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Induction melting of steel - melting time calculation

Discussion in 'Electronic Design' started by James H., Dec 18, 2005.

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  1. James H.

    James H. Guest


    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

    How long would it take to melt 3.5 lbs of 4140 Steel (2"Dia. x 4"L bar at
    1500 deg. C, 7.8 g/cc) with a 10 KW medium frequency (10-30Khz) water-cooled
    induction melting machine using the proper crucible (350 cc capacity)? Are
    there any other factors missed here that would influence this calculation?

    Thanks in advance,
  2. Tim Williams

    Tim Williams Guest

    Err, insulation? Open to radiation or is it in a dewar?

    I would ballpark 5 to 20 minutes, or never if uninsulated.

  3. James H.

    James H. Guest

    How long would it take to melt 3.5 lbs of 4140 Steel (2"Dia. x 4"L bar
    Is it not assumed that a commercial furnace designed for melting would be
    properly insulated?

    Thanks for your response though.
  4. Sjouke Burry

    Sjouke Burry Guest

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

    Tim Williams Guest

    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? ;-)

  6. 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
    1.5kWh, or about 9 minutes for 3.5/2000 ton of good ol' 4140.

    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
  7. Spehro Pefhany wrote...
    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?
  8. 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
  9. James H.

    James H. Guest

  10. Tim Williams

    Tim Williams Guest

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

    Silver and copper and fucking 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).

  11. James H.

    James H. Guest

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

  12. Tim Williams

    Tim Williams Guest

    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.)

  13. RobS

    RobS Guest

    I got a similar result to James. You need about ~1.1MJ of energy to
    melt the bar. At 100% furnace efficiency that takes about 70 seconds to
    melt. I pulled an induction furnace efficiency of 3.6% to 17% (for
    aluminium) from the web - presumably for a practical design with all
    factors taken into account.
  14. James H.

    James H. Guest

    How did you derive the 1.1 MJ value?

  15. Spehro Pefhany wrote...
    Nice work Spef.
  16. James H.

    James H. Guest

    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
  17. newsgroup: alt.binaries.schematics.electronic

    Message ID: <>

    Best regards,
    Spehro Pefhany
  18. James H.

    James H. Guest

    Spehro, Thanks for this.

  19. theJackal

    theJackal Guest

    The answer can only be a very very approximate figure for many
    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"

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