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I am trying to determine the density of a gas (oxygen). As the gas is a compressible fluid a modified version of the ideal gas equation is required. I have been given the following equation and I am struggling to interpret it. I would like advice on the unit conversions and whether this is the correct equation at all.

$$\rho=\frac{(P+1.01325).10^5.MW}{z.8314.(T+273)}$$

where:

$\rho$ = density (Kg/m3)

P = pressure (bar(g))

MW = molecular weight (in kg/kmol)

z = compressibility factor (dimensionless)

T = temperature (deg. C)

I think that the 1.01325 converts the bar.g. to bar.a. then to pascals using the 100,000 multiplier?

Has the universal gas constant $8.314 J/(k.mol)$ been converted to $J/(K.kmol)$ ?

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  • $\begingroup$ I can't see why you bother with it. It's simply pV=zRT transformed so you could easily calculate density with data given. $\endgroup$ – Mithoron Feb 4 '16 at 17:21
  • $\begingroup$ Z been compressibility in your equation? Apparently I need to consider compressibility. Something to do with pressure drop over pipe lengths. I will comment further soon $\endgroup$ – LiamH Feb 4 '16 at 17:27
  • $\begingroup$ This has nothing to do with flow in pipes. The z in this equation is the compressibility (correction) factor for real gases compared to ideal gases (not the distance along a pipe). You need to determine the reduced temperature and pressure of oxygen, scaling by the critical properties. You have a graph that gives z as a function of reduced temperature and pressure. Once you know z, you can calculate the density. Isn't this being covered in your course? $\endgroup$ – Chet Miller Feb 4 '16 at 22:29
  • $\begingroup$ I am not on a course, unfortunately, and I am having to learn this as I go. My intention was to use this equation to obtain the density, which is used in the Darcy equation to calculate pressure drop per unit length of straight pipe. $\endgroup$ – LiamH Feb 5 '16 at 7:51
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    $\begingroup$ See any Thermodynamics book, but the best one that I am aware of is Fundamentals of Engineering Thermodynamics by Moran et al. To get an extended treatment of chemical thermodynamics (chemical reactions and phase equilibrium), I suggest Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness. $\endgroup$ – Chet Miller Feb 5 '16 at 13:02
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What an awkward, ugly equation that is! Mithoron is correct about the traditional, elegant and understandable form.

Yes, (P+1.01325)*10^5 converts gauge pressure in atmospheres to absolute pressure in Pascals.

z is compressibility - look up the Nelson-Obert generalized compressibility charts.

As for the 8314, the units here should be Pa*m^3/kmol-K. Dividing it by the molecular weight in kg/kmol will give you the specific gas constant for oxygen in Pa-m^3/kg-K. That's already been done in this equation.

Look up compressibility factor on wikipedia then find the Nelson Obert charts online, also critical data for oxygen, and use them to determine whether your oxygen is ideal in the P, T range you are dealing with (z factor between 1.01 and 0.99). The specific volume of the gas depends on z as well as on P and T, but you can neglect the z correction if it's within a percent or two of the ideal value of 1.

Responding to your subsequent comment:

P, T and z determine the gas's specific volume in m^3/kg, which along with mass flow rate (kg/s) and pipe cross-sectional area (m^2) determine flow velocity (m/s), which along with pipe roughness will determine friction factor, which along with gas viscosity, pipe length and the presence of fittings like elbows and open valves, will determine pressure drop.

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