Re: ATM - Kiln Design

[Anthony Stillman <atmer@flash.net>]

The power dissipation Q through the walls of a kiln can be described by the
equation:

  Q = K * A * deltaT / L

Where K is the wall thermal conductivity
      L the wall thickness
      A the surface area
      deltaT the temperature difference across the wall


A kiln with a void one meter in diameter and half a meter tall, has pi
square meters of surface area.  Assuming an internal temperature of
1100C.(1) and kiln walls 50mm thick made of alumina(2), the power
dissipation through the walls is 15KW.  As the thermal conductivity of
alumina is a function of temperature(3), dropping T to 400C cuts the power
loss to 2.8KW.  A smaller kiln with a void a half meter in diameter and a
third of a meter high only dumps 4.4KW at 1100 C and 800 Watts at 400C.  Of
course the heat loss can be halved by doubling the wall thickness.  And,
there are other refractory materials with better insulating properties.
For very high temperatures, an onion skin approach can keep kiln
construction costs down.

A search on google turned up a number of interesting hits and I found this
page particularly fun(4):

http://www.zircar.com/Sample/products/furnmod/furnmodp.htm

Anthony



1) And for the sake of simplicity a room temperature of zero.
2) At 1100C Alumina's coef. of thermal conducitivity is 0.22 Watts/meterKelvin
3) At 400C Alumina's coef. of thermal conducitivity is 0.11 Watts/meterKelvin
4) I do not want to be around when they power up the one in the lower left. ;)