# 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.