# Engineering Hydrology

 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).

 A 3.8 B 4.6 C 8.5 D 12.5
GATE CE 2023 SET-2      Evaporation, Transpiration and Stream Flow Measurement
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}
$\begin{array}{|c|c|c|c|} \hline \begin{array}{l}\text { Avg. width } \\ \text { (m) } \\ \text { (1) }\end{array} & \begin{array}{l}\text { Avg. Depth } \\ \text { (m) } \\ \text { (2) }\end{array} & \begin{array}{l}\text { Avg. Velocity } \\ \text { (m/s) } \\ \text { (3) }\end{array} & \begin{array}{c}\text { Segmental } \\ \text { discharge } \\ \left(\mathrm{m}^{3} / \mathrm{s}\right)=(1) \times(2) \times(3)\end{array} \\ \hline 1.6875 & 0.40 & 0.40 & 0.270 \\ \hline 1.5 & 0.70 & \frac{0.76+0.70}{2}=0.73 & 0.767 \\ \hline 1.5 & 1.20 & \frac{1.19+1.13}{2}=1.16 & 2.088 \\ \hline 1.5 & 1.40 & \frac{1.25+1.10}{2}=1.175 & 2.467 \\ \hline 1.5 & 1.10 & \frac{1.13+1.09}{2}=1.11 & 1.831 \\ \hline 1.5 & 0.80 & \frac{0.69+0.65}{2}=0.67 & 0.804 \\ \hline 1.6875 & 0.45 & 0.42 & 0.319 \\ \hline & & & \Sigma=8.546 \\ \hline \end{array}$
 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 ? ___

 A 912.56 B 521.32 C 5463.25 D 472.36
GATE CE 2023 SET-2      Precipitation and General Aspects of Hydrology
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 $=\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4}$ $=912.56 \mathrm{~mm}$

 Question 3
Which one of the following options provides the correct match of the terms listed in Column-I and Column-2 ?
$\begin{array}{|l|l|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{P : Horton equation} & \text{I: Precipitation} \\ \hline \text{Q : Muskingum method} & \text{II: Flood frequency} \\ \hline \text{R : Penman method} & \text{III: Evapotranspiration} \\ \hline & \text{IV : Infiltration} \\ \hline & \text{V : Channel routing} \\ \hline \end{array}$
 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      Infiltration, Runoff and Hydrographs
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.
 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)
 A 25.3 B 58.2 C 70 D 78.5
GATE CE 2023 SET-1      Infiltration, Runoff and Hydrographs
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}
 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)_____
 A 2.25 B 3.6 C 4.15 D 5.25
GATE CE 2023 SET-1      Infiltration, Runoff and Hydrographs
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]
$\Rightarrow \quad \phi=3.6 \mathrm{~mm} / \mathrm{hr}$.

There are 5 questions to complete.