I have been posed by many young entry level engineers about gas density calculations which I do in a jiffy using a calculator. I will try to summarize these below:

As a basic we are all supposed to know that gas density is calculated as:

rho = P*M / R*T*Z ------------- 1

or

rho = M / ((R*T*Z) / P) ----------- 2

where:

rho = gas density

M = Molecular weight of gas

R = Universal gas constant

T = Absolute Temperature

P = Absolute Pressure

For different standard conditions the denominator in equation 2, i.e. (R*T*Z/P) varies and below are a few examples:

Standard Conditions 1:

P = 1.01325 bar (abs), T = 15.6 deg C = 288.75 K, Z =1

R = 0.0831447 bara.m

^{3}/kmole.K

**rho = M / 23.7 kg/m**

^{3}Standard Conditions 2:

P = 1.01325 bar (abs), T = 0 deg C = 273.15 K, Z =1

R = 0.0831447 bara.m

^{3}/kmole.K

**rho = M / 22.414 kg/m**

^{3}Note: These are also called normal conditions and the volume defined as Nm

^{3}. The term

**normal**is archaic in today's definition of standard conditions of temperature and pressure.

Standard Conditions 3:

P = 1.01325 bar (abs), T = 15 deg C = 288.15 K, Z =1

R = 0.0831447 bara.m

^{3}/kmole.K

**rho = M / 23.64 kg/m**

^{3}Standard Conditions 4:

P = 14.696 psia, T = 60 deg F = 519.67 R, Z =1

R = 10.7316 psia.ft

^{3}/lbmole.R

**rho = M / 379.5 lb/ft**

^{3}**Note: The compressibility factor Z is practically unity at all standard conditions mentioned above.**

The utility of the standard density is to convert standard volume flow in Sm

^{3}or Scf to mass flow by multiplying it with the standard density, given the standard conditions. It is always good engineering practice to specify the standard conditions in the "Design Basis" document of the particular project being executed. For more details on standard conditions of temperature and pressure please refer the following link:

http://en.wikipedia....re_and_pressure

Regards,

Ankur.