---
title: "Daily lake evaporation by Penman, 30 N in April"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-lake-30n-april"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-lake-30n-april
site: BCT Engineering
---

# Daily lake evaporation by Penman, 30 N in April

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2068 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-lake-30n-april
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Estimate daily evaporation from a lake at 30$^\circ$N for April by Penman method with the following mean monthly data.

| $T_a$ (Kelvin) | RH (%) | n (hrs) | $u_2$ (m/s) | $H_a$ (mm/day) | N (hrs) |
|---|---|---|---|---|---|
| 293 | 65 | 10 | 1.2 | 14.8 | 12.9 |

## Short answer

Applying Penman's method with net radiation and aerodynamic terms to the given data yields a daily lake evaporation of 5.41 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
- $T = 20\,^\circ$C $\Rightarrow T_a = 293$ K; RH $= 65\%$; $n = 10$ h; $N = 12.9$ h; $u_2 = 104$ km/day
- $H_a = 14.8$ mm/day; $r = 0.05$; $\varphi = 30.00^\circ$; $e_w = 17.54$ mm Hg; $A = 1.09$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- $T_a = 293$ K $= 20^\circ$C; $u_2 = 1.2$ m/s $= 103.7$ km/day; $r = 0.05$ for open water. From standard tables at $20^\circ$C: $e_w = 17.54$ mm Hg and $A = 1.09$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.65\times17.54 = 11.40\ \text{mm Hg}
$$

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

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.8\times0.950\times(0.2511 + 0.52\times0.7752) = 9.199\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(293)^4 = 14.814\\
\text{Back radiation} &= 14.814\times(0.56 - 0.092\sqrt{11.40})\times(0.1 + 0.9\times0.7752) = 2.947\\
H_n &= 9.199 - 2.947 = 6.252\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{104}{160}\right)(17.54 - 11.40) = 3.541\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.09\times6.252 + 3.541\times0.49}{1.09 + 0.49} = \frac{6.815 + 1.735}{1.580} = 5.41\ \text{mm/day}
$$

**Answer: Daily lake evaporation = 5.41 mm/day.**

## Check yourself

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

1. Ta = 293 K means T is
   A. 293°C
   B. 30°C
   C. 566°C
   D. 20°C
   Answer: D. 293 - 273 = 20°C, used to read ew and A. (Key point: Temperature)

2. u2 = 1.2 m/s is
   A. 4.3 km/day
   B. 43 km/day
   C. 1.2 km/day
   D. 103.7 km/day
   Answer: D. 1.2 x 86400 / 1000 = 103.7. (Key point: Wind)

3. r = 0.05 as it is
   A. Desert
   B. Green crop
   C. Snow
   D. Lake (open water)
   Answer: D. Open water has low albedo. (Key point: r)

4. Daily lake evaporation is
   A. 5.41 mm/day
   B. 0.5 mm/day
   C. 12 mm/day
   D. 1.1 mm/day
   Answer: A. (1.09x6.252 + 3.541x0.49)/1.58 = 5.41. (Key point: PET)

## Related in Engineering Hydrology

- [Lake loss from monthly pan evaporation](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apan-evap-kathmandu-lake) — Engineering Hydrology, chapter 3
- [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
- [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
- [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
- [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
- [Free water evaporation by Penman method, June](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apenman-june-33n) — Engineering Hydrology, chapter 3

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