# Evaporation, Transpiration and Stream Flow Measurement

 Question 1
The average surface area of a reservoir in the month of June is 20 $km^{2}$. In the same month, the average rate of inflow is 10 $m^{3}$/s, outflow rate is 15 $m^{3}$/s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million $m^{3}$. The evaporation (in cm) in that month is
 A 46.8 B 136 C 13.6 D 23.4
GATE CE 2015 SET-2   Engineering Hydrology
Question 1 Explanation:
Let 'x' cm evaporation takes place in month of June.
Total inflow
$\begin{array}{l} =\left(\frac{10 \times 30 \times 24 \times 60 \times 60}{20 \times 10^{6}} \times 100\right)+10 \\ =139.6 \mathrm{cm} \end{array}$
Total outflow
$\begin{array}{l} =\left(\frac{15 \times 30 \times 24 \times 60 \times 60}{20 \times 10^{6}} \times 100\right)+1.8+x\\ =196.2+x \mathrm{cm} \end{array}$
As total outflow is more than total inflow, therefore depression in storage takes place.
Depression in storage
\begin{aligned} =-\frac{16 \times 10^{6}}{20 \times 10^{6}} \times 100&=-80 \mathrm{cm} \\ \Rightarrow 139.6-(196.2+x)&=-80 \\ -x &=-80+56.6\\ \therefore \quad x&=23.4 \mathrm{cm} \end{aligned}
 Question 2
The top width and the depth of flow in a triangular channel were measured as 4 m and 1 m, respectively. The measured velocities on the centre line at the water surface, 0.2 m and 0.8 m below the surface are 0.7 m/s, 0.6 m/s and 0.4 m/s, respectively. Using two-point method of velocity measurement, the discharge (in $m^3$/s) in the channel is
 A 1.4 B 1.2 C 1 D 0.8
GATE CE 2012   Engineering Hydrology
Question 2 Explanation:

\begin{aligned} V_{0.2}&=0.6m/s \\ V_{0.8}&=0.4 m/s \\ V&=\frac{V_{0.2}+V_{0.8}}{2} \\ &=\frac{0.6+0.4}{2}=0.5 m/s\\ Q&=AV \\ &= \frac{1}{2}\times 1\times 4\times 0.5=1.0m^{3}/s. \end{aligned}
 Question 3
The ratio of actual evapo-transpiration to potential evapo-transpiration is in the range
 A 0.0 to 0.4 B 0.6 to 0.9 C 0.0 to 1.0 D 1.0 to 2.0
GATE CE 2012   Engineering Hydrology
 Question 4
The plan area of a reservoir is 1 $km^{2}$. The water level in the reservoir is observed to decline by 20 cm in certain period. During this period the reservoir receives a surface inflow of 10 hectare-meters, and 20 hectare-meters are abstracted from the reservoir for irrigation and power. The plan evaporation and rainflow recorded during the same period at a nearby meteorological station are 12 cm and 3 cm respectively. The calibrated pan factor is 0.7. The seepage loss from the reservoir during the this period in hectare-meters is
 A 0 B 1 C 2.4 D 4.6
GATE CE 2003   Engineering Hydrology
Question 4 Explanation:
Inflow to reservoir,
I= 10 ha-m
Outflow from reservoir,
Q=20 ha-m
Evaporation loss,
$E=1\times 10^{6}\times \frac{12}{100}\times 0.7 =8.4 ha-m$
Rainfall, $P=1\times 10^{6}\times \frac{3}{100} =3 ha-m$
Change in storage,
$\Delta S=1\times 10^{6}\times \frac{20}{100} =-20 ha-m$
We know that,
$(I+P)-(E+Q+\text{Seepage})=\Delta S$
$\Rightarrow \; (10+3)-(8.4+20+\text{Seepage})=-20$
$\Rightarrow \; 13-28.4-\text{Seepage} = -20$
$\Rightarrow \; \text{Seepage} =4.6 ha-m$
 Question 5
Isopleths are lines on a map through points having equal depth of
 A Rainfall B Infiltration C Evapotranspiration D Total runoof
GATE CE 2001   Engineering Hydrology
Question 5 Explanation:
The mean annual PET (in cm) over various parts of the country is shown in the form of isopleths. Isopleths are the line on a map through places having equal depth of "evapotranspiration".
There are 5 questions to complete.