This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Evaporimeters – Set 4”.
1. To maintain the water in an evaporation pan at a certain level, the water needs to be removed from the pan in which case?
a) Precipitation = Evaporation
b) Precipitation > Evaporation
c) Precipitation < Evaporation
d) In case of no precipitation
View Answer
Explanation: The pan evaporimeter is used on the principle that the water level needs to be maintained at the initial level after the study period. In the case that the amount of precipitation is more than the evaporation, the level of water will be higher than the initial level and must be removed in order to bring it to the initial reading.
2. After an observation period of 24 hours, the water level in an evaporation pan of capacity 250 litres and depth 25 cm, dropped by 9 mm. How many litres of water needs to be added/removed to maintain the initial water level?
a) Add 0.9 L
b) Remove 0.9 L
c) Add 9 L
d) Remove 9 L
View Answer
Explanation: Volume of the pan = 250 L = 2,50,000 cm3
Surface area of the pan=\(\)\frac{Volume}{Depth}={250000 cm^3}{25 cm}=10,000 cm2
The amount of water that needs to be added = 9 mm = 0.9 cm
∴ Volume of water to be added=0.9 cm*10,000 cm2=9000 cm3=9 L
3. The following table shows the level of water in an evaporation pan, under study for 8 hours, during different times of the day.
| Time | 8 am | 10 am | 12 pm | 2 pm | 4 pm |
| Depth of water in pan (in cm) | 20 (initial) | 19.6 | 19.1 | 18.2 | 19.8 (final) |
A rainfall of uniform intensity 2 cm/hr was observed from 3 pm onwards. What is the total depth of water evaporated from the pan between 8 am and 4 pm?
a) 2 mm
b) 18 mm
c) 22 mm
d) 34 mm
View Answer
Explanation: 8 am to 10 am→ initial level = 20 cm, no rainfall, evaporation = 20 – 19.6 = 0.4 cm
10 am to 12 pm→ initial level = 19.6 cm, no rainfall, evaporation = 19.6 – 19.1 = 0.5 cm
12 pm to 2 pm → initial level = 19.1 cm, no rainfall, evaporation = 19.1 – 18.2 = 0.9 cm
2 pm to 4 pm→ initial level = 18.2 cm, total rainfall = 2 cm, evaporation = 18.2 + 2 – 19.8 = 0.4 cm
Therefore, total evaporation = 0.4 + 0.5 + 0.9 + 0.4 = 2.2 cm = 22 mm
4. After a week of observation, 7.6 litres of water is added to an evaporation pan of 1.22 m diameter, to bring the water to the level it was at the start of the week. A nearby raingauge records a total rainfall of 5.1 mm during the week. What is the pan evaporation during the week?
a) 1.4 mm
b) 6.7 mm
c) 11.6 mm
d) 15.7 mm
View Answer
Explanation: Since water is added to the pan, the evaporation should be more than the rainfall.
Depth of water added=\(\frac{Volume added}{Area of pan}=\frac{(7.6*10^{-3})m^3}{(\frac{π}{4}*1.22^2)m^2} \))=0.0065 m=6.5 mm
∴ Total evaporation=water added+rainfall depth=6.5+5.1=11.6 mm
5. The evaporation pan readings and rainfall data for a 12 hour period is given below.
| Time | 6 am | 9 am | 12 pm | 3 pm | 6 pm |
| Rainfall intensity (in mm/hr) | 4 | 0 | 7 | 8 | |
| Volume of water in pan (in L) | 250 (initial) | 245 | 229 | 242 | 260 (final) |
In which time interval did the most evaporation take place? Assume pan surface area as 1 m2.
a) 6 am to 9 am
b) 9 am to 12 pm
c) 12 pm to 3 pm
d) 3 pm to 6 pm
View Answer
Explanation: Volume = ‘V’ litres=(V*10-3) m3=\(\frac{(V*10^{-3})m^3}{1 m^2}\) =(V*10-3)m = \(\frac{V}{10}\) cm depth
Let IL = initial level, FL = final level, R = rainfall and E = evaporation. Now,
6 am to 9 am→ IL = 25 cm, FL = 24.5 cm, R = 0.4 * 3 = 1.2 cm, E = 25 – 24.5 + 1.2 = 1.7 cm
9 am to 12 pm→ IL = 24.5 cm, FL = 22.9 cm, R = 0 * 3 = 0 cm, E = 24.5 – 22.9 + 0 = 1.6 cm
12 pm to 3 pm→ IL = 22.9 cm, FL = 24.2 cm, R = 0.7 * 3 = 2.1 cm, E = 22.9 – 24.2 + 2.1 = 0.8 cm
3 pm to 6 pm→ IL = 24.2 cm, FL = 26 cm, R = 0.8 * 3 = 2.4 cm, E = 24.2 – 26 + 2.4 = 0.6 cm
Therefore, the maximum evaporation of 1.7 cm takes place between 6 am to 9 am.
6. After a 72 hour observation period, 3.24 litres of water is emptied from a square evaporation pan of side 90 cm, to bring the water to the initial level. A nearby raingauge records a total rainfall of 10.9 mm during the observation period. What is the pan evaporation during this period?
a) 6.9 mm
b) 9.3 mm
c) 11.7 mm
d) 14.9 mm
View Answer
Explanation: Since water is removed from the pan, the evaporation should be lesser than the rainfall.
Depth of water removed=\(\frac{Volume \, removed}{Area \, of \, pan}=\frac{(3.24*10^{-3})m^3}{(0.9^2)m^2}\)=0.004 m=4 mm
∴ Total evaporation=rainfall depth-water removed=10.9-4=6.9 mm
7. An ISI standard pan is installed near a lake of surface area 6 hectares. The evaporation data of the pan is given for the third quarter of 2020.
| Month | July | August | September |
| Pan evaporation (in cm) | 21.7 | 20.9 | 21.1 |
What is the approximate amount of water (in million litres) that evaporated from the lake in the month of July? Assume any data suitably.
a) 7.8
b) 10.4
c) 13
d) 15.6
View Answer
Explanation: Let us assume the pan coefficient to be 0.8.
For the month of July, Lake evaporation=Cp*pan evaporation=0.8*21.7=17.36 cm
∴ Volume of lake evaporation=lake evaporation*area=\(\frac{17.36}{100}\) m*(6*104)m2
= 10416 m3=10416000 litres=10.416 million litres≅10.4 Ml
8. A river has an average surface width of 30 m. The evaporation at a nearby point measured by a class A pan is 4.5 mm/day. What is the volume of water (in million m3) evaporated in an 80 km stretch of the river in the month of March? Assume pan coefficient = 0.75.
a) 0.19
b) 0.22
c) 0.25
d) 0.30
View Answer
Explanation: Total pan evaporation in March = 4.5 * 31 = 139.5 mm = 13.95 cm
Actual evaporation from river in March = 0.75 * 13.95 = 10.46 cm
∴ Volume of water evaporated=\(\frac{10.46}{100}\) m*30 m*(80*1000) m=251040 m3≅0.25 Mm3
Sanfoundry Global Education & Learning Series – Engineering Hydrology.
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