Daily lake evaporation by Penman, 30 N in April

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.