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Alternator Stator Outputs Shown as Center Tapped Transformers.

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  • Harvey D Norris
    A question that comes up is the difference between a delta and wye configuration. These things are usually looked at with respect to how a 3 phase load is
    Message 1 of 1 , Jan 12, 2004
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      A question that comes up is the difference between a delta and wye
      configuration. These things are usually looked at with respect to how
      a 3 phase load is configured with respect to the three available
      wires of input. However the 3 phase source of emf itself has two
      options of how it can be connected, and is always connected in WYE.
      The WYE connection from the stator coils, is theoretically 1.7 times
      the voltage present as if the voltages were instead extracted as
      Delta. If those stator coils themselves were arranged as center
      tapped trasformers typically used on the "outside" application from
      power sources, then the resultant voltage of the ending coils would
      be twice that found on each coil alone, because of the 180 phased
      voltages on each coil, but here it assumed that because the three
      phase motional emf voltages are 120 degrees out of phase, instead of
      180, we then get only 1.7 times the series coil voltage instead of 2
      times that individual phases voltage. As output forms then we
      generally think of delta and wye as the only two possible expressions
      of stator wiring output. There is however a third possibility where
      diodes could be used to switch the stator output between a delta and
      a wye expression, depending on the polarity or the currents found in
      the AC. It gets a little hard to imagine how that signal would look?

      Heres a letter I recieved some time ago on the subject.

      > I was reading you message on yahoo about Delta and
      > WYE stator
      > windings. I am just an automotive student and our
      > instructor asked us
      > why a Delta stator carries more current
      Right off the bat I would think it does NOT contain
      more current, but;
      These are somewhat complicated issues when both
      voltage and amperage issues are mentioned, but I would
      hypothesise that the output in delta could only
      contain more current if the delta phase were obtained
      totally in isolation, meaning that we would then have
      6 terminals, where each phase would have two seperate
      terminals obtained for its output. This is a confusing
      issue I have not yet mentally resolved, and I could be
      wrong on this specific point.
      and a WYE
      > stator carries more
      > voltage.
      The wye stator does contain more voltage because it is
      the result of two stator coils in series, rather then
      just one. This is why it puzzles me that your
      instructor would say that the Delta output should
      contain more current, which ordinarily I would think
      is untrue, and the only possible way I can think to
      say that it might be true is the above reservation
      where that MIGHT occur if the phases were derived in
      isolation, with no interphasal connections. If we had
      three delta phases in isolation, each outputing one
      amp, that would be 6 wires holding 6 amps outwards
      from the source. If we allowed each of those delivery
      wires to be joined with the adjacent delta phase then
      we would instead have three wires, no longer holding 6
      total amps, but 1.7 amps times 3 or 5.1 amps.

      My instructor is a 30 year ASE certified
      > technition and he
      > did not know the answer. If your could help me, not
      > to show my
      > instructor up, but to educate our class with an
      > answer I would
      > sincerely appreciate it.
      > Thank You,
      > Mike
      Its rather simple, they simply dont make the stator
      windings in Delta, they make them in Wye instead. If
      they made then in Delta, each coil encompassing the
      flux from the motional magnetic field would be
      recieving that quantity, and a measurement of that
      quantity would show a lesser amount of total flux
      change converted to voltage. Say we have 1 volts emf
      for a single delta stator winding. We have three of
      them doing this. Now we take those same windings and
      reconvert them to WYE. Formerly we only had one stator
      winding per phase for output. Picture the delta
      triangle being pushed inwards from the middle of each
      delta pathway, we can only push it so far inwards
      until the collapsed triangle becomes a WYE. But when
      we have done this we have allowed one delta branch to
      share the pathways with each of its neighbors delta
      pathways. This however may only be a mental picture to
      understand the analogy later on. If the analogy
      applied then now the currents available on those new
      pathways would be the vector sum of each two delta
      stator currents, which is also what occurs when we
      join 6 wires in delta isolation to delivery with only
      three wires. If we look at the problem from the
      reverse direction, from the currents found on the
      delta load vs the line that carries those shared
      currents to the load, then the answer instead becomes
      a vector subtraction with respect to those currents.
      Each action carries the inverse aspect; depending if
      we are working from the inside-out, the more common
      circumstance, or the outside-in, in which case the
      laws are reversed.

      Now picture instead the actual change of stator
      connections. Now each output phase; (OBTAINED AS A
      ARE VIEWING THE PROBLEM); it now has TWO stator coils
      in series to make that output, instead of merely one
      obtained in the Delta to Delta case. But just because
      we formerly had one volt in the delta to delta case,
      we do not expect that now because now we have two
      sources of motional emf in series, that we would now
      obtain 2 volts output. Rather we obtain the VECTOR SUM
      of those two motional emf's, which is 1.7 volts,
      because the AC waveforms on each of these stator
      windings are 120 degrees out of phase. So there is a
      trade off here, even though we are not recieving
      double the amount by using twice the amount of stator
      coils for delivery, we are still getting more then if
      we drew on merely one stator coil alone.

      One may wonder why in the stuff I do, that I never
      wire the outputs in WYE. This is because then each of
      those WYE based outputs would then be recieving a
      lesser amount of voltage. Twice the amount of load
      would be present, just like in the analogy where we
      pressed the delta triangle into a wye formation, since
      we now have two loads on that former delta path,
      shouldnt that 1.7 volts be cut in half for the new
      voltage division? If this were done for just a single
      phase that would be true. But once we do it for two
      delta phases, now each of those former phases is
      sharing a half a pathway with its neighboring phase,
      and when we do it for all three phases, the total
      pathways on the wye become all pathways where the
      current is being shared by the adjacent phase. Thus
      the 1.7 volts doesnt get chopped into half, because
      the adjacent phase also contributes some voltage,
      albeit a voltage that is not in phase. Suppose I had
      that outside 1.7 volts being extracted from the WYE to
      DELTA conversion. If I made it instead a WYE stator
      input, (which never changes, because thats how they
      are manufactured), to a WYE output then another
      question comes into being, then we are back to the
      original problem. For that output conversion,
      ordinarily we use a fourth return wire, and this
      fourth return wire connects back to the WYE stator
      junction. We have returned to the original problem,
      because a DELTA TO DELTA conversion is conceptially
      the same thing as connecting the alternator output to
      a WYE load, because that is a WYE to WYE conversion.
      Both of these things ARE the same thing, with the
      exception of how may lines are being employed that
      share currents from different phases. So in the WYE to
      WYE conversion, once again we are deriving a source of
      motional emf from only a single stator coil, and in
      this conversion, only the return fourth wire back to
      the stator is the line that shares currents from all
      three phases, and if all the currents on all the
      phases are equal, that wire will hold complete
      cancellation of current, equalling zero return
      current, Thus for balanced WYE delivery, all of the
      returning current returns on the adjacent phases. The
      only difference then from a delta to delta and a wye
      to wye is that for the delta case we do not have one
      wire sharing current for all three phases, rather we
      have three (output) wires that each share currents
      with the adjacent phase, but in both cases the
      motional emf is derived from only a single motional
      emf obtained by only one stator phase. Since a wye to
      wye is the same as a delta to delta, there is no
      reason to specifically make the stator combinations
      into exclusively a delta output, because practically
      the same effect takes place in procuring the outputs
      in wye, from the wye based stator.

      I am still in confusion regarding the statement that
      the delta output will be an output where MORE current
      would be obtained. Again if we wish to be totally
      specific and talk only about what is being done to to
      the stator connections; we can say the following for
      ALL cases of shared line formations.

      The DELTA wired stator delivers LESS voltage, but it
      does not deliver MORE current, it delivers LESS
      current because it is being sourced from only a single
      stator coil winding.

      The WYE wired stator delivers both MORE current and
      MORE voltage because then the motional emf stator
      winds use two of these windings in series, but it does
      not deliver twice the voltage or amperage found on
      each stator winding because of a consequence of the
      120 degree phasing differences.

      A 1 amp delivery for these stator lines attached to a
      wye load will cause 1 amp delivery to those loads for
      the balanced case.

      A 1 amp delivery for these stator lines to a load in
      delta will contain 70% less current to the delta
      phase, but it will be delivered at 70% more voltage.

      I would suppose then if each of the loads were equal
      for both wye and delta extractions, the 70% higher
      voltage would then mean that actually more current was
      extracted for the equal resistive load values placed
      in delta rather then wye. Perhaps this is what is
      meant by saying that more current is obtained IN A

      This is where things get totally confusing if we do
      not specifically state whether the aspect is internal,
      as the question seems to be expounded; or external as
      the load configuration. The laws are exactly opposite
      for each case, where the voltages divide for a wye
      load, and the voltages sum if we are intead speaking
      of how the source itself is wired.

      Perhaps we are making the problem appear needlessly
      more complicated. Perhaps it needs to be rephrased
      with the initial condition that we are using EQUAL
      resistive loads for both cases. Since we obtain more
      voltage in the delta output from an alternator, which
      is always manufactured with the stator construed in
      wye, when we obtain that output in Delta, it is a
      higher voltage, so for equal loads in both cases, we
      also obtain more current in that delta extraction. For
      the Wye extraction, we obtain less voltage on each
      individual load, thereby less current on those
      branches when obtained in WYE.

      Are we confused yet?

      Take this into consideration;

      When we take the outputs before the field is turned
      on, (parametric), at my rpm I can draw 1.5 amps out
      for a single Delta short. Adding the second short
      reduces both of these values to slightly under 1.5
      amps, a total just under three amps. In this
      condition, two thirds of the stator outputs are
      exclusively made for use on the two shorts, and one of
      the stator outputs is shared by both phases. When we
      add the third phase short, the total amperages on all
      three phases drops to .75 A, and now all the stator
      outputs are being shared with all three phases. But we
      actually obtained more total current out when we drew
      on just two phases. I do not think such a thing would
      happen in real operating conditions where the field
      was actually energized, but it is something to wonder
      about for future comparisons. It would seem that the
      three shorts has placed the machine in overload
      operation, where the voltage drops down below the
      effective values for maximum power transfer, or
      something similar to that scenario.

      Sincerely HDN
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