Q2GATE 2019NAT

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 _______

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📖 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}

Q3GATE 2019MCQ

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

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📖 Explanation

Irrigation canal below the bed level of a natural stream $\rightarrow$ Super passage

Q4GATE 2019NAT

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______

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📖 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}

Q5GATE 2017MCQ

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

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Q6GATE 2016MCQ

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)?

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📖 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}

Q7GATE 2008MCQ

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

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📖 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}

Q8GATE 2008MCQ

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

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📖 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}

Q9GATE 2005MCQ

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

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📖 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}

Q10GATE 2005MCQ

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.

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📖 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}

Q11GATE 2003MCQ

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

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📖 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}

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