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}

There are 5 questions to complete.