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

  1. #301
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    Quote Originally Posted by Sahib View Post
    ... As for your equation for power factor, it holds good for linear loads, but not for non linear loads.
    If the equation is essentially the definition, how can it not hold for nonlinear loads???
    I will have achieved my life's goal if I die with a smile on my face.

  2. #302
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    Quote Originally Posted by Smart $ View Post
    If the equation is essentially the definition, how can it not hold for nonlinear loads???
    The definition does not take into account the distortion introduced by non linear loads and so not applicable in case of non linear loads.

  3. #303
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    Quote Originally Posted by Sahib View Post
    The definition does not take into account the distortion introduced by non linear loads and so not applicable in case of non linear loads.
    As I mentioned earlier. Power factor is just a number to correct VRMS*ARMS to equal P. You only run into a problem when you make the errant assumption that the power triangle applies to all scenarios.
    I will have achieved my life's goal if I die with a smile on my face.

  4. #304
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    Quote Originally Posted by Smart $ View Post
    As I mentioned earlier. Power factor is just a number to correct VRMS*ARMS to equal P. You only run into a problem when you make the errant assumption that the power triangle applies to all scenarios.
    Correct. I was misled by Gar's use of term 'real power' in his equation. If by real power is meant total actual power including that due to distortion loss, then his equation is okay.

  5. #305
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    170618-1047 EDT

    Smart $:

    Knowing that you have a calculus background makes our discussions much easier.

    I would like to treat the complete VFD as two black boxes. One is the DC supply, and for simplicity assume it has constant output voltage and zero output impedance, and can absorb energy. A large capacitor with appropriate trickle charge fits this criteria. Any DC source that meets that criteria can be used. The other box is the converter that takes a DC input and provides whatever voltage and current the end load requires.

    The said other box (the converter) will always have a constant input voltage, and its input current will be whatever is required. I expect the input current to vary with time. The instantaneous power is V*i. The average power over some time period T is V*Iave over the time period T.

    The power factors looking into said other box, or into its load can be whatever is existent. We don't care.

    The only requirement on the said first box is that its average input power over T equals its average output power over T, such that excessive output ripple does not occur. The instantaneous input current does not need to track the instantaneous output current. The instantaneous input current will be whatever the said first box circuitry requires. For constant load power on the DC supply (first box output) the said first box will have the same input power factor which is defined by its circuitry independent of any changes in final load power factor.

    .

  6. #306
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    Quote Originally Posted by gar View Post
    170618-1047 EDT

    Smart $:

    Knowing that you have a calculus background makes our discussions much easier.

    I would like to treat the complete VFD as two black boxes. One is the DC supply, and for simplicity assume it has constant output voltage and zero output impedance, and can absorb energy. A large capacitor with appropriate trickle charge fits this criteria. Any DC source that meets that criteria can be used. The other box is the converter that takes a DC input and provides whatever voltage and current the end load requires.

    The said other box (the converter) will always have a constant input voltage, and its input current will be whatever is required. I expect the input current to vary with time. The instantaneous power is V*i. The average power over some time period T is V*Iave over the time period T.

    The power factors looking into said other box, or into its load can be whatever is existent. We don't care.

    The only requirement on the said first box is that its average input power over T equals its average output power over T, such that excessive output ripple does not occur. The instantaneous input current does not need to track the instantaneous output current. The instantaneous input current will be whatever the said first box circuitry requires. For constant load power on the DC supply (first box output) the said first box will have the same input power factor which is defined by its circuitry independent of any changes in final load power factor.

    .



    Better yet...
    I will have achieved my life's goal if I die with a smile on my face.

  7. #307
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    170618-1313 EDT

    Sahib:

    The equation I provided is a definition. It applies to any two terminal device, linear or nonlinear. And the Vrms and Irms are really RMS values no matter what are the waveforms. Is the equation always useful or meaningful? Most likely not.

    The wording "real power" is because some people do not read the word "power" as meaning heat or physical work. And it is the average real power over the same time period as that of the RMS measurements.

    This is the power read on a true wattmeter.

    .

  8. #308
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    170618-2226 EDT

    This post is about the line current of one phase of a three phase source to one of our HAAS CNC mills. This is one wire to the input to the HAAS vector spindle drive. There are only three power wires to the drive.

    Our power source is two pole mounted transformers wired as a 240 V open delta. There are 300 to 400 feet of lines from the transformers to the machinre. I judge the voltage and phase balance may not be real good based upon th current plot below.

    It was most convenient to run the spindle at 3500 RPM and no additional mechanical load. The machine is an advertized 20 HP, and under no machining load the HAAS load meter read 15%

    A Fluke Hall device current probe was used to measure current on the 2 V per 20 A range.

    The plot is a single shot. Repetitive plotting shows considerable variation in the peaks, but the unbalance was still evident. This is nothing close a sine wave current signal and thus a low power factor.
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    Last edited by gar; 06-19-17 at 12:02 AM.

  9. #309
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    Quote Originally Posted by gar View Post
    .... This is nothing close a sine wave current signal and thus a low power factor.
    ...
    Interesting...

    I hear your claim of a low power factor but see no proof. And if you say the proof is in the plot, I'll ask where? I see no rms current determination. You can't determine power factor without it.

    Now even if you do manage to show proof, I will note that you took your measurements while the motor is relatively unloaded (15% of 20HP... 3HP to turn a spindle???) when the motor power factor is at its worst. Don't forget the debate is whether or not a VFD improves power factor (at least I think it is... its been kind of hard to keep track of what's actually being debated ). To adequately compare, you must determine the power factor of both the input and output for the same period of time... and preferably several examples depicting the same (to demonstrate repeatability) and also at differing amounts of motor loading.


    While you're at it, and for a fair overall comparison, determine the power factor atof the same motor at the same differing amounts of loading when the motor is run across the line, i.e. connected without the VFD.

    What it amounts to is, we'd need an electrical testing lab and all the necessary equipment to even come close to settling the debate.
    I will have achieved my life's goal if I die with a smile on my face.

  10. #310
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    Quote Originally Posted by Smart $ View Post
    Interesting...

    I hear your claim of a low power factor but see no proof. And if you say the proof is in the plot, I'll ask where? I see no rms current determination. You can't determine power factor without it.

    Now even if you do manage to show proof, I will note that you took your measurements while the motor is relatively unloaded (15% of 20HP... 3HP to turn a spindle???) when the motor power factor is at its worst. Don't forget the debate is whether or not a VFD improves power factor (at least I think it is... its been kind of hard to keep track of what's actually being debated ). To adequately compare, you must determine the power factor of both the input and output for the same period of time... and preferably several examples depicting the same (to demonstrate repeatability) and also at differing amounts of motor loading.


    While you're at it, and for a fair overall comparison, determine the power factor atof the same motor at the same differing amounts of loading when the motor is run across the line, i.e. connected without the VFD.

    What it amounts to is, we'd need an electrical testing lab and all the necessary equipment to even come close to settling the debate.
    Here ya go....

    GRAPHIC HARMONIC ANALYSIS V(av) 270.1
    I(rms) 340 1.220208208
    Supply Voltage (V) 267 I(h) 326 367 0.959301243
    Delay angle (deg) 42 95.9%
    DC Choke (mH) 2
    Initial Current (A) 380.0
    Choke ripple (A) 326.43 102
    Choke ripple (pk-pk) (A) 53.9 13%
    Average Current (A) 415.2 338.8


    Angle Rect 1 Vmean I inst pk-pk Iac A1 B1 An Bn R.M.S Rads
    378 270 380 54 380 1 1
    0 369 99 383 0 0 0 0 0 0 -4 0 -8 2460 A/ms
    1 368 98 385 0 0 0 0 0 0 -12 0 -1 0.055555556
    2 366 96 388 0 0 0 0 0 0 -20 0 -2 0.154471545
    3 365 95 391 0 0 0 0 0 0 -28 0 -2 2.780487805
    4 363 93 393 0 0 0 0 0 0 -36 0 -2
    5 361 91 396 0 0 0 0 0 0 -44 0 -2
    6 359 89 398 0 0 0 0 0 0 -52 0 -2
    7 357 87 401 0 0 0 0 0 0 -60 0 -2
    8 355 85 403 0 0 0 0 0 0 -68 0 -2
    9 353 82 405 0 0 0 0 0 0 -76 0 -2
    10 350 80 408 0 0 0 0 0 0 -84 0 -2
    11 348 78 410 0 0 0 0 0 0 -92 0 -3
    12 345 75 412 0 0 0 0 0 0 -99 0 -3
    13 342 72 414 0 0 0 0 0 0 -107 0 -3
    14 339 69 416 0 0 0 0 0 0 -115 0 -3

    On the other hand, you could just take my word for it.
    Si hoc legere scis nimium eruditionis habes.

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