---
title: "PET by Penman method, Kathmandu in March"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-kathmandu-march"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-kathmandu-march
site: BCT Engineering
---

# PET by Penman method, Kathmandu in March

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2071 Chaitra, 14 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-kathmandu-march
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate the potential evapotranspiration for an area over Kathmandu in the month of March by Penman Method. The following data is available:

Mean monthly temp: 10.0$^\circ$C
Mean RH: 60%
Mean sunshine hours: 9 h
Potential sunshine hours: 12.9 h
Wind velocity at 2 m height: 5 km/hour
Albedo: 0.25
Upper terrestrial solar radiation = 11 mm of Hg/day

Other values:
Latitude: 28.5$^\circ$
Longitude: 84.5$^\circ$
Saturated vapor pressure at 10.0$^\circ$C = 9.2 mm of Hg
Slope of saturated vapor pressure = 1.24 mm/$^\circ$C
Psychrometric constant = 0.49 mm/$^\circ$C
Boltzmann constant = $2.01 \times 10^{-9}$ mm/day

## Short answer

PET = 1.98 mm/day (about 61 mm for March).

## 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
- $T = 10.0\,^\circ$C $\Rightarrow T_a = 283.0$ K; RH $= 60\%$; $n = 9$ h; $N = 12.9$ h; $u_2 = 120$ km/day
- $H_a = 11$ mm/day; $r = 0.25$; $\varphi = 28.50^\circ$; $e_w = 9.2$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- $H_a = 11$ is taken as mm of evaporable water per day, wind $5\ \text{km/h} = 120$ km/day, and albedo is used as $r$. $b = 0.52$.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.60\times9.2 = 5.52\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(28.50^\circ) = 0.2549$, and $n/N = 0.6977$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 11\times0.750\times(0.2549 + 0.52\times0.6977) = 5.096\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(283.0)^4 = 12.893\\
\text{Back radiation} &= 12.893\times(0.56 - 0.092\sqrt{5.52})\times(0.1 + 0.9\times0.6977) = 3.227\\
H_n &= 5.096 - 3.227 = 1.869\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{120}{160}\right)(9.2 - 5.52) = 2.254\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.24\times1.869 + 2.254\times0.49}{1.24 + 0.49} = \frac{2.317 + 1.104}{1.730} = 1.98\ \text{mm/day}
$$

**Answer: PET = 1.98 mm/day, which is about 61 mm for the 31 days of March.**

## Check yourself

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

1. Wind 5 km/h used in Ea as
   A. 50 km/day
   B. 120 km/day
   C. 5 km/day
   D. 12 km/day
   Answer: B. 5 km/h x 24 = 120 km/day, as u2 must be in km/day. (Key point: Unit conversion)

2. RH = 60%, ew = 9.2 mm Hg gives ea =
   A. 3.68 mm Hg
   B. 15.3 mm Hg
   C. 9.2 mm Hg
   D. 5.52 mm Hg
   Answer: D. 0.60 x 9.2 = 5.52. (Key point: Actual vapour pressure)

3. n/N for n = 9 h and N = 12.9 h is
   A. 0.698
   B. 1.43
   C. 0.58
   D. 0.9
   Answer: A. 9/12.9 = 0.6977. (Key point: Sunshine ratio)

4. Final PET is about
   A. 1.98 mm/day
   B. 5.4 mm/day
   C. 0.5 mm/day
   D. 12 mm/day
   Answer: A. PET = (1.24x1.869 + 2.254x0.49)/1.73 = 1.98 mm/day. (Key point: Result)

## Close variants set in other exams

- [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×

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## Related in other subjects

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