How would one calculate pressure increase in cooker

  • #1
If you have a cooker with a volume of 3000 cm^3, then you add 200 ml of water, and let it turn into vapor at 100 C over time, how would you calculate the pressure at any point during the evaporation? It's not homework, just an example so you get what I am saying.
 

Answers and Replies

  • #3
Is the vessel sealed so that gases can't escape?
Yes that would be the case
 
  • #4
Nugatory
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Under those conditions the ideal gas law ##PV=nRT## will give you your answer.
 
  • #5
Under those conditions the ideal gas law ##PV=nRT## will give you your answer.
I got 114.7 bars by putting in all the values, but how do I know the pressure after a given amount of time? For example, it will not be the same at 5 minutes as it was at 2 minutes? When was it 114.7 bars, is that when it completely evaporates?

By the way ,114.7 bars sounds like a lot
 
  • #6
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I got 114.7 bars by putting in all the values, but how do I know the pressure after a given amount of time? For example, it will not be the same at 5 minutes as it was at 2 minutes?
Once all the water boils into steam, the pressure will stabilize at whatever level the ideal gas law gives you (aside from small corrections because steam isn't quite an ideal gas, but that's a rounding error here).

However, you can't calculate how quickly the pressure ramps up to that level unless you also specify the rate at which heat is entering the pressure vessel - it whether it heats up to 100 degrees in a year, a day, or a minute makes a big difference in when any given pressure is reached.

If you do have that rate, you'll be able to calculate the pressure reasonably well from the initial temperature of the water and the pressure vessel, the specific heat of water, and the heat of vaporization of water. Depending on how accurate you need your results to be and how interested you are in the time before the water starts boiling, you will also need to consider the contribution to the the pressure from the air already in the vessel; this contributes to the value of ##n## in the gas law.
 
  • #7
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When the vessel reaches equilibrium at 100 C, only a tiny fraction of the water will have evaporated into the head space. I suggest that you simplify this problem a little to begin with by assuming that the heating is done slowly so that, at any time, the system is basically at thermodynamic equilibrium at the existing temperature (which is assumed uniform throughout the vessel). You can then calculate at each temperature the
  • Fraction of liquid
  • Vapor pressure of water
  • The partial pressure of the air
  • The total pressure
  • The internal energy of the vessel contents
Based on the internal energy change, and assuming a constant heating rate, you can calculate the amount of time required to reach each specified temperature.

If you try to extend this to temperatures much higher than 100 C, you will need to consider non-ideal behavior of the air-vapor mixture.

I would start out even simpler than this by assuming there is no air in the vessel. Why? If you can't solve the problem with no air present, you certainly won't be able to solve for the case with air present. And it will give you a better mechanistic picture of what is happening. And you can solve this simply using the Steam Tables.
 
  • #8
However, you can't calculate how quickly the pressure ramps up to that level unless you also specify the rate at which heat is entering the pressure vessel - it whether it heats up to 100 degrees in a year, a day, or a minute makes a big difference in when any given pressure is reached.

If you try to extend this to temperatures much higher than 100 C, you will need to consider non-ideal behavior of the air-vapor mixture.

It would be a constant 100 C from beginning to end. By the way, does 114 bars sound reasonable for the example I gave?
 
  • #9
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It would be a constant 100 C from beginning to end. By the way, does 114 bars sound reasonable for the example I gave?
No way. Only a tiny fraction of the 200 ml of water would evaporate at 100 C. Maybe about 10 ml. Then, the water vapor in the head space would be in equilibrium with the liquid water. The pressure would be no more than 3 bars.
 
  • #10
No way. Only a tiny fraction of the 200 ml of water would evaporate at 100 C. Maybe about 10 ml. Then, the water vapor in the head space would be in equilibrium with the liquid water. The pressure would be no more than 3 bars.
How do you figure out the temperature needed to evaporate a specific amount?
 
  • #12
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@Chestermiller is right.... I'd been reading the problem as if you were planning to heat the vessel past 100C to whatever temperature was needed to boil all the water into vapor, not to heat it to 100C and leave it there.
 

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