proof of full load formula

Discussion in 'Homework Help' started by newfara, Jul 23, 2013.

  1. newfara

    Thread Starter New Member

    Jul 23, 2013
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    hi all,

    please provide proof of this approximate full load current formula for induction motor. I can't find it anywhere. Thanks.

    I = 600 * P(h) / E

    I = full-load current [A]
    P(h) = output power in HorsePower
    E = rated line voltage [V]
    600 = empirical constant
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    How do you expect someone to provide a "proof" of a formula that you indicate involves and "imperical" constant? The very concept of "imperical" means that it was determined by experiment and observation.

    This post should probably be in the Homework Help forum. It definitely should NOT be in the Feedback and Suggestions forum. I'll alert a moderator so that they can move it to an appropriate place.
     
  3. newfara

    Thread Starter New Member

    Jul 23, 2013
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    0
    Hey thanks. Sorry if i phrase the question wrongly. I going through a textbook and this formula came out, i don't know how it came to this conclusion. I'm wondering if this is a common formula so asking around here.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's possible this relates to a "rule of thumb" that states the DOL (direct on line) starting current per phase may typically be ~600% of the motor's rated per phase current.

    Otherwise it makes little sense to me.
     
  5. Ramussons

    Active Member

    May 3, 2013
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    1 Electrical HP = 746 Watts, 1 Metric HP = 735.5 watts.

    Logically, I = [ (746 or 735.5) * HP ] / E

    How did that '600' come about is not clear.

    Ramesh
     
  6. newfara

    Thread Starter New Member

    Jul 23, 2013
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    Guess its not a common formula then. Haha. Anyway i highly recommend theodore wildi's electrical machines, power drives and systems, a highly organized and clear book, the best of related books i read. This formula popped up here, but to be fair a fair share of the examples and practice question used this formula.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Is this formula for single-phase or three-phase systems?

    Assuming it is for a three phase system, is the I the per-phase current? Is the E the phase-phase voltage? Assuming a balanced load, what is the power being delivered to the load in terms of I and E?
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If it is a 3-phase motor with full load power factor of pf then the rated load current per phase would be

    \text{ I= \frac{746x Rating[HP]x\sqrt{3}}{3xpfxE}}

    E would be the line-to-line voltage.

    For the aforementioned empirical factor of 600 a power factor of 0.718 would be required. This seems a rather arbitrary assumption. A modern well designed induction motor would most likely have a full load power factor substantially better than that.
     
    Last edited: Jul 23, 2013
  9. WBahn

    Moderator

    Mar 31, 2012
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    But the 600 may include a cushion for the purpose of sizing the wire.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Could well be the case.

    I recollect when I used to deal with induction motor installations in mining operations, we often had quite long cable runs from distribution boards to the actual motor location. In that situation the cable voltage drop at full load was also an important factor in conductor sizing.
     
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