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 ______.
\frac{\pi}{4}D^2V | |
\frac{\pi}{4}D^2\sqrt{2gh_2} | |
\frac{\pi}{4}D^2\sqrt{2g(h_1+h_2)} | |
\frac{\pi}{4}D^2\sqrt{V^2-2gh_2} |
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.
350 \pi ^2 | |
750 \pi | |
1000 \pi | |
3000 |
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).
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).
11.3 | |
18.5 | |
8.6 | |
14.2 |
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).
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).
5 | |
10 | |
15 | |
20 |
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 _____
21.4 mm | |
22.4 mm | |
20.4 mm | |
23.4 mm |
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}
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}
There are 5 questions to complete.