Pressure equalization in cartriges and related safety risks

[Matthew99]

Dear community!
I've recently been pondering about an intriguing thought experiment that can have serious consequences in the real world. Imagine you've got a box filled with a liquid J with a partial pressure p_J_eq(20°C) > 1 bar (higher than the air surrounding the box). If the box is closed, the remaining volume which is not occupied by J_liquid will be filled with J_gas at a pressure of p_J_eq(T_box,initial = 20°C). Now you pierce a small hole in the box, lets say 2mm x 2mm. According to common sense, gaseous J will escape and therefore more J_liquid will vaporize in order to increase p_J up to p_J_eq(T_box).

Here, lets consider two scenarios:

Scenario 1:
Heat transfer from the surrounding air to the box and within the box is immediate (alpha & lambda = infinity), thus the temperature of the box and J within the box will not change (T_box=T_box,initial=20°C). Following that assumption, p_J_eq is not a function of time and will be constant at p_J_eq = p_J_eq(20°C) > 1bar. Here it gets interesting: Considering entropy and the fact that particles in ambient air move quite fast, it will be highly likely that at least some air particles will manage to get inside the box. My assumption is that the partial pressure of air particles inside the box p_Air,box will look something like the following graph. As far as I know, the equilibrium pressure of air molecules inside the box p_Air,box,eq should be equal to the air pressure surrounding the box (1 bar). My question here is, firstly, is this assumption right, and secondly, can anyone think of an easy way to estimate how long this will take? The only thing that comes to my mind is some CPU intensive particle modelling.

Scenario 2 (real world):
Heat transfer is not immediate, and the box will cool down until the partial pressure of J is approximately 1bar (the speed of which gaseous J particles escape is such that the rate of vaporization and the thereby induced cooling of the box is equal to the rate of heat transfer to the box). Therefore, p_J after some time will be p_J_eq(T_box,eq). Let's consider a real world example of where this might happen: Think of a gas cartridge filled with J=butane (p_Butane(20°C)=2bar, T(p_Butane=1bar)=-10°C (ball park numbers). Assume you unintentionally open the outlet of the cartridge without any safety mechanism so that the outlet is directly exposed to the surrounding air. In this case, butane will escape & air will get inside, and eventually the box will cool down to -10°C. According to previous reasoning the situation inside the box after some time will be following: p_Butane = 1bar, p_Air,box = 1bar. Now assume you close the cartridge again for later usage. First thing that comes to my mind is that the cartridge could theoretically burst as it heats up again as it is now filled with p_Butane(20°C) = 2 bar and some additional air with a pressure that is definitely above 1bar (it was 1bar at -10°C). Lets say it does not burst, and you use the cartridge again in a gas burner. Could this theoretically be dangerous? I reckon that the explosion limit of butane should not be an issue as the concentration of oxygen must be quite high. Whats your assessment of this?

Edited May 16, 2021 by Matthew99
Troubleshooting

[exchemist]

1 hour ago, Matthew99 said:

Dear community!
I've recently been pondering about an intriguing thought experiment that can have serious consequences in the real world. Imagine you've got a box filled with a liquid J with a partial pressure p_J_eq(20°C) > 1 bar (higher than the air surrounding the box). If the box is closed, the remaining volume which is not occupied by J_liquid will be filled with J_gas at a pressure of p_J_eq(T_box,initial = 20°C). Now you pierce a small hole in the box, lets say 2mm x 2mm. According to common sense, gaseous J will escape and therefore more J_liquid will vaporize in order to increase p_J up to p_J_eq(T_box).

Here, lets consider two scenarios:

Scenario 1:
Heat transfer from the surrounding air to the box and within the box is immediate (alpha & lambda = infinity), thus the temperature of the box and J within the box will not change (T_box=T_box,initial=20°C). Following that assumption, p_J_eq is not a function of time and will be constant at p_J_eq = p_J_eq(20°C) > 1bar. Here it gets interesting: Considering entropy and the fact that particles in ambient air move quite fast, it will be highly likely that at least some air particles will manage to get inside the box. My assumption is that the partial pressure of air particles inside the box p_Air,box will look something like the following graph. As far as I know, the equilibrium pressure of air molecules inside the box p_Air,box,eq should be equal to the air pressure surrounding the box (1 bar). My question here is, firstly, is this assumption right, and secondly, can anyone think of an easy way to estimate how long this will take? The only thing that comes to my mind is some CPU intensive particle modelling.

Scenario 2 (real world):
Heat transfer is not immediate, and the box will cool down until the partial pressure of J is approximately 1bar (the speed of which gaseous J particles escape is such that the rate of vaporization and the thereby induced cooling of the box is equal to the rate of heat transfer to the box). Therefore, p_J after some time will be p_J_eq(T_box,eq). Let's consider a real world example of where this might happen: Think of a gas cartridge filled with J=butane (p_Butane(20°C)=2bar, T(p_Butane=1bar)=-10°C (ball park numbers). Assume you unintentionally open the outlet of the cartridge without any safety mechanism so that the outlet is directly exposed to the surrounding air. In this case, butane will escape & air will get inside, and eventually the box will cool down to -10°C. According to previous reasoning the situation inside the box after some time will be following: p_Butane = 1bar, p_Air,box = 1bar. Now assume you close the cartridge again for later usage. First thing that comes to my mind is that the cartridge could theoretically burst as it heats up again as it is now filled with p_Butane(20°C) = 2 bar and some additional air with a pressure that is definitely above 1bar (it was 1bar at -10°C). Lets say it does not burst, and you use the cartridge again in a gas burner. Could this theoretically be dangerous? I reckon that the explosion limit of butane should not be an issue as the concentration of oxygen must be quite high. Whats your assessment of this?

There is one assumption you are making that I would question. I don't think air will get into the box. The issue is not so much the speed of the molecules as their diffusion rate. Don't forget that while they move fast, the mean free path is very short as they keep rebounding off other molecules in random directions. For air to get into the box, the speed of diffusion, which is the net rate of progress, taking into account all these random collisions,  has to exceed the flow rate of gas out of the hole. Doesn't it?  

[Matthew99]

@exchemist

Thank's for your comment! That's probably true - scenario 1 is incorrect, as the flow rate of gas out of the hole will very likely be far greater than rate of diffusion. However, that does not apply to scenario 2 (real world example). When T_box is such that p_J_eq(T_box) = ~1bar , there won't be a pressure gradient driven flow of J and therefore diffusion of air into the box should happen, shouldn't it?

[exchemist]

1 hour ago, Matthew99 said:

@exchemist

Thank's for your comment! That's probably true - scenario 1 is incorrect, as the flow rate of gas out of the hole will very likely be far greater than rate of diffusion. However, that does not apply to scenario 2 (real world example). When T_box is such that p_J_eq(T_box) = ~1bar , there won't be a pressure gradient driven flow of J and therefore diffusion of air into the box should happen, shouldn't it?

It seems to me that, in practice, with a real metal butane cartridge, so long as some liquid butane remains, there is likely to continue to be a +ve pressure sufficient to prevent air from diffusing against the flow into the cartridge. Unless you insulate the cartridge, the temperature difference between the air at 20C and the colder liquid will most likely keep it above -10C. But supposing that one way or another it does get down to -10C, the pressure equalises and then air diffuses in, then I suppose you eventually end up with a mixture in which the rate of diffusion of butane from the liquid is in competition with the rate of diffusion of air in through the hole. I would imagine that would become a matter of the relative areas of the liquid surface and that of the hole.  

If that is right, then I should imagine the mixture inside the cartridge would remain principally butane, the excess pressure on closing the hole would be marginal and the risk of reaching the explosion limit would be negligible.  

Edited May 16, 2021 by exchemist

[Matthew99]

@exchemist

Thank you very much for sharing your thoughts on this! It seems reasonable that the temperature of butane in the box will be always be slightly above -10, because if it would be exactly -10 °C, there wouldn't be net-vaporization except to account for diffusion losses and therefore no net cooling thus the temperature would increase. Good point. So, just to clarify, you're saying that because of the temperature of butane being slightly higher than -10° the diffusion of air into the box would be negligible because even though the flow of butane out of the box (which is driven only by a slight pressure gradient) is quite slow, the 'speed' of diffusion of air is even smaller?

Edited May 17, 2021 by Matthew99

[exchemist]

52 minutes ago, Matthew99 said:

@exchemist

Thank you very much for sharing your thoughts on this! It seems reasonable that the temperature of butane in the box will be always be slightly above -10, because if it would be exactly -10 °C, there wouldn't be net-vaporization except to account for diffusion losses and therefore no net cooling thus the temperature would increase. Good point. So, just to clarify, you're saying that because of the temperature of butane being slightly higher than -10° the diffusion of air into the box would be negligible because even though the flow of butane out of the box (which is driven only by a slight pressure gradient) is quite slow, the 'speed' of diffusion of air is even smaller?

That's part of it. But to my mind the more significant part what would happen assuming the temperature of the liquid butane did drop to -10C. In that case I think the relative rates of diffusion would be determined chiefly by the surface ares of the hole compared to  the surface area of the liquid.

I've now added chiefly because of one aspect my earlier post did not take into account, which is that as I recall the rate of diffusion is inversely proportional to the square root of the molecular weight. The MW of butane is 48 +8+2 = 58 and that of air is 29 average, exactly half. So the air will diffuse 1.414 time as fast as the butane, per unit area, but the liquid area will be a lot greater than the area of the hole.

I actually think there may be another effect too, thinking more about it. As soon as the butane vapour become significantly diluted with air, the partial pressure of butane will drop lower than 1bar and so more butane will evaporate. So there will actually be still a net flow of butane out of the hole, I think.