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]
[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]
\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}} | |
\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}^{2}} | |
\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}^{2}} | |
\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}} |
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}}
\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
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
3 | |
2 | |
1 | |
4 |
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) ____
0.25 | |
0.38 | |
0.51 | |
0.62 |
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}
\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}
\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}
P-VI, Q-I, R-II, S-VII | |
P-VI, Q-III, R-IV, S-V | |
P-VI, Q-I, R-II, S-V | |
P-VI, Q-III, R-IV, S-VII |
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.
(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 ?
M_2 and S_2 | |
M_2 only | |
S_2 only | |
S_2 and hydraulic jump |
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.