PET by Penman formula, Simara 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

• $T = 23\,^\circ$C $\Rightarrow T_a = 296$ K; RH $= 75\%$; $n = 10$ h; $N = 12.5$ h; $u_2 = 85$ km/day
• $H_a = 15.0$ mm/day; $r = 0.2$; $\varphi = 27.00^\circ$; $e_w = 21.04$ mm Hg; $A = 1.27$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• $r = 0.20$ is taken for a closed green crop (range 0.15-0.25) and $\gamma = 0.49$ mm Hg/$^\circ$C is assumed, since they are not given. Elevation (107 m) has a negligible effect and is neglected. $b = 0.52$.

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.75\times21.04 = 15.78\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(27.00^\circ) = 0.2584$, and $n/N = 0.8000$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 15.0\times0.800\times(0.2584 + 0.52\times0.8000) = 8.093\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(296)^4 = 15.430\\ \text{Back radiation} &= 15.430\times(0.56 - 0.092\sqrt{15.78})\times(0.1 + 0.9\times0.8000) = 2.461\\ H_n &= 8.093 - 2.461 = 5.631\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{85}{160}\right)(21.04 - 15.78) = 2.819\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.27\times5.631 + 2.819\times0.49}{1.27 + 0.49} = \frac{7.152 + 1.381}{1.760} = 4.85\ \text{mm/day}$$

Answer: PET = 4.85 mm/day, about 145 mm for April.