Furnace heating element

John Ings · 2004-01-13T23:44:58+00:00

After perusing posts about the Gingery furnace and visiting' have bought myself a nice little thermocouple and controller kit one-bay with the intention of constructing a small tempering andannealing furnace and maybe doing some enameling. Nothing so ambitious as a melting furnace for metal casting.Anyhow, the question of what to use as a heating element arises. Wheredoes one buy suitable heating coils, and how do you calculate how muchyou need with respect to the volume of the furnace? Would a stoveelement do? Or some other clever substitute for expensive purposebuilt oven elements? I have both 115 and 240 volts available at up to50 amps.

Jan 13, 2004 23 Replies

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Don Foreman 22 years ago

Yes, if the insulation's conductivity can be expressed as a function of temperature, either as an equation or as a set of tabular values.

I'd take as many datapoints as are available for insulation conductivity at various temps and then fit a curve to those points to get an empirical equation.

The area also increases with distance from the inside surfaces but I'd ignore that. Calculations like this are approximations at best anyway, because they don't take into account radiation or convection.

Ron Reil reports that coating the inside with some ceramic stuff that reflects radiation helps quite a bit.

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dave martin 22 years ago

The effective area of a wall that increases in size from the inside to the outside is close to:

EffectiveArea=SquareRoot(OutsideArea*InsideArea)

The heat flow resistance of a particular wall i is: R_i=thickness_i/(ThermalConductivity_i*EffectiveArea_i)

where ThermalConductivity_i is evaluated at wall i's average temperature.

The total resistance of a wall with n layers is close to: (using units of BTU/hr, squarefeet for area, inches for thickness) R_total= 2/OutsideArea + R_1+R_2+...+R_n

The first term on the RHS accounts for the so-called "film effect" at the furnace's outside wall. The film effect on the inside can be ignored for a hot furnace.

The heat loss from a furnace wall is TotalTemperatureDrop/R_total

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Don Foreman 22 years ago

Wouldn't you have to do this iteratively until the various intermediate temps converged to the corresponding Rvalues used?

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dave martin 22 years ago

......

You are correct. The iterations will quickly converge.

Use the fact that just as in a series electrical circuit:

deltaT_i/deltaT_total = R_i/R-total

ie. the temperature drop across a particular layer is proportional to the resistance of that layer. Restimate a layer's thermal conductivity using the temperatures indicated by the last iteration. It will converge rapidly.

Don't bother with more than a few iterations. The thermal conductivity data isn't all that good & neither are the construction details.

Iterate a few times to find the heat loss. This is the absolute minimum amount of power to hold the system at temperature.

You might want to add some power so things will heat up at a reasonable speed.

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