Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}.
Question 4 |
The ordinates of a one-hour unit hydrograph for a catchment are given below :
\begin{array}{|l|l|l|l|l|l|l|l|l|} \hline \mathrm{t} (hour) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{Q}\left(\mathrm{m}^{3} / \mathrm{s}\right) & 0 & 9 & 21 & 18 & 12 & 5 & 2 & 0 \\ \hline \end{array}
Using the principle of superposition, a D-hour unit hydrograph for the catchment was derived from the one-hour unit hydrograph. The ordinate of the D-hour unit hydrograph were obtained as 3 \mathrm{~m}^{3} / \mathrm{s} at t=1 hour and 10 \mathrm{~m}^{3} / \mathrm{s} at t=2 hour. the value of D (in integer) is _____
\begin{array}{|l|l|l|l|l|l|l|l|l|} \hline \mathrm{t} (hour) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{Q}\left(\mathrm{m}^{3} / \mathrm{s}\right) & 0 & 9 & 21 & 18 & 12 & 5 & 2 & 0 \\ \hline \end{array}
Using the principle of superposition, a D-hour unit hydrograph for the catchment was derived from the one-hour unit hydrograph. The ordinate of the D-hour unit hydrograph were obtained as 3 \mathrm{~m}^{3} / \mathrm{s} at t=1 hour and 10 \mathrm{~m}^{3} / \mathrm{s} at t=2 hour. the value of D (in integer) is _____
2 | |
3 | |
4 | |
5 |
Question 4 Explanation:
Clearly seen
The duration D=3 hours
Question 5 |
A two-hour duration storm event with uniform excess rainfall of 3 cm occurred
on a watershed. The ordinates of streamflow hydrograph resulting from this event
are given in the table.
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Time (hours)}& 0&1 &2 &3 &4 &5 &6 &7 \\ \hline \text{Streamflow}(m^3/s)& 10 &16 & 34 &40 & 31 & 25 &16&10 \\ \hline \end{array}
Considering a constant baseflow of 10 m^3/s , the peak flow ordinate (in m^3/s ) of one-hour unit hydrograph for the watershed is ________ . (in integer)
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Time (hours)}& 0&1 &2 &3 &4 &5 &6 &7 \\ \hline \text{Streamflow}(m^3/s)& 10 &16 & 34 &40 & 31 & 25 &16&10 \\ \hline \end{array}
Considering a constant baseflow of 10 m^3/s , the peak flow ordinate (in m^3/s ) of one-hour unit hydrograph for the watershed is ________ . (in integer)
12 | |
14 | |
22 | |
18 |
Question 5 Explanation:
C_1 = Time
C_2 = Ordinates of 2hr DRH
C_3 = Ordinates of 2hr UH=(Ordinates of 2hr DRH)/(Rianfall excess of 3cm)
C_4 = S-curve lag by 2hr
C_5 = S-curve ordinates S_2
C_6 = S_1 curve
C_7 = Ordinates of 1hr UH = \frac{S_2-S-1}{1/2}
\begin{array}{|c|c|c|c|c|c|c|} \hline C_1& C_2& C_3& C_4& C_5& C_6&C_7 \\ \hline 0&0 &0 &- &0 &- & 0\\ \hline 1&6 & 2& -&2 & 0& 4\\ \hline 2&24 & 8&0 &8 & 2&12 \\ \hline 3&30 &10 &2 &12 &8 &8 \\ \hline 4&21 & 7& 8& 15& 12& 6\\ \hline 5&15 &5 & 12& 17& 15& 4\\ \hline 6&6 & 2& 15& 17&17& 0\\ \hline 7 &0 & 0& 17& 17& 17&0\\ \hline & & & 17& 17& 17&0\\ \hline \end{array}
C_2 = Ordinates of 2hr DRH
C_3 = Ordinates of 2hr UH=(Ordinates of 2hr DRH)/(Rianfall excess of 3cm)
C_4 = S-curve lag by 2hr
C_5 = S-curve ordinates S_2
C_6 = S_1 curve
C_7 = Ordinates of 1hr UH = \frac{S_2-S-1}{1/2}
\begin{array}{|c|c|c|c|c|c|c|} \hline C_1& C_2& C_3& C_4& C_5& C_6&C_7 \\ \hline 0&0 &0 &- &0 &- & 0\\ \hline 1&6 & 2& -&2 & 0& 4\\ \hline 2&24 & 8&0 &8 & 2&12 \\ \hline 3&30 &10 &2 &12 &8 &8 \\ \hline 4&21 & 7& 8& 15& 12& 6\\ \hline 5&15 &5 & 12& 17& 15& 4\\ \hline 6&6 & 2& 15& 17&17& 0\\ \hline 7 &0 & 0& 17& 17& 17&0\\ \hline & & & 17& 17& 17&0\\ \hline \end{array}
There are 5 questions to complete.