0
$\begingroup$

We have a building with a large propane cylinder that distributes propane to different sources within the building. When the cylinder is filled, the propane is compressed to a liquid, and our bill is therefore in liters of liquid propane. It is then converted to a vapor in the pipes due to lower pressure.

After the cylinder there is a pressure regulator that regulates the pressure in the line down to 10 PSI. In the line there is a gas meter that measures consumption in units of 100 cubic feet (2831.68 liters). Finally before the appliance is another regulator which regulates the final output pressure to 0.5 PSI for the appliance.

enter image description here

I am trying to calculate the consumption of liquid propane by a tenant relative to the consumption measured on the gas meter. In several sources online (for example), I see that the ratio of liquid to gas propane is 1:270, however I would assume that this is probably at STP. If the temperature is 25°C and the pressure is 10 PSI, how would I determine this ratio in order to accurately charge the tenants for their consumption?

Note, I am unfamiliar with gas regulator terminology, but I would assume that this would mean pressure above atmospheric pressure, because if it was only 10 PSI, gas would flow into the pipe instead of out of it since atmospheric pressure is 14.7 PSI. So 10 PSI would actually mean 24.7 PSI. Please correct me if I'm wrong.

Improve this question
$\endgroup$
4
  • $\begingroup$ Just fit an extra meter - much easier. $\endgroup$
    – user207455
    Apr 26, 2019 at 19:02
  • $\begingroup$ @SolarMike I don't see how that would solve the problem $\endgroup$
    – Mike
    Apr 26, 2019 at 19:05
  • $\begingroup$ If all the pipework losses for each tenant were identical then you can find one value that you could use for calculating each tenant’s use. As that probably is not true, unless the building is circular and the propane tank at the centre, then using a meter to record the use per tenant is probably more accurate. That is, of course, assuming that that is what you want to achieve. $\endgroup$
    – user207455
    Apr 26, 2019 at 19:12
  • $\begingroup$ I am using one meter per tenant. See updated question with a diagram. $\endgroup$
    – Mike
    Apr 26, 2019 at 19:19

2 Answers 2

Reset to default
1
$\begingroup$

According to http://dotnetdesigner.com/Tools/GasDensityCalculator.aspx, the difference between the 0.5psig vapor and the 10psig vapor is

3.03 kg/m^3 (0.19 lb/ft^3) @10psig

1.86 kg/m^3 (0.12 lb/ft^3) @0.5psig

So a cubic foot of liquid becomes 270 cubic feet of low-pressure vapor, becomes 270 *(1.86/3.03) or ~165 cubic feet of 10psig vapor.

Also, is this 1:270 ratio at 0°C or 25°C?

I'm assuming it's a rule of thumb used for simple calculations of "room-temperature" installations. If you need greater precision, look up the density of liquid propane directly.

My usual source is engineering-toolbox, but in this case, the data is a bit sparse, with nothing super close to room temperature.

519.2 kg/m^3 (32.41 lb/ft^3) @6.9C

473.0 kg/m^3 (29.53 lb/ft^3) @36.9C

I would think interpolation between those items would get you something useful, and probably much less than the measurement errors.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Is the 1:270 ratio with 0 psig or 0.5 psig? I would assume 0. $\endgroup$
    – Mike
    Apr 26, 2019 at 22:48
  • $\begingroup$ I would assume 0 as well. I just ignored that particular conversion. Looks like about 3% difference between 0 and 0.5psig. That said, it's about the same error as dropping from 25C to 15C. So probably close to the errors in these devices. $\endgroup$
    – BowlOfRed
    Apr 26, 2019 at 23:12
  • $\begingroup$ Ok, doing a bit of investigation, this number is slightly off. According to to here the density of propane at STP is 1.882 Kg/m3 and according to Google, the density of propane liquid is 493 kg/m3, which gives a ratio of 1:261.96. $\endgroup$
    – Mike
    Apr 26, 2019 at 23:21
0
$\begingroup$

BowlOfRed's answer really pointed me in the right direction here, but I managed to simplify it greatly instead of using numbers from unreliable sources.

We know the density of liquid propane at 25°C to be 493 kg/m³ [source].

We can use this calculator to determine that the density at 10 psig and 25°C is 3.02937 Kg/m³

Therefore, 1 L liquid = 493/3.02937 = 162.74 L vapor @ 10 psig and 25°C.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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