# Use of reflector to contain heat in kiln

My understanding is that most of the heat in a kiln is transmitted by radiation, not convection. This would suggest the use of a reflective wall for the kiln to contain the heat.

So, for example, imagine we line the kiln with polished silver. Will the silver reflect the heat away until it is absorbed by the bolus, or will the air in the kiln just get so hot that it will melt the silver?If the latter is true, could we contain the kiln in a thin wall of a refractory ceramic interposed between the reflective wall and the heating area, then evacuate the air between the two?

Will the silver absorb some portion of the infrared radiation rather than reflect it? How can I model how much energy will be absorbed versus reflected?

How can I model something like this to explore what happens in a simulation? I have Mathematica. Is it as simple as entering in a few equations, or will it be very complicated to model?

Ultimately, I imagine there are 4 main variables: the temperature of the bolus, of the air in the chamber, of the heating elements and of the wall of the kiln.

• Not exactly sure what the objective is here. You're trying to limit the escape of heat from the kiln (oven) as a whole? Or you're trying to limit the escape of heat from the object inside the kiln to the walls of the kiln? But doing this doesn't make sense because anything that would block the flow of heat from the heated object to the walls would also block the flow of heat from the kiln walls to the object. May help if you provided a diagram to illustrate your thinking. – Samuel Weir Nov 13 '16 at 1:26

There's a reason why metals are rarely (if ever) used as refractory materials in kiln design: they're prone to either oxidation and/or melting. Silver has an MP of only $962\:\mathrm{Degrees}$ (Celsius) and therefore very prone to melting in a high temperature kiln.

From the point of reflectivity, there's relatively little lost by designing with more traditional and throroughly tested refractory materials, with respect to highly reflective metals like silver or aluminium.

My understanding is that most of the heat in a kiln is transmitted by radiation, not convection.

True as that may be, the kiln strives towards uniform temperature distribution, as heat always flows from hot to cold. Even modest amounts of natural convection would heat up any refractory parts quickly (depending on kiln operating temperature, of course).

How can I model something like this to explore what happens in a simulation? I have Mathematica. Is it as simple as entering in a few equations, or will it be very complicated to model?

It would be hard to do and not a question of 'entering in a few equations'. But almost certainly not impossible either: specialist literature would probably give important clues as to what kind of engineering equations, know how and approaches to use.

Even a relatively 'crude' model should allow rough estimates of refractory temperature (even if based on radiation only), to see if 'safe' temperatures will be exceeded or not.

• I am confused by the claim that a kiln will head towards a "uniform temperature distribution". Kilns I have seen are always much hotter on the inside than on the outside, so there must large differences of temperature within. – Ambrose Swasey Nov 13 '16 at 14:02
• The FACT that a kiln strives towards uniform temperature doesn't mean it succeeds at achieving that. It's a universal principle that heat flows from hot to cold, so that temperature gardients are diminished, See Fourier's Law. – Gert Nov 13 '16 at 15:25

The objective of the kiln is to heat whatever is inside. A conventional (thermal, non-RF) kiln does this by heating the walls, after which you correctly state that thermal radiation transfers the power to the bolus (charge of the kiln).

Now the emitted power scales with $\epsilon T_0^4$ where $\epsilon$ is the emissivity of the kiln, and $T_0$ the temperature of the walls. On the other hand, the charge will emit power in proportion to the fourth power of its own temperature. If the walls are at 1000 °C, and the bolus is at 800 °C, then the rate at which heat goes into the bolus is 2x greater than the rate at which the bolus emits - $\left(\frac{1273}{1073}\right)^4 = 1.98$ .

If you make the walls of the kiln from a material that is more reflecting, the emissivity will go down (although that is a strong function of wavelength, depending on the material). It is likely, then, that the rate at which heat initially is transferred to the charge will be lower.

The only thing you can do to increase the efficiency of the kiln is to make sure it is well insulated on the outside (multiple layers of insulation, including reflective shields) - on the inside, it is probably best to make it black.