Question about converting propane vapor volume to liquid gallons

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SUMMARY

This discussion focuses on calculating the volume of liquid propane recovered from rail cars during offloading. The ideal gas law is applied to determine the density of propane vapor at specific conditions, revealing that at 53°F and 45 PSI, the density is approximately 0.36 lb/ft³. Given a rail car capacity of 34,010 gallons, it is estimated that around 350 gallons of liquid propane remains after offloading. The conversation highlights the need for accurate calculations to address excessive shrinkage, which currently exceeds the industry standard of 2%.

PREREQUISITES
  • Understanding of the ideal gas law and its application in gas density calculations
  • Knowledge of propane properties, including specific gravity and vapor-liquid relationships
  • Familiarity with pressure and temperature conversions in gas measurements
  • Experience with propane rail car operations and offloading procedures
NEXT STEPS
  • Research the ideal gas law and its formula: ρ = (pM)/(RT)
  • Explore propane density charts and tables for various temperatures and pressures
  • Investigate methods to minimize propane vapor loss during offloading
  • Learn about the impact of specific gravity on liquid volume calculations in propane
USEFUL FOR

Propane company employees, logistics coordinators, and anyone involved in the offloading and measurement of propane from rail cars will benefit from this discussion.

My05stang
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Hello, I work for a propane company where i unload propane rail cars. I am trying to figure out how much propane is being shipped back to the refinery after I off load the rail car. We are experiencing a large amount of shink and the accountant want to get to the bottom of it. With propane the industry acceptable shrink is 2% and we are over 4% currently.

Depending on air temperature my cars range from 150 PSI summer to 70 PSI winter +\-. During the off load I pump the rail car dry of all liquid propane then revers the compressor and begin to recover the vapor. I typically draw the rail cars down to 40 PSI because at that point it gets difficult on the compressor and it begins to build up heat. Typical capacity of a rail cars is 34,000 gallons.

Here is my question... I will use today as my example. Rail car capacity was 34,010 gallons
propane temperature was 53 f.
Pressure was 95 psi when i began to recover vapor. I ran my compressor till pressure was down to 45 PSI
not sure if this is needed or not but specific gravity of the load was .501. Thats the liquid as stated on Bill of lading. correct me if I'm wrong its been a long time since all my propane training but I think Propane vapor is 1.5 specific gravity.

how much liquid propane was recovered from the car and how much was left inside the rail car?

I really hope I was able to give enough information if not please let me know. I have been goolging for the better part of the afternoon trying to figure this out with no luck. I'm hoping once i figure out the equation I can put it as part of my spread sheet to be sent in with the paperwork,

thank you very much
Michael
 
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From the ideal gas law, the density of propane vapor at 53 F and 45 psi (I assume absolute pressure) is 0.36 lb/ft^3. 34000 gallons is 4545 ft^3. So the remaining mass is about 1600 lb. If converted to liquid, that would be about 350 gallons.
 
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I had always figured it to be around 400 gallons or so but had not scientific proof to back it up! I have been trying to google a chart to give me the the density of propane vapor at a specific temp but can't come up with anything or... or..more likely I probably don't understand what I'm looking at

Is there a simple formula or chart that shows how to get to the 0.36 lb/ft^3.

thanks again sorry to be a bother.
Michael
 
My05stang said:
I had always figured it to be around 400 gallons or so but had not scientific proof to back it up! I have been trying to google a chart to give me the the density of propane vapor at a specific temp but can't come up with anything or... or..more likely I probably don't understand what I'm looking at

Is there a simple formula or chart that shows how to get to the 0.36 lb/ft^3.

thanks again sorry to be a bother.
Michael
Well, the ideal gas law is useful. From the ideal gas law, the density of the gas is given by:
$$\rho=\frac{pM}{RT}$$
where p is the absolute pressure (psi), M is the molecular weight (44), T is the absolute Rankine temperature (460 + F), F is the Fahrenheit temperature, and R is the universal gas constant 10.7 in corresponding units. ρ is the density in lb/ft^3.

Chet
 
thanks you very much for taking the time to help me out with this!
 
I know it has been more than a year since the conversation started so I am jumping in a little late. I too work for a propane company and have a similar issue. The question I have is how could we ensure that all of the propane on the car is offloaded.
 
DM1924 said:
I know it has been more than a year since the conversation started so I am jumping in a little late. I too work for a propane company and have a similar issue. The question I have is how could we ensure that all of the propane on the car is offloaded.
It doesn't seem practical to try to offload all the vapors.
 

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