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

# PET by Penman formula, Simara in April

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2074 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-simara-april
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

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

Latitude: 27$^\circ$N
Elevation (from msl): 107 m
Mean monthly temperature: 23$^\circ$C
Mean relative humidity: 75%
Mean observed sunshine hour: 10
Wind velocity at 2 m height: 85 km/day
Nature of sunshine cover: closed ground green crop

Given: A = 1.27 mm/$^\circ$C; $H_a$ = 15.00 mm of evaporable water per day; mean monthly value of possible sunshine hour (N): 12.5 hours; saturated vapour pressure at 23$^\circ$C = 21.04 mm of Hg.

## Short answer

PET = 4.85 mm/day (about 145 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
- $T = 23\,^\circ$C $\Rightarrow T_a = 296$ K; RH $= 75\%$; $n = 10$ h; $N = 12.5$ h; $u_2 = 85$ km/day
- $H_a = 15.0$ mm/day; $r = 0.2$; $\varphi = 27.00^\circ$; $e_w = 21.04$ mm Hg; $A = 1.27$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- $r = 0.20$ is taken for a closed green crop (range 0.15-0.25) and $\gamma = 0.49$ mm Hg/$^\circ$C is assumed, since they are not given. Elevation (107 m) has a negligible effect and is neglected. $b = 0.52$.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.75\times21.04 = 15.78\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(27.00^\circ) = 0.2584$, and $n/N = 0.8000$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 15.0\times0.800\times(0.2584 + 0.52\times0.8000) = 8.093\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(296)^4 = 15.430\\
\text{Back radiation} &= 15.430\times(0.56 - 0.092\sqrt{15.78})\times(0.1 + 0.9\times0.8000) = 2.461\\
H_n &= 8.093 - 2.461 = 5.631\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{85}{160}\right)(21.04 - 15.78) = 2.819\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.27\times5.631 + 2.819\times0.49}{1.27 + 0.49} = \frac{7.152 + 1.381}{1.760} = 4.85\ \text{mm/day}
$$

**Answer: PET = 4.85 mm/day, about 145 mm for April.**

## Check yourself

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

1. In the Penman equation PET = (A Hn + Ea γ)/(A + γ), what do the two terms combine?
   A. Infiltration and interception terms
   B. Energy-balance (net radiation) and aerodynamic (drying power of air) terms
   C. Radiation and runoff terms
   D. Temperature and rainfall terms
   Answer: B. Penman combines the energy-balance term Hn with the aerodynamic term Ea, weighted by A and γ. (Key point: Penman form)

2. With RH = 75% and saturated vapour pressure 21.04 mm Hg, what is ea?
   A. 28.05 mm Hg
   B. 21.04 mm Hg
   C. 5.26 mm Hg
   D. 15.78 mm Hg
   Answer: D. ea = RH x ew = 0.75 x 21.04 = 15.78 mm Hg. (Key point: Actual vapour pressure)

3. In Hn, what is subtracted from the incoming short-wave part to get net radiation?
   A. Albedo only
   B. Depression storage
   C. Aerodynamic term Ea
   D. Back (long-wave) radiation σTa^4(0.56 - 0.092√ea)(0.1 + 0.9 n/N)
   Answer: D. Hn = short-wave absorbed minus back radiation, which is 8.093 - 2.461 = 5.631 mm/day. (Key point: Net radiation)

## Related in Engineering Hydrology

- [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
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## 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