Question 1 |

At the foot of a spillway, water flows at a depth of 23 cm with a velocity of 8.1 m/s, as shown in the figure.

The flow enters as an M-3 profile in the long wide rectangular channel with bed slope = \frac{1}{1800} and Manning's n= 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 m/s^2. Just before the hydraulic jump, the depth of flow y_1 (in m, round off to 2 decimal places) is _______

The flow enters as an M-3 profile in the long wide rectangular channel with bed slope = \frac{1}{1800} and Manning's n= 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 m/s^2. Just before the hydraulic jump, the depth of flow y_1 (in m, round off to 2 decimal places) is _______

0.24 | |

0.42 | |

0.82 | |

0.96 |

Question 1 Explanation:

\begin{aligned} y &=0.23m \\ V&=8.1m/s \\ q&=Vy=0.23 \times 8.1 \\ &= 1.863 m^3/s-m\\ S_0&=\frac{1}{800} \\ D&=0.015 \end{aligned}

y_n= Normal depth of flow

R = y for wide rectangular channel

By Manning's equation

\begin{aligned} q&=\frac{y_n}{n}R^{2/3}S_0^{1/2} \\ 1.863&=\frac{y_n^{5/3}}{0.015} \times \left ( \frac{1}{1800} \right )^{1/2} \\ y_n&= 1.108m \end{aligned}

y_1 is conjugate depth of y_0,

\begin{aligned} \frac{y_1}{y_n}&=\frac{-1+\sqrt{1+8Fr_n^2}}{2} \\ \because &\;\;\left ( Fr_n^2=\frac{q^2}{gy_n^3} \right ) \\ y_1&=\frac{-1+\sqrt{1+8 \times \frac{1.863^2}{9.81 \times 1.108^3}}}{2} \times 1.108\\ &=0.418m\simeq 0.42m \end{aligned}

Question 2 |

A confined aquifer of 15 m constant thickness is sandwiched between two aquicludes as shown in the figure.

The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days,round off to 1 decimal place) is______

The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days,round off to 1 decimal place) is______

925.6 | |

654.4 | |

369.6 | |

254.2 |

Question 2 Explanation:

Discharge velocity,

\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}

\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}

Question 3 |

If the path of an irrigation canal is below the bed level of a natural stream, the type of cross-drainage structure provided is

Aqueduct | |

Level crossing | |

Sluice gate | |

Super passage |

Question 3 Explanation:

Irrigation canal below the bed level of a natural stream

\rightarrow Super passage

\rightarrow Super passage

Question 4 |

Group I contains three broad classes of irrigation supply canal outlets. Group II presents hydraulic performance attributes.

The correct match of the items in Group I with the items in Group II is

The correct match of the items in Group I with the items in Group II is

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

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

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

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

Question 4 Explanation:

Non-modular outlet :These are the outlets whose
discharge depends on the difference in water
levels in the distributing channel and the water <
course. The discharge of such outlets, therefore,
varies with the variation of the water levels in the
distributing channel and the water course.

Semi-modular outlet : These are the outlets whose discharge varies with the variation of the water level in the distribution channel but it is independent of the water level in the water course, so long as the minimum working head required for their working is available.

Modular outlet : These are the outlets whose discharge is independent of the water levels in the distributing channel and the water course, within reasonable working limits. In other words modular outlets maintain a constant discharge irrespective of variation of the water levels in the distributing channel and the water course.

Semi-modular outlet : These are the outlets whose discharge varies with the variation of the water level in the distribution channel but it is independent of the water level in the water course, so long as the minimum working head required for their working is available.

Modular outlet : These are the outlets whose discharge is independent of the water levels in the distributing channel and the water course, within reasonable working limits. In other words modular outlets maintain a constant discharge irrespective of variation of the water levels in the distributing channel and the water course.

Question 5 |

Profile of a weir on permeable foundation is shown in figure I and an elementary profile of 'upstream pile only case' according to Khosla's theory is shown in figure II. The uplift pressure heads at key points Q, R and S are 3.14 m, 2.75 m and 0 m, respectively (refer figure II).

What is the uplift pressure head at point P downstream of the weir (junction of floor and pile as shown in the figure I)?

What is the uplift pressure head at point P downstream of the weir (junction of floor and pile as shown in the figure I)?

2.75 m | |

1.25 m | |

0.8 m | |

data not sufficient |

Question 5 Explanation:

\begin{aligned} \phi_{\mathrm{R}} &=\frac{2.75}{4} \times 100=68.75 \% \\ \phi_{\mathrm{P}} &=100-\phi_{\mathrm{R}}=31.25 \% \\ \text { Now, } \phi_{p} &=\frac{\text { Pressure head at point } \mathrm{P}}{\text { Total head }} \times 100 \\ 31.25 &=\frac{h}{4} \times 100 \\ \therefore \quad h &=1.25 \mathrm{m} \end{aligned}

Question 6 |

Water emerges from an ogee spilway with velocity = 13.72 m/s and depth = 0.3
m at its toe. The tail water depth required to form a hydraulic jump at the toe is

6.48m | |

5.24m | |

3.24m | |

2.24m |

Question 6 Explanation:

Water depth required to form a hydraulic jump may be given as,

\begin{aligned} y_{2} &=\frac{y_{1}}{2}\left[-1+\sqrt{1+\frac{8 V_{1}^{2}}{g y_{1}}}\right] \\ &=\frac{0.3}{2} \times\left[-1+\sqrt{1+\frac{8 \times(13.72)^{2}}{9.81 \times 0.3}}\right] \\ &=3.24 \mathrm{m} \end{aligned}

\begin{aligned} y_{2} &=\frac{y_{1}}{2}\left[-1+\sqrt{1+\frac{8 V_{1}^{2}}{g y_{1}}}\right] \\ &=\frac{0.3}{2} \times\left[-1+\sqrt{1+\frac{8 \times(13.72)^{2}}{9.81 \times 0.3}}\right] \\ &=3.24 \mathrm{m} \end{aligned}

Question 7 |

A weir on a permeable foundation with downstream sheet pile is shown in the
figure below. The exit gradient as per Khosla's method is

1 in 6.0 | |

1 in 5.0 | |

1 in 3.4 | |

1 in 2.5 |

Question 7 Explanation:

The exit gradient may be given as,

\begin{aligned} G_{E}&=\frac{H}{d} \times \frac{1}{\pi \sqrt{\lambda}}\\ \text{where }\lambda&=\frac{1+\sqrt{1+\alpha^{2}}}{2} \text{ and }\alpha=\frac{b}{d} \\ \text{Here }\quad b&=10 \mathrm{m}, d=4 \mathrm{m}, H=5 \mathrm{m} \\ \therefore \quad \alpha&=\frac{b}{d}=\frac{10}{4}=2.5 \\ \text{and }\quad \lambda&=\frac{1+\sqrt{1+(2.5)^{2}}}{2}=1.85 \\ \therefore \quad G_{E}&=\frac{5}{4} \times \frac{1}{\pi \times \sqrt{1.85}} \\ \Rightarrow \quad G_{E}&=\frac{1}{3.4} \end{aligned}

\begin{aligned} G_{E}&=\frac{H}{d} \times \frac{1}{\pi \sqrt{\lambda}}\\ \text{where }\lambda&=\frac{1+\sqrt{1+\alpha^{2}}}{2} \text{ and }\alpha=\frac{b}{d} \\ \text{Here }\quad b&=10 \mathrm{m}, d=4 \mathrm{m}, H=5 \mathrm{m} \\ \therefore \quad \alpha&=\frac{b}{d}=\frac{10}{4}=2.5 \\ \text{and }\quad \lambda&=\frac{1+\sqrt{1+(2.5)^{2}}}{2}=1.85 \\ \therefore \quad G_{E}&=\frac{5}{4} \times \frac{1}{\pi \times \sqrt{1.85}} \\ \Rightarrow \quad G_{E}&=\frac{1}{3.4} \end{aligned}

Question 8 |

Uplift pressure at point E and D (shown in figure ) of a straight horizonal floor of negligible
thickness with a sheet pile at downstream end are 28% and 20%, respectively. If the sheet pile is at upstream end of the floor, the uplift pressures at point D_{1} and C_{1} are

68% and 60% respectively | |

80% and 72% respectively | |

88% and 70% respectively | |

100% and zero respectively |

Question 8 Explanation:

\begin{aligned} \text{Uplift pressure at }\mathrm{D}_{1}, \phi_{D_{1}}&=100-\phi_{\mathrm{D}} \\ &=100-20=80 \% \\ \text{Uplift pressure at }C_{1}, \phi_{C_{1}}&=100-\phi_{E} \\ &=100-28=72 \% \end{aligned}

Question 9 |

Which one of the following equations represents the downstream profile of Ogee
spillway with vertical upstream face?

(x,y) are the co-ordinates of the point on the downstream profile with origin at the crest of the spillway and H_d is the design head.

(x,y) are the co-ordinates of the point on the downstream profile with origin at the crest of the spillway and H_d is the design head.

\frac{y}{H_{d}}= -0.5\left ( \frac{x}{H_{d}} \right )^{1.85} | |

\frac{y}{H_{d}}= -0.5\left ( \frac{x}{H_{d}} \right )^{\frac{1}{1.85}} | |

\frac{y}{H_{d}}= -2.0\left ( \frac{x}{H_{d}} \right )^{1.85} | |

\frac{y}{H_{d}}= -2.0\left ( \frac{x}{H_{d}} \right )^{\frac{1}{1.85}} |

Question 9 Explanation:

For a spillway having a vertical u/s face, the d/s crest is given by the equation,

\begin{aligned} x^{1.85}&=-2 H_{d}^{0.85} \times y\\ \Rightarrow \quad x^{1.85}&=-2 H_{d}^{0.85} y \times \frac{H_{d}}{H_{d}} \\ \Rightarrow \quad x^{1.85}&=-2 \frac{H_{d}^{1.85} y}{H_{d}} \\ \Rightarrow \quad \frac{x^{1.85}}{-2 H_{d}^{1.85}}&=\frac{y}{H_{d}} \\ \Rightarrow \quad \frac{y}{H_{d}}&=-0.5\left(\frac{\dot{x}}{H_{d}}\right)^{1.85} \end{aligned}

\begin{aligned} x^{1.85}&=-2 H_{d}^{0.85} \times y\\ \Rightarrow \quad x^{1.85}&=-2 H_{d}^{0.85} y \times \frac{H_{d}}{H_{d}} \\ \Rightarrow \quad x^{1.85}&=-2 \frac{H_{d}^{1.85} y}{H_{d}} \\ \Rightarrow \quad \frac{x^{1.85}}{-2 H_{d}^{1.85}}&=\frac{y}{H_{d}} \\ \Rightarrow \quad \frac{y}{H_{d}}&=-0.5\left(\frac{\dot{x}}{H_{d}}\right)^{1.85} \end{aligned}

Question 10 |

While designing a hydraulic structure, the piezometric head at bottom of the
floor is computed as 10 m. The datum is 3m below floor bottom. The assured
standing water depth above the floor is 2 m. The specific gravity of the floor
material is 2.5. The floor thickness should be

2m | |

3.3m | |

4.4m | |

6m |

Question 10 Explanation:

G=2.5

Pressure head =10-3=7 \mathrm{m}

At critical condition,

\begin{aligned} \text { Uplift }&=\text { Downward pressure } \\ 7 \times \gamma_{w}&=\left(\gamma_{w} \times 2\right)+\left(\gamma_{\text {floor }} \times t\right) \\ \frac{5 \gamma_{w}}{\gamma_{\text {floor }}}&=t=\frac{5}{G}=\frac{5}{2.5}\\ \therefore \quad t&=2 \mathrm{m} \end{aligned}

There are 10 questions to complete.