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

# PET by Penman method, Kathmandu in February

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2066 Magh (old course), 12 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-kathmandu-february
- 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 February by Penman Method. The following data is available.

Mean Monthly Temperature: 12.5$^\circ$C
Mean RH: 70%
Mean Sunshine Hours: 7 h
Potential Sunshine Hours: 11.9 h
Wind Velocity at 2 m Height = 120 km/day
Albedo: 0.15
Upper Terrestrial Solar Radiation: 9 mm of water/day

Other values:
Latitude: 28.50
Longitude: 84.50
Saturated Vapor Pressure at 12.5$^\circ$C: 11.4 mm of Hg
Slope of Saturated Vapor Pressure: 1.24 mm/$^\circ$C
Psychrometric Constant: 0.49 mm/$^\circ$C
Boltzman Constant: $2.0 \times 10^{-9}$ mm/day

## Short answer

Actual vapour pressure is 7.98 mm Hg; using a = 0.29 cos 28.5° and n/N = 0.588 gives net radiation H_n = 1.780 mm/day and E_a = 2.095 mm/day. Penman's equation gives PET = 1.87 mm/day, about 52 mm for February.

## Answer

**Penman's method** combines the energy-balance and aerodynamic (mass-transfer) approaches:

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

where $A$ = slope of the saturation vapour pressure curve, $\gamma$ = psychrometric constant, $H_n$ = net radiation (mm of water/day) and $E_a$ = drying power of air (mm/day).

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

#### Step 1: Vapour pressure
$$
e_a = RH \times e_s = 0.70 \times 11.4 = 7.98\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
Sunshine ratio $n/N = 7/11.9 = 0.588$, and $a = 0.29\cos(28.5^\circ) = 0.2549$.

Short-wave part:
$$
9 \times (1-0.15)\times(0.2549 + 0.52\times0.588) = 7.65 \times 0.5607 = 4.290\ \text{mm/day}
$$

Long-wave part, with $T_a = 12.5 + 273 = 285.5$ K:
$$
\sigma T_a^4 = 2.0\times10^{-9}\times 285.5^4 = 13.29\ \text{mm/day}
$$
$$
13.29\times(0.56 - 0.092\sqrt{7.98})\times(0.1 + 0.9\times0.588) = 13.29\times0.3001\times0.6294 = 2.510\ \text{mm/day}
$$

$$
H_n = 4.290 - 2.510 = 1.780\ \text{mm/day}
$$

#### Step 3: Drying power of air
$$
E_a = 0.35\left(1 + \frac{u_2}{160}\right)(e_s - e_a) = 0.35\times\left(1+\frac{120}{160}\right)\times(11.4-7.98) = 2.095\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.24\times1.780 + 0.49\times2.095}{1.24+0.49} = \frac{2.207+1.027}{1.73} = 1.87\ \text{mm/day}
$$

For February (28 days): $1.869\times28 \approx 52.3$ mm.

**Answer: PET = 1.87 mm/day, about 52 mm for February.**

(The latitude is used only to find $a = 0.29\cos\varphi$; the longitude is not needed. The extraterrestrial radiation is already given.)

## Check yourself

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

1. Latitude not needed because
   A. Wind
   B. Not important
   C. Ha is already given
   D. Cancels
   Answer: C. Extraterrestrial radiation is given. (Key point: Lat/long)

2. E a with u2 = 120, ew - ea = 3.42 gives
   A. 2.095 mm/day
   B. 5 mm/day
   C. 0.5 mm/day
   D. 3.4 mm/day
   Answer: A. 0.35 x 1.75 x 3.42 = 2.095. (Key point: Ea)

3. PET is
   A. 3 mm/day
   B. 10 mm/day
   C. 1.56 mm/day
   D. 0.2 mm/day
   Answer: C. (1.664 + 1.027)/1.73 = 1.56. (Key point: PET)

## Close variants set in other exams

- [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×

## 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
- [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
- [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 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
- [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