# Engineering Hydrology Questions and Answers – Analytical Methods of Evaporation Estimation – Set 2

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Analytical Methods of Evaporation Estimation – Set 2”.

1. The energy budget method for evaporation estimation is based on which of the following laws of physics?
a) Law of conservation of charge
b) Law of conservation of energy
c) Law of conservation of mass
d) Law of conservation of momentum

Explanation: As the name suggests, the energy budget method is an application of the energy conservation principle. The energy entering, stored in and leaving the system over a time period is considered to evaluate the energy required for evaporation.

2. Which of the following units is not applicable to the terms of the energy budget equation?
a) Watts/m2
b) Calories/mm2/day
c) kJ/min/hectare
d) Joule/kg

Explanation: All the terms of the energy budget equation should be in the units of energy absorbed or lost per unit time per unit area of the water body. Joule/kg is the unit for the latent heat of vapourization.

3. The incoming solar radiation onto a water surface is 0.5 kJ/s/m2/. If the reflection coefficient is 0.12, what is the net heat energy (in watt/m2/) received by the water body. Assume back radiation as 15% of incoming radiation.
a) 365
b) 374
c) 425
d) 440

Explanation: Incoming radiation, H=0.5 kJ/s/m2=500 watts/m2
Therefore, Hnet=Hn-Hb=440-(15% of H)=440-(0.15*500)=365 watts/m2

4. Which of the following pair of terms can be neglected from the water budget equation if the time period is short?
a) Back radiation and heat flux into ground
b) Stored heat and advected energy
c) Advected energy and heat flux into ground
d) Back radiation and stored heat

Explanation: According to the energy budget equation, the net heat received is equal to the sum of back radiation, heat lost to air, heat for evaporation, heat flux into the ground, heat stored and advected energy. If the time period considered is short, the stored heat and advected energy are negligibly small and can be ignored.

5. The heat loss from water to atmosphere cannot be measured directly.
a) True
b) False

Explanation: All the terms in the energy budget equation, except heat lost from water to air, can be measured or evaluated. The sensible heat term is instead indicated using a term called Bowen ratio.

6. The total heat energy used up for evaporation per second by a water body (surface area = 150 ha), is 40000 kJ. Find the amount of water evaporated (in mm) from the water body in one day. Assume the latent heat of evaporation as 2300 kJ/kg and density of water as 998 kg/m3.
a) 0.6
b) 0.8
c) 1.0
d) 1.2

Explanation: Heat energy used up for evaporation, He = ρw*L*E⇒E = $$\frac{H_e}{ρ_w*L}$$
Now, He = 36000 kJ/s = $$\frac{40000*10^3}{150*10^4}$$ J/s/m2=26.67 watts/m2
∴ E = $$\frac{26.67}{998*2300*10^3}$$ = 1.162*10-8m/s = 1.162*10-8*103*86400=1 mm/day

7. Which of the following equations represents the correct equation for evaporation rate (E) as per energy budget equation? Hn = net heat energy received; Hc = sum of heat stored, heat flux to ground and advected energy; ρ = density of water; L = latent heat of vapourization of water; β = Bowen ratio.
a) E=$$\frac{H_n-H_c}{βρL}$$
b) E=$$\frac{H_n-H_c}{(β+1)ρL}$$
c) E=$$\frac{β(H_n-H_(c))}{ρL}$$
d) E=$$\frac{(β+1).(H_n-H_c)}{ρL}$$

Explanation: From the energy budget equation, Hn = Ha+He+Hc ⋯⋯⋯(1), where
He=energy used in evaporation=ρ*L*E ⋯⋯⋯(2) and Ha = heat lost from water to air.
Now from the definition of Bowen ratio,
β = $$\frac{H_a}{H_e} = \frac{H_a}{ρ.L.E}$$ ⇒ Ha = β.ρ.L.E ⋯⋯⋯(3)
Substituting (2) and (3) in (1),
Hn = (β.ρ.L.E) + (ρ.L.E) + Hc
⇒ Hn=(ρLE)(β+1)+Hc
⇒ ρLE = $$\frac{H_n-H_c}{β+1}$$ ⇒ E = $$\frac{H_n-H_c}{(β+1)ρL}$$

8. Find the evaporation rate if actual heat lost to air is 70 watts/m2, density of water is 998 kg/m3, latent heat of vapourization is 2405 kJ/kg and Bowen ratio = 0.21.
a) 0.5 mm/hour
b) 12 cm/hour
c) 0.5 cm/day
d) 1.2 mm/day

Explanation: E = $$\frac{H_a}{βρL} = \frac{70}{0.21*998*2405*10^3}$$=1.388*10-7 m/s
=1.388*10-7*103*3600≅0.5 mm/hour
=1.388*10-7*103*86400≅12 mm/day

9. Find the amount of water (in L) evaporated from a pond of area 260 m2 in one hour for the following parameters. Atmospheric pressure = 760 mm of Hg, water temperature = 15°C, air temperature = 13°C, relative humidity = 30%, saturated vapour pressure = 12.8 mm of Hg, density of water = 1 g/cm3, latent heat of evaporation = 2260 kJ/kg, sensible heat transfer from water to air = 50 watt/m2.
a) 86
b) 200
c) 770
d) 4800

Explanation: Bowen ratio can be given as,
β=6.1*10-4*pa*$$\frac{T_w-T_a}{e_w-e_a}$$ = 6.1*10-4*760*$$\frac{15-13}{12.8-0.3*12.8}$$=0.1035
Now, β = $$\frac{H_a}{ρLE}$$ = 0.1035
⇒ E = $$\frac{H_a}{0.1035*ρ*L} = \frac{50}{0.1035*1000*2260*10^3}$$ = 2.1376*10-7 m/s
=2.1376*10-7*103*3600=0.77 mm/hour
∴ Amount evaporated in one hour=0.77*10-3*260=0.2 m3=200 litres

10. If the product of density and latent heat evaporation of water is 2.4×106 kJ/m3 and Bowen ratio is 0.1, what is the evaporation rate (in mm/day) for a lake with net radiation of 100 watt/m2. Ignore all other energy budget terms.
a) 4.36
b) 2.22
c) 4.85
d) 3.27

Explanation: The evaporation rate is given as,
E = $$\frac{H_n}{ρ.L.(β+1)} = \frac{100}{2.4*10^6*10^3*(0.1+1)}$$ = 3.788*10-8 m/s
= 3.788*10-8*103*86400
= 3.27 mm/day

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