Pressure Decay from a Fixed Volume

[PacificSnapper]

If an ideal gas is leaking from a fixed volume, does the temperature of the air in the fixed volume go down, or does it remain constant? We do pressure decay measurements under choked leakage flow conditions, but I wonder if our assumption of constant temperature for the gas remaining in the fixed volume is valid. Assuming an adiabatic wall, of course. Any solid guidance on that would be helpful! Thanks.

 

[boneh3ad]

The answer is going to depend on what sort of boundary conditions you have. Is your fixed volume adiabatic? Is there some means by which heat can move into the system? How fast is it discharging? All of these will play a role.

 

[PacificSnapper]

Let's assume adiabatic wall, no other means for heat to flow into or out of the system. Air or some other ideal gas is leaking to outside ambient. In practice, the flow is usually choked, through a small crack or defect. So, the speed of discharge is sonic at the exit. The typical case might be a 2 ft3 rigid walled tank, pressurized to 100 psia, leaking to ambient for one minute. Pressure might reduce to, say, 80 psia in one minute.
Thanks for responding!

 

[boneh3ad]

You could get a pretty good approximation just using the relationship between total pressure, total temperature, throat area, and mass flow rate that you can find in any compressible flow textbook (usually discussed near the same point as nozzles). The pressures you cite should remain choked the entire time. The temperature of the reservoir will change as it discharges since it is dropping in pressure but not gaining any heat from the outside, which will be a fun wrinkle in the problem.

 

[Nidum]

In a real situation the flow rate often depends on the geometry of the leakage passageway .

Where leakage is taking place through a defect such as a crack in or thin gap between components made from material of significant thickness the calculations can become very difficult and sometimes they are unsolvable by purely analytic means .

 

[JBA]

As I view it, it depends upon whether you are interested in plotting the values during flow or simply the final volume at the a given finished vessel pressure. For a zero heat flow ideal adiabatic final gas volume calculation, the method, rate and time of discharge are irrelevant to the final gas volume at a given reduced tank pressure. The temperature in the tank can be determined by the classical isentropic ΔT vs ΔP calculation; and, then calculating the gas volume volume with the final P & calculated T in the tank is straightforward.

 

[boneh3ad]

I don't know where you are coming up with a changing volume. The original question had to do with the temperature of gas as it discharges from a fixed volume, meaning the volume is constant while the pressure, temperature, and density drop. The nice thing about making an adiabatic assumption is that you can treat the problem isentropically, which simplifies certain aspects of the problem like the temperature's relationship with the other variables. It does complicate the mass flow rate equation somewhat, though.

Basically, the ideal mass flow through a choked opening in a tank or other similar pressure vessel is
\dot{m} = \dfrac{p_t A^*}{\sqrt{T_t}}\sqrt{\dfrac{\gamma}{R}}\left( \dfrac{2}{\gamma + 1} \right)^{\frac{\gamma + 1}{2(\gamma - 1)}}.
So really you just need to account for the temperature change in order to solve for the pressure as a function of time. In the adiabatic case, you can treat it as isentropic and use isentropic relations. Then you can solve the equation for the time history of the pressure and use that to then backsolve for the temperature as a function of time.

This is a pretty good approximation provided that viscous effects are fairly small. In other words, as long as the hole's size is suitably large compared to the wall thickness, then viscous effects are likely to be pretty small and this equation would give you a very good approximation. What constitutes "suitably large" is not necessarily simple, though.

 

[PacificSnapper]

Thanks for the good input. First, to reiterate my confusion, I am only thinking about the temperature of the ideal gas remaining in the fixed volume rigid container, after some period of choked-flow leakage. No heat flows in or out of the volume, and it's a slow process so we can call it isentropic. The pressure would go down of course, if only because there are fewer molecules remaining to bounce off of the walls. I think my issue with this is: has the kinetic energy of the molecules remaining in the container been reduced, which equates to a reduction in temperature? Or I guess I am asking, did that gas remaining in the container do work on the outside world so that the energy per unit mass of remaining gas has been reduced?

 

[JBA]

A simple measure of a contained gas energy for an isothermal condition is E (lb-in) = P (psi) x V (in^3) and the difference between these values at the start and end of the venting cycle represents the difference in potential energy of the tank's contained volume; therefore E lost = (P1 - P2) x V (tank volume).

In the real world cases, the long term condition, lacking any other external heat sources or sinks, is isothermal (i.e. a given container will over time recover to the surrounding ambient temperature). Adiabatic conditions only exist only for very short period flow periods and will become polytropic due to combined temperature loss due to gas expansion cooling and heat flow input from the surrounding ambient temperature.

The result, from my view, is that if you want to know the mass of the original gas in the vessel you have lost then assuming a constant isotropic (constant temperature) approach is a good choice. On the other hand, if you want to know the mass flow rate being discharged at any given instantaneous point in time over a short flow period ; then, an adiabatic solution to the determine the instantaneous temperature of the contained gas at the pressure at that point, as I described above, is a good method.

 

[JBA]

With relation to my last post, I think I have deviated into a theoretical discussion; where, in fact there is a better answer regarding your initial post. In order to uniformly compare the amount of gas within a given vessel volume it is necessary to establish a base pressure and temperature reference condition for your comparisons.

For industrial purposes in the USA that reference is SCF (Standard Cubic Feet) at 14.7 psig and 60°F and SCFM (Standard Cubic Feet per Minute) at those conditions; and, are used in sizing pneumatic systems because this universal can be used to quickly and easily compare the performance of any piece of equipment from any source.

The only way to accurately determine and relate the amount of gas that has been discharged up to a point or being discharged at a point in the test is to have the appropriate pressure and internal gas temperature probes that allow you to simultaneously determine both the gas P & T at that point of interest during the leak test.
In order to then standardize those values the amount of gas contained at the measured P & T at the start of a test would be converted to SCF; and, to determine the relative amount of gas discharged up to any point, the measured P & T at that point is used to convert the quantity of gas lost to the equivalent SCF. Similarly, the instantaneous discharge rate calculated at any point based upon the the existing P & T at that point is also converted to the equivalent SCFM conditions.

In this manner, you are then capable of comparing the relative quantity of gas lost and flow rate at any point of interest directly regardless of the P & T at that point.

 

[boneh3ad]

I don't think you necessarily deviated too far. The original question was a bit vague, so the answers are going to be quite broad. My response was assuming you wanted to know the temperature at a certain time after the draining started. I also accepted the adiabatic assumption because the more realistic isothermal kind of defeats the original purpose of his question, though I agree with your previous post that it is more realistic in many situations.

That said, if all @PacificSnapper wants is the gas temperature given a certain amount of pressure drop, that's just algebra and he only needs to decide whether he wants to stick with the isentropic assumption or use a different model. That's extremely straightforward. If he wants temperature (and pressure, etc.) as a function of time, he needs to use my method above.

 

[JBA]

Assuming @PacificSnapper is satisfied with an assumed adiabatic assumption and given his stated 100 psia tank pressure, you are correct and the equation should give reasonably accurate results.

My purpose was to give him the general procedure for insuring good accuracy if there is a high exhaust rate and resulting significant gas temperature drop during his test with the possibility/probability of heat transfer into the vessel from the surrounding ambient temperature. As you are aware, under those conditions this becomes a polytropic process with no defined Gamma value.

I learned about this the hard way when asked to write a program to more "accurately" predict, than from currently available data, how many scuba tanks can be "filled" to 2000 psig from a specified number of air storage tanks at 6000 psig. Even given temperature decline rates vs pressure drop from storage tank tests; in the end, part of the answer became "at what rate are you filling the scuba tanks and/or how much time is required to switch between tanks for sequential tank filling"; and, with no clear consistent answer to those elements, I had the program give two results, one at a given estimated filling rate; and, a second, if the storage tanks were to remain or be allowed to return to their original pre-fill ambient temperature by a prolonged filling cycle. At the same time, the program still did not deal with the required or estimated cooling time for the scuba tanks to return to ambient as well. (Standard filling procedure is to assume that there will be about a 200 psi pressure drop in a scuba tank after cooling so this was considered acceptable by the user). Of course, at the pressures used the Z gas compression factor had to be incorporated into the program as well.
On a side note, if anyone else has ever wondered, this is when I also learned the reason scuba tanks are immersed in water during their filling cycle is to assist with cooling from compression heating during filling. Also, filling from a compressor automatically deals with this issue because it can simply continue to cycle until fully isothermal stability is reached in the filled tank.