PENMAN EQUATION EMPIRICAL METHOD

Mat Jasper Basil Penman Equation to Estimate Potential (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 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- 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 = 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