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PET by Penman's method, Dang in May

Short answer

Penman's equation gave PET = 3.48 mm/day, about 108 mm for May. Net radiation was 4.40 mm/day, aerodynamic term 1.16 mm/day; weighting by slope A and psychrometric constant yielded the result.

Calculate the potential evapotranspiration for an area over Dang in the month of May by Penman's method using the following data: Latitude = 26∘^\circN; Mean temperature = 10∘^\circC; Mean relative humidity = 65%; Mean monthly solar radiation at the top of atmosphere = 15.88 mm of evaporable water/day; Wind velocity at 2 m height = 5 km/day; Psychrometric constant = 0.49 mm/∘^\circC; Mean sunshine hour = 10.1 hr; Potential sunshine hours = 13.5 hr; Reflection coefficient = 0.25; Slope of saturated vapor pressure at 10∘^\circC = 1.24 mm of Hg.

Penman's equation

PET=AHn+EaγA+γPET = \frac{A H_n + E_a \gamma}{A + \gamma} Hn=Ha(1−r)(a+bnN)−σTa4(0.56−0.092ea)(0.1+0.9nN),a=0.29cos⁡φ, b=0.52H_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 pressures

The saturation vapour pressure at 10 °C is taken from the standard table: es=9.21e_s = 9.21 mm Hg (assumed, since it is not given).

ea=0.65×9.21=5.99 mm Hge_a = 0.65\times 9.21 = 5.99\ \text{mm Hg}

Step 2: Net radiation

n/N=10.1/13.5=0.748n/N = 10.1/13.5 = 0.748, and a=0.29cos⁡(26∘)=0.2607a = 0.29\cos(26^\circ) = 0.2607.

Short-wave part:

Ha(1−r)(a+bnN)=15.88×0.75×(0.2607+0.52×0.748)=7.738 mm/dayH_a(1-r)\left(a+b\frac{n}{N}\right) = 15.88\times0.75\times(0.2607 + 0.52\times0.748) = 7.738\ \text{mm/day}

Long-wave part, with Ta=283T_a = 283 K and σ=2.01×10−9\sigma = 2.01\times10^{-9} mm/day/K4^4:

σTa4=12.89 mm/day\sigma T_a^4 = 12.89\ \text{mm/day} 12.89×(0.56−0.0925.99)×(0.1+0.9×0.748)=12.89×0.335×0.773=3.339 mm/day12.89\times(0.56 - 0.092\sqrt{5.99})\times(0.1+0.9\times0.748) = 12.89\times0.335\times0.773 = 3.339\ \text{mm/day} Hn=7.738−3.339=4.399 mm/dayH_n = 7.738 - 3.339 = 4.399\ \text{mm/day}

Step 3: Aerodynamic term

Ea=0.35(1+5160)(9.21−5.99)=0.35×1.031×3.22=1.163 mm/dayE_a = 0.35\left(1+\frac{5}{160}\right)(9.21 - 5.99) = 0.35\times1.031\times3.22 = 1.163\ \text{mm/day}

Step 4: PET

PET=1.24×4.399+0.49×1.1631.24+0.49=5.455+0.5701.73=3.48 mm/dayPET = \frac{1.24\times4.399 + 0.49\times1.163}{1.24+0.49} = \frac{5.455+0.570}{1.73} = 3.48\ \text{mm/day}

For May (31 days): 3.482×31≈1083.482\times 31 \approx 108 mm.

Answer: PET = 3.48 mm/day, about 108 mm for May.

✓ Answer checked by Claude Sonnet 5.5

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