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Mechanical Engineering
INDUSTRIAL ENGINEERING
TOPICWISE GATE SOLUTION
1991-2013
F-108, Katwaria Sarai, Near Mother Dairy Booth, New Delhi-16
Ph. +91-011-64551144, 9810758209
www.drona.org
DRONACHARYA INSTITUTE OF ENGINEERS
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Fluid Mechanics:Fluid properties; fluid statics, manometry, buoyancy; control-volume analysis of mass,
momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoullis
equation; viscous flow of incompressible fluids; boundary layer; elementary turbulent flow; flow through
pipes, head losses in pipes, bends etc. Pelton-wheel, Francis and Kaplan turbines - impulse and reaction
principles, velocity diagrams.
Syl labus of GATE Examination
S.NO. TOPIC PAGE NO.
Fluid Mechanics 69 - 156
1. ............... Properties of Fluids................................................................................ 71- 75
2. ............... Pressure and its Measurement ................................................................. 76 - 80
3. ............... Hydrostatic Forces on Surfaces .............................................................. 81 - 84
4. ............... Buoyancy and Flotation .......................................................................... 85 - 86
5. ............... Fluid Kinematics .................................................................................... 87 - 97
6. ............... Fluid Dynamics...................................................................................... 98 - 111
7. ............... Dimensional and Model Analysis ............................................................. 112 - 115
8. ............... Boundary Layer Theory ......................................................................... 116 - 1249. ............... Laminar and Turbulent Flow ................................................................... 125 - 131
10. ............. Flow Through Pipes............................................................................... 132 - 139
11. .............. Hydraulic Turbines ................................................................................. 140 - 147
12. ............. Centrifugal Pump ................................................................................... 148 - 153
13. ............. Compressible Flow................................................................................ 154 - 155
NOMENCLATURE OF CHAPTERS
Fluid MechanicFluid Mechanic
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Conclusion
S.No. Chapter Name 03 04 05 06 07 08 09 10 11 12 13
1 Properties of Fluids 1 1 2
2Pressure and its
Measurement2 2
3Hydrostatic Forces on
Surfaces2 2 2
4Buoyancy and
Flotation1 1
5 Fluid Kinematics 2 1 1 3 5 2 1
6 Fluid Dynamics 2 4 4 2 2 2 2 2 2 2
7Dimensional and
Model Analysis1 3
8Boundary Layer
Theory2 4 5 2
59Laminar and Turbulent
Flow2 1 2 1 1
10 Flow Through Pipes 4 4 1 2 1
11 Hydraulic Turbines 4 3 4 2 1 1
12 Centrifugal Pump 2 2 2
13. Compressible Flow
Total 13 18 7 17 16 11 8 8 3 5 6
Statistical Analysis
1. Fluid Mechanics has approximate 6 to 8% weightage in GATE.
2. From analysis it is clear that one should focus on Kinematics and Dynamics of Flow, Boundry Layer
Theory, Francis Turbine, Flow through Pipes, Laminar Flow, and Centrifugal Pump.
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Properties of Fluids
1
Year 2008
1. A journal bearing has shaft diameter of 40 mmand a length of 40 mm. The shaft is rotating at 20
rad/s and the viscosity of the lubricant is 20 mPa-
s. The clearance is 0.020 mm. The loss of torque
due to the viscosity of the lubricant is
approximately
(a) 0.040 Nm (b) 0.252 Nm
(c) 0.400 Nm (d) 0.652 Nm
Year 2006
2. For a Newtonian fluid (a) Shear stress is proportional to shear strain
(b) Rate of shear stress is proportional to shear
strain
(c) Shear stress is proportional to rate of shear
strain
(d) Rate of shear stress is proportional to rate of
shear strain
Year 2004
3. An incompressible fluid (kinematic viscosity, 7.4x 107 m2/s, specific gravity, 0.88) is held between
two parallel plates. If the top plate is moved with
a velocity of 0.5 m/s while the bottom one is held
stationary, the fluid attains a linear velocity profile
in the gap of 0.5 mm between these plates; the
shear stress in Pascals on the surface of top
plate is:
(a) 0.651 x 103 (b) 0.651
(c) 6.51 (d) 0.651x103
Year 2001
4. The SI unit of kinematic viscosity (v) is(a) m2/sec (b) kg/(m-sec)
(c) m/sec2 (d) m3/sec2
5. A static fluid can have
(a) non-zero normal and shear stress
(b) negative normal stress and zero shear stress
(c) positive normal stress and zero shear stress
(d) zero normal stress and non-zero shear stress
Year 1999
6. Kinematic viscosity of air at 20oC is given to be
1.6105m2/s. It kinematic viscosity at 70oC will
be vary approximately
(a) 2.2105m2/s (b) 1.6105m2/s
(c) 1.2105m2/s (d) 105m2/s
Year 1996
7. The dimension of surface tension is
(a) ML1 (b) L2 T1
(c) ML1T1 (d) None of these
Year 1995
8. A fluid is said to be Newtonian when the shear
stress is
(a) directly proportional to the velocity gradient
(b) inversely proportional to the velocity gradient
(c) independent of the velocity gradient
(d) none of the above
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Chapter-1Answers
1. Ans. (a) 2. Ans. (c) 3. Ans. (b) 4. Ans. (a) 5. Ans. (c)6. Ans. (a) 7. Ans. (d) 8. Ans. (a)
Space for Rough work
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Chapter-1Answer & Explanations
Q.1 Ans. (a)
Given: Shaft diameter, d = 40 mm
Shaft length, L = 40 mm
Speed, = 20 rad/s Viscosity, = 20 mPa-s Clearance, y = 0.020 mm
40 mm
40 mm
0.02 mm = 20 Pasm
Shear stress given by Newtons law of viscosity
=du
dy
Here, u = r = 20 0.02 = 0.4 m/s
= 33
0.420 10
0.02 10
= 400 N/m2
Shear force, F = A = 400 d L= 400 0.04 0.04 = 2.0106 N
Torque loss, T = F r = 2.0106 0.02 = 0.0402 NmReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-22, Example-
1.10.
Q.2 Ans. (c)
Exp. Consider a fluid element in a real flow. In a real flow there exist a velocity gradient in the perpendicular
direction of the flow. The change in velocity in two conscutive layer of fluid flow is shown in the figure.
dy
u
du.dtu + du
d
Shear strain, tan d =du dt
dy
If dis small, then tan d ~ d .
Therefore, d =du dt
dy
d
dt
=
du
dt
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75
From Newtons law of viscosity
=du
dy =
d
dt
Hence, for a Newtonian Fluid, the shear stress is directly proportional to rate of shear strain.
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 6, 1.3.3.
or
For a Newtonian fluid, Shear stress, du
dy
du
dy, where
du
dy= velocity gradient
dy
yu
u + du
x
dxdt
dy
dx
dy
dt
, where
dx
dyis shear strain of fluid
Thus
dx
dy
dt
is rate of shear strain
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, 1.8.
Q.3 Ans. (b)
Given, Kinematic viscosity, = 7.4107m2/secSpecific gravity, S = 0.88
Density of fluid, = 0.88 1000 kg/m3
Dynamic viscosity, = = 0.88 103 7.4 107= 0.6512103Pa.s
V = 0.5 m/s
0.5 mm
Now, from Newtons law of viscosity
=3
3
.du 0.6512 10 0.50.6512
dy 0.5 10
N/m2
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.
Q.4 Ans. (a)
The SI unit of kinematic viscosity () is m2/s whereas CGS unit is cm2/s which is also known as Stoke.1 m2/s = 104stoke
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.
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Q.5 Ans. (c)
Static fluid has normal stress only. Since fluid starts flowing under the action of shear stress irrespective of
its magnitude. In static fluid, there is no flow. Therefore, there is no shear stress.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-34, Equation-2.1.
Q.6 Ans. (a)
The viscosity of liquid decreases with increase in temperature due to decrease in intermoleculer force of
attraction while the viscosity of gas increases with increase in temperature due to increase in random motion
of the molecules.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.
Q.7 Ans. (d)
Surface tension () is defined as force per unit length. It is also equivalent to surface energy per unit surfacearea. It is mainly due to force of cohesion.
Dimension of =2MLT
L
= MT2
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-14, Equation-
1.11.
Q.8 Ans. (a)
A fluid is said to be Newtonian fluid when it obeys the Newtons law of viscosity. For such fluids the viscosity
is independent from the rate of shear strain. For example water, air etc.The other types of fluid is shown in the following figure:
Ideal solid
Bingham
plasticflu
id
Pseudopl
asticfluid
Newtonia
nfluid
Ideal Fluid
Velocity gradient,du
dy
Shearstress,
Dilata
ntflu
id
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-11, Equation-1.8.
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Pressure and its Measurement
2
Year 2005
1. A U-tube manometer with a small quantity of
mercury is used to measure the static pressure
difference between two locations A and B in a
conical section through which an incompressible
fluid flows. At a particular flow rate, the mercury
column appears as shown in the figure. The
density of mercury is 13 600 kg/m3and g = 9.81
m/s2. Which of the following is correct?
A
B
150 mm
(a) Flow direction is A to B and pA p
B= 20 kPa
(b) Flow direction is B to A and pA p
B=1.4 kPa
(c) Flow direction is A to B and pB p
A=20 kPa
(d) Flow direction is B to A and pB p
A=1.4 kPa
Year 2004
2. The pressure gauges G1and G
2installed on the
system show pressures of PG1
= 5.00 bar and
PG2
= 1.00 bar. The value of unknown pressure
P is
G1
G2
P
Atmospheric pressure
1.01 bar
(a) 1.01 bar (b) 2.01 bar
(c) 5.00 bar (d) 7.01 bar
Year 2000
3. In figure if the pressure of gas in bulb A is 50 cm
Hg vaccum and patm= 76 cm Hg, the height ofcolumn H is equal to
A
HPatm
Hg
(a) 26 cm (b) 50 cm
(c) 76 cm (d) 126 cm
Year 1999
4. If p is the gauge pressure within a spherical
droplet, the gauge pressure within a bubble of the
same fluid and of same size will be
(a)p
4(b)
p
2
(c) p (d) 2p
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Year 1997
5. Refer to figure, the absolute pressure of gas A in
the bulb is
PA
10 cm
AB
C
D
F
E= 13.6 g/ml
2 cm
5 cm
(a) 771.2 mm Hg (b) 752.65 mm Hg
(c) 767.35 mm Hg (d) 748.8 mm Hg
Year 1996
6. A mercury manometer is used to measure the
static pressure at a point in a water pipe as shown
in Fig. The level difference of mercury in the two
limbs is 10 mm. The gauge pressure at the
point A is
10 mmH O2
Hg
WaterA
(a) 1236 Pa (b) 1333 Pa
(c) zero (d) 98 Pa
Year 1994
7. Net force on a control volume due to uniform
normal pressure alone
(a) depends upon the shape of the control volume(b) translation and rotation
(c) translation and deformation
(d) deformation only
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Chapter-2Answers
1. Ans. (a) 2. Ans. (d) 3. Ans. (b) 4. Ans. (d) 5. Ans. (a)6. Ans. (b) 7. Ans. (a)
Space for Rough work
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Chapter-2Answer & Explanations
Q.1 Ans. (a)
AB
150 mm
Writing the pressure balance equation,
pA
= pB+ gh
pA p
B=
150136000 9.81 20.012kPa
1000
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-36, Equation-2.2.
Q.2 Ans. (d)
G1
G2
P
Atmospheric pressure1.01 bar
Absolute pressure at 2
Pabs2
= PG2
+P
atm
= 1 + 1.01 = 2.01 bar
Absolute pressure at 1 Pabs1
= PG1
+ Patm
(Atmospheric pressure for G1becomes 2.01 bar)
= 5 + 2.01 = 7.01 bar
Q.3 Ans. (b)
A
HPatm
Hg
Applying pressure balancing equation at free surface
PA+ P
H= P
atm
Patm
= PA P
H
Taking Patm
= 0
Therefore, PH
= 50 cm
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-847, Equation-19.7.
Q.4 Ans. (d)
Pressure inside spherical droplet =4
d
Pressure inside soap bubble =8
d
, where is surface tension force and d is diameter..
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-15,
Equation-1.11.
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Q.5 Ans. (a)
PA
17 cm AB
C
D
F
E = 13.6 g/mldatum
5 cm
2 cm
Gauge pressure at A,P
A+
1gh
1=
2gh
2+
1gh
3(Taking the unknown liquid as water)
PA+
171000 9.81
100 =
2 513600 9.81 1000 9.81
100 100
PA
= 2668.32 + 490.5 1667.2 = 1491.12 N/m2
Pab s
= Patm
+ PA
= 1.013105+ 1491.12 = 102791.12 N/m2
Pab s = mghm
m
m
density of mercuryh mercury column height
hm
=102791.12
0.77055m13600 9.81
= 771 mm
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 40, 2.6.2.Q.6 Ans. (b)
10 mm
H O2
Hg
WaterA
Neglecting the depth of water column, gauge pressure is given as
Pguage
= gh =10
13600 9.811000
= 1334.16 N/m2
Q.7 Ans. (a)
Exp. Pressure =Force
Area
Net force = PressureArea.
Therefore, area defined by the shape of the control volume.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-34, Equation-2.1.
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Hydrostatic Forces on Surfaces
3Year 2013
1. A hinged gate of length 5 m,inclined at 30 with
the horizontal and with mass on its left, is shown
in the figure below. Density of water is 1000 kg/
m3. The minimum mass of the gate in kg per unit
width (perpendicular to the plane of paper),
required to keep it closed is
5m
(a) 5000 (b) 6600(c) 7546 (d) 9623
Year 2003
2. A water container is kept on a weighing balance.
Water from a tap is falling vertically into the
container with a volume flow rate of Q; the velocity
of the water when it hits the water surface is U.
At a particularly instant of time the total mass of
the container and water is m. The force registered
by the weighing balance at this instant of time is
(a) mg + QU (b) mg + 2QU(c) mg + QU2/2 (d) QU2/2
Year 2001
3. The horizontal and vertical hydrostatic forces Fx
and Fyon the semi-circular gate, having a width
w into the plane of figure, are
(a) Fx= ghrw and F
y= 0
(b) Fx= 2ghrw and Fy= 0(c) Fx= 2ghrw and F
y= gwr2/2
(d) Fx= 2ghrw and F
y= gwr2/2
Year 1992
4. A 3.6 m square gate provided in an oil tank is
hinged at its top edge (Figure). The tank contains
gasoline (sp. gr. = 0.7) upto a height of 1.8 m
above the top edge of the plate. The space above
the oil is subjected to a negative pressure of 8250
N/m2. Determine the necessary vertical pull to
be applied at the lower edge to open the gate.
45
Negative pressure (8250 N/m )2
Hinge
Gate
Gasoline surface
1.8m
P
Gasoline (S = 0.7)
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Chapter-3Answers
1. Ans. (d) 2. Ans. (a) 3. Ans. (d) 4. Ans. (144.5 kN)
Space for Rough work
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Chapter-3Answer & Explanations
Q.1 Ans. (d)
mg
mg cos 30
30
30
5m
b=1m
G
F
2.5mx
B
h
Depth of centre of gravity from free surface of water,
x = 2.5 sin 30o= 1.25 m
Hydrostatic force, F = gAx = 1000 9.81 5 1 1.25 = 61312.5 NDepth of centre of pressure,
h =2GIx sin
Ax
= 1.25 +
3
2o
11 5
12 sin305 1 1.25
= 1.67 m
For gate to be closed, moment of all forces about the hinge point must be zero. Therefore, taking moment of
all forces about hinge point.
mg cos 30o 2.5 = F h /sin 30o
Therefore, m =0 0
F h
g cos30 2.5 sin30
= 061312.5 1.67
9.81 cos30 1.25
= 9641 kg
Therefore, the nearest possible value is 9623 kg.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-132, Example-
3.22.
Q.2 Ans. a)
Mass of water strike = AV = QForce on weighing balance due to strike of water = Initial momentum final momentum
= QU Q.0 = QUSince weight of water and container = mg
Total force on weighing balance = mg + QUQ.3 Ans. (d)
Horizontal component of hydrostatic force, Fx= gAx where = density of the liquid, A = surface area, x
= depth of centre of pressure from free surface of liquid
Hence, Fx
= g Ax where projected area, A = w 2rTherefore, F
x= 2 gwrh Projected area (ABCD),
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r + r
2r
r r
w
A
BC
D
Vertical component of hydrostatic force,
Fy
= Weight of water supported by the curved surfaceF
y= g Volume of curved portion
= 2g r w2
where,
2r
2
= Area of semi-circle
=2
g w r
2
where, w is the width of the gate.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-94, Equation-3.4.
Q.4 Ans. ( 144.5 kN)
45
Negative pressure (8250 N/m )2
Hinge
Gate
Gasoline surface
1.8m
P
Gasoline (S = 0.7)
Head of oil equivalent to negative pressure 238 N/m2, h =p 8250
1.2 mw 0.7 9810
This negative pressure will reduce the oil head above the top edge of the gate from 1.8 - 1.2 = 0.6 m of oil.
Calculations for the magnitude and location of the pressure force are thus to be made corresponding to 0.6 mof oil.
Now, x = 0.6 +3.6
sin452
= 1.873 m
Area, A = 3.6 3.6 = 12.96 m2
Pressure, P = wAx = 0.7 9810 12.96 1.873 = 166690 N
Centre of pressure, h =
2
GI sin
xAx
=
3 21 3.6 (3.6) (sin 45 )12 1.873 2.16m
12.96 1.873
Vertical distance of centre of pressure below top edge of the gate = 2.16 0.6 = 1.56 mTaking moments about the hinge.
F sin 45 3.6 = P 1.56
sin45
Hence, vertical force, F =2
P 1.56
3.6 (sin45)
= 2166690 1.56
3.6 (sin45)
= 144465 N = 144.5 kN
Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-101.
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Buoyancy and Flotation
4
Year 2010
1. For the stability of a floating body, under theinfluence of gravity alone, which of the following
is TRUE?(a) Metacentre should be below centre of gravity.(b) Metacentre should be above centre of gravity.
(c) Metacentre and centre of gravity must lie on
the same horizontal line.
(d) Metacentre and centre of gravity must lie on
the same vertical line.
Year 2003
2. A cylindrical body of cross-sectional area A,
height H and density s, is immersed to a depth h
in a liquid of density , and tied to the bottom witha string. The tension in the string is
h
(a) ghA (b) (s ) ghA(c) (
s) ghA (d) (h
sH) gA
Year 1994
3. Bodies in flotation to be in stable equilibrium, the
necessary and sufficient condition is that the
centre of gravity is located below the...........
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Chapter-4Answer & 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 < 03. For neutral equilibrium, MG = 0
M
Meta centre
Centre ofbuoyancy
Centre ofgravityB
G
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-151, 4.3.Q.2 Ans. (d)
h
Free body diagram of the cylindrical body will be
WG
T
FBB
At equilibrium condition
T + weight of body = Buoyancy force
T + Mg = h AgT + (sHA)g = h Ag
T = (h sH) gAQ.3 Ans. metacentre
For floating body the equlibrium conditionds are as follows:-
1. For stable equilibrium, metacentre should be above the centre of gravity.
2. For unstable equilibrium, metacentre should be below the centre of gravity.3. For neutral equilibrium, metacentre should coincide the centre of gravity.
Reference: Fluid Mechanics, R. K. Rajput, Edition 2005, Page-129, 4.3.
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Fluid Kinematics
5
Year 2011
1. A streamline and an equipotential line in a flowfield
(a) are parallel to each other
(b) are perpendicular to each other
(c) intersect at an acute angle
(d) are identical
Year 2009
2. You are asked to evaluate assorted fluid flows
for their suitability in a given labortory application.
The following three flow choices expressed interms of the two dimensional velocity field in the
xy plane are made available
P. u = 2y, v = 3x
Q. u = 3xy, v = 0
R. u = 2x. v = 2y
Which flow (s) should be recommended when
the application requires the flow to be
incompressible and irrota tional?
(a) P and R (b) Q
(c) Q and R (d) R
Year 2008
3. For the continuity equation given by V
to be
valid, where V
is the velocity vector, which one
of the following is a necessary condition?
(a) steady flow
(b) irrotational flow
(c) inviscid flow
(d) incompressible flow
Statement for linked answer questions
4 and 5
The gap between a moving circular plate and astationary surface is being continously reduced,
as the circular plate comes down at a uniform
speed V towards the stationary bottom surface,
as shown in the figure. In the process, the fluid
contained between the two plate flows out
radially. The fluid is assumed to be incompressible
and inviscid.
R
r
Stationarysurface
Movingcircular plate
Vh
4. The radial velocity vrat any radius r, when the
gap width is h, is
(a) vr =Vr
2h(b) vr =
Vr
h
(c) vr =2 V h
r(d) vr =
Vh
r
5. The radial component of the fluid acceleration at
r = R is
(a)2
R
2
3V
4h(b)
2R
2
V
4h
(c)2
R
2
V
2h
(d)
2R
2
V
4h
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Year 2006
6. In a two-dimensional velocity field with velocities
u and v along the x and y directions respectively,
the convective acceleration along the x-direction
is given by
(a)u u
u vx y
(b)
u vu v
x y
(c)v u
u vx y
(d)
u uv u
x y
7. A two-dimensional flow field has velocities along
thexandydirections given by u=x2tand v=
2xytrespectively, where tis time. The equation
of streamlines is
(a) x2y = constant
(b) x y2= constant
(c) x y = constant
(d) not possible to determine
Year 2005
8. The velocity components in the x and and y
directions of a two dimensional potential flow are
u and v, respectively. Then ux
is equal to
(a)v
x
(b)v
x
(c)v
y
(d)v
y
Year 2004
9. A fluid flow is represented by the velocity field
V x i y j a a , where a is a constant. The
equation of stream line passing through a point
(1, 2) is:
(a) x - 2y = 0 (b) 2x + y = 0
(c) 2x - y = 0 (d) x + 2y = 0
Year 2003
10. The vector field F xi yj
(where i
andj
are
unit vectors), is:
(a) divergence free, but not irrotational
(b) irrotational, but not divergence free
(c) divergence free and irrotational(d) neither divergence free nor irrotational
Year 2001
11. The 2-D flow with velocity
V x 2y 2 i 4 y j
, is
(a) compressible and irrotational
(b) compressible and not irrotational
(c) incompressible and irrotational
(d) incompressible and not irrotational
Year 1999
12. For the function f = ax2y y3 to represent the
velocity potential of an ideal fluid. D2f should be
equal to zero. In that case, the value of a has to
be:
(a) 1 (b) 1
(c) 3 (d) 3
13. If the velocity vector in 2-D flow field is given by
2 2V = 2xyi + (2y - x )j
, the vorticity vector, curl
V
will be
(a) 22y j
(b) 6yk
(c) zero (d) -4xk
Year 1995
14. The velocity components in the x and y directions
are given by
u xy x y v xy y 3 2 2 43
4,
The value of for a possible flow field involvingan incompressible fluid is
(a) 3
4(b)
4
3
(c)4
3(d) 3
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15. The force F needed to support the liquid of density
d and the vessel on top (Fig) is
(a) gd[ha (h H) A] (b) gdHA
(c) gdHa (d) gd (H h) A
Year 1994
16. Stream lines, path lines and streak lines are virtuallyidentical for
(a) Uniform flow
(b) Flow of ideal fluids
(c) Steady flow
(d) Non uniform flow
17. In a flow field, the streamlines and equipotential
lines
(a) are parallel
(b) are orthogonal everywhere in the flow field
(b) cut at any angle
(d) cut orthogonally except at the stagnation points
18. For a fluid element in a two dimensional flow field
(x-y plane), if it will undergo
(a) translation only
(b) translation and rotation
(c) translation and deformation
(d) deformation only
Year 1992
19. Existence of velocity potential implies that(a) Fluid is in continuum
(b) Fluid is irrotational
(c) Fluid is ideal
(d) Fluid is compressible
20. Circulation is defined as line integral of tangential
component of velocity about a..........
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Chapter-5Answers
1. Ans. (b) 2. Ans. (d) 3. Ans. (d) 4. Ans. (a) 5. Ans. (c)6. Ans. (a) 7. Ans. (a) 8. Ans. (d) 9. Ans. (c) 10.Ans. (c)11.
Ans. (d) 12.
Ans. (d) 13.
Ans. (d) 14.
Ans. (d) 15.
Ans. (a)
16.Ans. (c) 17.Ans. (b) 18.Ans. (c) 19.Ans. (b) 20.Ans.(closedcontour in afluid flow)
Space for Rough work
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Chapter-5Answer & Explanations
Q.1 Ans. (b)
If and are the stream function and potential function respectively representing the possible flow field.
Slope of stream line represented by is given by
slope (m1) =dy
dx=
d
dx
d
dy
=v
-u......(i)
Slope of potential line represented by is given by
slope (m2) =dy
dx=
d
dx
d
dy
= -u u=-v v
......(ii)
Now, product of the slopes,
m1 m2 =v u
u v
= 1
Since the product of the slope of these two lines at the point of intersection is 1, which indicates that thesetwo lines are prependicular to each other.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-244, Sec. 6.12.
Q.2 Ans. (d)For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations
u v
x y
= 0 ......(i)
z =
1 v u
2 x y
= 0 ......(ii)
For P,Given u = 2y and v = 3x
u
x
= (2y)x
= 0 andu
y
=
(2y)y
= 2
v
x
= 3x 3x
and
v
y
= 3x 0y
From equation (i)u v
x y
= 0 + 0 = 0
From equation (ii) z =1 v u
-2 x y
= 1
3 - 22
0
Since the given velocity field is satifying the equation (i) only, therefore it is a possible case of steady,incompressible and rotational flow.
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For Q, Given u = 3xy and v = 0
u
x
= (3xy)x
= 3y andu
y
=
(3xy)y
= 3x
v
x
= 0 0x
and
v
y
= 0 0y
From equation (i)u v
x y
= 3y 0
From equation (ii) z =1 v u
-2 x y
=-3x
2 0
Given velocity field is neither satisfying the equation (i) nor (ii), therefore the flow is neither steady norirrotational.For R, Given u = 2x and v = 2y
u
x
= ( 2x)x
= -2 and
u
y
=
( 2x)y
= 0
v
x
=
2y 0
x
and
v
y
= 2y 2
y
From equation (i)u v
x y
= 0
From equation (ii) z =1 v u
-2 x y
= 0
Given velocity field is satisfying the equation (i) and (ii), therefore, the flow is a possible case of steady,incompressible and irrotational flow.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn-6.33 a.
Q.3 Ans. (d)
Given that V
= 0
i.e. i j k ui vj wk 0x y z
i.e.u v w
x y z
= 0
which represents the three dimensional continuity equation of steady, and incompressible flow.Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page -146, Eqn. 5.4.
Q.4 Ans. (a)At radius r, volume of fluid moving out radially is equal to the volume of fluid displaced by moving plate withinradius r.Given that V = downward velocity of circular plate in m/s
vr
= radial velocity at radius rR
V
Vrr
h
So volume displaced by moving plate= Velocity Area= V r2
Now, volume flow out at radius,
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r = vr 2rhFrom above stated condition
vr 2rh = r2 V
Therefore, vr =V
2
rh
Q.5 Ans. (c)Radial component of the fluid acceleration at r = R
aR =R
Vd
dt=
VR
2d
h
dt
=
VR
2d
dhh
dh dt
(ve as h is reducing with time)
= 2VR 1
( V)2 h
as V
dh
dt
Therefore, aR =
2
2
V R
2hQ.6 Ans. (a)Acceleration of fluid particle alongx-axis is given by
ax =u u u u
u + v + w +x y z t
......(i)
For 2-D flowu
z
= 0
Thus, from equation (i), ax =
Temporalor localConvective
accelerationacceleration
u u uu + v +
x y t
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-232, Equation6.27.
Q.7 Ans. (a)Given: u = x2t and v = 2xytStream line equation is given as
dx
u=
dy
v
2dx
x t=
dy
-2xyt
dx
x=
1 dy-
2 y
Integrating both sidedx
x =1
2
dy
y
ln x =1
ln y2
c
ln x2+ ln y = c x2y = ConstantReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219,Equation 6.2.
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Q.8 Ans. (d)Exp. For two dimensional potential flow, the continuity equation is given as
u v
x y
= 0
Therefore,u
x
= v
y
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-223, Eqn. 6.5.Q.9 Ans. (c)The velocity field is given as,
V
= ax i ay j
= ui vj
The equation of stream line
dx
u=
dy
v......(i)
from equation (i),dx
ax=
dy
ay
Integrating both side,dx
x =dy
yln x = ln y + c
xn
y = ln c
x
y= c ......(ii)
Since this stream line passes through point (1, 2) hence c = 1/2Therefore, equation of stream line is (from equation (ii))
2x y = 0Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn. 6.2.
Q.10 Ans. (c)Given vector filed F
= xi yj
The divergence of V
is defined as V
It can also be written as x y zx y z
i j k i j k
=u v w
x y z
=0
x y z
x y
= 1-1 = 0
Rotational component, z
=1 v u
2 x y
= 0
Hence, the vector field is divergence free and irrotational.Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page-156, Sec. 5.8.
Q.11 Ans. (d)
The 2-D flow with velocity, V
= x 2y 2 i 4 y j
= ui vj
For incompressible and irrotational flow, the velocity field should satisfy the following equations
u v
x y
= 0 ......(i)
z =
1 v u
2 x y
= 0 ......(ii)
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Here,u
x
= (x 2y 2)x
= 1 and
u
y
=
(x 2y 2)y
= 2
v
x
= 4 yx
= 0 and
v
y
= 4 y 1y
From equation (i),u v
x y
= 1 1 = 0
For irrotational flow from equation (ii),
z
=1 v u
2 x y
= 1
0 22
0
Hence, this flow is steady, incompressible and rotational.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-219, Eqn 6.2.
Q.12 Ans. (d)Velocity Potential, f = ax2y y3
Nowf
x
= 2axy &
2
2
f2ay
x
......(i)
andf
y
= 2 2ax 3y &
2
2
f6yy
......(ii)
As D2f should equal to 0
or D2(f) = 2 (f ) 0 2 2
2 2
f f
x y
= 0
From euation (i) and (ii),2ay 6y = 02y (a 3) = 0
Therefore, a = 3Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-243, Eqn. 6.45.
Q.13 Ans. (d)The curl of V
is defined as V
.
curl of V
=
i j k
x y z
u v w
Flow field, V
= 2 22xyi + (2y -x )j + 0k
It can also be written asw v u w v u
i j ky z z x x y
= v u
0 0 0 0x y
i j k
= -2x - 2x k
= - 4xk
Reference: Fluid Mechanics, R. K. Rajput, Edition 2005, Page-172, Equation 5.32.Q.14 Ans. (d)
The velocity components in the x and y directions are given by
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u xy x y v xy y 3 2 2 43
4,
Continuity equation for steady, incompressible and irrotataional flow is
u v
x y
= 0 ......(i)
u
x
=3
y 2xy &
v
y
=3
2xy 3yPut these value in equation (i),
3 3y 2xy 2xy 3y = 0 y3 3y3 = 0 y3 ( 3) = 0 3 = 0 = 3Reference: Fluid Mechanics, S.K. Aggarwal, Page No. 104.
Q.15 Ans. (a)Let Free body diagram of liquid columns due to symmetry
Here A1
= a and A2= A
3=
A a
2
A1
A2 A2
a
a
A
h
H-h
H
(Aa)2
Aa2
Now F is equal to the weight of water supported by the piston.W = Mg
or M.g = d.g.V where d is the density of the liquidF = d.g.V ......(i)Now V = A
1H + 2 (A
1(Hh))
= aH + 2A a
(H h)2
= aH + A (Hh) aH + ahV = ah + A(H h)
= ah A (h H) ......(ii)Put Value of V in equation (i)
F = dg [ah A(hH)]Q.16 Ans. (c)
In steady and uniform flow stream line, path line and streak line are same.
In the given problem steady flow and uniform flow are separate option. Hence option (a) & (c) both arecorrect but most appropriate single answer is (c).Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page No. 160, 5.4.4.
Q.17 Ans. (b)In a flow field, the streamlines and equipotential lines are always orthogonal to each other.
= stream lines
= equipotential lines
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Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-245, Fig. 6.18.Q.18 Ans. (c)
For 2-D flow, irrotational component,
z =
1 v u
2 x y
= 0
Therefore, there is no variation in velocity in z-direction.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn 6.34c.
Q.19 Ans. (b)For steady, incompressible and irrotational flow, the velocity field should satisfy the following equations
u v
x y
= 0 ......(i)
z =
1 v u
2 x y
= 0 ......(ii)
If is the potential function representing the possible flow field. Then from definition of potential function
u =
-x
and v =
-y
From equation (i),u v
x y
=2 2
2 2x y
which is known as Laplace equation.
From equation (ii),1 v u
2 x y
=
2 2
x y y x
= 0
The velocity potential of the flow denoted by if satisfies the continuity/Laplace equation, then it will be apossible case of irrotational flow.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-238, Eqn 6.34c.
Q.20 Ans. (closed contour in a fluid flow)Circulation is defined as the line integral of the tangential component of the velocity taken around a closedcontour. Mathematically, the circulation is obtained if the product of the velocity component along the curveat any point and the length of the small element containing that point is integrated around the curve.
ds
C
V
X
Y
Mathematically, circulation = V cos .ds = (udx vdy)
Area of closedcurve
= Vorcitity along the axis perpendicular to the plane containing the closed
curve.
= Vorticity area = 2 z area
=1 v u
2 x y2 x y
For irrotational flow in xy plane,
z= 0 hence vorticity which leads to circulation also equal to zero.
Reference: Fluid Mechanics & Hydraulic Machines, K. Subramanya, Edition 2012, 108,3.5.1.
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Fluid Dynamics
6
Year 2013
1. Water is coming out from a tap and falls verticallydownwards. At the tap opening, the stream
diameter is 20mmwith uniform velocity of 2m/s.
Acceralation due to gravity is 9.81 m/s 2.
Assuming steady, inviscid flow, constant atmo
spheric pressure everywhere and neglecting
curvature and surface tension effects, the
diameter in mmof the stream 0.5 m below the
tap is approximately
(a) 10 (b) 15
(c) 20 (d) 25
Year 2012
2. A large tank with a nozzle attached contains three
immiscible, inviscid fluids as shown. Assuming
that the changes in h1, h2and h3are negligible,
the instantaneous discharge velocity is
h1
h2
h3
1
2
3
(a)1 1 2 2
3
3 3 3 3
2 1h h
gh
h h
(b)1 2 32 ( )g h h h
(c)1 1 2 2 3 3
1 2 3
2h h h
g
(d)1 2 3 2 3 1 3 1 2
1 1 2 2 3 3
2h h h h h h
gh h h
Year 2011
3. Figure shows the schematic for the measurement
of velocity of air (density = 1.2 kg/m3) through a
constant-area duct using a pitot tube and a water-
tube manometer. The differential head of water
(density = 1000 kg/m3) in the two columns of the
manometer is 10 mm. Take acceleration due to
gravity as 9.8 m/s2. The velocity of air in m/s is
Flow
10 mm
(a) 6.4 (b) 9.0
(c) 12.8 (d) 25.6
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Year 2010
4. A smooth pipe of diameter 200 mm carries water.
The pressure in the pipe at section S1(elevation
: 10 m) is 50 kPa. At section S2(elevation : 12 m)
the pressure is 20 kPa and velocity is 2 ms1.
Density of water is 1000 kgm3and accelerationdue to gravity is 9.8 ms2. Which of the following
is TRUE
(a) flow is from S1to S2 and head loss is
0.53 m
(b) flow is from S2to S1and head loss is 0.53 m
(c) flow is from S1 to S2 and head loss is
1.06 m
(d) flow is from S2to S1and head loss is 1.06 m
Year 2009
5. Consider steady, incompressible and irrotational
flow through a reducer in a horizontal pipe where
the diameter is reduced from 20 cm to 10 cm.
The pressure in the 20 cm pipe just upstream of
the reducer is 150 kPa. The fluid has a vapour
pressure of 50 kPa and a specific weight of 5
kN/m3. Neglecting frictional effects, the maxi-
mum discharge (in m3/s) that can pass through
the reducer without causing cavitation is(a) 0.05 (b) 0.16
(c) 0.27 (d) 0.38
Year 2007
6. Which combination of the following statements
about steady incompressible forced vortex flow is
correct ?
P : Shear stress is zero at all points in the flow.
Q : Vorticity is zero at all points in the flow.R : Velocity is directly proportional to the radius
from the centre of the vortex.
S : Total mechanical energy per unit mass is con-
stant in the entire flow field.
Select the correct answer using the codes given
bewlow:
(a) P and Q (b) R and S
(c) P and R (d) P and S
Year 2006
7. A siphon draws water from a reservoir and
discharges it out at atmospheric pressure.
Assuming ideal fluid and the reservoir is large,
the velocity at point P in the siphon tube is
h2
h1
P
(a) 12gh (b) 22gh
(c) 2 12 ( )g h h (d) 2 12 ( )g h h
Year 2005
8. A venturimeter of 20 mm throat diameter is used
to measure the velocity of water in a horizontal
pipe of 40 mm diameter. If the pressure differ-
ence between the pipe and throat sections is found
to be 30 kPa then, neglecting frictional losses, theflow velocity is
(a) 0. 2 m/s (b) 1. 0 m/s
(c) 1. 4 m/s (d) 2. 0 m/s
9. A leaf caught in a whirlpool. At a given instant
the leaf is at a distance of 120 m from the centre
of the whirlpool. The whirlpool can be described
by the following velocity distribution;
Vr =
360 10/2 m sr
& VV=
3300 10
/ ,2 m sr
angular Velocity V where r (in metres) is the
distance from the centre of the whirlpool. What
will be the distance of the leaf from the centre
when it has moved through half a revolution
(a) 48 m (b) 64 m
(c) 120 m (d) 142 m
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Year 2004
10. A closed cylinder having a radius R and height H
is filled with oil of density . If the cylinder isrotated about its axis at an angular velocity of ,the thrust at the bottom of the cylinder is:
(a) R gH2
(b)
RR2
2 2
4
(c) R R gH2 2 2 j
(d)
RR
gH2
2 2
4HG KJ
11. A centrifugal pump is required to pump water to
an open water tank situated 4 km away from the
location of the pump through a pipe of diameter
0.2 m having Darcys friction factor of 0.01. Theaverage speed of water in the pipe is 2 m/s. If it
is to maintain a constant head of 5 m in the tank,
neglecting other minor losses, then absolutedischarge pressure at the pump exit is
(a) 0.449 bar (b) 5.503 bar (c) 44.911 bar (d) 55.203 bar
Year 2003
12. Air flows through a venturi and into atmosphere.Air density is; atmospheric pressure is P
a; throat
diameter is Dt; exit diameter is D and exit velocity
is U. The throat is connected to a cylinder
containing a frictionless piston attached to aspring. The spring constant is k. The bottom
surface of the piston is exposed to atmosphere.
Due to the flow, the piston moves by distance x.
Assuming incompressible frictionless flow,x is
xDs
k
Dt
D U
Pa
(a) U k Ds2 2
2/ j
(b) U kD
DD
t
s2
2
2
28 1/e j HG KJ
(c) U kD
DD
t
s2
2
2
22 1/e j HG KJ
(d) U kD
DD
t
s2
4
4
28 1/e j HG KJ
Year 1999
13. Water flows through a vertical contraction from
a pipe of diameter d to another of diameter d/2
(see Fig.). The flow velocity at the inlet to the
contraction is 2 m/s and pressure 200 kN/m2. If
the height of the contraction measures 2 m, the
pressure at the exit of the contraction will be
very nearly
2m
d
d/2
(a) 168 kN/m2 (b) 192 kN/m2
(c) 150 kN/m2 (d) 174 kN/m2
Year 1996
14. A venturimeter (throat diameter = 10.5 cm) isfitted to a water pipe line (internal diameter
= 21.0 cm) in order to monitor flow rate. To
improve accuracy of measurement, pressure
difference across the venturimeter is measured
with the help of an inclined tube manometer, the
angle of inclination being 30 (Figure). For a
manometer reading of 9.5 cm of mercury, find
the flow rate. Discharge coefficient of venturi is
0.984.
30
9.5cmy
From
venturi
Water
Hg
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Chapter-6Answers
1. Ans. (b) 2. Ans. (a) 3. Ans. (c) 4. Ans. (c) 5. Ans. (b)6. Ans. (c) 7. Ans. (c) 8. Ans. (d) 9. Ans. (b) 10.Ans. (d)11.
Ans. (b) 12.
Ans. (d) 13.
Ans. (c) 14.
Ans.(0.0302
m3/s)
Space for Rough work
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Chapter-6Answer & Explanations
Q.1 Ans (b)
0.5m
1
2
1
2
Applying Bernoullis equation at section (1-1) & (2-2)
2
1 11
P VZ
g 2g
=2
2 22
P VZ
g 2g
......(i)
P1
= P2= P
atm.(taking section 2-2 as datum)
From equation (i)
220.5
2 9.81
=
2
2V
2g
V2
= 3.716 m/sec.
From Continuity equation, 1A1V1 = 2A1V2 (since flow is incompressible, i.e. 1= 2)A
1V
1= A
2V
2
A2
=1 1
2
A V
V
2
2d
4
=
2
1 1
2
d V
4 V
Therefore, d2
=
2
1 1
2
d V
V
=
11
2
Vd
V
=2
0.023.716
= 0.01467 m 15 mm
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-306, Ex.7.3.
Q.2 Ans (a)
Applying Bernoullis equation, just before the exit from the tank and just after entry in the atmosphere
21 1
1
3
P VZ
g 2g
=2
2 22
3
P VZ
g 2g
......(i)
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h1
h2
h3
1
2
3
From the above figure, it is clear thatZ
1= Z
2, V
1= 0 and P
2= Atmospheric pressure = 0;
Then the Bernoullis equation reduces to :
1
3
P
=
2
2V
2
V2
=1
3
2P ......(ii)
From given figure we can find pressure P1
P1
= 1gh
1+
2gh
2+
3gh
3
Substitute this value of P1in eqution (ii), we get
V2
= 1 1 2 2 3 33
2gh h h
=1 1 2 2
3
3 3 3 3
h h2gh 1
h h
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-277, Eqn 7.24.Q.3 Ans. (c)
10 mm
Flow a
w
Given that
Density of air, a = 1.2 kg/m3,
Density of water, w = 1000 kg/m3
x = 10 mm, g = 9.8 m/s2
Now h =
2v
2g
v = 2gh
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where, h = 1
w
a
x
=3 100010 10 1 8.32 m
1.2
Velocity of air, v = 2 9.81 8.32 = 12.8 m/s
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-294, 295.
Q.4 Ans. (c)
Given:
At section S1: P1 = 50 kPa At section S2: P2 = 20 kPa
Z1 = 10 m Z2 = 12 m
V1 = 2 m/s V2 = 2 m/s
10 m12 m
S1
S2
P = 50 KPa1
P = 20 KPa2
Datum line
Since diameter of the pipe is constant hence velocity of the flow will be same through out the length of
the pipe. Therefore V1= V2= 2 m/s. Since velocity of flow is constant throughout the pipe, hence direction
of flow is decided by the piezometric head only.
Total piezometric head at S1,
H1
=1
1
PZ
g
=
350 1010
1000 9.81
= 15.096 m
Total piezometric head at S2, H2 =2
2
PZ
g
=
320 1012
1000 9.81
= 14.038 m
Since H1> H2therefore flow direction is from S1to S2.
Therefore, head loss = H1 H
2= 15.096 14.038 = 1.06 m
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-285, 7.10.
Q.5 Ans. (b) Given,
Inlet diameter, d1 = 0.2 m
Inlet pressure, P1 = 150 kPa
Exit diameter, d2 = 0.1 m
Specific weight, w (g) = 5 kN/m3
Vapour pressure, P2 = Pv= 50 kPa(To avoid cavitation, pressure at exit should not be allowed tofall below the vapour pressure of the liquid)
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106
12
2
1
From continuity equationa1V1 = a2V2
2
1 1d V
4
=
2
2 2d V
4
V2 =
2
112
2
dV
d=
2
12
0.2V
0.1= 4V1
Applying Bernoullis equation between at section (1-1) and (2-2)
2
1 11
P VZ
2
g g =2
2 22
P VZ
2
g g
2
1V150
5 2g =
2
2V50
5 2g
=
2
116 V102g
Therefore, V1 = 5.114 m/s
Therefore, maximum discharge, Q = 21 1r V
= 0.12 5.114 = 0.161 m3/sReference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-347, Ex. 8.7.
Q.6 Ans. (c)
In forced vortex flow when steady state is reached the liquid attains equilibrium condition in this position andit rotates as a solid mass with the container at the same angular velocity. The liquid is then at rest with respect
to its container and therefore no shear stress will exist in the liquid mass.In the forced vortex flow the stream lines are concentric circles and the velocity v of any liquid particle ata distance r from the axis of rotation may be expressed as v = r.Therefore, v r. In forced votex flow, velocity is directly propotional to distance from the axis of rotation.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, Page-185, 5.5 (a), Page-301, 7.62.
Q.7 Ans. (c)
2
1
P
z1
h1
h2
vP
Applying Bernoullis equation between section (1) and (2)
2
1 11
P VZ
2
g g =2
2 22
P VZ
2
g g
Now P1 = P2= Patmand Z2= 0, (taking point 2 as datum)
Z1 = (h2 h1), V1= 0
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Thus from Bernoullis equation
atm
2 1
P0 ( )
h h
g=
2
atm 2P V
02
g g
h2 h1 =2
2V
2g
V2 = 2 12 ( )g h hAs area of siphon is constant, therefore velocity of flow is same
Hence, VP = V2= 2 12 ( )g h h
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-274, 7.17.
Q.8 Ans. (d)
Given that
D1 = 40 mm
D2 = 20 mm
40 mm 20 mmD1 D2
1
12
2
From continuity equation, A1V1 = A2V2
V2 =1
1
2
AV
A
=
2
11
2
DV
D
=
2
1
40V
20
= 4V1
Now applying Bernoullis equation in between the sections 1-1 and 2-22
1 11
P Vz
g 2g
=2
2 22
P Vz
g 2g
( Since pipe is horizontal, hence Z2= Z1)
1 2P P
g
=
2 22 1V V
2g
=2 2 2
1 1 1(4V ) V 15V
2 2
Since, P1 P2= 30 kPa
330 10
1000
=
2115V
2
V12 =
360 10
41000 15
Therefore, flow velocity V1 = 2 m/sec
Or
Theoretical discharge (Qth) through a venturimeter is given by
Qth =1 2
2 2
1 2
A A2gh
A A
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A1V1 =1 2
2 2
1 2
A A 2gh
A A
V1 = 21
2
2gh
A1
A
=4
1
2
2gh
d1
d
=4
2 9.81 (30 / 9.81)
0.041
0.01
= 2 m/sec.
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 241, 6.7.1.
Q.9 Ans. (b)
Given, Vr =
360 10m/sec
2 r
and V=
3300 10m/sec
2 r
NowrV
V= 60 1
300 5
vrV
A
B
C
D
r
r
Therefore, Vr = V
5
Also V = r.= r.d
dt
and Vr=
dr
dt
dr
dt=
r.d
5dt
r
120
dr
r =
0
d
5
r120[ n r]l = 1( 0)5
r
ln120
=5
r
120= 0.5336
Therefore, r = 120 0.5336 = 64.03 m
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 147, 5.6.1.
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Q.10 Ans. (d)
Given :
Radius of cylinder = R
Height of cylinder = H
Angular speed = Density of oil =
As the cylinder is closed and completely filled with oil, the rise of oil level at the ends and depression of oil at
the centre due to rotation of the vessel, will be prevented. Thus the oil will exert force on the complete top ofthe vessel. Also the pressure will be exerted at the bottom of the cylinder.
Thrust at the bottom of cylinder = Weight of oil in cylinder + total force on the top of the cylinder
Now Weight of water = V.g= R2 H g = gR2H ......(i)
H
R
R
r
dr
Now lets consider an elementary ring of radius r and thickness dr. Then pressure gradient in the elementary
ring in free as well as in forced vortex flow is given as
p
r
=2V
r
=
22(r )
rr
Integrating the above equation,
p = 2 r r
p =
2 2w r
2
......(ii)
Now Force on elementary ring is = pressure intensity area of elementary circular ring
dF = p 2rdr
Total force on the top of the cylinder, FT =
2 2R R
0 0
w rdF .2 rdr
2
. (p from equation (ii)
FT = w2
R4
0
r
4
=2 4R
4
......(iii)
Now, total thrust at bottom of cylinder is given by adding the equation (iii) and (i)
=4
2 2Rw g R H
4
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=
2 22 w R
R g H4
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 180, 5.30.
Q.11 Ans. (b)
0.2 m 5mPump
4 km
1
1
2
2
Applying Bernoulliss equation at the section (1-1) and section (2-2)
21 1
1
P VZ
g 2g
=2
2 22 f
P VZ h
g 2g
......(i)
Since pipe is horizontal, therefore Z1= Z2From question, V2 = 0, V1= 2 m/s
Head loss due to friction in the pipe is given as hf=
2f LV
2g dwhere, f, V, L are the friction factor, mean velocity and length of the pipe respectively.
=
20.01 4000 2
2 9.81 0.2
= 40.774 m
from equation (i),1
P
g =2
2 1f
P Vh
g 2g
= 5 + 40.774
22
2 9.81 = 45.57 m
Therefore, P1 = 45.57 1000 9.81 N/m2 = 447.04 kPa
Therefore, absolute discharge pressure at the pump exit = P1+ Patm.= 447.04 + 101.325 = 548.365 kPa = 5.5 bar
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 420, Eqn. 11.1.Q.12 Ans. (d)
xDs
k
Dt
D U
Pa
From continuity equation at the throat and at the exit of the venturimeterA1V1 = A2V2
V2 =1 1
2
A V
A=
2
12t
D.V
D
......(i)
Now applying Bernoulliss equation at the throat and at the exit of the venturimeter
21 1
1
P VZ
g 2g
=2
2 22
P VZ
g 2g
Since venturi is horizontal, therefore Z1= Z2
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1 2P P
g
=
2 22 1V V
2g
P1 P2 =
22 1
2 22
VV 1
2 V
P1 Patm. =
42
4t
DU 12 D
At throat velocity is greater than U, hance pressure will be less than atmospheric
P1 =
42
4t
DU 1
2 D
=4
2
4t
DU 1
2 D
Now spring is elongated due to lower pressure at throat.Therefore, in equilibrium,
Spring force = Pressure Force
Hence, kx =2sD
4
(P1)
=
2 2 4s
4t
D U D1
4 2 D
Hence, x =
2 42s4
t
U D1 D
8k D
Reference: Fluid Mechanics & Hydraulic Machines, Dr. R.K. Bansal, Page No. 242, 6.6.
Q.13 Ans. (c)
2m
d
d/2
2 2
11
From continuity equation, A1V1 = A2V2
2
d .24
=
2
2
d
.V4 2
V2 =
2
2
2d4 8 m /sec
d
Now Applying Bernoullis theorem at section (1-1) & (2-2)
21 1
1
P Vz
g 2g
=2
2 22
P Vz
g 2g
......(i)
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From question, P1 = 200 kN/m2= 200103N/m2
Taking section (1-1) as datum surface
V1 = 200 m/sec, z1= 0, V2 = 8 m/sec, z2= 2 m
Putting all the above values in equation (i), we get
23 2200 10
01000 9.81 2 9.8
=
2P 8 8 21000 9.81 2 9.81
20.39 + 0.204 =2P 3.26 2
9810
P2 = 9810 (20.594 5.26) = 150426.5 N/m2
Therefore, pressure at the exit of the contraction,
P2 = 150.4 kN/m2
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-274, 17.16.
Q.14 Ans. ( 0.0302 m3/s)
Internal diameter, D1 = 21.0 cm = 0.2 m;
Area of inlet, A1 =2
2D
4
D12 =
2 2(0.21) 0.0346 m4
Throat diameter, D2 = 10.5 cm = 0.105 m
Area at throat, A2 =2
2D
4
=
4
(0.105)2 = 0.0087 m2
Coefficient of discharge of venturi,
Cd = 0.984
Pressure head, h = yHg
water
S1
S
= (9.5 sin 30)13.6 1
1
= 59.85 cm = 0.5985 m
Discharge (Q) through a venturimeter is given by:
Q = 1 2d2 2
1 2
A AC 2gh
A A
= 0.984 2 2
0.0346 0.00872 9.81 0.5985
(0.0346) (0.0087)
= 0.984 0.008945 3.427 = 0.0302 m3/s
Reference: Fundamentals of Fluid Mechanics, B. R. Munson, Edition 2010, Page-441, 8.37.
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Dimensional and Model Analysis
7
Year 2010
1. A phenomenon is modeled using n dimensionalvariables with k primary dimensions. The number
of non-dimensional variables is
(a) k (b) n
(c) n k (d) n + k
2. Match the following
P :Compressibleflow U : Reynoldsnumber
Q: Freesurfaceflow V :Nusseltnumber
R : Boundarylayerflow W : Weber number
S: Pipeflow X : Froudenumber
T : Heat convection Y: Machnumber
Z :Skinfrictioncoefficient
(a) P-U; Q-X; R-V; S-Z; T-W
(b) P-W; Q-X; R-Z; S-U; T-V
(c) P-Y; Q-W; R-Z; S-U; T-X
(d) P-Y; Q-W; R-Z; S-U; T-V
Year 2007
3. Consider steady laminar incompressible axi-sym-
metric developed viscous flow through a straightcircular pipe of constant cross-sectional area at
a Reynolds number of 5. The ratio of inertia force
to viscous force on a fluid particle is
(a) 5 (b)1
5
(c) 0 (d)
Year 2002
4. If there are m physical quantities and nfundamental dimensions in a particular process,
the number of non-dimensional parameters is
(a) m + n (b) m n(c) m n (d) m/n
Year 1997
5. The Reynolds number for flow of a certain fluid
in a circular tube is specified as 2500. What will
be the Reynolds number when the tube diameter
is increased by 20% and the fluid velocity isdecreased by 40% keeping fluid the same?
(a) 1200 (b) 1800
(c) 3600 (d) 200
Year 1994
6. The ratio of inertia forces to gravity forces may
be expressed as square of non-dimensional group
known as.........
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Chapter-7Answers
1. Ans. (c) 2. Ans. (d) 3. Ans. (a) 4. Ans. (c) 5. Ans. (b)Ans. (Froude Number)
Space for Rough work
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Chapter-7Answer & Explanations
Q.1 Ans. (c)
Buckinghams -theorem states that if there are ntotal dimensional variables (dependent as well as independentvariables) involved in a phenomenon which can be completely described by mfundamental dimensions (such
as mass, length, time etc.), and are related by a dimensionally homogeneous equation, then the relationshipamong the nquantities can be expressed in terms of exactly (n m) dimensionless and independentterms.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b).
Q.2 Ans. (d)
P. Compressible flow Mach Number
Q. Free surface flow Weber Number
R. Boundary layer flow Skin friction coefficient
S. Pipe flow Reynolds Number
T. Heat convection Nusselt Number
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, 769 & 455, , 17.11e &
11.2.
Q.3 Ans. (a)
Reynolds number is defined as the ratio of inertia force and viscous force.
Re =Inertia force
5Viscous force
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-767, 17.11 (b).
Q.4 Ans. (c)
Buckinghams -theorem states that if there are n total dimensional variables involved in a phenomenonwhich can be completely described by mfundamental dimensions (such as mass, length, time etc.), and are
related by a dimensionally homogeneous equation, then the relationship among thenquantities can be expressed
in terms of exactly (n m) dimensionless and independentterms.Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-757, 17.4 (b).
Q.5 Ans. (b)
Exp. Reynolds number, Re =vd
2500 =vd ......(i)
when new diameter = 1.2 d then new velocity = 0.6v
Reynolds number, Renew =1.2d 0.6v
= 0.72 Re
= 0.72 2500 = 1800Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-325, Equation 7.15.
Q.6 Ans. (Froude Number)
Reynolds number, Re =Inertia force
Viscous force=
VL or
Vd
Froude number, Fr =Inertia force
Gravity force=
V
Lg
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Euler number, Eu =Inertia force
Pressure force=
V
p /
Weber number, We =Inertia force
Surface tension force=
V
/ L
Mach number, M =Inertia force
Elastic force =
V
K /
Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357.
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Boundary Layer Theory
8
Year 2012
1. An incompressible fluid flows over a flat platewith zero pressure gradient. The boundary layer
thickness is 1mm at a location where the Reynolds
number is 1000. If the velocity of the fluid alone
is increased by a factor of 4, then the boundary
layer thickness at the same location, in mm will
be
(a) 4 (b) 2
(c) 0.5 (d) 0.25
Year 2007
2. Consider an incompressible laminar boundary
layer flow over a flat plate of length L, aligned
with the direction of an oncoming uniform free
stream. If F is the ratio of the drag force on the
front half of the plate to the drag force on the
rear half, then
(a) F 1/ 2 (b) F = 1/2
(c) F = 1 (d) F > 1
Linked Answer Questions : Q. 3 - Q. 4
Consider a steady incompressible flow through a
channel as shown below.
A B
H
x
y
uo
Vmu0
The velocity profile is uniform with a value of u0at the inlet section A. The velocity profile at
section B downstream is
m
m
m
yV , 0 y
u V , y H
H yV , H y H
3. The ratio0
Vm
uis
(a) 1
1 2H
(b) 1
(c)1
1H
(d)1
1H
4. The ratio A B20
p p
1u
2
(where pA and p
B are the
pressures at section A and B, respectively, and is the density of the fluid) is
(a) 21
1
1H
(b)2
1
1H
(c)2
11
21
H
(d)1
1H
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Year 2006
Linked Questions 5 and 6
A smooth flat plate with a sharp leading edge is
placed along a gas stream flowing at U = 10 m/s.
The thickness of the boundarylayer at section
r-s is 10 mm, the breadth of the plate is 1 m (into
the paper) and the density of the gas = 1.0kg/m3. Assume that the boundary layer is thin,
two-dimensional, and follows a linear velocity
distribution, u= U ,y
at the section r-s,where
yis the height from plate.
r U
u
flat plate
q
p
U
s
5. The mass flow rate (in kg/s) across the section
q-r is
(a) zero (b) 0.05
(c) 0.10 (d) 0.15
6. The integrated drag force (in N) on the plate,
between p-s, is
(a) 0.67 (b) 0.33
(c) 0.17 (d) zero
Year 2004
7. For air flow over a flat plate, velocity (U) and
boundary layer thickenss () can be expressedrespectively, as
3
x
U 3 y 1 y 4.64x;
U 2 2 Re
If the free stream velocity is 2 m/s, and air has
kinetmatic viscosity of 1.5 105m2/s and density
of 1.23 kg/m3, the wall stress at x = 1m, is
(a) 2.36 102N/m2 (b) 43.6 103N/m2
(c) 4.36 103N/m2 (d) 2.18 103N/m2
8. If x is the distance measured from the leading
edge of a flat plate, the laminar boundary layer
thickness varies as
(a) 1/x (b) x4/ 5
(c) x2 (d) x1/ 2
9. Flow separation in flow past a solid object is caused
by
(a) a reduction of pressure to vapour pressure
(b) a negative pressure gradient
(c) a positive pressure gradient
(d) the boundary layer thickness reducing to zero
Year 1994
10. For air near atmosphere conditions flowing over
a flat plate, the laminar thermal boundary layer is
thicker than the hydrodynamic boundary layer.
(True/false)
Year 1993
11. The predominant forces acting on an element of
fluid in the boundary layer over a flat plate in a
uniform parallel stream are :
(a) Viscous and pressure forces
(b) Viscous and inertia forces
(c) Viscous and body forces
(d) Inertia and pressure forces
Year 1991
12. A streamlined body is defined as a body about
which
(a) The flow is laminar
(b) The flow is along the sreamlines
(c) The flow separation is suppressed
(d) The drag is zero
Year 2002
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Chapter-8Answers
1. Ans. (c) 2. Ans. (d) 3. Ans. (c) 4. Ans. (a) 5. Ans. (b)6. Ans. (c) 7. Ans. (c) 8. Ans. (d) 9. Ans. (c) 10.Ans. (False)11.
Ans. (b) 12.
Ans. (c)
Space for Rough work
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Chapter-8Answer & Explanations
Q.1 Ans. (c)
As per Blasius result thickness of laminar boundary layer is given as
x
5x
Re
Hence, =5x
vx
Therefore,1
v
1 1
24
Therefore, if the velocity of fluid is increase by four times then boundary layer thickness reduces by
1/2.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-525, 12.6.
Q.2 Ans. (d)V
L/2 L/2
As we know that local drag coefficient is given by
Cf = 20.664
V Re
2
o
x
or o = 0.332 V2(Rex)
1/2
Now drag force on the front half is given by
F1 =
L
2
0
Bo dx (B = width of Plate)
=
L
122 2
0
0.332 V (Re )
x dx
Reynolds number is given as, Rex =
V
x
=
L1
122
2 2
0
V0.332. V
x dx
=
L
12 2
2
0
0.332 V
V
x dx
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121
=
L1 122 2
0
LK 2K
1 2
2
x
......(i)
where, K =
20.332 V
V
Similarly, drag force on the rear half,
F2 =
L
L
2
Bo
dx
=
L 1
2
L
2
K x dx
=1
12
2L
2K (L)2
Now required ratio, F =
1
2
1
112 22
L2K
F 21
FL
2K L2
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-540, Ex. 12.6.
Q.3 Ans. (c)
Given:
A B
H
x
y
uo
Vmu0
u =
V , 0
V , H
HV , H H
m
m
m
yy
y
yy
Assuming width of channel as unity
Applying mass conservation at section A and B. Taking density of liquid constant, the conservation of mass
principle becomes volume flow equation.
Volume flow rate incoming at section A = Volume flow rate outgoing from section B
Therefore, total volume flow rate inlet
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122
entryQ
= uo H B = uoH
Total volume flow rate leaving,
exitQ
= Volume flow rate from boundary layer + Volume flow rate from mid section
u
ydy
Volume flow rate from mid section = Vm(H 2)For boundary layer 0y
Bd Q = u . dy
Bd Q
= Vmy
dy
Integrating the above equation
BQ
=0
Vm y dy
=2
0
V y
2
m
=V .
2
m
By symmetry for H yH, Volume flow rate =V .
2
m
Therefore, entryQ
=exit
Q
uoH =V .
V (H 2 ) 22
m
m
uoH = Vm(H )
Vm
ou=
H 1
H1
H
Reference: Fluid Mechanics, R.K. Bansal, 4th Edition, Page-655, 13.3.
Q.4 Ans. (a)
A B
H
x
y
uo
Vmu0
Applying Bernoullis equation at section A and B
2
A Ap v+g 2g
=2
B Bp +g 2g
(Since
A Bp - p
=
2 2
B A -
2
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A B
p - p
=
2 2
m oV - u
2
A B2
o
p - p
1 u
2
=
2V
1m
ou
A B2
o
p - p
1 u
2
= 21
1
1H
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-280, Equation
7.29.
Q.5 Ans. (b)
Given:Free stream velocity,
U = 10 m/s
Boundry layer thickness, = 10 mmBreadth of plate, B = 1 m
Density of air, = 1.0 kg/m3
velocity distribution, u = Uy
r U
u
flat plate
q
p
U
s
Applying mass conservation:
Mass rate entering section q p= Mass leaving section q r+ mass leaving section r s
Mass rate entering q p = Density Volume flow rate
= B U= 1.0 1 10 103 10 = 0.1 kg/s
Mass flow rate through the element dy at section r s
dm = u B dy
dm = BU
ydy
Integrating the above equation gives,
m =0
BU
y
dy
=
2BU
2
=
BU
2
=
31 1 10 10 10
2
= 0.05 kg/s
Thus mass flow rate leaving across the section, q r = 0.1 0.05 = 0.05 kg/s
Q.6 Ans. (c)
Drag froce on the plate will be the rate of change of momentum of control volume qprs
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Thus, momentum rate entering section q p = Um
= 0.1 10 = 1N
Momentum rate leaving through section rs
=0
u Bd y u
=2
2
0
BU
y
dy
=
2 3
2
BU
3
=
2BU
3
=
2 31.0 1.0 10 10 10
3
= 0.33 N
Momentum rate leaving through section q r = 0.05 10 = 0.5 N
Drag force, F = Change in momentum rate= 1 0.33 0.5 = 0.17 N
Reference: Fluid Mechanics and Hydraulic Machines, R.K. Bansal, 4th Edition, Page-655, 13.3.
Q.7 Ans (c)
Reynolds number at x,
Rex =Ux
= -5
2 1
1.5 10
= 1.33 105
Boundry layer thickness, =5
x
4.64x 4.64 1
Re 1.33 10
= 0.0127
Since the velocity profile is given as,
U
U=
33 y 1 y
2 2
Velocity gradient,dU
dy=
2
3
3 1 1 3yU .
2 2
Therefore,dU
at y 0dy
=3 1
. .U2
From Newtons law of viscosity,
o =dU
at y 0dy
=3U
2
=5 3 21.5 10 1.23
2 0.0127
= 435.82 105N/m2 = 4.36 103N/m2
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-537, Ex. 12.3.
Q.8 Ans. (d)
As per Blasius result thickness of laminar boundary layer is given asx
5x
Re
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Hence, =5x
vx
Therefore, 1
12x
1
2x
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-525, 12.6.
Q.9 Ans. (c)
In direction of flow usually pressure gradient is negative i.e.P
x
= -ve, supports the fluid flow. But in case of
flow separation, pressure gradient is positive i.e.P
x
= +ve, supports the fluid separation &P
x
= 0, it means
that the fluid is on the verge of separation.
Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-662, Equation-13.7.
Q.10 Ans. (False)Q.11 Ans. (b)
In a fluid flow over a flat plate, the dominant forces are inertia force and viscous force. Therefore Reynolds
number decides the nature of the flow.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-770, 17.13.
Q.12 Ans. (c)
A body where flow separation is suppresed is called streamlined body. For a well stream lined body the
separation occurs only at the down stream end.
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition , Page-805, 18.2.
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126
Laminar and Turbulent Flow
9
Year 2013
1. For steady, fully developed flow inside a straightpipe of diameter D, neglecting gravity effects,the pressure drop p over a lengthLand the wallshear stress
ware related by
(a) wpD
4L
(b)
2
w 2
p D
4L
(c) wpD
2L
(d) w
4 pL
D
Year 2010
2. The maximum velocity of a one-dimensionalincompressible fully developed viscous flow,
between two fixed parallel plates, is 6 ms1. Themean velocity (in ms1) of the flow is(a) 2 (b) 3(c) 4 (d) 5
Year 2009
3. The velocity profile of a fully developed laminarflow in a straight circular pipe, as shown in thefigure, is given by the expression
u(r) =2 2
2
R dp r 14 dx R
wheredp
dxis a constant
u(r)Rr
x
The average velocity of fluid in the pipe is
(a)
2R dp
8 dx
(b)
2R dp
4 dx
(c)
2R dp
2 dx
(d)
2R dp
dx
Year 2007
4. Consider steady laminar incompressible axi-sym-
metric developed viscous flow through a straight
circular pipe of constant cross-sectional area at
a Reynolds number of 5. The ratio of inertia forceto viscous force on a fluid particle is
(a) 5 (b)1
5
(c) 0 (d)
Year 2006
5. The velocity profile in fully developed laminar flow
in a pipe of diameter D is given by
u =
2
0 24ru 1 ,D
where r is the radial distance
from the center. If the viscosity of the fluid is ,the pressure drop across a length L of the pipe is
(a)0
2
u L
D
(b)
0
2
4 u L
D
(c)0
2
8 u L
D
(d)
0
2
16 u L
D
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127
Year 1996
6. In flow through a pipe, the transition from laminar
to turbulent flow does not depend on
(a) velocity of the fluid
(b) density of the fluid
(c) diameter of the pipe
(d) length of the pipe
7. For laminar flow through a long pipe, the pressure
drop per unit length increases
(a) in linear proportion to the cross-sectional area
(b) in proportion to the diameter of the pipe
(c) in inverse proportion to the cross-sectional
area
(d) in inverse proportion to the square of cross-
sectional area
Year 1995
8. In fully developed laminar flow in a circular pipe,
the head loss due to friction is directly proportional
to........
(mean velocity/square of the mean velocity)
Year 1994
9. For a fully developed viscous flow through a pipe,
the ratio of the maximum velocity to the average
velocity is.......
10. Prandtls mixing length in turbulent flow signifies
(a) the average distance perpendicular to the
mean flow covered by the mixing particles
(b) the ratio of mean free path to characteristic
length of the flow field
(c) the wavelength corresponding to the lowest
frequency present in the flow field
(d) the magnitude of turbulent kinetic energy
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128
Chapter-9Answers
1. Ans. (a) 2. Ans. (c) 3. Ans. (a) 4. Ans. (a) 5. Ans. (d)6. Ans. (d) 7. Ans. (c) 8. Ans.
(meanvelocity)
9. Ans. (two) 10.Ans. (a)
Space for Rough work
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129
Chapter-9Answer & Explanations
Q.1 Ans. (a)
In a pipe flow, =dp r
dx 2
=dp d
dx 4
Therefore, shear stress at wall, w
=dp D
dx 4
Above equation can be written as, w=
p D
4L
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-547, Eqn.-13.3.
Q.2 Ans. (c)
For the flow of fully developed between two fixed parallel plates
Vmax
B
x
y
The velocity distribution for laminar flow between fixed parallel plates is given as,
V =1
2
p
x(By y2)
Flow velocity is maximum when y =B
2
Therefore, Vmax =
2B
8
p
x
Average flow velocity is obtained by dividing the total discharge with cross sectional area.
Hence, Vavg. =
2B
12
p
x
Therefore,max
avg
V
V =3
2
So mean velocity, Vavg = max.2
V3
=2
6 4m/s3
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-557, Equation13.389.
Q.3 Ans. (a)
x
rr
Ru(r)
dr
Consider an element ring of thickness dr at a radius of r.
Therefore, element discharge from this ring, dQ = (2r).dr. u(r)
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130
Therefore, total discharge Q = Qd =R 2 2
2
0
R P(2 ) . 1 .
4 R
rr dr
x
=
R2 3
2
0
R P2
4 R
rr dr
x
=
R2 2 4
20
R P r r 24 x 2 4R
=
2 2 4
2R P R R 24 x 2 4R
Total discharge, Q =
4R P
8 x
Now, Q = Area Average velocity
Area Vavg. =
4R P
8 x
R2 Vavg =4R P
8 x
Therefore,Vavg =2R P
8
x
=
2R P
8
x
Reference: Hydraulics and Fluid Mechanics, Modi and Seth, 17th Edition, Page-550, Eqn-13.14.
Q.4 Ans. (a)
Reynolds number, Re =Inertia force vd
Viscous force
Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357.
Q.5 Ans. (d)
Given:
Fully developed laminar flow velocity profile in a circular pipe is given by
u =
2
o 2
4ru 1
D
......(i)
From Hagen-Poiseuille equation the pressure loss in fully developed laminar flow across the length of the
pipe is given as
p1 p2 = 232u L
d......(ii)
The ratio of maximum velocity and average velocity in case of fully developed laminar flow through a circular
pipe is 2 i.e.ou
u= 2
Therefore, from equation (ii), p1 p2 =
o
2
u32 L
2
d=
o
2
16u L
d
Or
Fully developed laminar flow velocity profile in a circular pipe is given by
u =
2
o 2
4ru 1
D
......(i)
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131
Rr
Now =P r
x 2
......(ii)
From Newtons law viscosity, =du
dy
Here y = R r
Therefore, dy = dr
Putting the value of dy in equation (i), we get
=u
r
......(iii)
From equation (ii) and (iii),u
r
=
P r.
x 2
u
r
=1 P r
x 2
From equation (i) ,
2
o 2
4ru 1
r D
=
1 P r
x 2
o28u
rD
=1 P r
x 2
P
x
= o
2
16 u
D
P =o
2
16 ux
D
Integrating over a length of L,
2
1
P
P
dp =L
o
2
o
16 udx
D
p2 p1 =o
2
16 u L
D
Therefore, p2 p1 =o
2
16 u L
D
Reference: Hydraulics and Fluid mechanics, Modi and Seth, 17 Edition, Page-548. Equation-13.3,
13.6, 13.7.
Q.6 Ans. (d)
In flow through pipe the transition from laminar to turbulent depends upon Reynolds number which is given as
Re =vd where, d is the characteristic dimension of the pipe.
Reference: Fluid Mechanics, R.K. Rajput, Edition 2005, Page-357.
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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 p2 = 232u L
d
1 2p p
L
= 2
32u
dTherefore, for laminar flow through a pipe, the pressure drop per unit length increases in inverse proportion
to the cross-sectional area.
Reference: Fluid Mechanics, R.K. Rajput, 2005 Edition, Page-442, Equation-10.11.
Q.8 Ans. (mean velocity)
The head loss due to friction in fully developed laminar flow in a circular pipe is given as
hf
=1 2
P P
g
=avg.
2
32 u L
gd
Head loss in laminar flow over a length L of circular pipe varies as the first power of the mean velocity of the
flow.
Reference: Hydraulics and Fl
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