Penman Equation Empirical Method

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PENMAN EQUATION EMPIRICAL METHOD Mat Jasper Basil Penman Equation to Estimate Potential Evapotranspiration (PET) Penman’s equation is based on sound theoretical reasoning and is obtained by a combination of the energy – balance and mass – transfer approach. - It is used to estimate evaporation from water, and land Penman Equation to Estimate Potential Evapotranspiration (PET) Eep – potential evapotranspiration(mm/day) Rn – net radiation(MJ/m/d) G – heat flux density to the ground ll –– latentlatent heatheat ofof vaporization(MJ/kg)vaporization(MJ/kg) U2 – wind speed(m/sec) ∆ - slope of the saturation vapor pressure- temperature curve(kpa/⁰ C) g - psychometric constant(kpa/⁰ C) P – atmospheric pressure es-ed - vapor pressure deficit ∆ = 0.2(0.00738T + 0.8072)7 – 0.000116 P = 101.3 – 0.01055H g = Cp P 0.622l l = 2.501 – 2.361x10-3 T (Ti+1 – Ti-1) G = 4.2 -------------∆ t Example: Estimate the Eep for Aug 2, 1992 from a field close to Hoytville, OH, using Penman’s method. Radiation data, wind speed and pan evaporation were measured at the site. The other weather data are taken from the Hoytville daily summaries published by NOAA: Tavg(Aug 2)=18.6ᵒ C; Tmax(Aug 2) = 24.4ᵒ C; Tmin(Aug 2) = 12.8ᵒ C; Tavg(Aug 1) = 15ᵒ C; Tavg(Aug 3) = 20.6ᵒ C; ed(8am) = 1.1759kPa; U2 = 1.83m/sec; 2 R1 = 19.33MJ/m /d; ELEV = 213.4m and lat = 41ᵒ N Solution Calcute: for G (20.6 -15.0) G = 4.2 ------------------- = 11.76MJ/m2/d 2 Calculate l l = (2.501 – 2.361x10-3)(18.6) = 2.4571MJ/Kg Calculate for P P = 101.3 – 0.01055(213.4) = 99.05kPa 0.001013(99.05) g =----------------------- =0.0657 0.622(2.4571) Calculate ∆ ∆ = 0.2[0.00738(18.6) + 0.8072]7 – 0.000116 = 0.134 Rn = 10.63MJ/m2/d es-ed = 1.0927 kPa 0.134(10.63 – 11.76) 0.657 + 0.134+0.0657 0.134+0.0657 (6.43)(1+0.53(1.83))(1.0927) 2.4571 Eep = Eep = 1.54mm/d Empirical Method Empirical equations are available to estimate lake evaporation using commonly available meteorological data. Empirical Evaporation Equation Meyer’s Formula: EL = KM (ew-ea)(1+u9/16) Meyer’s formula EL = KM (ew-ea)(1+u9/16) EL – lake evaporation in mm/day Ew - saturated vapor pressure at the water –surface temperature in mm of mercury Ea – actual vapor pressure of overlying air at a specific height in mm of mercury u9 – monthly mean wind velocity in km/h at about 9m above ground KM – coefficient accounting for various other factors with a value of 0.36 for large deep waters and 0.50 for small, shallow waters Example A reservoir with a surface area of 250 hectares had the following average values of climate parameters during a week: Water temperature = 20⁰ C Relative humidity = 40% Wind velocity at 1m above ground surface = 16km/h. saturated vapor pressure = 17.54mm Actual vapor pressure = 7.02mm Estimate the average daily evaporation from the lake. Solution: Ew = 17.54mm Ea = 7.02mm 1/7 U9 = U1 x (9) = 21.9km/h EL = 0.36(17.53 – 7.02)(1 + 21.9/16) EL = 8.97mm/day.

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Homework No. 4 - Solution
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