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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

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  • $\begingroup$ But humid air is lighter than dry air.... $\endgroup$ – Pieter Mar 12 at 13:24
  • $\begingroup$ Why are you bothering with the van der waals equation when water and air are pretty much ideal gases under these conditions? What is the mole fraction water? What is the molar average molecular weight of the mixture? $\endgroup$ – Chet Miller Mar 12 at 13:42
  • $\begingroup$ i know only themperature and pressure, i want to calculate specific weight of air of my room... i hope can you help me $\endgroup$ – Borja Mar 12 at 13:47
  • $\begingroup$ @ChesterMiller where is the problem for you ? $\endgroup$ – Borja Mar 12 at 13:53
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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 = $\frac{pM}{RT}=\frac{(101325)(28.8)}{(8.314)(293)}=1198\ \frac{gm}{m^3}=1.20\ \frac{kg}{m^3}$

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  • $\begingroup$ Oh yes is more simpler :D but i don't understand why in previous comment an user said that humid air is lighter than dry air. My data are correct because dry air is calculate with 10% oh humidity. Right ? Now i have doubt only about "specific weight of steam" $\endgroup$ – Borja Mar 12 at 15:03
  • $\begingroup$ What you did is correct, although your calculated saturation vapor pressure of water is low by a factor of 100. Regarding what the other user said, dry air at the same total pressure as moist air has a higher density. If you redo you calculation for purely dry air, your calculation will confirm this. $\endgroup$ – Chet Miller Mar 12 at 15:27
  • $\begingroup$ thanks a lot :) $\endgroup$ – Borja Mar 12 at 15:58

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