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.