Saturation Vapor Pressure

Given T in Kelvin these equations will give es in milliBar.

from Fleagle and Businger, Vol.5, pg. 62 (QC880.F59)

The first law can be written as

L = T · (S2-S1) = U2 - U1 + Ps · (alpha2 - alpha1)

where alpha1 ident 1/rho1 and index 1 refers to the liquid phase and index 2 refers to the gas phase.

For an isothermal change of phase, the Clausius-Clapeyron equation has the form

d Ps/d T = L/[T · (alpha2-alpha1)]

Water vapor behaves like an ideal gas and alpha2 gg alpha1 for a change in state.

L approx 2.5 · 103 Joules/gm

L approx 2.824 · 103 Joules/gm over ice

Rw = R*/<mw>w = 8.3143/18.016 = 0.4615 Joules/gm/K

Ps = rho2 · Rw T

d Ps/d T = L/[T · (alpha2-alpha1)] appeq (L · rho2)/T = (L · Ps) /(Rw · T2)

d log(es) = d Ps/Ps = (L/Rw) · (dT/T2)

loge(es) = integ (L/Rw) · (dT/T2) = -(L/Rw)|T->T0 + C = -L/(Rw · T0) + L/(Rw· T) + C

at triple point all 3 phases can exist in equilibrium, 0.0098° C and Ps = 6.11 mB

es(T=T0) = 6.11

L/Rw = 5417.12

L/(Rw T0) = 19.8313

6.11 · exp[ L/(Rw T0) ] = 2.504 · 109

es(T) = 6.11 · exp[ 5417( 1/T0 - 1/T) ] = 2.504 · 109 · exp[ 5417/T ]

Undocumented fit is used in the program watsat.F (over liquid)

es = 2.229 · 109 · exp[ -5385/T ]

Note this is the same equation as above, except that it assumes L = 2485.2 Joules/gm and T0 = 273.15 ° K

Another undocumented fit is given (but not used) in the program watsat.F

es = 0.001 · exp[ a/T + b + c log(T) + d · T + e · T2 ]



coefover iceover water
a-5631.1206-2313.0338
b-8.363602-164.03307
c8.231238.053682
d-3.861449 · 10-2-1.3844344 · 10-1
e2.77494 · 10-57.4465367 · 10-5


From Rogers and Yau, pg. 16

es = 6.112 · exp[ a · (T-273.16)/(T-b) ]



coefRogers & Yau
over water
a17.67
b29.66

Murray, F.W. 1966. ``On the computation of Saturation Vapor Pressure'' J. Appl. Meteor. 6 p.204

es = 6.1078 · exp[ a · (T-273.16)/ (T-b) ]



coefMurray
over ice
Murray
over water
a21.874558417.2693882
b7.6635.86

Saucier, W.J. 1883. ``Principles of Meteorological Analysis'' Dover pg. 9 who uses values of Tetens (1930). Note, he used 10[(a'T)/(T-b')] with T in Centigrade so to convert into the form above a = log 10 · a' and b = 273.16-b'.



coefover iceover water
a21.87517.27
b7.6635.86