Dear all,

I have been posed by many young entry level engineers about standard 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 Condition 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 Condition 2 (NTP):

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/Nm**

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

^{3}.

Standard Condition 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 and volume flow at normal conditions to mass flow by multiplying it with the density, given the standard conditions as mentioned above. For the upstream oil and gas the most common standard conditions that are utilized are the standard condition 3 when using SI units and standard condition 4 when using English units. As per the Wikipedia reference provided below, the standard conditions ot temperature utilized by "Organization Of Petroleum Expostring Countries" (OPEC) are 14.73 psia and 60°F. However, this only seems to be applied for natural gas transmission applications in in the United States and Canada. 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

Let me hear comments from some fresh chemical engineers who have just graduated from college.

Regards,

Ankur.

Source: Short Cut Formulas For Gas Densities At Various Std.conditions

Α.Let us try to make density calculations handy, at least for ideal gas conditions as above, remembering that quantity of 1 kmol of gas(α) has mass equal to gas molecular weight M

(β) has volume equal to molar volume Vm

As we remember from chemistry, Vm = 22.4 lt/gmol = 22.4 m3/kgmol at 0 oC and 1 Atm a (*). A more precise value is 22.414 m3/kgmol, though difference can be insignificant in practical calculations.

To calculate gas density at some standard condition (or even non standard), we have to find molar volume at this condition by ideal gas law, then divide (α)/(β). Molar gas constant R need not be used in the calculation.

Β. Examples of gas density at several standard conditions P, T.1. P= 1 Atm a, T=15 oC = 288.15 oK

Vm=22.414*(288.15/273.15)=23.64 m3/kgmol, ρ=Μ/23.64 kg/m3

2. P=1.0 Bar a, T=25 oC=298.15 oK

Vm=22.414*(1.01325/1.0)*(298.15/273.15)=24.79 m3/kgmol, ρ=M/24.79 kg/m3

3. P= 14.696 psi a, T=60 oF = 519.67 oR, density is required in lb/ft3.

We can first convert Vm = 22.414 m3/kgmol to ft3/lbmol, taking into account 1 ft = 0.3048 m and 1 lb = 0.4536 kg; so Vm = 22.414/0.3048^3 ft3/(1/0.4536)lbmol = 359.04 ft3/lbmol (at 1 Atm a =14.696 psi a and 0 oC = 491.67 oR).

Thus for considered standard condition Vm=359.04*(14.696/14.696)*(519.67/491.67) = 379.5 m3/lbmol, ρ=Μ/379.5 lb/ft3.

If conversions looks difficult in this specific case due to English units, we can find density in metric units, then convert it:

T=60 oF = 15.56 oC=288.71 oK, P = 14.696 psi a = 1 Atm a

Vm=22.414*(288.71/273.15)=23.691 m3/kgmol, ρ=Μ/23.691 kg/m3 = (M/0.4536)/(23.691/.3048^3) lb/ft3 = (M*0.3048^3)/23.691*0.4536) lb/ft3 = M/379.5 lb/ft3.

Γ. Concerning ground level concentrations estimated in dispersion models, conversion of ppm (v) to mg/m3 of a substance in air at some standard condition is similarly facilitated by consideration of air molar volume. This is indicated in http://www.cheresour...r-mgm3-to-ppm-v '> http://www.cheresour...r-mgm3-to-ppm-v . Air volume Vm contains 1 kgmole of air, and 1*ppm(v)*1E-6*M kg of the substance. Finding Vm at "standard" condition, mg/m3 is then specified.Δ.With simple thinking and some skill in unit conversions (rather necessary for a Chemical Engineer) we can do a lot using a calculator, even out of the office.Note (*): Chemistry courses in USA schools seem to give Vm in metric units and at 0 oC, 1 Atm a, see Schaum's Colledge Chemistry, Chapter 8, Molar Volume.