PET by Penman method for May

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 $= 75\%$; $n = 10$ h; $N = 13.5$ h; $u_2 = 192$ km/day
• $H_a = 14.4$ mm/day; $r = 0.028$; $\varphi = 27.00^\circ$; $e_w = 11$ mm Hg; $A = 1.42$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• $H_a = 14.4$ is taken as mm of evaporable water per day. $\gamma = 0.49$ is assumed (not given). $e_w$ and $A$ are taken as given.

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.75\times11 = 8.25\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(27.00^\circ) = 0.2584$, and $n/N = 0.7407$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.4\times0.972\times(0.2584 + 0.52\times0.7407) = 9.008\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(293)^4 = 14.814\\ \text{Back radiation} &= 14.814\times(0.56 - 0.092\sqrt{8.25})\times(0.1 + 0.9\times0.7407) = 3.359\\ H_n &= 9.008 - 3.359 = 5.649\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{192}{160}\right)(11 - 8.25) = 2.118\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.42\times5.649 + 2.118\times0.49}{1.42 + 0.49} = \frac{8.022 + 1.038}{1.910} = 4.74\ \text{mm/day}$$

Answer: PET = 4.74 mm/day, about 147 mm for the 31 days of May.