Suppose we have a pot of known dimensions filled to level L with boiling water on a stove, covered by a lid with a vent of known dimensions. Given the steady-state temperature of the stove (bottom of pot's surface), how to quantitatively estimate the ambient pressure and temperature of steam in the pot, at 1 atm pressure outside the pot?

All I can think of is that the pressure will be a function of the vent size (monotonically decreasing) and stove temperature (monotonically increasing). That's because more steam will be produced at higher stove temperature and the velocity of steam escaping through the vent is a monotonically increasing function of pressure; at equilibrium the amount of steam escaping == the amount of steam produced per unit time. But is it only the pressure that's increasing as stove temperature increases? In the limiting case of no water and zero-sized vent, the temperature of steam will be increasing until it reaches the temperature of the stove. Does the presence of 100°C water affect the temperature of steam?

One way to attempt to solve this problem might be to consider the amount of energy entering and exiting the pot (the rate of mass lost due to steam escaping is low, so assume no mass is lost).

I expect the answer to the whole problem would be a set of simultaneous equations to solve. But what equation to use to find the rate of energy input? It would be to do with the temperature of the stove and the temperature of the water, 100°C, but I can't think of how to find it.

  • $\begingroup$ Actually, since the pot is metal and a good conductor of heat, it will be at the same temperature as the water. It's a gas stove, so next to a flame temperature of around 1000°C en.wikipedia.org/wiki/Flame#Common_flame_temperatures it will appear cold (in the typical case where the water temperature is around 100°C), so most of the heat will flow to the water. This makes the power supplied to the pot the useful independent variable, which is related to the rate of gas burning (let's assume linearly). $\endgroup$ Oct 6 '13 at 10:58

There are too many unknowns to model the situation with any accuracy.

If you know the power being pumped into your pan then you can easily calculate the amount of steam generated from the latent heat of vaporisation of water. You can calculate the power being generated by your cooker from the flow rate of the gas, or the current if it's an electric cooker, but it's anyone's guess what percentage of this is lost to the environment and how much ends up in the pan.

For any significant flow rate of steam I would guess that turbulent mixing will ensure the steam temperature is roughly constant throughout the pan. The temperature will be whatever the boiling point of water is at the internal pressure. The steam flow rate through the hole in the pan is fairly straightforward to calculate from first principles, though since steam is so important industrially I'd guess some Googling will find tables and empirical equations for flow rate.

Response to comment:

Let the mass of water lost per second be $m$, then the power applied to the water in the pan is just:

$$ W = mL $$

where $L$ is the latent heat of vaporisation of water. This will be equal to the power generated by your cooker times some unknown factor less than unity to allow for heat loss to the environment.

Response to second comment:

The temperature of the water will be close to the boiling point because any water hotter than the boiling point turns to steam, and the latent heat required will cool the water again. The steam in the layer immediately above the water will be at the same temperature as the water because it's in thermal contact with it.

If the steam above the water is hotter than the water you have to ask what is heating it. The only things I can think of that could heat the steam are the pan walls and lid. However the pan is only being heated from the bottom, and heat flow by conduction though the pan walls is a lot slower than heating/cooling by convection between the pan walls and the water in the pan. Therefore I would guess that the pan walls and lid are also close to the boiling point of water - actually they will probably be slightly cooler because they will lose heat to the surrounding air.

So I would guess that as long as there is enough water in the pan the steam in the pan will be close to the temperature of the water.

  • $\begingroup$ All unknowns can be expressed as variables. $\endgroup$ Oct 6 '13 at 10:30
  • $\begingroup$ I don't know the power, I know only the temperature at the bottom of the pot. How much power is being pumped in? $\endgroup$ Oct 6 '13 at 10:33
  • $\begingroup$ You're saying that the steam temperature will be equal to the water temperature, right? I still can't think of a compelling first-principles explanation of why this should be the case... Does temperature have to be a continuous function of space? So at the interface of water and steam there won't be a discontinuity? If so, why? The state of matter, whatever that is, is discontinuous. So perhaps what I really want to understand is the distributions of kinetic energies of H2O molecules in the liquid and gaseous forms in the pot. $\endgroup$ Oct 6 '13 at 10:40
  • $\begingroup$ For any temperature, there is a distribution of kinetic energies, which would be interesting, but I guess it does make sense that the water and the steam will be at the same temperature: turbulence will cause each domain to be at an isotemperature, and any steam hotter than the water will transfer the excess energy to the water, partially (is that possible?) providing the latent heat that is required to convert (a small part of) the water to steam. (Reminder: this is the energy put into a sample to achieve phase transition while it remains at the same temperature.) $\endgroup$ Oct 6 '13 at 11:44
  • $\begingroup$ Given all of the above, the final piece of the puzzle seems to be the constraint of Clausius-Clapeyron, as I've read in this gentle discussion on the subject geosci.uchicago.edu/~moyer/GEOS24705/Notes/SteamNotes.pdf‎ This is a function where pressure is exponential in (1/T), (ha! I would never have guessed) and is valid while there are both water and steam in the pot. I think this, and the above discussion, is enough to finally draw a set of curves of both the pressure and temperature of the steam as a function of theta, the angle of the gas handle. $\endgroup$ Oct 6 '13 at 11:53

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