RH CGS-240 - Page 9

Theory of operation
CGS-240 Manual 6/15/2021
Page 9 of 148
a
2
= –4.1935419·10
–6
a
2
= –3.5912529·10
–6
a
3
= 6.2038841·10
–9
a
3
= 5.0194210·10
–9
b
0
= –1.0385289·10
2
b
0
= -8.2308868·10
1
b
1
= 8.5753626·10
–1
b
1
= 5.6519110·10
–1
b
2
= –2.8578612·10
–3
b
2
= –1.5304505·10
–3
b
3
= 3.5499292·10
–6
b
3
= 1.5395086·10
–6
2.3.4 TEMPERATURE FROM SATURATION VAPOR PRESSURE
Equations 1 and 2 are easily solved for saturation vapor pressure over water or ice for a given saturation
temperature. However, if vapor pressure is known and temperature is the unknown desired quantity, the solution
immediately becomes complicated and must be solved by iteration. For ease of computation, the following inverse
equation is provided. This equation is generally used to find the dew point or frost point temperature when the
vapor pressure of a gas has been determined. When vapor pressure is known, use the water coefficients to obtain
the dew point and use the ice coefficients to obtain the frost point.
( )
( )


=
=
=
3
0
3
0
ln
ln
i
i
i
i
i
i
ed
ec
T
(7)
where T is the temperature in kelvin
and e is the saturation vapor pressure in pascals
with coefficients
for water: for ice:
c
0
= 2.0798233·10
2
c
0
= 2.1257969·10
2
c
1
= –2.0156028·10
1
c
1
= –1.0264612·10
1
c
2
= 4.6778925·10
–1
c
2
= 1.4354796·10
–1
c
3
= –9.2288067·10
–6
c
3
= 0
d
0
= 1 d
0
= 1
d
1
= –1.3319669·10
–1
d
1
= –8.2871619·10
–2
d
2
= 5.6577518·10
–3
d
2
= 2.3540411·10
–3
d
3
= –7.5172865·10
–5
d
3
= –2.4363951·10
–5