Daily PET by Penman, latitude 30 N, 300 m

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

• $T = 15\,^\circ$C $\Rightarrow T_a = 288$ K; RH $= 70\%$; $n = 10$ h; $N = 12.9$ h; $u_2 = 50$ km/day
• $H_a = 14.8$ mm/day; $r = 0.05$; $\varphi = 30.00^\circ$; $e_w = 12.79$ mm Hg; $A = 0.82$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• Month not given: April at $30^\circ$N is assumed, giving $H_a = 14.8$ mm/day and $N = 12.9$ h. At $15^\circ$C, $e_w = 12.79$ mm Hg and $A = 0.82$ mm Hg/$^\circ$C (standard tables). $\gamma = 0.49$ mm Hg/$^\circ$C and $\sigma = 2.01\times10^{-9}$ mm/day/K⁴ are used. The elevation (300 m) changes $\gamma$ by only about 3% and is neglected.

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.70\times12.79 = 8.95\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(30.00^\circ) = 0.2511$, and $n/N = 0.7752$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 14.8\times0.950\times(0.2511 + 0.52\times0.7752) = 9.199\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(288)^4 = 13.828\\ \text{Back radiation} &= 13.828\times(0.56 - 0.092\sqrt{8.95})\times(0.1 + 0.9\times0.7752) = 3.141\\ H_n &= 9.199 - 3.141 = 6.058\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{50}{160}\right)(12.79 - 8.95) = 1.763\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{0.82\times6.058 + 1.763\times0.49}{0.82 + 0.49} = \frac{4.968 + 0.864}{1.310} = 4.45\ \text{mm/day}$$

Answer: Daily PET = 4.45 mm/day.