Induction melting of steel - melting time calculation

Hi,

We're trying to size a small induction melting furnace (similar to those used in the jewelry making industry) for our application and I was wondering if anyone with some experience in this subject can help answer the following question:

How long would it take to melt 3.5 lbs of 4140 Steel (2"Dia. x 4"L bar at

Thanks in advance, James

Is it not assumed that a commercial furnace designed for melting would be properly insulated?

Thanks for your response though.

You'd be suprised -- I could assume it has heavy duty castable on the order of 6 BTU-in/hr-ft^2-°F (that unit needs a name, typing it is too long!) conductivity value, but, it could also be a standard crucible inside the coils, with or without a high [insulating] value ceramic fiber insulator, or so on. I've heard of inconel melted bare inside a vacuum chamber, for casting standard quality (by which I mean, fricking high quality) jet engine combustor parts.

A dewar is perhaps a bit far off and I haven't heard of one able to hold molten steel... but if you could get one, it would rock, wouldn't it? ;-)

Tim

-- Deep Fryer: a very philosophical monk. Website:

The steel might block your induction current partly, as iron core,increasing your coil value very much, hysteresis losses might not melt anything maybe.

How does your figure change if you account for the radiation heat loss during the latter part of the nine minutes? What's the equilibrium temperature?

My copy of "Standard Handbook for Electrical Engineers" (C) 1907-1948 gives the energy *consumption* of an induction furnace at 600kWh/ton for melting steel, and remarks that the numbers will be higher for small furnaces. If we add 50% for the latter, I come up with about

I note that this (and RobS's numbers) are well within the ballpark off-the-cuff estimate that Tim gave. Which fits nicely with the thesis of this cute little (somewhat overpriced) book:

Best regards, Spehro Pefhany

I don't know because I just read the empirical data out of the handbook. Presumably it accounts for all the real-world losses required to bring a 'melt' up to a useful temperature. My concern is that my own +50% figure could be way off, since losses are related to surface area (proportional to melt size squared) and volume is proportional to melt size cubed, so a big melt that is roughly spherical has WAY less surface area per unit mass. The range they cover begins at 100 lbs. Still, they say "somewhat affected" by size.

I'll post a PDF with the relevant four pages in abpse- there are sketches of the construction, some graphs, etc.

Best regards, Spehro Pefhany

KW):

Hum, I wonder what the SiC crucible's contribution is, to heating.

Silver and copper and f****ng hard to melt with induction...sorry to put it that way but that's the best way to say it! The low resistivity just reflects back the induced power, burning roughly the same heat in the coil as the work, not to mention requiring a lot of volt-amp capacity in the coil and capacitor. Something like 4140 should have no problem coupling to a furnace so equipped.

BTW, is your crucible SiC? If it is, it's incompatible with ferrous alloys (both silicon and carbon are soluble).

Tim

-- Deep Fryer: a very philosophical monk. Website:

Ok, good. I was wondering if you were going to use what might already be installed, or what.

Hmm, the linked crucible is awfully thin, and even more ouch on the price! Why not a cheaper standard #2 or so foundry crucible? (Clay-graphite crucibles are okay for ferrous melting AFAIK, since the graphite burns out of the surface.)

Tim

-- Deep Fryer: a very philosophical monk. Website:

Maybe this abbreviated link works.

Thanks for all your responses. Here's a typical furnace (this one is 9.5 KW):

James

As far as I know we will be using Alumina crucibles, unless they can't be used with this process for some reason. See

James

Nice work Spef.

How did you derive the 1.1 MJ value?

James

newsgroup: alt.binaries.schematics.electronic

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Best regards, Spehro Pefhany

Call me ignorant but I don't know anything about "abpse". Me being the original poster would be very interested in this PDF file. Is it possible for you to post this somewhere that I can access?

Thanks, James

Spehro, Thanks for this.

Regards, James

The answer can only be a very very approximate figure for many reasons. First of all they are plenty of induction furnaces each which work on entirely different principles each of which has different heat transfer mechanisms like magnetic ,arc etc. So more information would be needed on that . Second heat transfer is a complex field . So many factors will be missed in the calculation . Just to consider one form of heat loss like radiation. To get an idea of how much heat is lost you'd have to know the Radiosity , Irradiation, radiation shape factor of all the bodies involved then form a radiation network to solve the problem. And I won't mention conduction or convection.

"Go easy with the whisky"

theJackal