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# Re: [femm] Power Factor

- To: femm@egroups.com
- Subject: Re: [femm] Power Factor
- From: David Meeker <dmeeker@foster-miller.com>
- Date: Mon, 16 Oct 2000 09:25:11 -0400
- Organization: Foster-Miller, Inc.
- References: <8sacf9+ki6l@eGroups.com>

    Bruce McGee wrote:

    > Hi
    >
    > I am working on an induction heating problem. I induce currents from
    > a source into a metal material for heating purposes. From the
    > solution calculated by femm30 is it possible to determine the power
    > factor of the load relative to the source.

    Specify a purely real current in your conductor a "circuit" property. Then,
    solve the problem and integrate A . J over the conductor. For convenience,
    we could denote the result of this integral as X. The power factor would
    then be:
    -Im[X]/Abs[X].

    This works because you can use the A.J integral to get sort of a
    frequency-dependent inductance by dividing X by i^2, where i is the per-turn
    current in the coil:

    L=X/(i^2)

    The voltage that's needed to drive the current is:

    v = j*omega * L * i

    just like a normal inductor. If you split L into real and imaginary parts (
    L=Lr + j Li) and expand the above,

    v=j*omega*Lr*i - omega Li*i

    so Lr is associated with reactive power, and Li with real power.

    Dave.

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    org:Foster-Miller, Inc.;Electrical and Electronic Systems Group
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    email;internet:dmeeker@foster-miller.com
    title:Senior Engineer
    adr;quoted-printable:;;350 Second Avenue=0D=0A;Waltham;MA;02451-1196;USA
    fn:David Meeker, Ph.D.
    end:vcard

- Follow-Ups:
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- References:
  - Power Factor
    - From: "Bruce McGee" <mcgee@mcmillan-mcgee.com>

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