# How to estimate power consumption for an oven given volume?

Is there a good way to estimate the power consumption of an oven/heating device given the volume of the oven? I'm trying to do really, really dirty/quick math - I know that the average oven uses probably between 2-4000 Watts, but I'd be looking for a similar estimate with a volume of about 1.3 cubic inches, so it would be far, far less than a normal oven.

I'm under the assumptions that heating up and insulating a smaller space would be significantly easier (better than just the linear ratio in size) than a regular oven. Is there a way to figure out how much power input would be needed to heat up that much volume in ~20 minutes (or any time)? I'm not sure of all of the factors that would need to come into play (I'm sure insulation/heat retention is important, but I don't know a typical number for that). I'm not sure if convection or conduction would be better, but presumably the one that would use less power? The maximum temperature would be ~400F.

This is for a project I just thought of, which is why everything is somewhat vague. I'm basically just looking for the worst case ceiling, and the best case ideal to have something to start from. Thank you!

• The heat transfer properties of the oven walls will be at least, if not more important than the volume. Do you have a guess as to what material(s) will be used in the walls? Commented Feb 29, 2016 at 15:56
• Unfortunately, I wouldn't be sure, at least at this point. The goal would be to have a portable device, so the walls wouldn't be extremely thick. I would assume it should be possible to have similar insulation characteristics to a regular, electric oven, though? My brother approached me with the project just today, but heat transfer kinds of problems aren't something I have much experience with other than at a high level. I could possibly do some extended research, but I was hoping to use "typical characteristics" of an oven to get an estimate Commented Feb 29, 2016 at 16:07
• If you'd want to have similar insulation characteristics, then you'd want to have similar insulation, not scaled-down insulation - even if the "inside" is ten times smaller, you'd need a comparable wall thickness (inch? half-inch?) to get comparable insulation; in which case the volume of insulation would be far larger than the inside volume of your oven. Commented Feb 29, 2016 at 20:31
• It can also be interesting to consider the electrical power consumed. Commented Mar 1, 2016 at 8:29

If you can make a guess at the interior and exterior surface temperatures, then you can use the following simplified steady-state equation for conduction through an arbitrary shape. It is from A.F. Mills, "Heat Transfer 2nd Edition", equation 3.32:

$$\dot{Q} = k \Delta T \sum_i S_i$$

where $\dot Q$ is the rate of heat transfer, $k$ is the thermal conductivity of the material, $\Delta T$ is the temperature difference across the wall of the material, and $S_i$ is a "shape factor" for surface $i$, which is a function of the size and shape of the material through which the heat is flowing. If your oven will be a cube with interior width $W$ and uniform insulation thickness, $L$, on all sides, then it can be decomposed into two basic shapes; 6 walls and 12 edges. I'm ignoring the corners because the amount of heat lost there will be small. According to table 3.2 from the same book, the shape factors for these surfaces will be:

$$\sum_i S_i = 6S_{wall} + 12S_{edge} = 6\frac{W^2}{L} + 12(0.54W)$$

For example, I'll assume the interior of your oven will be a cube, ~1 inch (0.0254 meters) on all interior dimensions, with insulation thickness of ~1 inch. So, $W=L=0.0254m$. The total shape factor will be

$$S_{tot} = 0.152+0.165 = 0.317m$$

If the interior temperature is 400F (477K) and the outer surface is at 120F (322K), then the temperature difference will be $477K-322K = 155K$

Assuming fiberglass insulation, like the kind on a common kitchen stove, the thermal conductivity will be $\approx 0.11 \mathrm{\frac{W}{m\cdot K}}$.

Putting it all together:

$$\dot{Q} = (0.11\mathrm{\frac{W}{m\cdot K}})(155K)(0.317m) = 5.4 W$$

So, after the oven is heated up and running at steady state, you'll need about 6 watts to keep it at operating temperature. Like I said, this solution depends on knowledge of the surface temperature, so if it's way off from my guess of 120F, you'll need to re-do the calculation.

• Thank you (both)! I made some painfully quick and dirty estimates of my own and got something similar to these (with a much less scientific approach). It's good to know that I was in the right order of magnitude, and that the idea is viable. Commented Mar 1, 2016 at 0:45

The volume, in itself, doesn't make much of a difference as this just determines the mass of air that you need to heat and air has quite a small specific heat capacity.

However the volume is directly related to the external surface area, which will have a significant effect. Assuming that the walls of your oven are reasonably comparable to those of your base for comparison in terms of heat transfer coefficient then the relative power requirements will be in the same ratio as surface area.

For example if both ovens are cubes and your reference oven is a 10 inch cube it will have a surface area of 10x10 inches per side (with six sides) so 600 square inches.

A 1 inch cube will have a surface area of 6 square inches.

So the ratio of surface areas is 600/6 = 100 so a reasonable approximation would be that the 1 inch oven requires 1/100th of the power to maintain a stable temperature.

So as a first approximation you are looking at tens of watts rather than thousands. This is about the same order of magnitude as a candle and so seems fairly reasonable to me.

There is also the consideration that for such a small volume it is fairly easy to be a bit more generous with the insulation than with a larger oven.

In terms time to warm up the key factor is the thermal mass of the oven structure and insulation which is proportional to volume ie a cubic relationship so you get even more benefit from reducing the size. Here I'm ignoring the scaling effects of the thickness of the insulation to have comparable heat flux per surface area but in terms of the practicalities of building the thing this is fairly trivial for a general case.

A simple way to make a very efficient oven on the scale of a few inches is simply to carve it out of soft fire bricks and, if necessary line it with a short length of stainless steel box section.