Work done by expansion of liquidfied natural gas to atmosphere

[casts_by_fly]

Hi All,

Been try at this for a week now and I know I'm missing something. I'm too far out of university to remember how to do it, but just close enough to know that I should remember.

Anyway, I have a steel can containing a mixture of liquidfied hydrocarbon gasses (butane/i-butane/propane) at 40 psi internal pressure. Most of the hydrocarbons are liquified, though there is some headspace. Additionally there is another fluid in the container that is being expelled by the propellant (they are mixed, but not miscible if it matters). They are released through a valve to the atmosphere in short bursts with time between the bursts for the can to return to room temperature (nominally 300K). How much work does the compressed hydrocarbon do on the other liquid when released to the atmostphere? If I were to replace the propellant with a battery, how much energy would the battery need to hold to be able to get the same work into the other liquid?

I feel like it is an expansion problem, but can't decide between isothermal (because both the internal gas and the released gas return to room temp) or adiabatic (because the expansion is rapid and there is immediate cooling from both the expansion of the gas and the evaporation of liquid to gas in the container). Neither seems to fit though.

I tried to work it out from a kinetics standpoint by figuring out the velocity of the ejecta and the mass ejected, but the number was very low for what I was expecting.

Any help in setting this up?

Thanks,
Rick

 

[RonL]

Hi All,

Been try at this for a week now and I know I'm missing something. I'm too far out of university to remember how to do it, but just close enough to know that I should remember.

Anyway, I have a steel can containing a mixture of liquidfied hydrocarbon gasses (butane/i-butane/propane) at 40 psi internal pressure. Most of the hydrocarbons are liquified, though there is some headspace. Additionally there is another fluid in the container that is being expelled by the propellant (they are mixed, but not miscible if it matters). They are released through a valve to the atmosphere in short bursts with time between the bursts for the can to return to room temperature (nominally 300K). How much work does the compressed hydrocarbon do on the other liquid when released to the atmostphere? If I were to replace the propellant with a battery, how much energy would the battery need to hold to be able to get the same work into the other liquid?

I feel like it is an expansion problem, but can't decide between isothermal (because both the internal gas and the released gas return to room temp) or adiabatic (because the expansion is rapid and there is immediate cooling from both the expansion of the gas and the evaporation of liquid to gas in the container). Neither seems to fit though.

I tried to work it out from a kinetics standpoint by figuring out the velocity of the ejecta and the mass ejected, but the number was very low for what I was expecting.

Any help in setting this up?

Thanks,
Rick

I'm a little surprised someone has not responded to this, It sounds a lot like some things I have brought up, my math sucks so I won't be of much help, but there are some very sharp people here, just hope I have not worn them down too much.

Bump!

RonL

 

[astrokreng]

Hi Rick,

It seems like you are trying to calculate the work done by the expansion of the liquidfied natural gas to the atmosphere. This is indeed an expansion problem, and the type of expansion depends on the conditions of the system.

If the expansion is happening slowly enough that the system can reach thermal equilibrium with the surroundings, then it can be considered an isothermal process. In this case, the work done is given by the equation W = -PΔV, where P is the external pressure and ΔV is the change in volume of the system. In your case, the external pressure is atmospheric pressure and the change in volume will depend on the amount of gas that is being released.

If the expansion is happening rapidly and there is not enough time for the system to reach thermal equilibrium, then it can be considered an adiabatic process. In this case, the work done is given by the equation W = -ΔE, where ΔE is the change in internal energy of the system. This change in internal energy can be calculated using the ideal gas law, where the initial and final temperatures and volumes of the system are known.

To calculate the work done by the expansion of the liquidfied natural gas, you will need to know the amount of gas being released, the initial and final temperatures and volumes of the system, and the type of expansion (isothermal or adiabatic). From there, you can use the appropriate equations to calculate the work done.

If you were to replace the propellant with a battery, the amount of energy needed would depend on the efficiency of the system. In an ideal system, the work done by the expansion of the gas would be equal to the energy stored in the battery. However, in a real system, there may be losses due to friction or other factors, so the energy needed from the battery may be slightly higher.

I hope this helps. Good luck with your calculations!