Roskam in book "Airplane Aerodynamics and Performance" explains (in paragraph "1.1 ATMOSPHERIC FUNDAMENTALS"):

To determne how water vapor affects air density, consider a given volume filled with a dry air and water vapor mixture. Dalton's law of partial pressures states that the obvious pressure, $p$, of the mixture equals the sum of the dry air pressure, $p_a$ and the water vapor pressure, $p_v$:

$$p = p_a + p_v \qquad \mbox{(1.2)}$$

Also, since the total mass of air inside the volume is equal to the sum of the dry air mass and the water vapor mass, the observed density of the mixture is given by:

$$\rho = \rho_a + \rho_v \qquad \mbox{(1.3)}$$

In Eqn. (1.3) it is also assumed that the distribution of masses of dry air and water vapor is uniform. By using the perfect gas law, $p=\rho g R T$, it is possible to show that the density of the mixture is given by:

$$\rho=\frac{p-p_v}{gRT}+\frac{p_v}{gR_vT} = \frac{1}{gRT}\left( p-p_v\left( 1 - \frac{R}{R_v}\right)\right) \qquad \mbox{(1.4)}$$

For water vapor or steam, $R_v$=85.89 ft/°R. It follows that $R < R_v$ so that second term in equation (1.4) is negative. Therefore, the density of a mixture of dry air and water vapor is less than that of dry air. Normally the effect of water vapor on air density is small. To illustrate this fact, assume that the observed temperature and pressure of a mixture of dry air and water vapor are 90 °F and 2116.2 lb/ft^2 respectively. If the relative humidity is 100%, the water vapor pressure can be found to be 100.6 lbs/ft^2. The density of the mixture then follows from Eqn (1.4) as:

$$\rho =\frac{1}{53.35*32.17*549.7}(2116.2-100.6*0.379)=0.002203 \mbox{slug}/ft^3 \qquad \mbox{(1.5)}$$

The corresponding dry air density is 0.002243 slugs/ft^3. The density reduction in this case is therefore only 1.8%.

Granted that in the equations above the term "R" is referred to dry air, in my opinion the sentence "The corresponding dry air density is 0.002243 slugs/ft^3" is not correct because in my calculation I found 0.002130 with the following calculation: $$\begin{array}{} \rho_a & =\frac{p_a}{gRT} \\ \rho_a & =\frac{p-p_v}{gRT} \\ \rho_a & =\frac{2116.2-100.6}{32.17*53.5*549.7} \\ \rho_a & =0.002130 \end{array} $$

So, can someone explain if I'm wrong?

  • $\begingroup$ You seem to have calculated the density of something like "100% water vapor and 0% air" at 90F. That's not what "Relative humidity of 100%" means. (But I'm not going to attempt to check what you did in any more detail using a crazy units system that only survives because one country doesn't want to be consistent with the rest of the world.) $\endgroup$ – alephzero Nov 16 '16 at 2:30
  • $\begingroup$ I think I used exactly the same formula used by Roskam when he writes in Eqn (1.4) $$\rho=\frac{p-p_v}{gRT}+\frac{p_v}{gR_vT} $$ ; it intended, in my opinion, that the first addendum is $$\rho_a$$ $\endgroup$ – d.pensopositivo Nov 16 '16 at 16:23
  • $\begingroup$ Does someone know if an email address exists in order to write to someone from Professor Roskam's staff in order to have an answer? $\endgroup$ – d.pensopositivo Nov 21 '16 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.