Pressure rate increase between two vessels

• j117
j117
TL;DR Summary
generating pressure between vessels to check for structural integrity
I have to cylindrical vessels using air as a medium. I want to pressurise vessel one, release the pressure into vessel to and register a .525mbar pressure in vessel 2 - within .3s.

My understanding so far,
dp/dt = 175000 Pa/s required

P1V1/n1 = p2v2/n2
52500 * 0.088627484 * 0.0041846 = 52500 * 0.088627484 * 0.0041846

for me if I have the required pressure in the same volume, and release this within 0.3 seconds, then i will get the exact same pressure on the opposite side.
But, once the valve is opened, the volume essentially doubles.

would I double the volume on V2 but i don't want the final pressure in V2 to change

My aim is to generate the pressure on the bottom section and see if it remains structurally sound.

Welcome, @j117 !

For structural test of recipients, use water and a pump rather than compressed air.
It is not safe to use any compressible gas.

russ_watters and berkeman
the test has to be air, it is to replicate a delayed ignition in gas appliances, all safety is taken care of. I used a small tube to fill, the rate was not fast enough as I need am immediate impact as would be experienced in real terms

I don't see how the equations you posted are relevant.
I believe you're trying to apply the ideal gas equations.

Initially, everything is known

After the expansion, we can find the equilibrium state assuming some things

The final volume I assume is known.
Then, if no mass is lost ##m=constant## and the process is adiabatic ##Q=0## we can conclude the internal energy will be the same as in the beginning because there is no work coming out ##W=0##.
$$\Delta U = Q-W \rightarrow U_f-U_i=0 \rightarrow U_i = U_f$$
Since the internal energy is only a function of the temperature in ideal gases, then the temperature must be the same before and after the expansion.
$$U_i = U_f \rightarrow T_i=T_f$$
https://en.wikipedia.org/wiki/Joule_expansion

Anyway, if you want to know something like ##dp/dt## you will need to characterize so many things that I believe it's easier to just use CFD. The gas will accelerate through the valve and then collide with the chamber walls. Something similar to a water hammer. I'm not certain if the pressure on the walls could be at some instant higher than the equilibrium pressure it will end up reaching as time goes on because of the mentioned collision.

Maybe if you check more info about water hammers and their gaseous version you can find something that works for you. If that's the case, post it back. It'd be interesting to read.

Thread is in Moderation...

Lnewqban said:
Welcome, @j117 !

For structural test of recipients, use water and a pump rather than compressed air.
It is not safe to use any compressible gas.
j117 said:
the test has to be air, it is to replicate a delayed ignition in gas appliances, all safety is taken care of. I used a small tube to fill, the rate was not fast enough as I need am immediate impact as would be experienced in real terms
After Mentor review, this thread will remain closed per the "dangerous discussion" prohibition in the PF rules.

Delayed ignition is a lot like a potato gun without the potato in that the temperatures and pressures are very similar. Peak pressure on the order of 35-40 PSI is a number that sticks in my mind, and is consistent with what I observed when I watched a friend firing tennis balls from a potato gun.

The larger concern is that a test with compressed air does not properly test the system. If I was running this testing, I would set it up for the worst delayed ignition possible, then program it to repeat until stopped. My passing criteria would be about 100,000 cycles with no catastrophic failures. Allowable failures, such as the flame sensor, would shut the system down.

Lnewqban

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