4
$\begingroup$

I am trying to calculate a cost per hour use of a gas we buy in a cylinder. The details I have been working with are:

Gas used CP-grade N2 material number: 110628-L can

  • Gas: N, density = 1.251 g/L @ 0.101 MPa M (molar mas) = 14.007
  • Gas cylinder: Pressure = 200 bar or 20 MPa Cylinder volume 9.45 m3 or 9450 L
  • Operating pressure of gas from cylinder = 4.5 bar or 0.45 MPa and flow rate = 60 cL/min

I have been applying the ideal gas law to PV=nRT to figure out how many total litres of N are in the can at 4.5 bar pressure to then get a rough estimate of how many hours of flowing gas I get. However I seem to be getting unreasonable numbers, potentially from me using the equation or reasoning incorrectly:

When I apply the gas law under isothermal V2=V1/(P2/P1) conditions to see how much the internal gas volume expands from the 20 MPa pressure, as a sanity check I wanted to see how many L at 1 atm came out. The result was 1.89 ML which makes little sense as the conversion of 1.89 ML N2 to mass = 2.364 Mg.

Similarly it equates to 420 kL @ 0.42 MPa.

Where am I going wrong?

Maybe the specifications mean that its 9.45m3 or 9450 L of compressed gas to 200 bar?

$\endgroup$
  • 1
    $\begingroup$ 9450 L is about 10 cubic meters, so unless your cylinder is really big -- say like 1 meter in diameter and 12 meters high, it isn't 9450 L big. So you are probably right that what you are receiving is 9450 L of gas at STP. A further note: nitrogen gas is $\textrm{N}_2$, a diatomic molecule, so its molecular weight is 28 g/mol, not 14. $\endgroup$ – Curt F. Apr 15 '15 at 14:23
  • $\begingroup$ I worked it out in the end. The specifications given of 9450 L are indeed the volume of the uncompressed gas, which compresses to 20 MPa results in about 47 L which fits the dimensions of the L gas cylinder. I simply understood the information backwards. Noted for the 28 g/mol. Thank you. $\endgroup$ – Fiztban Apr 15 '15 at 14:27
  • $\begingroup$ Have a look at this video. This Should you help you out. youtube.com/watch?v=Yo9IZ2RJCCU $\endgroup$ – Anup Dec 31 '16 at 17:38
3
$\begingroup$

Turns out my assumptions were wrong about the stated $9\,450\ \mathrm L$ of $\ce{N2}$ gas. The $9\,450\ \mathrm L$ is the total volume of standard pressure gas compressed into the cylinder to $20\ \mathrm{MPa}$.

Using $V_2=V_1/(p_2/p_1)$ having $p_1=0.101\ \mathrm{MPa}$, $p_2=20\ \mathrm{MPa}$, $V_1=9\,450\ \mathrm L$. The total volume of gas within the cylinder is about $47\ \mathrm L$ a reasonably fitting size for the cylinder dimensions.

Hence to answer my own question, at the pressure of usage ($0.45\ \mathrm{MPa}$) the volume of gas (using same equation as above) is $2\,121\ \mathrm L$ which at $60\ \mathrm{cL/min}$ equates to $59\ \mathrm h$ of flow roughly.

$\endgroup$

protected by Community Sep 28 at 16:20

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.