H2 gas venting from a pressure vessel 1

[hoff16]

Hi,

I have been presented with this problem, hopefully someone can help.

I have a 4000 ft^3 pressure vessel filled with Hydrogen gas at 60 PSIG. The initial temperature is 300K. At one end of the tank there is a 1-1/2" ball valve with CV=125 and KV=111.

I need to determine the time it would take for the pressure in the tank to drop to 5 PSIG if the valve were opened to 100%.

I greatly appreciate anyone's help.

Thanks,
Kyle

 

[dcasto]

Never enough information.

I'll lead you through somewhat. You know the initial pressure and temperatur and the valve Cv, so calculate the flow rate. Now take a time period of 1 minute. How many pounds were lost? When that many pounds are lost from a volume, what is the new pressure?
Here the tough part. When the pressure drops x psi, what is the corresponding temperature reduction due to JT effect.
OK, How much heat enters from the surroundings, need Q=UAdTlnm.

Repeat the same steps again.

If you put the equations in a spread sheet, you can replicate and use the initial pressure temperature from the previous calculations. Once done you can copy the formulas 10,000 times and set the time interval to .1 second if you want.

It's called digital Integration.

 

[MortenA]

I would like to expand a little on dcastos fine walk through:

From initial P/T comes initial density comes intial total mass

For each time step you calculate your loss of mass - and thus know your end-of-timestep mass - since vessel volume dosnt change you now know your end-of-timestep density.

Knowing your inital stage P/T and your final density and assuming adiabatic expansion you can now calceulate your final P/T - this is however trial and error/itteration where you change pressure until the resulting density is the same as your known end-of-timestep density. Use of a process simulation tool such as HYSYS may make it easier. HYSYS (and simlar programs) has a function for this type of calculation.

Best regards

Morten

 

[israelkk]

To make it a little bit more complex you have to take into account heat transfer and cooling of the gas inside the vessel. If the process is very fast you may assume adiabatic process. If it is quite slow then you may consider it as an isothermal process and constant gas temperature. In between it is a polytropic process.

 

[sailoday28]

israelkk (Aerospace)I agree with everything stated except for the use of polytropic. Usually the polytropic exponent is a constant and I doubt that the constant will account for the heat transfer.

Regards

 

[israelkk]

sailoday28

Completely agree with you, this is why I mentioned the heat transfer analysis/check.

 

[iainuts]

Hi dcasto,
One minor nit pick on your post.

When the pressure drops x psi, what is the corresponding temperature reduction due to JT effect.

The JT effect is described as:

A thermodynamic process (also called a throttling process) in which enthalpy is conserved

Ref:

http://scienceworld.wolfram.com/physics/Joule-ThomsonProcess.html

For the gas exiting the tank through the valve, enthalpy is conserved (ie: the process is aproximated by an isenthalpic process). However, for the gas inside the vessel which is expanding, the process is not isenthalpic, it is isentropic (ie: dS = 0). You can also apply the first law to the gas inside the vessel and note that dU = Hout.

Other than that minor nit pick, I agree with what you have.

One other point however is worth raising. For most pressure vessels, the thermal mass of the steel will be significant, such that dQ is mostly due to the convective heat transfer of the gas with the inner surface of the vessel and the subsequent cooling of the metal. This means that the heat transfer to atmosphere might effectively be overlooked which means we don't look at convection on the outer surface being a limitation on the total heat transfer. The limitation may only be the convective heat transfer of the inner vessel surface. When the thermal mass of the vessel is considered, the gas inside the vessel is seen to have a higher heat flux than for the case where the tank's thermal mass is neglected.