# Open Channel Flow

 Question 1
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]
 A $\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}}$ B $\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}^{2}}$ C $\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}^{2}}$ D $\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}}$
GATE CE 2023 SET-2   Fluid Mechanics and Hydraulics
Question 1 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, $A=A_{C}$;
$\mathrm{T}=\mathrm{T}_{\mathrm{C}}$
$\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}$
 Question 2
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
 A 3 B 2 C 1 D 4
GATE CE 2023 SET-2   Fluid Mechanics and Hydraulics
Question 2 Explanation: Question 3
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) ____
 A 0.25 B 0.38 C 0.51 D 0.62
GATE CE 2023 SET-1   Fluid Mechanics and Hydraulics
Question 3 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 $=0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}$
 Question 4
Which of the following options match the test reporting conventions with the given material tests in the table?
$\begin{array}{|l|l|} \hline \text{Test reporting convention} & \text{Material test} \\ \hline \text{(P) Reported as ratio} & \text{(I) Solubility of bitumen} \\ \hline \text{(Q) Reported as percentage} & \text{(II) Softening point of bitumen }\\ \hline \text{(R) Reported in temperature} & \text{(III) Los Angeles abrasion test} \\ \hline \text{(S) Reported in length} & \text{(IV) Flash point of bitumen} \\ \hline & \text{(V) Ductility of bitumen} \\ \hline & \text{(VI) Specific gravity of bitumen} \\ \hline & \text{(VII) Thin film oven test} \\ \hline \end{array}$
 A P-VI, Q-I, R-II, S-VII B P-VI, Q-III, R-IV, S-V C P-VI, Q-I, R-II, S-V D P-VI, Q-III, R-IV, S-VII
GATE CE 2023 SET-1   Fluid Mechanics and Hydraulics
Question 4 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.
 Question 5
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 $P$ ? A $M_2$ and $S_2$ B $M_2$ only C $S_2$ only D $S_2$ and hydraulic jump
GATE CE 2023 SET-1   Fluid Mechanics and Hydraulics
Question 5 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 $S_{2}$.

There are 5 questions to complete.