Q61GATE 2006MCQ

A hydraulic jump occurs in a rectangular, horizontal, frictonless channel. What would be the pre-jump depth if the discharge per unit width is 2 $m^{3}/s/m$ and the energy loss is 1 m ?

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📖 Explanation

\begin{aligned} \frac{2 q^{2}}{g} &=y_{1} y_{2}\left(y_{1}+y_{2}\right) \quad \ldots(i)\\ H_{L} &=\frac{\left(y_{2}-y_{1}\right)^{3}}{4 y_{1} y_{2}} \quad\ldots(ii)\\ \text{Given,}\\ H_{L} &=1 \mathrm{m} \\ \left(y_{2}-y_{1}\right)^{3} &=4 \mathrm{y}_{1} \cdot \mathrm{y}_{2}\\ \text{Put,}\quad y_{2}&=m y_{1} \text { \in equation (ii) } \\ y_{1}&=\frac{4 m}{(m-1)^{3}} \end{aligned}

Similarly, put $y_{2}=m y_{1}$ in equation (i), we get

\begin{array}{l} m(m+1) y_{1}^{3}=\frac{2 \times 2^{2}}{9.81}=0.8155 \\ m(m+1)\left\{\frac{4 m}{(m-1)^{3}}\right\}^{3}=0.8155\\ \frac{m(m+1)\left(m^{3}\right)}{(m-1)^{9}}=\frac{0.8155}{4^{3}}\\ \text{By trial and error,} m=5.09 \\ y_{1}=0.3 \mathrm{m} \\ y_{2}=1.527 \mathrm{m} \end{array}

Q63GATE 2005MCQ

A triangular irrigation lined canal carries a discharge of 25 $m^{3}$ /s at bed slope = $\frac{1}{6000}$ . If the side slopes of the canal are 1:1 and Manning's coefficient is 0.018, the central depth of flow is equal to

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📖 Explanation

By Manning's equation, $Q=\frac{1}{n} R^{2 / 3} S^{1 / 2} A$

\begin{aligned} \Rightarrow \quad 25&=\frac{1}{0.018} \times \left(\frac{y}{2 \sqrt{2}}\right)^{2 / 3} \times \left(\frac{1}{6000}\right)^{1 / 2} \times (y)^{2} \\ \Rightarrow \quad 25&=\frac{1}{0.018} \times (y)^{8 / 3} \times \frac{1}{154.92} \\ \Rightarrow \quad y&=4.91 \mathrm{m} \end{aligned}

Q64GATE 2005MCQ

A horizontal bed channel is followed by a steep bed channel as shown in the figure. The gradually-varied profiles over the horizontal and sleep beds are

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Q65GATE 2005MCQ

Critical depth at a section of a rectangular channel is 1.5 m. The specific energy at that section is

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📖 Explanation

For Rectanguler channel, $E_{c}= 1.5y_{c}=1.5\times 1.5=2.25m$

Q66GATE 2005MCQ

A partially open sluice gate discharge water into a rectangular channel. The tail water depth in the channel is 3 m and Froude number is $\frac{1}{2\sqrt{2}}$ . If a free hydraulic jump is to be formed at downstream of the sluice gate after the vena contracta of the jet coming out from the sluice gate, the sluice gate opening should be (coefficient of contraction $C_c$ =0.9)

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📖 Explanation

For hydraulic jump,

\begin{aligned} \frac{y_{1}}{y_{2}}&=\frac{1}{2}\left[-1+\sqrt{1+8 F_{2}^{2}}\right]\$\\ \text{Given},\quad y_{2}&=3 \mathrm{m} \\ F_{2} &=\frac{1}{2 \sqrt{2}} \\ \therefore \quad y_{1} &=\frac{3}{2}\left[-1+\sqrt{1+8 \times \frac{1}{4 \times 2}}\right]\\ \Rightarrow \quad y_{1}&=0.62 \mathrm{m} \\ \end{aligned}

Sluice gate opening $=\frac{y_{1}}{C_{c}}=\frac{0.62}{0.9}=0.69 \mathrm{m}$

Q67GATE 2004MCQ

For a pipe of radius, r, flowing half full under the action of gravity, the hydraulic depth is

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📖 Explanation

Area, $A=\frac{\pi r^{2}}{2}$ Top width, $T=2 r$ $\therefore$ Hydraulic depth $=\frac{A}{T}=\frac{\pi r^{2}}{2 \times 2 r}=\frac{\pi r}{4}$

Q68GATE 2004MCQ

A steep wide rectangular channel takes off from a reservoir having an elevation of 101.2 m. At the entrance, the bottom elevation of the channel is 100 m. If the slope of the channel is increased by 4%, the discharge per unit length in the channel will be

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📖 Explanation

It is a example of super critical canal intake, the reservoir water surface falls to the critical depth at the head of the canal and then towards the water surface follows the $S_{2}$ Curve. The critical depth occurring at the u/s end of the channel is the control for this flow. And critical depth only depends upon channel geometry not the slope.

\begin{aligned} E_{c} &=101.2-100 \\ &=1.2 m \\ y_{c} &=\frac{2}{3} E_{c}=0.8 m \\ y_{c} &=\left(\frac{q^{2}}{g}\right)^{1 / 3} \\ q &=2.24 m^{2} / s e c \end{aligned}

Note: On changing the slope discharge will not change; but velocity and depth vary with slope.

Q69GATE 2004MCQ

A wide channel is 1 m deep and has a velocity of flow, V, as 2.13 m/s. If a disturbance is caused, an elementary wave can travel upstream with a velocity of

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📖 Explanation

$F_{\mathrm{r}}=\frac{V}{C_{0}}=\frac{\text { Flow speed }}{\text { Wave speed }}$ At low speed, $F_{t} \lt 1;$ a small disturbance travels u/s (with a velocity $C_{0}-V$ relative to a stationary observer) and affects the u/s conditions. At high flow velocities, $F_{r} \gt 1$ , a small disturbance travels d/s at a velocity of $\mathrm{V}-\mathrm{C}_{0}$ relative to a stationary observer. (As $\mathrm{V} \gt \mathrm{C}_{0}$ small disturbance can not travel u/s) and thus the u/s conditions can not be influenced by the d/s conditions and the flow is controlled by the u/s conditions. Therefore, we can say that the region which remains unaffected controls the flow. $F_{r}=\frac{2.13}{\sqrt{9.81 \times 1}}=0.68 \lt 1$ (Subcritical flow) $C_{0}=\sqrt{9.81 \times 1}=3.13 \mathrm{m} / \mathrm{s}$ An elementary wave travels u/s with a velocity $C_{0}-V$ relative to a stationary observer $=3.13-2.13=1 \mathrm{m} / \mathrm{s}$

Q70GATE 2004MCQ

A steady flow occurs in an open channel with lateral inflow of $q \; m^{3}/s$ per unit width as shown in the figure. The mass conservation equation is

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📖 Explanation

Considering an element of width dx The mass conservation equation will be

\begin{aligned} Q+q d x-\left(Q+\frac{\partial Q}{\partial x} d x\right)&=0 \\ \therefore \quad Q+q d x-Q-\frac{\partial Q}{\partial x} d x &=0 \\ \therefore \quad \frac{\partial Q}{\partial x}-q &=0 \end{aligned}

Q71GATE 2003MCQ

A very wide rectangular channel carries a discharge of 8 $m^3/s$ per m width. The channel has a bed slope of 0.004 and Manning's roughness coefficient, n = 0.015. At a certain section of the channel, the flow depth is 1 m. At what distance from this section the flow depth will be 0.9 m ? (use the direct step method employing a single step.)

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📖 Explanation

Depth at midpoint,

\begin{aligned} y&=\frac{y_{1}+y_{2}}{2}=\frac{0.9+1}{2}=0.95\\ \text{But,}\quad \Delta x&=\frac{E_{2}-E_{1}}{S_{0}-S_{e}} \\ \text{Now,}\quad q&=\frac{1}{n}(y)^{5 / 3} \sqrt{S_{0}} \\ \Rightarrow \quad 8&=\frac{1}{0.015} \times (0.95)^{5 / 3} \times \sqrt{S_{e}} \\ \Rightarrow \quad S_{e} &=0.0171 \\ E_{2} &=y_{2}+\frac{V_{2}^{2}}{2 g}=1+\frac{q^{2}}{2 g y_{2}^{2}} \\ &=1+\frac{8^{2}}{2 \times 9.81 \times 1^{2}} \\ &=4.26 \\ E_{1}&=y_{1}+\frac{V_{1}^{2}}{2 g}=0.9+\frac{q^{2}}{2 g y_{1}^{2}}\\ &=0.9+\frac{8^{2}}{2 \times 9.81 \times (0.9)^{2}}=4.927 \\ \Delta x &=\frac{E_{2}-E_{1}}{S_{0}-S_{0}}=\frac{4.26-4.927}{0.004-0.017} \\ &=51.31 \mathrm{m} (upstream) \end{aligned}

Q72GATE 2003MCQ

A hydraulic jump takes place in a triangular channel of vertex angle 90 $^{\circ}$ , as shown in figure. The discharge is 1 $m^{3}/s$ and the pre-jump depth is 0.5 m. What will be the post-jump depth ? (Take g = 981 $m/s^{2}$ )

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📖 Explanation

Specific force will remain constant on both sides of hydraulic jump.

\begin{aligned} \therefore \frac{Q^{2}}{g A_{1}}+A_{1} \bar{Z}_{1} &=\frac{Q^{2}}{g A_{2}}+A_{2} \bar{Z}_{2} \\ A_{1} &=y_{1}^{2} ; A_{2}=y_{2}^{2} \\ \bar{Z}_{1} &=\left(\frac{y_{1}}{3}\right) ; \quad \bar{Z}_{2}=\left(\frac{y_{2}}{3}\right) \\ \text{Given}\quad Q &=1 \mathrm{m}^{3 / \mathrm{s}} ; y_{1}=0.5 \mathrm{m}\\ \therefore \frac{(1)^{2}}{9.81 \times (0.5)^{2}}&+(0.5)^{2} \times \left(\frac{0.5}{3}\right) \\ &=\frac{(1)^{2}}{9.81 \times y_{2}^{2}}+\left(y_{2}\right)^{2} \times \left(\frac{y_{2}}{3}\right) \\ \Rightarrow \quad 0.4494&=\frac{0.1019}{y_{2}^{2}}+\frac{y_{2}^{3}}{3} \\ \end{aligned}

By hit and trial, we get $y_{2}=1.02 \mathrm{m}$

Q73GATE 2002MCQ

In a lined rectangular canal, the Froude number of incoming flow is 3. A hydraulic jump forms when it meets the pool of water. The depth of flow after the jump formation is 1.51 m. Froude number of flow after the hydraulic jump is

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📖 Explanation

\begin{array}{l} \frac{y_{2}}{y_{1}}=\frac{1}{2}\left[-1+\sqrt{1+8 F_{1}^{2}}\right]\$=3.77 \\ \frac{y_{1}}{y_{2}}=\frac{1}{2}\left[-1+\sqrt{1+8 F_{2}^{2}}\right]\$=\frac{1}{3.77} \\ F_{2}=0.41 \end{array}

Q74GATE 2001MCQ

A trapezoidal channel with bottom width of 3 m and side slope of IV:1.5H carries a discharge of 8.0 $m^3/sec$ with the flow depth of 1.5 m. The Froude number of the flow is

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📖 Explanation

\begin{aligned} Q&=8 \mathrm{m}^{3} / \mathrm{s} \\ A&=\frac{1}{2}(3+7.5) \times 1.5=7.875 \mathrm{m}^{2} \\ V&=\frac{Q}{\mathrm{A}}=\frac{8}{7.875}=1.015 \mathrm{m} / \mathrm{s}\\ \text{Now}\quad \frac{1}{1.5}&=\frac{1.5}{x} \\ x&=2.25\\ T&=3+2 \times 2.25=7.5 \mathrm{m} \\ D&=\frac{A}{T}=\frac{7.875}{7.5}=1.05 \\ F&=\frac{V}{\sqrt{\mathrm{g} \mathrm{D}}}=\frac{1.05}{\sqrt{9.81 \times 1.05}}=0.316 \end{aligned}

Hence no any option is correct.