[engineerscott]

ty1295 is on the right track. If you halve the voltage you need to double the current to supply a given amount of power (which is measured in watts). A 220VAC oven that draws 50 amps would require 100 amps at 110VAC. Other down sides are that you will have higher waste power lost in your supply feed and you will have to use a significantly larger guage wire at the increased current level. This is why the voltage requiremens for industrial machinery generally starts at 220V, and 440V and 660V is not at all uncommon for high power equipment. Large industrial motors like to run off of three phase at 240V and above.

I do question ty1295's estimate that you need 350W per cubic ft to heat your oven. This seems awful high. The example he cites is of a 410 cubic ft oven (9'x7'x6.5'). I have a really difficult time believing that you need 143.5 kW to heat an oven of that size. Just to give you a point of reference most moderate size homes are wired with a 200 amp service. At full load such a service could deliver 22 kW (in reality you would never come close to drawing that kind of load for any significant amount of time). The amount of power you need per cubic ft is a function of how good your insulation is and what kind of temperature ramp you want to support (i.e. how fast the temperature needs to rise in degrees/second). With good insulation and a reasonable ramp rate I would think you could cut that 350W/ft^3 from 1/2 to 1/5 that number.

Lets say you have a commercial operation using ty's 143.5 kW oven. If we assume $0.09 per kilowatt - hour then it would cost $12.91 per hour to operate. Lets assume that the oven is in operation 20 hrs a week (i.e. 4 hrs per day). At that usage rate that oven is going to cost $1033 per month to run (or about $13,000 per year). You'd need to push a lot of product through that 9'x7'x6.5' oven just to pay the electric bill.

In spite of the fact that I think his power number is too high, for an oven of the size you are talking about I think running it off of 110V is really unrealistic. The only way you might even come close to making it work is to have extremely effective insulation ($$$) and to be willing to tolerate a very slow ramp rate (i.e. warm up time to operating temp).

I did look at the largest walk-in oven that Caswell sells. It is 190 cubic ft and requires 15 kW. That works out to about 79 W per cubic ft. I have no idea what the insulation factor is and the temp ramp rate it can achieve but considering that this oven is a little under half the volume you are considering and it runs off of 240V three phase tell me that 110V just ain't going to cut it.

If I was a little more sober I could calculate the require power for a given oven volume, insulation factor and temperature ramp rate, but it is getting late and my math ability is fading fast.