Fluid Dynamics

Question 1
A tube of uniform diameter D is immersed in a steady flowing inviscid liquid stream of velocity V, as shown in the figure. Gravitational acceleration is represented by g. The volume flow rate through the tube is ______.

A
\frac{\pi}{4}D^2V
B
\frac{\pi}{4}D^2\sqrt{2gh_2}
C
\frac{\pi}{4}D^2\sqrt{2g(h_1+h_2)}
D
\frac{\pi}{4}D^2\sqrt{V^2-2gh_2}
GATE ME 2022 SET-2   Fluid Mechanics
Question 1 Explanation: 


By applying Bernoulli's equation between (1) and (2)
\begin{aligned} \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1&=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \frac{P_{atm}+\rho gh_1}{\rho g}+\frac{V_1^2}{2g}+0&=\frac{P_{atm}}{\rho g}+\frac{V_2^2}{2g}+h_1+h_2 \\ &[\because \;\; V_1=V]\\ \frac{V_1^2}{2g}&=\frac{V}{2g}+h_2-(h_1+h_2)\\ \therefore \; V_2&=\sqrt{V^2-2gh_1}\\ \therefore \; Q&=A_2V_2=\frac{\pi d^2}{4} \times \sqrt{V^2-2gh_1} \end{aligned}
Question 2
Consider steady, one-dimensional compressible flow of a gas in a pipe of diameter 1 m. At one location in the pipe, the density and velocity are 1 kg/m^3 and 100 m/s, respectively. At a downstream location in the pipe, the velocity is 170 m/s. If the pressure drop between these two locations is 10 kPa, the force exerted by the gas on the pipe between these two locations is ____________ N.
A
350 \pi ^2
B
750 \pi
C
1000 \pi
D
3000
GATE ME 2022 SET-1   Fluid Mechanics
Question 2 Explanation: 


Applying momentum equation to the pipe flow,
\Sigma \vec{F}=(\dot{m}\vec{V})_{out}-(\dot{m}\vec{V})_{in}+\frac{\partial }{\partial t}(m\vec{V})_{e,v}
P_1A-P_2A+F=\dot{m}(V_2)-\dot{m}(V_1)
\begin{aligned} \therefore \; F&=\dot{m}(V_2-V_1)-(P_1-P_2)A\\ &=\rho _1AV_1(V_2-V_1)-(P_1-P_2)A\\ &=\frac{ \pi d^2}{4}\left [ \rho (V_1V_2-V_1^2)-(P_1-P_2) \right ]\\ &=\frac{\pi \times 1^2}{4}\left [ 1 \times (100 \times 170 -100^2)-(10 \times 10^3) \right ]\\ &=-750 \pi N \end{aligned}
Negative sign shows that assumed direction of force is opposite of actual direction.
Question 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 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 8
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 8 Explanation: 
Within boundary layer, Bernoulli's equation is not valid because due to friction, flow is irrotional
Question 9
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 9 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 10
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 10 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}
There are 10 questions to complete.

Leave a Comment