Induction power estimator

A first-pass kilowatt figure from the three things that actually determine it: how much metal, how hot, how fast.

Sets a typical target temperature. Override it below.

kg

Only the mass you actually have to heat, not the whole component.

/hr

Parts per hour, at the rate the line must sustain.

s

Enter cycle time or production rate; we keep both in sync.

°C
°C
%

Coil, converter and radiation losses combined. 55 to 70 percent is realistic for well-coupled steel work. It is an input here rather than hidden inside the answer.

Estimated power required

0kW

Useful energy per part
0 kJ
Throughput
0 kg/hr
Cycle time available
0 s
Temperature rise
0 K

How this is calculated

The useful energy to raise a mass through a temperature difference is the standard sensible-heat relation:

Q = m · cp · ΔT

Multiply by the number of parts per hour to get useful energy per hour, convert to kilowatts, and divide by the overall system efficiency to get the power the supply must actually deliver.

P = (m · cp · ΔT · n) / (3600 · η)

Efficiency is exposed as an input rather than buried, because it is the single largest source of uncertainty in the answer and it depends on how well your coil couples to your part.

Check the estimate against a real machine.

Send the part, the material, the temperature you need and the rate. You will get a specification with a frequency, a coil concept and a handling arrangement, which is what a number on its own cannot give you.