UntitledFluid Properties
FM & HM
1. When subjected to shear force, a fluid a) deforms continuously no matter
how small the shear stress may be b) deforms continuously only for large
shear forces c) undergoes static deformation d) deforms continuously only for small
shear stresses Ref : K Subramanya
2. Match List -I (Rheological equation) with List - II(Types of fluids) and select the cor- rect answer using the codes given below the lists: (IES -ME 2003) List -I
A. n


List -II 1) Bingham plastic 2) Dilatant fluid 3) Newtonian fluid 4) Pseudo - plastic fluid
(IES -ME 2003) Codes :
A B C D a) 3 2 4 1 b) 4 1 2 3 c) 3 4 2 1 d) 4 2 1 3
Ref : Class Room Teaching (AEE) 3. Match List -I (Curve identification in
figure) with List - II (Nature of fluid) and select the correct answer using the codes given below the lists (IES -CE 2010)
Rate of Shear deformation
A
B
C
D
List -I List-II A. Curve A 1) Newtonian B. Curve B 2) Dilatant C. Curve C 3) Ideal bingham plastic D. Curve D 4) Pseudo-plastic Codes :
A B C D a) 3 4 1 2 b) 2 4 1 3 c) 3 1 4 2 d) 2 1 4 3
Ref : Class Room Teaching (AEE) 4. Match List - I with List - II and select the
correct answer using the codes given below the lists : (CIVIL SERVICE 94) List - I (Description) I. Property which explains the spherical shape of the drop of a liquid II. Property which explains the phenomenon of cavitation in a fluid flow III. Property which explains the rise of sap in a tree IV. Property which explains the flow of a jet of oil in an unbroken stream List - II (Property of fluid) A. Viscosity B. Surface tension C. Compressibility D. Vapour pressure
E. Capillarity
Codes : I II III IV
a) A B D E b) B D E A c) D B E A d) A B C D
Ref : Class Room Teaching (AEE) 5. If cohesion > adhesion, then
(TSPSC AE -2015) a) Capillary rise occurs b) Depression occurs c) Remain plane d) Either rise or fall
Ref : Class Room Teaching (AEE) 6. The dimensions of the coefficient of
dynamic viscosity in [F, L, T] notation system are a) F T L–2 b) FL–1T-1
c) FL2T–1 d) FT–2L Ref : K Subramanya 7. At a liquid-air-solid interface the contact
angle measured in the liquid is less than 900. The liquid is, a) wetting b) non-wetting c) ideal d) does not form a stable bubble
Ref : K Subramanya 8. A liquid at 200C has a relative density of
0.80 and a kinematic viscosity of 2.3 centistoke. Determine its (i) unit weight in (KN/m3)________ (ii) dynamic viscosity in (Pa.s.) ________
Ref : K Subramanya Ans : i) 7.832 ii) 1.836×10–3
Sol : i) S=0.8
= 2.3×10–6 m2/sec
liquid 7832.3



9. The velocity distribution near the solid wall at a section in a laminar flow is given by u=5.0 sin (5 y) for y0.10 m. Compute the shear stress at a section at (i) y=0, (ii) y= 0.05 m and (iii) y= 0.10m. The dynamic viscosity of the fluid is 5 poise
Ref : K Subramanya Ans : i) 39.27 N/m2 ii) 27.76 N/m2 iii) 0
Sol: u=5 sin (5 y)
5 poise
39.26 cos 5 y
= 39.26 cos (0)
= 27.6 N/m2
= 0
10. A 10 kg block slides down on a smooth inclined plate. A gap of 0.1 mm between the block and plate contains oil having viscosity 0.4 N–s/m2. If the velocity distribution in the gap is linear and the area of the block is contact with the oil is 0.4 m2, the terminal velocity of the block is
V
300
o
a) 0.03125 m/s b) 0.3125 m/s c) 3.125 m/s d) 0.03125 mm/s
Ref : Noida & Company 10. Ans : (a) Sol:
W
Wcos300
2


V 0.030 m/sec
11. A flat plate of 0.15 m2 is pulled at 20 cm/s relative to another plate, fixed at a distance of 0.02 cm from it with a fluid having =0.0014 N-s/m2 separating them. What is the power required to maintain the motion? a) 0.014 W b) 0.021 W c) 0.035 W d) 0.042 W Ref : (IES 2006 Civil)
11. Ans : (d) Sol: Area (A) =0.15m2
V = 20 cm/s
P 0.042 W
12. A high-pressure reactor contains a fluid at 5 atm pressure. If the pressure in the reactor increases to 30 atm and the temperature is controlled to remain constant, the percentage reduction in volume is found to be 0.18%. Estimate the average bulk modulus of elasticity(KPa) of the fluid over the process. What is the coefficient of compressibility ( c ) of the fluidj (in m2/KN)?
Ref : K Subramanya Ans : K = 1.407×106 KPa cβ = 7.106×10–7 m2/KN
Sol: P1 = 5 atm
= 506.625 kPa

7 2 C 7.106 10 m /kN
13. A solid cylindrical needle of diameter 1.6 mm and density 7824 k g/ m 3 may float on a liquid surface. Neglect buoyancy and assume a contact angle of 00. What will be the surface tension ? a) 0.0772 N/m b) 0.154 N/m c) 0.772 N/m d) 0.0154 N/m
Ref : Noida & Company 13. Ans : (a) Sol: diameter (d) = 1.6 mm = 1.6×10–3m
density ( ) = 7824 kg/m3


xL cos xL
14. A glass tube of 3.8mm diameter is inserted into milk and milk rises upto 2.5mm in the tube. If the density of milk is 960 kg/ m3 and contact angle is 150, the surface tension of milk is a) 0.2315 N/m b) 0.025 N/m c) 0.0236 N/m d) 0.02315 N/m
Ref : Noida & Company
14. Ans : (d) Sol: d = 3.8 mm = 3.8×10–3 m
h = 2.5 mm
= 2.5×10–3m
960 9.81 3.8 10


Pressure and its measurements
1. Match List-I with List-II and select the correct answer using the codes given below the lists: (IAS-1999) List-I (Device) A. Barometer B. Hydrometer C. U-tube manometer D. Bourdon gauge List-II (Use) 1. Gauge pressure 2. Local atmospheric pressure 3. Relative density 4. Pressure differential
Codes: A B C D
a) 1 3 1 4 b) 3 2 1 4 c) 3 2 4 1 d) 2 3 4 1
Ref: ACE Gate P.B. 2. In inverted U- tube manometer, the mano-
metric liquid is____the flowing liquid (CRT AE -1993) a) heavier than b) lighter than c) same as d) none of these
3. Small pressure differences are measured by (CRT APPSC -2008 L.R.) a) pressure gauge b) anemometer c) Inverted U - tube manometer d) piezometer
4. A Piezometer is a device used to measure (PREV-DYSY -2011)
a) low pressures b) differential pressures c) vacuum pressures d) None of these
5. A force F of 800 N is applied as shown. What maximum weight W can be supported? Neglect the weight of the pistons. (NTPC)
WF
8 cm dia Cylinder
80 cm dia Cylinder 80 cm dia Cylinder 80 cm dia Cylinder
2 cm dia Pipe
a) 800 N b) 8000 N
c) 40000 N d) 80000 N Ref: AEE CRT 5. Ans : (d) Sol:
W F
Given
from Pascals Law


W 80,000 N
6. Three rigid buckets, as shown in the figure (1), (2) and (3) are of identical heights and base areas. Further, assume that each of these buckets have negligible mass and are full of water. The weights of water in these buckets are denoted as W1, W2 and W3 respectively. Also, let the force of water on
Class Room Teaching
GATE MASTER’S ACADEMY
GATE M ASTER’S
Pressure and its measurements
the base of the bucket be denoted as F1, F2 and F3 respectively. The option giving an accurate description of the system physics is (GATE (C) 2014 Set-I)
h h h
(3)(2)(1) All three buckets have the same base area
a) 2 1 3 2 1 3W W W andF F F
b) 2 1 3 2 1 3W W W andF F F
c) 2 1 3 1 2 3W W W andF F F
d) 2 1 3 1 2 3W W W andF F F
Ref: Gate Civil 7. The force F needed to support the liquid
of density d and the vessel on top (fig) is [1995 : 2Marks]
Frictionless piston of area=A
Vessel
F
a) gd[ha-(H-h)]A b) gdHa c) gdHA d) gd(H-h)A
Ref: AEE CRT 8. The ratio of pressures between two points
A and B located respectively at depths 0.5 m and 2m below a constant level of water in a tank is (CIVIL SERVICES 95)
a) 1: 2 b) 1: 2 c) 1: 4 d) 1 : 16
Ref: AEE CRT
8. Ans : (c)
9. Find the pressure represented by a column of
i) 10 cm of water. ii) 5 cm of oil of relative denstiy 0.75 iii) 2 cm of mercury. Ans : a) 981 N/m2
b) 367.13 N/m2
c) 2668.32 N/m2
Ref : K. Subramayan
i) p= 1000×9.81×0.1
2P = 981 N/m
P = 750×9.81 × 0.05
P = 13600 × 9.81×0.02
2P 2668.32 N/m
10. An open tank contains water upto a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity
GATE MASTER’S ACADEMY
GATE M ASTER’S
Pressure and its measurements
a) at the interface of the two liquids and b) at the bottom of the tank.
Ans : a) 9000 N/m2 b) 29000 N/m2
Sol : (a) aP gh
1m
(b) bottom water waterP 9000 g h
Pbottom = 9000+1000×10×2
Pbottom = 29,000 N/m2
11. The intensity of pressure required to suck milk by a straw through a height of 200mm from an open vessel (Sp. Gravity of milk is 1.08) (ACE GATE PB) a) 2.12kPa absolute b) 99.18kPa vacuum c) 99.18kPa absolute d) none
Ref: Ace Gate 11. Ans : (c) Sol: Given h= 200 mm = 0.2 m
SP. Gravity of milk = 1.08
3 milk 1.08 1000 1080 kg /m
P gh
= 101.325-2.118
= 99.18 KPa (Absolute) 12. Refer to figure, the absolute pressure of
gas A in the bulb is [GATE(M)1997 : 2Mark]
PA
13.6g/ml
a) 771.2mm Hgb) 752.65mm Hg c) 767.35mm Hg d) 748.8mm Hg
Ref: GATE ME 1997 12. (a) Sol:
h =171 =5
Absolute pressure at point D= Absolute pressure at Point E
A w 1 Hg 3 w 2 atmP gh gh gh P
PA+1000×9.81×0.17=13600×9.81×0.02+ 1000 × 9.81×0.05+1.01325×105
PA = 102816.12 Pa (absolute)
A Hg AP gh
GATE MASTER’S ACADEMY
GATE M ASTER’S
770.6mmof Hg
13. The cross-sectional area of one limb of a U-tube manometer (figure shown below) is made 500 times larger than the other, so that the pressure difference between the two limbs can be determined by measuring ‘h’ in one limb of the manometer. The percentage error involved is [GATE(M) 1990: 2Mark]
Initial Level
a) 1.0 b) 0.5 c) 0.2 d) 0.05
13. Ans (c) Sol: Volume drop in larger limb = Volume rise
in smaller limb
2
1


14. An inclined manometer contains a liquid of relative density 0.8 and has an inclination of 30o to the horizontal. For a certain pres- sure the column length was 10cm. If there is an uncertainty of 1o in the measurement of the angle of inclination, the calculated pressure would have an uncertainty of
(TSPSC -Tribal Welfare -2017) a) 1% b) 0.28% c) 1.75% d) 3.33%
Ref: AEE CRT 14. Ans : (d) Sol: S = 0.8
30.8 1000 800 kg /m
030
dp kcos .d




15. The pressure difference measured by a mercury oil differential manometer on the two sides of the orifice meter gives a reading of 50cm of mercury. Determine the differential pressure head in orifice. Take the specific gravity of oil as 0.9. a) 7.05m of oil b) 5.05m of oil c) 4.75m of oil d) 3.35m of oil
Ref: ISRO-2008, 45 Q
GATE MASTER’S ACADEMY
GATE M ASTER’S
30.9 1000 900 kg /m
x = 50 cm= 0.5m
h = 7.05 m of oil

Ref:Nodia & company 16. Ans : (d) Sol:
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
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... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
... ... .
..................................................................................................................................................................................................................
...............
................................................
..................................................................................................................................................................................................................................................................................................................................
............................
...............................................................
....... ..................................................... .............
PA=PB
1 w 2 m w 1P gh P P gh gh
1 2 m w 1 wP P gh gh gh
3 m w w 134.4 10 h g g gh
=h(13600×9.81–1000×9.81)+1000×9.81×2
h 12cm
17. Determine the pressure difference between points A and B, for the set up shown in Fig
B
A
B m oil A oilP g (0.08) g(0.13) P g(0.31)
A B m oil oilP P g(0.08) g(0.13) g(0.31)
=13600(9.81)(0.08)+800×9.81(0.13)–
PA–PB = 9564.75 N/m2
18. An inverted U-tube manometer is used to measure the pressure difference between two pipes A and B, as shown in the figure. Pipe A is carrying oil (specific gravity=0.8) and pipe B is carrying water. The densities of air and water are 1.16kg/m3 and 1000kg/m3, respectively. The pressure difference between pipes A and B is (Acceleration due to gravity: g=10m/s2) Ans : -2.199KPa.
[GATE (M) 2016 : 1Mark, Set-1]
A Oil
Pc=PD
A oil 2 air 1 B w 1 2 3P gh gh P g(h h h )
A B oil 2 air 1 w 1 2 3P P gh gh g(h h h )
= 800 × 10 × 0.2 + 1.16 × 10 × 0.08 –1000× 10 × 0.38
PA-PB= –2199.07 Pa
Hydrostatic forces
LEVEL - I 1. The centre of pressure for a vertically im-
mersed surface lies at a distance equal to ....... the centre of gravity
(AEE -2001) CRT
GI AX above
AX I above
2. A lamina “l ” units long and “h” units deep is placed in a liquid vertically so that its upper edge lies at the free liquid sur- face. The distance of the centre of pres- sure from the centroid of the lamina is
CRT(AEE-1996)


3. A stepped cylindrical container is filled with a liquid as shown in the figure below.
d
h
h
2d
The container with its axis vertical, is first placed with its larger diameter downward and then upward. The ratio of the forces at the bottom in the two cases will be
CRT (IES-ME 1998) a) 1/2 b) 1
c) 2 d) 4



4. The specific weight of the liquid in the container whose cross section is shown below is and the container is W units wide. Find the magnitude of the upward force on the inclined wall. CRT (DRDO)
450
h



5. The hydrostatic pressure, in kgf, exerted on one side of an annular area enclosed by concentric circles of radii 2m and 1m and having its centroid 4m below the free wa- ter surface, is
CRT (AEE CE/ME & 1984 & 2012) a) 12000 b) 11000
c) 10000 d) none of these
6. The water total pressure per meter length on a vertical masonary wall of dam is
CRT (APPSC -2011) a) WH/2 b) WH
c) WH2/2 d) 2WH / 4
LEVEL - II 1. A rectangular plate 0.6 m wide and 1.2m
deep lies within a water body such that its plane is inclined at 450 to the horizon- tal and the top edge is 0.70 m below the water surface. Determine the total pres- sure force on one side of the plate and the location of the centre of pressure
(KSBN -2.19) a) 7.925 kN, 1.665 m b) 6.926 kN, 2.666 m
c) 8.925 kN, 1.665 m
d) 9.925 kN, 2.665 m
Sol :
G
C
Hydrostatic forces
h = depth of CG of the plate = 0.7+0.6 sin 450
= 1.1243 m Total pressure force
F Ah
= 9790×(0.6×1.2)×1.1243 = 7924.9N = 7.925 kN
Centre of pressure : Because of symmetry xP= x , i.e. C lies on the axis Y1Y1 passing through the CG of the area.
P ggy y Ay
l
l

= 1.665 m 2. A door in a tank is in the form a quadrant
of a cylinder of 1.5 m radius and 1.8 m wide as shown in below figure. Calculate the resultant force on the door and its lo- cation on the gate. (KSBN -2.38)
1.5 m
2.0 m
Sol : Horizontal component of the force
FH = (Projected area) × h

FH = (Volume TNMS)
21 (1.5) 1.8
= 2 2(72.69) (61.37) = 95.13 kN
If is the inclination of FR to vertical
tan= H

yR = vertical height of resultant on the gate.
= R cos =1.5 cos (57.1730) = 0.813 m
3. The force ‘F’ required at equilibrium on the semi - cylindrical gate shown below is
(GATE (C) 2002 : 1 Mark)
2 m Water
Hinge
F
a) 9.81 kN b) 0.0 kN c) 19.62 kN d) None of these
GATE MASTER’S ACADEMY
GATE M ASTER’S
Hydrostatic forces
Sol : Depth of water (H) = 2 m. Consider unit width, of cylinder,
H hF ghA

2R1000 9.81 1
* 2 4h H m 3 3

4R 4 1x 0.424m 3 3


419.62 1 15.41 0.424 F 1 3

4. A triangular gate with a base width of 2m
and a height of 1.5m lies in a vertical
plane. The top vertex of the gate is 1.5m
below the surface of a tank which contains
oil of specific gravity 0.8. Considering the
density of water and acceleration due to
gravity to be 31000 kg /m and 29.81 m/s , respectively, the hydrostatic
force (in kN) exerted by the oil on the gate
is ...... 29.43 kW
1. 5
m 1.
5 m
2 m

Buoyancy and Floatation
Level -I 1. The line of action of the buoyant force acts
through the CRT (AEE -1984) a) centroid of the volume of the floating
body
c) centroid of the horizontal projection of the body
d) centroid of the vertical projection of the body
2. Which principle does not apply to a balloon lifting in air ? CRT (MT) a) Archemedes principle b) Principle of buoyancy c) Law of gravitation
d) Continuity equation 3. A solid body sinks in a fluid when
CRT (AEE -1984) a) the specific gravity of its material is
greater than unity
b) the buoyancy force does not pass through the metacentre
c) the weight of the fluid displaced is less than the weight of the body
d) the metacentre lies below the C.G. 4. A body floats in stable equilibrium
CRT(AEE -1984) a) when its metacentric height is zero b) only when its centre of gravity below
its centre of buoyancy c) when the metacentre is above its
centre of gravity d) none of these
5. A rectangular floating body of 20m long 5m wide and 2m height is floating in wa- ter. The water line is 1.5m above the bot- tom, then the metacentre height will be
approximately CRT (AEE -1984)
c) 0.34 m d) 0.30 m
Sol :
GM = 1.13 m
6. The metacentric height of a ship is 0.6m and the radius of gyration is 4m. The time of rolling of a ship is
CRT (AEE CE/ME - 2009)
Class Room Teaching
T = 10.35 sec
7. A wooden cube with 2m×2m×2m of spe- cific gravity 0.5 floats in water. The vol- ume of the cube submerged under water, in m3 is CRT (AEE -1984) a) 4 b) 2 c) 1 d) None of these
C.G. and Meta centre (GM) lies at same point.
Sol:
2m
2m
Given
Sbody = 0.5
V = 4m3
Level -II 1. A body weighs 30N and 15N when weighed
under submerged conditions in liquids of relative densities 0.8 and 1.2 respectively. What is the volume of the body in litres?
CRT (IES-ME 2009)
1BW 30 F 2BW 15 F
1BW F 30 2BW F 15
1f W V 30
2f W V 15
W-(0.8×9810)V=30 ......(1)
V = 3.82×10–3m3 ( 3 31 litre 10 m )
V = 3.82 litres
2. In an iceberg, 15% of the volume projects above the sea surface. If the specific weight of sea water is 10.5 kN/m3 ,the specific weight of iceberge in kN/m3 is
CRT (GATE 1999) a) 12.52 b) 9.81 c) 8.93 d) 7.83
Sol: 15%
ice berg s.w s.wV V
ice berg 1 10.5 0.85
3 ice berg 8.95 kN/m
3. A 15cm length of steel rod with relative density of 7.4 is submerged in a two layer fluid. The bottom layer is mercury and the top layer is water. The height of top surface of the rod above the liquid inter- face in ‘cm’ is CRT (GATE 2001) a) 8.24 b) 7.82
c) 7.64 d) 7.38
GATE MASTER’S ACADEMY
GATE M ASTER’S
7.4×15=13.6×(15-x)+1(x)
x = 7.38 cm
4. A wooden plank (sp gr.0.5) 1m x 1mx 0.5m floats in water with 1.5kN load on it with 1m x 1m surface horizontal. The depth of plank lying below water surface shall be CRT (CIVIL SERVICES 95) a) 0.178 m b) 0.250 m
c) 0.403 m d) 0.500 m
Sol:
h
bodyW ( ) F.B( )
gv +1.5 kN = 9810× (1×1×h)
(500×9.81)×(1×1×0.5)+1.5=9810×h
h = 0.403 m
nozzle of length L is given by 2xV 2t 1
2L
where V= velocity in m/s,
t=time in seconds from commencement of flow, x=distance from inlet to the nozzle. Find (KSBR, 3.12)
i) convective acceleration,
ii) local acceleration and the total accel- eration when t=3s, x=0.5 and L=0.8m
Ans: i) -14.623m/s2
ii) 0.945m/s2 & -13.68m/s2
1. Given that x = 0.5, t = 3 sec, L = 0.8 m
Sol: 2xV 2t 1

2L x 2L
2L 2L 2L
2. A streamline is a line (KSBR, 3.4)
a) which is normal to the velocity vector at every point
b) which represents lines of constant ve- locity potential
c) which is normal to the lines of con- stant stream function
d) which is tangential to the velocity vector everywhere at a given instant
3. A flow has parallel curved streamlines and is steady. This flow has (KSBR, 3.16) a) tangential convective acceleration b) local acceleration c) normal convective as well as local ac- celeration d) normal convective acceleration.
4. A flow is said to be rotational when (KSBR, 3.21)
a) the streamlines are curved b) a velocity gradient in the normal direc- tion to flow exists. c) every fluid element has finite angu- lar velocity about its mass centre d) every fluid element has an angular ve- locity about a common axis
5. A stream function 3 3x y is observed for a two-dimensional flow field. What is the magnitude of the velocity at point (1,- 1)? (KSBR, 3.39) a) 4.24 b) 2.83 c) 0 d) -2.83
Class Room Teaching
5. Fluid Kinematics
v = ? (1,–1)



6. Velocity potential exists for (KSBR, 3.32)
a) Flow of a perfect fluid only b) Steady, irrotational flow only c) All irrotational flows d) All 3-D flows
7. X-component of velocity in a 2-D incom- pressible flow is given by u=y2+4xy. If Y- component of velocity ‘v’ equals zero at y=0, the expression for ‘v’ is given by
(GATE (C), 1996-1M) a) 4y b) 2y2
c) -2y2 d) 2xy 7. Ans : (c) Sol: u = y2 + 4xy
v = 0 at y = 0
u v 0 x y


2
24yv
2
v = –2y2
8. The circulation ‘ ’ around a circle of ra- dius 2 units for the velocity field u=2x+3y and v=-2y is (GATE (C), 2005-2M) a) -6 units b) -12 units c) -18 units d) -24 units
Ans : (b) Sol:
r= 2 units
u = 2x+3y


= -12
9. A two dimensional flow field has veloci- ties along the x and y directions given by u=x2t and v=-2xyt respectively, where t is time. The equation of streamline is
(GATE (M), 2006-2M) a) x2y= constant b) xy2=constant c) xy=constant d) not possible to determine
9. Ans : () Sol: u=x2t
v = –2xyt


1 12 .dx .dy Const. x y

2x y= constant
Level-II 1. The velocity field for a flow is given by:
ˆ ˆV 5x 6y 7z i 6x 5y 9z j
ˆ3x 2y z k and the density varies as
o exp (-2t). In order that the mass is conserved, the value of should be
(GATE (C), 2006-2M) a) -12 b) -10 c) -8 d) 10
1. Ans : (c)
Sol: v (5x 6y 7z) i (6x 5y 9z) j
(3x 2y z)k

2t oe (8 ) 0
8
8
Common Data for Question 2 : The laminar flow takes place between closely spaced parallel plates as shown figure below.
The velocity profile is given by yu V h
. The gap
height, ‘h’ is 5mm and the space filled with oil (specific gravity=0.86, viscosity =2×10-4Ns/ m2). The bottom plate is stationary and the top plate moves with a steady velocity of V=5cm/s. The area of the plate is 0.25m2.
FIGURE DRAW 2. The rate of rotation of a fluid particle is
given by (GATE (C), 2004-2M)
a) y z V0; 2h
b) y z V0; h

d) y z V ; 0 h



3. You are asked to evaluate assorted fluid flows for their suitability in a given labo- ratory application. The following three flow choices, expressed in terms of the two- dimensional velocity fields in the xy-plane, are made available.
(GATE (M), 2009-2M) P. u=2y, v=-3x Q. u=3xy, v=0 R. u=-2x, v=2y
GATE MASTER’S ACADEMY
GATE M ASTER’S
4
Which flow(s) should be recommended when the application requires flow to be incompressible and irrotational? a) P and R b) Q c) Q and R d) R
3. Ans : () Sol: (P) u = 2y, v = –3x
z 1 v u 2 x y


z 1 v u 2 x y


z 1 v u 2 x y




4. A flow is described by the stream func-
tion 2 3xy . Locate the point at which the velocity vector has a magnitude of 4 units and makes an angle of 150o with the x-axis. (KSBR, 3.28) a) (2.00, 0.5774) b) (1.00, 0.5774) c) (2.00, 0.6774) d) (3.00, 0.5774)
Sol : The velocity vector is diagrammatically shown in figure.
2 3 xy



22 24.0 4 3 x (0.5774 x )
216x 4x
5
The required point (x, y) is (1.00, 0.5774). 5. A velocity potential for a two-dimensional
flow is given by =(x2-y2)+3xy. Calculate (i) the stream function and (ii) the flow rate between the streamlines passing through points (1,1) and (1, 2) (KSBR, 3.34)
a) 2 233xy x y 2
, 7.5uits
, 8.5units
, 6.5units
, 5.5units

23f(x) x 2



Flow rate between the stream lines pass- ing through (1, 1) and (1, 2)
2 1 (8.5 2.0) 6.5 units
6. In a flow the velocity vector is given by V=3xi+4yj-7zk. Determine the equation of the streamline passing through a point M=(1, 4, 5) (KSBR, 3.9)
Ans: y=4x4/2 & z=5/x7/2
Sol : The equation of the streamline is
dx dy dz u w

Hence dx dy dz 3x 4y 7z

1 1 1inx iny InC 3 4

or y = C1x4/3
where C1 is another constant. (i) Similarly, by considering equations with x and z and on integration
2 1 1inx inz InC 3 7

2 7/3
C z
x (ii)
where C2 is another constant. Putting the coordinates of the point M (1, 4, 5)
from Eq. (i) 1 4/3
4C 4 (1)
The streamline passing through M is given by
y = 4x4/3 and z= 5/x7/3
GATE MASTER’S ACADEMY
GATE M ASTER’S
2
constant along
a stream line holds true for (GATE (C) 1996 : 1 Mark)
a) steady, frictionless, compressible fluid b) steady, uniform, incompressible fluid c) steady, frictionless, incompressible fluid d) unsteady incompressible fluid
2. Match List -I (Devices) with List - II (Uses) and select the answer using the codes given below the lists :
(GATE (C) 2010 : 1 Mark) List - I A. Pitot tube B. Manometer C. Venturimeter D. Anemometer List - II 1. Measuring pressure in a pipe 2. Measuring velocity of flow in a pipe 3. Measuring air and gas velocity 4. Measuring discharge in a pipe Codes :
A B C D a) 1 2 4 3 b) 2 1 3 4 c) 2 1 4 3 d) 4 1 3 2
3. The reading of differential manometer of a Venturimeter, placed at 045 to the hori- zontal is 11 cm. If the venturimeter is turned to horizontal position, the manom- eter reading will be
(GATE (C) 2005 : 1 Mark)
a) zero b) 11 cm 2
c) 11cm d) 11 2cm 4. In a venturimeter, the angle of the diverging
section is more than that of converging section. State: (T/F)
Ans : F (GATE (M) 1994 : 1 Mark] 5. The Bernoulli equation is written with
usual notation as p/ +V2/2g+Z = con- stant. In this equation each of the terms represents (SBRM 4.1, 168) a) energy in kg.m/kg mass of fluid b) energy in N.m/kg mass of fluid c) energy in N.m/N weight of fluid d) power in k.W/kg mass of fluid
6. The height of the water columns in a piezometer and a pitot tube are measured to be 35 cm and 50 cm respectively (Both measured from the top surface of the pipe). Both are tapped into a horizontal water pipe. What will be the velocity at the center of pipe? (NODIA, 3.19) a) 17.2 m/s b) 1.72 m/s c) 0.172 m/s d) 15.6 m/s
Ans (b) :
35 cm
50 cm
The application of the Bernoulli’s equa- tion between points (1) and (2) gives
22
p Vp V Z Z g 2g g 2g

2 2 2 1 1 2p p V V
g 2g

Since point 2 is a stagnation point and thus V2=0
2 pitot piezo2 11


1 pitot piezoV 2g h h 2 9.81 (0.50 0.35)
= 1.715 1.72 m/s 7. A Prandtl tube (Pitot-static tube with C=1)
is used to measure the velocity of water. The
differential manometer reading is 10mm of
liquid column with a relative density of 10.
Assuming g=9.8m/s2, the velocity of water
(in m/s) is [GATE (M) 2015: 2 Mark, Set-3]
a) 2.428m/s b) 1.328m/s
c) 3.428m/s d) 5.328m/s
Sol : Given data : C = 1 x = 10 mm of liquid = 0.01 m of liquid S = 10
mano =10×1000 = 10000 kg/m3
g = 9.8 m/s2

= 0.09 m of water Velocity of water,
V = C 2gh 1 2 9.8 0.09
= 1.328 m/s KEY : Level - I 1. c 2. c 3. c 4. False 5. c 6. b 7. b
Level - II 1. A pipeline is 15 cm in diameter and is at
an elevation of 100.00 m at section A. At section B it is at an elevation of 107.00 m and has a diameter of 30 cm. When a discharge of 50 L/s of water is passed through this pipe the pressure at section A is observed to be 30 kPa. The energy loss in the pipe is 2 m. Calculate the pressure at B when the flow is (i) from A to B and (ii) from B to A.
(SBRM 4.2, 144)
Sol: d1=15 cm 21A 0.15 4


hL = 2m
Q = A1V1
4
4
2 2 A A B B

3 2 2

2 2 A A B B

3 2 2

BP =-15.21 kPa
2.Water flows through a pipe of diameter 0.30m. What would be the velocity V for the conditions shown in the figure below? (GATE (M) 1988 : 2Mark]
0.3 m
Water
a) 2.084 m/s b) 1.084 m/s c) 3.084 m/s d) 4.084 m/s
2. Ans : (b)
V 1.084m/s
3. A fire hose nozzle has a diameter of 3 cm. The nozzle must be capable of delivering at least 0.016 m3/s. If the nozzle is at- tached to a 7.5 cm diameter hose, what pressure must be maintained just up- stream of the nozzle to deliver this flow rate? (NODIA 3.20) a) 500 kPa b) 125 Pa c) 250 kPa d) 2.50 kPa
Ans : (c) Sol: Q=0.016 m3/sec
d1 = 7.5 cm = 7.5×10–2 m
22 3 2 1A 7.5 10 4.41 10 m
4
4


g 2g g 2g
9810 2 9.81 2 9.81

1P 250 kpa
4.Water 31000kg /m flows through a venturimeter with inlet diameter 80mm and throat diameter 40mm. The inlet and throat gauge pressures are measured to be 400kPa and 130kPa respectively. Assuming the venturimeter to be horizontal and neglecting friction, the inlet velocity (in m/s) is
[(GATE (M) 2015: 2Mark, Set-1] a) 6m/s b) 9m/s c) 3m/s d) 12m/s
4. Ans : (a) Sol: d1 = 80 mm
= 0.08 m
v2 = 4.01v1
g 2g g 2g


22 3 3 1 1V 4.01V 130 10 400 10
2 9.81 1000 9.81
1V 5.98 m/sec
5. A venturimeter,having a diameter of 7.5 cm at the throat and 15cm at the enlarged end, is installed in a horizontal pipeline of 15 cm diameter. The pipe carries an incompressible fluid at a steady rate of 30 litres per second. The difference of pres- sure head measured in terms of the mov- ing fluid in between the enlarged and the throat of the venturimeter is observed to be 2.45m. Taking the acceleration due to gravity as 29.81m/s , the coefficient of discharge of the venturimeter (correct up to two places of decimal) is ......
(GATE (C) 2014 : 2 Marks, Set-I) a) 0.75 b) 0.85 c) 0.65 d) 0.95
5. Ans : (d) Sol: d1 = 15 cm = 0.15m
21a 0.15 4
h = 2.45 m
1 2
a a




Level - I 1. When the whole fluid mass rotates either
due to gravity or rotation imparted previ- ously, the flow is known as
CRT (AEE -2004)
a) free vortex
b) forced vortex
c) rotational vortex
d) none-potential vortex
2. Flow in a whirl pool in row river is an ex- ample of the following vortex
CRT (AEE -2012)
a) Free b) Forced
c) Spiral d) Radial 3. For a forced vertex flow the height of pa-
raboloid formed is equal to CRT (AEE -2011)
a) 2P v+
2r 2g
4. In a free Cylindrical vortex of water, the tangential velocity at a radius of 12cm from axis of rotation is 7.2 m/sec and in- tensity of pressure is 2.5 kg/sq.cm. The velocity at a radius of 24 cm from the axis (in m/sec) is (AEE -1987) a) 2.4 b)3.2 c) 3.6 d) 4.8
Ans : C Free cylindrical vortex Radius = 12 cm Velocity = 7.2 m/s Intensity of pressure = 2.5 kg/cm2
Velocity at a radius of 24 cm = ? For free vortex flow = v.r = constant v1r1 = v2r2 7.2 12 = v2 24
v2 = 3.6 m/s 5. An open circular tank of 20 cm diameter
and 100 cm long contains water upto a height of 60 cm. The tank is rotated about its vertical axis at 300 r.p.m. find the depth of parabola formed at the free surface of water. (RK Bansal) a) 50.28 cm b) 55.28 cm c) 60.28 cm d) 65.28 cm
Ans : a Given Diameter of cylinder = 20 cm
Radius (R) = Diameter 20
2 2
Radius (R) = 10 cm Height of liquid (H) = 60 cm Speed (N) = 300 r.p.m
Angular velocity () = 2 N 60
2 300 60
2 2 2rZ
Z = 50.28 cm
6. An open circular cylinder 1.2 m high is filled with a liquid to its top. The liquid is given a rigid body rotation about the axis of the cylinder and the pressure at the centre line at the bottom surface is found to be 0.6 m of liquid. What is the ratio of volume of liquid spilled out of the cylin- der to the original volume? [IES(M) 2007]
a) 1/4 b) 3/8
c) 1/2 d) 3/4


Level - II 1. An open circular cylinder of 15 cm diam-
eter and 100 cm long contains water upto a height of 80 cm. Find the maximum speed at which the cylinder is to be rotate about its vertical axis so that no water spills (RK Bansal)
Ans : 356.66 r.p.m Sol. Given :
Diameter of cylinder = 15 m
Radius (R) = Diameter 15
2 2
Radius (R) = 7.5 cm Length of cylinder (L) = 100 cm Initial height of water = 80 cm Let the cylinder is rotated at an angular speed of rad/sec, when the water is about to spill then using Rise of liquid at ends = Fall of liquid at centre height Rise of liquid at ends = Length - Initial height
Fall of liquid at centre = 100–80 = 20 cm Height of parablola = 20+20 = 40 cm
Z = 40 cm 2 2RZ 2g



N = 356.66 r.p.m 2. An open circular cylinder of 15 cm diam-
eter and 100 cm long contains water upto a height of 70 cm. Find the speed at which the cylinder is to be rotated about its ver- tical axis, so that the axial depth becomes zero. (RK Bansal)
Ans : 563.88 r.p.m Sol.
Radius (R) = 15 7.5 2
cm
Length of cylinder= 100 cm Initial height of water = 70 cm When axial depth is zero, the depth of paraboloid = 100 cm
GATE MASTER’S ACADEMY
GATE M ASTER’S


N = 563.88 r.p.m 3. A vessel is cylindrical in shape and closed
at the top and bottom, contains water upto a height of 80 cm. The diameter of the vessel is 20 cm and length of vessel is 120 cm. The vessel is rotated at a speed of 400 r.p.m. about its vertical axis. Find the height of paraboloid formed.
(RK Bansal) Ans : 84.56 cm Sol.
1.2m
0.6m
Given : Initial height of water = 80 cm Diameter of vessel = 20 cm
Radius (R) = Diameter 20
2 2
Radius (R) = 10 cm Length of vessel = 120 cm Speed (N) = 400 r.p.m
2 N 2 400 60 60

= 41.88 rad/sec when the vessel is rotated, let Z
= Height of paraboloid formed r = Radius of paraboloid at the top of the vessel This is the case of closed vessel volume of air before rotation
= volume of air after rotation



2 4000 2r Z 8000
..... (i)



Z Z 8000 0.894
GATE MASTER’S ACADEMY
GATE M ASTER’S
Level-I 1. Consider fully developed flow in a circu-
lar pipe with negligible entrance length effects. Assuming the mass flow rate, den- sity and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of
[GATE(M)-2015] a) 4 b) 16 c) 32 d) 64
Sol: Ans (d)
2
2
2
f
f
h
2. Due to aging of a pipeline, its carrying ca- pacity has decreased by 25%. The corre- sponding increase in the Darcy Weisbach friction factor, f is 77 % [GATE(C)-1995]
Sol: Q1=Q Q2=0.75Q f =?
* 2
f2=1.77 f1
3. If a single pipe of length L and diameter D is to be replaced by three pipes of same material, same length and equal diameter d (d<D), to convey the same total discharge under the same head loss, then d and D related by [GATE(C)-1997]
a) 1 5
L LLL D d d d

1 2 3


4. The head loss in a sudden expansion from 6 cm diameter pipe to 12 cm diameter pipe, in terms of velocity V1 in the 6 cm pipe is
[K SBRM 11.2]
4. Ans : (d) Sol: d1= 6 cm d2=12 cm
A1= 28.27 A2 = 113.09
2 1V9
16 2g
5. Three pipes are connected in series. Then [K SBRM 11.7]
a) The head loss in each pipe is the same b) The total discharge is the sum of the
discharge in the individual pipes c) The discharge through each pipe is
the same d) the Reynolds number for each pipe is
the same 6. Two pipelines of equal length and diam-
eter of 20cm and 30cm respectively are connected in parallel between two reser- voirs. If the friction factor f is the same for both the pipes, the ratio of the discharges in the smaller to the largest size of the pipe is [K SBRM 11.16] a) 0.363 b) 0.444 c) 0.667 d) 0.137
6. Ans : (a) Sol: d1=20 cm d2= 30 cm
L1=L2
f1= f2
22
f L Qf L Q 12.1d 12.1d

= 0.363
7. In a sudden contraction, the velocity head changes from 0.5m to 1.25m. The coeffi- cient contraction is 0.66. The head loss in this contraction is [K SBRM 11.4] a) 0.133m b) 0.332m c) 0.644m d) 0.750m
7. Ans : (b)




ch 0.331 m
8. If two reservoirs whose water levels differ by 2m are connected by 2 C.I. pipes laid in sense whose lengths and diameters are 100m, 40cm and 200m, 20cm respectively
[AEE CRT 15] a) the loss of head in the system is 2m
of water b) the loss of head cannot be determined c) there is no flow from one reservoir to
the other d) none of the above is true
GATE MASTER’S ACADEMY
GATE M ASTER’S
8. Ans : (a) Sol: L1=100 m L2=200 m
d1 = 40 cm d2 = 20 cm
z1=2 m z2 = 0 m
2m
g 2g g 2g

fh 2m
9. In a siphon, the summit 4m above the water level in the tank. If the head loss from the inlet of the siphon to the sum- mit is 2m and velocity head is 0.5m, then the pressure at the summit is
[AEE PREV 48] a) -63.6kPa b) -9 of H2O c) -40.5kPa d) -8.5cm of Hg
9. Ans : (a) Sol:
g 2g g 2g

2P 63.7kPa
10. Water (density=1000kg/m3) at ambient temperature flows through a horizontal pipe of uniform cross-section at the rate of 1kg/s. If the pressure drop across the pipe is 100kPa, the minimum power re- quired to pump the water across the pipe, in watts, is_________ [GATE(M)-2017]
ans: 100W 10. Ans : () Sol: mass/sec=1 kg/sec
P1–P2 = 100 kPa

P = 100W
Level-II 1. A siphon is used to drain water from a
large tank as shown in figure below. As- sume that the level of water is maintained constant. Ignore frictional effect due to viscosity and losses at entry and exit. At the exit of the siphon, the velocity of wa- ter is [GATE(M)-2014]
GATE MASTER’S ACADEMY
GATE M ASTER’S
a) Q R2g Z Z b) P R2g Z Z
c) O R2g Z Z d) Q2gZ
Sol : Ans (b) Applying Bernoulli’s equation between P and R,
2 2 atm atmP R
P R P PV VZ Z
g 2g 2g g

R P RV 2g (Z Z )
2. A 6 cm diameter pipe has a discharge of 450 L/min. At a section the pipe has a sudden expansion to a size of 9cm diam- eter. If the pressure just upstream of the expansion is 20kN/m2, calculate the pres- sure just after the expansion. Assume the pipe to be horizontal at the expansion re- gion. [K SBRM 11.2]
a) 21.83kPa b) 21.73kPa c) 22.73kPa d) 31.73kPa
Sol : Ans (b)
2
2 2 1

21 2 L
V V H
2g 2g
2g 2g


p2 = 2.22×9.79 = 21.73 kPa 3. A 2 km pipe of 0.2m diameter connects
two reservoirs. The difference between the water levels in the reservoir is 8m. The Darcy Weisbach friction factor of the pipe is 0.04. Accounting for frictional entry and exit losses. The velocity in the pipe in (m/sec) is_____ [GATE(C)-2013] a) 0.63 b) 0.35 c) 2.52 d) 1.25
Sol : Ans (a) Frictional loss,
2
L 0.5Vh
C Vh 2g


8 = 20.3873 V2 +0.07645 V2
V = 0.625 m/s 0.63 m/s
4. A pipe of 0.7 m diameter has a length of 6 km and connects two reservoirs A and
B. The water level in reservoir A is at an elevation 30m above the water level in res- ervoir B. Halfway along the pipeline, there is a branch through which water can be supplied to a third reservoir C. The fric- tion factor of the pipe is 0.024. The quan- tity of water discharged into reservoir C is 0.15m3/s. Considering the accelera- tion due to gravity as 9.81 m/s2 and ne- glecting minor losses, the discharge (in m3/s) into the reservoir B is______
[GATE(C)-2015] Sol : Ans:(0 .572m 3/s)
A
B
C
Q
Q-0.15
0.15
3
2
1
f 5
or, 2



Q = 0.722 m3/s QB = Q – 0.15 = 0.572 m3/s
GATE MASTER’S ACADEMY
GATE M ASTER’S
Level-I 1. Laminar flow takes place in a circular
tube. At what distance from the bound- ary does the local velocity equal the aver- age velocity? [KSBR 7.3] a) 0.193 R b) 0.293 R c) 0.493 R d) 0.393 R
1. Ans : (a)




=d.707R
4 u x 2

From boundary 0.293R
2. The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression
2 2

r x

2
maxU 2 u
2 2
Laminar Flow
3. An oil of density 917 kg/m3 is being pumped in a pipe of diameter 15cm. The discharge is measured as 850 L/min. The drop in pressure in a stretch of 800 m of pipeline, both ends of which are at the same elevation, is measured as 95 kPa. Estimate the absolute viscosity of the oil.
[KSBR 7.13] a) 1.1041Pa.s b) 0.2041Pa.s c) 2.1041Pa.s d) 0.1041Pa.s
Ans : (d)
Q AV

Level-II 1. A water tank containing 5m depth of wa-
ter has a horizontal crack of thickness 0.12 mm extending all the way through the 50 cm thickness of the wall at that location. The water seeps through this crack to atmosphere. Estimate the leak- age rate per minute per metre width of the crack. [Kinematic viscosity of water=0.01 stoke]
[KSBR 7.19] a) 0.848L/min b) 0.648L/min c) 0.748L/min d) 0.948L/min
Ans : (a) Sol: h = 5m
B = 0.12mm = 0.12×10–3m
L = 50 cm = 50×10–2 m
0.01 stokes = 0.01cm2/sec
= 0.01×10–4m2/sec




Q = A×V B 0.12 mm
L=50 cum
= 1.412×10–5 m3/sec

Laminar Flow
2. Oil of viscosity 1.5 Pa.s and relative den- sity 0.9 flows through a circular pipe a diameter 5cm with a mean velocity of 1.2m/s. The shear stress at the wall in pa is [KSBR 7.6] a) 360 b) 288 c) 180 d) 144
Ans : (a)
R = 2.5×10–2m



= 2.4 m/s



max 288 Pa
3. Oil of viscosity 0.97 poise and relative density 0.9 is flowing in a horizontal cir- cular pipe of diameter 100 mm and of length 10m. If 100 kg of oil is collected at the outlet of the pipe in 30sec. The differ- ence in pressure at the two ends of the pipe is [KSBR 7.11] a) 2467 Pa b) 3467 Pa c) 1467 Pa d) 5467 Pa
Ans : (c)
Sol: 2
100Mass /sec kg /sec 3.33kg /sec 30

3AV 3.7 10

1 2P P 1458 Pa
4. Oil of relative density 0.92 and dynamic viscosity 1.05 poise flows between two fixed parallel plates kept 1.2cm apart. If the mean velocity is 1.40m/s calculate (a) the maximum velocity, (b) the boundary
GATE MASTER’S ACADEMY
GATE M ASTER’S
Laminar Flow
shear stress, (c) velocity and shear stress at a distance of 0.2cm from the plates, and (d) head loss in a distance of 25m.
[KSBR 7.21] Ans : a) 2.10m/s
b) 73.5Pa c) 1.167m/s, 49Pa d) 34.0m
Sol: S=0.92, 3920kg /m
=1.05 poise
2 2

V = 1.40 m/sec.
221 P2.1 1.2 10 8 1.05 0.1 x


2 y


Vy =1.167 m/s


fh 34m
Common Data for Questions 5 and 6 : An upward flow of oil (mass density 800 kg/ m3, dynamic viscosity 0.8 kg/m-s) takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown in the figure. The pressures at sections 1 and 2 are measured as p1=435 kN/m2 and p2=200 kN/m2.
1 2
5 m
450
5. The discharge in the pipe is equal to (GATE (C/M)2006 : 2 Marks)
a) 0.100 m3/s b) 0.127 m3/s c) 0.144 m3/s d) 0.161 m3/s
Sol: =800 kg/m3
22 2 21 1
g 2g g 2g
2 3


800 9.81 0.1
3Q 0.127 m /sec
6. If the flow is reversed, keeping the same discharge, and the pressure at section 1 is maintained as 435 kN/m2, the pressure at section 2 is equal to
(GATE (C/M) 2006 : 2 Marks) a) 488 kN/m2 b) 549 kN/m2
c) 586 kN/m2 d) 614 kN/m2
6. Ans : (d) Sol:
22 2 21 1
g 2g g 2g


Turblent flow
LEVEL - I 1. In pipe flow the Prandtl mixing length l is
assumed as equal to Ref : 10.5 KSB a) k y2 b) k y c) k log y d) –k/y
2. A liquid flows in a 30 cm diameter pipe at a Reynolds number of 106. If the friction factor is 0.025, the thickness of laminar sublayer in mm is Ref : 10.19 KSB a) 0.025 b) 1.00 c) 0.062 d) 0.0031
2. Ans : (c) Sol: dia of pipe (D) = 30 cm = 300 mm
R = 150 mm
0.062mm
3. In a turbulent flow through a pipe of ra- dius r0, the radial distance at which the local velocity is equal to the mean velocity is Ref : 10.21 KSB a) 0.1 r0 b) 0.223 r0
c) 0.777 r0 d) 0.5 r0
4. The head loss in 100m length of a 0.1 m diameter pipe (f=0.02) carrying water is 10 m. The boundary shear stress, in kPa, is
Ref : 10.24 KSB a) 9.79 b) 0.298 c) 0.1958 d) 0.0245
4. Ans : (d) Sol: L = 100 m
d = 0.1 m


Turblent flow
LEVEL - II 1. In turbulent flow in a pipe of radius R the
friction factor f was found to be 0.02. Estimate a) the shear velocity u*
b) the ratio of the local velocity to mean velocity at a radial distance of 0.3R from the axis of the pipe
ans: a) 0.05V b) 1.143 Ref : 10.3 KSB
Sol: f = 0.02 R y
r=0.3 R
*
l
u 1.143 u
2. A 30 cm diameter pipe conveys water in turbulent flow regime. The friction factor has been estimated as 0.02 for this flow. If the centreline velocity is measured as 3.57m/s, determine the discharge in the pipe. Ref : 10.5 KSB a) 0.2138 m3/s b) 0.312m3/s c) 0.412m3/s d) 0.512m3/s Ans : (a)
Sol: D = 30 cm = 0.3 m
2Area (A) 0.3 4


3Q 0.2138m /sec
3. A smooth pipe 200 mm in diameter carries crude oil at a velocity of 2.5 m/s. Estimate the loss of head in 100 m and the power required to maintain the flow. Assume = 0.40 cm2/s and relative density =0.90.
Ref : 10.6 KSB hf = 4.762 m P = 3.30 kW
GATE MASTER’S ACADEMY
GATE M ASTER’S
s = 0.90


P = 3299W
4. A discharge of 300L/s of petrol ( 3680m /s and 42.9 10 Pa.s) is passing through a 30cm diameter pipe in smooth turbulent flow regime. The friction factor f of this flow estimated as 0.0097.
Determine the maximum velocity in the pipe. Ref : 10.8 KSB a) 4.8m/s b) 5.8m/s c) 7.8m/s d) 6.8m/s
Ans : (a) Sol: Q = 300 L/S
Q = 0.3 m3/s
maxu 4.8m/sec
5. If the velocities in a 60cm pipe carrying oil are 4.50m/s and 4.20m/s on the centreline and at a radial distance of 10cm from the axis, calculate the discharge in the pipe. Ref : 10.11 KSB a) 1.958m3/s b) 0.758m3/s c) 0.958m3/s d) 0.858m3/s Ans : (c)
Sol: D = 60 cm = 0.6m
GATE MASTER’S ACADEMY
GATE M ASTER’S
Q = ?
*
3Q 0.958 m /s
6. A pipe of 25cm diameter carries air (=1.22kg/m3, =1.5×10-5m2/s) at an av- erage velocity of 8m/s. The equivalent sand grain roughness of the pipe is 0.50mm. Calculate the friction factor f by assuming the flow to be fully rough-tur- bulent. What is the shear stress at the boundary? Ref : 10.12 KSB a) 0.3285Pa b) 0.4285Pa c) 0.2285Pa d) 0.1285Pa Ans : (c)
Sol: D = 25 cm = 0.25 m
R = 0.125m
31.22kg /m
*
*
LEVEL - I 1. Given that the velocity distribution in a
laminar boundary layer due to flow over a
flat plate is 3u 3 1
U 2 2
momentum thicknesses in terms of the nominal boundary layer thickness .
Ref : 8.2 KSB
a) * 3 8
0
1d dy
3
0

0




0


3 3 3 1 3 3 4 4 20 8 20 28
39 280

2. The correct relationship among displacement thickness ‘d’, momentum thickness ‘m’ and energy thickness ‘e’ is Ref : MCQ CE 361 page a) d > m > e b) d > e > m c) e > m > d d) e > d > m
3. The concept of boundary layer decides the flow fluid into two zones, these zones are : Ref : MCQ CE 361 page a) the region in which eddies formation
takes place b) the region in which viscous effect are
negligible and flow is analysed by potential flow theory
c) the region in which turbulence and viscous effect both takes place
d) the region in which viscous effect are predominant
Class Room Teaching
Boundary Layer Theory
4. The predominant forces acting on an element of fluid in the boundary layer over a flat plate in a uniform parallel stream are [GATE ME 1990 : 1 Mark] a) Viscous and pressure forces b) Viscous and inertia forces c) Viscous and body forces d) Inertial and pressure forces
5. If x is the distance measured from the leading edge of a flat plate, then laminar boundary layer thickness varies as [GATE ME 2002 : 1 Mark] a) 1/x b) x4/5
c) x2 d) x1/2
x
1 2x
6. Velocity distribution in a boundary layer flow over a plate is given by
m(u/u ) 1.5
Where, y / ; y is the distance measured normal to the plate; is the boundary layer thickness; and um is the maximum velocity at y= . If the shear stress , acting on the plate is given by
mK( u )/
Where, is the dynamic viscosity of the fluid, K takes the value of [GATE CE 2002 : 1 Mark] a) 0 b) 1 c) 1.5 d) None of these
Ans : (c)
K = 1.5 Level - II
1. Assuming that the shear stress distribution in a laminar boundary layer
is such that 0 y1
. Calculate the
displacement and momentum thickness of this boundary layer in terms of Ref : 8.3 KSB Page No 261
a) * 2, 3 15
b) * 12, 3 15
d) * 2, 13 15






2 3



Linked Answer Questions 2 and 3: An automobile with projected area 2.6 m2
is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are 1.2 kg/m3
and 1.5×10–5 m2/s, respectively. The drag coefficient is 0.30.
2. The drag force on the automobile is Ref : [GATE CE 2008 : 2 Marks]
a) 620 N b) 600 N c) 580 N d) 520 N
Ans : (d) Sol. The drag force on the automobile may be
given as, 2

V = 120 kmph
2 (60 60)
FD = 520 N 3. The metric horse power required to
overcome the drag force is Ref : [GATE CE 2008 : 2 Marks] a) 33.23 b) 31.23 c) 23.23 d) 20.23
Ans : (c) Sol. Power required to overcome the drag
= Drag force Velocity
= 17333.33
735.5 metric horse power
= 23.567 metric horse power 4. For air flow over a flat plate, velocity (U)
and boundary layer thickness ( ) can be expressed respectively, as
3
x

If the free stream velocity is 2 m/s, and air has kinematic viscosity of 1.5×10–5
m2/s and density of 1.23 kg/m3, the wall shear stress at x=1m, is [GATE ME 2004 : 2 Marks, set-1] a) 2.36×102 N/m2 b) 43.6×10-3 N/m2
c) 4.36×10-3 N/m2 d) 2.18 ×10-3 N/m2
Ans : (c)








2 0.0127
0 = 4.36 10–3 N/m2
5. A steady laminar boundary layer is formed over a flat plate as shown in the figure. The free stream velocity of the fluid is U0. The velocity profile at the inlet a-b is uni- form, while that at a downstream location
c-d is given by 2
0 y yu U 2
.
a c
d bdm
The ratio of the mass flow rate, bdm , leav- ing through the horizontal section b-d to that entering through the vertical section a-b is______
[GATE ME 2016 : 2 Marks, set-1] a) 3.33 b) 0.333 c) 333 d) 33.3
Ans : (b)
bd m.
U0 = Velocity of free stream d = Boundary layer thickness at section c-d b = Width of the plate
Velocity profile u = 2
0 y yU 2
0mab = b U Mass flow rate through section c-d,
cd 0
m bdyU

=

Mass flow rate through section b-d
bd ab cdm m m
= 0 0 2b U b U 3
= 0b U 3


Level - I 1. The number of parameters needed to
express the function F (A, V, t, , L) = 0 are (GATE (C) 1994 : 1 Mark) a) 5 b) 4 c) 3 d) 2
Ans : (d) Sol. f (A, V, t, , L) = 0
where, Number of variables, n = 0 number of fundamentals dimension, m=0 Number of -terms = n – m
= 5–3 = 2
2. The Euler number En is written as En= (KSBM 6.4 page 226)
a)
m p
r rV L
3. A dimensionless combination of pressure drop p , dynamic viscosity , velocity V and length L is (KSBM 6.6 page 226)
a) p L V
p V L
4. Dynamic similarity exists when the model and prototype have the same
(KSBM 6.7 page 226) a) length scale ratio and time scale ratio b) length scale ratio and velocity scale
ratio c) length scale ratio, time scale ratio and
velocity scale ratio d) length scale ratio, velocity scale ra-
tio and force scale ratio 5. The time scale ratio for a model based on
Froude law criterion in terms of length scale ratio Lr is (KSBM 6.10 page 226)
a) Lr b) rL c) r
1 L d) 1.5
rL
6. In Reynolds law of similitude the discharge ratio Qr= (KSBM 6.15 page 227)
a) r r
r
L
7. Match List - I (Predominant force) with List - II (Dimensionless numbers) and se- lect the correct answer using the codes given below the lists
(ESE (M) : 1996) List - I A. Compressibility force B. Gravity force C. Surface tension force D. Viscous force List - II 1. Euler number 2. Froude number 3. Mach number 4. Reynolds number 5. Weber number Codes :
A B C D a) 1 2 3 4 b) 3 2 5 4 c) 3 1 4 2 d) 2 3 5 1
Class Room Teaching
13. DIMENSIONAL ANALYSIS
Dimensional Analysis
8. Match List-I (Flow/Wave) with List-II (Di- mensionless Number) and select the cor- rect answer using the codes given below the lists : (ESE : 2003) List-I A. Capillary waves in channel B. Testing of aerofoils C. Flow around bridge piers D. Turbulent flow through pipes List-II 1. Reynolds number 2. Froude number 3. Weber number 4. Euler number 5. Mach number
Codes : A B C D
a) 5 4 3 2 b) 3 5 4 1 c) 5 4 2 1 d) 3 5 2 1
9. The relationship between the length scale
ratio rL and the velocity scale ratio rV in hydraulic models, in which Froude dy- namic similarity is maintained, is
(GATE (C) 2015 : 2 Marks, Set-II)
a) r rV L b) rrL V
c) 1.5 r rV L d) r rV L
Level - II 1. An 1 : 50 model of an ogee spillway crest
records an acceleration of 1.5 m/sec2 at a certain location. The homologous value of acceleration in the prototype is ______ (ap=1.5 m/s2)
(GATE (C) 1991 : 2 Marks) Ans : ap = 1.5 m/s2
Sol. (Fr)m = (Fr)p

Ta V T V a V T V T
p
....... (i)



ap = 1.5 m/sec2
2. A laboratory model of a river is built to a geometric scale of 1:100. The fluid used in the model is oil of mass density 900 kg/m3. The highest flood in the river is 10,000 m3/s. The corresponding dis- charge in the model shall be
(GATE (C) 2003 : 1 Mark) a) 0.095 m3/s b) 0.100 m3/s c) 0.105 m3/s d) 10.5 m3/s
Ans : (b) Sol. From froude's model law
3
m m
r rV L
Dimensional Analysis
Qm = 0.1 m3/sec 3. The model of a propeller, 3m in diameter,
cruising at 10 m/s in air, is tested in a wind tunnel on a 1:10 scale model. If a thrust of 50N is measured on the model at 5m/s wind speed, then the thrust on the prototype will be (ESE (M) : 1995) a) 20,000 N b) 2,000 N c) 500 N d) 200 N
Ans : (a)



Fp = 20,000 N 4. A 1:50 spillway model has a discharge of
1.25 m3/s. What is the corresponding pro- totype discharge? If a flood phenomenon takes 12 hours to occur in the prototype, how long should it take in the model?
(KSBR) a) 11097 m3/s, 3.697 hr b) 33097 m3/s, 2.697 hr c) 22097 m3/s, 1.697 hr d) 44097 m3/s, 4.697 hr
Ans : (c) Sol. From Froude's model law
pm r rm p
Discharge ratio (Qr) = 2 5/2 r r rV L L
r 1L 50

Time ratio (Tr) = r r
r r r


tm = 1.697 hours
5. A pipe of diameter 1.5 m is required to transport an oil of relative density 0.9 and kinematic viscosity=3×10-2 stoke at a rate of 3.0 m3/s. If a 15 cm diameter pipe with water at 20oC ( =0.01 stoke) is used to model the above flow, find the velocity and discharge in the model. (KSBR)
a) 5.659m/s, 0.1m3/s b) 6.659m/s, 0.1m3/s
c) 3.659m/s, 0.2m3/s
d) 2.659m/s, 0.1m3/s
Ans : (a) Sol. The Reynolds number must be the same
in the model and prototype for similar pipe flows
p p m m

4
Level-I 1. A circular jet of water emanating from a
nozzle strikes a stationary vertical plate normally. The force exerted is measured as 635 N. If the discharge from the nozzle is measured as 42.4 litres/s estimate the diameter of the nozzle at the exit.
[KSBR 5.6] a) 7cm b) 5cm c) 4cm d) 6cm
Ans : (d) Sol. Force on the stationary plate due to Jet
impingement F = Qv
m3/sec
F = 635 N 635 = 998 0.0424 v v = 15 m/sec
2 0.0424D 15
D 0.0036 D = 0.06m D = 6 cm
2. A two-dimensional jet of water of thick- ness 10 cm and issuing with a velocity of 10 m/s strikes a stationary plate at an angle of 30o to the normal of the plate. Calculate the force on the plate.
[KSBR 5.3] a) 8.65kN b) 9.65kN c) 7.65kN d) 6.65kN
3. A horizontal nozzle of 30mm diameter dis- charges a steady jet of water into the at- mosphere at a rate of 15 litres per sec- ond. The diameter of inlet to the nozzle is 100 mm. The jet impinges normal to a flat stationary plate held close to the nozzle end. Neglecting air friction and considering the density of water as 1000
kg/m3, the force exerted by the jet (in N) on the plate is_____ (318.3N)
(GATE (C) 2014 - 2M)
Level-II 1. A 5 cm diameter jet of water having a ve-
locity of 20 m/s impinges normally on a flat vertical plate that is moving away from the jet with a velocity of 8.0 m/s. Find (a) force on the plate, (b) work done per sec- ond and (c) efficiency of power transfer by jet impingement. [KSBR 5.4]
Ans: a) 282.15N b) 2.26kW c) 28.8%
Vr = (v–u) = 20–8 = 12m/sec
V V-u
V-u Relative steady ?ow
(a) Quantity of fluid mass that strikes the plate per second = relative discharge
2 rQ D (v u)
4
Rx = 998 0.02356 12 Rx = 282.15N
Force on the plate (Fx) = Rx in opposite direction
Fx = 282.15 N (b) Workdone per second
p = Fxu p = 282.15 8 p = 2257.2 W p = 2.26 kw
Class Room Teaching
impingement = workdone per second
Kinetic energy flux of the jet = 2 1
1 v(Av ) 2
4 2
KE = 7838.3W
2257.2Efficiency ( ) 7838.3
= 0.288 = 28.8%
2. A square plate is suspended vertically from one of its edges using a hinge support as shown in figure. A water jet of 20 mm di- ameter having a velocity of 10 m/s strikes the plate at its midpoint, at an angle of 300
with the vertical. Consider ‘g’ as 9.81 m/s2
and neglect the self-weight of the plate. The force ‘F’ (expressed in N) required to keep the plate in its vertical position is________ (F = 7.85 N)
Water Jet
300
Force exerted by jet in x-direction Fx = m [Vsin – 0]
= Q V sin = AV vsin
= 2 21000 (0.02) (10) sin30 4

x 0.2F F 0.2 2

F = 7.85 N 3. A 60mm diameter water jet strikes a plate
containing a hole of 40mm diameter as shown in the figure. Part of the jet passes through the hole horizontally, and the remaining is deflected vertically. The den- sity of water is 1000 kg/m3. If velocities are as indicated in the figure, the magni- tude of horizontal force (in N) required to hold the plate is______.
(GATE (M) 2014 - 2M)
20 m/s 20 m/s
Ans : F = 628.32N Sol. = 1000 kg/m3
Force (F) = momentum inlet – momentumoutlet
in out F mv mv
2 2 2 2 1F D D v
4
F = 628.32 N
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 Ans : (c) Sol. discharge (Q) = 3 m3/sec
Head (H1) = 15 m Head (H2) = 12 m Rotational speed (N1) = 500 rpm Rotational speed (N2) = ?
21

N2 = 447.21 rpm 2. A hydraulic turbine has a discharge of 3
m3/s when operating under a head of 15 m and a speed of 500 rpm. If it is to operate under 12 m of head, the rotational speed will be [KSRB-16.1, 588P]
a) 600 rpm b) 559 rpm c) 447 rpm d) 400 rpm
3. A turbine works under a head of 20m, has a speed of 375 rpm and develops 400 kW of power. Its specific speed is
[KSRB-16.7, 588P]
a) 375 b) 83
c) 177 d) 1677 Ans : (c) Sol. Head (H) = 20 m
Speed (N) = 375 rpm Power (P) = 400 KW
Specific speed (Ns) = N P
5(H) 4
Ns = 177.32 4. A turbine develops 500 kW power under
a net head of 30 m. If the overall efficiency of the turbine is 0.83, the discharge through the turbine, in m3/s, is
[KSRB-16.14, 589P] a) 20.5 b) 2.05 c) 1.41 d) 1.51
Ans : (b) Sol. Power (P) = 500 KW = 500 103 W
Head (H) = 30m Overall efficiency (0) = 0.83 Discharge (Q) = ? Power (P) = 0 gQH
Discharge (Q) = 0

Q = 2.046 2.05 m3/sec
5. A Francis turbine has runner of outer and inner diameters 4 m and 2m respectively. The breadth at inlet and also at outlet is 0.8 m and the velocity of flow is constant at 3.0 m/s. The discharge through the turbine, in m3/s, is
[KSRB-16.25, 590P] a) 28.28 b) 37.70 c) 30.16 d) 15.08
Ans : D
Discharge (Q) =D1B1 1f V
Q = 2 0.8 3 Q = 15.079 15.08 m3/sec Q =D2B2 2f
V
Q = 4 0.8 3 Q = 30.159 Q 30.16 m3/sec
6. In a general sense, water turbines may be put in the following decreasing order of specific speeds, as :
[KSRB-16.22, 590P]
b) Pelton wheel, Francis turbine, Kaplan turbine
c) Kaplan turbine, impulse turbine, Francis turbine
d) Francis turbine, Kaplan turbine, Pelton wheel
7. Which one of the following statements is TRUE? [GATE 2017 : Set -II 1 mark]
a) Both Pelton and Francis turbines are impulse turbines.
b) Francis turbine is a reaction turbine but Kaplan turbine is an impulse turbine.
c) Francis turbine is an axial- flow reaction turbine.
d) Kaplan turbine is an axial - flow reaction turbine.
8. In order to have maximum power from a pelton turbine, the bucket speed must be
[GATE(M)-2013 : 1Mark]
b) equal to half of the jet speed
c) equal to twice the jet speed
d) independent of the jet speed Ans : (b)
Sol. Condition for maximum power from a pelton turbine Bucket speed = Half of the jet speed
Vu 2

9. A hydraulic turbine develops 1000kW power for a head of 40m. If the head is reduced to 20m, the power developed (in kW) is ________ [GATE(M)-2010 : 1Mark]
a) 177 b) 354 c) 500 d) 707
Ans : (b) Sol. Power (P1) = 1000 KW
Head (H1) = 40 m Head (H2) = 20 m Power (P2) = ?
Unit Power (Pu) = 3 2
P
1 2
P2 = 353.55 KW 354 KW
LEVEL - II 1. A penstock of 1 m diameter and 5 km
length is used to supply water from a reservoir to an impulse turbine. A nozzle of 15 cm diameter is fixed at the end of the penstock. The elevation difference between the turbine and water level in the reservoir is 500 m. Consider the head loss due to friction as 5% of the velocity head
GATE MASTER’S ACADEMY
GATE M ASTER’S
Hydraulic Turbines
available at the jet. Assume unit weight of water =10 kN/m3 and acceleration due to gravity (g)= 10 m/s2. If the overall efficiency is 80%, power generated (expressed in kW and rounded to a nearest integer) is_______
(GATE(C) - 2016 : 2 Marks, Set-II) Ans : 6570 KW Sol. Given : Length (L) = 5 km = 5 103 m
Penstock Diameter (D) = 1m Nozzle Diameter (d) = 15 cm = 0.15 m Elevation difference between the turbine and water level in the reservior (H) = 500m Head loss due to friction (hf) = 5% of the velocity head
2
Acceleration due to gravity (g)= 10m/sec2
Overall efficiency (0) = 80% = 0.80 Applying energy equation
22 2 21 1
g 2g g 2g
2g 2g V = 97.59 m/sec
Overall efficiency (0) = Shaft power Water power
3



Shaft power (S.P) = 6569.74 103W Shaft power (S.P) = 6569.74 KW
6570 KW
2. A Kaplan turbine develops 15000 kW power at a head of 30 m. The diameter of the boss is 0.35 times the diameter of the runner. Assuming a speed ratio of 2.0, a flow ratio of 0.65 and an overall efficiency of 90% calculate the (i) diameter of the runner, (ii) rotational speed and (iii) specific speed. [KSRB-16.11, 548P] Ans :
i) D= 2.285 m
ii) N = 405.5 rpm
iii) Ns = 707.4 rpm
2gH
Diameter of boss (Db) = 0.35D
Flow ratio = 1f
1f V 0.65 2 9.81 30
1f V 15.77 m/sec Power (P) = gQH 1500 103 = 0.90 998 9.81 Q 30 Q = 56.745 m3/sec
Discharge (Q) = 1
4
(ii) Speed = N rpm
5(H) 4
s 5/4

Ns = 707.4 3. A conical draft tube of a reaction turbine
has diameters of 1.5 m and 2.0 m at its ends. It is set 7.0 m above the tailrace level. When the discharge from the tur- bine is 12 m3/s, calculate (i) pressure h ead at the entrance of the draft tube and (ii) efficiency of the draft tube. Assume loss of energy in the draft tube as 0.35 times the velocity head at the draft tube exit. [KSRB-16.14, 550P]
Ans :
2g 2 9.81
2g 2 9.81
2V 0.35
Applying Bernoulli's equation sections (1) and (2)
22 2 21 1
g 2g g 2g

Take the bottom of the draft tube as datum Let the depth of draft tube below tail Water level = y Taking atmospheric pressure head
= 10.3 m of water
1P = 10.3+0.744+0.260–2.351–7
= 1.953 m (abs) (ii) Efficiency of the draft tube
L d 22
d = 83.8% 4. A double overhung 1.5 m diameter im-
pulse turbine installation is to develop 3000 kW at 400 rpm under a net head of 270 m. If the overall efficiency is 0.90, determine the (i) diameter of the jet (ii) speed ratio and (iii) specific speed. (Take
GATE MASTER’S ACADEMY
GATE M ASTER’S
2
Power (P) = gQH 1500 103 = 0.90 1000 9.81 270 Q
31500 10Q 0.90 1000 9.81 270

2 v0.629 d C 2gH
4


N P (H)
ing in series and in parallel is
[GATE(C) 1998, 1 M 4 Q]
a) Pumps operating in series boost the discharge, whereas pumps operating in parallel boost the head
b) Pumps operating in parallel boost the discharge, whereas pumps op- erating in series boost the head
c) In both cases there would be a boost in discharge only
d) In both cases there would be a boost in head only
2. A pump can lift water at a discharge of 0.15m3/s to a head of 25 m. The critical cavitation number c for the pump is found to be 0.144. The pump is to be in- stalled at a location where the barometric pressure is 9.8m of water and the vapour pressure of water is 0.30m of water. The intake pipe friction loss is 0.40 m. Using the minimum value of NPSH (Net Positive Suction Head), the maximum allowable elevation above the sump water surface at which the pump can be located is___
[GATE(C) 2002 : 2 M Q.6 ]
a) 9.80 m b) 6.20 m
c) 5.50 m d) None of these
Ans : (c) Sol. c = NPSH/H
NPSH0.144 25
NPSH = 3.6m
NPSH = Hatm – Hv – hs– hL 3.6 = 9.8 – 0.3 – hs – 0.4 hs = 5.5 m
3. A centrifugal pump is started with its de- livery valve kept CRT [ESE(M) 1997]
a) fully open b) fully closed
c) partially open d) 50% open
4. Consider the following statements:
Cavitation in hydraulic machines occurs at
P.Q. [ESE(M) 1995]
a) 1 and 2 b) 1 and 3
c) 1,2 and 3 d) 2 and 3
5. A centrifugal pump driven by a directly coupled 3kW motor of 1450 rpm speed, is proposed to be connected to another motor of 2900 rpm speed. The power of the motor should be
P.Q. [ESE(M) 2002]
a) 6 kW b) 12 kW
c) 18 kW d) 24 kW Ans : (d) Sol. P N3 (for constant diameter)
3 3 2 2

P2 = 3 8 = 24 KW 6. Two centrifugal pumps A and B operate
at their maximum efficiencies at 1000 rpm and 500 rpm respectively. Against the
Class Room Teaching
16. CENTRIFUGAL PUMPS
Centrifugal Pumps
same delivery head, pump A discharges 1m3/s and pump B discharges 4m3/s respectively. What is the ratio of specific
speeds s sA BN : N ?
P.Q. (ESE (M) : 2004) a) 1 : 2 b) 1 : 1
c) 1 : 4 d) 4 : 1 Ans : (b)
Sol. s 3/4
N Q N
N 1000 1 1:1 N 500 4
7. Water is pumped to a height of 10m at the rate of 30.1m /sec . Frictional and other losses are 5m. The power of pump required will be CRT (AEE CE/ME-2004) a) 1000 Kg m/sec b) 1500 Kg m/sec c) 500 Kg m/sec d) 2000 Kg m/sec
Ans : (b) Sol. Power of pump (P) = gQ (H+hf)
P = gQ (H+hf) = 1000 9.81 0.1 (10+5)
= 1500 9.81 N-m/sec = 1500 kg-m/sec
Level- II 1. A centrifugal pump is to discharge 0.11
m3/s at speed of 1450 rpm against a head of 25 m. The impeller diameter is 25 cm, its width at outlet is 5 cm and manomet- ric efficiency is 75%. The vane angle at outer periphery of impeller is
(ESE (M) : 2013) a) 56.770 b) 59.770
c) 61.770 d) 48.770
Outlet velocity triangle
Outlet velocity triangle Given, Q = 0.118 m3/s N = 1450 rpm Hm = 25m D2 = 0.25m b2 = 0.05 m mano = 0.75
2f 2 2
QV D b

Centrifugal Pumps
2. Water at 200C is delivered by a pump with 1500 L/min against a pressure rise of 270 kN/m2. The driving motor supplies 9 kW of power. If the change in kinetic and po- tential energies are negligible, the overall efficiency of the pump is
[Nodia 9.5, 329P] a) 68% b) 50% c) 71% d) 75%
Ans : (d) Sol. Pwater = gVH [P = gH)
= Vp
= 1500 270
1000 60
= 6.75 KW

= 75% 3. Consider the test pump as shown in the
figure below. The data are :
1
2
65 cm
Inlet pressure p1= 100 mm Hg (Vacuum) Outlet pressure p2 = 500 mm Hg (gauge) Inlet diameter D1= 12 cm Outlet diameter D2= 5 cm
Flow rate =0.01136 m3/s Fluid is light oil S.G. = 0.91 Efficiency = 75% What will be the input power?
[Nodia 9.24, 332P]
a) 1525 W b) 858 W
c) 760 W d) 1715 W Ans : (a) Sol. P1 = 100 mm Hg = 13332 Pa
P2 = 500 mm Hg = 66661 Pa oil = 0.91 9790
= 8909 N/m3
4
4
2 1 P V P VH Z Z
2g 2g
= 11.3m
= 1525 W 4. A pump running at 1000 rpm consumes
1 kW and generates head of 10 m of wa- ter. When it is operated at 2000 rpm, its power consumption and head generated would be
(ESE (M) : 2002)
a) 4 kW, 50 m of water
b) 6 kW, 20 m of water
c) 3 kW, 30 m of water
d) 8 kW, 40 m of water Ans : (d) Sol. Given data
Condition - 1 N1 = 100 rpm P1 = 1 KW H1 = 10 m Condition-2 N2 = 2000 rpm P2 = ? H2 = ? Head coefficient,
H 2 2
gHC N D
gHgH N D N D

HH N N
2
p 3 5
PC N D
1 1 3 5 3 5
1 1 2 2