This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Empirical Evaporation Equations – Set 2”.
1. Estimate the weekly evaporation for a lake of water temperature 25°C, surrounding relative humidity 35% and wind speed 10 km/hr. Saturation vapour pressure data is given below. Use Fitzgerald equation.
| Water temperature (in °C) | Saturation vapour pressure (in mm of Hg) |
| 20 | 17.5 |
| 25 | 23.7 |
| 30 | 31.8 |
a) 17.68 mm
b) 17.68 cm
c) 25.26 mm
d) 25.26 cm
View Answer
Explanation: Saturation vapour pressure = ew = 23.7 mmHg
Actual vapour pressure = ea = 0.35*23.7 = 8.3 mmHg
Now, rate of evaporation, E=(0.4+0.124V)(ew-ea)
=(0.4+0.124*10)(23.7-8.3)=25.26 mm/day
∴ Evaporation in one week=25.26*7=176.82 mm=17.68 cm
2. What is the per day evaporation from a pond if the saturation vapour pressure = 12.79 mm of Hg and relative humidity = 60%. The average wind speed at a height of 9 m above the ground is 22 kmph. Use Meyer’s formula.
a) 4.37
b) 6.08
c) 6.56
d) 9.11
View Answer
Explanation: For a pond or shallow water, Km = 0.5.
Evaporation=Km.(ew-ea).\(\left(1+\frac{u_9}{16}\right)\)=0.5*(12.79-0.6*12.79)*\(\left(1+\frac{22}{16}\right)\) =6.075 mm/day
≅6.08 mm/day
3. A 300 hectare lake of surface temperature 30°C (saturation vapour pressure = 31.82 mm of Hg) underwent an evaporation of 94 mm in a given week. The wind velocity measured at a height of 9 m from the ground level is 3 m/s. Estimate the average relative humidity of the surrounding area during the given week.
a) 30%
b) 37.5%
c) 62%
d) 70%
View Answer
Explanation: Using Meyer’s equation and assuming the lake as a large water body, Km = 0.36.
Evaporation per day of the given week = \(\frac{94}{7}\)=13.43 mm/day
Evaporation per day=Km*(ew-ea)*\(\left(1+\frac{u_9}{16}\right)\)
⇒13.43=0.36*(31.82-(humidity*31.82))*\(\left(1+\frac{3*3.6}{16}\right)\)
⇒13.43=0.36*31.82*(1-humidity)*1.675
⇒(1-humidity)=\(\frac{13.43}{0.36*31.82*1.675}\)=0.6999
∴ Humidity=1-0.6999=0.3001=30%
4. If the actual vapour pressure and humidity (average values for the month of April) are 9 mm of Hg and 60% respectively, find the evaporation (in cm) from a shallow pond for the month of April. The wind speed is 8 kmph at 9 m from ground level.
a) 5.8
b) 9.7
c) 8.1
d) 13.5
View Answer
Explanation: Saturation vapour pressure, ew=\(\frac{e_a}{humidity}=\frac{9}{0.6}\)=15 mm of Hg
From Meyer’s formula, E=0.5*(15-9)*\(\left(1+\frac{8}{16}\right)\)=0.5*6*1.5=4.5 mm/day
∴ Total evaporation in April=4.5*30=135 mm=13.5 cm
5. An ISI pan (Cp = 0.78) recorded an evaporation of 43.7 mm in a given week. During the same week, the average saturation vapour pressure, humidity and wind speed (at 9 m from ground) were found out to be 9.21 mm of Hg, 16% and 13 kmph, respectively. Calculate the absolute error in evaporation during the week by Meyer’s equation (take Km = 0.36) as compared to the value obtained as per evaporimeter data.
a) 0.47 mm
b) 0.85 mm
c) 1.18 mm
d) 1.25 mm
View Answer
Explanation: From Meyer’s equation, E1=0.36*(9.21-0.16*9.21)*\(\left(1+\frac{13}{16}\right)\) = 5.048 mm/day
Evaporation during the week as per Meyer’s equation, E1 = 5.048 * 7 = 35.336 mm
Evaporation during the week as per evaporimeter, E2 = 43.7 * 0.78 = 34.086 mm
∴ Erorr in weekly evaporation= |E1-E2|=|35.336-34.086|=1.25 mm
6. Find the percentage error in the daily evaporation on a particular day estimated by Meyer’s formula (Km = 0.5), if the expected lake evaporation value for that day as per evaporimeter (Cp = 0.8) records is 5.6 mm. The climate parameters for that day are as follows: saturation vapour pressure = 12.8 mm of Hg, relative humidity = 50%, wind speed at 5 m from ground level = 12 kmph.
a) 0%
b) 1.04%
c) 3.6 %
d) 3.75%
View Answer
Explanation: For Meyer’s formula, the wind speed at 9 m from ground needs to be computed.
⇒ \(\frac{u_9}{u_5}=\frac{(9)^{\frac{1}{7}}}{(5)^{\frac{1}{7}}}\) ⇒ u9 = \(\frac{12*(9)^{\frac{1}{7}}}{(5)^{\frac{1}{7}}}\) = 13.05 kmph
Now, E=0.5*(12.8-0.5*12.8)*\(\left(1+\frac{13.05}{16}\right)\)=5.81 mm/day
∴ Percentage error=\(\frac{|Calculated \, value-Expected \, value|}{Expected \,
value}\)*100=\(\frac{5.81-5.6}{5.6}\)*100=3.75%
7. The total lake evaporation in a week as estimated by Meyer’s formula (Km = 0.36) is 90 mm. The humidity of the region is 34% and the wind speed measured at 2 m from the ground level is 7 kmph. Find the average temperature (in °C) of the water surface during the week. The saturation vapour pressure data is given below.
| Water temperature (in °C) | Saturation vapour pressure (in mm of Hg) | |
| 30.0 | 31.82 | |
| 32.5 | 36.68 | |
| 35.0 | 42.81 |
a) 30.3
b) 31.7
c) 32.9
d) 34.1
View Answer
Explanation: The daily evaporation = \(\frac{90}{7}\)=12.857 mm
The wind speed at 9 m from ground level,
⇒ \(\frac{u_9}{u_2} = \frac{(9)^{\frac{1}{7}}}{(2)^{\frac{1}{7}}}\) ⇒ u9 = \(\frac{7*(9)^{\frac{1}{7}}}{(2)^{\frac{1}{7}}}\) = 8.678 kmph
So from Meyer’s equation, 12.857=0.36*(ew-0.34*ew)*\(\left(1+\frac{8.678}{16}\right)\)
⇒ 12.857=0.36*ew*0.66*1.54
⇒ ew=\(\frac{12.857}{0.36*0.66*1.54}\) = 35.14 mm of Hg
From the given table, interpolating the value of temperature,
∴ Water temperature=30+\([\frac{(32.5-30)}{(36.68-31.82)}*(35.14-31.82)]\) = 31.7°C
8. Using Rohwer’s formula, estimate the per day lake evaporation (in mm) for a saturation vapour pressure of 25.5 mm of Hg, humidity 65% and ground wind velocity 6 kmph. Take mean barometer reading as 1 atm.
a) 5.5
b) 6.6
c) 7.7
d) 8.8
View Answer
Explanation: As per Rohwer’s formula,
E=0.771*(1.465-0.000732pa)*(0.44+0.0733u)*(ew-ea)
⇒ E=0.771*(1.465-0.000732*760)*(0.44+0.0733*6)*(25.5-0.65*25.5)
⇒ E=0.771*0.9087*0.8798*8.925
∴ E=5.5 mm/day
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