Question

Using Penman’s method, estimate the Etp for July 15, 1994, from a field at an elevation of 200 m. The average temperature for July 15 was 16° C, with a maximum T of 22° C and a minimum of 10° C. The previous day’s average temperature was 14° C, and the following day’s average temperature was 21° C. The early morning actual vapor pressure on July 15 was 1.3 kPa: the wind speed was 2 m/sec, and Rn was 15 MJ/m2/d. (heat index was found to be 40 and saturated vapor pressure was found to be 2.0 kPa)

Answer 1

Tmean = (Tmax+Tmin)/2 = (22+10)/2 = 16 deg C

Wind speed at 100 m elevation, U100 = 2 m/s

Required Wind speed at Height 2m, U2=Uh*4.87/(ln(67.8*h-5.42)) = 2*4.87/(ln(67.8*100-5.42)) = 1.1 m/s

Slope of Saturation vapour pressure curve,

Therefore, Delta = 0.11

Atmospheric pressure, P = 101.3*(293-0.0065*z)/293)^5.26

Z=200 m

Therefore, P = 98.95 Kpa

Psychometric constant, γ=0.00065*P = .065

Delta Term for radiation, DT =Delta/(delta+γ*(1+.34*U2) = 0.55

PSI term for radiation, PT = γ/(Delta+γ(1+0.34*U2) = 0.32

Temperature term for wind, TT = (900/(Tmean+273))*U2 = 3.42

Saturation Vapor Pressure, e(T) = 0.61*exp(17.27*T/(T+237.3)) = 1.78

Similarly e(Tmax) = 2.64

e(Tmin) = 1.22

Mean Saturation vapor pressure, e(s) = (e(Tmax)+e(Tmin))/2=1.92

Actual vapor pressure, e(a) = e(Tmin) = 1.22

ETP(wind) = PT*TT*(e(s)-E(a))) = 0.32*3.42*(1.92-1.22) = 0.76 mm/day

Rng = 0.48*Rn

Given Rn = 15 MJ/m2/d

Therfore, Rng = 7.2 MJ/m2/d

ETP radiation = Rng*DT = 7.2*0.55 = 3.96 mm/day

ETP = ETP radiation + ETP wind = 3.96+.76 = 4.72 mm/day