Gate 1991-2013 Topic Wise Solution.pdf 30213o

Mechanical Engineering

INDUSTRIAL ENGINEERING

TOPICWISE GATE SOLUTION 1991-2013

DRONACHARYA INSTITUTE OF ENGINEERS F-108, Katwaria Sarai, Near Mother Dairy Booth, New Delhi-16 Ph. +91-011-64551144, 9810758209 www.drona.org

Fluid Mechanics Syllabus of GATE Examination 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. Pelton-wheel, Francis and Kaplan turbines - impulse and reaction principles, velocity diagrams.

NOMENCLATURE OF CHAPTERS 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 - 124 9. ............... 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

Statistical Analysis S.No.

Chapter Name

03

04

05

06

1

Properties of Fluids

1

2

Pressure and its Measurement

2

3

Hydrostatic Forces on Surfaces

2

4

Buoyancy and Flotation

1

5

Fluid Kinematics

2

1

1

3

6

Fluid Dynamics

2

4

4

2

7

Dimensional and Model Analysis

8

Boundary Layer Theory

59

Laminar and Turbulent Flow

10

Flow Through Pipes

11

Hydraulic Turbines

12

Centrifugal Pump

13.

Compressible Flow Total

07

1

08

09

12

13

2 2

2 1

2

5 2

2 2

5

2

1

2

1

2

2

18

7

3

4

2

2

17

16

1 2

2

2

2

3

4

4 4

13

11

2

1

4

10

2

2

11

1

1 1

1

8

8

1

3

5

6

Conclusion 1. 2.

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

1 Properties of Fluids

Year 2008

Year 2001

1.

4.

The SI unit of kinematic viscosity (v) is (a) m2 /sec (b) kg/(m-sec) (c) m/sec2 (d) m3 /sec 2

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

A journal bearing has shaft diameter of 40 mm and a length of 40 mm. The shaft is rotating at 20 rad/s and the viscosity of the lubricant is 20 mPas. 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, 7.4 x 10–7 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 10–3 (b) 0.651 (c) 6.51 (d) 0.651x103

Year 1999 6.

Kinematic viscosity of air at 20oC is given to be 1.6×10–5 m2/s. It kinematic viscosity at 70oC will be vary approximately (a) 2.2×10–5 m2/s (b) 1.6×10–5 m2/s (c) 1.2×10–5 m2/s (d) 10–5 m2/s

Year 1996 7.

The dimension of surface tension is (a) ML–1 (b) L2 T–1 –1 1 (c) ML T (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



73

Chapter-1 Answers 1. Ans. (a) 6. Ans. (a)

Space for Rough work

2. Ans. (c) 7. Ans. (d)

3. Ans. (b) 8. Ans. (a)

4. Ans. (a)

5. Ans. (c)

74

Chapter-1 Answer & Explanations Q.1

Ans. (a) Given:

Shaft diameter, d Shaft length, L Speed,  Viscosity,  Clearance, y

= = = = =

40 mm 40 mm 20 rad/s 20 mPa-s 0.020 mm 0.02 mm

µ = 20 mPas 40 mm

40 mm

Shear stress given by Newton’s law of viscosity  = 

du dy

Here,

u = × r = 20 × 0.02 = 0.4 m/s



 = 20  103 

0.4 = 400 N/m2 0.02  103

F = × A = 400 × d L = 400 ×  × 0.04 × 0.04 = 2.0106 N  Torque loss, T = F × r = 2.0106 × 0.02 = 0.0402 Nm Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-22, Example1.10. 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. Shear force,

Q.2

u + du

d

dy u

Shear strain, tan d =

du  dt dy

If d is small, then tan d ~ d . Therefore,



d =

du.dt

du  dt dy

d du = dt dt

75

From Newton’s law of viscosity

du d  =  dy =  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 du , where = velocity gradient dy dy

dy y

u + du u x

 dx       dt  dy

 dx   dy  dx     , where is shear strain of fluid dt dy

Q.3

 dx   dy  Thus   is rate of shear strain dt Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, 1.8. Ans. (b) Given, Kinematic viscosity,  = 7.4×10–7 m2/sec Specific gravity, S = 0.88 Density of fluid,  = 0.88 × 1000 kg/m3 Dynamic viscosity,  = ×  = 0.88 × 103 × 7.4 × 10–7 = 0.6512×10–3 Pa.s V = 0.5 m/s 0.5 mm

Now, from Newton’s law of viscosity  =

Q.4

.du 0.6512  10 3  0.5   0.6512 N/m2 dy 0.5  10 3

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8. 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 = 104 stoke Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.

76

Q.5

Q.6

Q.7

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. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-34, Equation-2.1. 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. Ans. (d) Surface tension () is defined as force per unit length. It is also equivalent to surface energy per unit surface area. It is mainly due to force of cohesion. MLT 2 = MT–2 L Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-14, Equation1.11. Ans. (a) A fluid is said to be Newtonian fluid when it obeys the Newton’s 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:

Dimension of  =

Ideal solid Shear stress, 

Q.8

fluid astic l p ham Bing fluid stic a l p do fluid Pseu nian o t w Ne id t flu n a at Dil

Ideal Fluid

 du    dy 

Velocity gradient, 

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



77

2 Pressure and its Measurement

Year 2005 G2

Atmospheric pressure 1.01 bar

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/m3 and g = 9.81 m/s2. Which of the following is correct?

G1

P

(a) 1.01 bar (c) 5.00 bar

(b) 2.01 bar (d) 7.01 bar

Year 2000 3.

B

In figure if the pressure of gas in bulb A is 50 cm Hg vaccum and patm = 76 cm Hg, the height of column H is equal to

A 150 mm

Patm

A H Hg

(a) (b) (c) (d)

Flow direction is A to B and pA – pB = 20 kPa Flow direction is B to A and pA – pB =1.4 kPa Flow direction is A to B and pB – pA =20 kPa Flow direction is B to A and pB – pA =1.4 kPa

Year 2004 2.

The pressure gauges G1 and G2 installed on the system show pressures of PG1 = 5.00 bar and PG2 = 1.00 bar. The value of unknown pressure P is

(a) 26 cm (c) 76 cm

(b) 50 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

(c) p

(b)

p 2

(d) 2p

Year 1997 5.

Refer to figure, the absolute pressure of gas A in the bulb is PA C 10 cm BA F D

E

5 cm 2 cm

 = 13.6 g/ml

(a) 771.2 mm Hg (c) 767.35 mm Hg

(b) 752.65 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 Water

A

H2O

10 mm

Hg

(a) 1236 Pa (c) zero

(b) 1333 Pa (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



79

Chapter-2 Answers 1. Ans. (a) 6. Ans. (b)

Space for Rough work

2. Ans. (d) 7. Ans. (a)

3. Ans. (b)

4. Ans. (d)

5. Ans. (a)

80

Chapter-2 Answer & Explanations Q.1

Ans. (a)

B A 150 mm

Writing the pressure balance equation, pA = pB + gh

150  20.012 kPa 1000 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-36, Equation-2.2. Ans. (d) pA– pB = 136000  9.81 

Q.2

Atmospheric pressure 1.01 bar

G2

G1

P

Absolute pressure at 2

Q.3

P abs2 = = Absolute pressure at 1 Pabs1 = = Ans. (b)

PG2 + Patm 1 + 1.01 = 2.01 bar PG1 + Patm (Atmospheric pressure for G1 becomes 2.01 bar) 5 + 2.01 = 7.01 bar Patm

A H Hg

Q.4

Applying pressure balancing equation at free surface –PA + PH = P atm P atm = PA – PH Taking P atm = 0 Therefore, P H = 50 cm Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-847, Equation19.7. Ans. (d) Pressure inside spherical droplet =

4 d

8 , where  is surface tension force and d is diameter.. d Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-15, Equation-1.11. Pressure inside soap bubble =

Q.5

Ans. (a) PA C 17 cm

BA F

datum D

5 cm 2 cm

E

 = 13.6 g/ml

Gauge pressure at A, PA + 1gh1 = 2gh2 + 1gh3 (Taking the unknown liquid as water) PA + 1000  9.81



2 5 17  1000  9.81  = 13600  9.81  100 100 100 PA = 2668.32 + 490.5 – 1667.2 = 1491.12 N/m2 P abs = Patm + PA = 1.013×105 + 1491.12 = 102791.12 N/m2 P abs = mghm m  density of mercury   h  mercury column height   m 

102791.12  0.77055m = 771 mm 13600  9.81 Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 40, 2.6.2. Ans. (b) hm =

Q.6

Water

A

H2O

10 mm

Hg

Neglecting the depth of water column, gauge pressure is given as P gua ge = gh = 13600  9.81  Q.7

10 = 1334.16 N/m2 1000

Ans. (a) Exp. 

Force Area Net force = Pressure×Area. Pressure =

Therefore, area defined by the shape of the control volume. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-34, Equation2.1.



82

3 Hydrostatic Forces on Surfaces Year 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

(a) (b) (c) (d)

Fx = ghrw and Fy = 0 Fx = 2ghrw and Fy = 0 Fx = 2ghrw and Fy = gwr2/2 Fx = 2ghrw and Fy = gwr2/2

Year 1992 5m

(a) 5000 (c) 7546

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.

(b) 6600 (d) 9623

Gasoline surface

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 ed by the weighing balance at this instant of time is (a) mg + QU (b) mg + 2 QU 2 (c) mg + QU /2 (d) QU2 /2

Negative pressure (8250 N/m2) 1.8m

Gasoline (S = 0.7)

The horizontal and vertical hydrostatic forces Fx and Fy on the semi-circular gate, having a width w into the plane of figure, are

Gate 45º

Year 2001 3.

Hinge P



83

Chapter-3 Answers 1. Ans. (d)

Space for Rough work

2. Ans. (a)

3. Ans. (d)

4. Ans. (144.5 kN)

84

Chapter-3 Answer & Explanations Q.1

Ans. (d)

5m

m 2.5

x

h

G B

mg cos 30º 30º F

1m b=

30º mg

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 N Depth of centre of pressure,

h = x

IG sin 2  Ax

1  1 53 2 12 sin 30o  = 1.67 m  = 1.25 +  5  1  1.25

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,

Q.2

Q.3

m =

61312.5  1.67 Fh = = 9641 kg 0 9.81  cos300  1.25 g  cos30  2.5  sin 30 0

Therefore, the nearest possible value is 9623 kg. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-132, Example3.22. Ans. a) Mass of water strike = AV = Q Force on weighing balance due to strike of water = Initial momentum – final momentum = QU – Q.0 = QU Since weight of water and container = mg Total force on weighing balance = mg + QU 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 A x where projected area, A = w × 2r Therefore, Fx = 2 gwrh Projected area (ABCD),

r+r B

C w

A

r

r 2r

D

Vertical component of hydrostatic force, F y = Weight of water ed by the curved surface F y = g × Volume of curved portion  r2  =  g r 2  w where, = Area of semi-circle 2 2  g w r 2 where, w is the width of the gate. 2 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-94, Equation-3.4.

=

Q.4

Ans. (144.5 kN) Gasoline surface Negative pressure (8250 N/m2) 1.8m

Gasoline (S = 0.7)

Hinge P

Gate 45º

p 8250   1.2 m w 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 m of oil. Head of oil equivalent to negative pressure 238 N/m2, h =

Area, Pressure,

3.6 sin 45º = 1.873 m 2 A = 3.6 × 3.6 = 12.96 m2 P = wAx = 0.7 × 9810 × 12.96 × 1.873 = 166690 N

Centre of pressure,

h =

x = 0.6 +

Now,

I G sin 2  x Ax

1  3.6  (3.6)3  (sin 45º ) 2 12  1.873  2.16 m = 12.96  1.873  Vertical distance of centre of pressure below top edge of the gate = 2.16 – 0.6 = 1.56 m Taking moments about the hinge. F sin 45º × 3.6 = P ×

1.56 sin 45º

166690  1.56 P  1.56 = 3.6  (sin 45º ) 2 = 144465 N = 144.5 kN 3.6  (sin 45º ) 2 Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-101. Hence, vertical force,

F =



86

4 Buoyancy and Flotation

Year 2010 1.

For the stability of a floating body, under the influence 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 with a string. The tension in the string is

h

(a) ghA (c) ( – s) ghA

(b) (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...........



Chapter-4 Answer & Explanations 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 < 0 3. For neutral equilibrium, MG = 0

Meta centre M G B

Q.2

Centre of gravity Centre of buoyancy

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

h

Free body diagram of the cylindrical body will be G B W FB T

Q.3

At equilibrium condition T + weight of body = Buoyancy force T + Mg =  h Ag T + (s HA)g =  h Ag T = (h – sH) gA 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.



88

5 Fluid Kinematics

Statement for linked answer questions 4 and 5 The gap between a moving circular plate and a stationary 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.

Year 2011 1.

A streamline and an equipotential line in a flow field (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 in 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 app lication r equ ir es the flow to be incompressible and irrotational? (a) P and R (b) Q (c) Q and R (d) R

R r

Moving circular plate

h V

Stationary surface

4.

The radial velocity vr at any radius r, when the gap width is h, is Vr

(a) vr =

2h 2V h

(c) vr =

r

(b) vr = (d) vr =

Vr h Vh r

Year 2008 3.

  For the continuity equation given by   V to be

5.

The radial component of the fluid acceleration at r = R is



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

2

(a)

3V R 4h

2

2

(b)

4h

(c)

2h

2

2

2

2

V R

V R

(d)

V R 4h

2

89

Year 2006 6.

Year 2003

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 v x y

v u v (c) u x y 7.

(b) u

10.

u v v x y

u u u (d) v x y

A two-dimensional flow field has velocities along the x and y directions given by u = x2t and v = – 2xyt respectively, where t is time. The equation of streamlines is (a) x2y = constant (b) x y2 = constant (c) x y = constant (d) not possible to determine

Year 2001 11.

u is equal to x

(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  ax i  ay j , where a is a constant. The equation of stream line ing through a point (1, 2) is: (a) x - 2y = 0 (b) 2x + y = 0 (c) 2x - y = 0 (d) x + 2y = 0

compressible and irrotational compressible and not irrotational incompressible and irrotational incompressible and not irrotational

Year 1999 12.

For the function f = ax2 y – y3 to represent the velocity potential of an ideal fluid. D2 f 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

The velocity components in the x and and y directions of a two dimensional potential flow are u and v, respectively. Then

The 2-D flow with velocity    V   x  2y  2  i   4  y  j , is (a) (b) (c) (d)

Year 2005 8.

     The vector field F  xi  yj (where i and j 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

   V = 2xyi + (2y 2 - x 2 )j , the vorticity vector, curl  V will be   (a) 2y 2 j (b) 6yk  (c) zero (d) -4xk

Year 1995 14.

The velocity components in the x and y directions are given by u  xy 3  x 2 y, v  xy 2 

3 4 y 4

The value of  for a possible flow field involving an incompressible fluid is (a)  (c)

4 3

3 4

(b)  (d) 3

4 3

90

15.

The force F needed to 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 virtually identical 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..........



91

Chapter-5 Answers 1. 6. 11. 16.

Ans. (b) Ans. (a) Ans. (d) Ans. (c)

Space for Rough work

2. 7. 12. 17.

Ans. (d) Ans. (a) Ans. (d) Ans. (b)

3. 8. 13. 18.

Ans. (d) Ans. (d) Ans. (d) Ans. (c)

4. 9. 14. 19.

Ans. (a) Ans. (c) Ans. (d) Ans. (b)

5. 10. 15. 20.

Ans. (c) Ans. (c) Ans. (a) Ans.(closed contour in a fluid flow)

92

Chapter-5 Answer & Explanations Q.1

Ans. (b) If ψ and φ are the stream function and potential function respectively representing the possible flow field. Slope of stream line represented by ψ is given by  d     dx  dy v slope (m1) = =  d  = dx -u  dy   

......(i)

Slope of potential line represented by φ is given by  d     dx  dy slope (m2) = =  d   dy  dx  

=

-u u = -v v

......(ii)

Now, product of the slopes,

v u  = –1 u v Since the product of the slope of these two lines at the point of intersection is –1, which indicates that these two lines are prependicular to each other. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-244, Sec. 6.12. Ans. (d) For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations u v  ......(i) x y = 0 m1 × m2 =

Q.2

z =

1  v u     2  x y 

= 0

......(ii)

For P, Given u = 2y and v = –3x

u  (2y) = 0 = x x  v =  3x    3 x x From equation (i)

and and

u  (2y) = 2 = y y  v  3x   0 = y y

u v  x y = 0 + 0 = 0

1  v u  1 -  =  3 - 2   0  2  x y  2 Since the given velocity field is satifying the equation (i) only, therefore it is a possible case of steady, incompressible and rotational flow.

From equation (ii)

z =

93

For Q, Given

u = 3xy and v = 0

u  u  (3xy) = 3x (3xy) = 3y and = = y y x x  v  v =  0   0 and y = y  0   0 x x From equation (i)

u v  x y = 3y  0

1  v u  -3x -  = 0  2  x y  2 Given velocity field is neither satisfying the equation (i) nor (ii), therefore the flow is neither steady nor irrotational. For R, Given u = – 2x and v = 2y

From equation (ii)

z =

u  u  (2x) = 0 ( 2x) = -2 and = = y y x x  v  v =  2y   0 and y = y  2y   2 x x From equation (i)

1  v u  =0 2  x y  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. Ans. (d)   Given that V = 0

From equation (ii)

Q.3

u v  x y = 0 z =

     j  k    ui  vj  wk   0 i.e.  i   x  y  z  

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. Ans. (a) At radius r, volume of fluid moving out radially is equal to the volume of fluid displaced by moving plate within radius r. Given that V = downward velocity of circular plate in m/s vr = radial velocity at radius ‘r’ R i.e.

Q.4

V h

So volume displaced by moving plate = Velocity × Area = V × r2 Now, volume flow out at radius,

r

Vr

94

r = vr × 2rh From above stated condition vr × 2rh = r2 × V Therefore, Q.5

vr =

V r 2h

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

aR

d VR = = dt

 VR  d   2h  dt

 VR  d   2h    dh  =   (–ve as h is reducing with time) dh  dt  =





VR  1  dh V   2   (  V) as  2  h  dt

V2R Therefore, aR = 2h 2

Q.6

Ans. (a) Acceleration of fluid particle along x-axis is given by ax = u

u u u u +v +w + x y z t

u = 0 z

For 2-D flow

u u u +v + ax = u x y t   Temporal or local

Thus, from equation (i),

Convective acceleration

Q.7

......(i)

acceleration

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-232, Equation 6.27. Ans. (a) Given: u = x2t and v = –2xyt Stream line equation is given as



dx dy = u v dy dx = 2 -2xyt x t

1 dy dx = 2 y x

 Integrating both side



dx 1 =  x 2



dy y

1 ln x =  ln y  c 2  ln x2 + ln y = c  x2y = Constant Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Equation 6.2. 

95

Q.8

Ans. (d) Exp. For two dimensional potential flow, the continuity equation is given as u v  x y = 0

u v = – x y Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-223, Eqn. 6.5. Ans. (c) The velocity field is given as,      ui  vj = = ax i  ay j V Therefore,

Q.9

The equation of stream line dx dy = u v dx dy from equation (i), = ax ay Integrating both side,

Q.10

......(i)

dx dy =  x y ln x = ln y + c x n y = ln c



x ......(ii) y = c Since this stream line es through point (1, 2) hence c = 1/2 Therefore, equation of stream line is (from equation (ii)) 2x – y = 0 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn. 6.2. Ans. (c)    Given vector filed F = x i  yj   The divergence of V is defined as   V

          i  j  k  x i  yj  zk It can also be written as   y z   x u v w x   y 0     = 1-1 = 0 = = x y z x y z



1  v u    =0 2  x y  Hence, the vector field is divergence free and irrotational. Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page-156, Sec. 5.8. 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  ......(i) x y = 0

Rotational component,

Q.11



z =

z =

1  v u     = 0 2  x y 

......(ii)

96

Here,

u  u  (x  2y  2) = 2 (x  2y  2) = 1 and = = y y x x  v  v =  4  y  = 0 and y = y  4  y   1 x x

u v  = 1–1=0 x y For irrotational flow from equation (ii), From equation (i),

1 1  v u    =  0  2  0  2  x y  2 Hence, this flow is steady, incompressible and rotational. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn 6.2. Ans. (d) Velocity Potential, f = ax2y – y3 z =

Q.12

Now

f  2f  2ay = 2axy & x x 2

......(i)

and

f  2f  6y ax 2  3y 2 & = y y 2

......(ii)

As D2f should equal to 0 or D2 (f) =  2 (f )  0

Q.13

 2 f  2f  = 0 x 2 y 2 From euation (i) and (ii), 2ay – 6y = 0 2y (a– 3) = 0 Therefore, a = 3 Reference: Hydraulics and Fluid Ans. (d)   The curl of V is defined as  V .  i   curl of V = x

u

Flow field,

Mechanics, Modi and Seth, 17th Edition, Page-243, Eqn. 6.45.

 j  y

 k  z

v w

    2 2 V = 2xyi + (2y -x )j + 0k

 w v    u w    v u    i    It can also be written as   j   k  y z   z x   x y     v u    =  0  0  i  0  0 j   k  x  y    =  -2x - 2x  k = - 4xk Q.14

Reference: Fluid Mechanics, R. K. Rajput, Edition 2005, Page-172, Equation 5.32. Ans. (d) The velocity components in the x and y directions are given by

97

u  xy 3  x 2 y, v  xy 2 

3 4 y 4

Continuity equation for steady, incompressible and irrotataional flow is u v  = 0 x y

Q.15

......(i)

v u = y3  2xy & y = 2xy  3y3 x Put these value in equation (i), y3  2xy  2xy  3y3 = 0  y3 – 3y3 = 0  y3 ( – 3) = 0  – 3 = 0   = 3 Reference: Fluid Mechanics, S.K. Aggarwal, Page No. 104. Ans. (a) Let Free body diagram of liquid columns due to symmetry Here

 Aa  A 1 = a and A2 = A3 =    2  a

A–a 2

A1

A2

h

(A–a) 2 A2

H

H-h

a A

Now F is equal to the weight of water ed by the piston. W = Mg or M.g = d.g.V where d is the density of the liquid F = d.g.V Now V = A1H + 2 (A1 (H–h))

Q.16

Q.17

......(i)

Aa = aH + 2   (H  h)  2  = aH + A (H–h) – aH + ah V = ah + A(H – h) = ah – A (h – H) ......(ii) Put Value of V in equation (i) F = dg [ah – A(h–H)] 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 are correct but most appropriate single answer is (c). Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page No. 160, 5.4.4. Ans. (b) In a flow field, the streamlines and equipotential lines are always orthogonal to each other. = equipotential lines

= stream lines

98

Q.18

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-245, Fig. 6.18. Ans. (c) For 2-D flow, irrotational component, 1  v u     =0 2  x y  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. Ans. (b) For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations

z =

Q.19

u v  x y = 0

......(i)

1  v u  ......(ii)    = 0 2  x y  If φ is the potential function representing the possible flow field. Then from definition of potential function

z =

u = From equation (i),

Q.20

φ φ and v = y x

u v  2  2   = which is known as Laplace equation. x y x 2 y 2

2 2  1  v u        =0 From equation (ii),  =  2  x y   xy yx  The velocity potential of the flow denoted by ‘’ if satisfies the continuity/Laplace equation, then it will be a possible case of irrotational flow. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn 6.34c. Ans. (closed contour in a fluid flow) Circulation is defined as the line integral of the tangential component of the velocity taken around a closed contour. Mathematically, the circulation is obtained if the product of the velocity component along the curve at any point and the length of the small element containing that point is integrated around the curve.

Y C ds



V X

Mathematically, circulation  =  V cos .ds =

 (udx  vdy)

 Area of closed curve = Vorcitity along the axis perpendicular to the plane containing the closed curve.  = Vorticity × area = 2 × z × area 1  v u  = 2     x y 2  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.



99

6 Fluid Dynamics

Year 2013 1.

Water is coming out from a tap and falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/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 mm of the stream 0.5 m below the tap is approximately (a) 10 (b) 15 (c) 20 (d) 25

3. A large tank with a nozzle attached contains three immiscible, inviscid fluids as shown. Assuming that the changes in h1, h2 and h3 are negligible, the instantaneous discharge velocity is

h1

 h  h   2 gh3  1  1 1  2 2  3 h3 3 h3  

(b)

2 g ( h1  h2  h3 )

(c)

  h  2 h2  3 h3  2g  1 1   1  2  3 

(d)

  h h  2 h3 h1  3 h1h2  2g  1 2 3   1h1  2 h2  3 h3 

Year 2011

Year 2012 2.

(a)

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 watertube 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

1 Flow

h2

2

h3

3

10 mm

(a) 6.4 (c) 12.8

(b) 9.0 (d) 25.6

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 ms–1. Density of water is 1000 kgm–3 and acceleration due to gravity is 9.8 ms–2. Which of the following is TRUE (a) flow is from S 1 to S 2 and head loss is 0.53 m (b) flow is from S2 to S1 and head loss is 0.53 m (c) flow is from S 1 to S 2 and head loss is 1.06 m (d) flow is from S2 to S1 and head loss is 1.06 m

Year 2006 7.

P

h1 h2

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 maximum discharge (in m3/s) that can through the reducer without causing cavitation is (a) 0.05 (b) 0.16 (c) 0.27 (d) 0.38

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

(a)

2gh1

(b)

2gh2

(c)

2 g (h2  h1 )

(d)

2 g (h2  h1 )

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 difference between the pipe and throat sections is found to be 30 kPa then, neglecting frictional losses, the flow 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;

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 constant 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

 60  103 

Vr = –   m / s & V =  2 r 

300  103 m / s, 2 r

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

101

Year 2004 10.

A closed cylinder having a radius R and height H is filled with oil of density . If the cylinder is rotated about its axis at an angular velocity of , the thrust at the bottom of the cylinder is:  2 R 2 4

(b) R 2

e

F  R GH 4 2

2

(d)

13.

2 2 2 (c) R  R  gH

11.

j

2

 gH

I JK

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 Darcy’s friction factor of 0.01. The average 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 absolute discharge pressure at the pump exit is (a) 0.449 bar (b) 5.503 bar (c) 44.911 bar (d) 55.203 bar

2 t

4

2

4 t

2 s

2 s

Air flows through a venturi and into atmosphere. Air density is ; atmospheric pressure is Pa; 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 a spring. 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 D

U

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 d/2

2m

d

(a) 168 kN/m2 (c) 150 kN/m2

Year 2003 12.

2

Year 1999

(a) R 2 gH

(d) R

j FGH DD  1IJK  D eU / 8kj FGH DD  1IJK  D e

2 (c) U / 2 k

(b) 192 kN/m2 (d) 174 kN/m2

Year 1996 14.

A venturimeter (throat diameter = 10.5 cm) is fitted 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 30º (Figure). For a manometer reading of 9.5 cm of mercury, find the flow rate. Discharge coefficient of venturi is 0.984.

Dt

From venturi Ds

x

Pa

k

e

2

(a) U / 2 k (b)

e

U 2 / 8k

j

y

 D 2s

j FGH DD  1IJK  D 2 2 t

9.5 cm

Water Hg

30º 2 s



102

Chapter-6 Answers 1. Ans. (b) 6. Ans. (c) 11. Ans. (b)

Space for Rough work

2. Ans. (a) 7. Ans. (c) 12. Ans. (d)

3. Ans. (c) 8. Ans. (d) 13. Ans. (c)

4. Ans. (c) 9. Ans. (b) 14. Ans. (0.0302 m3 /s)

5. Ans. (b) 10. Ans. (d)

103

Chapter-6 Answer & Explanations Q.1

Ans (b)

1

1 0.5m

2

2

Applying Bernoulli’s equation at section (1-1) & (2-2) P1 V12 P V2   Z1 = 2  2  Z2 g 2g g 2g

......(i)

P 1 = P2 = Patm. (taking section 2-2 as datum) From equation (i)

22 V2  0.5 = 2 2  9.81 2g

V 2 = 3.716 m/sec. From Continuity equation, 1A1V1 = 2A1V2 (since flow is incompressible, i.e. 1 = 2) A1V 1 = A2V 2 A2 =

A1V1 V2

 d12  V1  2 d2 = 4 V2 4

Therefore, d2 =

V d12  V1 d1 1 = V2 V2

= 0.02

Q.2

2 = 0.01467 m  15 mm 3.716

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-306, Ex.7.3. Ans (a) Applying Bernoulli’s equation, just before the exit from the tank and just after entry in the atmosphere P1 V12 P V2   Z1 = 2  2  Z2 3 g 2g 3g 2g

......(i)

104

h1

1

h2

2

h3

3

From the above figure, it is clear that Z 1 = Z2, V1 = 0 and P2 = Atmospheric pressure = 0; Then the Bernoulli’s equation reduces to :

P1 V2 = 2 3 2 V2 =

2P1 3

......(ii)

From given figure we can find pressure P1 P 1 = 1 gh1 + 2 gh2 + 3 gh3 Substitute this value of P1 in eqution (ii), we get V2 =

2g 1h1  2 h 2  3h 3  3

 h  h  2gh 3  1  1 1  2 2   3 h 3 3 h 3  Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-277, Eqn 7.24. Ans. (c)

=

Q.3

a

Flow

10 mm w Given that Density of air, Density of water,

 a = 1.2 kg/m3,  w = 1000 kg/m3 x = 10 mm, g = 9.8 m/s2

Now

v2 h = 2g



v =

2gh

105

where,

 w   1 h = x  a 

 3  1000  1  8.32 m = 10  10   1.2 

 Velocity of air, v = Q.4

2  9.81  8.32 = 12.8 m/s

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-294, 295. Ans. (c) Given: At section S1 : P 1 = 50 kPa At section S2: P2 = 20 kPa Z1 = 10 m Z2 = 12 m V1 = 2 m/s V2 = 2 m/s S2 S1 P2 = 20 KPa P1 = 50 KPa

12 m

10 m

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.

P1 Total piezometric head at S1,H1 = g  Z1 50  103  10 = 15.096 m = 1000  9.81

P2 Total piezometric head at S2 , H2 = g  Z2 =

Q.5

20  103  12 = 14.038 m 1000  9.81

Since H1 > H2 therefore flow direction is from S1 to S2. Therefore, head loss = H1 – H2 = 15.096 – 14.038 = 1.06 m Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-285, 7.10. 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 to fall below the vapour pressure of the liquid)

106

1 2

2 1

From continuity equation a1 V1 = a2 V2  2  d1 V1 = d 2 2 V2 4 4 2 d12 V1 = 0.2 V1 = 4V V2 = 2 1 d2 0.12 Applying Bernoulli’s equation between at section (1-1) and (2-2)

P1 V2 P V2  1  Z1 = 2  2  Z2 g 2g g 2 g 2

50 V22 150 V1   = 5 2g 5 2g 16 V12 2g Therefore, V1 = 5.114 m/s Therefore, maximum discharge, Q =  r12  V1 =  × 0.12 × 5.114 = 0.161 m3/s Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-347, Ex. 8.7. Ans. (c) In forced vortex flow when steady state is reached the liquid attains equilibrium condition in this position and it 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 at a 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. Ans. (c) P vP = 10 

Q.6

Q.7

h1

1 h2 z1 2

Applying Bernoulli’s equation between section (1) and (2)

Now

P1 V2 P V2  1  Z1 = 2  2  Z2 g 2g g 2 g P 1 = P2 = Patm and Z2 = 0, (taking point 2 as datum) Z1 = (h2 – h1), V1 = 0

107

Thus from Bernoulli’s equation

Patm P V2  0  (h2  h1 ) = atm  2  0 g g 2 g 

h2 – h1 =

V22 2g

 V2 = 2 g (h2  h1 ) As area of siphon is constant, therefore velocity of flow is same

Q.8

Hence, VP Reference: Hydraulics and Ans. (d) Given that D1 D2

= V2 = 2 g (h2  h1 ) Fluid Mechanics, Modi and Seth, 17th Edition, Page-274, 7.17.

= 40 mm = 20 mm 1 2

40 mm

D2

D1

20 mm

2 1

From continuity equation,

A1 V1 = A2 V2 V2 =

A1 V1 A2 2

2  D1  40 =   V1 =    V1 = 4V1  D2   20  Now applying Bernoulli’s equation in between the sections 1-1 and 2-2

P1 V12 P V2   z1 = 2  2  z 2 ( Since pipe is horizontal, hence Z = Z 2 1) g 2g g 2g

P1  P2 g

=

V22  V12 2g

=

(4V1 ) 2  V12 15V12  2 2

Since, P1 – P2 = 30 kPa 15V12 30  103  = 1000 2 60  103 4 1000  15 Therefore, flow velocity V1 = 2 m/sec Or Theoretical discharge (Qth) through a venturimeter is given by

V1 2 =

Q th =

A1A 2 A12  A 22

 2gh

108

A1 V1 =

A1A 2 2gh A12  A 2 2 2gh

V1 =

2gh

=

2

4

 A1    1  A2 

 d1    1  d2 

2  9.81  (30 / 9.81)

=

4

= 2 m/sec.

 0.04    1  0.01 

Q.9

Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 241, 6.7.1. Ans. (b) Given,

Vr = –

60  103 300  103 m / sec and V = m / sec  2 r 2 r

Vr 1  60    V =  300  5

Now

D

V



vr

A C B r  r

V 5

Therefore,

Vr = –

Also

V = r. = r.



dr r.d =  dt 5dt



 d dr =  5 120 r 0

d dr and Vr = dt dt

r

 



r = [ln r] 120

1 (   0) 5

 r   ln   =  120 5  

r = 0.5336 120 Therefore, r = 120 × 0.5336 = 64.03 m Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 147, 5.6.1.



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 of the 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 = gR2 H ......(i) R

H

dr R r

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 V 2 (r)2  r2 = = r r r Integrating the above equation, 2  p =    r  r

w 2  r 2 2 Now Force on elementary ring is = pressure intensity × area of elementary circular ring dF = p × 2rdr

p =

R

R

Total force on the top of the cylinder, FT =  dF    0

0

......(ii)

w 2r2 .2rdr . (p from equation (ii) 2

R

 r4   2 4 2 F T =  w   =   R 4 4  0 Now, total thrust at bottom of cylinder is given by adding the equation (iii) and (i) 2 = w

R 4  g  R2H 4

......(iii)

110

Q.11

2 2  2  w R  R   g H  =  4  Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 180, 5.30. Ans. (b)

Pump

1 0.2 m 1

5m

2 2

4 km

Applying Bernoullis’s equation at the section (1-1) and section (2-2) P1 V12 P V2   Z1 = 2  2  Z2  h f g 2g g 2g Since pipe is horizontal, therefore Z1 = Z2 From question, V2 = 0, V1 = 2 m/s

......(i)

f LV 2 Head loss due to friction in the pipe is given as hf = 2g  d where, f, V, L are the friction factor, mean velocity and length of the pipe respectively. 0.01  4000  22 = = 40.774 m 2  9.81  0.2

from equation (i),

P1 P2 V12  h  f g = g 2g

22 = 45.57 m 2  9.81 Therefore, P 1 = 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. Ans. (d)

= 5 + 40.774 –

Q.12

D

U

Dt

Ds

x

Pa

k

From continuity equation at the throat and at the exit of the venturimeter A1 V1 = A2 V2  D2  A1V1   .V1 V2 = A 2 =  D 2t  Now applying Bernoullis’s equation at the throat and at the exit of the venturimeter P1 V12 P V2   Z1 = 2  2  Z2 g 2g g 2g Since venturi is horizontal, therefore Z1 = Z2

......(i)

111

P1  P2 g

V22  V12 = 2g

 2 V12  V 1  2  P1 – P2 = 2  2  V2   2  D4  P1 – Patm. = U 1  4  2  Dt  At throat velocity is greater than U, hance pressure will be less than atmospheric

P1 = –

 2  D4  U 1  4  2  Dt 

  2  D4 U  4  1 2  Dt  Now spring is elongated due to lower pressure at throat. Therefore, in equilibrium, Spring force = Pressure Force

=

Hence,

kx =

=

Q.13

 2 Ds (P1) 4   Ds2    U 2  D 4  4  1 4 2  Dt 

 U 2  D 4 2  4  1  Ds Hence, x = 8k  D t  Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 242, 6.6. Ans. (c) d/2 2

2

1

2m 1 d

From continuity equation, A1 V1 = A2 V2 2

d  2 d .2 =   .V2 42 4

2d 2  4  8 m / sec d2 Now Applying Bernoulli’s theorem at section (1-1) & (2-2) V2 =

P1 V12 P V2   z1 = 2  2  z 2 g 2g g 2g

......(i)

112

From question, P 1 = 200 kN/m2 = 200×103 N/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 2

P2 88  2  0 = 200  103  2  1000  9.81 2  9.81 1000  9.81 2  9.8 P2  3.26  2 9810 P 2 = 9810 (20.594 – 5.26) = 150426.5 N/m2 Therefore, pressure at the exit of the contraction, P 2 = 150.4 kN/m2 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-274, 17.16. Ans. (0.0302 m3/s) Internal diameter, D1 = 21.0 cm = 0.2 m; 20.39 + 0.204 =

Q.14

Area of inlet,

A1 =

  D 22 4

  (0.21) 2  0.0346 m 2 4 = 10.5 cm = 0.105 m

D1 2 = Throat diameter,

D2

   D 22 = × (0.105)2 = 0.0087 m2 4 4 Coefficient of discharge of venturi, Cd = 0.984  Area at throat,

Pressure head,

A2 =

 SHg   1 h = y   Swater 

13.6   1 = 59.85 cm = 0.5985 m = (9.5 sin 30º)   1  Discharge (Q) through a venturimeter is given by: Q = Cd

A1A 2 A12  A 22

= 0.984 ×

 2gh

0.0346  0.0087 (0.0346)2  (0.0087)2

 2  9.81  0.5985

= 0.984 × 0.008945 × 3.427 = 0.0302 m3/s Reference: Fundamentals of Fluid Mechanics, B. R. Munson, Edition 2010, Page-441, 8.37.



113

7 Dimensional and Model Analysis

Year 2010

Year 2002

1.

A phenomenon is modeled using ‘n’ dimensional variables with ‘k’ primary dimensions. The number of non-dimensional variables is (a) k (b) n (c) n – k (d) n + k

4.

2.

Match the following

Year 1997

P : Compressibleflow Q : Freesurfaceflow R : Boundary layer flow

U : Re ynolds number V : Nusselt number W : Weber number

S : Pipeflow T : Heat convection

X : Froude number Y : Mach number Z : Skin friction coefficient

(a) (b) (c) (d)

P-U; Q-X; R-V; S-Z; T-W P-W; Q-X; R-Z; S-U; T-V P-Y; Q-W; R-Z; S-U; T-X P-Y; Q-W; R-Z; S-U; T-V

5.

3.

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 is decreased by 40% keeping fluid the same? (a) 1200 (b) 1800 (c) 3600 (d) 200

Year 1994 6.

Year 2007

If there are ‘m’ physical quantities and ‘n’ fundamental dimensions in a particular process, the number of non-dimensional parameters is (a) m + n (b) m  n (c) m – n (d) m/n

The ratio of inertia forces to gravity forces may be expressed as square of non-dimensional group known as.........

Consider steady laminar incompressible axi-symmetric developed viscous flow through a straight circular 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 1 5

(a) 5

(b)

(c) 0

(d) 



Chapter-7 Answers 1. Ans. (c) 2. Ans. (d) Ans. (Froude Number)

Space for Rough work

3. Ans. (a)

4. Ans. (c)

5. Ans. (b)

115

Chapter-7 Answer & Explanations Q.1

Q.2

Q.3

Ans. (c) Buckingham’s -theorem states that if there are n total dimensional variables (dependent as well as independent variables) involved in a phenomenon which can be completely described by m fundamental dimensions (such as mass, length, time etc.), and are related by a dimensionally homogeneous equation, then the relationship among the n quantities can be expressed in of exactly (n – m) dimensionless and independent . Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b). 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. Ans. (a) Reynolds number is defined as the ratio of inertia force and viscous force. Re =

Q.4

Q.5

Inertia force 5 Viscous force

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-767, 17.11 (b). Ans. (c) Buckingham’s -theorem states that if there are n total dimensional variables involved in a phenomenon which can be completely described by m fundamental dimensions (such as mass, length, time etc.), and are related by a dimensionally homogeneous equation, then the relationship among the n quantities can be expressed in of exactly (n – m) dimensionless and independent . Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b). Ans. (b) Exp. Reynold’s number, Re =

vd 

vd  when new diameter = 1.2 d then new velocity = 0.6v

2500 =

  1.2d  0.6v  = 0.72× Re = 0.72 × 2500 = 1800 Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-325, Equation 7.15. Ans. (Froude Number)

Reynold’s number, Renew =

Q.6

......(i)

Reynolds number, Re =

Froude number, Fr =

VL Vd Inertia force = or   Viscous force

Inertia force = Gravity force

V Lg

116

Euler number, Eu =

Weber number, We =

Inertia force = Pressure force

V p/

Inertia force = Surface tension force

V  / L

V Inertia force = K / Elastic force Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357. Mach number, M =



117

8 Boundary Layer Theory

The velocity profile is uniform with a value of u0 at the inlet section A. The velocity profile at section B downstream is

Year 2012 1.

An incompressible fluid flows over a flat plate with 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

y  Vm , 0 y    u Vm ,   y  H    Hy Vm , H yH  

3.

The ratio

Year 2007 2.

(a)

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

(c)

(d) F > 1 4.

Linked Answer Questions : Q. 3 - Q. 4 Consider a steady incompressible flow through a channel as shown below. y

A

B

1  1  H

1    1    H

H  x

(d)

1  1   H

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

Vm

(b) 1

1 2 u 0 2

(c)

uo

1  1  2  H

The ratio p A  p B (where pA and pB are the

 u0

Vm is u0

  

2

1   2   1   H     

2

1

(b)

1     1   H     

1

(d)

1  1   H

2

Year 2002

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  = 1.0 kg/m3. Assume that the boundary layer is thin, two-dimensional, and follows a linear velocity

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 2 (c) x (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

 y

distribution, u = U   , at the section r-s, where  y is the height from plate. r

U q

Year 1994

U u



10.

s

p

flat plate

5.

6.

The mass flow rate (in kg/s) across the section q-r is (a) zero (b) 0.05 (c) 0.10 (d) 0.15

7.

Year 1993 11.

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

For air near atmosphere conditions flowing over a flat plate, the laminar thermal boundary layer is thicker than the hydrodynamic boundary layer. (True/false)

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

For air flow over a flat plate, velocity (U) and boundary layer thickenss () can be expressed respectively, as 3

U 3 y 1 y 4.64x     ;  U 2  2    Re x

If the free stream velocity is 2 m/s, and air has kinetmatic viscosity of 1.5 × 10–5 m2/s and density of 1.23 kg/m3, the wall stress at x = 1m, is (a) 2.36 × 102 N/m2 (b) 43.6 × 10–3 N/m2 –3 2 (c) 4.36 × 10 N/m (d) 2.18 × 10–3 N/m2

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



119

Chapter-8 Answers 1. Ans. (c) 6. Ans. (c) 11. Ans. (b)

Space for Rough work

2. Ans. (d) 7. Ans. (c) 12. Ans. (c)

3. Ans. (c) 8. Ans. (d)

4. Ans. (a) 9. Ans. (c)

5. Ans. (b) 10. Ans. (False)

120

Chapter-8 Answer & Explanations Q.1

Ans. (c) As per Blasius result thickness of laminar boundary layer is given as  

Hence,

5x

 =

vx 

1  1 1 4 2 v Therefore, if the velocity of fluid is increase by four times then boundary layer thickness reduces by 1/2. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-525, 12.6. Ans. (d)

Therefore,  

Q.2

5x Re x

V

L/2

L/2

As we know that local drag coefficient is given by o 0.664  2 V Re x 2 or  o = 0.332 V2(Rex)–1/2 Now drag force on the front half is given by Cf =

L 2

F1 =



o

(B = width of Plate)

B dx

0 L 2

=

1

2  0.332  V (Re x ) 2 dx 0

Reynolds number is given as, Rex =

V x 

 V  = 0.332. V 2     

=

0.332  V 2 V 

L 2

L 1 2 2

1

 x 2 dx 0

1

 x 2 dx 0

121

L 1  12  2 x   L 2  = K    2K  1 2    2 0

where,

......(i)

0.332  V 2 V 

K =

Similarly, drag force on the rear half, L

F2 =



o

B dx

L 2

L

= K x L 2

1 2

1  1  2 L   2 dx = 2K  (L)       2  

1

Now required ratio,

Q.3

 L 2 2K  F1 2  1 F = 1 F2  1  2 L 2 K  L2       2  

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-540, Ex. 12.6. Ans. (c) Given: y

A

B 

u0

Vm

H 

uo

x

y  Vm , 0 y     Vm ,   y  H   u =   H y  Vm  , H    y  H    

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

122 

Qentry = u × H × B = u H o o Total volume flow rate leaving, 

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

y dy

u

Volume flow rate from mid section = Vm(H – 2) For boundary layer 0y   d QB = u . dy  d QB = Vm

y dy 

Integrating the above equation 



QB

Vm  Vm  y 2  Vm .  y dy = =   =   0   2 0 2

By symmetry for H –  y  H, Volume flow rate =

Vm .  2



Therefore,

Qentry = Q exit

uo H = Vm (H  2)  2  

uo H = Vm(H – )

Vm uo

H 1  H   1  H Reference: Fluid Mechanics, R.K. Bansal, 4th Edition, Page-655, 13.3. Ans. (a)



Q.4

Vm . 2

=

y

A

B 

u0

Vm

H 

uo

Applying Bernoulli’s equation at section A and B p ν2 p A vA2 + = B + B (Since  ρg 2g ρg 2g

p A - pB ν2 - ν2 = B A ρ 2

x

123



p A - pB Vm2 - u o2 = ρ 2 2





Q.5

 Vm  p A - pB 1 =  uo  1 2  ρ uo 2 1 p A - pB 1 = 2 1 2   ρ uo 1  H  2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-280, Equation 7.29. Ans. (b) Given:Free stream velocity, U = 10 m/s Boundry layer thickness,  = 10 mm Breadth of plate, B = 1m Density of air, = 1.0 kg/m3  y velocity distribution, u = U     r

U q

U u

p



s flat plate

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 × 10–3 × 10 = 0.1 kg/s Mass flow rate through the element dy at section r – s dm =  u B dy  y dm = BU   dy  Integrating the above equation gives, 

m =

=

BU BU  2  = 2  2

1  1  10  10  103 = 0.05 kg/s 2 Thus mass flow rate leaving across the section, q – r = 0.1 – 0.05 = 0.05 kg/s Ans. (c) Drag froce on the plate will be the rate of change of momentum of control volume qprs

=

Q.6

 BU y dy  0



124



Thus, momentum rate entering section q – p = mU = 0.1 × 10 = 1N Momentum rate leaving through section r–s 2





y =   u B dy × u =   BU   dy  0 0 2

=

 BU 2 3  BU 2   = 2 3 3

1.0  1.0  102  10  103 = 0.33 N 3 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) Reynold’s number at x,

Boundry layer thickness,

Rex =

Ux 2 1 = = 1.33 × 105 υ 1.5  10-5

 =

4.64x 4.64  1  = 0.0127 Re x 1.33  105

Since the velocity profile is given as,

U U

3 y 1 y    = 2  2  

3

 3 1 1 3y 2  dU U Velocity gradient, =  .   3  dy 2  2   Therefore,

3 1 dU at y  0 = . .U  2  dy

From Newton’s law of viscosity, o = 

dU at y  0 dy

3U  3 2 5 = 1.5  10  1.23  2 2  0.0127 = 435.82 ×10–5 N/m2 = 4.36 × 10–3 N/m2 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-537, Ex. 12.3. Ans. (d) = 

Q.8

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

5x Re x

Hence,

5x

 =

vx  1

Therefore,

  x 

Q.9

1 2

1 x2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-525, 12.6. Ans. (c) In direction of flow usually pressure gradient is negative i.e.

flow separation, pressure gradient is positive i.e.

Q.10 Q.11

P = -ve, s the fluid flow. But in case of x

P P = +ve, s the fluid separation & = 0, it means x x

that the fluid is on the verge of separation. Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-662, Equation-13.7. Ans. (False) Ans. (b) In a fluid flow over a flat plate, the dominant forces are inertia force and viscous force. Therefore Reynold’s number decides the nature of the flow.

Q.12

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-770, 17.13. 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.



126

9 Laminar and Turbulent Flow

Year 2013

The average velocity of fluid in the pipe is

1.

(a) 

R 2  dp    8  dx 

(b) 

R 2  dp    4  dx 

(c) 

R 2  dp    2  dx 

(d) 

R 2  dp      dx 

For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, the pressure drop p over a length L and the wall shear stress w are related by (a) w 

pD 4L

(b) w 

p D 2 4L2

(c) w 

p D 2L

(d) w 

4p L D

Year 2007 4.

Year 2010 2.

The maximum velocity of a one-dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is 6 ms–1. The mean velocity (in ms–1) of the flow is (a) 2 (b) 3 (c) 4 (d) 5

Consider steady laminar incompressible axi-symmetric developed viscous flow through a straight circular pipe of constant cross-sectional area at a Reynold’s number of 5. The ratio of inertia force to viscous force on a fluid particle is (b)

(c) 0

(d) 

Year 2009

Year 2006

3.

5.

The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression

where

x

4r 2 

u = u 0 1  2  , where r is the radial distance  D  from the center. If the viscosity of the fluid is , the pressure drop across a length L of the pipe is

dp is a constant dx

r

The velocity profile in fully developed laminar flow in a pipe of diameter D is given by 

R 2  dp   r2  1  u(r) =     4  dx   R 2 

1 5

(a) 5

u(r)

(a)

 u0 L D2

(b)

4 u 0 L D2

(c)

8 u 0 L D2

(d)

16 u 0 L D2

R

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 crosssectional 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.

Prandtl’s 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



128

Chapter-9 Answers 1. Ans. (a) 6. Ans. (d)

Space for Rough work

2. Ans. (c) 7. Ans. (c)

3. Ans. (a) 8. Ans. (mean velocity)

4. Ans. (a) 9. Ans. (two)

5. Ans. (d) 10. Ans. (a)

129

Chapter-9 Answer & Explanations Q.1

Ans. (a)

 dp  r In a pipe flow,  =    =  dx  2

 dp  d –   dx  4

 dp  D Therefore, shear stress at wall, w =  –   dx  4 Above equation can be written as, w = Q.2

pD 4L

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-547, Eqn.-13.3. Ans. (c) For the flow of fully developed between two fixed parallel plates

y B Vmax x The velocity distribution for laminar flow between fixed parallel plates is given as,

Flow velocity is maximum when

V =

1  p     (By – y2) 2  x 

y =

B 2

B 2  p  Therefore, V max =   8  x  Average flow velocity is obtained by dividing the total discharge with cross sectional area. Hence,

Therefore,

Vmax 3 = Vavg 2

2 2  Vmax. =  6  4 m/s 3 3 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-557, Equation 13.389. Ans. (a) So mean velocity,

Q.3

B2  p  Vavg. =   12  x 

Vavg =

R

r

u(r)

x

Consider an element ring of thickness dr at a radius of r. Therefore, element discharge from this ring, dQ = (2r).dr. u(r)

dr r

130

R

Therefore, total discharge Q =

 d Q =  (2r ) . 0

= 2

R2   P   r2     1  2  .dr 4  x   R 

R R2   P   r3      r  2  dr 4  x  0  R 

R

R2   P  R2 R4  R 2   P   r2 r4  2    = 2 =         4  x   2 4 R 2  4  x   2 4 R 2  0 Total discharge, Now,

R4   P    Q = 8  x  Q = Area × Average velocity

Area × Vavg. R2

× Vavg

R4   P    = 8  x  R4   P  =   8  x 

R2   P  R2   P    =    8  x  8  x  Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-550, Eqn-13.14. Ans. (a) Therefore,Vavg =

Q.4

Inertia force vd  Viscous force  Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357. Ans. (d) Given: Fully developed laminar flow velocity profile in a circular pipe is given by

Reynold’s number, Q.5

Re =

 4r 2  u = u o 1  2  ......(i)  D  From Hagen-Poiseuille equation the pressure loss in fully developed laminar flow across the length of the pipe is given as 32μu L ......(ii) d2 The ratio of maximum velocity and average velocity in case of fully developed laminar flow through a circular p1 – p 2 =

pipe is 2 i.e.

uo =2 u

Therefore, from equation (ii), p1 – p2 =

uo L 16μu o L 2 = d2 d2

32μ

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

 4r 2  u u = o 1  2   D 

......(i)

131

R r

 P  r  =    x  2

Now

From Newtons law viscosity,  = 

......(ii)

du dy

Here y = R–r Therefore, dy = – dr Putting the value of dy in equation (i), we get  u   =     r 

From equation (ii) and (iii),





= 

P r . x 2

1P r u  =    x  2 r



From equation (i) ,

u r

......(iii)

1  P  r    4r 2     u o 1  2   =    x  2 r   D  

1  P  r 8u o r =   2   x  2 D



P 16 u o =  x D2



P = 

16u o x D2

P2

Integrating over a length of L,

L

16 u o dx D2 o

 dp =  

P1

p2 – p 1 = 

16 u o L D2

16 u o L D2 Reference: Hydraulics and Fluid mechanics, Modi and Seth, 17 Edition, Page-548. Equation-13.3, 13.6, 13.7. Ans. (d) In flow through pipe the transition from laminar to turbulent depends upon Reynolds number which is given as Therefore, p2 – p1 =

Q.6

vd where, d is the characteristic dimension of the pipe.  Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357.

Re =

132

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 – p 2 =

Q.8

p1  p 2 32μu = L d2 Therefore, 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. Ans. (mean velocity) The head loss due to friction in fully developed laminar flow in a circular pipe is given as hf =

=

Q.9

P1  P2 g 32 u avg.L  g d2

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 Fluid Mechanics, Modi and Seth, 17th Edition, Page-550, Equation13.18. Ans. (two) In fully developed laminar flow through circular pipe of radius R mean velocity and maximum velocity are given as

Therefore,

Q.10

32μu L d2

Vmean =

1  p  2  R 8  x 

Vmax. =

1  p  2   R 4  x 

Vmax Vmean = 2

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-550, Equation13.15. Ans (a) From Prandtl’s hypothesis Prandtl mixing length, l = y where  = a constant of proportionality known as ‘Karman universal constant’ and y is the distance from the wall. Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-507, Equation-11.5.



133

10 Flow Through Pipes

Year 2012 Year 2004 1.

Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m. The volumetric flow rate is 0.2 m3/s. The head loss (in m) due to friction is (assume g = 9.81 m/s2) (a) 116.18 (b) 0.116 (c) 18.22 (d) 232.36

4.

(a)

Year 2009 2.

Water at 25°C is flowing through a 1.0 km long G.I. pipe of 200 mm diameter at the rate of 0.07 m3/s. If value of Darcy friction factor for this pipe is 0.02 and density of water is 1000 kg/m3. The pumping power ( in kW) required to maintain the flow is (a) 1.8 (b) 17.4 (c) 20.5 (d) 41.0

Year 2007 3.

In a steady flow through a nozzle, the flow velocity

3x on the nozzle axis is given by v  u 0 1   i, L  where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is L (a) u0 (c)

L 4u 0

For a fluid flow through a divergent pipe of length L having inlet and outlet radii of R 1 and R 2 respectively and a constant flow rate of Q, assuming the velocity to be axial and uniform at any cross-section, the acceleration at the exit is:

L In (4) (b) 3u 0 (d)

L 2. 5u 0

(c)

5.

b

2Q R1  R 2

g

(b)

LR 32

b

2Q 2 R1  R 2 

2

LR 52

g

(d)

b

2Q 2 R1  R 2

g

LR 32

b

2 Q 2 R 2  R1 

2

g

LR 52

The following data about the flow of liquid was observed in a continuous chemical process plant

Flow rate 7.5 7.7 7.9 8.1 8.3 8.5 (litres / sec.) to to to to to to 7.7 7.9 8.1 8.3 8.5 8.7 Frequency 1 5 35 17 12 10 Mean flow rate of the liquid is (a) 8.00 litres/sec (b) 8.06 litres/sec (c) 8.16 litres/sec (d) 8.26 litres/sec

Year 2003 Linked Data Question 6 and 7 A syringe with a frictionless plunger contains water and has its end a 100 mm long needle of 1 mm diameter. The internal diameter of syringe is 10 mm. Water density is 1000 kg/m3. The plunger

is pushed in at 10 mm/sec and the water comes out as jet 10 mm

10 mm/sec F

needle

water jet

1 mm Syringe

100 mm

6.

Assuming ideal flow, the force F in newtons required on the plunger to push out the water is (a) 0 (b) 0.04 (c) 0.13 (d) 1.15

7.

Neglect losses in the cylinder and assume fully developed laminar viscous flow throughout the needle; the Darcy friction factor is 64/Re, where Re is the Reynolds number. Given that the viscosity of water is 1.0  10–3 kg/s m, the force F is newtons required on the plunger is (a) 0.13 (b) 0.16 (c) 0.3 (d) 4.4

Year 1998 8.

The dicharge velocity at the pipe exit in figure is:

(a)

2 gH

(c)

g Hh

b

(b)

g

2 gh

(d) 0

Year 1994 9.

Fluid is flowing with an average velocity of V through a pipe of diameter d. Over a length of L, the “head” loss is given by

fLV 2 . The friction 2gd

factor ‘ f ’ for laminar flow in of Reynolds number (Re) is.......



135

Chapter-10 Answers 1. Ans. (a) 6. Ans. (b)

Space for Rough work

2. Ans. (b) 7. Ans. (c)

3. Ans. (b) 8. Ans. (b)

4. Ans. (c) 64 9. Ans.( ) Re

5. Ans. (c)

136

Chapter-10 Answer & Explanations Q.1

Ans. (a) Head loss due to friction in flow through pipe is given as, head loss hL = where,

L d V f hL

Therefore,

Q.2

= = = = =

Length of the pipe Diameter of the pipe Mean velocity of flow Friction factor Head Loss due to friction

0.2   0.0225  500    2    0.2 / 4  hL = 2  9.81 0.4

2

= 116.18 m

Reference: Fluid Mechanics, R.K. Rajput, 2005 Edition, Page-536, Equation-12.3.1. Ans. (b) Given: Pipe length, L = 1.0 km = 1000 m Pipe diameter, D = 200 mm = 0.2 m Flow rate, Q = 0.07 m3/s Friction factor, f = 0.02 Density of water,  = 1000 kg/m3 From Darcy Weisbach equation, head loss due to friction in pipe is given by  Q  f L  2  D  2 f LV 4   hf = 2g D 2g D

=

Q.3

fLV 2 2gd

2

8  0.02  0.07 2  1000 8 f Q2L = = 25.304 m 2  9.81  0.25 2 g D5

Therefore, pumping power to overcome this loss, P = w.Q.hf = g.Q.hf = 1000 × 9.81 × 0.07 × 25.304 = 17376.26 W = 17.4 kW Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-459, Equation11.2 Ans. (b) Given :

x t=0 x=0  3x  Equation of motion, v = u 0 1    L

t=T x=L

137

dx 3x   = u 0 1    L dt

or

dt =

Integrating both side,

 3x  u 0 1    L

T

L

 dt =



0

T

 t 0

0

dx  3x  u 0 1    L

L = 3u 0

T = Q.4

dx

L

  3x    ln  1  L    0

L ln (4) 3u 0

Ans. (c)

R1

R2 Flow direction

L

Inlet velocity,

V1 =

Q  R12

Outlet velocity,

V2 =

Q  R 22

Acceleration =

Since,

dv dv dx vdv × = = dt dx dt dx

dv V2 -V1 Q  1 1  = =  -  dx L π L  R 22 R12 

dv Q Q  R12 -R 22  =  Acceleration at exit = V2 dx π R 2 L π L  R 2 R 2  2  1 2 Q2  (R1 -R 2 )(R1 + R 2 )   = π2 R 2 L  R12 R 22 2   Assuming velocity axial and uniform at any cross section i.e. V1 = V2 Therefore, R1 = R2 Hence acceleration at the exit =

2Q 2 (R 1 – R 2 ) π 2 L R 52

138

Q.5

Ans (c)

Flow rate

 Q.6

Mean value

frequency

fx

of flow rate (x) (f )

7.5  7.7 7.7  7.9

7.6 7.8

1 5

7.6 39

7.9  8.1 8.1  8.3

8.0 8.2

35 17

280 139.4

8.3  8.5 8.5  8.7

8.4 8.6

12 10

100.8 86

f  80

fx  652.8

Mean flow rate =

fx 652.8  8.16 litres/sec. = f 80

Ans. (b) 1 2

10 mm/sec

2 1

At section 1–1 & 2–2 from continuity equation for incompressible flow A1V 1 = A2V 2 A1 d12 .V  .V1 V2 = 1 A2 d 22 2

V2 =

 0.01  0.01  1m / sec 2  0.001

Now applying Bernoulli’s theorem at sections (1-1) and (2-2) P1 V12 P V2   z1 = 2  2  z 2 g 2g g 2g

P1 =





 (0.01) 2 = 0.04 N 4 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-343, Ex. 8.3. Ans. (c) = 499.95 ×

Q.7

(Also P2 = 0, atmospheric pressure)

1000 2 1  0.012  499.95 N / m 2 2 = P 1 .A 1

P1 = Force on plunger required

 2 (V2  V12 ) 2

Viscosity,

µ = 1 × 10–3 kg/s.m

139

m  s2 1  kg. = 1 × 10  . . s 2  m s.m  –3

= 1 × 10–3 Ns/m2 We know, Reynold’s number

Re =

Darcy friction factor, f = Head loss in needle,

V2 d 2 1000  1  0.001  1000 =  1  103

64 64   0.064 Re 1000

f .L.V22 0.064  0.1  12 hf = = 2gd 2  9.81  0.001

= 0.326 m of water 1 2

10 mm/sec

2 1

Applying Bernoulli’s equation at sections (1-1) and (2-2) P1 V12 P2 V22     hf = g 2g g 2g

P1 =

 2 V2  V12  g.h f 2





1000 2 (1  0.012 )  1000  9.81 0.326 2 = 3702.9 N/m2 = P1 × A1 =

Now Force required

 (0.01)2 = 0.3 N 4 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-347, Ex. 8.7. Ans. (b) = 3702.9 ×

Q.8

Applying Bernoulli’s equation just before the entry into the pipe and just after exit from the pipe P1 V12 P V2   z1 = 2  2  z 2 g 2g g 2g Here, V 1 = 0, P2 = Patm. = 0, z1 = 0, z2 = H – h (taking 1-1 as datum) Therefore, the above equation reduces to H =

V22 Hh 2g

140

V2 =

2gh

Or At exit, figure shows that height of liquid level above the centre of pipe is h. Total potential energy = Total kinetic energy m.g.h =

1 2 mv 2

V =

2gh

Reference: Fluid Mechanics, R.K. Rajput, 2005 Edition, Page-234, Section-6.6.3. Q.9

Ans.

64 Re

64 Re Head loss in laminar flow through a circular pipe is given by

In laminar flow friction factor, f =

f Lv 2 (a) Darcy weisbach equation = 2gd (b) Hagen - Poisullie equation =

32v L gd

By equating the above two equations, we get,

64 Re whereas, in turbulent flow friction factor, f =

0.3164 f = (Re)1/4 Reference: Fluid Mechanics, R.K. Rajput, 2005 Edition, Page-536, Equation-12.3.1.



141

11 Hydraulic Turbines

Year 2013 1.

Deflected jet

In order to have maximum power from a Pelton turbine, the bucket speed must be (a) equal to the jet speed (b) equal to half of the jet speed (c) equal to twice the jet speed (d) independent of the jet speed

120° Incoming jet

120°

Deflected jet

Year 2010 2.

A hydraulic turbine develops 1000 kW power for a head of 40 m. If the head is reduced to 20 m, the power developed (in kW) is (a) 177 (b) 354 (c) 500 (d) 707

(a) 0 (N.m)/(Kg/s) (c) 2.5 (N.m)/(Kg/s)

Year 2007 4.

Year 2008 3.

Water having a density of 1000 Kg/m3, issues from a nozzle with a velocity of 10 m/s and the jet strikes a bucket mounted on a Pelton wheel. The wheel rotates at 10 rad/s. The mean diameter of the wheel is 1m. The jet is split into two equal streams by the bucket, such that each stream is deflected by 120° as shown in the figure. Friction in the bucket may be neglected. Magnitude of the torque exerted by the water on the wheel, per unit mass flow rate of the incoming jet, is

(b) 1.25 (N.m)/(Kg/s) (d) 3.75 (N.m)/(Kg/s)

5.

The inlet angle of runner blade of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant throughout the blade age and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is (a) 25 % (b) 50 % (c) 80 % (d) 89 % A model of a hydraulic turbine is tested at a head of

1 th of that under which the full scale turbine 4

works. The diameter of the model is half of that of the full scale turbine. If N is the RPM of the full scale turbine, then the RPM of the model will be (a)

N 4

(c) N

(b)

N 2

(d) 2N

Year 2006

Year 1997

6.

In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0.1 m3/s. If the jet deflection angle is 120º and the flow is ideal, the power developed is (a) 7.5 kW (b) 15.0 kW (c) 22.5 kW (d) 37.5 kW

10.

7.

A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1 : 4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW) will be (a) 2.34 (b) 4.68 (c) 9.38 (d) 18.75

Year 2004 8.

At a hydroelectric power plant site, available head and flow rate are 24.5 m and 10.1m 3 /s respectively. If the turbine to be installed is required to run at 4.0 revolution per second (rps) with an overall efficiency of 90%, the suitable type of turbine for this site is (a) Francis (b) Kaplan (c) Pelton (d) Propeller

9.

Match the following P. Reciprocating pump Q. Axial flow pump R. Microhydel plant S. Backward curved vanes 1. Plant with power output below 100 kW 2. Plant with power output between 100 kW to 1 MW 3. Positive displacement 4. Draft tube 5. High flow rate, low pressure ratio 6. Centrifugal pump impeller (a) P-3, Q-5, R-6, S-2 (b) P-3, Q-5, R-2, S-6 (c) P-3, Q-5, R-1, S-6 (d) P-4, Q-5, R-1, S-6

Kaplan turbine is (a) a high head, mixed flow turbine (b) a low head, axial flow turbine (c) an outward flow reaction turbine (d) an impulse inward flow turbine



143

Chapter-11 Answers 1. Ans. (b) 6. Ans. (c)

Space for Rough work

2. Ans. (b) 7. Ans. (a)

3. Ans. (d) 8. Ans. (a)

4. Ans. (c) 9. Ans.(c)

5. Ans. (c) 10. Ans. (b)

144

Chapter-11 Answer & Explanations Q.1

Ans. (b) Pelton wheel is a tangential flow impulse turbine. It was invented by Lester A. Pelton, an American engineer in 1870. It is preferably used in case of high head and low volume flow rate. V1 = Vw1 u1

Vr1 u

Vr2

Power developed

=

V2





u2

Vw2

Vf2

wQ  Vw1  Vw 2   u g 

......(i)

From exit velocity triangle, Vw2 = Vr2 × cos  – u2 = Vr1 × cos  – u2 (If friction in the runner bucket is neglelected, then Vr2 = Vr1 ) = (V1 – u1) cos  – u2 (since in Pelton turbine u1 = u2 = u ) From equation (i), Power developed

wQ  V1  (V1  u)cos   u   u g

=

=

wQ wQ (V1 .u  u 2 )(1  cos )   (V1  u)(1  cos )  u = g g 

2 = K.(V1 .u  u ) where K =

Therefore,

Q.2

......(ii)

d (Power developed) =0 du

For maximum power,

From equation (ii),

wQ (1  cos ) g



 =0

d K(V1u  u 2 ) du V1–2u = 0 u =

V1 2

For maximum power from a Pelton turbine the bucket speed ‘u’ must be equal to half of the jet speed, V1 . Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-924, Equation21.13. Ans. (b) Given : P 1 = 1000 kW, H1 = 40 m, H2 = 20 m The power developed by a turbine, working under a head of one meter is called unit power of the turbine.

145

Unit power is defined as, Pu =

P

......(i)

3

H2 3

P  H1  2 Therefore from equation (i), 1 =   P2  H2  3

3

 H2  2  20  2 P2 =    P1 =    1000 = 353.55 W  40   H1 

Q.3

Hence, power developed, P2 = 354 kW. Reference: Hydraulic Machines, Dr. Jagdish Lal, Edition 2004, Page-69, Equation 3.3. Ans. (d) Given: Water jet velocity, V1 = 10 m/s Diameter of wheel, D = 1m Radius of wheel, R = 0.5 m Angular speed,  = 10 rad/s Density,  = 1000 kg/m3 Clearance angle180o - 120o = 60o Tangential velocity, u1 = u2 =  × R = 10 × 0.5 = 5 m/s V1 = Vw1 u1

60 = 

Vr1 u

120°  Vr2

V2





u2

Vw2

Vf2

Power developed by Pelton wheel means the shaft power. Since there is no mechanical loss, hence shaft power is equal to the runner power. Runner power is equal to the work done per second by water on runner. Power developed = Q  Vw1  Vw 2   u = Q  Vw1  Vw 2     R     It can be written as power developed = torque × 

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

From equation (i) and (ii), torque = Q  Vw1  Vw 2   R

Therefore, torque exerted by water per unit mass flow rate =

Q  Vw1  Vw 2   R Q

=  Vw1  Vw 2   R From exit velocity triangle, Vw2 = Vr2 × cos  – u2 = Vr1 × cos  – u2 = (V1 – u1) cos  – u2 (since in Pelton turbine u1 = u2 = u ) = (10 – 5) × cos 60º – 5 = –2.5 m/s

......(iii)

146

From equation (iii), torque developed = 10  2.5 0.5 = 3.75 (N.m)/(kg/s) Q.4

Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-924, Eq. 21.10. Ans. (c) Given: Inlet and outlet velocity triangle of Francis turbine is shown in the figure below: Vw1 = u1   Vf1 = Vr1 V 1

u

Vr2 



V2 = Vf2

u2

V12 V22  actual conversion of kinetic head 2g 2g Blade efficiency = = kinetic head available at inlet of the turbine V12 2g

V12  V22 = V12

......(i)

From question it is given that Vf1 = Vf2 = u1 2 2

From inlet velocity triangle

V1

2

5 2  u1  = Vw1 + Vf1 = u1 +   =  u1 4  2 2

2

V  From equation (i) blade efficiency, = 1   2   V1  2

Q.5

2

 u1    2 = 1 –   = 0.8 = 80% 5 2  u1 4 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, Page-895, Equation-20.45, Fig. 20.9 Ans. (c) For complete similarity to exist between the model and prototype turbines, the following conditions must be satisfied.

 gH   gH   2 2 =  2 2  N D m  N D P Hm g HP 4 = 2 N 2P D 2P 2  Dm  Nm    2  g



147

Q.6

Therefore, NP = Nm Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-990, 22.17. Ans. (c) Given: Peripheral or tangential speed, u1 = u2 = 10 m/s Water jet velocity, V1 = 25 m/s Flow rate, Q = 0.1 m3/s Density,  = 1000 kg/m3 V1 = Vw1 u1

60 = 

Vr1 u

120°  Vr2

V2





u2

Vw2

Vf2

Power developed by Pelton wheel means the shaft power. Since there is no mechanical loss, hence shaft power is equal to the runner power. Runner power is equal to the work done per second by water on runner.

wQ  Vw1  Vw 2   u ......(i) g  = Vr2 × cos  – u2 (Vr2 = Vr1, since there is no friction) = Vr1 × cos  – u2 = (V1 – u1) cos  – u2 = (25 – 10) × cos 60º – 10 = –2.5 m/s

Therefore, power developed = From exit velocity triangle, Vw2

9.81  0.1  25  2.510 = 22.5 kW 9.81 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-924, Eq. 21.10. Ans. (a) Given: Power generated, Pp = 300 kW N p = 1000 rpm H p = 40 m Model testing head, Hm = 10 m, Scale is 1 : 4 Specific power is same for similar turbines

from equation (i), power developed = Q.7



 P Specific Power =  3  D2 H 2 

  P 3     D2 H 2 P  2

   m

3/ 2

2

3

 1   10  2 D  H  P m = PP   m    m  = 300      4   40   DP   HP  P m = 2.34 kW Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-990, Eq. 22.189.

Q.8

Ans. (a) Given, Available head, H Discharge, Q Rotation, N Overall effeiciency,0

= = = =

24.5 m 10.1 m3/s 4 rps = 4 × 60 rpm = 240 rpm 90% Shaft power Overall effeiciency,0 = Water power Shaft power = 0 × Water power

148

= 0 × (W × Q × H) = 0.9 × 9.81 × 10.1 × 24.5 = 2184.74 kW Specific speed plays an important role for selecting the type of the turbine. Also the performance of a turbine can be predicted by knowing the specific speed of the turbine. Specific speed is defined as the speed of the turbine which is identical in shape, geometrical dimensions etc. with actual turbine but of such size that it will developed unit power working under unit head. Therefore, specific speed Ns =

N P H5/ 4

......(i)

240  2184.74 = 205.80  206. (24.5)5 / 4 In equation (i), if P is taken in horse power (1 HP = 746 Watt) the specific speed is obtained in M.K.S. units. But if P is taken in kilowatts, the specific speed is obtained in S.I. unit. The type of turbine for different specific speed is given in following Table

=

S.No. 1. 2. 3. 4.

Specific Speed

(M.K.S.) 10 to 35 35 to 60 60 to 300 300 to 1000

(S.I.) 8.5 to 30 30 to 51 51 to 255 255 to 860

Type of Turbine

Pelton wheel with single jet Pelton wheel with two or more jets Francis turbine Kaplan or Propeller turbine

Since the specific speed of the turbine is 206, therefore suitbale turbine is Francis turbine. Q.9

Q.10

Reference: Fluid Mechanics and Hydraulic Machines, Dr. R.K. Bansal, Page-832, 18.1. Ans. (c) Reciprocating pump - It is most common positive displacement pump. The positive displacement pumps are those pumps in which the liquid is sucked and pushed due to the thrust exerted on it by a moving member, which results in lifting the liquid to the required height. Mycro-hydel Plant - The hydro plant producing up to 100 kW of power using natural flow of water. More frequently Pelton wheel is used in micro hydel power plant. Axial flow pump - An axial-flow pump consists of a propeller (an axial impeller) in a pipe. The main advantage of an AFP is that it has a relatively high discharge at a relative low head. For example, it can pump up to 3 times more water and other fluids at lifts of less than 4 meters as compared to the more common centrifugal pump. These pumps have the smallest of the dimensions among many of the conventional pumps and are more suited for low heads and higher discharges. In India, millions of smaller horsepower (6-15 HP) mobile units powered mostly by single cylinder Diesel and Petrol engines are used by smaller farmers for crop irrigation, drainage and fisheries. Backward curved vanes - A series of backward curved vanes/blades is mounted on the impeller of the centrifugal pump. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-1016, 1061, Section-23.1, 24.3. Ans. (b) It is an axial flow reaction turbine, which is suitable for relatively low head and hence requires a large quantity of water. It was developed by Austrian professor Victor Kaplan in 1913. The head ranges from 10 to 70 meters and used where power developed ranges from 5 to 120 MW. The difference of pressure or pressure drop between the inlet and the outlet of the runner is called reaction pressure, and hence thses turbines are known as reaction turbines. Thomson, Francis, Propeller and Kaplan are some important reaction turbines. Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-935, Section-22.18.



149

12 Centrifugal Pump

(a) 24 (c) 28

Year 2007 1.

Match the items in columns I and II Column I Column II P : Centrifugal 1 : Axial flow compressor Q : Centrifugal pump 2 : Surging R : Pelton wheel 3 : Priming S : Kaplan turbine 4 : Pure impulse (a) P – 2, Q – 3, R – 4, S – 1 (b) P – 2, Q – 3, R – 1, S – 4 (c) P – 3, Q – 4, R – 1, S – 2 (d) P – 1, Q – 2, R – 3, S – 4

(b) 26 (d) 30

Year 2003 3.

A centrifugal pump running at 500 rpm and at its maximum efficiency is delivering a head of 30 m at a flow rate of 60 litres per minute. If the rpm is changed to 1000, then the head H in metres and flow rate Q in litres per minute at maximum efficinecy are estimated to be (a) H = 60, Q = 120 (b) H = 120, Q = 120 (c) H = 60, Q = 480 (d) H = 120, Q = 30

Year 2006 Year 2002 2.

A horizontal-shaft centrifugal pump lifts water at 65ºC. The suction nozzle is one meter below pump centerline. The pressure at this point equals 200 kPa gauge and velocity is 3 m/s. Steam tables show saturation pressure at 65ºC is 25 kPa and specific volume of the saturated liquid is 0.001020 m3 /kg. The pump Net Positive Suction Head (NPSH) in meters is

1m

Common Data Question No. 4 & 5 A centrifugal pump has an efficiency of 80%. The specifications of the pump are : Discharge = 70 m3/hr, head = 7 m, speed = 1450 rpm and diameter = 2000 mm. If the speed of this pump is increased to 1750 rpm. 4.

Discharge and head developed are given respectively: (a) 84.48 m3/hr and 10.2 m (b) 48.8 m3/hr and 20 m (c) 48.8 m3/hr and 10.2 m (d) 58.4 m3/hr and 12 m

5.

Power input required is given by (a) 1.066 kW (b) 1.96 kW (c) 2.12 kW (d) 20 kW

Year 2000 6.

When the speed of a centrifugal pump is doubled, the power required to drive the pump will (a) increase 8 times (b) increase 4 times (c) double (d) remain the same

Year 1994 7.

In of speed of rotation of the impeller (N), discharge (Q) and change in total head through the machine, the specific speed for a pump is.........



151

Chapter-12 Answer 1. Ans. (a) 6. Ans. (a)

Space for Rough work

2. Ans. (c) N Q 7. Ans .( 3 / 4 ) Hm

3. Ans. (b)

4. Ans. (a)

5. Ans. (a)

152

Chapter-12 Answer & Explanations Q.1

Q.2

Ans. (a) Column I P. Centrifugal compressor Q. Centrifugal pump R. Pelton wheel S. Kaplan turbine Reference: Hydraulics and & 24.4. Ans. (c)

Column II 2. Surging 3. Priming 4. Pure impulse 1. Axial flow Fluid Mechanics, Modi and Seth, 17th Edition, Page-919 & 1061, 21.4

2 1m 1

NPSH: Net positive suction head is defined as the absolute pressure head at the inlet to the pump minus the vapour pressure head (in absolute units) corresponding to the temperature of the liquid pumped, plus the velocity head at this point Or NPSH may be defined as the head required to make the liquid to flow through the suction pipe to the impeller. 2  P2 Pa  PV VS    Thus, NPSH =  ...................... (i)   g g  g 2 g

where, P2 Pv Pa Vs

= = = =

Gauge pressure at the inlet of pump Vapour pressure of the liquid in absolute unit Atmospheric pressure (101.325 kPa) Velocity of flow in suction pipe = 3 m/s

At 65º C, Vapour pressure, P v = 25 kPa Specific volume, w = 0.001020 m3/kg Therefore, mass density of water at 65ºC

1 1  = 980.39 kg/m3 w 0.001020 Now given that at suction nozzle (at point 1) P 1 = 200 kPa (Gauge pressure) V s = 3 m/s Applying Bernoulli’s equation at section (1) and (2)  =

P1 V12 P V2   Z1 = 2  2  Z2 g 2 g g 2 g But

V1 = V2 = VS and Z1 = 0

(Taking (1) as datum)

153

Z2 = 1 m P2 VS2 200  103 VS2  1  0 = g 2 g g 2g



200  103 P2 1 = g g

So from equation (i) NPSH =

P P V2 200  103 1 a  V  S g g g 2 g

(200  101.325  25)  103 32  1  NPSH = 980.39  9.81 2  9.81

Q.3

= 28.73 – 1 + 0.4587 = 28.1 m Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-1086, 24.18. Ans. (b) In a centrifugal pump for model testing there are following two conditions,  Hm   Hm    =    DN     1  DN  2

Q1 Q  32 3 D1 N1 D 2 N 2 For given pump,

......(i)

......(ii)

D1 = D2, from equation (i) 2

2  N2   1000   H  H2 =  1 =    30 =120 m  500   N1 

From equation (ii)

Q1 Q  32 3 D1 N1 D 2 N 2

N2 1000  60 = 120 lit/minute Q2 = N  Q 2 = 500 1 Reference: Fluid Mechanics and Hydraulic Machines, Dr. R.K. Bansal, Page-881, 19.22. Ans. (a) Given, Efficiency,  = 80% Discharge, Q1 = 70 m3/hr = 0.0194 m3/s Monometric head, Hm1 = 7 m Diameter, D1 = 2000 mm Speed, N1 = 1450 rPm Increased speed, N2 = 1750 rpm Therefore,

Q.4

Q2 Q1 = 3 3 D 2 N3 D1 N1 From question, diameter is constant, i.e. D1 = D2 For Discharge from model analysis,

From equation (i),

Q1 Q2 = N1 N2 70 Q2 = 1450 1750

......(i)

154

Therefore, discharge Q2 = 84.48 m3/hr = 0.0235 m3/s For head from model analysis of pump

N1 Q1 3/ 4

(Hm1 )

=

N 2 Q2

......(ii)

(Hm2 )3/ 4

1750 0.0235 1450 0.0194 = 3/ 4 (H m2 )3/ 4 (7)

Q.5

Therefore, head H m2 = 10.2 m Reference: Fluid Mechanics and Hydraulic Machines, Dr. R.K. Bansal, Page-881, 19.22. Ans. (a)

WQ H m Shaft Power WQ H m Shaft Power = 

Overall effeiciecy of pump,  =

9.81  0.0235  10.2 0.8 Shaft or Input Power = 2.94 kW Reference: Fluid Mechanics and Hydraulic Machines, Dr. R.K. Bansal, Page-858, 19.8. Ans. (a) =

Q.6

 P   P  In a centrifugal pump for model testing,  5 3  =  5 3   D N m  D N m From question,

 P1   5 3 = D N 

 P  5 2 3  D  2N  

   

Therefore, P2 = 8P1 Reference: Fluid Mechanics and Hydraulic Machines, Dr. R.K. Bansal, Page-881, 19.22. Q.7

N Q H m 3/ 4 Speed at which a geometrically similar centrifugal pump runs discharging 1 m3 of volume running under a head of one meter is called specific speed of the pump.

Ans.

N Q H m 3/ 4 where, N = Speed of pump in rpm Q = Discharge from the pump in m3/s Hm= Manometric head of the pump in ‘m’ Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-1078, 24.36.

Specific speed, Ns =



155

13 Compressible Flow Year 2000 1.

For a compressible fluid, sonic velocity is (a) a property of the fluid (b) always given by (RT)1/2 where , R and T are respectively the ratio of specific heats, gas constant and temperature in K

b

g

(c) always given by p / 

1/ 2 s

where p,  and s

are respectively pressure, density and entropy (d) always greater than the velocity of fluid at any location

Year 1999 2.

An aeroplane is cruising at a speed of 800 km/hr at an altitude, where the air temperature is 0oC. The flight Mach number at this speed is nearly (a) 1.5 (b) 0.254 (c) 0.67 (d) 2.04



Chapter-13 Answer & Explanations Q.1

Ans. (c) Exp. In case of compressible fluid, Sonic velocity is denoted by ‘C’ which is given by

 p      s

where,

p = pressure intensity  = desntiy of the liquid Reference: Fluid Mechanis, R. K. Rajput, Edition 2005, Page-723, Equation 15.29. Q.2

Ans. (c) Speed of aeroplane,

V = 800 km/hr

800  1000 m/s 60  60 = 222.22 m/s. = Temperature = 0º C = 273 K. Sonic speed,

C =

RT (assuming process as Adiabatic process)

= 1.4  287  273 = 331.196 m/s.

V 222.22  = 0.67 C 331.196 Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-847, Equation-19.7. Mach number =

