Question 1 |
The cross-section of a small river is sub-divided into seven segments of width 1.5 \mathrm{~m} each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is ____ \mathrm{m}^{3} / \mathrm{s} (round off to one decimal place).
3.8 | |
4.6 | |
8.5 | |
12.5 |
Question 1 Explanation:
Average width for 1 st land last sections,
\begin{aligned} \bar{\omega}_{1}=\bar{\omega}_{\mathrm{n}} & =\frac{\left(\omega_{1}+\frac{\omega_{2}}{2}\right)^{2}}{2 \omega_{1}} \\ & =\frac{\left(1.5+\frac{1.5}{2}\right)^{2}}{2 \times 1.5} \\ & =1.6875 \mathrm{~m} \end{aligned}
\begin{array}{|c|c|c|c|} \hline \begin{array}{l}\text { Avg. width } \\ \text { (m) } \\ \text { (1) }\end{array} & \begin{array}{l}\text { Avg. Depth } \\ \text { (m) } \\ \text { (2) }\end{array} & \begin{array}{l}\text { Avg. Velocity } \\ \text { (m/s) } \\ \text { (3) }\end{array} & \begin{array}{c}\text { Segmental } \\ \text { discharge } \\ \left(\mathrm{m}^{3} / \mathrm{s}\right)=(1) \times(2) \times(3)\end{array} \\ \hline 1.6875 & 0.40 & 0.40 & 0.270 \\ \hline 1.5 & 0.70 & \frac{0.76+0.70}{2}=0.73 & 0.767 \\ \hline 1.5 & 1.20 & \frac{1.19+1.13}{2}=1.16 & 2.088 \\ \hline 1.5 & 1.40 & \frac{1.25+1.10}{2}=1.175 & 2.467 \\ \hline 1.5 & 1.10 & \frac{1.13+1.09}{2}=1.11 & 1.831 \\ \hline 1.5 & 0.80 & \frac{0.69+0.65}{2}=0.67 & 0.804 \\ \hline 1.6875 & 0.45 & 0.42 & 0.319 \\ \hline & & & \Sigma=8.546 \\ \hline \end{array}
\begin{aligned} \bar{\omega}_{1}=\bar{\omega}_{\mathrm{n}} & =\frac{\left(\omega_{1}+\frac{\omega_{2}}{2}\right)^{2}}{2 \omega_{1}} \\ & =\frac{\left(1.5+\frac{1.5}{2}\right)^{2}}{2 \times 1.5} \\ & =1.6875 \mathrm{~m} \end{aligned}
\begin{array}{|c|c|c|c|} \hline \begin{array}{l}\text { Avg. width } \\ \text { (m) } \\ \text { (1) }\end{array} & \begin{array}{l}\text { Avg. Depth } \\ \text { (m) } \\ \text { (2) }\end{array} & \begin{array}{l}\text { Avg. Velocity } \\ \text { (m/s) } \\ \text { (3) }\end{array} & \begin{array}{c}\text { Segmental } \\ \text { discharge } \\ \left(\mathrm{m}^{3} / \mathrm{s}\right)=(1) \times(2) \times(3)\end{array} \\ \hline 1.6875 & 0.40 & 0.40 & 0.270 \\ \hline 1.5 & 0.70 & \frac{0.76+0.70}{2}=0.73 & 0.767 \\ \hline 1.5 & 1.20 & \frac{1.19+1.13}{2}=1.16 & 2.088 \\ \hline 1.5 & 1.40 & \frac{1.25+1.10}{2}=1.175 & 2.467 \\ \hline 1.5 & 1.10 & \frac{1.13+1.09}{2}=1.11 & 1.831 \\ \hline 1.5 & 0.80 & \frac{0.69+0.65}{2}=0.67 & 0.804 \\ \hline 1.6875 & 0.45 & 0.42 & 0.319 \\ \hline & & & \Sigma=8.546 \\ \hline \end{array}
Question 2 |
A catchment may be idealized as a circle of radius 30 \mathrm{~km}. There are five rain gauges, one at the center of the catchment and four on the boundary (equi-spaced), as shown in the figure (not to scale).
The annual rainfall recorded at these gauges in a particular year are given below.
\begin{array}{|c|c|c|c|c|c|} \hline Gauge & \mathrm{G}_{1} & \mathrm{G}_{2} & \mathrm{G}_{3} & \mathrm{G}_{4} & \mathrm{G}_{5} \\ \hline Rainfall (mm) & 910 & 930 & 925 & 895 & 905 \\ \hline \end{array}
Using the Thiessen polygon method, what is the average rainfall (in \mathrm{mm}, rounded off to two decimal places) over the catchment in that year ? ___
The annual rainfall recorded at these gauges in a particular year are given below.
\begin{array}{|c|c|c|c|c|c|} \hline Gauge & \mathrm{G}_{1} & \mathrm{G}_{2} & \mathrm{G}_{3} & \mathrm{G}_{4} & \mathrm{G}_{5} \\ \hline Rainfall (mm) & 910 & 930 & 925 & 895 & 905 \\ \hline \end{array}
Using the Thiessen polygon method, what is the average rainfall (in \mathrm{mm}, rounded off to two decimal places) over the catchment in that year ? ___
912.56 | |
521.32 | |
5463.25 | |
472.36 |
Question 2 Explanation:
Radius of basin =30 \mathrm{~km}
We know that in Theissen polygon method, the polygons are obtained by joining the perpendicular bisectors of triangles formed, when various rain gauge stations are joined.
The Theissen polygon in the given basin can be drawn as:
Area under influence of station GI
=30 \times 30=900 \mathrm{~km}^{2}
Area under influence of stations
G2, G3, G4, G5
=\frac{\pi \times 30^{2}-900}{4}
=481.86 \mathrm{~km}^{2}
mean rainfall =\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4} =912.56 \mathrm{~mm}
We know that in Theissen polygon method, the polygons are obtained by joining the perpendicular bisectors of triangles formed, when various rain gauge stations are joined.
The Theissen polygon in the given basin can be drawn as:
Area under influence of station GI
=30 \times 30=900 \mathrm{~km}^{2}
Area under influence of stations
G2, G3, G4, G5
=\frac{\pi \times 30^{2}-900}{4}
=481.86 \mathrm{~km}^{2}
mean rainfall =\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4} =912.56 \mathrm{~mm}
Question 3 |
Which one of the following options provides the correct match of the terms listed in Column-I and Column-2 ?
\begin{array}{|l|l|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{P : Horton equation} & \text{I: Precipitation} \\ \hline \text{Q : Muskingum method} & \text{II: Flood frequency} \\ \hline \text{R : Penman method} & \text{III: Evapotranspiration} \\ \hline & \text{IV : Infiltration} \\ \hline & \text{V : Channel routing} \\ \hline \end{array}
\begin{array}{|l|l|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{P : Horton equation} & \text{I: Precipitation} \\ \hline \text{Q : Muskingum method} & \text{II: Flood frequency} \\ \hline \text{R : Penman method} & \text{III: Evapotranspiration} \\ \hline & \text{IV : Infiltration} \\ \hline & \text{V : Channel routing} \\ \hline \end{array}
P-IV, Q-V, R-III | |
P-III, Q-IV, R-I | |
P-IV, Q-III, R-II | |
P-III, Q-I, R-IV |
Question 3 Explanation:
Horton's equation is used to calculate total infiltration.
Muskingum method is used in flood response analysis in channels.
Penman equation is used to calculate potential evapotranspiration.
Muskingum method is used in flood response analysis in channels.
Penman equation is used to calculate potential evapotranspiration.
Question 4 |
In Horton's equation fitted to the infiltration data for a soil, the initial infiltration capacity is 10 \mathrm{~mm} / \mathrm{h}; final infiltration capacity is 5 \mathrm{~mm} / \mathrm{h}; and the exponential decay constant is 0.5 \mathrm{~h}. Assuming that the infiltration takes place at capacity rates, the total infiltration depth (in \mathrm{mm} ) from a uniform storm of duration 12 \mathrm{~h} is ____. (round off to one decimal place)
25.3 | |
58.2 | |
70 | |
78.5 |
Question 4 Explanation:
\begin{aligned}
f_{0} & =10 \text { meter } \\
f_{C} & =5 \text { meter } \\
K & =0.5 h r^{-1} \\
f & =f_{c}+\left(f_{0}-f_{c}\right) e^{-k t} \\
f & =5+5 e^{-0.5 \times t}
\end{aligned}
At capacity rates i.e. t \rightarrow \infty
\begin{aligned} f & =5+5 e^{-0.5 \times \infty} \\ & =5 \text { meters } \end{aligned}
Infiltration in 12 \mathrm{hr} .=5 \times 12=60 \mathrm{~mm}
If above is not the case :
then
\begin{aligned} \text { Infiltration } & =\int \mathrm{fdt} \\ & =\int_{0}^{12} 5+5 \mathrm{e}^{-0.5 \mathrm{t}} \mathrm{dt} \\ & =5 \times(\mathrm{t})_{0}^{12}+\left.\frac{5}{-0.5} \mathrm{e}^{-0.5 \mathrm{t}}\right|_{0} ^{2} \\ & =60+10\left(1-\mathrm{e}^{-0.5 \times 12}\right) \\ & =69.975 \mathrm{~mm}=70 \mathrm{~mm} \end{aligned}
At capacity rates i.e. t \rightarrow \infty
\begin{aligned} f & =5+5 e^{-0.5 \times \infty} \\ & =5 \text { meters } \end{aligned}
Infiltration in 12 \mathrm{hr} .=5 \times 12=60 \mathrm{~mm}
If above is not the case :
then
\begin{aligned} \text { Infiltration } & =\int \mathrm{fdt} \\ & =\int_{0}^{12} 5+5 \mathrm{e}^{-0.5 \mathrm{t}} \mathrm{dt} \\ & =5 \times(\mathrm{t})_{0}^{12}+\left.\frac{5}{-0.5} \mathrm{e}^{-0.5 \mathrm{t}}\right|_{0} ^{2} \\ & =60+10\left(1-\mathrm{e}^{-0.5 \times 12}\right) \\ & =69.975 \mathrm{~mm}=70 \mathrm{~mm} \end{aligned}
Question 5 |
A 12-hour storm occurs over a catchment and results in a direct runoff depth of 100 \mathrm{~mm}. The time-distribution of the rainfall intensity is shown in the figure (not to scale). The \phi-index of the storm is (in \mathrm{mm}, rounded off to two decimal places)_____
2.25 | |
3.6 | |
4.15 | |
5.25 |
Question 5 Explanation:
D=12 hrs.
Direct runoff (R)=100 \mathrm{~mm}=10 \mathrm{~cm}
\mathrm{P}= total rainfall
P= Area under above diagram
P=\frac{1}{2} \times(12+2) \times 20
=14 \times 10=140 \mathrm{~mm}=14 \mathrm{~cm}
Total Infiltration = Total rainfall - runoff
=140-100=40 \mathrm{~mm}
Assuming infiltration rate as x\; \mathrm{ mm} / \mathrm{hr}.
\frac{x}{y_{1}}=\frac{20}{4}
y_{1}=\frac{x}{5}
\begin{aligned} & y_{2}=\frac{x}{y_{2}}=\frac{20}{6} \\ & y_{2}=\frac{3 x}{10} \end{aligned}
Total Infiltration
= Area under rainfall intensity duration curve Area of same curve below \phi-index line.
\Rightarrow \frac{1}{2} \times \frac{x}{5} \times x+\frac{1}{2} \times \frac{3 x}{10} \times x+\left(12-\frac{5 x}{10}\right) \times x=40
\Rightarrow \frac{x^{2}}{10}+\frac{3 x^{2}}{20}+12 x-\frac{5 x^{2}}{10}=40
\Rightarrow-\frac{5 x^{2}}{20}+12 x=40
\Rightarrow \mathrm{x}^{2}-48 \mathrm{x}+160=0
\Rightarrow \quad \mathrm{x}=3.6 and \mathrm{x}=44.39 [Discarded]
\Rightarrow \quad \phi=3.6 \mathrm{~mm} / \mathrm{hr}.
Direct runoff (R)=100 \mathrm{~mm}=10 \mathrm{~cm}
\mathrm{P}= total rainfall
P= Area under above diagram
P=\frac{1}{2} \times(12+2) \times 20
=14 \times 10=140 \mathrm{~mm}=14 \mathrm{~cm}
Total Infiltration = Total rainfall - runoff
=140-100=40 \mathrm{~mm}
Assuming infiltration rate as x\; \mathrm{ mm} / \mathrm{hr}.
\frac{x}{y_{1}}=\frac{20}{4}
y_{1}=\frac{x}{5}
\begin{aligned} & y_{2}=\frac{x}{y_{2}}=\frac{20}{6} \\ & y_{2}=\frac{3 x}{10} \end{aligned}
Total Infiltration
= Area under rainfall intensity duration curve Area of same curve below \phi-index line.
\Rightarrow \frac{1}{2} \times \frac{x}{5} \times x+\frac{1}{2} \times \frac{3 x}{10} \times x+\left(12-\frac{5 x}{10}\right) \times x=40
\Rightarrow \frac{x^{2}}{10}+\frac{3 x^{2}}{20}+12 x-\frac{5 x^{2}}{10}=40
\Rightarrow-\frac{5 x^{2}}{20}+12 x=40
\Rightarrow \mathrm{x}^{2}-48 \mathrm{x}+160=0
\Rightarrow \quad \mathrm{x}=3.6 and \mathrm{x}=44.39 [Discarded]
\Rightarrow \quad \phi=3.6 \mathrm{~mm} / \mathrm{hr}.
There are 5 questions to complete.