Q41GATE 2013MCQ

The normal depth in a wide rectangular channel is increased by 10%. The percentage increase in the discharge in the channel is:

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📖 Explanation

The discharge in the channel is given by

\begin{aligned} Q&=\frac{1}{n} A R^{2 / 3} S^{1 / 2}\\ \Rightarrow \quad Q&=\frac{1}{n}(B y)\left(\frac{B y}{B+2 y}\right)^{2 / 3} S^{1 / 2} \end{aligned}

For a wide rectangular channel

\begin{aligned} B &\gg y \\ \therefore\quad Q&=\frac{1}{n}(B y)(y)^{2 / 3} S^{1 / 2} \\ \therefore\quad Q &\propto y^{5 / 3} \\ \text{let} Q_{1}&=k(y)^{5 / 3} \\ \therefore\quad Q_{2}&=k(1.1 y)^{5 / 3} \\ \therefore\quad \frac{Q_{2}-Q_{1}}{Q_{1}} \times 100&=\frac{(1.1 \mathrm{y})^{5 / 3}-(\mathrm{y})^{5 / 3}}{(\mathrm{y})^{5 / 3}} \times 100 \\ &= 17.21 \% \end{aligned}

Q43GATE 2012MCQ

A rectangular open channel of width 5.0 m is carrying a discharge of 100 $m^3$ /s. The Froude number of the flow is 0.8. The depth of flow (in m) in the channel is

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📖 Explanation

Froude Number

\begin{aligned} F&=\frac{V}{\sqrt{g y}} \\ \Rightarrow 0.8&=\frac{V}{\sqrt{9.81 \mathrm{y}}} &\ldots(i)\\ \because \quad Q&=A V \\ \Rightarrow \quad 100&=(5 \mathrm{y}) \mathrm{V} \\ \Rightarrow \quad v&=\frac{100}{5 y}=\frac{20}{y} \\ \therefore \text{From eq. (i)}\\ 0.8&=\frac{20}{\sqrt{9.81} \cdot(y)^{3 / 2}} \\ \Rightarrow \quad y^{3 / 2}&=\frac{20}{0.8 \sqrt{9.81}} \\ \therefore \quad y&=3.99 \simeq 4 \mathrm{m} \end{aligned}

Q44GATE 2012MCQ

A trapezoidal channel is 10.0 m wide at the base and has a side slope of 4 horizontal to 3 vertical. The bed slope is 0.002. The channel is lined with smooth concrete (Manning's n = 0.012). The hydraulic radius (in m) for a depth of flow of 3.0 m is

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📖 Explanation

\begin{aligned} \because \quad A &=(B+n y) y=\left(10+\frac{4}{3} \times 3\right) \times 3 \\ &=(10+4) 3=42 m^{2} \\ P &=B+2 y \sqrt{n^{2}+1} \\ &=10+6 \sqrt{(4 / 3)^{2}+1} \\ &=10+6 \times \frac{5}{3}=20 m \\ \therefore \quad R &=\frac{A}{P}=\frac{42}{20}=2.1 \end{aligned}

Q45GATE 2011MCQ

For a given discharge, the critical flow depth in an open channel depends on

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Q46GATE 2011MCQ

The flow in a horizontal, frictionless rectangular open channel is supercritical. A smooth hump is built on the channel floor. As the height of hump is increased, choked condition is attained. With further increase in the height of the hump, the water surface will

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Q47GATE 2011MCQ

A spillway discharges flood flow at a rate of 9 $m^3$ /s per metre width. If the depth of flow on the horizontal apron at the toe of the spillway is 46 cm, the tail water depth needed to form a hydraulic jump is approximately given by which of the following options?

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📖 Explanation

\begin{aligned} \mathrm{q}&=9 \mathrm{m}^{3} / \mathrm{s} / \mathrm{m}\\ \text{and }y_{1}&=46 \mathrm{cm}=0.46 \mathrm{m} \\ \therefore \quad V_{1}&=\frac{q}{y_{1}}=\frac{9}{0.46}=19.565 \mathrm{m} / \mathrm{s} \\ F_{1}&=\frac{V_{1}}{\sqrt{g y_{1}}}=\frac{19.565}{\sqrt{9.81 \times 0.46}}=9.21 \end{aligned}

Thus, flow is super -critical. Thus the sequent depth will be

\begin{aligned} \frac{y_{2}}{y_{1}}&=\frac{1}{2}\left[-1+\sqrt{1+8 F_{1}^{2}}\right]\$\\ \Rightarrow \quad y_{2}&=\frac{0.46}{2}\left[-1+\sqrt{1+8 \times (9.21)^{2}}\right]\$ \\ \Rightarrow \quad y_{2}&=0.23 \times 25.07 \\ \Rightarrow \quad y_{2}&=5.77 \mathrm{m} \end{aligned}

Q48GATE 2010MCQ

For a rectangular channel seetion, Group I lists geometrical elements and Group II gives proportions for hydraulically efficient section $y_e$ is the flow depth corresponding to hydraulically efficient section. The correct match of Group I with Group II is

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📖 Explanation

Let the width of rectangular channel be B and depth of flow be y. Area of flow, A=By Wetted perimeter, $P=B+2 y=\frac{A}{y}+2 y$ For hydraulically most efficient section, P should be minimum.

\begin{aligned} \therefore \quad \frac{d P}{d y}&=0 \\ \Rightarrow \quad \frac{-A}{y^{2}}+2&=0 \\ \Rightarrow \quad A_{\theta}&=2 y_{e}^{2} \\ \Rightarrow \quad B_{e} y_{e}&=2 y_{\theta}^{2} \\ \Rightarrow \quad B_{\theta}&=2 y_{e} \end{aligned}

Hydraulic radius, $R=\frac{A}{P}=\frac{2 y_{e}^{2}}{2 y_{e}+2 y_{e}}=\frac{y_{e}}{2}$ Top width, $T=B=2 y_{e}$ Perimeter, $P=B+2 y_{e}=2 y_{e}+2 y_{e}=4 y_{e}$ Hydraulic depth, $D=\frac{A}{T}=\frac{2 y_{e}^{2}}{2 y_{e}}=y_{e}$

Q49GATE 2010MCQ

The flow in a rectangular channel is subcritical. If width of the channel is reduced at a certain section, the water surface under no-choke condition will

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Q50GATE 2010MCQ

The Froude number of flow in a rectangular channel is 0.8. If te depth of flow is 1.5m, the critical depth is

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📖 Explanation

Froude Number for a rectangular channel is given by

\begin{aligned} F &=\frac{V}{\sqrt{g y}} \\ \Rightarrow \quad V &=0.8 \times \sqrt{9.81 \times 1.5} \\ &=3.07 \mathrm{m} / \mathrm{s} \end{aligned}

$\therefore$ Discharge per unit width of rectangular channel is given by $q=V y=3.07 \times 1.5=4.6 \mathrm{m}^{3} / \mathrm{s} / \mathrm{m}$ $\therefore$ Critical depth, $y_{c}=\left(\frac{q^{2}}{g}\right)^{1 / 3}=\left(\frac{4.6^{2}}{9.81}\right)^{1 / 3}=1.29 m$

Q51GATE 2010MCQ

A mild-sloped channel is followed by a steep-sloped channel. The profiles of gradually varied flow in the channel are

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Q52GATE 2009MCQ

A rectangular open channel of width 4.5 m is carrying a discharge of 100 $m^{3}/ sec$ . The critical depth of the channel is

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📖 Explanation

The critical depth for a rectangular channel is given by

\begin{aligned} y_{c}^{3}&=\frac{q^{2}}{g}\\ \text{But }\quad q&=\frac{Q}{B}=\frac{100}{4.5}=22.22 \mathrm{m}^{2} / \mathrm{sec} \\ \therefore \quad y_{c}^{3}&=\frac{(22.22)^{2}}{9.81} \\ \Rightarrow \quad y_{c}&=3.69 m \end{aligned}

Q53GATE 2009MCQ

Direct step method of computation for gradually varied flow is

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Q54GATE 2008MCQ

A person standing on the bank of a canal drops a stone on the water surface. He notices that the disturbance on the water surface is not travelling upstream. This is because the flow in the canal is

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📖 Explanation

For super-critical flow, $F_{r} \gt 1$

\begin{aligned} \frac{V}{\sqrt{g y}} & \gt 1 \\ V & \gt \sqrt{g y} \\ V & \gt C \end{aligned}

Hence, on upstream, the waves will be washed away and we will not be able to see them but they can be seen on downstream side. Hence, the flow here is super-critical.

Q55GATE 2008MCQ

A rectangular channel 6.0 m wide carries a discharge of 16.0 $m^{3}/s$ under uniform flow condition with normal depth of 1.60 m. Minning's n is 0.015. The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition is

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📖 Explanation

If the channel is contracted, the depth of flow at this location decreases. Thus width of the channel can be decreased till the specific energy becomes minimum. If the channel is contracted further, the depth of flow starts decreasing.

\begin{aligned} \therefore \quad y&+\frac{Q^{2}}{2 g A^{2}}=E_{\min } \\ \Rightarrow 1.6&+\frac{(16)^{2}}{2 \times 9.81 \times (9.6)^{2}}=\frac{3}{2} \mathrm{y}_{\mathrm{c}} \\ \Rightarrow \quad 1.74&=\frac{3}{2} \times \left(\frac{\mathrm{Q}^{2} / \mathrm{b}^{2}}{\mathrm{g}}\right)^{1 / 3} \\ \Rightarrow \quad 1.74&=\frac{3}{2} \times \left[\frac{(16)^{2}}{9.81}\right]^{1 / 3} \times \frac{1}{(b)^{2 / 3}} \\ \Rightarrow \quad b^{2 / 3}&=2.557 \\ \Rightarrow \quad b&=4.1 \mathrm{m} \end{aligned}

Q56GATE 2007MCQ

A triangle open channel has a vertex angle of $90^{\circ}$ and carries flow at a critical depth of 0.30 m. The discharge in the channel is

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📖 Explanation

The side slope of triangle open channel is given as z horizontal to 1 vertical When vertex angle is $90^{\circ},z=1$ Now, $y_{c}=\left(\frac{2 Q^{2}}{g z^{2}}\right)^{1 / 5}$ for a triangular channel $\Rightarrow \quad Q^{2}=\frac{g z^{2}\left(y_{c}\right)^{5}}{2}$ $\Rightarrow \quad Q^{2}=\frac{9.81 \times (1)^{2} \times (0.30)^{5}}{2}$ $\Rightarrow \quad Q=0.11 \mathrm{m}^{3} / \mathrm{s}$

Q57GATE 2007MCQ

The flow rate in a wide rectangular open channel is 2.0 $m^{3}/s$ per metre width. The channel bed slope is 0.002. The Manning's roughness coefficient is 0.012. The slope of the channel is classified as

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📖 Explanation

The critical depth for a rectangular channel is given by $y_{c}=\left(\frac{q^{2}}{g}\right)^{1 / 3}=\left(\frac{(2)^{2}}{9.81}\right)^{1 / 3}=0.74 m$ Using Manning's equation, we get

\begin{aligned} & q=\frac{1}{n}\left(y_{n}\right)^{5 / 3} \sqrt{S} \\ \Rightarrow \quad 2&=\frac{1}{0.012} \times \left(y_{n}\right)^{5 / 3} \times \sqrt{0.002}\\ \Rightarrow \quad y_{n}&=0.69 \mathrm{m} \\ \end{aligned}

For steep slopes $y_{n} \lt y_{c}$

Q58GATE 2007MCQ

A rectangular open channel needs to be designed to carry a flow of 2.0 $m^{3}/s$ under uniform flow conditions. The Manning's roughness coefficient is 0.018. The channel should be such that the flow depth is equal to half the width, and the Froude number is equal to 0.5 Keeping the width, flow depth and roughness the same, if the bed slope of the above channel is double, the average boundary shear stress under uniform flow conditions is

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📖 Explanation

The average shear stress, $\tau_{0}=\gamma \mathrm{RS}_{0}$ Bed slope is doubled i.e.

\begin{aligned} S_{0} &=2 \times 0.0021=0.0042 \\ \therefore \quad \tau &=9810 \times \frac{0.836}{2} \times 0.0042 \\ \Rightarrow \quad \tau_{0} &=17.22 \mathrm{N} / \mathrm{m}^{2} \end{aligned}

Q59GATE 2007MCQ

There is a free overfall at the end of a long open channel. For a given flow rate, the critical depth is less than the normal depth. What gradually varied flow profile will occur in the channel for this flow rate ?

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📖 Explanation

All free fall curves are zone II profiles. since the critical depth $\left(\mathrm{y}_{\mathrm{c}}\right)$ is less than the normal depth $\left(y_{n}\right),$ therefore the slope is mild. Thus, the gradually varied profile is $\mathrm{M}_{2}$

Q60GATE 2006MCQ

A channel with a mild slope is followed by a horizontal channel and then by a steep channel. What gradually varied flow profiles will occur?

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