PET by Penman method, Kathmandu in March

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

• $T = 10.0\,^\circ$C $\Rightarrow T_a = 283.0$ K; RH $= 60\%$; $n = 9$ h; $N = 12.9$ h; $u_2 = 120$ km/day
• $H_a = 11$ mm/day; $r = 0.25$; $\varphi = 28.50^\circ$; $e_w = 9.2$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• $H_a = 11$ is taken as mm of evaporable water per day, wind $5\ \text{km/h} = 120$ km/day, and albedo is used as $r$. $b = 0.52$.

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.60\times9.2 = 5.52\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(28.50^\circ) = 0.2549$, and $n/N = 0.6977$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 11\times0.750\times(0.2549 + 0.52\times0.6977) = 5.096\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(283.0)^4 = 12.893\\ \text{Back radiation} &= 12.893\times(0.56 - 0.092\sqrt{5.52})\times(0.1 + 0.9\times0.6977) = 3.227\\ H_n &= 5.096 - 3.227 = 1.869\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{120}{160}\right)(9.2 - 5.52) = 2.254\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.24\times1.869 + 2.254\times0.49}{1.24 + 0.49} = \frac{2.317 + 1.104}{1.730} = 1.98\ \text{mm/day}$$

Answer: PET = 1.98 mm/day, which is about 61 mm for the 31 days of March.