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VFD Input I less than Motor FLC? SKM Analysis kVA Out > kVA In?

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    VFD Input I less than Motor FLC? SKM Analysis kVA Out > kVA In?

    Hey Guys, I would like to get some of your thoughts. It is a 2-parter question:
    1) Looking at large 480VAC, 3PH, VFDs the input drive current rating is < motor FLC? - See attachment. What's the deal? I've tried to explain this by assuming a high pf on line side and lower pf on output but after doing the calcs this does not explain it. I've read on these forums someone claim the drive does not need to supply reactive power kvar on the output which explains the lower rating. I'm not sure I buy that.
    2) I've ran a model in SKM to see what it says. As I had anticipated, load flow study still shows kVAR to motor but unexpected was that load side kVA is greater than line side kVA. What gives?

    Fairly new to SKM so please excuse my ignorance if this is an obvious one.

    3 attachments to illustrate my question.

    Thanks
    Attached Files

    #2
    VFDs generally have a PF on the line of close to unity.

    the load PF is typically much lower, as motor PFs tend to be.

    this leads to a lower line current than load current, and thus lower line KVA then load KVA.

    for the purposes of this exercise, assume a load has a PF of 0.8 and uses 1 kW. What would the current be? What would the current be for a 1 kW load with a PF of 1? assume 480V.
    Last edited by petersonra; 08-27-14, 05:58 PM.
    Bob

    Comment


      #3
      Originally posted by smoothops10 View Post
      HI've read on these forums someone claim the drive does not need to supply reactive power kvar on the output which explains the lower rating. I'm not sure I buy that
      Without knowing just what posts in what threads you were looking at, I would hope that what you saw was in fact that the drive inputdoes not need to supply the reactive power kVAR on the output because that reactive power is generated within the drive itself, working from the DC bus. Since the reactive energy flows into the load and returns, there is no net drain on the DC bus when averaged over a full cycle.

      Comment


        #4
        On the load side, motor current is inclusive of reactive current, the current that makes the motor a motor instead of a boat anchor. The VFD is providing that current. As far as that goes, the LOAD side of the VFD is almost like (can be considered as) a NEW power source for the motor, think of it as a generator that is dedicated to that one motor.

        On the LINE side of the VFD, the VFD is just using the source as the "fuel" for that generator. It is pulling current IN PHASE WITH the voltage sine wave (because that's how diodes work). So the PF it presents to the source of that line is at around .95, all of the time. The only thing that changes is the AMOUNT of current, based on the load, but from the perspective of the line source, that current is always still drawn at a .95 PF.

        So... 100HP 460V motor, 124A FLC. But that is 74.6kW, / 460V / 1.732 = 93.6 A, then assuming 95% efficiency = 98.6A, so that means 25.4A of the current going to that motor is reactive current. That means the VFD components (mainly the transistors) must be sized to safely deliver all 124A if necessary, but the input side draws current consistently without regard to PF on the load.

        If you look at that 750 series chart for a 100HP 480V drive, the output current is rated for 125A, as it should be, but the input current is only 117.4A. The difference between the 98.6A and the 117.4A is representative of the 5% PF (reactive current) still used by the drive mostly due to what is called distortion power factor (due to harmonics mostly) and the drive's internal efficiency losses (about 3%). Still, lower than the output amp rating however.
        Last edited by Jraef; 08-27-14, 06:13 PM.
        __________________________________________________ ____________________________
        Many people are shocked when they discover I am not a good electrician...

        I'm in California, ergo I am still stuck on the 2014 NEC... We'll get around to the 2017 code in around 2021.

        Comment


          #5
          Thanks for the responses guys. I'm thinking on them but initially I believe kVA is kVA. Apparent power magnitude is not dependent on power factor. Real power kW and reactive power kVAR are but not apparent power in kVA. So using the 100HP motor example:


          VFD in
          Smax=97.6kVA (from AB catalog)

          Motor side
          http://www.ab.com/support/abdrives/d...on/fb/1018.pdf
          S=V*I*3^.5/1000
          460*124*3^.5/1000=98.8kVA

          The drive appears undersized to me.

          Comment


            #6
            Originally posted by smoothops10 View Post
            Apparent power magnitude is not dependent on power factor. ....
            Quite to the contrary, if the real power is held constant, the apparent power magnitude is exactly real power/PF. That comes directly from the definition of PF.
            Similarly if reactive power, real power and apparent power are represented by Pr, P and Pa respectively, then for any situation with only linear components Pr + P = Pa.
            That is also consistent with Ir2 + I2 = Ia2 since I is a vector and P is not.....

            Comment


              #7
              Golddigger you are correct but real power is not a constant in this scenario. Let's use what the mfr gives us: a drive efficiency of 97.5%. Drive pf of 0.97 to 0.98. Let's assume a motor pf ranging anywhere from .8 to .93. Using these in any range of the motor pf, the drive power, real or apparent, never meets or exceeds the motor power. I have a spreadsheet I can post tomorrow I used to check this but I still feel drive is undersized albeit only a little but the point remains.

              Comment


                #8
                Originally posted by smoothops10 View Post
                ...Let's assume a motor pf ranging anywhere from .8 to .93.
                Do you mean for the same motor under different loading conditions? Or substituting different design motors with the same HP?

                If you change the PF of a motor (whether by changing the load or the motor) you will generally change the apparent power value. You cannot say as a generalization that the real power and the reactive power will both change but the apparent power remains the same. Just not going to happen.
                More likely is that the reactive power will remain close to constant but both the real and apparent power will change as the load (and PF) together change.

                Comment


                  #9
                  No just saying a range of pf for a typ 100hp motor. We never know the exact motor going to be used til after bid.
                  Last edited by smoothops10; 08-27-14, 11:20 PM. Reason: Spelling

                  Comment


                    #10
                    Originally posted by smoothops10 View Post
                    No just saying a range of pf for a typ 100hp motor. We never know the exact motor going to be used til after bid.
                    And if you do not know what motor is being used, it makes no sense to claim that PF will not affect the apparent power. Sorry....
                    It may not affect the amperage that the NEC requires you to use in wire ampacity calculations based on the motor HP, but that is not the same as saying that it does not affect the actual apparent power.

                    Comment


                      #11
                      Originally posted by smoothops10 View Post
                      Golddigger you are correct but real power is not a constant in this scenario. Let's use what the mfr gives us: a drive efficiency of 97.5%. Drive pf of 0.97 to 0.98. Let's assume a motor pf ranging anywhere from .8 to .93. Using these in any range of the motor pf, the drive power, real or apparent, never meets or exceeds the motor power. I have a spreadsheet I can post tomorrow I used to check this but I still feel drive is undersized albeit only a little but the point remains.
                      For a variable frequency drive, the input kVA is almost invariably lower than the output kVA at rated load.
                      It's simply a result of the lower output power factor. Nothing more complicated than that.
                      Si hoc legere scis nimium eruditionis habes.

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