The original amount was 36,000 m^3 not 3,600 m^3; supposing it's for gaseous propane.
For gas the specific volume is roughly 0.552 m^3/kg; so 1.8 kg/m^3 so 6500 kg for 3600 m^3 or 65,000 kg for 36,000 m^3
What's missing is the allowable temperature rise in the liquid propane, or is the idea that there is a delta T of 2C when the liquid propane is sprayed or otherwise introduced into the tank to drop the final temp to -42C.
It seems like the goal is to get the gas to the saturation temperature by allowing the liquid to expand/boil to reach the saturation temperature. I'd say that the thermal mass of the tank will also have to be considered in this as well as thermal gains through the tank walls.
You are looking at a delta T of 62 C for a material that has a specific heat, c sub p, of 1630 J/kg/K; so
3600 kg * 62 * 1630 -> 3.6E8 Joules.
Heat of vaporization is 428000 Joules/kg, so 3.6E8/428000 = 850 kg of liquid propane. (or 8,500 for 36,000 m^3 as originally stated.)
I'm ignoring the heat required to warm the liquid from -44 to -42; that's all the calcs I'm interested in.
I'm also ignoring the potential that propane is not a perfect gas and that the specific heat may vary - I don't have access to tables showing what the value is across the range of interest. This calculation might be far off if that's the case.
Values of constants from
so check to see these match what you expect.
