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.

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.

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