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I got this project to make, to anneal brass pipe edges (to be more precise, empty brass gun-shells) of diameter about 8mm. I would like to raise the temperature up to about 450-500°C in few seconds, not minutes. I was working with 120W induction module (as seen on photo below) which was able to properly raise the temperature, but in few minutes instead of seconds. I don't expect it to be hot in 2 seconds, but not more than 15 should be per cartridge. I'm not sure if induction goes linear in comparison with time (power x2 = 1/2 time) so therefor I'm asking, how am I able to calculate the power needed for induction?

Current setup:

Power supply: U=12V I=9.5-10A (current draw slightly varies)

Coil: Number of turns: 10 turns Diameter of wire: 2.1mm 20mm is inner diameter of copper wire Lenght of coil is 25mm

enter image description here

Brass, that I want to heat up is marked with red color. The end bullet is not present (obviously). If there is few millimeters more to the left heated up, there's no problem at all. Just the red part need to reach 450-500°C. Wall thickness is 0.4mm.

enter image description here

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  • \$\begingroup\$ Twice the power for half the time basically.. \$\endgroup\$
    – Andy aka
    Commented Oct 18, 2021 at 14:56
  • \$\begingroup\$ @Andyaka: You're trying to say, it's going by square up? \$\endgroup\$
    – Jakey
    Commented Oct 18, 2021 at 15:21
  • \$\begingroup\$ I'm wondering if a slug of brass might absorb more power than a ring of brass, which is actually a one-turn coil? Orienting a ring at right-angles inside your coil might make a difference? \$\endgroup\$
    – glen_geek
    Commented Oct 18, 2021 at 17:14

1 Answer 1

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The power is relatively easy to calculate. I estimate the mass of the shell be about 10g (you can just weight it) and the heat capacity of brass is 920 Joules per kg per Kelvin. So you need about 4000 J to heat 10 g of brass by 450 degrees.

Since brass is a very good thermal conductor, we can assume that the whole shell will be more or less at the same temperature.

If you want to deliver this in 15 seconds the required power is 4000J/15s = 266W. That is the power you need to be delivered into the shell. If your total efficiency is, say, 50% you will need to feed it 520 Watts,

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  • \$\begingroup\$ That would sound true, but with 120W that I was currently working, didn't do it in either 30 neither 60 seconds (as with your equation it should). \$\endgroup\$
    – Jakey
    Commented Oct 18, 2021 at 15:29
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    \$\begingroup\$ @Jakey Hilmar made an assumption about the efficiency of the power transfer, but the equation they use is correct. From what you state in your post, your few minutes would be 120s or 4000J/120s = 33W => 33W/120W = 27.5%. I've seen worse... Those losses are spread between the components in the circuit itself and the transfer efficiency. In a calculation like this, the hardest part is going to be that efficiency number as it varies with a lot of factors including the switching frequency of the heater and the size/shape of the output coil. \$\endgroup\$
    – Stiddily
    Commented Oct 18, 2021 at 16:02
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    \$\begingroup\$ @Jakey: that just means that your efficiency is significant less than 50%. Efficiency is not particularly linear: the longer it takes the more heat you lose to the environment, fixture, etc. If you can be REALLY fast you can get away with less energy since you don't heat the entire shell \$\endgroup\$
    – Hilmar
    Commented Oct 18, 2021 at 16:16

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