---
title: "Reservoir evaporation by Meyer's formula"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:meyer-reservoir-evap-2km2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/meyer-reservoir-evap-2km2
site: BCT Engineering
---

# Reservoir evaporation by Meyer's formula

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2078 Bhadra, 5 marks
- Video: [2. Estimate Reservoir Evaporation Using Meyer's Formula](https://www.youtube.com/watch?v=sKAYDqtbxyU) — Dhrubo Raj Roy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/meyer-reservoir-evap-2km2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

A reservoir of a hydropower project has a surface area of 2 km$^2$. Estimate the volume of water evaporated from the reservoir in March (30 days), if temperature = 25$^\circ$C, relative humidity = 70%, wind speed at 2 m above the ground surface is 12 km/h, and saturation vapor pressure at 25$^\circ$C is 23.76 mm of Hg. Take Meyer's coefficient as 0.36.

## Short answer

Wind speed at 9 m is 14.88 km/h; Meyer's formula gives 4.95 mm/day. Over 30 days on 2 km², evaporation loss is about 2.97×10⁵ m³.

## Answer

Meyer's formula:

$$
E_L = K_M (e_w - e_a)\left(1 + \frac{u_9}{16}\right)\ \text{mm/day}
$$

where $u_9$ is the wind speed in km/h at 9 m above the ground. The given wind is at 2 m, so it is converted by the 1/7 power law.

#### Step 1: Wind speed at 9 m
$$
u_9 = u_2\left(\frac{9}{2}\right)^{1/7} = 12\times1.2397 = 14.88\ \text{km/h}
$$

#### Step 2: Vapour pressures
$e_w = 23.76$ mm Hg; $e_a = RH\times e_w = 0.70\times23.76 = 16.632$ mm Hg; $e_w - e_a = 7.128$ mm Hg

#### Step 3: Evaporation rate
$$
E_L = 0.36\times7.128\times\left(1 + \frac{14.88}{16}\right) = 4.952\ \text{mm/day}
$$

#### Step 4: Volume in March (30 days)
Depth $= 4.952\times30 = 148.6$ mm $= 0.1486$ m

$$
V = 0.1486\times 2\times10^6 = 297,117\ \text{m}^3
$$

**Answer: Evaporation = 4.95 mm/day, so the volume evaporated in March is about 297,117 m³ (≈ 2.97 × 10⁵ m³).**

(If the 2 m wind is used directly without correction, the answer is 2.69 × 10⁵ m³.)

## Check yourself

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

1. The 2 m wind speed is converted to 9 m using:
   A. Multiply by 9/2
   B. The 1/7 power law
   C. Subtract 2
   D. No change
   Answer: B. u9 = u2 (9/2)^(1/7) = 14.88 km/h. (Key point: Wind correction)

2. ea in this problem is:
   A. 23.76 mm Hg
   B. 16.632 mm Hg
   C. 7.128 mm Hg
   D. 33.9 mm Hg
   Answer: B. ea = 0.70 x 23.76 = 16.632, so ew - ea = 7.128. (Key point: Vapour pressure)

3. Meyer's evaporation rate is:
   A. 4.95 mm/day
   B. 7.13 mm/day
   C. 0.36 mm/day
   D. 14.9 mm/day
   Answer: A. 0.36 x 7.128 x (1 + 14.88/16) = 4.95 mm/day. (Key point: Rate)

4. The March volume is about:
   A. 297,117 m3
   B. 4952 m3
   C. 148.6 m3
   D. 2.97 x 10^7 m3
   Answer: A. 0.1486 m x 2 x 10^6 m2. (Key point: Volume)

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