Reservoir evaporation by Meyer's formula

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.)