Skip to main content

Free water evaporation by Penman method, June

Short answer

Free water surface evaporation = 9.2 mm/day (about 275 mm in June).

Calculate the free water surface evaporation in June using the Penman method from an area whose latitude is approximately 33∘^\circN. The available data include air temperature = 30∘^\circC, wind speed at 2 m height = 10 km/h, relative humidity = 60%, mean observed sun shine hours = 12 and reflection coefficient = 0.05.

Free-water evaporation is obtained from Penman with r=0.05r = 0.05 for a water surface.

Penman equation (potential evapotranspiration):

PET=A Hn+Ea γA+γPET = \frac{A\,H_n + E_a\,\gamma}{A + \gamma}

with

Hn=Ha(1−r)(a+bnN)−σTa4 (0.56−0.092ea)(0.1+0.9nN)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) Ea=0.35(1+u2160)(ew−ea),a=0.29cos⁡φ, b=0.52E_a = 0.35\left(1 + \frac{u_2}{160}\right)(e_w - e_a), \qquad a = 0.29\cos\varphi,\ b = 0.52

where HnH_n is the net radiation in mm of evaporable water per day, EaE_a the aerodynamic term in mm/day, u2u_2 the wind speed at 2 m in km/day, and ew,eae_w, e_a in mm of Hg.

Data and constants used

  • T=30 ∘T = 30\,^\circC ⇒Ta=303\Rightarrow T_a = 303 K; RH =60%= 60\%; n=12n = 12 h; N=14.2N = 14.2 h; u2=240u_2 = 240 km/day
  • Ha=16.9H_a = 16.9 mm/day; r=0.05r = 0.05; φ=33.00∘\varphi = 33.00^\circ; ew=31.82e_w = 31.82 mm Hg; A=1.83A = 1.83 mm Hg/∘^\circC; γ=0.49\gamma = 0.49 mm Hg/∘^\circC; σ=2.01×10−9\sigma = 2.01\times10^{-9} mm/day/K4^4
  • Assumed values (not given): ewe_w at 30∘30^\circC =31.82= 31.82 mm Hg and slope A=1.83A = 1.83 mm Hg/∘^\circC; for 33∘33^\circN in June, Ha≈16.9H_a \approx 16.9 mm/day and possible sunshine N≈14.2N \approx 14.2 h (standard tables / solar geometry); wind 10 km/h=24010\ \text{km/h} = 240 km/day; γ=0.49\gamma = 0.49.

Step 1: Actual vapour pressure

ea=RH100 ew=0.60×31.82=19.09 mm Hge_a = \frac{RH}{100}\,e_w = 0.60\times31.82 = 19.09\ \text{mm Hg}

Step 2: Net radiation HnH_n

a=0.29cos⁡(33.00∘)=0.2432a = 0.29\cos(33.00^\circ) = 0.2432, and n/N=0.8451n/N = 0.8451

Ha(1−r)(a+bnN)=16.9×0.950×(0.2432+0.52×0.8451)=10.960σTa4=2.01×10−9×(303)4=16.942Back radiation=16.942×(0.56−0.09219.09)×(0.1+0.9×0.8451)=2.304Hn=10.960−2.304=8.656 mm/day\begin{aligned} H_a(1-r)\left(a + b\tfrac{n}{N}\right) &= 16.9\times0.950\times(0.2432 + 0.52\times0.8451) = 10.960\\ \sigma T_a^4 &= 2.01\times10^{-9}\times(303)^4 = 16.942\\ \text{Back radiation} &= 16.942\times(0.56 - 0.092\sqrt{19.09})\times(0.1 + 0.9\times0.8451) = 2.304\\ H_n &= 10.960 - 2.304 = 8.656\ \text{mm/day} \end{aligned}

Step 3: Aerodynamic term

Ea=0.35(1+240160)(31.82−19.09)=11.137 mm/dayE_a = 0.35\left(1 + \frac{240}{160}\right)(31.82 - 19.09) = 11.137\ \text{mm/day}

Step 4: PET

PET=1.83×8.656+11.137×0.491.83+0.49=15.841+5.4572.320=9.18 mm/dayPET = \frac{1.83\times8.656 + 11.137\times0.49}{1.83 + 0.49} = \frac{15.841 + 5.457}{2.320} = 9.18\ \text{mm/day}

Answer: Free water surface evaporation = 9.2 mm/day, so about 275 mm in June (30 days).

✓ Answer checked by Claude Sonnet 5.5

All Engineering Hydrology chapter 3 questions →