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    Calculating Phase Currents for Unbalanced Loads

    Hi All,

    I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

    I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

    Cheers, W

    #2
    You should be able to write a KCL equation for each node of the delta (relating the two phase currents to the line current) and assuming you know BOTH the magnitude and phase of each of the line currents, it should be straightforward phasor math. This sure sounds like a test question of some sort?
    Dennis L. Karst, P.E.

    Comment


      #3
      This may not be quite as straightforward as I indicated... the solution may have to be split into two problems. Could you post the real situation with values, please.
      Dennis L. Karst, P.E.

      Comment


        #4
        Originally posted by mea03wjb View Post
        Hi All,

        I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

        I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

        Cheers, W
        Would be glad to help but can you provide some numbers for us to crank?

        Comment


          #5
          Originally posted by mea03wjb View Post
          Hi All,

          I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

          I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

          Cheers, W


          ..........................
          It depends on what the meaning of "is" is.

          Comment


            #6
            If we have the magnitudes of the line currents, it is easy enough to obtain the angles with a graphical solution. After that, I am stumped. I can write 3 equations, but there are 6 unknowns--the phase currents and their angles. And too, the determinant equals zero.

            Help!
            Don't mess with B+!
            (Signal Corps. Motto)

            Comment


              #7
              Thinking out loud: Connect test wye impedances and solve for impedance values. Then make a wye-delta impedance transformation.
              BB+/BB=?

              Comment


                #8
                I have looked at the equations and don't think you can solve them using the magnitudes only, I think some of you may be coming to the same conclusion - it looks like it should be simple but I don't think there is enough information!

                It is for a system that samples the currents and voltages at regular intervals so I can give you some instantaneous values as an example (recorded from a test rig):

                v1 (phase-neutral) = 308V
                v2 (phase-neutral) = -275V
                v3 (phase-neutral) = -35V

                i1 (line) = 17.7A
                i2 (line) = -2.95A
                i3 (line) = -15.5A

                I want it to be an online (during operation) calculation so this rules out connecting in Y, thanks anyway Mivey!

                Thanks, W.

                Comment


                  #9
                  Originally posted by rattus View Post
                  If we have the magnitudes of the line currents, it is easy enough to obtain the angles with a graphical solution. After that, I am stumped. I can write 3 equations, but there are 6 unknowns--the phase currents and their angles. And too, the determinant equals zero.

                  Help!
                  Well I figured out a geometric solution (assuming each line-to-line load has the same power factor)... see below. First step is arrange the line current vectors tail to head. Of course we only know the magnitudes, so they'll be arranged relative to each other but not to voltage.

                  Next, let each line current vector be the base of a new 30-120-30 isoceles triangle to the outside of the original triangle. The apex of each triangle is the center of a circle which passes through the its base vector endpoints. The intersection of the three circles is the point where a vector to each tail-head intersection represents the phase (line-to-line) current of the delta load.

                  Now the math I'll leave as an exercise for you other guys

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

                  Comment


                    #10
                    Originally posted by Smart $ View Post
                    ....

                    Now the math I'll leave as an exercise for you other guys

                    ...
                    FWIW, my solution appears to be an offshoot of (read: essentially the same as) a First Fernat Point.
                    I will have achieved my life's goal if I die with a smile on my face.

                    Comment


                      #11
                      Okay, here it is:
                      Vab = 533.47 /30 deg
                      Vbc = 476.31 /-90 deg
                      Vca = 60.62 /150 deg

                      Iab = 2.95 / 22.64 deg amperes
                      Ibc = 0
                      Ica = 15.5 / 157.36 deg amperes

                      Comment


                        #12
                        Originally posted by topgone View Post
                        Okay, here it is:
                        Vab = 533.47 /30 deg
                        Vbc = 476.31 /-90 deg
                        Vca = 60.62 /150 deg

                        Iab = 2.95 / 22.64 deg amperes
                        Ibc = 0
                        Ica = 15.5 / 157.36 deg amperes
                        Don't forget he said instantaneous values (actually, just instant values, being of a single point in time). Instant voltages and currents have no phase angle. The line-to-line voltages are simply the arithmetic difference in potential, and currents evaluate as if DC.

                        So now we're back sort of to what Mivey suggested, but doing it theoretically and mathematically. Google "wye-delta transform" or "...transformation". The following link takes you to the wikipedia page on the topic:

                        http://en.wikipedia.org/wiki/Y-%CE%94_transform

                        Using such we can mathematically determine an equivalent wye-configured resistor circuit, then transform it into a delta-configured resistor circuit and obtain our "phase" instant currents...
                        I will have achieved my life's goal if I die with a smile on my face.

                        Comment


                          #13
                          Not very elegant, but:

                          Originally posted by Smart $ View Post
                          FWIW, my solution appears to be an offshoot of (read: essentially the same as) a First Fernat Point.
                          After finding the angles geometrically, one can simply draw a 120 degree wye on a sheet of tracing paper, then lay this over the line current delta sketch and wiggle it around until each leg of the wye intersects a vertex of the delta. Now scale the load current phasors and measure the angles if you wish. Probably good to 5% or so.
                          Don't mess with B+!
                          (Signal Corps. Motto)

                          Comment


                            #14
                            Originally posted by rattus View Post
                            After finding the angles geometrically, one can simply draw a 120 degree wye on a sheet of tracing paper, then lay this over the line current delta sketch and wiggle it around until each leg of the wye intersects a vertex of the delta. Now scale the load current phasors and measure the angles if you wish. Probably good to 5% or so.
                            A caveman could do it

                            Seriously though, with scale, straight edge, a drafting compass, and a sheet of paper, it can be done rather easily. Six lines, six arcs... pictured below, but I included some extra lines just for the heck of it, I guess

                            Any way one goes about it, this way is definitely easier than doing the math

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

                            Comment


                              #15
                              Ok thanks Smart - I will have a look at your link and see if I can calculate a solution.

                              Is assuming the PF is equal in each phase a valid assumption for the induction motor load (I am not challenging your original statement, I am a novice and I just want to understand more!) ?

                              Thanks.

                              Comment

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