Question to settle a bet… if you took a sealed container containing some water, put it in a pressure cooker also containing some water, and heated it to a temp well above boiling, what would happen? Would there be any pressure differential causing the sealed inner container to explode, implode, ec?
I believe the pressure would be equal inside and outside the container, and its integrity would not be in jeopardy.
It would depend on whether the amount of water in the cooker and in the container were equal relative to the size of the respective spaces. If there was significantly less water (relatively) in the cooker, for example, the expansion of the water in the container would create a lot more pression inside than there would be on the outside to balance it.
Ok, it’s been a while, but let me see if I can remember some high school stuff and information from my job of about 20 years ago (selling temperature and pressure controls, gauges, and valves).
In an ideal saturated system (a closed system in which there exists both water and steam and where any heat applied is distributed evenly through the system) the pressure and temperature are tied together in a specific way.
A given temperature above boiling will always give you the same pressure, and a given pressure always means the system is at the same temperature as any other saturated system at the same pressure.
For example, saturated steam at 20 psia (or about 5 psig at STP) will always be about 228 degrees Fahrenheit. Doesn’t matter how much water you have, is the container 10% or 90% full, big container like a boiler, small container like a pressure cooker, it’s going to be about 228 degrees and about 5 psig.
I think this means that assuming the heat applied to the pressure cooker raised the temperature of the interior vessel the same amount that it raised the temperature of the water in the pressure cooker, the pressure would have to be the same inside and outside of the smaller vessel.
In reality, I’d expect the water in the pressure cooker that was outside of the smaller vessel would heat slightly faster than the water in the smaller vessel, and so there would be some pressure pushing in on the smaller vessel. How much? Hard to say, it would depend on many variables.
But given time for the heat to equalize, the pressure should also equalize.
This means no explosion or distortion, the pressure is (or will come to be, given time) the same inside and outside of the smaller vessel.
At least I think. It’s been a long time.
Yup, at equilibrium.
Here’s a table: Properties of Saturated Steam - Pressure in Bar
With a real world pressure cooker sitting a stove, and equipped with a pressure relief valve, you might build up a little excess pressure in a can sitting on the bottom of the cooker during the steady state. A can will mess with the heat flow a little bit.
Assuming this is a reasonable experiment and we are not going to do anything stupid such as going too far in any direction:
Put a pint of water in a quart canning jar and seal the lid tight.
Place the jar in a pressure cooker and add water to the level in the jar or little higher.
Aplly heat from stove top and bring pressure cooker up to normal operating pressure.
The jar will heat up slower, and the pressure will be lower than that in p.c. but since the jar is in compression it will not fail.
When, if, you remove the pressure control weight on the p.c. the jar may explode or may buldge the lid insert.
If you raise the heat and hold the p.c.w. down the safety disc will blow letting the interior pressure to suddenly fall to atmospheric with greatest possibility of jar exploding.
Oh, no doubt about it. I think it’s quite likely that there will be small temperature differences between the inside and outside of the smaller vessel, and possibly fairly large ones for short times.
But given time for the system temp to stabilize, I’d think the differences would be pretty small, and the pressure difference would be similarly small and (I think) pretty much negligible.
Aha… so important safety tip, we don’t release the pressure valve, we allow the entire system to cool to room temp prior to opening?
Correct me if I’m wrong but ISTM that the previous answers perhaps don’t emphasise a particular point enough: if you assume that temperature throughout the system are even, then the pressures in the inner can and the pressure vessel will remain equal as long as we are talking about saturated steam. That is, steam at a temperature equivalent to the boiling point of water at the pressure in the vessel. What that means in practice is that as long as any liquid water remains in both containers, equilibrium will reign.
In effect, the presence of water moderates the system. For any given temperature, only a certain pressure can be reached because boiling will stop.
However, if one or both containers boil dry, as I understand it the situation changes completely. At that point, gas pressure becomes a function of (a) container volume (b) temperature and © the amount of moles of gas in the container.
So if the temperature goes high enough to boil off all the water, and if the ratio of gas to container volume as between the pressure cooker and the inner can is not equal by a sufficient margin, the can will collapse/explode.
What this implies to me is that as long as there’s an excess of water in the outer container, then pressure differentials would not occur (since saturated steam could not go beyond 228F, how would the inner container ever exceed that temperature?)
Yes, absolutely, the pressure/temperature relationship holds for saturated steam only, exactly as you say.
I began my original post with “In an ideal saturated system”, but did not emphasize the point that all the rest of my post relied on that. Good catch.
It’s not that saturated steam can’t go beyond 228F, but that saturated steam at about 5 psig will always be about 228 F. Saturated steam at about 10 psig will always be about 239F, etc. As the temperature goes up and down, so does the pressure, and vice-versa. You can change either one by changing the other, and a given pressure will always equal a given temperature, as long as you’re talking about a saturated system (one where even a little liquid water is still available to change to steam).
So as long the temperature and pressure don’t get to the point where all the water has changed to steam, either inside or outside the smaller vessel, the pressure should be about equal inside and outside of the smaller vessel.