GATE CE · Fluid Mechanics And Hydraulics

Open Channel Flow – GATE CE Fluid Mechanics And HydraulicsPractice Questions & PYQs

Generate GATE-level questions on Open Channel Flow in Fluid Mechanics And Hydraulics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

74 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025MCQ

Consider flow in a long and very wide rectangular open channel. Width of the channel can be considered as infinity compared to the depth of flow. Uniform flow depth is 1.0 m. The bed slope of the channel is 0.0001. The Manning roughness coefficient value is 0.02. Acceleration due to gravity, $g$ can be taken as $9.81\ \text{m/s}^2$ . The critical depth (in m) corresponding to the flow rate resulting from the above conditions is ______ (round off to one decimal place).

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📖 Explanation

Given: Uniform depth, $y = 1.0\ \text{m}$ Slope, $S = 0.0001$ Manning's $n = 0.02$ $g = 9.81\ \text{m/s}^2$ Since the channel is very wide, hydraulic radius $R \approx y = 1.0\ \text{m}$ Discharge per unit width ( $q$ ) using Manning's equation for wide channel: $q = \frac{1}{n} y^{5/3} S^{1/2}$ $q = \frac{1}{0.02} (1.0)^{5/3} (0.0001)^{1/2} = 50 \times 0.01 = 0.5\ \text{m}^2/\text{s}$ Now, to find critical depth $y_c$ , use the critical flow condition: $q^2 = \frac{g y_c^3}{1}$ for a wide rectangular channel $y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{0.5^2}{9.81}\right)^{1/3}$ $y_c = \left(\frac{0.25}{9.81}\right)^{1/3} = (0.02549)^{1/3} \approx 0.295\ \text{m}$ Rounded to one decimal place: $\boxed{0.3\ \text{m}}$

Q2GATE 2025MCQ

A hydraulic jump occurs in an open channel when the slope of the channel changes from ______.

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Q3GATE 2025NAT

A hydraulic jump is formed in a 5 m wide rectangular channel, which has a horizontal bed and is carrying a discharge of $15 \mathrm{~m}^{3} / \mathrm{s}$ . The depth of water upstream of the jump is 0.5 m . The power dissipated by the jump (in kW) is _________ (rounded off to the nearest integer). Note: Acceleration due to gravity $=9.81 \mathrm{~m} / \mathrm{s}^{2}$ Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ Kinetic energy correction factor $=1.0$

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📖 Explanation

Given: rectangular channel.

\begin{aligned} \text { width } B & =5 \mathrm{~m} \\ \text { Discharge } Q & =15 \mathrm{~m}^{3} / \mathrm{sec} \\ \text { Prejump depth } y_{1} & =0.5 \mathrm{~m} \\ \text { Power dissipated } & =? \\ \Delta P & =\rho g Q \Delta E \\ \Delta E & =\frac{\left(y_{2}-y_{1}\right)^{3}}{4 y_{1} y_{2}} \end{aligned}

For Froude number at upstream

\begin{aligned} F r_{1}^{2} & =\frac{V_{1}^{2}}{g y_{1}}=\frac{q^{2}}{g y_{1}^{3}} \\ & =\frac{3^{2}}{9.81 \times 0.5^{3}} \\ F r_{1}^{2} & =7.34 \\ \frac{y_{2}}{y_{1}} & =\frac{-1+\sqrt{1+8 F_{1}^{2}}}{2} \\ \frac{y_{2}}{0.5} & =\frac{-1+\sqrt{1+8 \times 7.34}}{2} \\ y_{2} & =1.68 \mathrm{~m} \mathrm{~m} \end{aligned}

Now, $\Delta E =\frac{(1.68-0.5)^{3}}{4 \times 1.68 \times 0.5}=0.49$ From above equation

\begin{aligned} \Delta P & =10^{3} \times 9.81 \times 15 \times 0.49 \\ & =72.10 \mathrm{kWatt} \simeq 72 \mathrm{~kW} \quad \text { (Nearest integer) } . \end{aligned}

Q4GATE 2024NAT

A $2 \mathrm{~m}$ wide rectangular channel is carrying a discharge of $30 \mathrm{~m}^{3} / \mathrm{s}$ at a bed slope of 1 in 300. Assuming the energy correction factor as 1.1 and acceleration due to gravity as $10 \mathrm{~m} / \mathrm{s}^{2}$ , the critical depth of flow (in meters) is _____ (rounded off to 2 decimal places)

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📖 Explanation

Discharge $(\mathrm{Q})=30 \mathrm{~m}^{3} / \mathrm{s}$ Width of channel $(B)=2 \mathrm{~m}$ Bed slope $=1$ in 300 Energy correction factor $(\alpha)=1.1$ $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Discharge per unit width $(q)=\frac{Q}{B} =\frac{30}{2}=15 \mathrm{~m}^{3} / \mathrm{sec} / \mathrm{m}$ Critical depth does not depend on slope or the roughness of channel. Critical depth $\left(y_{c}\right)=\left(\frac{q^{2} \cdot \alpha}{g}\right)^{1 / 3} =\left(\frac{1.1 \times 15^{2}}{10}\right)^{1 / 3}=2.914 \mathrm{~m} \simeq 2.91 \mathrm{~m}$

Q5GATE 2024NAT

A spillway has unit discharge of $7.5 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m}$ . The flow depth at the downstream horizontal apron is $0.5 \mathrm{~m}$ . The tail water depth (in meters) required to form a hydraulic jump is _____ (rounded off to 2 decimal places).

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📖 Explanation

Unit discharge $(q)=7.5 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m}$

\begin{aligned} \frac{q^{2}}{g} & =\frac{y_{1} y_{2}\left(y_{1}+y_{2}\right)}{2} \\ \frac{7.5^{2}}{9.81} & =0.5 y_{2}\left(\frac{0.5+y_{2}}{2}\right) \end{aligned}

Tail water depth, $y_{2}=4.54 \mathrm{~m}$

Q6GATE 2024NAT

A rectangular channel is $4.0 \mathrm{~m}$ wide and carries a discharge of $2.0 \mathrm{~m}^{3} / \mathrm{s}$ with a depth of $0.4 \mathrm{~m}$ . The channel transitions to a maximum width contraction at a downstream location, without influencing the upstream flow condtions. The width (in meters) at the maximum contraction is _____ (rounded off to 2 decimal places)

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📖 Explanation

width of channel $(B)=4 \mathrm{~m}$ discharge $(Q)=2 \mathrm{~m}^{3} / \mathrm{s}$ depth of $\operatorname{flow}(\mathrm{y})=0.4 \mathrm{~m}$ Specific energy at section (1) - (1) (E)

\begin{aligned} & =\mathrm{y}_{1}+\frac{\mathrm{V}_{1}^{2}}{2 \mathrm{~g}} \\ & =0.4+\frac{\mathrm{Q}^{2}}{\mathrm{~B}_{1}^{2} \times 2 \mathrm{~g} \times \mathrm{y}_{1}^{2}} \\ & =0.4+\frac{2^{2}}{4^{2} \times 2 \times 9.81 \times 0.4^{2}} \\ & =0.4796 \mathrm{~m} \end{aligned}

When channel section is contracted to minimum width and for constant discharge Q. The flow over contracted section will be critical flow.

\begin{aligned} E & =E_{c}=\frac{3}{2} \cdot y_{c} \\ \therefore \quad 0.4796 & =\frac{3}{2} \times y_{c} \\ \Rightarrow \quad \quad \quad y_{c} & =0.4796 \times \frac{2}{3}=0.3197 \mathrm{~m} \end{aligned}

For critical flow condition, $\frac{\mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA}^{3}}=1$ $\frac{\mathrm{Q}^{2} \mathrm{~B}_{\min }}{\mathrm{gB}_{\min \mathrm{y}_{\mathrm{c}}^{3}}^{3}}=1$

\begin{aligned} \Rightarrow \quad \mathrm{B}_{\min }^{2} & =\frac{\mathrm{Q}^{2}}{9 \mathrm{y}_{\mathrm{c}}^{3}}=1 \\ \mathrm{~B}_{\min } & =\left(\frac{\mathrm{Q}^{2}}{\mathrm{gy_{c } ^ { 3 }}}\right)^{1 / 3} \\ & =\left(\frac{2^{2}}{9.81 \times 0.3197^{3}}\right)^{1 / 2} \\ & =3.5325 \mathrm{~m} \simeq 3.53 \mathrm{~m} \end{aligned}

Explanation Figure 1

Explanation Figure 2

Q7GATE 2024MCQ

The number of degrees of freedom for a natural open channel flow with a mobile bed is

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Q8GATE 2023MSQ

Which of the following options match the test reporting conventions with the given material tests in the table?

\begin{array}{|l|l|} \hline \text{Test reporting convention} & \text{Material test} \\ \hline \text{(P) Reported as ratio} & \text{(I) Solubility of bitumen} \\ \hline \text{(Q) Reported as percentage} & \text{(II) Softening point of bitumen }\\ \hline \text{(R) Reported \in temperature} & \text{(III) Los Angeles abrasion test} \\ \hline \text{(S) Reported \in length} & \text{(IV) Flash point of bitumen} \\ \hline & \text{(V) Ductility of bitumen} \\ \hline & \text{(VI) Specific gravity of bitumen} \\ \hline & \text{(VII) Thin film oven test} \\ \hline \end{array}

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📖 Explanation

(I) Solubility of bitumen: It is defined as weight of bitumen soluble in $100 \mathrm{ml}$ of carbon disulphide. (II) Softening point of bitumen: It is the temperature at which bitumen attains a particular degree of softneess. (III) Los angles abrasion test: It is reported as percentage of material finer than $1.7 \mathrm{~mm}$ with respect to total weight. (IV) Flash point of bitumen: It is the lowest temperature at which bitumen catch five momentarily. (V) Ductility of bitumen: Ductility is measured by stretching a standard briquette of bitumen. The distance in $\mathrm{cm}$ that the briquette can be stretched before breaking is the ductility. (VI) Specific gravity of bitumen: It is the ratio of unit weight of bitumen to unit weight of water. (VII) Thin film oven test: It measure mass charge a sample as a percent of initial mass.

Q9GATE 2023NAT

A hydraulic jump occurs in a $1.0 \mathrm{~m}$ wide horizontal, frictionless, rectangular channel, with a pre-jump depth of $0.2 \mathrm{~m}$ and a post-jump depth of $1.0 \mathrm{~m}$ . The value of $g$ may be taken as $10 \mathrm{~m} / \mathrm{s}^{2}$ . The values of the specific force at the pre-jump and post-jump sections are same and are equal to (in $\mathrm{m}^{3}$ , rounded off to two decimal places) ____

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📖 Explanation

Given,

\begin{aligned} & B=1.0 \mathrm{~m} \\ & y_{1}=0.2 \mathrm{~m} \\ & y_{2}=1.0 \mathrm{~m} \\ & g=10 \mathrm{~m} / \mathrm{s} \end{aligned}

The value of specific force

\begin{aligned} & =\frac{\mathrm{P}+\mathrm{M}}{\gamma_{\mathrm{w}}}=\text { constant } \\ & =\frac{\mathrm{P}_{1}+\mathrm{M}_{1}}{\gamma_{\mathrm{w}}}=\frac{\mathrm{P}_{2}+\mathrm{M}_{2}}{\gamma_{\mathrm{w}}} \\ & =A \overline{\mathrm{y}}_{1}+\frac{\mathrm{Q}^{2}}{\mathrm{Ag}} \end{aligned}

For horizontal, frictionless rectangular channel $\therefore \quad \frac{2 q^{2}}{g}=y_{1} y_{2}\left(y_{1}+y_{2}\right)$

\begin{aligned} Q & =\sqrt{\frac{1 \times 0.2 \times (1.2) \times 10}{2}} \quad(B=1 \mathrm{~m}) \\ & =1.095 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m} \\ Q & =1.095 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Now, specific force $=0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}$

Q10GATE 2023MCQ

When a supercritical stream enters a mild-sloped (M) channel section, the type of flow profile would become ____

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📖 Explanation

When super critical stream enters a mild (m) channel section, the type of flow. Pofile would be $M_{3}$ or $S_{1}$ .

Explanation Figure 1

Q11GATE 2023MCQ

A very wide rectangular channel carries a discharge (Q) of $70 \mathrm{~m}^{3} / \mathrm{s}$ per meter width. Its bed slope changes from $0.0001$ to $0.0009$ at a point $P$ , as shown in the figure (not to scale). The Manning's roughness coefficient of the channel is $0.01$ . What water surface profile(s) exist(s) near the point $P$ ?

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📖 Explanation

$\eta=0.01$ Critical depth, $y_{C}=\left(\frac{q^{3}}{g}\right)^{1 / 3}=\left(\frac{70^{2}}{9.81}\right)^{1 / 3}=7.934 \mathrm{~m}$ For normal depth of flow for wide rectangular channel. For bed slope $(0.001)$

\begin{aligned} Q & =\left(\frac{1}{\eta} R^{2 / 3} \sqrt{S}\right) \times A \\ R & =\frac{A}{P}=\frac{\beta y}{B+2 y}=y \\ Q & =\frac{1}{n}(y)^{2 / 3} \sqrt{S} \times B y \\ q & =\frac{1}{n} y^{5 / 3} \sqrt{S} \\ y_{1} & =\left(\frac{q n}{\sqrt{S}}\right)^{3 / 5}\left(\frac{70 \times 0.01}{\sqrt{0.0001}}\right)^{3 / 5} \\ & =12.796 \mathrm{~m} \gt y_{c} \text { (Mild slope) } \\ y_{2} & =6.619 \mathrm{~m} \lt y_{c} \text { (Steep slope) } \end{aligned}

The profile would be $\mathrm{M}_{2}$ and $S_{2}$ .

Explanation Figure 2

Q12GATE 2023MCQ

A compound symmetrical open channel section as shown in the figure has a maximum of depth(s). $B_{m}-$ Bottom width of main channel $B_{\mathrm{f}}-$ Bottom width of flood channel $\mathrm{y}_{\mathrm{m}}$ - Depth of main channel $y$ - Total depth of the channel $\mathrm{n}_{\mathrm{ma}}$ -Manning's roughness of the main channel $\mathrm{n}_{f}$ - Manning's roughness of the flood channel

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Q13GATE 2023MCQ

The critical flow condition in a channel is given by ____ [Note: $\alpha -$ kinetic energy correction factor; $Q -$ discharge; $A_{C} -$ cross-sectional area of flow at critical flow condition; $T_{C}-$ top width of flow at critical flow condition; $g-$ acceleration due to gravity]

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📖 Explanation

Specific Energy $(E)=y+\frac{v^{2}}{2 g}$ $\Rightarrow \quad E=y+\alpha \cdot \frac{Q^{2}}{2 g A^{2}}\left[\right.$ as $\left.\frac{Q}{A}=v\right]$ For critical flow condition,

\begin{aligned} \frac{d E}{d y} & =0 \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g} \frac{d}{d y}\left(A^{-2}\right) \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g}\left(-\frac{2}{A^{3}}\right) \frac{d A}{d y} \end{aligned}

'dA' can be written as 'Tdy' $\Rightarrow \frac{d A}{d y}=T$ $\Rightarrow \quad \frac{\mathrm{dE}}{\mathrm{dy}}=1-\frac{\alpha \mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA} \mathrm{A}^{3}}$ $\Rightarrow 1-\frac{\alpha Q^{2} \mathrm{~T}}{g A^{3}}=0$ $\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{\mathrm{A}^{3}}{\mathrm{~T}}$ At critical flow, $A=A_{C}$ ; $\mathrm{T}=\mathrm{T}_{\mathrm{C}}$ $\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}$

Explanation Figure 1

Q14GATE 2022NAT

A hydraulic jump takes place in a 6 m wide rectangular channel at a point where the upstream depth is 0.5 m (just before the jump). If the discharge in the channel is 30 $m^3/s$ and the energy loss in the jump is 1.6 m, then the Froude number computed at the end of the jump is ___________. (round off to two decimal places) (Consider the acceleration due to gravity as 10 $m/s^2$ .)

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📖 Explanation

$Q=30m^3/sec$ $B=6m$ $y_1=0.5m$ $E_L=1.6m$ $q= \frac{Q}{B} =\frac{30}{6}=5m^3/sec/m$ We know that,

\begin{aligned} E_L=\frac{(y_2-y_1)^3}{4y_1y_2}&=1.6m\\ \frac{(y_2-0.5)^3}{4 \times 0.5 \times y_2}&=1.6\\ y_2^{3-}1.5y_2^{2+}0.75y_2-0.125&=3.2y_2\\ y_2=2.5m,-0.0527m,&-0.0947m \end{aligned}

Hence, $y_2=2.5m$ Post jump Froude's No. $(F_2)=\frac{V_2}{\sqrt{\sqrt{gy_2}}}=\frac{\left ( \frac{30}{6 \times 2.5} \right )}{\sqrt{10 \times 2.5}}=0.4$ $$

Q15GATE 2022MSQ

Water is flowing in a horizontal, frictionless, rectangular channel. A smooth hump is built on the channel floor at a section and its height is gradually increased to reach choked condition in the channel. The depth of water at this section is $y_2$ and that at its upstream section is $y_1$ . The correct statement(s) for the choked and unchoked conditions in the channel is/are

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Q16GATE 2022MCQ

A rectangular channel with Gradually Varied Flow (GVF) has a changing bed slope. If the change is from a steeper slope to a steep slope, the resulting GVF profile is

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Q17GATE 2022NAT

Depth of water flowing in a 3 m wide rectangular channel is 2 m. The channel carries a discharge of 12 $m^3/s$ . Take g = 9.8 $m/s ^2$ . The bed width (in m) at contraction, which just causes the critical flow, is _________ without changing the upstream water level. (round off to two decimal places)

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📖 Explanation

Given: $B = 3m, y = 2m, Q = 12 m^3 /sec$ $Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec$ Specific energy at section (1-1) $E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m$ When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place. $E=E_c=\frac{108}{49}$ We have that, $E_c=\frac{3}{2}y_c$ (for rectangular crosssection) $y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}$ For critical flow condition,

\begin{aligned} \frac{Q^2T}{gA^3}&=1\\ \frac{Q^2B_{min}}{g(B_{min} \times y_c)^3}&=1\\ (B_{min})^2&=\frac{Q^2}{gy_c^3}\\ B_{min}&=\left ( \frac{(12)^2}{9.8 \times \left ( \frac{72}{49} \right )^3} \right )^{1/2}\\ &=2.152m \end{aligned}

Explanation Figure 1

Q18GATE 2021MCQ

A rectangular open channel of 6 m width is carrying a discharge of $20 \mathrm{~m}^{3} / \mathrm{s}$ . Consider the acceleration due to gravity as $9.81 \mathrm{~m} / \mathrm{s}^{2}$ and assume water as incompressible and inviscid. The depth of flow in the channel at which the specific energy of the flowing water is minimum for the given discharge will then be

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📖 Explanation

Minimum specific energy will correspond to a critical flow condition. The critical depth

\begin{aligned} \left(Y_{C}\right) &=\left[\frac{q^{2}}{g}\right]^{1 / 3} \\ Y_{C} &=\left[\frac{(20 / 6)^{2}}{9.81}\right]^{1 / 3}=1.042 \mathrm{~m} \end{aligned}

Explanation Figure 1

Q19GATE 2021MCQ

If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is

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📖 Explanation

Efficient channel section

\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}

Explanation Figure 1

Q20GATE 2020MCQ

Water flows at the rate of 12 $m^3/s$ in a 6 m wide rectangular channel. A hydraulic jump is formed in the channel at a point where the upstream depth is 30 cm (just before the jump). Considering acceleration due to gravity as 9.81 $m/s^2$ and density of water as 1000 $kg/m^3$ , the energy loss in the jump is

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📖 Explanation

Assuming channle bed to be horizontal and frictionless. $q=\frac{12}{6}=2 m^3/s/m$ Initial Froude No.

\begin{aligned} (F_r) &=\left ( \frac{q^2}{gY_1^3} \right )^{1/2} \\ &= \left ( \frac{2^2}{9.81 \times 0.3^3} \right )^{1/2}\\ &=3.88 \end{aligned}

From Belenger's Momentum equation for a rectangular channel

\begin{aligned} \frac{Y_2}{Y_1}&=\frac{1}{2}(-1+\sqrt{1+8F_1^2}) \\ &=\frac{1}{2}(-1+\sqrt{1+8\times 3.88^2}) \\ &= 5.018\\ Y_2&=5.018 \times 0.3=1.505m \end{aligned}

Heal loss in the jump

\begin{aligned} (h_L) &=\frac{(Y_2-Y_1)^3}{4Y_1Y_2} \\ &= \frac{(1.505-0.3)^3}{4 \times 1.505 \times 0.3}\\ &= 0.968m \end{aligned}

Power lost in the jump

\begin{aligned} &=\gamma _wQh_L \\ &= (9.81 \times 12 \times 0.968)kW\\ &=114.04 kW \end{aligned}

Explanation Figure 1

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