---
title: "PET by Penman formula, Dharan in April"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-dharan-april"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-dharan-april
site: BCT Engineering
---

# PET by Penman formula, Dharan in April

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2075 Asoj, 10 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-dharan-april
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate the potential evapotranspiration from an area near Dharan, Sunsari in the month of April by Penman's formula. The following data are available.

Latitude: 26$^\circ$49'N; Elevation (from msl): 250.00 m
Mean monthly temperature: 22.5$^\circ$C; Mean relative humidity: 75%
Mean observed sunshine hour: 10 hr; Wind velocity at 2 m height: 80 km/day
Psychrometric constant: 0.49 mm of Hg/$^\circ$C; Reflection coefficient: 0.20
$e_w$: 20.4 mm of Hg; A: 1.24 mm/$^\circ$C; b = 0.52; $H_a$ = 14.9 mm of evaporable water per day
Mean monthly value of possible sunshine hour (N): 12.7 hours
Nature of sunshine cover: closed ground green crop, where the symbols carry their usual meanings.

## Short answer

Using Penman’s method: net radiation 5.49 mm/day, aerodynamic term 2.68 mm/day; PET = 4.69 mm/day, approximately 141 mm for April.

## Answer

**Penman equation (potential evapotranspiration):**

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

with
$$
H_n = H_a(1-r)\left(a + b\frac{n}{N}\right) - \sigma T_a^4\,(0.56 - 0.092\sqrt{e_a})\left(0.1 + 0.9\frac{n}{N}\right)
$$
$$
E_a = 0.35\left(1 + \frac{u_2}{160}\right)(e_w - e_a), \qquad a = 0.29\cos\varphi,\ b = 0.52
$$

where $H_n$ is the net radiation in mm of evaporable water per day, $E_a$ the aerodynamic term in mm/day, $u_2$ the wind speed at 2 m in km/day, and $e_w, e_a$ in mm of Hg.

#### Data and constants used
- $T = 22.5\,^\circ$C $\Rightarrow T_a = 295.5$ K; RH $= 75\%$; $n = 10$ h; $N = 12.7$ h; $u_2 = 80$ km/day
- $H_a = 14.9$ mm/day; $r = 0.2$; $\varphi = 26.82^\circ$; $e_w = 20.4$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- Latitude $26^\circ49' = 26.82^\circ$. Elevation (250 m) is not needed because $\gamma$ is given. The closed green crop is represented by $r = 0.20$ (given).

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.75\times20.4 = 15.30\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(26.82^\circ) = 0.2588$, and $n/N = 0.7874$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.9\times0.800\times(0.2588 + 0.52\times0.7874) = 7.966\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(295.5)^4 = 15.326\\
\text{Back radiation} &= 15.326\times(0.56 - 0.092\sqrt{15.30})\times(0.1 + 0.9\times0.7874) = 2.480\\
H_n &= 7.966 - 2.480 = 5.485\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{80}{160}\right)(20.4 - 15.30) = 2.677\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.24\times5.485 + 2.677\times0.49}{1.24 + 0.49} = \frac{6.802 + 1.312}{1.730} = 4.69\ \text{mm/day}
$$

**Answer: PET = 4.69 mm/day, so about 141 mm for April (30 days).**

## Check yourself

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

1. Why is elevation not needed?
   A. It is irrelevant
   B. gamma is given
   C. Temperature is given
   D. Wind is given
   Answer: B. Elevation affects gamma, which is provided. (Key point: Elevation)

2. ea is:
   A. 15.30 mm Hg
   B. 20.4 mm Hg
   C. 27.2 mm Hg
   D. 5.1 mm Hg
   Answer: A. 0.75 x 20.4. (Key point: ea)

3. Ea is:
   A. 2.677 mm/day
   B. 5.485 mm/day
   C. 1.3 mm/day
   D. 4.69 mm/day
   Answer: A. 0.35 x 1.5 x 5.1. (Key point: Ea)

4. PET is:
   A. 4.69 mm/day
   B. 5.49 mm/day
   C. 2.68 mm/day
   D. 8.1 mm/day
   Answer: A. (1.24x5.485 + 2.677x0.49)/1.73 = 4.69. (Key point: PET)

## Related in Engineering Hydrology

- [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
- [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
- [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's method, Dang in May](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-dang-may) — Engineering Hydrology, chapter 3
- [PET by Penman method, Kathmandu in March](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-kathmandu-march) — Engineering Hydrology, chapter 3, asked 2×
- [Daily PET by Penman, 27.5 N, 1400 m](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-1400m-27-5n) — 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