PureBasic Forums - English

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<math>R_{dry air} = 287.05 \frac{\mbox{J}}{\mbox{kg} \cdot \mbox{K}}<math> 

The pressure at altitude h is given by:

    <math>p = p_0 \cdot (1 + \frac{L \cdot h}{T_0})^\frac{g \cdot M}{R \cdot -L}<math> 

Density can then be calculated according to a molar form of the original formula:

    <math>\rho = \frac{p \cdot M}{R \cdot T}<math> 


    $ \rho_{\,\mathrm{humid~air}} = \frac{p_{d}}{R_{d} \cdot T} + \frac{p_{v}}{R_{v} \cdot T} \, $[1] 

where:

    $ \rho_{\,\mathrm{humid~air}} = $ Density of the humid air (kg/m³) 
    $ p_{d} = $ Partial pressure of dry air (Pa) 
    $ s_{d} = $ Partial sound of dry air (Pa) 
    $ R_{d} = $ Specific gas constant for dry air, 287.05 J/(kg·K) 
    $ T = $ Temperature (K) 
    $ p_{v} = $ Pressure of water vapor (Pa) 
    $ s_{v} = $ Sound of water vapor (dB) 
    $ R_{v} = $ Specific gas constant for water vapor, 461.495 J/(kg·K) 

Code: Select all

R_{dry air} = 287.05 \frac{\mbox{J}}{\mbox{kg} \cdot \mbox{K}}