PET by Penman method, Kathmandu in February

Penman's method combines the energy-balance and aerodynamic (mass-transfer) approaches:

$$PET = \frac{A H_n + E_a \gamma}{A + \gamma}$$

where $A$ = slope of the saturation vapour pressure curve, $\gamma$ = psychrometric constant, $H_n$ = net radiation (mm of water/day) and $E_a$ = drying power of air (mm/day).

$$H_n = H_a(1-r)\left(a + b\frac{n}{N}\right) - \sigma T_a^4\left(0.56 - 0.092\sqrt{e_a}\right)\left(0.1 + 0.9\frac{n}{N}\right), \qquad a = 0.29\cos\varphi,\ b = 0.52$$

Step 1: Vapour pressure

$$e_a = RH \times e_s = 0.70 \times 11.4 = 7.98\ \text{mm Hg}$$

Step 2: Net radiation $H_n$

Sunshine ratio $n/N = 7/11.9 = 0.588$, and $a = 0.29\cos(28.5^\circ) = 0.2549$.

Short-wave part:

$$9 \times (1-0.15)\times(0.2549 + 0.52\times0.588) = 7.65 \times 0.5607 = 4.290\ \text{mm/day}$$

Long-wave part, with $T_a = 12.5 + 273 = 285.5$ K:

$$\sigma T_a^4 = 2.0\times10^{-9}\times 285.5^4 = 13.29\ \text{mm/day}$$ $$13.29\times(0.56 - 0.092\sqrt{7.98})\times(0.1 + 0.9\times0.588) = 13.29\times0.3001\times0.6294 = 2.510\ \text{mm/day}$$ $$H_n = 4.290 - 2.510 = 1.780\ \text{mm/day}$$

Step 3: Drying power of air

$$E_a = 0.35\left(1 + \frac{u_2}{160}\right)(e_s - e_a) = 0.35\times\left(1+\frac{120}{160}\right)\times(11.4-7.98) = 2.095\ \text{mm/day}$$

Step 4: PET

$$PET = \frac{1.24\times1.780 + 0.49\times2.095}{1.24+0.49} = \frac{2.207+1.027}{1.73} = 1.87\ \text{mm/day}$$

For February (28 days): $1.869\times28 \approx 52.3$ mm.

Answer: PET = 1.87 mm/day, about 52 mm for February.

(The latitude is used only to find $a = 0.29\cos\varphi$; the longitude is not needed. The extraterrestrial radiation is already given.)