First of all sorry for my bad English …
I have to calculate the specific weight of air knowing that in the air there is actually a small percentage of water vapor.
Therefore the specific weight of the humid air is:
dry air + water vapor
I want to know specific weight of air with temperature 293.15 K, pressure 101325 Pascal and RH 10% (relative humidity)
I found formulas but I don't know if they are correct…
1) I have to calculate the saturation pressure of the steam:
"Psat"= 6.0178*(10^(((7.5*T)-2048.25)/(T-35.85)))
.
.
2) I have to calculate the partial pressure of the steam:
"Pv"= ("RH"/100)*"Psat"
.
.
3) Specific weight of dry air:
"Pd"=("P"-"Pv")/(287.058*T)
.
.
.It would be the ideal gas formula as a function of Temperature (T), gas constant (R), molecular weight (PM)
4) Specific weight of steam
"Pv2"="Pv"/(461.495*T)
.
.
Where data of humid air are:
a= 0,1358 costant van der waals
b= 0,000036 costant van der waals
PM= 28,962788 molecular weight
R= 8,314472 costant
P= 101325 Pa pressure
T= 293,15 K temperature
RH= 10% realtive humidity
Where data of water are:
a= 0,548 costant van der waals
b= 0,000031 costant van der waals
PM= 18,0153 molecular weight
R= 8,314472 costant
P= 230,36 Pa pressure
T= 293,15 K temperature
My doubts are also about the constants and the molecular weight of humid air and water.
Anyway with these formulas and these data we have:
Saturation pressure of steam: 23,113848 Pa
Partial pressure of the steam: 2,3113848 Pa
Specific weight of dry air: 1,204057292 kg/m3
Specific weight of steam: 0,000017085 kg/m3
Final specific weight of humid air is: 1,20407437 kg/m3
So formulas are correct?
a and b constant (van der Walls) e molecular weight are correct ?
Thanks a lot and sorry for my english 🙂
Giovanni
Best Answer
What you did is fine. But here is a simpler approach.
Based on 21% oxygen and 79% nitrogen, the molecular weight of air is 28.8. At 20 C, the saturation vapor pressure of water is 2333 Pa. So the partial pressure of the water vapor in the air is 233 Pa.
Partial pressure of air is 101325 - 233 = 101092 Pa.
Mole fraction of water = 233/101325 = 0.0023
Mole fraction of air = 1 - 0.0023 = 0.9977
Molecular weight of humid air = (28.8)(0.9977)+(18)(0.0023)= 28.8
Density of moist air = $\frac{pM}{RT}=\frac{(101325)(28.8)}{(8.314)(293)}=1198\ \frac{gm}{m^3}=1.20\ \frac{kg}{m^3}$