---
title: "Daily PET by Penman, latitude 30 N, 300 m"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-daily-lat30-elev300"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-daily-lat30-elev300
site: BCT Engineering
---

# Daily PET by Penman, latitude 30 N, 300 m

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2070 Chaitra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-daily-lat30-elev300
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate the daily potential evapotranspiration by the Penman method from an area having the following characteristics: latitude = 30$^\circ$N, elevation = 300 m above mean sea level, mean monthly temperature = 15$^\circ$C, mean relative humidity = 70%, mean observed sunshine hours = 10, wind velocity at 2 m height = 50 km/day and reflection coefficient is 0.05.

## Short answer

Using Penman's method that combines net radiation and aerodynamic terms with the given data, the daily potential evapotranspiration is 4.45 mm/day.

## 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
- $T = 15\,^\circ$C $\Rightarrow T_a = 288$ K; RH $= 70\%$; $n = 10$ h; $N = 12.9$ h; $u_2 = 50$ km/day
- $H_a = 14.8$ mm/day; $r = 0.05$; $\varphi = 30.00^\circ$; $e_w = 12.79$ mm Hg; $A = 0.82$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- **Month not given:** April at $30^\circ$N is assumed, giving $H_a = 14.8$ mm/day and $N = 12.9$ h. At $15^\circ$C, $e_w = 12.79$ mm Hg and $A = 0.82$ mm Hg/$^\circ$C (standard tables). $\gamma = 0.49$ mm Hg/$^\circ$C and $\sigma = 2.01\times10^{-9}$ mm/day/K⁴ are used. The elevation (300 m) changes $\gamma$ by only about 3% and is neglected.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.70\times12.79 = 8.95\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(30.00^\circ) = 0.2511$, and $n/N = 0.7752$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.8\times0.950\times(0.2511 + 0.52\times0.7752) = 9.199\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(288)^4 = 13.828\\
\text{Back radiation} &= 13.828\times(0.56 - 0.092\sqrt{8.95})\times(0.1 + 0.9\times0.7752) = 3.141\\
H_n &= 9.199 - 3.141 = 6.058\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{50}{160}\right)(12.79 - 8.95) = 1.763\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{0.82\times6.058 + 1.763\times0.49}{0.82 + 0.49} = \frac{4.968 + 0.864}{1.310} = 4.45\ \text{mm/day}
$$

**Answer: Daily PET = 4.45 mm/day.**

## Check yourself

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

1. Elevation 300 m was
   A. Neglected as it changes γ by about 3%
   B. Used for wind
   C. Added to Ha
   D. Used for RH
   Answer: A. The effect on γ is small. (Key point: Elevation)

2. RH = 70%, ew = 12.79 gives ea =
   A. 18.3 mm Hg
   B. 12.79 mm Hg
   C. 5.6 mm Hg
   D. 8.95 mm Hg
   Answer: D. 0.7 x 12.79 = 8.95. (Key point: ea)

3. Final daily PET is
   A. 4.45 mm/day
   B. 1.2 mm/day
   C. 9 mm/day
   D. 0.44 mm/day
   Answer: A. (0.82x6.058 + 1.763x0.49)/1.31 = 4.45. (Key point: PET)

## Related in Engineering Hydrology

- [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
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