Humid air – ravnik.eu

Problem

We consider humid air at the temperature of $20^\circ C$ and pressure $1~bar$ . Humidity expressed with the mixing ratio is $x=0.014$ . Estimate absolute humidity $\rho_w$ , relative humidity $\phi$ , mass fraction of water vapour $\xi_w$ and molar fraction of water vapour $\psi_w$ in air!

Data

$x=0.014$
$T=293~K$
$p=10^5~Pa$
$M_w=18~kg/kmol$
$M_a=29~kg/kmol$
$\rho_w=?$
$\phi=?$
$\xi_w=?$
$\psi_w=?$

Solution

Mass fraction (mass of water vapour / mass of humid air) is connected to mixing ratio (mass of water vapour / mass of dry air) by

$\xi_v=\frac{x}{1+x}=0.0138.$

Humid air is a mixture of air and water vapour, so the mass fraction of air is $\xi_z=1-\xi_v=0.9862$ . Since mass fractions are now know, we can estimate the molar fraction of water vapur as

$\psi_v=\frac{\xi_w/ M_w}{\sum_j\xi_j/M_j}=\frac{\xi_w/ M_w}{\xi_w/M_w+\xi_a/M_a}=0.022,$

where $M_w$ and $M_a$ are the molar masses of water and air, respectively. Molar fraction can be used to estimate partial pressure of water vapour:

$p_w=\psi_wp=2205~Pa.$

Using an engineering handbook we find that the saturated partial pressure of water vapour at temperature $20^\circ C$ and pressure $1~bar$ is $p_{w,s}=2337~Pa$ . Using this, we calculate the relative humidity as

$\varphi=\frac{p_v}{p_{v,s}}=0.94.$

Finally, absolute humidity (i.e. the mass concentration of water vapour in the air) is calculated by assumig that air is an ideal gas

$\rho_v=\frac{p_wM_w}{RT}=0.016~kg/m^3.$

Here $R=8314~J/(kmolK)$ .

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