Flowing Gas Density
The density of a gas is given by:

G is the specific gravity of the gas, which is given by

P is the absolute pressure (PSIA) which is equal to line pressure + local barometric pressure

Z is the gas compressibility factor

T is the absolute temperature (ºR) which is equal to line temperature + 459.67ºF

The molecular weight of air is 28.9625, the molecular weight of any given gas can be found from many reference sources. (e.g.) Flow Measurement Engineering Handbook by Richard W. Miller or Perry and Chilton's Chemical Engineer's Handbook.

The compressibility factor for any given gas may be calculated using various methods or can be looked up in standard reference sources. (e.g.) Flow Measurement Engineering Handbook by Richard W. Miller or Matheson Gas Data Book.

Base Gas Density
The gas density equation remains the same but is referenced to what are known as standard temperature and pressure conditions (STP). The reference conditions used in general industry are 60ºF and 14.696 PSIA. The reference conditions used vary from country to country and from industry to industry. In the U.S. gas industry typical reference conditions are 60ºF and 14.73 PSIA.

Volumetric and Mass Flow
A flow rate expressed in volume units, (e.g.)ft³ per minute, is known as volumetric flow. A flow rate expressed in mass units, (e.g.) lb per minute, is known as mass flow. These flow units are linked together in the mass conservation equation.

Converting Between Flowing and Base Units
By using the conservation of mass equation shown above we can easily convert between flowing and base conditions. The following example illustrates the procedure.

Example
Carbon dioxide is flowing through a pipe at a rate of 100 ft³ per minute (ACFM), the line pressure is 80 PSI gauge and the line temperature is 50ºF. The installation is situated at sea level. Calculate the flow rate in standard ft³ per minute (SCFM)

From the Flow Measurement Engineering Handbook appendix D we find that the molecular weight of air is 28.9625 and that of carbon dioxide is 44.01. The barometric pressure at sea level is 14.696 PSIA.

From equation 2 we find that the S.G. of the gas is

The absolute pressure P = 80 + 14.696 = 94.696

The absolute temperature T = 50 + 459.67 = 509.67

The compressibility at pressures below 150 PSIA can be assumed to be 1

Substituting the above values into equation 1 we get

Standard conditions are defined as 14.696 PSIA and 60ºF. To find the density of carbon dioxide at these conditions we substitute the values into equation 1. This gives a density of 0.12 lb per ft³.

By substituting flow and density values into equation 3, the conservation of mass equation, we have the following.

Molecular Weights for Gas Mixtures
When gas mixtures are being considered there are no readily available reference tables, it is then necessary to calculate the molecular weight by using the simple combination procedure shown below.

Example
Calculate the molecular weight of the following gas mixture.

Hydrogen 10%
Nitrogen 20%
Carbon dioxide 30%
Oxygen 40%

Gas Component	Gas Fraction	Component Molecular weight	Fraction x MW (MW)
Hydrogen	0.1	2.016	0.2016
Nitrogen	0.2	28.013	5.6026
Carbon dioxide	0.3	44.010	13.2030
Oxygen	0.4	31.998	12.7992
	Sum of MW products		31.8064

Specific Heat of a Gas (Cp)
Sometimes it is necessary to calculate the heat content of a flowing gas stream, in order to do this the specific heat of the gas must be known. This data is easily obtained for common gasses by referring to the reference books mentioned above. The units of specific heat are btu per lb ºF

The heat of the gas stream is given by:

Specific Heat of a Gas Mixture (Cp mix)
The specific heat of a gas mixture is calculated as follows:

Specific heat of mixture (Cp mix) =
(fraction of component 1 x Cp of component 1) + (fraction of component 2 x Cp of component 2) + .......etc.

Example
Calculate the specific heat of the gas mixture given earlier.

Cp of Hydrogen = 7.06 Btu/lb ºF
Cp of Nitrogen = 7.00 Btu/lb ºF
Cp of Carbon dioxide = 8.80 Btu/lb ºF
Cp of Oxygen = 7.04 Btu/lb ºF

Cp of mixture = (0.1 x 7.06) + (0.2 x 7.00) + (0.3 x 8.80) + (0.4 x 7.04) = 7.562 btu per lb ºF