Question 1 |

Match Column X with Column Y:

\begin{array}{|l|l|}\hline \text{Column X}&\text{Column Y} \\ \hline \text{(P) Horton equation} & \text{((I) Design of alluvial channel} \\ \hline \text{(Q) Penman method} & \text{(II) Maximum flood discharge}\\ \hline \text{(R) Chezys formula}& \text{(III) Evapotranspiration}\\ \hline \text{(S) Lacey's theory}& \text{(IV) Infiltration}\\ \hline \text{(T) Dicken's formula}& \text{(V) Flow velocity}\\ \hline \end{array}

Which one of the following combinations is correct?

\begin{array}{|l|l|}\hline \text{Column X}&\text{Column Y} \\ \hline \text{(P) Horton equation} & \text{((I) Design of alluvial channel} \\ \hline \text{(Q) Penman method} & \text{(II) Maximum flood discharge}\\ \hline \text{(R) Chezys formula}& \text{(III) Evapotranspiration}\\ \hline \text{(S) Lacey's theory}& \text{(IV) Infiltration}\\ \hline \text{(T) Dicken's formula}& \text{(V) Flow velocity}\\ \hline \end{array}

Which one of the following combinations is correct?

(P)-(IV), (Q)-(III), (R)-(V), (S)-(I), (T)-(II) | |

(P)-(III), (Q)-(IV), (R)-(V), (S)-(I), (T)-(II) | |

(P)-(IV), (Q)-(III), (R)-(II), (S)-(I), (T)-(V) | |

(P)-(III), (Q)-(IV), (R)-(I), (S)-(V), (T)-(II) |

Question 2 |

Two reservoirs are connected by two parallel pipes of equal length and of
diameters 20 cm and 10 cm, as shown in the figure (not drawn to scale). When
the difference in the water levels of the reservoirs is 5 m, the ratio of discharge in
the larger diameter pipe to the discharge in the smaller diameter pipe is
____________. (round off to two decimal places)

(Consider only loss due to friction and neglect all other losses. Assume the friction factor to be the same for both the pipes)

(Consider only loss due to friction and neglect all other losses. Assume the friction factor to be the same for both the pipes)

2.25 | |

6.32 | |

4.22 | |

5.66 |

Question 2 Explanation:

\begin{aligned}
h_{f1} &=h_{f2} \\
\frac{8Q_1^2}{\pi ^2 g} \times \frac{fl}{D_1^5}&= \frac{8Q_2^2}{\pi ^2 g} \times \frac{fl}{D_2^5} \\
\left ( \frac{Q_1}{Q_2} \right )^2&= \left ( \frac{D_1}{D_2} \right )^5\\
\frac{Q_1}{Q_2} &= \left ( \frac{D_1}{D_2} \right )^{5/2}\\
&= \left ( \frac{0.2}{0.1} \right )^{5/2}\\&=5.66
\end{aligned}

Question 3 |

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 3 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 4 |

Two water reservoirs are connected by a siphon (running full) of total length 5000 m and diameter of 0.10m, as shown below

The inlet leg length of the siphon to its summit is 2000m. The difference in the water surface levels of the two reservoirs is 5 m. Assume the permissible minimum absolute pressure at the summit of siphon to be 2.5 m of water when running full. Given:

friction factor f=0.02 throughout,

atmospheric pressure= 10.3 m of water, and

acceleration due to gravity g= 9.81 m/s^2.

Considering only major loss using Darcy-Weisbach equation, the maximum height of the summit of siphon from the water level of upper reservoir,h (in m, round off to 1 decimal place)is _____

The inlet leg length of the siphon to its summit is 2000m. The difference in the water surface levels of the two reservoirs is 5 m. Assume the permissible minimum absolute pressure at the summit of siphon to be 2.5 m of water when running full. Given:

friction factor f=0.02 throughout,

atmospheric pressure= 10.3 m of water, and

acceleration due to gravity g= 9.81 m/s^2.

Considering only major loss using Darcy-Weisbach equation, the maximum height of the summit of siphon from the water level of upper reservoir,h (in m, round off to 1 decimal place)is _____

2.8 | |

6.9 | |

7.8 | |

5.8 |

Question 4 Explanation:

Applying Bernoulli's equation between 1 & 3

\begin{aligned} \frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1&= \frac{P_3}{\rho g}+\frac{v_3^2}{2g}+z_3+h_{f13}\\ z_1-z_3&=h_{f13}\;\;\;(\because \; P_1=P_3=P_{atm}\; and \; v_1=v_3=0) \\ 5&=\frac{fLv^2}{2gd} \\ v^2&=\frac{5 \times 2\times 9.81\times 0.1}{0.02\times 5000} \\ v&=0.313m/s \end{aligned}

Applying Bernoulli's equation between 1 & 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_{f12}\\ 10.3+0+0&=2.5+\frac{0.313^2}{2 \times 9.81}+h+\frac{0.02 \times 2000 \times 0.313^2}{2 \times 9.81 \times 0.1}\\ h&=5.798m \end{aligned}

Question 5 |

Water flows through a 90^{\circ} bend in a horizontal plane as depicted in the figure.

A pressure of 140 kPa is measured at section 1-1. The inlet diameter marked at section 1-1 is \frac{27}{\sqrt{\pi }}cm, while the nozzle diameter marked at section 2-2 is \frac{14}{\sqrt{\pi }}cm. Assume the following:

(i) Acceleration due to gravity = 10 m/s^{2}

(ii) Weights of both the bent pipe segment as well as water are negligible.

(iii) Friction across the bend is negligible.

The magnitude of the force (in kN, up to two decimal places) that would be required to hold the pipe section is ______

A pressure of 140 kPa is measured at section 1-1. The inlet diameter marked at section 1-1 is \frac{27}{\sqrt{\pi }}cm, while the nozzle diameter marked at section 2-2 is \frac{14}{\sqrt{\pi }}cm. Assume the following:

(i) Acceleration due to gravity = 10 m/s^{2}

(ii) Weights of both the bent pipe segment as well as water are negligible.

(iii) Friction across the bend is negligible.

The magnitude of the force (in kN, up to two decimal places) that would be required to hold the pipe section is ______

1.85 | |

2.60 | |

4.45 | |

6.58 |

Question 5 Explanation:

\begin{aligned} P_{1} A_{1} &=F_{x} \\ F_{x} &=\left(140 \times 10^{3}\right)\left[\frac{\pi}{4}\left(\frac{0.27}{\sqrt{\pi}}\right)^{2}\right] \\ &=2551.5 \mathrm{N}\\ F_{y} &=P_{2} A_{2} \\ &=\left(1.013 \times 10^{5}\right)\left[\frac{\pi}{4}\left(\frac{0.14}{\sqrt{\pi}}\right)^{2}\right] \\ &=49637 \mathrm{N} \\ \text { Resultant force } &=\sqrt{F_{x}^{2}+F_{y}^{2}} \\ &=\sqrt{(2551.5)^{2}+(49637)^{2}} \\ &=259933 \mathrm{N} \\ &=2.60 \mathrm{kN} \end{aligned}

Question 6 |

A venturimeter having a throat diameter of 0.1 m is used to estimate the flow rate of a horizontal pipe having a diameter of 0.2 m. For an observed pressure difference of 2 m of water head and coefficient of discharge equal to unity, assuming that the energy losses are negligible, the flow rate (in m^3/s) through the pipe is approximately equal to

0.5 | |

0.15 | |

0.05 | |

0.015 |

Question 6 Explanation:

Diameter of throat,

d=0.1 \mathrm{m}

Diameter of pipe,

D=0.2 \mathrm{m}

Pressure difference,

\frac{P_{1}-P_{2}}{w}=2 m=h

Coefficient of discharge

C_{D}=1

Discharge,

\begin{aligned} Q&=\frac{C_{D} A_{1} A_{2} \sqrt{2 g h}}{\sqrt{A_{1}^{2}-A_{2}^{2}}}\\ A_{1} &=\frac{\pi}{4}(0.2)^{2} \\ \text { and } A_{2} &=\frac{\pi}{4}(0.1)^{2} \\ \Rightarrow \quad Q &=\frac{1 \times\left(\frac{\pi}{4}\right)^{2} \times(0.2)^{2}(0.1)^{2} \sqrt{2 \times 9.81 \times 2}}{\left(\frac{\pi}{4}\right) \sqrt{(0.2)^{4}-(0.1)^{4}}} \\ &=0.0508=0.050 \mathrm{m}^{3} / \mathrm{sec} \end{aligned}

d=0.1 \mathrm{m}

Diameter of pipe,

D=0.2 \mathrm{m}

Pressure difference,

\frac{P_{1}-P_{2}}{w}=2 m=h

Coefficient of discharge

C_{D}=1

Discharge,

\begin{aligned} Q&=\frac{C_{D} A_{1} A_{2} \sqrt{2 g h}}{\sqrt{A_{1}^{2}-A_{2}^{2}}}\\ A_{1} &=\frac{\pi}{4}(0.2)^{2} \\ \text { and } A_{2} &=\frac{\pi}{4}(0.1)^{2} \\ \Rightarrow \quad Q &=\frac{1 \times\left(\frac{\pi}{4}\right)^{2} \times(0.2)^{2}(0.1)^{2} \sqrt{2 \times 9.81 \times 2}}{\left(\frac{\pi}{4}\right) \sqrt{(0.2)^{4}-(0.1)^{4}}} \\ &=0.0508=0.050 \mathrm{m}^{3} / \mathrm{sec} \end{aligned}

Question 7 |

Group I lists a few devices while Group II provides information about their uses. Match the devices with their corresponding use.

P-1; Q-2; R-3; S-4 | |

P-2; Q-1; R-4; S-3 | |

P-4; Q-2; R-1; S-3 | |

P-4; Q-3; R-2; S-1 |

Question 8 |

A venturimeter, having a diameter of 7.5 cm at the throat and 15 cm at the enlarged end, is installed in a horizontal pipeline of 15 cm diameter. The pipe carries an incompressible fluid at a steady rate of 30 litres per second. The difference of pressure head measured in terms of the moving fluid in between the enlarged and the throat of the venturimeter is observed to be 2.45 m. Taking the acceleration due to gravity as 9.81 m/s^{2}, the coefficient of discharge of the venturimeter (correct up to two places of decimal) is___

0.24 | |

0.68 | |

0.82 | |

0.95 |

Question 8 Explanation:

\begin{aligned} Q &=30 \; l / \mathrm{s} \\ \left(\frac{P_{1}}{w}+Z_{1}\right)&-\left(\frac{P_{2}}{w}+Z_{2}\right) =2.45 \mathrm{m}=h \\ g&=9.81 \\ Q&=C_{d} \frac{A_{1} A_{2}}{\sqrt{A_{1}^{2}-A_{2}^{2}}} \sqrt{2 g h} \\ C_{d} &=\frac{Q_{1} A_{2}}{\sqrt{A_{1}^{2}-A_{2}^{2}} \sqrt{2 g h}}\\ &= \frac{30 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s}}{\frac{\frac{\pi}{4}(0.15)^{2} \times(0.075)^{2} }{\sqrt{(0.15)^{2}-(0.075)^{2}}}\times \sqrt{2 \times 9.81 \times 2.45}} \\ C_{\mathrm{d}}&=0.95 \end{aligned}

Question 9 |

Group I gives a list of devices and Group II gives the list of uses

The correct match of Group I with Group II is

The correct match of Group I with Group II is

P-1, Q-2, R-4, S-3 | |

P-2, Q-1, R-3, S-4 | |

P-2, Q-1, R-4, S-3 | |

P-4, Q-1, R-3, S-2 |

Question 10 |

At two point 1 and 2 in a pipeline the velocities are V and 2V, respectively. Both
the points are at the same elevation. The fluid density is \rho. The flow can be
assumed to be incompressible, inviscid, steady and irrotational. The difference in
pressures P_{1}
and P_{2} at point 1 and 2 is

0.5\rho V^{2} | |

1.5\rho V^{2} | |

2\rho V^{2} | |

3\rho V^{2} |

Question 10 Explanation:

Applying Bernoulli's equation, we get

\begin{aligned} \frac{P_{1}}{\rho g}+\frac{(V)^{2}}{2 g}+Z&=\frac{P_{2}}{\rho g}+\frac{(2 V)^{2}}{2 g}+z \\ \Rightarrow \quad \frac{P_{1}-P_{2}}{\rho g}&=\frac{4 V^{2}-V^{2}}{2 g} \\ \Rightarrow \quad P_{1}-P_{2}&=\frac{3}{2} p V^{2} \\ \Rightarrow \quad P_{1}-P_{2}&=1.5 p V^{2} \end{aligned}

\begin{aligned} \frac{P_{1}}{\rho g}+\frac{(V)^{2}}{2 g}+Z&=\frac{P_{2}}{\rho g}+\frac{(2 V)^{2}}{2 g}+z \\ \Rightarrow \quad \frac{P_{1}-P_{2}}{\rho g}&=\frac{4 V^{2}-V^{2}}{2 g} \\ \Rightarrow \quad P_{1}-P_{2}&=\frac{3}{2} p V^{2} \\ \Rightarrow \quad P_{1}-P_{2}&=1.5 p V^{2} \end{aligned}

There are 10 questions to complete.