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    #16
    Originally posted by mea03wjb View Post
    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?
    Please define what you mean by "line current" and "phase current". By line current, do you mean the current in the line between the source and the load (each phase)? By phase current, do you mean the current in the delta load (between phases)?
    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
    You can't really do much with a single instantaneous value. Consider that any sinusoidal value will have a value of zero twice per cycle, so if you measure zero amps, the magnitude of the sinusoid could be zero or 10000000.

    If the values are sinusoidal (constant magnitude and phase with no distortion), you can define the magnitude and phase with two values 1/4 cycle apart.

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      #17
      Originally posted by mea03wjb View Post
      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.
      I will have achieved my life's goal if I die with a smile on my face.

      Comment


        #18
        I think we're getting there......

        Thanks everyone so far for their input!

        Firstly, people have asked for clarification on some of the terms I've used:

        Line current = current measured on each of the 3 lines (phases) between source and load.

        Phase current = current flowing through the 3 windings (phases) of the induction motor load (i.e. current flowing in the different sides of the delta load).

        Secondly, I was asked to provide example figures for people to have a look at - that is why I gave some instantaneous values. I am actually able to sample the data at up to 20kHz (5kHz realistically) through a dedicated sensor box.

        Therefore, each of the currents and voltages will be available as a discrete waveform (0.0002 secs between each data point). Does this make the calculation any easier?

        The graphical solution seems fine for one-offs but I wanted to implement a system to calculate the phase currents automatically in software based on the measured line quantities - that is why I'm trying to do it mathematically.

        --

        In terms of the Fermat solution, I think the idea is to calculate the centre point of the triangle made by the 'line' current vectors and use this with the original vector values to calculate the new 'phase' current values. This is what topgone has done in a previous post.

        If this was done at each time instant (every 0.0002 seconds) with the instantaneous 'line' values, then the instantaneous 'phase' values could be calculated at each time instant. Then if this process was repeated which each set of instantatneous (sampled) values over an extended period of time (say 1 second) the waveform for each of the 3 'phase' currents would be produced.

        Does anyone see a flaw in this approach?

        Cheers,
        W.
        Last edited by mea03wjb; 07-19-10, 03:05 PM. Reason: Corrected sampling time

        Comment


          #19
          Originally posted by mea03wjb View Post
          .... I am actually able to sample the data at up to 20kHz (5kHz realistically) through a dedicated sensor box.

          Therefore, each of the currents and voltages will be available as a discrete waveform (0.0002 secs between each data point). Does this make the calculation any easier?

          The graphical solution seems fine for one-offs but I wanted to implement a system to calculate the phase currents automatically in software based on the measured line quantities - that is why I'm trying to do it mathematically.

          --

          In terms of the Fermat solution, I think the idea is to calculate the centre point of the triangle made by the 'line' current vectors and use this with the original vector values to calculate the new 'phase' current values. This is what topgone has done in a previous post.

          If this was done at each time instant (every 0.0002 seconds) with the instantaneous 'line' values, then the instantaneous 'phase' values could be calculated at each time instant. Then if this process was repeated which each set of instantatneous (sampled) values over an extended period of time (say 1 second) the waveform for each of the 3 'phase' currents would be produced.

          Does anyone see a flaw in this approach?

          Cheers,
          W.
          First let's make a distinction here. The graphical solution (including any math solution associated thereto) is for RMS data, not instantaneous data.

          Values of an instant are processed with basic math. "Phase" voltage is the absolute difference of line-to-neutral values. For example, your earlier post gave the following values for a single instant:
          v1 (phase-neutral) = 308V
          v2 (phase-neutral) = -275V
          v3 (phase-neutral) = -35V
          Line-to-line voltages at that instant would be:
          I will have achieved my life's goal if I die with a smile on my face.

          Comment


            #20
            Originally posted by Smart $ View Post
            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.
            If the values were instantaneous, and the OP only mentioned "unbalanced load", then I must say the voltage values are unbalanced - because in a balanced three-phase voltage, the sum of the instantaneous voltages should be equal to zero!

            Comment


              #21

              Comment


                #22
                The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

                Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

                IA + IB + IC = 0
                IA = IAB - ICA
                IB = IBC - IAB
                IC = ICA - IBC

                Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

                The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.

                Comment


                  #23
                  Originally posted by jghrist View Post
                  The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

                  Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

                  IA + IB + IC = 0
                  IA = IAB - ICA
                  IB = IBC - IAB
                  IC = ICA - IBC

                  Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

                  The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.
                  Are you saying that this problem is impossible to solve for an exact single answer?

                  i.e. we can find 'a' solution problem mathematically but this may or may not correspond to the actual currents flowing in the motor windings.

                  Comment


                    #24
                    Originally posted by topgone View Post
                    If the values were instantaneous, and the OP only mentioned "unbalanced load", then I must say the voltage values are unbalanced - because in a balanced three-phase voltage, the sum of the instantaneous voltages should be equal to zero!
                    308v + (-275) + (-35) = -2V
                    ... which isn't quite zero due to rounding, and perhaps a slight imbalance, but well within a nominal tolerance.

                    However, that reminds me that he'll likely have to use non-absolute voltage values for the line-to-line calc's:and
                    583V + (-240V) + (-343V) = 0V
                    Where'd the 2V imbalance go??? Isn't it amazing
                    I will have achieved my life's goal if I die with a smile on my face.

                    Comment


                      #25
                      Originally posted by mea03wjb View Post
                      Are you saying that this problem is impossible to solve for an exact single answer?

                      i.e. we can find 'a' solution problem mathematically but this may or may not correspond to the actual currents flowing in the motor windings.
                      Yes. That's what I'm saying. It is similar to calculating the secondary load current in a delta-wye transformer when you know the primary line currents. You can calculate the line currents if you know the delta currents, but not vice-versa. In the case of a delta-wye transformer, if you add the same current (magnitude and angle) to each secondary phase current, the primary line currents do not change. The same current in each phase is a zero-sequence current and circulates in the primary delta winding.

                      It's a little more difficult to come up with an instance of adding zero-sequence current to a delta load, but the mathematical principle is the same.

                      Comment


                        #26
                        Originally posted by topgone View Post
                        If the values were instantaneous, and the OP only mentioned "unbalanced load", then I must say the voltage values are unbalanced - because in a balanced three-phase voltage, the sum of the instantaneous voltages should be equal to zero!
                        The delta phase voltages need not be unbalanced only the phase currents.

                        IMHO, the sum of the L-L voltages is always zero, instantaneous or RMS.
                        Don't mess with B+!
                        (Signal Corps. Motto)

                        Comment


                          #27
                          Originally posted by jghrist View Post
                          The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

                          Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

                          IA + IB + IC = 0
                          IA = IAB - ICA
                          IB = IBC - IAB
                          IC = ICA - IBC

                          Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

                          The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.
                          Quite correct, whether we're discussing instant or RMS values... just that for RMS values, we have to include the phase relationships
                          I will have achieved my life's goal if I die with a smile on my face.

                          Comment


                            #28
                            Originally posted by jghrist View Post
                            The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

                            Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

                            IA + IB + IC = 0
                            IA = IAB - ICA
                            IB = IBC - IAB
                            IC = ICA - IBC

                            Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.
                            I don't see circulating currents in the loads. These currents are defined by the L-L voltages and the load impedances. You may see these currents in the transformer though.

                            The thing is we have six unknowns--three complex numbers.

                            Now take a look at Smart's diagram which shows the line and load current phasors. Clearly, for the general case, the intersection of the load current phasors can be anywhere within the line current delta, and this amounts to an infinite number of solutions.
                            Don't mess with B+!
                            (Signal Corps. Motto)

                            Comment


                              #29
                              Balanced rms voltage value

                              First, it is necessary to have good data before we calculate.
                              Since the OP did not mention if the supply was a balanced 3-phase supply, I played around with possible values of balanced RMS voltages and simulated/plotted voltage wave swept from zero electrical degrees up to half a cyle (180 electrical degrees). I found out that the correct Line to Neutral values should have been:
                              V1 = 308.20 volts
                              V2 = -273.13 volts, and
                              V3 = -35.07 volts.
                              This values were computed using a balanced line to neutral RMS voltage of 238.62 volts (Vpeak = Vrms X 1.414 = 337.46 volts).
                              Above mentioned figures resolved at around 66 electrical degrees, i.e.:
                              V1(instantaneous) = Vpeak X sin(theta) = 337.46*sin(66) = 308.2
                              V2(instantaneous) = Vpeak X sin(theta-120) = 337.46*sin(66-120) = -273.0
                              V3(instantaneous) = Vpeak X sin(theta+120) = 337.46*sin(66+120)= -35.2
                              Now back to the OP.
                              For any phase to phase loads, it would be clear that finding the instantaneous phase current is easy (Vinst/load ohms). Bear in mind that only the magnitude of the instantaneous phase currents differ depending on the phase load of each), the angular displacement of the current wave will still be the same 120 electrical degrees. The complexity in getting the values of the instantaneous line current lies in the vectorial addition of the instantaneous currents at the delta corners of the load. Still, if you have good data, then everything can be computed as needed. I would leave the final computations to you in knowing the instantaneous line current.
                              Respectfully.

                              Comment


                                #30
                                Originally posted by rattus View Post
                                I don't see circulating currents in the loads. These currents are defined by the L-L voltages and the load impedances. You may see these currents in the transformer though.

                                The thing is we have six unknowns--three complex numbers.

                                Now take a look at Smart's diagram which shows the line and load current phasors. Clearly, for the general case, the intersection of the load current phasors can be anywhere within the line current delta, and this amounts to an infinite number of solutions.
                                I agree that at least for passive loads, I don't see how to get circulating currents in a delta load. Perhaps with an active load like a motor, it is possible. Mathematically, it is still possible which is why you can have an infinite number of solutions to the set of equations. The same as you can graphically in Smart's diagram as you point out. Only the one shown minimizes the load currents, however, and this is probably the solution with zero circulating current (although I can't prove it).

                                You can treat three complex numbers as six unknowns, but then you can double the number of equations by separating the real and imaginary components. Each equation would apply to both the real and the imaginary components. For instance, if for complex currents,

                                IA = IAB - ICA,

                                Then

                                IAreal = IABreal - ICAreal

                                and

                                IAimag = IABimag - ICAimag

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