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 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 3 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 4 |
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 4 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}
Question 5 |
Two identically sized primary settling tanks receive water for Type-I settling (discrete
particles in dilute suspension) under laminar flow conditions. The surface overflow rate
(SOR) maintained in the two tanks are 30 m^3/m^2\cdot d and 15 m^3/m^2\cdot d. The lowest diameters
of the particles, which shall be settled out completely under SORs of 30 m^3/m^2\cdot d and 15 m^3/m^2\cdot d are
designated as d_{30} and d_{15} respectively. The ratio \frac{d_{30}}{d_{15}}
(round off to two decimal places),
is __________.
2 | |
1.41 | |
0.85 | |
2.35 |
Question 5 Explanation:
For type-I setting, Stokes law is applicable.
\begin{aligned} V_s&\propto d^2 \\ \frac{d_{30}^2}{d_{15}^2}&=\frac{30}{15}=2 \\ \frac{d_{30}}{d_{15}}&=\sqrt{2}=1.41 \end{aligned}
\begin{aligned} V_s&\propto d^2 \\ \frac{d_{30}^2}{d_{15}^2}&=\frac{30}{15}=2 \\ \frac{d_{30}}{d_{15}}&=\sqrt{2}=1.41 \end{aligned}
Question 6 |
A circular water tank of 2 m diameter has a circular orifice of diameter 0.1 m at the bottom.
Water enters the tank steadily at a flow rate of 20 litre/s and escapes through the orifice.
The coefficient of discharge of the orifice is 0.8. Consider the acceleration due to gravity
as 9.81 m/s^2 and neglect frictional loses. The height of the water level (in m, round off
to two decimal places) in the tank at the steady state, is ______.
0.25 | |
0.34 | |
0.52 | |
0.78 |
Question 6 Explanation:
Assume H is the level of weter in the tank in steady condition.
For steady water level in the tank
Discharge through orifice = Water enters in the tank
\begin{aligned} c_d a \sqrt{2 gH}&=20 \times 10^{-3} \\ 0.8 \times \frac{\pi}{4} (0.1)^2 \sqrt{2gH}&=0.02 \\ H&=0.5164 \end{aligned}
Question 7 |
Three reservoir P, Q and R are interconnected by pipes as shown in the figure (not drawn
to the scale). Piezometric head at the junction S of the pipes is 100 m. Assume
acceleration due to gravity as 9.81 m/s^2 and density of water as 1000 kg/m^3. The length
of the pipe from junction S to the inlet of reservoir R is 180 m.
Considering head loss only due to friction (with friction factor of 0.03 for all the pipes), the height of water level in the lowermost reservoir R (in m, round off to one decimal places) with respect to the datum, is ________.
Considering head loss only due to friction (with friction factor of 0.03 for all the pipes), the height of water level in the lowermost reservoir R (in m, round off to one decimal places) with respect to the datum, is ________.
32.09 | |
97.51 | |
121.25 | |
78.45 |
Question 7 Explanation:
Apply conutinuity
\begin{aligned} Q_3&=Q-1+Q_2 \\ &= A_1V_1+A_2V_2\\ &= \frac{\pi}{4}(0.3)^2(2.56)+\frac{\pi}{4}(0.3)^2(1.98)\\ &= 0.3209 m^3/s \end{aligned}
Apply energy eq. between (S) and (R)
\begin{aligned} H_s&= H_r+h_f\\ 100&=+\frac{8Q_3^2}{\pi ^2 g} \times \frac{fL_3}{D_3^5} \\ 100 &=z+\frac{8(0.3209)^2}{\pi^2 g} \times \frac{(0.03)(180)}{(0.45)^5} \\ z&= 97.51 m \end{aligned}
\begin{aligned} Q_3&=Q-1+Q_2 \\ &= A_1V_1+A_2V_2\\ &= \frac{\pi}{4}(0.3)^2(2.56)+\frac{\pi}{4}(0.3)^2(1.98)\\ &= 0.3209 m^3/s \end{aligned}
Apply energy eq. between (S) and (R)
\begin{aligned} H_s&= H_r+h_f\\ 100&=+\frac{8Q_3^2}{\pi ^2 g} \times \frac{fL_3}{D_3^5} \\ 100 &=z+\frac{8(0.3209)^2}{\pi^2 g} \times \frac{(0.03)(180)}{(0.45)^5} \\ z&= 97.51 m \end{aligned}
Question 8 |
Two identical pipes (i.e., having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and discharge freely into the atmosphere. In the second case, they are attached in parallel and also discharge freely into the atmosphere. Neglecting all minor losses, and assuming that the friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is ______
1.45 | |
1.85 | |
2.83 | |
3.8 |
Question 8 Explanation:
Given: Two identical pipes of same length (L), diameter (D) and roughness (k_s).
First case (series)
Assume height of reservoir = H
\begin{aligned} H &=h_{f1} +h_{f2}\;\;\;...(i)\\ H&=2h_{f1} \;\;\;(\because h_{f1}=h_{f2})\\ H&=2\left ( \frac{8Q_1^2}{\pi^2 g} \times \frac{fL}{D^5} \right ) \end{aligned}
Second case (Parallel)
H =\frac{8}{\pi^2 g}\left ( \frac{Q_2}{2} \right )^2 \times \frac{fL}{D^5}\;\;\;...(ii)
By eq. (i) and (ii)
\begin{aligned} 2\left ( \frac{8Q_1^2}{\pi^2 g} \times \frac{fL}{D^5} \right )&=\frac{8}{\pi^2 g}\left ( \frac{Q_2}{2} \right )^2 \times \frac{fL}{D^5}\\ \frac{Q_2}{Q_1}&=\sqrt{8}=2.83 \end{aligned}
First case (series)
Assume height of reservoir = H
\begin{aligned} H &=h_{f1} +h_{f2}\;\;\;...(i)\\ H&=2h_{f1} \;\;\;(\because h_{f1}=h_{f2})\\ H&=2\left ( \frac{8Q_1^2}{\pi^2 g} \times \frac{fL}{D^5} \right ) \end{aligned}
Second case (Parallel)
H =\frac{8}{\pi^2 g}\left ( \frac{Q_2}{2} \right )^2 \times \frac{fL}{D^5}\;\;\;...(ii)
By eq. (i) and (ii)
\begin{aligned} 2\left ( \frac{8Q_1^2}{\pi^2 g} \times \frac{fL}{D^5} \right )&=\frac{8}{\pi^2 g}\left ( \frac{Q_2}{2} \right )^2 \times \frac{fL}{D^5}\\ \frac{Q_2}{Q_1}&=\sqrt{8}=2.83 \end{aligned}
Question 9 |
A circular duct carrying water gradually contracts from a diameter of 30 cm to 15 cm. The figure shows the arrangement of differential manometer attached to the duct
When the water flows, the differential manometer shows a deflection of 8 cm of mercury (Hg). The values of specific gravity of mercury and water are 13.6 and 1.0, respectively. Consider the acceleration due to gravity, g=9.81m/s^2. Assuming frictionless flow, the flow rate (in m^3/s, round off to 3 decimal places) through the duct is ____
When the water flows, the differential manometer shows a deflection of 8 cm of mercury (Hg). The values of specific gravity of mercury and water are 13.6 and 1.0, respectively. Consider the acceleration due to gravity, g=9.81m/s^2. Assuming frictionless flow, the flow rate (in m^3/s, round off to 3 decimal places) through the duct is ____
10.012 | |
0.025 | |
0.081 | |
0.096 |
Question 9 Explanation:
\begin{aligned} Q&=\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh} \\ h&=x\left [ \frac{\rho H_g}{\rho _w}-1 \right ] \\ &=0.08(13.6-1)=1.008m \;\;(A_1=4A_2) \\ Q&=\frac{4A_2 \times A_2}{\sqrt{16A_2^2-A_2^2}}\times \sqrt{2 \times 9.81 \times 1.008} \\ &= \frac{4}{\sqrt{15}} \times A_2 \times \sqrt{2 \times 9.81 \times 1.008}\\ &= \frac{4}{\sqrt{15}} \times \frac{\pi}{4}(0.15)^2 \times \sqrt{2 \times 9.81 \times 1.008} \\ Q&=0.081m^3/s \end{aligned}
Question 10 |
Water is pumped at a steady uniform flow rate of 0.01 m^{3}/s through a horizontal smooth circular pipe of 100 mm diameter. Given that the Reynolds number is 800 and g is 9.81 m/s^{2}, the head loss (in meters, up to one decimal place) per km length due to friction would be_____
66.1 | |
133.5 | |
33.3 | |
76.7 |
Question 10 Explanation:
Head loss due to frictions,
\begin{aligned} h_{L}&=\frac{n O^{2}}{12.1 d^{5}}\\ \text{Here. }f&= \text{ friction factor}\\ &=\frac{64}{\text { Re }}=\frac{64}{800}=0.08\\ \end{aligned}
Head loss per metre length
\begin{aligned} \frac{h_{L}}{L}&=\frac{fQ^{2}}{12.1d^{5}}=\frac{0.08\times(0.01)^{3}}{12.1\times(1.0)^{5}}\\ \frac{h_{L}}{L}&=\frac{0.8}{12.1} = 0.06611 \text{meter/meter}\\ &=66.11\text{meter/km} \end{aligned}
There are 10 questions to complete.
