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I made an electric furnace that consume ~18 kW (3 phase) of power. I want to know how much energy is lost when its used to heat only air. The only sensing tool available as of now are thermocouple installed inside and amperemeter. It took approximately 250s to reach 500°C from 25°C. It's not fully sealed so there are some holes for air to go in and out.

Cant seem to make up my mind to decide which factor I should consider for the heat loss.

To add, it's an electric furnace and has temperature limit set to certain degree (500°C in my case, can be changed). At that point the contactor will start turn on and off to keep temperature constant. All I know is that it consume constant 27.3 A when its powered. That's where I got 18 kW from.

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    $\begingroup$ Wait for thermal equilibrium then measure the energy input. $\endgroup$
    – Solar Mike
    Jun 18 at 19:12
  • $\begingroup$ Can you elaborate on thermal equilibrium? Since its and electric furnace and has temperature limit set to certain degree (500°C in my case). At that point the contactor will start turn on and off to keep temperature constant. All I know is that it consume constant 27.3 A when its powered. Thats where I got 18 kW from. Leaving it powered with no limit could risk overheating and breaking component inside. $\endgroup$
    – Bramble
    Jun 18 at 19:21
  • $\begingroup$ Who said remove the limit? $\endgroup$
    – Solar Mike
    Jun 18 at 20:22
  • $\begingroup$ There are lots of equations to calculate heat loss. This is a good portion of a heat transfer course. Though most ovens don't have holes in them becaue usually you only want to heat the inside. $\endgroup$
    – Tiger Guy
    Jun 19 at 23:00

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Since its and electric furnace and has temperature limit set to certain degree (500°C in my case). At that point the contactor will start turn on and off to keep temperature constant. All I know is that it consume constant 27.3 A when its powered. Thats where I got 18 kW from.

That's fine. Run it to 500°C and let it stabilise for a while. Then record the duty cycle of the contactor. (That's the on-time divided by the total cycle time. You might want to average this over many cycles.)

Heat loss at 500°C = average power in = 18 kW × duty cycle.

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  • $\begingroup$ I see, yeah, that could be a good way to calculate it $\endgroup$
    – Bramble
    Jun 19 at 3:19

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