Question 1 |
In a certain month, the reference crop evapotranspiration at a location is
6 mm/day. If the crop coefficient and soil coefficient are 1.2 and 0.8, respectively,
the actual evapotranspiration in mm/day is
5.76 | |
7.2 | |
6.8 | |
8 |
Question 1 Explanation:
Actual evapotranspiration (ET_C)
=K_S \times K_C \times Reference evapotranspiration (ET_0)
=0.8 \times 1.2 \times 6=5.76mm
=K_S \times K_C \times Reference evapotranspiration (ET_0)
=0.8 \times 1.2 \times 6=5.76mm
Question 2 |
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
46.8 | |
136 | |
13.6 | |
23.4 |
Question 2 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}
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 3 |
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
1.4 | |
1.2 | |
1 | |
0.8 |
Question 3 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 4 |
The ratio of actual evapo-transpiration to potential evapo-transpiration is in the range
0.0 to 0.4 | |
0.6 to 0.9 | |
0.0 to 1.0 | |
1.0 to 2.0 |
Question 5 |
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
0 | |
1 | |
2.4 | |
4.6 |
Question 5 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
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 6 |
Isopleths are lines on a map through points having equal depth of
Rainfall | |
Infiltration | |
Evapotranspiration | |
Total runoof |
Question 6 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 6 questions to complete.