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).
3.8 | |
4.6 | |
8.5 | |
12.5 |
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}
\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 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 |
There are 5 questions to complete.