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Thread: Single Phase current draw for a 3 phase output VFD: technical discussion

  1. #21
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    Quote Originally Posted by Smart $ View Post
    You're saying it's probably not a lot different... then you're saying it may very well be different.

    It's simple physics If you put 10.5kW 3Ø in and you 10kW out, the process has a 95% efficiency (i.e. 5% loss). Then everyone (but Jraef) is saying it'll likely take significantly more 1Ø power in to get the same 10kW 3Ø power out. Logic dictates a lesser efficiency (greater loss). How can this be "not a lot different"?
    No, I'm not saying that at all.
    In both cases, the DC current conducts through two diodes at any one time. There's just more of them taking part when it's three phase.
    The issue is how hard you can push the input rectifier when not using all of its capacity. The efficiency is a bit of a red herring.
    Si hoc legere scis nimium eruditionis habes.

  2. #22
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    Quote Originally Posted by Jraef View Post
    If I can't tell them "1.732x the motor current, because you are feeding the same kW load from a single phase source", then what DO I say?
    Consult the manufacturer.
    Si hoc legere scis nimium eruditionis habes.

  3. #23
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    There seem to be two different discussions that have been going on.

    One was whether an existing system designed for 3 phase input could be used with a single phase source.

    The other seems to have something to do with input to output power, and input current related to something.

    I am going to address current and power for the moment. Assume the three phase source is a 208 wye. The load is a balanced wye resistor. For example each leg of the load consists of two 20 ohm resistors in parallel making the resistance of each leg = 10 ohms.

    Each 3 phase line current is 12 A. Power into each load leg is 1440 W. Total power is 3 * 1440 = 4320 W. You can do this with DC or AC supplies.

    Now take all 6 resistors and put them in parallel and apply 120 V. Each resistor is worth 6 A and 720 W, or a total of 36 A and 4320 W. The current is 3 times the current on one 3 phase line. As expected.

    Next organize the 6 resistors to be balanced across 240 V. Now the line current is 6+6+6 = 18 A, or 1.5 times the 3 phase line current. Total power is 18 * 240 = 4320 W.

    .

  4. #24
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    OK, but again, look at the current formulae for three phase vs single phase that I posted in the first place.
    How does that relate to what you just said?

    So as a form of backup, take a 10HP 230V 3 phase motor, FLA = 28A
    Now look at a comparable 10HP 230V single phase motor, FLA = 50A
    50/28 = 1.78; efficiency would be different, but it is a lot closer to 1.732 than it is to 1.5.
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  5. #25
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    Quote Originally Posted by Jraef View Post
    OK, but again, look at the current formulae for three phase vs single phase that I posted in the first place.
    How does that relate to what you just said?

    So as a form of backup, take a 10HP 230V 3 phase motor, FLA = 28A
    Now look at a comparable 10HP 230V single phase motor, FLA = 50A
    50/28 = 1.78; efficiency would be different, but it is a lot closer to 1.732 than it is to 1.5.
    But the topic is not about single phase v three phase motors.
    It is about 3-phase VFD output powered from either a single phase or three phase input.
    Can we stick to that?
    Si hoc legere scis nimium eruditionis habes.

  6. #26
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    Quote Originally Posted by Besoeker View Post
    But the topic is not about single phase v three phase motors.
    It is about 3-phase VFD output powered from either a single phase or three phase input.
    Can we stick to that?
    Ok sure.

    So if i send 10kW of power out of my VFD to a 3 phase motor, the input power cannot be LESS than 10kW, correct?

    So if my input power is 10kW, what is the current that is represented within that value, an how it that NOT represented by the formula:

    Three phase; I = W/V x PF x 1.732 ergo I = 10,000/230 x .95 x 1.732 = 26.42A

    Single phase; I = W/V x PF ergo I = 10,000/230 x .95 = 45.76A

    45.76/26.42 = 1.732
    ???

    If Isp = 1.5 x Itp as purported, then is the PF the difference? So is the PF for the single phase input only .6 then?
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  7. #27
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    Much of this has to do with standard sizing
    usually that will give you some headroon over the 1.732
    in addition the motor is seldom loaded to 100%
    most mfgs now give the 1 ph ratings anyways

    sanity check I use
    calc 3 ph fla at sf, 1.15 for example
    mult x 1.732
    find a 3 ph drive rated for that i, it will usually be larger than the calculated
    slop
    sf
    assumed 100%+ loaded
    margin to next standard size
    usually 2x plus



  8. #28
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    Jraef:

    In response to your post #24. You are comparing three phase line current to three phase line to line load voltage for your three phase motor input, and not three phase motor winding curtrent (a voltage to current phase angle problem), to the single phase winding voltage and its winding current (no phase angle problem).

    Change your three phase thinking to a wye motor. Now the voltage to current phase angle problem is corrected by using the voltagre magnitude from line to neutral.

    Line to neutral is 230/1.732 = 132.8 V. Then 123.8*28 = 3466.4 VA per winding or 10,399 VA for the entire motor.

    For the single phase motor we have 230*50 = 11,500 VA.

    The two values are adequatly close to indicate a close comparison.

    .

  9. #29
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    Gar, you are mixing up heating in diodes with power thru them. The current on 1ph is exactly 1.73 x higher on 1ph into the vfd compared to 3ph into it for the same load.

    Your 1a vs 50a scenarios are not realistic to this discussion.

    Your idea that the diode drop is constant is no different than the 208v input voltage - it too is constant.

    You cannot supply the same 3ph load from anything less than exactly 1.73x more current. And don't bring in efficiency or pf, neither is changing enough to discuss.

    If you want to argue how much heat the diode junction will see on 1ph vs 3ph load, be it resistor or motor, ok, and your 1.5 vs avg vs rms discussion has merit. But not regarding 1.73x more current for the same load, be it resistor or motor.


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  10. #30
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    Quote Originally Posted by Besoeker View Post
    Consult the manufacturer.
    There is a simpler way--> conduct a load test!

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