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I am working on determining the specs needed for an induction heater I plan on buying or building. However, I’m having trouble calculating the potential current draw and wattage needed for my application.

The use will be for brazing copper to brass with silver solder. The melting point is around 1200 F. The copper is a 3” diameter hollow pipe with about 0.079 wall thickness. And the brass is about 0.5” thick and a 5” diameter.

I understand the resistivity of copper to be 1.7*10^-8. But with that value I am coming with extremely small numbers for resistance of my target hearing area and it just doesn’t seem correct.

How can I accurately determine my power and current needs for this application. The plan is to use 48VDC supply but that can be changed.

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  • \$\begingroup\$ Talk to an induction heating manufacturer to help you with your project. The combined electrical and thermal design is not trivial. It wouldn't surprise me if you need 50 kW and 50 kHz induction heater. \$\endgroup\$ – Marla Oct 19 '18 at 18:55
  • \$\begingroup\$ accidently edited your question, so I removed the edit. Apology. \$\endgroup\$ – Marla Oct 19 '18 at 23:15
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I had commented, "Talk to an induction heating manufacturer to help you with your project. The combined electrical and thermal design is not trivial. It wouldn't surprise me if you need 50 kW and 50 kHz induction heater".

Here is further information just to show that the thermal and mechanical considerations are a big part of the design.

The brass is about, "0.5” thick and a 5” diameter." Also, the copper is about 0.079 wall thickness. Both parts need to be heated to about 1200 degrees. So, you will need to apply about 5 times the power in the brass as you will apply to the copper. This will require a special induction coil or coils.

Since copper and brass both have high thermal conductivity, as you apply heat to the pipes, the heat will be rapidly being removed from the heat zone. The heat conducting down the length of both pipes. Thus, you need to apply heat at a very high rate to over come the heat loss due to conduction away from the heat zone.

EDIT1 : Using your 48 volts, you will need 1050 amps for 50 kW.

EDIT2 : There is a minimum power level required which must be greater than the amount of heat flowing away from the work zone. This is a thermal calculation. Additional power then is required to heat the material in the work zone to 1200 degrees. Additional power is then required to offset the induction power supply losses.

The wattage required has nothing to do with the resistivity or resistance of the workpiece.

I came up with 50 kW ESTIMATE using my years of experience and observations.

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  • \$\begingroup\$ Actually, I do not need the brass and copper to get to 1200. The solder needs to, but the metals just need to be hot enough to allow the capillary action to flow the solder. Does that change any of your statements and do you have any suggestions as to calculate power requirements ? \$\endgroup\$ – TheValyreanGroup Oct 19 '18 at 20:36
  • \$\begingroup\$ Calculate the thermal requirement. Then assume the induction heat will be 20% efficient. \$\endgroup\$ – Marla Oct 19 '18 at 20:43
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    \$\begingroup\$ I don't think solder will flow and bond correctly onto a metal which is cooler than the solder melting point. So I think you DO need to heat the work piece to 1200 degrees. Have you ruled out using some type of gas torch? \$\endgroup\$ – mkeith Oct 19 '18 at 23:18
  • \$\begingroup\$ Regarding your edit, I know how to calculate that if the wattage is known. But I’m trying to figure how many watts it will use based on the resistance values of the metals. Not sure why or how you’re coming to with 50kW. @mkeith You may be correct, and yes, we currently use an Oxy-Acetylene setup now. But induction would be much much easier if feasible. \$\endgroup\$ – TheValyreanGroup Oct 20 '18 at 0:35
  • \$\begingroup\$ You used the word "pipe", which is why I have emphasized the heat flowing away from the heat zone. \$\endgroup\$ – Marla Oct 20 '18 at 0:59

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