This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Evapotranspiration Equations”.

1. Which of the following equations is not associated with estimating the evapotranspiration?

a) Penman equation

b) Thornthwaite equation

c) Blaney-Criddle equation

d) Stefan-Boltzmann equation

View Answer

Explanation: Penman equation is based on theoretical reasoning, whereas Blaney-Criddle equation and Thornthwaite equation are purely empirical. Stefan-Boltzmann equation is associated with the radiation emitted from a black body.

2. Penman equation is derived based on which of the following concepts?

a) Energy balance and mass transfer

b) Energy balance and water budget

c) Mass transfer and water budget

d) Energy balance, mass transfer and water budget

View Answer

Explanation: Penman equation is a formula based on theoretical concepts to find the daily potential evapotranspiration using data like temperature, wind velocity, pressure and radiation. It is derived using a combination of energy balance concept and mass transfer approach.

3. What is the unit of the slope term in Penman equation?

a) (mm of Hg)/°C

b) (mm of Hg).°C

c) (mm of Hg)^{-1}.(°C)^{-1}

d) °C/(mm of Hg)

View Answer

Explanation: The slope term in Penman equation represents the slope of the curve between saturation vapour pressure and temperature at a given air temperature. The unit should be in mm of Hg per °C.

4. In the Penman equation given as follows, what does the parameter E_{a} represent?

PET=\(\frac{AH_n+E_a γ}{A+γ} \)

a) Wind speed and air pressure

b) Radiation intensity and saturation deficit

c) Wind speed and saturation deficit

d) Radiation intensity and air pressure

View Answer

Explanation: The term E

_{a}is a parameter that includes wind speed and saturation deficit. It is given as,

E

_{a}=0.35\(\left(1+\frac{u_2}{160}\right)\)(e

_{w}-e

_{a})

where all terms have their usual meaning.

5. Penman’s equation for evapotranspiration can also be used to calculate the evaporation from a water surface.

a) True

b) False

View Answer

Explanation: The type of surface for which the PET is calculated is represented by the reflection coefficient (albedo) which is used in the equation to compute net radiation. For a pure water body, the albedo is around 0.05 and can be substituted as required to calculate evaporation only.

6. Which of the following data is not required for estimating PET using Penman equation?

a) Latitude of the region

b) Type of soil

c) Duration of sunshine

d) Mean air temperature

View Answer

Explanation: For calculating the net radiation to be used in the Penman equation, data regarding the latitude, mean air temperature, sunshine hours, actual vapour pressure and nature of the surface cover are required. Additional data like wind speed and saturation vapour pressure are required for E

_{a}calculation.

7. The net radiation for a region during a month is given as 2 mm of evaporable water per day. If the mean temperature for the month was 17.5°C, estimate the daily potential evapotranspiration for that month. The wind speed at 2 m height is 92 km/day. The vapour pressure data is given below.

Temperature (in °C) | Saturation vapour pressure (in mmHg) | Relative humidity (in %) | A (mmHg/°C) |

15 | 12.8 | 40 | 0.80 |

17.5 | 15.0 | 40 | 0.95 |

20 | 17.5 | 40 | 1.05 |

a) 3.00 mm/day

b) 3.47 mm/day

c) 4.22 mm/day

d) 4.96 mm/day

View Answer

Explanation: For 17.5°C, e

_{w}= 15 mmHg, e

_{a}= 0.4*15 = 6 mmHg, A = 0.95 mmHg/°C

So, E

_{a}=0.35\(\left(1+\frac{u_2}{160}\right)\)(e

_{w}-e

_{a})=0.35\(\left(1+\frac{92}{160}\right)\)(15-6)=4.96 mm/day

From Penman’s equation,

∴ PET=\(\frac{AH_n+E_a γ}{A+γ}=\frac{(0.95*2)+(4.96*0.49)}{0.95+0.49}\)=3 mm/day

8. Calculate the net radiation in mm of evaporable water per day, for a water surface exposed to a sunshine time of 12 hours with a maximum of 14.2 hours per day. Take air temperature = 36°C, mean solar radiation = 16 mm/day and actual vapour pressure = 11 mm Hg. Constants a = 0.25, b = 0.52.

a) 2.06

b) 3.31

c) 5.15

d) 6.46

View Answer

Explanation: For water surface r = 0.05, T

_{a}= 273+36 = 309 K

H

_{n}=H

_{a}.(1-r)\(\left(a+b.\frac{n}{N}\right)\) – σT

_{a}

^{4}.(0.56-0.092 √e

_{a}).\(\left(\frac{0.1+0.9n}{N}\right)\)

⇒ H

_{n}=16.(1-0.05)\(\left(0.25+\frac{0.52*12}{14.2}\right)\)-(2.01*10

^{-9}*(309)

^{4}).(0.56-0.092√11).\(\left(0.1+\frac{0.9*12}{14.2}\right)\)

⇒ H

_{n}=10.479-4.019=6.46 mm of water/day

9. Find the maximum monthly potential evapotranspiration for a closed ground crop for the following data: r = 0.25, net radiation = 1.66 mm/day, E_{a} = 1.94 mm/day, A = 1.85 mmHg/°C.

a) 1.72 cm

b) 49.9 mm

c) 5.33 cm

d) 71.8 mm

View Answer

Explanation: From Penman equation,

PET=\(\frac{AH_n+E_a γ}{A+γ}=\frac{(1.85*1.66)+(1.94*0.49)}{1.85+0.49}\)=1.7186 mm/day

Therefore, PET in one month = 1.7186*31= 53.27 mm = 5.33 cm

10. The saturation vapour pressure (in mm of Hg) is given for different temperatures t (in °C) as,

e_{w}=4.584*\(e^{\left(\frac{17.27t}{237.3+t}\right)}\)

For a mean temperature of 23.7°C and humidity of 44%, find the value of E_{a}. The wind speed at 2 m height from ground is given as 1.47 m/s.

a) 7.74 mm/day

b) 9.67 mm/day

c) 12.70 mm /day

d) 21.99 mm/day

View Answer

Explanation: For t = 23.7°C, the saturation vapour pressure is,

e

_{w}=4.584*\(e^{\left(\frac{17.27*23.7}{237.3+23.7}\right)}\)=4.584*e

^{1.5682}=21.99 mm of Hg

Since humidity = 44%, e

_{a}= 21.99 * 0.44 = 9.67 mm of Hg

Wind speed, u

_{2}= 1.47 m/s = 5.292 km/hr = 127.008 km/day

∴ E

_{a}=0.35\(\left(1+\frac{u_2}{160}\right)\)(e

_{w}-e

_{a}) = 0.35\(\left(1+\frac{127.008}{160}\right)\)(21.99-9.67)=7.74 mm/day

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