Question 1 |
A venturimeter as shown in the figure (not to scale) is connected to measure the flow of water in a vertical pipe of 20 cm diameter.
Assume g=9.8 \mathrm{~m} / \mathrm{s}^{2}. When the deflection in the mercury manometer is 15 cm, the flow rate (in lps, round off to two decimal places) considering no loss in the venturimeter is ___________
Assume g=9.8 \mathrm{~m} / \mathrm{s}^{2}. When the deflection in the mercury manometer is 15 cm, the flow rate (in lps, round off to two decimal places) considering no loss in the venturimeter is ___________
49.4 | |
23.36 | |
87.64 | |
68.22 |
Question 1 Explanation:
\begin{aligned} \text { Discharge }(Q) &=C_{d} \frac{A_{1} A_{2}}{\sqrt{A_{1}^{2}-A_{2}^{2}}} \sqrt{2 g h} \\ h &=X\left(\frac{\rho_{m}}{\rho}-1\right)=0.15\left(\frac{13.6 \times 10^{3}}{10^{3}}-1\right) \\ &=1.89 \mathrm{~m}\\ Q &=\frac{A_{1} A_{2}}{A_{2} \sqrt{\left(\frac{A_{1}}{A_{2}}\right)^{2}-1}} \times \sqrt{2 \times 9.8 \times 1.89} \\&= \frac{\frac{\pi}{4}(0.2)^{2}}{\sqrt{(2)^{4}-1}} \times \sqrt{2 \times 9.8 \times 1.89} \\&= 49.395 \mathrm{l} / \mathrm{s} \simeq 49.40 \mathrm{l} / \mathrm{s} \end{aligned}
Question 2 |
A fire hose nozzle directs a steady stream of water of velocity 50 m/s at an angle of 45^{\circ} above the horizontal. The stream rises initially but then eventually falls to the ground. Assume water as incompressible and inviscid. Consider the density of air and the air friction as negligible, and assume the acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2}. The maximum height (in m,round off to two decimal places) reached by the stream above the hose nozzle will then be _________
63.71 | |
56.21 | |
98.36 | |
88.24 |
Question 2 Explanation:
As we know that
\begin{aligned} {V}^{2}-u^{2}&=2 a S\\ \text{In vertical direction} (\uparrow)\\ 0^{2}-\left(50 \sin 45^{\circ}\right)^{2} &=2(-9.81) h_{\max } \\ h_{\max } &=\frac{\left(50 \sin 45^{\circ}\right)^{2}}{2(9.81)}=63.71 \mathrm{~m} \end{aligned}
Question 3 |
A rectangular open channel of 6 m width is carrying a discharge of 20 \mathrm{~m}^{3} / \mathrm{s}. Consider the acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and assume water as incompressible and inviscid. The depth of flow in the channel at which the specific energy of the flowing water is minimum for the given discharge will then be
0.82 m | |
1.04 m | |
2.56 m | |
3.18 m |
Question 3 Explanation:
Minimum specific energy will correspond to a critical flow condition.
The critical depth \begin{aligned} \left(Y_{C}\right) &=\left[\frac{q^{2}}{g}\right]^{1 / 3} \\ Y_{C} &=\left[\frac{(20 / 6)^{2}}{9.81}\right]^{1 / 3}=1.042 \mathrm{~m} \end{aligned}
Question 4 |
The ratio of the momentum correction factor to the energy correction factor for a laminar flow in a pipe is
\frac{1}{2} | |
\frac{2}{3} | |
1 | |
\frac{3}{2} |
Question 4 Explanation:
For laminar flow through pipe.
\frac{\text { Momentum correction factor }}{\text { Kinetic energy correction factor }}=\frac{4 / 3}{2}=\frac{4}{6}=\frac{2}{3}
\frac{\text { Momentum correction factor }}{\text { Kinetic energy correction factor }}=\frac{4 / 3}{2}=\frac{4}{6}=\frac{2}{3}
Question 5 |
A cylinder (2.0 m diameter, 3.0 m long and 25 kN weight) is acted upon by water on one side and oil (specific gravity=0.8) on other side as shown in the figure.
The absolute ratio of the net magnitude of vertical forces to the net magnitude of horizontal forces (round off to two decimal places) is _____________
The absolute ratio of the net magnitude of vertical forces to the net magnitude of horizontal forces (round off to two decimal places) is _____________
0.61 | |
1.22 | |
0.22 | |
0.94 |
Question 5 Explanation:
Net horizontal force \left(F_{H}\right) due to liquids
\begin{aligned} F_{H} &=F_{H 1}-F_{H 2} \\ &=\rho_{w} g \bar{h}_{1} A_{v_{1}}-\rho_{\mathrm{oil}} g \bar{h}_{2} A_{v_{2}} \\ &=\left(10^{3}\right)(9.81)\left(1+\frac{2}{2}\right)(2 \times 3)-(800)(9.81)\left(\frac{1}{2}\right)(1 \times 3) \\ F_{H} &=105.948 \mathrm{kN}(\rightarrow) \end{aligned}
Net vertical force \left(F_{v}\right) due to liquids
\begin{aligned} F_{v} &=F_{1}+F_{2} \\ &=\rho_{w} g \forall_{1}+\rho_{o i l} g \forall_{2} \\ &=\left(10^{3}\right)(9.81)\left(\frac{\pi(1)^{2} \times 3}{2}\right)+(800)(9.81)\left(\frac{\pi}{4}(1)^{2} \times 3\right) \\ &=64.7199 \mathrm{kN}(\uparrow) \\ \frac{F_{V}}{F_{H}} &=\frac{64.7199}{105.948}=0.61 \end{aligned}
Question 6 |
A fluid flowing steadily in a circular pipe of radius R has a velocity that is everywhere parallel to the axis (centerline) of the pipe. The velocity distribution along the radial direction is V_r=U\left ( 1-\frac{r^2}{R^2} \right ), where r is the radial distance as
measured from the pipe axis and U is the maximum velocity at r=0. The average velocity of the fluid in the pipe is
\frac{U}{2} | |
\frac{U}{3} | |
\frac{U}{4} | |
\frac{5}{6}U |
Question 6 Explanation:
\begin{aligned} u&=U\left ( 1-\frac{r^2}{R^2} \right ) \\ \dot{m}&=\int_{0}^{R} \rho (2 \pi r\; dr)u\\ &= 2 \pi \rho U\int_{0}^{R} \left ( 1-\frac{r^2}{R^2} \right ) r\; dr\\ \rho (\pi R^2)\bar{u} &= 2 \pi \rho U \left ( \frac{R^2}{2}-\frac{R^4}{R^2 \times 4} \right )\\ \bar{u}&= \frac{U}{2}\\ \bar{u} &= \text{Mean velocity} \\ U&=\text{Max velocity} \end{aligned}
Question 7 |
Kinematic viscosity' is dimensionally represented as
\frac{M}{LT} | |
\frac{M}{L^{2} T} | |
\frac{T^{2}}{L} | |
\frac{L^{2}}{T} |
Question 7 Explanation:
Kinematic viscosity
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
Question 8 |
If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is
\frac{1}{\sqrt{2}} | |
\sqrt{2} | |
1 | |
2 |
Question 8 Explanation:
Efficient channel section
\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}
\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}
Question 9 |
A hydraulic jump occurs, in a triangular (V-shaped) channel with side slopes 1:1 (vertical
to horizontal). The sequent depths are 0.5 m and 1.5 m. The flow rate (in m^3/s, round
off to two decimal places) in the channel is _________.
1.24 | |
1.68 | |
1.73 | |
2.14 |
Question 9 Explanation:
\begin{aligned} A &=\frac{1}{2} \times 2Y \times Y =Y^2 \\ \bar{Y} &=\frac{Y}{3} \end{aligned}
For a horizontal and frictionless channel
Specific Force (F) =A\bar{Y}+\frac{Q^2}{Ag}=constant
\Rightarrow \; Y^2\left ( \frac{Y}{3} \right )+\frac{Q^2}{(Y^2)g}=constant
\Rightarrow \; \frac{Y^3}{3}+\frac{Q^2}{gY^2}=constant
If Y_1 and Y_2 are conjugate depth
\begin{aligned} \frac{Y_1^3}{3}+\frac{Q^2}{gY_1^2} &= \frac{Y_2^3}{3}+\frac{Q^2}{gY_2}\\ \frac{0.5^3}{3}+\frac{Q^2}{g \times 0.5^2}&=\frac{1.5^3}{3}+\frac{Q^2}{g \times 1.5^2} \\ \frac{1.5^3}{3}-\frac{0.5^3}{3}&=\frac{Q^2}{g}\left ( \frac{1}{0.5^2}-\frac{1}{1.5^2} \right ) \\ Q&=1.728 m^3/sec \end{aligned}
Question 10 |
A cast iron pipe of diameter 600 mm and length 400 m carries water from a tank and
discharges freely into air at a point 4.5 m below the water surface in the tank. The friction
factor of the pipe is 0.018. Consider acceleration due to gravity as 9.81 m/s^2. The velocity
of the flow in pipe (in m/s, round off to two decimal places) is __________.
2.56 | |
1.52 | |
4.12 | |
6.82 |
Question 10 Explanation:
Apply energy 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+h_f\\ 4.5&=\frac{flV^2}{2gD}+0.5\frac{V^2}{2g}+\frac{V^2}{2g} \\ 4.5&=\frac{(0.018)(400)V^2}{2(9.81)(0.6)}+\frac{1.5V^2}{2g} \\ 4.5&=\frac{12V^2}{2g}+\frac{1.5V^2}{2g} \\ V^2&=6.54 \\ V&=2.557 m/s\simeq 2.56m/s \end{aligned}
There are 10 questions to complete.