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I am trying to calculate the density of natural gas to define the amount of natural gas flow through an orifice. I am coming up with 5.45 kg/m3 Is this correct?

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closed as too localized by David Z May 16 '13 at 21:33

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2 Answers 2

Natural gas doesn't have a standard composition. North Sea gas will differ from Russian gas, which will differ from shale gas. For this reason the density of natural gas is usually given as a range e.g. 0.7 - 0.9 kg/m$^3$.

Although natural gas is non-ideal, it does roughly obey a modified gas law:

$$ PV = ZnRT $$

where $Z$ is a constant that varies with gas composition (and is typically less than one). This means that to a good approximation density is proportional to pressure. So if we take North Sea gas, which is about 0.85 kg/m$^3$ then the density at 700kPa will be about:

$$ \rho_{700} = 0.85 \times \frac{700}{101.325} = 5.87 kg/m^3 $$

This is pretty close enough to your figure - presumably you started with a different source of gas and therefore a lower density at STP.

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Wolfram alpha makes this awfully easy, although I have not followed exactly what standard they use to give the answer. Put in the string:

methane at 293K and 700 kPa

Or just click this:


Density they give is $4.67 \frac{g}{cm^3}$. Since I don't know how reliable your figure is, I'm going to go with this number as the better one. Then again, natural gas is more than just methane. There are natural gas liquids, which may or may not be a part of the substance you're dealing with. Overwhelmingly, I would expect the other compounds to have a higher weight than the methane, increasing its density. So it would be consistent that the real density of your natural gas is slightly higher, consistent with your number. If I was given that by a petroleum engineer, I would be likely to believe it, but this is circumstantial.

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I think you meant $4.67 \frac{kg}{m^3}$, since that is what WolframAlpha is spitting out, correct? –  thewheelz May 15 '13 at 19:35
When I calculate the density at $600 kPa$, I also get $4.67\frac{kg}{m^3}$. So, is Wolfram calculating it with absolute pressure? –  thewheelz May 15 '13 at 20:03
@thewheelz I get 3.995 kg/m^3 when I put in 600 kPa. I don't see any indication that this would be the relative pressure. Maybe post what input you used and I'll retry it? –  AlanSE May 15 '13 at 20:41
what I meant is that Wolfram assumes you are inputing the gauge value and that is why it subtracts 100 kPa from the value you input. –  thewheelz May 16 '13 at 0:12
@thewheelz no, that can't be right because when you put in 100 kPa it puts you at STP. I think it's absolute. John's method is probably the better one for natural gas, these are probably correct numbers for just methane. –  AlanSE May 16 '13 at 0:40

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