Infiltration, Runoff and Hydrographs


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}
A
P-IV, Q-V, R-III
B
P-III, Q-IV, R-I
C
P-IV, Q-III, R-II
D
P-III, Q-I, R-IV
GATE CE 2023 SET-2   Engineering Hydrology
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.
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)
A
25.3
B
58.2
C
70
D
78.5
GATE CE 2023 SET-1   Engineering Hydrology
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}


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)_____
A
2.25
B
3.6
C
4.15
D
5.25
GATE CE 2023 SET-1   Engineering Hydrology
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}.
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 _____
A
2
B
3
C
4
D
5
GATE CE 2023 SET-1   Engineering Hydrology
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)
A
12
B
14
C
22
D
18
GATE CE 2022 SET-1   Engineering Hydrology
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}




There are 5 questions to complete.

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