---
title: "PET by Penman's method, Dang in May"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-dang-may"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-dang-may
site: BCT Engineering
---

# PET by Penman's method, Dang in May

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2082 Baisakh, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-dang-may
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate the potential evapotranspiration for an area over Dang in the month of May by Penman's method using the following data: Latitude = 26$^\circ$N; Mean temperature = 10$^\circ$C; Mean relative humidity = 65%; Mean monthly solar radiation at the top of atmosphere = 15.88 mm of evaporable water/day; Wind velocity at 2 m height = 5 km/day; Psychrometric constant = 0.49 mm/$^\circ$C; Mean sunshine hour = 10.1 hr; Potential sunshine hours = 13.5 hr; Reflection coefficient = 0.25; Slope of saturated vapor pressure at 10$^\circ$C = 1.24 mm of Hg.

## Short answer

Penman's equation gave PET = 3.48 mm/day, about 108 mm for May. Net radiation was 4.40 mm/day, aerodynamic term 1.16 mm/day; weighting by slope A and psychrometric constant yielded the result.

## Answer

**Penman's equation**

$$
PET = \frac{A H_n + E_a \gamma}{A + \gamma}
$$

$$
H_n = H_a(1-r)\left(a + b\frac{n}{N}\right) - \sigma T_a^4\left(0.56 - 0.092\sqrt{e_a}\right)\left(0.1 + 0.9\frac{n}{N}\right), \qquad a = 0.29\cos\varphi,\ b = 0.52
$$

#### Step 1: Vapour pressures
The saturation vapour pressure at 10 °C is taken from the standard table: $e_s = 9.21$ mm Hg (assumed, since it is not given).

$$
e_a = 0.65\times 9.21 = 5.99\ \text{mm Hg}
$$

#### Step 2: Net radiation
$n/N = 10.1/13.5 = 0.748$, and $a = 0.29\cos(26^\circ) = 0.2607$.

Short-wave part:
$$
H_a(1-r)\left(a+b\frac{n}{N}\right) = 15.88\times0.75\times(0.2607 + 0.52\times0.748) = 7.738\ \text{mm/day}
$$

Long-wave part, with $T_a = 283$ K and $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$:
$$
\sigma T_a^4 = 12.89\ \text{mm/day}
$$
$$
12.89\times(0.56 - 0.092\sqrt{5.99})\times(0.1+0.9\times0.748) = 12.89\times0.335\times0.773 = 3.339\ \text{mm/day}
$$

$$
H_n = 7.738 - 3.339 = 4.399\ \text{mm/day}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1+\frac{5}{160}\right)(9.21 - 5.99) = 0.35\times1.031\times3.22 = 1.163\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.24\times4.399 + 0.49\times1.163}{1.24+0.49} = \frac{5.455+0.570}{1.73} = 3.48\ \text{mm/day}
$$

For May (31 days): $3.482\times 31 \approx 108$ mm.

**Answer: PET = 3.48 mm/day, about 108 mm for May.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which form of Penman's equation is used?
   A. PET = (A*H_n + E_a*gamma)/(A + gamma)
   B. PET = (A*E_a + gamma*H_n)/(A - gamma)
   C. PET = A*H_n/(E_a*gamma)
   D. PET = (H_n + E_a)/(A + gamma)
   Answer: A. PET = (A H_n + E_a gamma)/(A + gamma), a weighted mean of radiation and aerodynamic terms. (Key point: Penman equation)

2. How is e_a obtained from the data?
   A. e_a = 0.35 x e_s
   B. e_a = e_s - 0.65
   C. e_a = RH x e_s = 0.65 x 9.21 = 5.99 mm Hg
   D. e_a = e_s / RH
   Answer: C. Actual vapour pressure is relative humidity times saturation vapour pressure. (Key point: Vapour pressure)

3. How is the net radiation H_n found?
   A. Short-wave (incoming x (1-r)) term minus the long-wave outgoing term
   B. Only the long-wave term
   C. Product of both terms
   D. Short-wave term plus long-wave term
   Answer: A. H_n = H_a(1-r)(0.18+0.55n/N) minus sigma Ta^4 (0.56-0.092 sqrt(e_a))(0.1+0.9n/N) = 7.045 - 3.339 = 3.705 mm/day. (Key point: Net radiation)

4. What is the final PET?
   A. About 12 mm/day
   B. About 29.9 mm/day
   C. About 0.99 mm/day
   D. About 2.99 mm/day, i.e. about 92.5 mm for May
   Answer: D. PET = (1.24x3.705 + 0.49x1.163)/1.73 = 2.99 mm/day, times 31 days = 92.5 mm. (Key point: Final result)

## Related in Engineering Hydrology

- [PET by Penman method for May](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-may-20c) — Engineering Hydrology, chapter 3
- [Daily PET by Penman, latitude 30 N, 300 m](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-daily-lat30-elev300) — Engineering Hydrology, chapter 3
- [Importance of solar radiation, humidity, evapotranspiration](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aimportance-solar-rh-et) — Engineering Hydrology, chapter 3
- [PET by Penman formula, Dharan in April](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-dharan-april) — Engineering Hydrology, chapter 3
- [PET by Penman method, Kathmandu in February](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-kathmandu-february) — Engineering Hydrology, chapter 3, asked 2×
- [PET by Penman formula, Simara in April](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-simara-april) — Engineering Hydrology, chapter 3

## Related in other subjects

- [Potential evapotranspiration by Penman method](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Apenman-pet-steps) — Irrigation and Drainage Engineering, chapter 2