---
title: "PET by Penman method for May"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-may-20c"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-may-20c
site: BCT Engineering
---

# PET by Penman method for May

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2076 Asoj, 8 marks
- Video: [Potential evapotranspiration using Penman's equation | Hydrology](https://www.youtube.com/watch?v=ifZ2sOisH34) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-may-20c
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate PET for May month by Penman method.
Mean monthly temperature = 20$^\circ$C
Mean RH = 75%
Mean sunshine hour = 10 hr
Potential sunshine hour = 13.5 hr
Wind velocity at 2 m height = 8 km/hr
Albedo = 0.028
Upper terrestrial solar radiation = 14.4 mm of Hg/day
Latitude = 27$^\circ$; Longitude = 86$^\circ$
Saturated vapour pressure at 20$^\circ$C = 11 mm of Hg
Slope of saturated vapour pressure = 1.42 mm/$^\circ$C

## Short answer

PET computed by Penman formula: net radiation 5.65 mm/day, aerodynamic term 2.12 mm/day; PET = 4.74 mm/day, about 147 mm for May.

## 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
- $T = 20\,^\circ$C $\Rightarrow T_a = 293$ K; RH $= 75\%$; $n = 10$ h; $N = 13.5$ h; $u_2 = 192$ km/day
- $H_a = 14.4$ mm/day; $r = 0.028$; $\varphi = 27.00^\circ$; $e_w = 11$ mm Hg; $A = 1.42$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- $H_a = 14.4$ is taken as mm of evaporable water per day. $\gamma = 0.49$ is assumed (not given). $e_w$ and $A$ are taken as given.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.75\times11 = 8.25\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(27.00^\circ) = 0.2584$, and $n/N = 0.7407$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.4\times0.972\times(0.2584 + 0.52\times0.7407) = 9.008\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(293)^4 = 14.814\\
\text{Back radiation} &= 14.814\times(0.56 - 0.092\sqrt{8.25})\times(0.1 + 0.9\times0.7407) = 3.359\\
H_n &= 9.008 - 3.359 = 5.649\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{192}{160}\right)(11 - 8.25) = 2.118\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.42\times5.649 + 2.118\times0.49}{1.42 + 0.49} = \frac{8.022 + 1.038}{1.910} = 4.74\ \text{mm/day}
$$

**Answer: PET = 4.74 mm/day, about 147 mm for the 31 days of May.**

## Check yourself

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

1. ea is:
   A. 8.25 mm Hg
   B. 11 mm Hg
   C. 14.67 mm Hg
   D. 2.75 mm Hg
   Answer: A. 0.75 x 11 = 8.25. (Key point: Actual vapour pressure)

2. 8 km/h converts to u2 of:
   A. 8 km/day
   B. 192 km/day
   C. 80 km/day
   D. 0.33 km/day
   Answer: B. 8 x 24 = 192 km/day. (Key point: Wind units)

3. Hn is:
   A. Short-wave term minus back radiation = 5.649 mm/day
   B. Sum of both terms
   C. 9.008 mm/day
   D. 3.359 mm/day
   Answer: A. 9.008 - 3.359 = 5.649 mm/day. (Key point: Net radiation)

4. The PET is:
   A. 4.74 mm/day
   B. 8.02 mm/day
   C. 1.04 mm/day
   D. 5.65 mm/day
   Answer: A. (1.42x5.649 + 2.118x0.49)/1.91 = 4.74 mm/day. (Key point: PET)

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