Pressure variations after rapid pressurization

  • #1
CCG
4
0
Summary:
Why pressure falls after rapid pressurization
Hi all,

A question on the physics behind rapid pressurization (relatively high pressures e.g. 300 bar) and perhaps transients in systems. For simplicity, imagine that we only have a piece of pipe with an initially closed valve on one side and a pressure indicator on the other side. The current pressure in the pipe is atmospheric (~1 bar, zero over pressure). Further, on the other side of the valve I have connected a high pressure line. If I open and within very short time close the valve, the pressure indicator would first quickly rise and indicate same pressure as in the high pressure line. Just after closing the valve, the pressure indicator drops to a value less than the pressure in the HP line.

I the valve would remain open for a longer period of time before being closed, the effect can not be observed.

Could someone explain the physics behind this? Do temperatur variations during the pressurization have to do with it?

Many thanks!

C
 

Answers and Replies

  • #3
CCG
4
0
Hi!

Upstream the valve, we have about 200 bar (very large volume compared to the volume downstream valve)
Downstream valve roughly 2 meters of approx 10 mm diamater piping. This piping is closed (confined volume) with a pressure indicator.
 
  • #4
CCG
4
0
And timing, valve may be open for 0.5 s or so. The longer the valve is open, the less the effect. The pressure drops and becomes stable after approx 10 s.

Does Joule - Thomson have anything to do with it? The gas can be said to be throttled when the valve is opened and after a while (10 seconds or so). If there was a temp increase in the pipe, this is then perhaps cooled down to the ambient temp after 10 s and the pressure consequently drops (by 10-15% in this case)
 
Last edited:
  • #5
21,154
4,671
Joule Thomson has something to do with this, but that is not the whole story. This effect would occur even if the air behaved like an ideal gas. And, for an ideal gas, the Joule Thomson coefficient is zero.

If we apply the open-system version of the first law of thermodynamics to this system, and assume that the initial transfer of gas takes place adiabatically, we obtain:
$$d(un)=h^0dn$$where u is the internal energy per mole of gas in the pipe, n is the number of moles of gas in the pipe and ##h^0## is the molar enthalpy of the gas in the high pressure line. If we integrate this equation (and neglect the initial contents of the pipe compared to the final contents), we obtain:
$$u=h^0=u^0+P^0v^0=u^0+RT^0$$where ##v^0##, ##P^0##, and ##T^0## are the molar volume, the pressure, and the absolute temperature, respectively, of the gas in the high pressure line. From the above equation, for an ideal gas, we have:
$$u-u^0=C_v(T-T^0)=RT^0$$where T is the temperature before final cooling begins. The solution to this equation is $$T=\gamma T^0$$where ##\gamma=C_p/C_v=1.4##. So, if ##T_0=293\ K##, T = 410 K = 137 C. And, if the high pressure line pressure is 200 bars, the pressure in the pipe after cooling would be 143 bars.

Of course, this is all for the case in which the air is treated as an ideal gas. The calculation would be a little more complicated when including the real-gas properties of air.
 
  • #6
tech99
Gold Member
2,123
782
Summary:: Why pressure falls after rapid pressurization

Hi all,

A question on the physics behind rapid pressurization (relatively high pressures e.g. 300 bar) and perhaps transients in systems. For simplicity, imagine that we only have a piece of pipe with an initially closed valve on one side and a pressure indicator on the other side. The current pressure in the pipe is atmospheric (~1 bar, zero over pressure). Further, on the other side of the valve I have connected a high pressure line. If I open and within very short time close the valve, the pressure indicator would first quickly rise and indicate same pressure as in the high pressure line. Just after closing the valve, the pressure indicator drops to a value less than the pressure in the HP line.

I the valve would remain open for a longer period of time before being closed, the effect can not be observed.

Could someone explain the physics behind this? Do temperatur variations during the pressurization have to do with it?

Many thanks!

C
It sesm analogous to an electrical transmission line, where wave will propagate down the pipe as a result of an impulse. When it bounces from the end, we can get a reversal of pressure. Once the initial wave dies down, due to friction, we see the steady pressure of the line.
 
  • #7
21,154
4,671
It sesm analogous to an electrical transmission line, where wave will propagate down the pipe as a result of an impulse. When it bounces from the end, we can get a reversal of pressure. Once the initial wave dies down, due to friction, we see the steady pressure of the line.
He mentioned that the effect was present after 0.5 sec., and took 10 sec. to die out. How long does it take a pressure wave to travel at the speed of sound for 2 meters?

Did you read my answer?
 
  • #8
tech99
Gold Member
2,123
782
He mentioned that the effect was present after 0.5 sec., and took 10 sec. to die out. How long does it take a pressure wave to travel at the speed of sound for 2 meters?

Did you read my answer?
Apologies, I missed that, it does look thermal.
 
  • #9
CCG
4
0
Thank you very much for the thorough explanation. I'll need some time to digest it and dust off my knowledge in physics from university which I haven't applied for quite some time but hopefully I'll understand it in detail.

And as pointed out by you, I think it is a 'thermal phenomenon' rather than wave and impuls related. I've tried it in other piping systems with different pressure indicators and pressure up to 300 bar as well. The pressure indicator(s) goes 'immediately' up to the high pressure upstream the valve and is very stable until the valve is closed (when it drops).

Again, thank you for the extensive answer. I might get back to you if there are details in the explanation which I don't manage to understand.
 
  • #10
21,154
4,671
Thank you very much for the thorough explanation. I'll need some time to digest it and dust off my knowledge in physics from university which I haven't applied for quite some time but hopefully I'll understand it in detail.

And as pointed out by you, I think it is a 'thermal phenomenon' rather than wave and impuls related. I've tried it in other piping systems with different pressure indicators and pressure up to 300 bar as well. The pressure indicator(s) goes 'immediately' up to the high pressure upstream the valve and is very stable until the valve is closed (when it drops).

Again, thank you for the extensive answer. I might get back to you if there are details in the explanation which I don't manage to understand.
How do your measurements compare quantitatively with this analysis for an ideal gas? Did you do any measurements of the gas temperature (or the pipe wall temperature)?
 

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