx=

d

dx

d

dy

=v

-u......(i)

Slope of potential line represented by is given by

slope (m2) =dy

dx=

d

dx

d

dy

= -u u=-v v

......(ii)

Now, product of the slopes,

m1 m2 =v u

u v

= 1

Since the product of the slope of these two lines at the point of intersection is 1, which indicates that thesetwo lines are prependicular to each other.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-244, Sec. 6.12.

Q.2 Ans. (d)For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations

u v

x y

= 0 ......(i)

z =

1 v u

2 x y

= 0 ......(ii)

For P,Given u = 2y and v = 3x

u

x

= (2y)x

= 0 andu

y

=

(2y)y

= 2

v

x

= 3x 3x

and

v

y

= 3x 0y

From equation (i)u v

x y

= 0 + 0 = 0

From equation (ii) z =1 v u

-2 x y

= 1

3 - 22

0

Since the given velocity field is satifying the equation (i) only, therefore it is a possible case of steady,incompressible and rotational flow.


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For Q, Given u = 3xy and v = 0

u

x

= (3xy)x

= 3y andu

y

=

(3xy)y

= 3x

v

x

= 0 0x

and

v

y

= 0 0y


x y

= 3y 0


-2 x y

=-3x

2 0

Given velocity field is neither satisfying the equation (i) nor (ii), therefore the flow is neither steady norirrotational.For R, Given u = 2x and v = 2y

u

x

= ( 2x)x

= -2 and

u

y

=

( 2x)y

= 0

v

x

=

2y 0

x

and

v

y

= 2y 2

y


x y

= 0


-2 x y

= 0

Given velocity field is satisfying the equation (i) and (ii), therefore, the flow is a possible case of steady,incompressible and irrotational flow.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn-6.33 a.

Q.3 Ans. (d)

Given that V

= 0

i.e. i j k ui vj wk 0x y z

i.e.u v w

x y z

= 0

which represents the three dimensional continuity equation of steady, and incompressible flow.Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page -146, Eqn. 5.4.

Q.4 Ans. (a)At radius r, volume of fluid moving out radially is equal to the volume of fluid displaced by moving plate withinradius r.Given that V = downward velocity of circular plate in m/s

vr

= radial velocity at radius rR

V

Vrr

h

So volume displaced by moving plate= Velocity Area= V r2

Now, volume flow out at radius,


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94

r = vr 2rhFrom above stated condition

vr 2rh = r2 V

Therefore, vr =V

2

rh

Q.5 Ans. (c)Radial component of the fluid acceleration at r = R

aR =R

Vd

dt=

VR

2d

h

dt

=

VR

2d

dhh

dh dt

(ve as h is reducing with time)

= 2VR 1

( V)2 h

as V

dh

dt

Therefore, aR =

2

2

V R

2hQ.6 Ans. (a)Acceleration of fluid particle alongx-axis is given by

ax =u u u u

u + v + w +x y z t

......(i)

For 2-D flowu

z

= 0

Thus, from equation (i), ax =

Temporalor localConvective

accelerationacceleration

u u uu + v +

x y t

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-232, Equation6.27.

Q.7 Ans. (a)Given: u = x2t and v = 2xytStream line equation is given as

dx

u=

dy

v

2dx

x t=

dy

-2xyt

dx

x=

1 dy-

2 y

Integrating both sidedx

x =1

2

dy

y

ln x =1

ln y2

c

ln x2+ ln y = c x2y = ConstantReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219,Equation 6.2.


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Q.8 Ans. (d)Exp. For two dimensional potential flow, the continuity equation is given as

u v

x y

= 0

Therefore,u

x

= v

y

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-223, Eqn. 6.5.Q.9 Ans. (c)The velocity field is given as,

V

= ax i ay j

= ui vj

The equation of stream line

dx

u=

dy

v......(i)

from equation (i),dx

ax=

dy

ay

Integrating both side,dx

x =dy

yln x = ln y + c

xn

y = ln c

x

y= c ......(ii)

Since this stream line passes through point (1, 2) hence c = 1/2Therefore, equation of stream line is (from equation (ii))

2x y = 0Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn. 6.2.

Q.10 Ans. (c)Given vector filed F

= xi yj

The divergence of V

is defined as V

It can also be written as x y zx y z

i j k i j k

=u v w

x y z

=0

x y z

x y

= 1-1 = 0

Rotational component, z

=1 v u

2 x y

= 0

Hence, the vector field is divergence free and irrotational.Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page-156, Sec. 5.8.

Q.11 Ans. (d)

The 2-D flow with velocity, V

= x 2y 2 i 4 y j

= ui vj

For incompressible and irrotational flow, the velocity field should satisfy the following equations

u v

x y

= 0 ......(i)

z =

1 v u

2 x y

= 0 ......(ii)


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Here,u

x

= (x 2y 2)x

= 1 and

u

y

=

(x 2y 2)y

= 2

v

x

= 4 yx

= 0 and

v

y

= 4 y 1y

From equation (i),u v

x y

= 1 1 = 0

For irrotational flow from equation (ii),

z

=1 v u

2 x y

= 1

0 22

0

Hence, this flow is steady, incompressible and rotational.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn 6.2.

Q.12 Ans. (d)Velocity Potential, f = ax2y y3

Nowf

x

= 2axy &

2

2

f2ay

x

......(i)

andf

y

= 2 2ax 3y &

2

2

f6yy

......(ii)

As D2f should equal to 0

or D2(f) = 2 (f ) 0 2 2

2 2

f f

x y

= 0

From euation (i) and (ii),2ay 6y = 02y (a 3) = 0

Therefore, a = 3Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-243, Eqn. 6.45.

Q.13 Ans. (d)The curl of V

is defined as V

.

curl of V

=

i j k

x y z

u v w

Flow field, V

= 2 22xyi + (2y -x )j + 0k

It can also be written asw v u w v u

i j ky z z x x y

= v u

0 0 0 0x y

i j k

= -2x - 2x k

= - 4xk

Reference: Fluid Mechanics, R. K. Rajput, Edition 2005, Page-172, Equation 5.32.Q.14 Ans. (d)

The velocity components in the x and y directions are given by


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97

u xy x y v xy y 3 2 2 43

4,

Continuity equation for steady, incompressible and irrotataional flow is

u v

x y

= 0 ......(i)

u

x

=3

y 2xy &

v

y

=3

2xy 3yPut these value in equation (i),

3 3y 2xy 2xy 3y = 0 y3 3y3 = 0 y3 ( 3) = 0 3 = 0 = 3Reference: Fluid Mechanics, S.K. Aggarwal, Page No. 104.

Q.15 Ans. (a)Let Free body diagram of liquid columns due to symmetry

Here A1

= a and A2= A

3=

A a

2

A1

A2 A2

a

a

A

h

H-h

H

(Aa)2

Aa2

Now F is equal to the weight of water supported by the piston.W = Mg

or M.g = d.g.V where d is the density of the liquidF = d.g.V ......(i)Now V = A

1H + 2 (A

1(Hh))

= aH + 2A a

(H h)2

= aH + A (Hh) aH + ahV = ah + A(H h)

= ah A (h H) ......(ii)Put Value of V in equation (i)

F = dg [ah A(hH)]Q.16 Ans. (c)

In steady and uniform flow stream line, path line and streak line are same.

In the given problem steady flow and uniform flow are separate option. Hence option (a) & (c) both arecorrect but most appropriate single answer is (c).Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page No. 160, 5.4.4.

Q.17 Ans. (b)In a flow field, the streamlines and equipotential lines are always orthogonal to each other.

= stream lines

= equipotential lines


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Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-245, Fig. 6.18.Q.18 Ans. (c)

For 2-D flow, irrotational component,

z =

1 v u

2 x y

= 0

Therefore, there is no variation in velocity in z-direction.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn 6.34c.

Q.19 Ans. (b)For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations

u v

x y

= 0 ......(i)

z =

1 v u

2 x y

= 0 ......(ii)

If is the potential function representing the possible flow field. Then from definition of potential function

u =

-x

and v =

-y

From equation (i),u v

x y

=2 2

2 2x y

which is known as Laplace equation.

From equation (ii),1 v u

2 x y

=

2 2

x y y x

= 0

The velocity potential of the flow denoted by if satisfies the continuity/Laplace equation, then it will be apossible case of irrotational flow.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn 6.34c.

Q.20 Ans. (closed contour in a fluid flow)Circulation is defined as the line integral of the tangential component of the velocity taken around a closedcontour. Mathematically, the circulation is obtained if the product of the velocity component along the curveat any point and the length of the small element containing that point is integrated around the curve.

ds

C

V

X

Y

Mathematically, circulation = V cos .ds = (udx vdy)

Area of closedcurve

= Vorcitity along the axis perpendicular to the plane containing the closed

curve.

= Vorticity area = 2 z area

=1 v u

2 x y2 x y

For irrotational flow in xy plane,

z= 0 hence vorticity which leads to circulation also equal to zero.

Reference: Fluid Mechanics & Hydraulic Machines, K. Subramanya, Edition 2012, 108,3.5.1.


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Fluid Dynamics

6

Year 2013

1. Water is coming out from a tap and falls verticallydownwards. At the tap opening, the stream

diameter is 20mmwith uniform velocity of 2m/s.

Acceralation due to gravity is 9.81 m/s 2.

Assuming steady, inviscid flow, constant atmo

spheric pressure everywhere and neglecting

curvature and surface tension effects, the

diameter in mmof the stream 0.5 m below the

tap is approximately

(a) 10 (b) 15

(c) 20 (d) 25

Year 2012

2. A large tank with a nozzle attached contains three

immiscible, inviscid fluids as shown. Assuming

that the changes in h1, h2and h3are negligible,

the instantaneous discharge velocity is

h1

h2

h3

1

2

3

(a)1 1 2 2

3

3 3 3 3

2 1h h

gh

h h

(b)1 2 32 ( )g h h h

(c)1 1 2 2 3 3

1 2 3

2h h h

g

(d)1 2 3 2 3 1 3 1 2

1 1 2 2 3 3

2h h h h h h

gh h h

Year 2011

3. Figure shows the schematic for the measurement

of velocity of air (density = 1.2 kg/m3) through a

constant-area duct using a pitot tube and a water-

tube manometer. The differential head of water

(density = 1000 kg/m3) in the two columns of the

manometer is 10 mm. Take acceleration due to

gravity as 9.8 m/s2. The velocity of air in m/s is

Flow

10 mm

(a) 6.4 (b) 9.0

(c) 12.8 (d) 25.6


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Year 2010

4. A smooth pipe of diameter 200 mm carries water.

The pressure in the pipe at section S1(elevation

: 10 m) is 50 kPa. At section S2(elevation : 12 m)

the pressure is 20 kPa and velocity is 2 ms1.

Density of water is 1000 kgm3and accelerationdue to gravity is 9.8 ms2. Which of the following

is TRUE

(a) flow is from S1to S2 and head loss is

0.53 m

(b) flow is from S2to S1and head loss is 0.53 m

(c) flow is from S1 to S2 and head loss is

1.06 m

(d) flow is from S2to S1and head loss is 1.06 m

Year 2009

5. Consider steady, incompressible and irrotational

flow through a reducer in a horizontal pipe where

the diameter is reduced from 20 cm to 10 cm.

The pressure in the 20 cm pipe just upstream of

the reducer is 150 kPa. The fluid has a vapour

pressure of 50 kPa and a specific weight of 5

kN/m3. Neglecting frictional effects, the maxi-

mum discharge (in m3/s) that can pass through

the reducer without causing cavitation is(a) 0.05 (b) 0.16

(c) 0.27 (d) 0.38

Year 2007

6. Which combination of the following statements

about steady incompressible forced vortex flow is

correct ?

P : Shear stress is zero at all points in the flow.

Q : Vorticity is zero at all points in the flow.R : Velocity is directly proportional to the radius

from the centre of the vortex.

S : Total mechanical energy per unit mass is con-

stant in the entire flow field.

Select the correct answer using the codes given

bewlow:

(a) P and Q (b) R and S

(c) P and R (d) P and S

Year 2006

7. A siphon draws water from a reservoir and

discharges it out at atmospheric pressure.

Assuming ideal fluid and the reservoir is large,

the velocity at point P in the siphon tube is

h2

h1

P

(a) 12gh (b) 22gh

(c) 2 12 ( )g h h (d) 2 12 ( )g h h

Year 2005

8. A venturimeter of 20 mm throat diameter is used

to measure the velocity of water in a horizontal

pipe of 40 mm diameter. If the pressure differ-

ence between the pipe and throat sections is found

to be 30 kPa then, neglecting frictional losses, theflow velocity is

(a) 0. 2 m/s (b) 1. 0 m/s

(c) 1. 4 m/s (d) 2. 0 m/s

9. A leaf caught in a whirlpool. At a given instant

the leaf is at a distance of 120 m from the centre

of the whirlpool. The whirlpool can be described

by the following velocity distribution;

Vr =

360 10/2 m sr

& VV=

3300 10

/ ,2 m sr

angular Velocity V where r (in metres) is the

distance from the centre of the whirlpool. What

will be the distance of the leaf from the centre

when it has moved through half a revolution

(a) 48 m (b) 64 m

(c) 120 m (d) 142 m


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101

Year 2004

10. A closed cylinder having a radius R and height H

is filled with oil of density . If the cylinder isrotated about its axis at an angular velocity of ,the thrust at the bottom of the cylinder is:

(a) R gH2

(b)

RR2

2 2

4

(c) R R gH2 2 2 j

(d)

RR

gH2

2 2

4HG KJ

11. A centrifugal pump is required to pump water to

an open water tank situated 4 km away from the

location of the pump through a pipe of diameter

0.2 m having Darcys friction factor of 0.01. Theaverage speed of water in the pipe is 2 m/s. If it

is to maintain a constant head of 5 m in the tank,

neglecting other minor losses, then absolutedischarge pressure at the pump exit is

(a) 0.449 bar (b) 5.503 bar (c) 44.911 bar (d) 55.203 bar

Year 2003

12. Air flows through a venturi and into atmosphere.Air density is; atmospheric pressure is P

a; throat

diameter is Dt; exit diameter is D and exit velocity

is U. The throat is connected to a cylinder

containing a frictionless piston attached to aspring. The spring constant is k. The bottom

surface of the piston is exposed to atmosphere.

Due to the flow, the piston moves by distance x.

Assuming incompressible frictionless flow,x is

xDs

k

Dt

D U

Pa

(a) U k Ds2 2

2/ j

(b) U kD

DD

t

s2

2

2

28 1/e j HG KJ

(c) U kD

DD

t

s2

2

2

22 1/e j HG KJ

(d) U kD

DD

t

s2

4

4

28 1/e j HG KJ

Year 1999

13. Water flows through a vertical contraction from

a pipe of diameter d to another of diameter d/2

(see Fig.). The flow velocity at the inlet to the

contraction is 2 m/s and pressure 200 kN/m2. If

the height of the contraction measures 2 m, the

pressure at the exit of the contraction will be

very nearly

2m

d

d/2

(a) 168 kN/m2 (b) 192 kN/m2

(c) 150 kN/m2 (d) 174 kN/m2

Year 1996

14. A venturimeter (throat diameter = 10.5 cm) isfitted to a water pipe line (internal diameter

= 21.0 cm) in order to monitor flow rate. To

improve accuracy of measurement, pressure

difference across the venturimeter is measured

with the help of an inclined tube manometer, the

angle of inclination being 30 (Figure). For a

manometer reading of 9.5 cm of mercury, find

the flow rate. Discharge coefficient of venturi is

0.984.

30

9.5cmy

From

venturi

Water

Hg


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Chapter-6Answers

1. Ans. (b) 2. Ans. (a) 3. Ans. (c) 4. Ans. (c) 5. Ans. (b)6. Ans. (c) 7. Ans. (c) 8. Ans. (d) 9. Ans. (b) 10.Ans. (d)11.

Ans. (b) 12.

Ans. (d) 13.

Ans. (c) 14.

Ans.(0.0302

m3/s)



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Q.1 Ans (b)

0.5m

1

2

1

2

Applying Bernoullis equation at section (1-1) & (2-2)

2

1 11

P VZ

g 2g

=2

2 22

P VZ

g 2g

......(i)

P1

= P2= P

atm.(taking section 2-2 as datum)

From equation (i)

220.5

2 9.81

=

2

2V

2g

V2

= 3.716 m/sec.

From Continuity equation, 1A1V1 = 2A1V2 (since flow is incompressible, i.e. 1= 2)A

1V

1= A

2V

2

A2

=1 1

2

A V

V

2

2d

4

=

2

1 1

2

d V

4 V

Therefore, d2

=

2

1 1

2

d V

V

=

11

2

Vd

V

=2

0.023.716

= 0.01467 m 15 mm

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-306, Ex.7.3.

Q.2 Ans (a)

Applying Bernoullis equation, just before the exit from the tank and just after entry in the atmosphere

21 1

1

3

P VZ

g 2g

=2

2 22

3

P VZ

g 2g

......(i)


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h1

h2

h3

1

2

3

From the above figure, it is clear thatZ

1= Z

2, V

1= 0 and P

2= Atmospheric pressure = 0;

Then the Bernoullis equation reduces to :

1

3

P

=

2

2V

2

V2

=1

3

2P ......(ii)

From given figure we can find pressure P1

P1

= 1gh

1+

2gh

2+

3gh

3

Substitute this value of P1in eqution (ii), we get

V2

= 1 1 2 2 3 33

2gh h h

=1 1 2 2

3

3 3 3 3

h h2gh 1

h h

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-277, Eqn 7.24.Q.3 Ans. (c)

10 mm

Flow a

w

Given that

Density of air, a = 1.2 kg/m3,

Density of water, w = 1000 kg/m3

x = 10 mm, g = 9.8 m/s2

Now h =

2v

2g

v = 2gh


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where, h = 1

w

a

x

=3 100010 10 1 8.32 m

1.2

Velocity of air, v = 2 9.81 8.32 = 12.8 m/s

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-294, 295.

Q.4 Ans. (c)

Given:

At section S1: P1 = 50 kPa At section S2: P2 = 20 kPa

Z1 = 10 m Z2 = 12 m

V1 = 2 m/s V2 = 2 m/s

10 m12 m

S1

S2

P = 50 KPa1

P = 20 KPa2

Datum line

Since diameter of the pipe is constant hence velocity of the flow will be same through out the length of

the pipe. Therefore V1= V2= 2 m/s. Since velocity of flow is constant throughout the pipe, hence direction

of flow is decided by the piezometric head only.

Total piezometric head at S1,

H1

=1

1

PZ

g

=

350 1010

1000 9.81

= 15.096 m

Total piezometric head at S2, H2 =2

2

PZ

g

=

320 1012

1000 9.81

= 14.038 m

Since H1> H2therefore flow direction is from S1to S2.

Therefore, head loss = H1 H

2= 15.096 14.038 = 1.06 m


Q.5 Ans. (b) Given,

Inlet diameter, d1 = 0.2 m

Inlet pressure, P1 = 150 kPa

Exit diameter, d2 = 0.1 m

Specific weight, w (g) = 5 kN/m3

Vapour pressure, P2 = Pv= 50 kPa(To avoid cavitation, pressure at exit should not be allowed tofall below the vapour pressure of the liquid)


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106

12

2

1

From continuity equationa1V1 = a2V2

2

1 1d V

4

=

2

2 2d V

4

V2 =

2

112

2

dV

d=

2

12

0.2V

0.1= 4V1

Applying Bernoullis equation between at section (1-1) and (2-2)

2

1 11

P VZ

2

g g =2

2 22

P VZ

2

g g

2

1V150

5 2g =

2

2V50

5 2g

=

2

116 V102g

Therefore, V1 = 5.114 m/s

Therefore, maximum discharge, Q = 21 1r V

= 0.12 5.114 = 0.161 m3/sReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-347, Ex. 8.7.

Q.6 Ans. (c)

In forced vortex flow when steady state is reached the liquid attains equilibrium condition in this position andit rotates as a solid mass with the container at the same angular velocity. The liquid is then at rest with respect

to its container and therefore no shear stress will exist in the liquid mass.In the forced vortex flow the stream lines are concentric circles and the velocity v of any liquid particle ata distance r from the axis of rotation may be expressed as v = r.Therefore, v r. In forced votex flow, velocity is directly propotional to distance from the axis of rotation.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, Page-185, 5.5 (a), Page-301, 7.62.

Q.7 Ans. (c)

2

1

P

z1

h1

h2

vP

Applying Bernoullis equation between section (1) and (2)

2

1 11

P VZ

2

g g =2

2 22

P VZ

2

g g

Now P1 = P2= Patmand Z2= 0, (taking point 2 as datum)

Z1 = (h2 h1), V1= 0


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Thus from Bernoullis equation

atm

2 1

P0 ( )

h h

g=

2

atm 2P V

02

g g

h2 h1 =2

2V

2g

V2 = 2 12 ( )g h hAs area of siphon is constant, therefore velocity of flow is same

Hence, VP = V2= 2 12 ( )g h h


Q.8 Ans. (d)

Given that

D1 = 40 mm

D2 = 20 mm

40 mm 20 mmD1 D2

1

12

2

From continuity equation, A1V1 = A2V2

V2 =1

1

2

AV

A

=

2

11

2

DV

D

=

2

1

40V

20

= 4V1

Now applying Bernoullis equation in between the sections 1-1 and 2-22

1 11

P Vz

g 2g

=2

2 22

P Vz

g 2g

( Since pipe is horizontal, hence Z2= Z1)

1 2P P

g

=

2 22 1V V

2g

=2 2 2

1 1 1(4V ) V 15V

2 2

Since, P1 P2= 30 kPa

330 10

1000

=

2115V

2

V12 =

360 10

41000 15

Therefore, flow velocity V1 = 2 m/sec

Or

Theoretical discharge (Qth) through a venturimeter is given by

Qth =1 2

2 2

1 2

A A2gh

A A


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108

A1V1 =1 2

2 2

1 2

A A 2gh

A A

V1 = 21

2

2gh

A1

A

=4

1

2

2gh

d1

d

=4

2 9.81 (30 / 9.81)

0.041

0.01

= 2 m/sec.


Q.9 Ans. (b)

Given, Vr =

360 10m/sec

2 r

and V=

3300 10m/sec

2 r

NowrV

V= 60 1

300 5

vrV

A

B

C

D

r

r

Therefore, Vr = V

5

Also V = r.= r.d

dt

and Vr=

dr

dt

dr

dt=

r.d

5dt

r

120

dr

r =

0

d

5

r120[ n r]l = 1( 0)5

r

ln120

=5

r

120= 0.5336

Therefore, r = 120 0.5336 = 64.03 m



41/89

109

Q.10 Ans. (d)

Given :

Radius of cylinder = R

Height of cylinder = H

Angular speed = Density of oil =

As the cylinder is closed and completely filled with oil, the rise of oil level at the ends and depression of oil at

the centre due to rotation of the vessel, will be prevented. Thus the oil will exert force on the complete top ofthe vessel. Also the pressure will be exerted at the bottom of the cylinder.

Thrust at the bottom of cylinder = Weight of oil in cylinder + total force on the top of the cylinder

Now Weight of water = V.g= R2 H g = gR2H ......(i)

H

R

R

r

dr

Now lets consider an elementary ring of radius r and thickness dr. Then pressure gradient in the elementary

ring in free as well as in forced vortex flow is given as

p

r

=2V

r

=

22(r )

rr

Integrating the above equation,

p = 2 r r

p =

2 2w r

2

......(ii)

Now Force on elementary ring is = pressure intensity area of elementary circular ring

dF = p 2rdr

Total force on the top of the cylinder, FT =

2 2R R

0 0

w rdF .2 rdr

2

. (p from equation (ii)

FT = w2

R4

0

r

4

=2 4R

4

......(iii)

Now, total thrust at bottom of cylinder is given by adding the equation (iii) and (i)

=4

2 2Rw g R H

4


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110

=

2 22 w R

R g H4

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 180, 5.30.

Q.11 Ans. (b)

0.2 m 5mPump

4 km

1

1

2

2

Applying Bernoulliss equation at the section (1-1) and section (2-2)

21 1

1

P VZ

g 2g

=2

2 22 f

P VZ h

g 2g

......(i)

Since pipe is horizontal, therefore Z1= Z2From question, V2 = 0, V1= 2 m/s

Head loss due to friction in the pipe is given as hf=

2f LV

2g dwhere, f, V, L are the friction factor, mean velocity and length of the pipe respectively.

=

20.01 4000 2

2 9.81 0.2

= 40.774 m

from equation (i),1

P

g =2

2 1f

P Vh

g 2g

= 5 + 40.774

22

2 9.81 = 45.57 m

Therefore, P1 = 45.57 1000 9.81 N/m2 = 447.04 kPa

Therefore, absolute discharge pressure at the pump exit = P1+ Patm.= 447.04 + 101.325 = 548.365 kPa = 5.5 bar

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 420, Eqn. 11.1.Q.12 Ans. (d)

xDs

k

Dt

D U

Pa

From continuity equation at the throat and at the exit of the venturimeterA1V1 = A2V2

V2 =1 1

2

A V

A=

2

12t

D.V

D

......(i)

Now applying Bernoulliss equation at the throat and at the exit of the venturimeter

21 1

1

P VZ

g 2g

=2

2 22

P VZ

g 2g

Since venturi is horizontal, therefore Z1= Z2


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111

1 2P P

g

=

2 22 1V V

2g

P1 P2 =

22 1

2 22

VV 1

2 V

P1 Patm. =

42

4t

DU 12 D

At throat velocity is greater than U, hance pressure will be less than atmospheric

P1 =

42

4t

DU 1

2 D

=4

2

4t

DU 1

2 D

Now spring is elongated due to lower pressure at throat.Therefore, in equilibrium,

Spring force = Pressure Force

Hence, kx =2sD

4

(P1)

=

2 2 4s

4t

D U D1

4 2 D

Hence, x =

2 42s4

t

U D1 D

8k D

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 242, 6.6.

Q.13 Ans. (c)

2m

d

d/2

2 2

11

From continuity equation, A1V1 = A2V2

2

d .24

=

2

2

d

.V4 2

V2 =

2

2

2d4 8 m /sec

d

Now Applying Bernoullis theorem at section (1-1) & (2-2)

21 1

1

P Vz

g 2g

=2

2 22

P Vz

g 2g

......(i)


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112

From question, P1 = 200 kN/m2= 200103N/m2

Taking section (1-1) as datum surface

V1 = 200 m/sec, z1= 0, V2 = 8 m/sec, z2= 2 m

Putting all the above values in equation (i), we get

23 2200 10

01000 9.81 2 9.8

=

2P 8 8 21000 9.81 2 9.81

20.39 + 0.204 =2P 3.26 2

9810

P2 = 9810 (20.594 5.26) = 150426.5 N/m2

Therefore, pressure at the exit of the contraction,

P2 = 150.4 kN/m2


Q.14 Ans. ( 0.0302 m3/s)

Internal diameter, D1 = 21.0 cm = 0.2 m;

Area of inlet, A1 =2

2D

4

D12 =

2 2(0.21) 0.0346 m4

Throat diameter, D2 = 10.5 cm = 0.105 m

Area at throat, A2 =2

2D

4

=

4

(0.105)2 = 0.0087 m2

Coefficient of discharge of venturi,

Cd = 0.984

Pressure head, h = yHg

water

S1

S

= (9.5 sin 30)13.6 1

1

= 59.85 cm = 0.5985 m

Discharge (Q) through a venturimeter is given by:

Q = 1 2d2 2

1 2

A AC 2gh

A A

= 0.984 2 2

0.0346 0.00872 9.81 0.5985

(0.0346) (0.0087)

= 0.984 0.008945 3.427 = 0.0302 m3/s

Reference: Fundamentals of Fluid Mechanics, B. R. Munson, Edition 2010, Page-441, 8.37.


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113

Dimensional and Model Analysis

7

Year 2010

1. A phenomenon is modeled using n dimensionalvariables with k primary dimensions. The number

of non-dimensional variables is

(a) k (b) n

(c) n k (d) n + k

2. Match the following

P :Compressibleflow U : Reynoldsnumber

Q: Freesurfaceflow V :Nusseltnumber

R : Boundarylayerflow W : Weber number

S: Pipeflow X : Froudenumber

T : Heat convection Y: Machnumber

Z :Skinfrictioncoefficient

(a) P-U; Q-X; R-V; S-Z; T-W

(b) P-W; Q-X; R-Z; S-U; T-V

(c) P-Y; Q-W; R-Z; S-U; T-X

(d) P-Y; Q-W; R-Z; S-U; T-V

Year 2007

3. Consider steady laminar incompressible axi-sym-

metric developed viscous flow through a straightcircular pipe of constant cross-sectional area at

a Reynolds number of 5. The ratio of inertia force

to viscous force on a fluid particle is

(a) 5 (b)1

5

(c) 0 (d)

Year 2002

4. If there are m physical quantities and nfundamental dimensions in a particular process,

the number of non-dimensional parameters is

(a) m + n (b) m n(c) m n (d) m/n

Year 1997

5. The Reynolds number for flow of a certain fluid

in a circular tube is specified as 2500. What will

be the Reynolds number when the tube diameter

is increased by 20% and the fluid velocity isdecreased by 40% keeping fluid the same?

(a) 1200 (b) 1800

(c) 3600 (d) 200

Year 1994

6. The ratio of inertia forces to gravity forces may

be expressed as square of non-dimensional group

known as.........


46/89

Chapter-7Answers

1. Ans. (c) 2. Ans. (d) 3. Ans. (a) 4. Ans. (c) 5. Ans. (b)Ans. (Froude Number)



47/89

115


Q.1 Ans. (c)

Buckinghams -theorem states that if there are ntotal dimensional variables (dependent as well as independentvariables) involved in a phenomenon which can be completely described by mfundamental dimensions (such

as mass, length, time etc.), and are related by a dimensionally homogeneous equation, then the relationshipamong the nquantities can be expressed in terms of exactly (n m) dimensionless and independentterms.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b).

Q.2 Ans. (d)

P. Compressible flow Mach Number

Q. Free surface flow Weber Number

R. Boundary layer flow Skin friction coefficient

S. Pipe flow Reynolds Number

T. Heat convection Nusselt Number

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, 769 & 455, , 17.11e &

11.2.

Q.3 Ans. (a)

Reynolds number is defined as the ratio of inertia force and viscous force.

Re =Inertia force

5Viscous force

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-767, 17.11 (b).

Q.4 Ans. (c)

Buckinghams -theorem states that if there are n total dimensional variables involved in a phenomenonwhich can be completely described by mfundamental dimensions (such as mass, length, time etc.), and are

related by a dimensionally homogeneous equation, then the relationship among thenquantities can be expressed

in terms of exactly (n m) dimensionless and independentterms.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b).

Q.5 Ans. (b)

Exp. Reynolds number, Re =vd

2500 =vd ......(i)

when new diameter = 1.2 d then new velocity = 0.6v

Reynolds number, Renew =1.2d 0.6v

= 0.72 Re

= 0.72 2500 = 1800Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-325, Equation 7.15.

Q.6 Ans. (Froude Number)

Reynolds number, Re =Inertia force

Viscous force=

VL or

Vd

Froude number, Fr =Inertia force

Gravity force=

V

Lg


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116

Euler number, Eu =Inertia force

Pressure force=

V

p /

Weber number, We =Inertia force

Surface tension force=

V

/ L

Mach number, M =Inertia force

Elastic force =

V

K /



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117

Boundary Layer Theory

8

Year 2012

1. An incompressible fluid flows over a flat platewith zero pressure gradient. The boundary layer

thickness is 1mm at a location where the Reynolds

number is 1000. If the velocity of the fluid alone

is increased by a factor of 4, then the boundary

layer thickness at the same location, in mm will

be

(a) 4 (b) 2

(c) 0.5 (d) 0.25

Year 2007

2. Consider an incompressible laminar boundary

layer flow over a flat plate of length L, aligned

with the direction of an oncoming uniform free

stream. If F is the ratio of the drag force on the

front half of the plate to the drag force on the

rear half, then

(a) F 1/ 2 (b) F = 1/2

(c) F = 1 (d) F > 1

Linked Answer Questions : Q. 3 - Q. 4

Consider a steady incompressible flow through a

channel as shown below.

A B

H

x

y

uo

Vmu0

The velocity profile is uniform with a value of u0at the inlet section A. The velocity profile at

section B downstream is

m

m

m

yV , 0 y

u V , y H

H yV , H y H

3. The ratio0

Vm

uis

(a) 1

1 2H

(b) 1

(c)1

1H

(d)1

1H

4. The ratio A B20

p p

1u

2

(where pA and p

B are the

pressures at section A and B, respectively, and is the density of the fluid) is

(a) 21

1

1H

(b)2

1

1H

(c)2

11

21

H

(d)1

1H


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Year 2006

Linked Questions 5 and 6

A smooth flat plate with a sharp leading edge is

placed along a gas stream flowing at U = 10 m/s.

The thickness of the boundarylayer at section

r-s is 10 mm, the breadth of the plate is 1 m (into

the paper) and the density of the gas = 1.0kg/m3. Assume that the boundary layer is thin,

two-dimensional, and follows a linear velocity

distribution, u= U ,y

at the section r-s,where

yis the height from plate.

r U

u

flat plate

q

p

U

s

5. The mass flow rate (in kg/s) across the section

q-r is

(a) zero (b) 0.05

(c) 0.10 (d) 0.15

6. The integrated drag force (in N) on the plate,

between p-s, is

(a) 0.67 (b) 0.33

(c) 0.17 (d) zero

Year 2004

7. For air flow over a flat plate, velocity (U) and

boundary layer thickenss () can be expressedrespectively, as

3

x

U 3 y 1 y 4.64x;

U 2 2 Re

If the free stream velocity is 2 m/s, and air has

kinetmatic viscosity of 1.5 105m2/s and density

of 1.23 kg/m3, the wall stress at x = 1m, is

(a) 2.36 102N/m2 (b) 43.6 103N/m2

(c) 4.36 103N/m2 (d) 2.18 103N/m2

8. If x is the distance measured from the leading

edge of a flat plate, the laminar boundary layer

thickness varies as

(a) 1/x (b) x4/ 5

(c) x2 (d) x1/ 2

9. Flow separation in flow past a solid object is caused

by

(a) a reduction of pressure to vapour pressure

(b) a negative pressure gradient

(c) a positive pressure gradient

(d) the boundary layer thickness reducing to zero

Year 1994

10. For air near atmosphere conditions flowing over

a flat plate, the laminar thermal boundary layer is

thicker than the hydrodynamic boundary layer.

(True/false)

Year 1993

11. The predominant forces acting on an element of

fluid in the boundary layer over a flat plate in a

uniform parallel stream are :

(a) Viscous and pressure forces

(b) Viscous and inertia forces

(c) Viscous and body forces

(d) Inertia and pressure forces

Year 1991

12. A streamlined body is defined as a body about

which

(a) The flow is laminar

(b) The flow is along the sreamlines

(c) The flow separation is suppressed

(d) The drag is zero

Year 2002


51/89

119

Chapter-8Answers

1. Ans. (c) 2. Ans. (d) 3. Ans. (c) 4. Ans. (a) 5. Ans. (b)6. Ans. (c) 7. Ans. (c) 8. Ans. (d) 9. Ans. (c) 10.Ans. (False)11.

Ans. (b) 12.

Ans. (c)



52/89

120


Q.1 Ans. (c)

As per Blasius result thickness of laminar boundary layer is given as

x

5x

Re

Hence, =5x

vx

Therefore,1

v

1 1

24

Therefore, if the velocity of fluid is increase by four times then boundary layer thickness reduces by

1/2.


Q.2 Ans. (d)V

L/2 L/2

As we know that local drag coefficient is given by

Cf = 20.664

V Re

2

o

x

or o = 0.332 V2(Rex)

1/2

Now drag force on the front half is given by

F1 =

L

2

0

Bo dx (B = width of Plate)

=

L

122 2

0

0.332 V (Re )

x dx

Reynolds number is given as, Rex =

V

x

=

L1

122

2 2

0

V0.332. V

x dx

=

L

12 2

2

0

0.332 V

V

x dx


53/89

121

=

L1 122 2

0

LK 2K

1 2

2

x

......(i)

where, K =

20.332 V

V

Similarly, drag force on the rear half,

F2 =

L

L

2

Bo

dx

=

L 1

2

L

2

K x dx

=1

12

2L

2K (L)2

Now required ratio, F =

1

2

1

112 22

L2K

F 21

FL

2K L2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-540, Ex. 12.6.

Q.3 Ans. (c)

Given:

A B

H

x

y

uo

Vmu0

u =

V , 0

V , H

HV , H H

m

m

m

yy

y

yy

Assuming width of channel as unity

Applying mass conservation at section A and B. Taking density of liquid constant, the conservation of mass

principle becomes volume flow equation.

Volume flow rate incoming at section A = Volume flow rate outgoing from section B

Therefore, total volume flow rate inlet


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122

entryQ

= uo H B = uoH

Total volume flow rate leaving,

exitQ

= Volume flow rate from boundary layer + Volume flow rate from mid section

u

ydy

Volume flow rate from mid section = Vm(H 2)For boundary layer 0y

Bd Q = u . dy

Bd Q

= Vmy

dy

Integrating the above equation

BQ

=0

Vm y dy

=2

0

V y

2

m

=V .

2

m

By symmetry for H yH, Volume flow rate =V .

2

m

Therefore, entryQ

=exit

Q

uoH =V .

V (H 2 ) 22

m

m

uoH = Vm(H )

Vm

ou=

H 1

H1

H

Reference: Fluid Mechanics, R.K. Bansal, 4th Edition, Page-655, 13.3.

Q.4 Ans. (a)

A B

H

x

y

uo

Vmu0

Applying Bernoullis equation at section A and B

2

A Ap v+g 2g

=2

B Bp +g 2g

(Since

A Bp - p

=

2 2

B A -

2


55/89

123

A B

p - p

=

2 2

m oV - u

2

A B2

o

p - p

1 u

2

=

2V

1m

ou

A B2

o

p - p

1 u

2

= 21

1

1H

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-280, Equation

7.29.

Q.5 Ans. (b)

Given:Free stream velocity,

U = 10 m/s

Boundry layer thickness, = 10 mmBreadth of plate, B = 1 m

Density of air, = 1.0 kg/m3

velocity distribution, u = Uy

r U

u

flat plate

q

p

U

s

Applying mass conservation:

Mass rate entering section q p= Mass leaving section q r+ mass leaving section r s

Mass rate entering q p = Density Volume flow rate

= B U= 1.0 1 10 103 10 = 0.1 kg/s

Mass flow rate through the element dy at section r s

dm = u B dy

dm = BU

ydy

Integrating the above equation gives,

m =0

BU

y

dy

=

2BU

2

=

BU

2

=

31 1 10 10 10

2

= 0.05 kg/s

Thus mass flow rate leaving across the section, q r = 0.1 0.05 = 0.05 kg/s

Q.6 Ans. (c)

Drag froce on the plate will be the rate of change of momentum of control volume qprs


56/89

124

Thus, momentum rate entering section q p = Um

= 0.1 10 = 1N

Momentum rate leaving through section rs

=0

u Bd y u

=2

2

0

BU

y

dy

=

2 3

2

BU

3

=

2BU

3

=

2 31.0 1.0 10 10 10

3

= 0.33 N

Momentum rate leaving through section q r = 0.05 10 = 0.5 N

Drag force, F = Change in momentum rate= 1 0.33 0.5 = 0.17 N

Reference: Fluid Mechanics and Hydraulic Machines, R.K. Bansal, 4th Edition, Page-655, 13.3.

Q.7 Ans (c)

Reynolds number at x,

Rex =Ux

= -5

2 1

1.5 10

= 1.33 105

Boundry layer thickness, =5

x

4.64x 4.64 1

Re 1.33 10

= 0.0127

Since the velocity profile is given as,

U

U=

33 y 1 y

2 2

Velocity gradient,dU

dy=

2

3

3 1 1 3yU .

2 2

Therefore,dU

at y 0dy

=3 1

. .U2

From Newtons law of viscosity,

o =dU

at y 0dy

=3U

2

=5 3 21.5 10 1.23

2 0.0127

= 435.82 105N/m2 = 4.36 103N/m2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-537, Ex. 12.3.

Q.8 Ans. (d)

As per Blasius result thickness of laminar boundary layer is given asx

5x

Re


57/89

Hence, =5x

vx

Therefore, 1

12x

1

2x


Q.9 Ans. (c)

In direction of flow usually pressure gradient is negative i.e.P

x

= -ve, supports the fluid flow. But in case of

flow separation, pressure gradient is positive i.e.P

x

= +ve, supports the fluid separation &P

x

= 0, it means

that the fluid is on the verge of separation.

Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-662, Equation-13.7.

Q.10 Ans. (False)Q.11 Ans. (b)

In a fluid flow over a flat plate, the dominant forces are inertia force and viscous force. Therefore Reynolds

number decides the nature of the flow.


Q.12 Ans. (c)

A body where flow separation is suppresed is called streamlined body. For a well stream lined body the

separation occurs only at the down stream end.

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition , Page-805, 18.2.


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126

Laminar and Turbulent Flow

9

Year 2013

1. For steady, fully developed flow inside a straightpipe of diameter D, neglecting gravity effects,the pressure drop p over a lengthLand the wallshear stress

ware related by

(a) wpD

4L

(b)

2

w 2

p D

4L

(c) wpD

2L

(d) w

4 pL

D

Year 2010

2. The maximum velocity of a one-dimensionalincompressible fully developed viscous flow,

between two fixed parallel plates, is 6 ms1. Themean velocity (in ms1) of the flow is(a) 2 (b) 3(c) 4 (d) 5

Year 2009

3. The velocity profile of a fully developed laminarflow in a straight circular pipe, as shown in thefigure, is given by the expression

u(r) =2 2

2

R dp r 14 dx R

wheredp

dxis a constant

u(r)Rr

x

The average velocity of fluid in the pipe is

(a)

2R dp

8 dx

(b)

2R dp

4 dx

(c)

2R dp

2 dx

(d)

2R dp

dx

Year 2007

4. Consider steady laminar incompressible axi-sym-

metric developed viscous flow through a straight

circular pipe of constant cross-sectional area at

a Reynolds number of 5. The ratio of inertia forceto viscous force on a fluid particle is

(a) 5 (b)1

5

(c) 0 (d)

Year 2006

5. The velocity profile in fully developed laminar flow

in a pipe of diameter D is given by

u =

2

0 24ru 1 ,D

where r is the radial distance

from the center. If the viscosity of the fluid is ,the pressure drop across a length L of the pipe is

(a)0

2

u L

D

(b)

0

2

4 u L

D

(c)0

2

8 u L

D

(d)

0

2

16 u L

D


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127

Year 1996

6. In flow through a pipe, the transition from laminar

to turbulent flow does not depend on

(a) velocity of the fluid

(b) density of the fluid

(c) diameter of the pipe

(d) length of the pipe

7. For laminar flow through a long pipe, the pressure

drop per unit length increases

(a) in linear proportion to the cross-sectional area

(b) in proportion to the diameter of the pipe

(c) in inverse proportion to the cross-sectional

area

(d) in inverse proportion to the square of cross-

sectional area

Year 1995

8. In fully developed laminar flow in a circular pipe,

the head loss due to friction is directly proportional

to........

(mean velocity/square of the mean velocity)

Year 1994

9. For a fully developed viscous flow through a pipe,

the ratio of the maximum velocity to the average

velocity is.......

10. Prandtls mixing length in turbulent flow signifies

(a) the average distance perpendicular to the

mean flow covered by the mixing particles

(b) the ratio of mean free path to characteristic

length of the flow field

(c) the wavelength corresponding to the lowest

frequency present in the flow field

(d) the magnitude of turbulent kinetic energy


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128

Chapter-9Answers

1. Ans. (a) 2. Ans. (c) 3. Ans. (a) 4. Ans. (a) 5. Ans. (d)6. Ans. (d) 7. Ans. (c) 8. Ans.

(meanvelocity)

9. Ans. (two) 10.Ans. (a)



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Q.1 Ans. (a)

In a pipe flow, =dp r

dx 2

=dp d

dx 4

Therefore, shear stress at wall, w

=dp D

dx 4

Above equation can be written as, w=

p D

4L

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-547, Eqn.-13.3.

Q.2 Ans. (c)

For the flow of fully developed between two fixed parallel plates

Vmax

B

x

y

The velocity distribution for laminar flow between fixed parallel plates is given as,

V =1

2

p

x(By y2)

Flow velocity is maximum when y =B

2

Therefore, Vmax =

2B

8

p

x

Average flow velocity is obtained by dividing the total discharge with cross sectional area.

Hence, Vavg. =

2B

12

p

x

Therefore,max

avg

V

V =3

2

So mean velocity, Vavg = max.2

V3

=2

6 4m/s3

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-557, Equation13.389.

Q.3 Ans. (a)

x

rr

Ru(r)

dr

Consider an element ring of thickness dr at a radius of r.

Therefore, element discharge from this ring, dQ = (2r).dr. u(r)


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Therefore, total discharge Q = Qd =R 2 2

2

0

R P(2 ) . 1 .

4 R

rr dr

x

=

R2 3

2

0

R P2

4 R

rr dr

x

=

R2 2 4

20

R P r r 24 x 2 4R

=

2 2 4

2R P R R 24 x 2 4R

Total discharge, Q =

4R P

8 x

Now, Q = Area Average velocity

Area Vavg. =

4R P

8 x

R2 Vavg =4R P

8 x

Therefore,Vavg =2R P

8

x

=

2R P

8

x

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-550, Eqn-13.14.

Q.4 Ans. (a)

Reynolds number, Re =Inertia force vd

Viscous force


Q.5 Ans. (d)

Given:

Fully developed laminar flow velocity profile in a circular pipe is given by

u =

2

o 2

4ru 1

D

......(i)

From Hagen-Poiseuille equation the pressure loss in fully developed laminar flow across the length of the

pipe is given as

p1 p2 = 232u L

d......(ii)

The ratio of maximum velocity and average velocity in case of fully developed laminar flow through a circular

pipe is 2 i.e.ou

u= 2

Therefore, from equation (ii), p1 p2 =

o

2

u32 L

2

d=

o

2

16u L

d

Or

Fully developed laminar flow velocity profile in a circular pipe is given by

u =

2

o 2

4ru 1

D

......(i)


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Rr

Now =P r

x 2

......(ii)

From Newtons law viscosity, =du

dy

Here y = R r

Therefore, dy = dr

Putting the value of dy in equation (i), we get

=u

r

......(iii)

From equation (ii) and (iii),u

r

=

P r.

x 2

u

r

=1 P r

x 2

From equation (i) ,

2

o 2

4ru 1

r D

=

1 P r

x 2

o28u

rD

=1 P r

x 2

P

x

= o

2

16 u

D

P =o

2

16 ux

D

Integrating over a length of L,

2

1

P

P

dp =L

o

2

o

16 udx

D

p2 p1 =o

2

16 u L

D

Therefore, p2 p1 =o

2

16 u L

D

Reference: Hydraulics and Fluid mechanics, Modi and Seth, 17 Edition, Page-548. Equation-13.3,

13.6, 13.7.

Q.6 Ans. (d)

In flow through pipe the transition from laminar to turbulent depends upon Reynolds number which is given as

Re =vd where, d is the characteristic dimension of the pipe.



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Q.7 Ans (c)

From Hagen-Poiseuille equation the pressure loss in fully developed laminar flow across the length of the

pipe is given as

p1 p2 = 232u L

d

1 2p p

L

= 2

32u

dTherefore, for laminar flow through a pipe, the pressure drop per unit length increases in inverse proportion

to the cross-sectional area.

Reference: Fluid Mechanics, R.K. Rajput, 2005 Edition, Page-442, Equation-10.11.

Q.8 Ans. (mean velocity)

The head loss due to friction in fully developed laminar flow in a circular pipe is given as

hf

=1 2

P P

g

=avg.

2

32 u L

gd

Head loss in laminar flow over a length L of circular pipe varies as the first power of the mean velocity of the

flow.

Reference: Hydraulics and Fl

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