PET by Penman formula, Dharan 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

• $T = 22.5\,^\circ$C $\Rightarrow T_a = 295.5$ K; RH $= 75\%$; $n = 10$ h; $N = 12.7$ h; $u_2 = 80$ km/day
• $H_a = 14.9$ mm/day; $r = 0.2$; $\varphi = 26.82^\circ$; $e_w = 20.4$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• Latitude $26^\circ49' = 26.82^\circ$. Elevation (250 m) is not needed because $\gamma$ is given. The closed green crop is represented by $r = 0.20$ (given).

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.75\times20.4 = 15.30\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(26.82^\circ) = 0.2588$, and $n/N = 0.7874$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.9\times0.800\times(0.2588 + 0.52\times0.7874) = 7.966\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(295.5)^4 = 15.326\\ \text{Back radiation} &= 15.326\times(0.56 - 0.092\sqrt{15.30})\times(0.1 + 0.9\times0.7874) = 2.480\\ H_n &= 7.966 - 2.480 = 5.485\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{80}{160}\right)(20.4 - 15.30) = 2.677\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.24\times5.485 + 2.677\times0.49}{1.24 + 0.49} = \frac{6.802 + 1.312}{1.730} = 4.69\ \text{mm/day}$$

Answer: PET = 4.69 mm/day, so about 141 mm for April (30 days).