# Calculating the rate of pressure increase in a water-filled boiler

I'm trying to think about how fast the rate of pressure will increase in an idealized boiler filled with water starting at 100°C, standard pressure, and supplied with 1000W of power. I'm assuming a constant volume process.

Initially, the boiling point of water at standard pressure is 100 °C, and the latent heat of vaporization is 2257 Joules/gram, and I assume I'm adding 1000W = 1000Joules/sec, so I'm converting 0.443 grams/second into steam. That works out 0.0246 moles/sec.

My naive assumption was that I could simply use the ideal gas law and solve for delta-P, but now I don't think that's correct. My reasoning is that as the pressure builds, the water is suddenly below the new boiling point of the water at the increased pressure, and thus the rate of vaporization will slow as the temperature of the water lags behind the increasing boiling point. Of course, the water is still evaporating at this sub-critical temperature, and heat energy is still being supplied at the rate of 1000W, so the pressure will continue to climb. But it's not at all obvious to me how quickly it will increase in this dynamic system.

This seems like it must be a solved problem, but I'm struggling to find an approach to developing even an approximately solution to estimate how quickly pressure will increase over time.

• Are you allowed to use the steam tables to solve this? Are you going to specify an initial volume fraction of liquid water? – Chet Miller Mar 24 '18 at 19:46
• This isn't a homework problem, it's just something I'm thinking about; so, sure, steam tables are fair game. I'd kind of like to end up with a generalized way to think about this at different volume fractions, but for sake of simplicity, let's assume equal volume fractions between liquid and vapor phases. – Chris Young Mar 24 '18 at 19:51

If you know the total volume of the container V and the total mass of water in the container m, then using the steam tables to get what you want is relatively straightforward. For any chosen temperature T, you look up in the saturated steam tables the specific volume of liquid water $v_L$, the specific volume of water vapor $v_V$, the specific internal energy of the liquid $u_L$, the specific internal energy of the vapor $u_V$, and, of course, the saturation pressure P. Then you have the two equations: $$m(1-x)v_L+mxv_V=V\tag{1}$$and$$m(1-x)u_L+mxu_V=U\tag{2}$$where x is the mass fraction of water that is vapor.
For a constant volume system, you know from the first law of thermodynamics that $Q=\Delta U$, where Q is the total amount of heat you have added to get from the initial temperature to temperature T. This is just the heater power times the time. So you have the time required to raise the container contents to temperature T and pressure P.