Temperature and pressure in sauna?

[AudriusR]

Imagine that I pour a bowl of water onto the stones in sauna. So, the problem is that I need to estimate how much did the pressure and the temperature change in cause of this action.
Detailed answer would be very appreciated.

My attempt : pressure and temperature would decrease, because heater should boil and evaporate the water,which I poured.

Any other ideas?

 

[russ_watters]

Welcome to PF!

Saunas are not sealed, so the pressure in them never differs from the atmospheric pressure outside by more than an immeasurably tiny fraction.

Pouring water on the hot rocks will cool the rocks, but it also efficiently transfers a great deal of heat into the water, which then goes into the air as the water boils. Injecting 100C steam into a 50C sauna will most definitely heat up the sauna.

 

[AudriusR]

Welcome to PF!

Saunas are not sealed, so the pressure in them never differs from the atmospheric pressure outside by more than an immeasurably tiny fraction.

So you're saying that after I had poured the water, pressure did not change?

 

[Hg00000]

A sauna is a rather complex system to take a look at like this. Most saunas are electrically heated, and therefore actively temperature controlled. So changing the temperature in the system will cause the control loop to add more heat to the system.

Saunas also must be well ventilated. A sauna that isn't will quickly start to asphyxiate its occupants. That is a bad thing.

Finally, there are a huge variety of saunas with different volumes, materials of construction, stove designs, amount of rocks in the stove, size of rocks in the stove and materials of rocks in the stove to store heat.

For your sake, let's look at a heat storage sauna. Let's say it has 10m³ of air inside and it has a 1000 kg pile of rocks that has been heated to 600°C. We'll hang an adiabatically insulated rigid container of water over the stove with a dosing valve on it (so that volume is conserved) and wait for a temperature sensor to come to equilibrium at 100°C at eye level on one of the benches at the wall.

At the moment the system equilibrates, we'll seal all the openings to the sauna. Assuming air is an ideal gas, that traps (101 kPa x 10 m³) / (373 K x 8.314 J/mol-K) = 327 mol of air inside the sauna.

Now, we'll dispense 1 kg of water or 55.6 mol from the bag.

The water will hit the hot rocks and vaporize. That absorbs 2270 Joules from the system.

We'll assume the rocks are mostly silica, so they have a specific heat of 0.7 kJ/kg-K. That water is going to lower the average temperature of the stove by 3.2K. For simplicity, let's assume that it lowers the temperature of the whole system by 3.2K.

With that, again assuming that the wet air is an ideal gas, we'll now have a sauna pressure of: [(327+56 mol) x 370 K x 8.314 J/mol-K]/10 m³ = 118 kPa.

So the temperature inside of our sauna reduced by 3K and the pressure increased by 17 kPa.

http://saunascape.com/2011/06/how-much-water-can-i-toss-on-the-sauna-rocks/