---
title: "Daily PET by Penman, 27.5 N, 1400 m"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:penman-1400m-27-5n"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-1400m-27-5n
site: BCT Engineering
---

# Daily PET by Penman, 27.5 N, 1400 m

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2067 Shrawan (old course), 10 marks
- Video: [Engineering Hydrology/ Penman's Equation/ Potential Evapotranspiration/ Daily consumptive use/ PET](https://www.youtube.com/watch?v=FCp5F0eUmpQ) — Rupak Deo
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/penman-1400m-27-5n
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Calculate the daily potential evaporation by Penman method using the following data: Latitude = 27.5$^\circ$N, Elevation = 1400 m above mean sea level, mean monthly temperature = 10$^\circ$C, relative humidity = 70%, mean observed sunshine hour = 7 h, wind velocity at 2 m height = 80 km/day, the ground surface is observed with green grass, Albedo = 0.15. Saturated vapor pressure at 10$^\circ$C = 11.4 mm of Hg, Slope of saturated vapor pressure = 1.24 mm/$^\circ$C, Psychrometric constant = 0.49 mm/$^\circ$C, and Boltzman constant = $2.01 \times 10^{-9}$ mm/day.

## Short answer

Compute actual vapour pressure as 0.70×11.4 = 7.98 mm Hg. Net radiation H_n = 3.895 mm/day and aerodynamic term E_a = 1.796 mm/day. Substituting into Penman's equation gives PET = 3.30 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
- $T = 10\,^\circ$C $\Rightarrow T_a = 283$ K; RH $= 70\%$; $n = 7$ h; $N = 11.8$ h; $u_2 = 80$ km/day
- $H_a = 13.2$ mm/day; $r = 0.15$; $\varphi = 27.50^\circ$; $e_w = 11.4$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
- **Month not given:** March is assumed (a cool, hill-station month at $27.5^\circ$N), for which standard tables or solar geometry give $H_a \approx 13.2$ mm/day and $N \approx 11.8$ h. $H_n$ and $E_a$ use the given $e_w$, $A$, $\gamma$, $\sigma$ and the albedo as $r = 0.15$. The elevation (1400 m) is not needed because $\gamma$ is given.

#### Step 1: Actual vapour pressure
$$
e_a = \frac{RH}{100}\,e_w = 0.70\times11.4 = 7.98\ \text{mm Hg}
$$

#### Step 2: Net radiation $H_n$
$a = 0.29\cos(27.50^\circ) = 0.2572$, and $n/N = 0.5932$

$$
\begin{aligned}
H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 13.2\times0.850\times(0.2572 + 0.52\times0.5932) = 6.347\\
\sigma T_a^4 &= 2.01\times10^{-9}\times(283)^4 = 12.893\\
\text{Back radiation} &= 12.893\times(0.56 - 0.092\sqrt{7.98})\times(0.1 + 0.9\times0.5932) = 2.453\\
H_n &= 6.347 - 2.453 = 3.895\ \text{mm/day}
\end{aligned}
$$

#### Step 3: Aerodynamic term
$$
E_a = 0.35\left(1 + \frac{80}{160}\right)(11.4 - 7.98) = 1.796\ \text{mm/day}
$$

#### Step 4: PET
$$
PET = \frac{1.24\times3.895 + 1.796\times0.49}{1.24 + 0.49} = \frac{4.829 + 0.880}{1.730} = 3.30\ \text{mm/day}
$$

**Answer: Daily PET = 3.30 mm/day.** (The value depends on the assumed $H_a$ and $N$.)

## Check yourself

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

1. n/N = 7/11.8 is
   A. 0.42
   B. 0.593
   C. 1.69
   D. 0.7
   Answer: B. 7/11.8 = 0.5932. (Key point: Sunshine ratio)

2. Daily PET is
   A. 0.3 mm/day
   B. 8 mm/day
   C. 1.5 mm/day
   D. 3.30 mm/day
   Answer: D. (1.24x3.895 + 1.796x0.49)/1.73 = 3.30. (Key point: PET)

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