Fluid Dynamics

Question 1
A cylindrical jet of water (density = 1000 kg/m^3) impinges at the center of a flat, circular plate and spreads radially outwards, as shown in the figure. The plate is resting on a linear spring with a spring constant k=1 \; kN/m. The incoming jet diameter is D=1 \; cm.

If the spring shows a steady deflection of 1 cm upon impingement of jet, then the velocity of the incoming jet is _______m/s (round off to one decimal place).
A
11.3
B
18.5
C
8.6
D
14.2
GATE ME 2021 SET-1   Fluid Mechanics
Question 1 Explanation: 


\begin{aligned} \delta &=1 \mathrm{~cm} \\ D &=1 \mathrm{~cm} \\ \rho &=1000 \mathrm{~kg} / \mathrm{m}^{3} \\ K &=1 \mathrm{kN}-\mathrm{m}\\ \text{Force due to jet }&=\text{ Spring force}\\ \rho A V^{2} &=k x \\ 10^{3} \times \frac{\pi}{4} \times(0.01)^{2} \times V^{2} &=1 \times 10^{3} \times 0.01 \\ V &=11.28 \mathrm{~m} / \mathrm{s} \simeq 11.3 \mathrm{~m} / \mathrm{s} \end{aligned}
Question 2
A sprinkler shown in the figure rotates about its hinge point in a horizontal plane due to water flow discharged through its two exit nozzles.

The total flow rate Q through the sprinkler is 1 litre/sec and the cross-sectional area of each exit nozzle is 1 cm^{2}. Assuming equal flow rate through both arms and a frictionless hinge, the steady state angular speed of rotation (in rad/s) of the sprinkler is ______ (correct to two decimal places).
A
5
B
10
C
15
D
20
GATE ME 2018 SET-1   Fluid Mechanics
Question 2 Explanation: 


Relative velocities of water with sprinkler
\begin{aligned} V_{A}&=\frac{Q / 2}{A}=\frac{1 \times 10^{-3}}{2 \times 10^{-4}}=5 \mathrm{m} / \mathrm{s} \\ V_{B}&=5 \mathrm{m} / \mathrm{s} \end{aligned}
Absolute velocity from B side
\begin{aligned} V_{\text {Abs }}^{\prime}-\left(+r_{B} \omega\right)&=V_{B}\\ V_{A b s}^{\prime} &=V_{B}+r_{B} \omega \\ &=5+0.1 \omega \end{aligned}
Absolute velocity from A side
\begin{aligned} V_{\mathrm{Abs}}-\left(-r_{A} \omega\right)&=V_{A}\\ V_{\mathrm{Abs}}&=V_{A}-r_{A} \omega \\ V_{\mathrm{Abs}}&=5-0.2 \mathrm{c} \end{aligned}
The external torque to the sprinkler is zero.
\begin{aligned} \text{So,}\qquad \Sigma T&=0\\ \dot{m}_{A} V_{A b s} r_{A}-\dot{m}_{B} V_{A b s}^{\prime} r_{B}&=0 \\ \rho\left(\frac{Q}{2}\right)\{5-0.2 \omega\} 0.2-\rho \frac{Q}{2}\{5+0.1 \omega\} 0.1&=0 \\ 1-0.04 \omega-0.5-0.01 \omega&=0 \\ 0.05 \omega&=0.5 \\ \omega&=10 \mathrm{rad} / \mathrm{s} \end{aligned}
Question 3
The arrangement shown in the figure measures the velocity V of a gas of density 1\, kg/m^{3} flowing through a pipe. The acceleration due to gravity is 9.81\, m/s^{2}.If the manometric fluid is water (density 1000 kg/s^{2}) and the velocity V is 20 m/s, the differential head h (in mm) between the two arms of the manometer is _____
A
21.4 mm
B
22.4 mm
C
20.4 mm
D
23.4 mm
GATE ME 2017 SET-2   Fluid Mechanics
Question 3 Explanation: 
Method: (i)
Given data: p_{\text {gas }}=1 \mathrm{kg} / \mathrm{m}^{3} ; \quad g=9.81 \mathrm{m} / \mathrm{s}^{2}
\rho_{m}=1000 \mathrm{kg} / \mathrm{m}^{3}, \quad V=20 \mathrm{m} / \mathrm{s}
Dynamic pressure of gas =(\rho g h)_{\text {water }}
\begin{aligned} \frac{1}{2} \rho_{\mathrm{gas}} V^{2} &=\rho_{m} \times 9.81 \times h \\ \frac{1}{2} \times 1 \times(20)^{2} &=1000 \times 9.81 \times h\\ \text{or }h &=0.02038 \mathrm{m} \text { of water } \\ &=20.38 \mathrm{mm} \text { of water } \end{aligned}
Method: (ii)
\begin{aligned} \rho_{m} &=1000 \mathrm{kg} / \mathrm{m}^{3}, \rho_{w}=1 \mathrm{kg} / \mathrm{m}^{3} \\ V &=20 \mathrm{m} / \mathrm{s} \\ V &=\sqrt{2 g h\left(\frac{\rho_{m}}{\rho_{w}}-1\right)} \\ 20 &=\sqrt{2 \times 9.81 \times h\left(\frac{1000}{1}-1\right)} \\ h &=20.40 \mathrm{mm} \text { of water } \end{aligned}
Question 4
The water jet exiting from a stationary tank through a circular opening of diameter 300 mm impinges on a rigid wall as shown in the figure. Neglect all minor losses and assume the water level in the tank to remain constant. The net horizontal force experienced by the wall is ___________ kN.
Density of water is 1000 kg/m^{3}.
Acceleration due to gravity g = 10 m/s^{2}.
A
8.76
B
6.56
C
9.24
D
8.36
GATE ME 2016 SET-3   Fluid Mechanics
Question 4 Explanation: 


Diameter of jet,
d = 300\mathrm{mm} = 0.30 \mathrm{m}
\therefore \quad Cross-section area,
\begin{aligned} a &=\frac{\pi}{4} d^{2}=\frac{3.14}{4} \times(0.30)^{2} \\ &=0.07065 \mathrm{m}^{2} \\ \text{Jet Velocity: }V \;&=\sqrt{2 g h}\\ &=\sqrt{2 \times 10 \times 6.2} = 11.135\mathrm{m}/\mathrm{s} \end{aligned}
Net horizontal force
\begin{aligned} F &=m[V-0] \\ &=\rho a V \times V \\ &=\rho a V^{2} \\ &=1000 \times 0.07065 \times(11.135)^{2} \\ &=8759.76 \mathrm{N} \\ &=8.76 \mathrm{kN} \end{aligned}
Question 5
A Prandtl tube (Pitot-static tube with C=1) is used to measure the velocity of water. The differential manometer reading is 10 mm of liquid column with a relative density of 10. Assuming g = 9.8 m/s^{2}, the velocity of water (in m/s) is ________
A
1.3m/s
B
5.2m/s
C
9.8m/s
D
8.6m/s
GATE ME 2015 SET-3   Fluid Mechanics
Question 5 Explanation: 
Given data:
\begin{aligned} C &=1 \\ x &=10 \mathrm{mm} \text { of liquid }=0.01 \mathrm{m} \text { of liquid } \\ S &=10 \\ \rho_{\text {mano }} &=10 \times 1000=10000 \mathrm{kg} / \mathrm{m}^{3} \\ g &=9.8 \mathrm{m} / \mathrm{s}^{2} \\ h &=x\left[\frac{\rho_{\operatorname{man}}}{\rho_{\text {water }}}-1\right]=0.01\left[\frac{1000}{1000}-1\right] \\ &=0.09 \mathrm{m} \text { of water } \end{aligned}
Velocity of water
\begin{aligned} V &=C \sqrt{2 g h}=1 \sqrt{2 \times 9.8 \times 0.09} \\ &=1.328 \mathrm{m} / \mathrm{s} \end{aligned}
Question 6
Within a boundary layer for a steady incompressible flow, the Bernoulli equation
A
holds because the flow is steady
B
holds because the flow is incompressible
C
holds because the flow is transitional
D
does not hold because the flow is frictional
GATE ME 2015 SET-2   Fluid Mechanics
Question 6 Explanation: 
Within boundary layer, Bernoulli's equation is not valid because due to friction, flow is irrotional
Question 7
Water (\rho =1000 kg/m^{3}) flows through a venturimeter with inlet diameter 80 mm and throat diameter 40 mm. The inlet and throat gauge pressures are measured to be 400 kPa and 130 kPa respectively. Assuming the venturimeter to be horizontal and neglecting friction, the inlet velocity (in m/s) is _______
A
6m/s
B
7m/s
C
8m/s
D
9m/s
GATE ME 2015 SET-1   Fluid Mechanics
Question 7 Explanation: 
Given data:
\begin{aligned} \rho &=1000 \mathrm{kg} / \mathrm{m}^{3} \\ d_{1} &=80 \mathrm{mm}=0.08 \mathrm{mm} \\ d_{2} &=40 \mathrm{mm}=0.04 \mathrm{m} \end{aligned}

\begin{aligned} p_{1}&=400 \mathrm{kPa}=400 \times 10^{3} \mathrm{Pa} \\ p_{2}&=130 \mathrm{kPa}=130 \times 10^{3} \mathrm{Pa} \end{aligned}
Applying Bernoullis equation between sections 1 and 2
\begin{aligned} \frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}&+z_{1}=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\ \text{where }z_{1}&=z_{1} \text{ for horizontal pipe}\\ \therefore \quad \frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}&=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g} \\ or \quad \frac{p_{1}-p_{2}}{\rho g}&=\frac{V_{2}^{2}-V_{1}^{2}}{2 g} \\ or \quad \frac{p_{1}-p_{2}}{\rho}&=\frac{V_{2}^{2}-V_{2}^{2}}{2} \quad\ldots(i)\\ \end{aligned}
Applying continuity equation between sections and 2
\begin{aligned} A_{1} V_{1}&=A_{2} V_{2} \\ \frac{\pi}{4} d_{1}^{2} V_{1}&=\frac{\pi}{2} d_{2}^{2} V_{2}\\ \text{or }\quad d_{1}^{2} V_{1}&=d_{2}^{2} V_{2}\\ \text{or }\quad V_{2}&=\left(\frac{d_{1}}{d_{2}}\right)^{2} V_{1}=\left(\frac{0.08}{0.04}\right)^{2} V_{1} \\ V_{2}&=4 V_{1}\\ \text{Substituting } V_{2}&=4 V_{1} \text{ in Eq. }( i ),\text{ we get}\\ \frac{400 \times 10^{3}-130 \times 10^{3}}{1000}&=\frac{\left(4 V_{1}\right)^{2}-V_{1}^{2}}{2} \\ 540&=16 \mathrm{V}_{1}^{2}-V_{1}^{2} \\ 540&=15 \mathrm{V}_{1}^{2}\\ or \quad V_{1}^{2}&=36\\ or \quad V_{1}&=6 \mathrm{m} / \mathrm{s}\\ \end{aligned}
Question 8
Water is coming out from a tap and falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/s^2. Assuming steady, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately
A
10
B
15
C
20
D
25
GATE ME 2013   Fluid Mechanics
Question 8 Explanation: 


Given data:
At section 1-1
Diameter: \quad d_{1}=20 \mathrm{mm}=0.02 \mathrm{m}
Velocity: \quad V_{1}=2 \mathrm{m} / \mathrm{s}
Acceleration due to gravity,
g=9.81 \mathrm{m} / \mathrm{s}^{2}
Applying Bernoulli's equation between sections 1 and 2 , we get
\begin{aligned} \frac{p_{1}}{p g}+\frac{V_{1}^{2}}{2 g}+z_{1} &=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\ 0+\frac{V_{1}^{2}}{\rho g}+0.5 &=0+\frac{V_{2}^{2}}{\rho g}+0 \\ \frac{V_{1}^{2}}{2 g}+0.5 &=\frac{V_{2}^{2}}{2 g}\\ \text{ or }\qquad V_{2}^{2}&=V_{1}^{2}+0.5 \times 2 g\\ \text{or }\qquad V_{2} &=\sqrt{V_{1}^{2}+0.5 \times 2 \times g} \\ &=\sqrt{(2)^{2}+0.5 \times 2 \times 9.81} \\ &=\sqrt{13.81}=3.716 \mathrm{m} / \mathrm{s} \end{aligned}
Applying continuity equation between sections and 2, we get
\begin{aligned} A_{1} V_{1} &=A_{2} V_{2} \\ \frac{\pi}{4} d_{1}^{2} V_{1} &=\frac{\pi}{4} d_{2}^{2} V_{2}\\ \text{or }\quad d_{1}^{2} V_{1}&=d_{2}^{2} V_{2} \\ (0.02)^{2} \times 2&=d_{2}^{2} \times 3.716\\ \text{or }\quad d_{2}^{2}&=2.1528 \times 10^{-4}\\ \text{or }\quad d_{2} &=0.01467 \mathrm{m} \\ &=14.67 \mathrm{mm} \approx 15 \mathrm{mm} \end{aligned}
Question 9
Figure shows the schematic for the measurement of velocity of air (density = 1.2kg /m^{3}) through a constant-area duct using a pitot tube and a water-tube manometer. The differential head of water (density = 1000 kg/m^{3} ) in the two columns of the manometer is 10mm. Take acceleration due to gravity as 9.8m/s^{2}. The velocity of air in m/s is
A
6.4
B
9
C
12.8
D
25.6
GATE ME 2011   Fluid Mechanics
Question 9 Explanation: 
Given data:

Density of air,
\rho_{a}=1.2 \mathrm{kg} / \mathrm{m}^{3}
Density of water,
\rho_{w}=1000 \mathrm{kg} / \mathrm{m}^{3}
Differential head in manometer
\begin{aligned} h &=10 \mathrm{mm} \text { of water } \\ &=0.10 \mathrm{m} \text { of water } \end{aligned}
This reading is the dynamic pressure head. Hence
dynamic pressure,
\begin{aligned} P_{\text {dyn }} &=(\rho g h)_{\text {water }}=\rho_{w} g h \\ &=1000 \times 9.81 \times 0.01=98.1 \mathrm{N} / \mathrm{m}^{2} \\ \text { also } p_{\text {dyn }} &=\left(\frac{1}{2} \rho V^{2}\right)_{\text {air }}=\frac{1}{2} \rho_{\mathrm{a}} V^{2}\\ 98.1 &=\frac{1}{2} \times 1.2 \times V^{2} \\ \text{or }\qquad V^{2} &=163.5\\ V&=12.78 \mathrm{m} / \mathrm{s} \\ \end{aligned}
Alternatively
V=\sqrt{2 g h}
h=x\left(\frac{\rho_{m}}{\rho}-1\right)
x= reading of differential manometer =10 \times 10^{-3} \mathrm{m}
\rho_{m}= density of manometric fluid =1000 \mathrm{kg} / \mathrm{m}^{3}
\rho= density of following fluid =1.2 \mathrm{kg} / \mathrm{m}^{3}
V=\sqrt{2 \times 9.81 \times 10 \times 10^{-3}\left(\frac{1000}{1.2}-1\right)}
V=12.779 \mathrm{m} / \mathrm{s} \approx 12.8 \mathrm{m} / \mathrm{s}
Question 10
Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20cm to 10cm. The pressure in the 20cm pipe just upstream of the reducer is 150kPa. The fluid has a vapour pressure of 50kPa and a specific weight of 5 kN/m^{3}. Neglecting frictional effects, the maximum discharge (in m^{3}/s) that can pass through the reducer without causing cavitation is
A
0.05
B
0.16
C
0.27
D
0.38
GATE ME 2009   Fluid Mechanics
Question 10 Explanation: 
Given data:
\begin{aligned} d_{1}&=20 \mathrm{cm}=0.2 \mathrm{m} \\ d_{2}&=10 \mathrm{cm}=0.1 \mathrm{mm} \\ p_{1}&=150 \mathrm{kP} \mathrm{a}=150000 \mathrm{Pa} \end{aligned}
Vapour pressure,
p_{v}=50 \mathrm{kPa}
Specific weight
w=5 \mathrm{kN} / \mathrm{m}^{3}=5000 \mathrm{N} / \mathrm{m}^{3}=\rho \mathrm{g}

Applying continuity equation at section (1)-(1) and
(2)-(2), we get
A_{1} V_{1}=A_{2} V_{2} \quad for incompressible flow
\begin{aligned} \frac{\pi}{4} d_{1}^{2} V_{1} &=\frac{\pi}{4} d_{2}^{2} V_{2} \\ d_{1}^{2} V_{1} &=d_{2}^{2} V_{2}\\ or \quad V_{1}&=\left(\frac{d_{2}}{d_{1}}\right)^{2} V_{2}=\left(\frac{10}{20}\right)^{2} V_{2} \\ V_{1}&=0.25 \mathrm{V}_{2} \end{aligned}
The maximum pressure in downstream of reducer should be greater or equal to vapour pressure \left(p_{v}\right) to avoid cavitation. Therefore for maximum discharge,
p_{2}=p_{v}=50 \mathrm{kPa}=50000 \mathrm{Pa}
Applying Bernoulli's equation a section (1)-(1) and (2)-(2), we get
\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2}
\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g} \quad \because z_{1}=z_{2}
\frac{150000}{5000}+\frac{(0.25)^{2} V_{2}^{2}}{2 \times 9.81}=\frac{50000}{5000}+\frac{V_{2}^{2}}{2 \times 9.81}
30+0.00318 V_{2}^{2}=10+0.0509 V_{2}^{2}
V_{2}^{2}=418.51
\text{or }\quad V_{2}=20.45 \mathrm{m} / \mathrm{s}
\therefore \quad Q=A_{2} V_{2}=\frac{\pi}{4} d_{2}^{2} V_{2}
=\frac{3.14}{4} \times(0.1)^{2} \times 20.45=0.160 \mathrm{m}^{3} / \mathrm{s}
There are 10 questions to complete.

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