Getting a sauna to a certain temperature - faster with more rocks?

[davidorn]

TL;DR Summary

Need help with a recent friendly debate: does increased mass of stones in a sauna heater make it easier to get to a higher temperature?

We've been discussing what it takes to get a sauna to heat to 240F. Most sauna heaters will have some stones to provide mass to hold heat. The question is whether lots of stone mass will make it easier to reach higher temperatures.

My argument is that it does. Imagine heating a room using a heater with no stones. Let's say that to raise the temperature in the room from 100 to 110 is going to require X units of energy. Then consider how much energy would be required to raise it from 110 to 120. My argument is that when you're trying to go from 110 to 120 you will require less energy if you have stone mass than if you don't because without stones the heater is the only thing transferring heat to the air but with the stones they help raise the temperature because they are hotter than the air. (While this example talks about stones versus no stones, it applies to more stones versus fewer stones, too).

It's not a closed system -- there will be some amount of cooling through the walls, etc. And because the cooling is at a relatively constant rate, the ability to hold the heat up higher because of the stones seems relevant.

Am I thinking about this correctly? I'm wondering if I'm missing some factors that mean this doesn't make any sense.

 

[Nugatory]

More stones makes it harder to change the temperature in both directions: slower to heat but once heated will hold the temperature longer.

The problem with your line of thinking (if I’m understanding you properly, and if not I apologize and you can ignore me) is that the stones have to be hotter than 110 if they’re going to raise the air temperature past 110. But to get them hotter than 110 we have to heat them, and we’d be better off just heating the air directly.

 

[davidorn]

Please don't apologize - I'm just trying to work this through, and I appreciate any help or clarification.

It seemed to me that in a fully closed system what you're saying would make sense. But it's not a closed system -- there is some amount of cooling (I don't know how to estimate that -- it's well insulated but still cools down if not being heated). In my real world sauna, it takes almost 2 hours to reach near 240F. So, I was thinking that during those two hours the heat/energy loss would be meaningfully less because the heat held in the stones will dissipate much more slowly than heat only held by the air. Does that make any sense?

 

[sophiecentaur]

The question is whether lots of stone mass will make it easier to reach higher temperatures.

I read a few discussions on saunas and the answers to most of the questions all started with "It depends". (OMG it's a rich vein for 'opinions' and BS) Your friend and you were probably arguing with different models in your heads so you could both have ben right.

More stones makes it harder to change the temperature in both directions: slower to heat but once heated will hold the temperature longer.

True but it's more complicated than that. It all depends on where you start in the process. If your 'time' starts at switch-on then more stones will mean it just takes longer to get going. If you start your timer when you get into the sauna and the stones have already reached a useful temperature then you can get more instant steam as the stones cool down to 100 (which they will, on the surface). So the user will get hot faster if there are a lot of stones (a sort of flywheel effect).

As for cooling down time, that will depend on details of how well insulated the stones are from the room and the relative heat loss out of the room.

Adding water to the stones will cool them down, by evaporation (heating turned off) but can increase the room effective temperature for a while as the steam condenses on the walls and the user. A low mass of stones will mean less 'flywheel mass' available when you turn off the mains supply.

Experiments: There are too many variables to come up with a definitive answer to this, I think. Some experimenting would be needed - or else you would just be in the hands of the salesperson (as always).
One thing you could do is to measure the temperature in the sauna after it has warmed up but with no added water. The steam will act as a heat pump, increasing the room and body surface temperature and there will be an optimum amount of added water to get the maximum desired sauna effect before the stones cool down. Try bigger and smaller quantities of water and stones, in turn and do some subjective testing.

 

[Mister T]

In my real world sauna, it takes almost 2 hours to reach near 240F. So, I was thinking that during those two hours the heat/energy loss would be meaningfully less because the heat held in the stones will dissipate much more slowly than heat only held by the air. Does that make any sense?

No. Heat is transferred from inside the sauna to outside the sauna. The rate of transfer is roughly proportional to the temperature difference between the inside and outside of your sauna. The rocks will keep the temperature inside the sauna higher for a longer time, thus more heat will be transferred out of the sauna, not less.

It takes a longer time to raise the the temperature inside the sauna because you have to raise the not only the temperature of the air in the sauna, but also the rocks.

 

[sophiecentaur]

No. Heat is transferred from inside the sauna to outside the sauna.

That, without expanding it, could be misleading. Whilst the rate of heat loss is nominally dependent on temperature difference between room walls and outside, the actual distribution of temperatures affects it. I'm not sure what the OP is getting at here, though. Heating up the mass of stones will account for a large proportion of the heat input ('flywheel effect'). Heat capacity of the wood walls will be relevant but I expect the choice of wood will be to have low conductivity and low specific heat.

Adding water to the stones will raise the wall temperature as vapour condenses and reduce the temperature of the stones. (Or is that just obvious?). The temperature distribution within the stones must also be relevant to how the sauna is operated.

 

[hutchphd]

No. Heat is transferred from inside the sauna to outside the sauna. The rate of transfer is roughly proportional to the temperature difference between the inside and outside of your sauna.

I disagree with this conclusion because the rocks, being on the heater, are always hotter than the ambient temperature of the Sauna. So fewer rocks will be faster. You do need a sufficiency of hot rocks to support instant water vaporization and the resultant moving "shock" wave of hot.
Of course the warmup time can be put to good use fortifying oneself against the eventual cold water plunge (or snowdrift roll in a pinch). I have been told that either beer or some very strange flavored Finnish vodkas will work but deny any direct knowledge.

 

[sophiecentaur]

I disagree with this conclusion because the rocks, being on the heater, are always hotter than the ambient temperature of the Sauna.

I still don't know what you are actually trying to say. The fact is that it's no good just talking about "temperature". What counts is temperature distribution within the room and within the rocks. If you have actual experience of the effects of using more and fewer rocks etc. etc. then that evidence would need explanation in terms of a more complicate model - or you could just enjoy your sauna and your beer and the fact that you have found a cheaper way to run the sauna.

 

[hutchphd]

The model is not very complicated. The stove heats the rocks which heat the room (faster if water is involved). I believe the rate limiting interface is rock/room and this will strongly depend upon rock temperature (radiant conductive and convective effects). A large rock thermal mass will take longer to get to a useful temperature. . KIPPIS !

 

[sophiecentaur]

The model is not very complicated.

I beg to differ - that is if you want a good answer. The air flow round the room (convection) is what carries the heat to the walls. That is not a linear process. Conduction through the rocks will probably be (until the water arrives) and so will conduction through the walls. But transfer from the air to the wall surface (a 'soft' wooden finish, iirc) will also depend on the air movement.

A very simple analogue would be resistors in series (potential divider) but I really don't think this situation is a simple as that although a potential divider would be good for hand waving.

 

[hutchphd]

I beg to differ

My statement is tautologically true. The situation may be more complicated (I think it is not really) but the model upon which I was basing it is as I described it (it is, after all, my model!) and the conclusions follow from it. Every situation is more complicated that the essential model.
I have built several woodstoves and one sauna (with a friend) and I frankly don't know what a "good" answer means to you. For this situation the mass of thermal rock will control the balance between speed of warmup and the pleasant operation of the sauna. I am certain these are competing requirements, and would determine the precise optimal amount experimentally. It would be a tough job but I would volunteer. Might take all winter.

 

[sophiecentaur]

My statement is tautologically true.

I can't argue with that.

However, if you want to be able to predict the behaviour of a system with some accuracy then you need a model that's complex enough. I may have missed the point of the conversation but I thought it was about having the right amount of rocks that balance economy with user experience. I really don't think the Potential Divider model can produce an optimum balance, somewhere between one small rock and 100kg of them. If you were to have forced convection with no steam then the simple model would be fine but a sauna is a self-driving system and natural convection is unpredictable.

At school, in the days when we did 'Heat' Experiments, I remember doing cooling and heating curves to eliminate errors. Perhaps something like that would be possible, using a range of rock masses. You would be forced to use the sauna many times before you found an optimum. Sounds good to me.

On the subject of wood stoves, I have used a few and have owned on for a few years now. I haven't found one that's stable when operating at suitable flue temperatures to avoid soot and condensation.

 

[hutchphd]

I may have missed the point of the conversation but I thought it was about having the right amount of rocks that balance economy with user experience.

Darn it you made me reread the original question. I have no idea what question I was actually answering! (I feel certain whatever question I answered was answered well!)
Truth be told I think I started to bliss out early in the process...I love both cold weather and the sauna and it affected my concentration.
In fact I don't really know what the original question relates to. It seems not a very useful concept. Generally the thermal mass of an object does not affect the steady-state temperature...only the time constant. And I agree that a cooldown profile and a warmup profile at known power give you the data you need; heat leakage..
So the answer to the original query

Summary:: Need help with a recent friendly debate: does increased mass of stones in a sauna heater make it easier to get to a higher temperature?

The question is whether lots of stone mass will make it easier to reach higher temperatures.

is a qualified no. In steady state there will be no difference. If you have a bigger pile of hot rocks you can produce a larger transient steam-induced delivery of heat to your skin.

 

[sophiecentaur]

Darn it you made me reread the original question.

That can be risky. When we do that, we can meet ourselves coming back.

 

[DaveE]

Y'all could argue for years about the best way to build a sauna. There are too many choices, too many variables, to come to any simple general solution to the original question.

The only simple, general, answer is that at equilibrium (steady state) the heat lost from the sauna must equal the heat input. How that heat is stored or distributed inside the system isn't relevant at the most basic "black box" level. For a limited amount of heat input, it is normal to assume that the equilibrium temperature is inversely proportional to the heat loss through the walls. That is the key parameter, the insulation of the walls. Internal heat flow to the walls is nearly irrelevant in equilibrium, all of the heat will get there eventually, somehow.

Sure, how you arrange stuff inside will have second order effects. But those questions aren't answerable without a great amount of detail.

 

[DaveE]

Darn it you made me reread the original question. I have no idea what question I was actually answering! (I feel certain whatever question I answered was answered well!)

Maybe because the title doesn't match the subject description. One asks "faster" the other asks "easier". Thus, inherent confusion...

Then the question and responses veer into what I consider second order poorly defined issues.

 

[sophiecentaur]

Internal heat flow to the walls is nearly irrelevant in equilibrium, all of the heat will get there eventually, somehow.

I suppose you are assuming that the surface of the rocks is a 'Constant Temperature source' (as with an electrical Voltage Source) and independent of the volume of air in the room and the wall area. Is that valid? A potential divider with one resistor much lower than the other, if you like.

 

[DaveE]

I suppose you are assuming that the surface of the rocks is a 'Constant Temperature source' (as with an electrical Voltage Source) and independent of the volume of air in the room and the wall area. Is that valid? A potential divider with one resistor much lower than the other, if you like.

No, I'm saying (to first order) nothing about the rocks inside the sauna matters. I am simply balancing the energy that goes into the box with what comes out, and then asking about the thermal resistance of the barrier between inside and outside. I don't necessarily need to know the temperature of the heater element.

So, in your electrical model, I would only care about the current that flows from the source branch. My version would represent heat flow as current, temperature as voltage, insulation as resistance, thermal capacity (mass) as capacitance, etc. Then if the question is what is the temperature difference across the wall (voltage), I only care about how much heat (current) must flow through that wall, and the wall's structure (resistance). Granted it is the simplest model possible. But that's really my point, simple questions usually only support simple models.

If you add structure to the model, and say, for example, the heat has to flow through the rocks, then you will have multiple temperatures and "inside the box" becomes poorly defined. As if there were multiple rooms; as an electrical network has multiple voltages, not "a voltage". This is certainly a better model of a sauna IRL, but requires a more detailed description and has no general answer.

People often focus on the solution, when the real work is in asking the question. In modelling it's all about building, verifying, and defining the limitations of the models. Then the solution is nearly automatic, with computers and such.

 

[sophiecentaur]

But the user is not concerned with the ‘energy’. It’s the temperature. That depends on the external rock temperature. This is where the potential divider is relevant.

 

[DaveE]

But the user is not concerned with the ‘energy’. It’s the temperature. That depends on the external rock temperature. This is where the potential divider is relevant.

If you ask about one temperature, my model has one thermal resistance. If you ask about 2 temperatures, I have two resistors. If you want a pretty picture, my FEA SW will have thousands of resistors.

The potential divider is of course relevant to any but the simplest models.

Anyway, "rock temperature" isn't an adequate quantification of thermal energy input. So you'll have to add stuff to your model to determine the heat flow into the system. Soon I'm afraid you'll find there isn't a general answer anymore, it becomes a specific, probably underspecified, problem.

 

[sophiecentaur]

All temperatures will be changing until equilibrium is reached.
But, if you don’t want steam, no rocks or very few would have least heat capacity. So that would raise the temperature quickest. Then it would depend on how much water needs to be added for operation. That would determine the size of heat ‘flywheel” needed (=mass of rocks). Heater power is presumed adequate?

 

[In the Net]

Test your replies with this little proof: If you fill the room with, say, 85% rocks, would it take longer to heat the air?

 

[hutchphd]

The sauna would be a lot less fun, so no one would ever fire it up. We would never know.

 

[sophiecentaur]

But it’s all more complicated than that. What is the function of the water that’s added?
It’s used to produce a rapid heat transfer when we want it. Evaporation and condensation together accelerate heat transfer (like in a Heat Pipe). It delivers a dollop of extra heat compared with what we would get at the ambient. I suspect the condensation works better on our skin than on the wall surface.

 

[hutchphd]

I suspect the condensation works better on our skin than on the wall surface.

And an ancillary question is how much does the temperature of the rocks affect the details of that delicious pulse of humid heat. Can any good Finn tell you the temperature of the rocks from the feel of the pulse? Clearly even more experiments are required than anyone knew.

 

[sophiecentaur]

But it’s all more complicated than that. What is the function of the water that’s added?
It’s used to produce a rapid heat transfer when we want it. Evaporation and condensation together accelerate heat transfer (like in a Heat Pipe). It delivers a dollop of extra heat compared with what we would get at the ambient.
The surface area of the rocks will affect the rate of evaporation and the mass will affect the length of the heat pulse on the body. It would be tempting to do a very lengthy investigation.