Daily PET by Penman, 27.5 N, 1400 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

• $T = 10\,^\circ$C $\Rightarrow T_a = 283$ K; RH $= 70\%$; $n = 7$ h; $N = 11.8$ h; $u_2 = 80$ km/day
• $H_a = 13.2$ mm/day; $r = 0.15$; $\varphi = 27.50^\circ$; $e_w = 11.4$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• Month not given: March is assumed (a cool, hill-station month at $27.5^\circ$N), for which standard tables or solar geometry give $H_a \approx 13.2$ mm/day and $N \approx 11.8$ h. $H_n$ and $E_a$ use the given $e_w$, $A$, $\gamma$, $\sigma$ and the albedo as $r = 0.15$. The elevation (1400 m) is not needed because $\gamma$ is given.

Step 1: Actual vapour pressure

$$e_a = \frac{RH}{100}\,e_w = 0.70\times11.4 = 7.98\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

$a = 0.29\cos(27.50^\circ) = 0.2572$, and $n/N = 0.5932$

$$\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 13.2\times0.850\times(0.2572 + 0.52\times0.5932) = 6.347\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(283)^4 = 12.893\\ \text{Back radiation} &= 12.893\times(0.56 - 0.092\sqrt{7.98})\times(0.1 + 0.9\times0.5932) = 2.453\\ H_n &= 6.347 - 2.453 = 3.895\ \text{mm/day} \end{aligned}$$

Step 3: Aerodynamic term

$$E_a = 0.35\left(1 + \frac{80}{160}\right)(11.4 - 7.98) = 1.796\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.24\times3.895 + 1.796\times0.49}{1.24 + 0.49} = \frac{4.829 + 0.880}{1.730} = 3.30\ \text{mm/day}$$

Answer: Daily PET = 3.30 mm/day. (The value depends on the assumed $H_a$ and $N$.)