Optimized insulation thickness - a theoretic apporach

[Petter]

Well, you can't insulate too much, but where is the point of diminishing return? I did a calculation assuming constant wall thickness and varying the mass to insulation ratio. For example 20 cm wall, 5 cm mass and 15 cm insulation. A wall of 20 cm mass will store a lot of heat, but loose it fast due to the thermal conductivity. A wall of 19.9 cm insulation and 0.1 cm mass would be well insulated but not able to store enough heat. Where is optimum?

I assumed the following in the calculations:
36" hemisphere
Under floor, side of floor and dome insulation were equally thick
Thermal conductivity of insulation was set to 0.1 W/mK
Temperature of hearth bricks and dome was 400 K above ambient.
Hearth bricks were 1.5"in all calculations. Hmmm, these should have been 2", but anyway, the difference is small.
2.30 g/cm3 for both wall mass and hearth bricks.
No heat loss through the oven opening, i.e. no opening...

I calculated the stored heat energy in floor and wall at 400 K above ambient and divided it with the total heat loss rate of the oven. This gives me: (kW x h) / (W / K) = h * K. The h (time) term is of interest since
the temperature is assumed to be constant (it will drop with time of cause, see it as a potential if you like). It will however explain the differences when changing ratio between insulation and mass.


By calculating for a total wall thickness of 10-20-30-40 cm I get the following results:


The higher the better (but the actual value has no practical application). For example, if your wall thickness is 20 cm and you wish to optimize retained heat cooking, your thermal mass should be 3-4", (7.5-10 cm thick). By doing so, the temperature will drop as slow as possible. Not too much mass or too much insulation.

If all these optimums are plotted against wall thickness, the following graph can be made:


All dots fall on the same line. The optimum was found to range from 40 % thermal mass at 10 cm wall thickness to 45 % at 40 cm wall thickness.


Was this understandable? Do you agree? Comments are welcome. Of cause, in a design situation, the actual values for thermal conductivity and densities should be used. However - I think this gives a good indication of what is reasonable and not overkill in either direction.

[david s]

Interesting work. Thermal conductivity of course does not remain constant with temperature change. As an example the tempesture vs time graphs of ovens is not a straight line. This applies both to heat up and cooling down. The old adage “the higher the temperature; the greater the heat loss” applies. As an example, when firing my kiln I’ll consume more gas from 800C -1100C than 20C-800C . And cooling down takes ages once the temp drops to around 300C. If kilns are too well insulated not only is the cost increased well beyond any gains from saving fuel, but the waiting to unload is extended beyond reasonable limits. A kiln is also a little different in that a large proportion of the thermal mass is due to the kiln furniture and wares inside whereas an oven chamber is nearly always empty apart from large cast iron pots that might affect the heat up and cooling down.

[Kvanbael]

I like your quest.

Allow me to generalize for a large room:

temperature drop = temperature difference * thermal conductivity of insulation / ( thickness of insulation * brick thickness * brick density * brick heat capacity )

The two thickness variables are just multiplied in the denominator. That actually means that, irrespective of the material properties, temperature drop will be least if the given total thickness is equality divided by brick and insulation.

Obviously, temperature will drop slower with lower thermal conductivity, higher heat capacity and higher total thickness. But when it comes to layer thickness, the best performance is with equal thickness.

I wonder why your calculation resulted in 40%. Maybe because you fixed the floor thickness.

In real life:
- you also want to optimize time and wood to heat up
- the bricks are never equal temperature all across.
- there is an opening

[david s]

I don’t agree that equal thickness of wall and insulation is the ideal. It is generally considered that for cast ovens a 50mm (2”) works pretty well, providing reasonable thermal mass for retained heat cooking at the same time as providing reasonable heat up time and fuel consumption. But, a 2” thick insulation over it would be considered absolute minimum, better with 75mm (3”) or more. Practice and theory don’t always agree.

Brick ovens are generally 100mm (4”) because only 2” of mortar depth is insufficient to bond units together for longevity.

[Petter]

Thanks for your input to the debate.

David s, What you refer to is that the Newton's law of cooling as is and in reverse? An object will suck up and loose heat in a rate proportional to the temperature difference, here between mass and ambient / mass and flame.

The thermal conductivity at RT has a neglectable radiation contribution while at 400 C, there is a small one that increase heat loss rate. I personally think that the initial sharp drop in temperature is attributed equalization in brick temperature once you stop fireing. See here for example.

https://community.fornobravo.com/for...-matters-to-us

Kvanbael, Intresting formula, did not know that one. Seems reasonable. I think of other reason as well why my calculation differs and that is that the dome mass thickness is spherical, not flat. Also, the insulation is assumed to have constant area, even though it expands radially over a dome. This is for The ease of calculation. I think we were fairly close anyhow from a simple Excel sheet to the idealized case. :-)

David, you fell in the trap :-) 3" over a 2" cast would equate 5". The optimum would thus be 2.5" + 2.5" for heat retention. The formula is only valid for a given wall thickness and does not encounter fuel consumption.
I do however agree that for practical purpouses, most of us are willing to sacrifice some heat retention för a large reduction in fuel consumption during heat up. You can easily take a ratio of 1:2 - 1:3 mass : insulation and still get very good heat retention for equal wall thickness but with much less fuel than a 1:1 ratio.

[david s]

Thanks Petter,

"when it comes to layer thickness, the best performance is with equal thickness."

So how much insulation would you expect to be ideal for a steel dome of 5mm thickness? surely not 5mm of insulation.

Just re-reading your post re thermal conductivity. I was always somewhat perplexed about the thermal conductivity/density of blanket I use as stated in their data sheet. Expecting the higher density blanket to be a poorer insulator. Re-visiting the data sheet I notice that is not the case and the lower density blanket at temperatures over 200C is as expected, the better insulator. They must have revised the error.
DATA Superwool 607 Blkt.pdf

[Petter]

So how much insulation would you expect to be ideal for a steel dome of 5mm thickness? surely not 5mm of insulation.

They must have revised the error.

Welcome,

Well, for a Wall thickness of 10 mm, your suggestion is correct. If the Wall thickness is variable, more insulation is allways better. You would need a minimum of 50 mm anyway to reduce heat losses during baking. Remember, the 50-50 rule is only valid for best heat retention for a given Wall thickness.

looking at the data sheet, it looks perfectly reasonable now.