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Q: 3 phase star voltage

Discussion in 'Electrical Engineering' started by Jonathan Barnes, Sep 5, 2004.

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  1. Please help a mechanical engineer with an electrical problem.

    If 3 equal heating elements are connected in star to a 400 V 3 phase supply
    then each element has 230 V across it.
    In this condition the star point will be at the neutral voltage,

    If one element blows each remaining element will have 200 V across it , and
    the star point will have a 100 V difference from neutral.

    I have a system where each leg consists of two heating elements in parallel.

    If a single element blows the remaining element on that leg will be subject
    to a voltage which must be less than 330 V, but more than 230 V.
    The other two legs will be between 200 V and 230 V each.
    The star point must have a voltage to neutral of less than 100 V.

    Can anyone tell me what the actual voltages will be, and how they are
    calculated... or a good reference.


    Barnes's theorem; for every foolproof device
    there is a fool greater than the proof.

    To reply remove AT
  5. Don Kelly

    Don Kelly Guest

    Here's an approach:
    Assume that the resistor connected between a and the star point (s) is Ra,
    ditto for b and c so we have a star Ra, Rb, Rc

    Taking a as a reference point and converting to current sources at a and b
    leads to Vba/Rb+Vca/Rc =Vsa(1/Ra +1/Rb +1/Rc)
    and the voltage at s with respect to the system neutral is Vsa -Vna
    Vca=-345+j200 Vba=-345-j200 Vna= -230

    If Ra=Rb=Rc then -690 =3Vsc so Vsa = -230V (rounded off to 3 figures which
    is all that one needs or should use). Vna =-230V so the voltage to neutral
    is 0.

    If Ra is infinite (open) then -690 =2Vsa and Vsa=-345V so Vsn = - 115V

    If Ra=2Rc=2Rb then -690 =2.5Vsa and Vsa = - 276V and Vsn = -46V

    In the above cases, the star point floats along a line through a and n
    bisecting the bc line o the voltage triangle

    This can be done for any combination of Ra,Rb, Rc but the star point will
    shift to anywhere within the line voltage triangle.

    Don Kelly

    remove the urine to answer

  6. Yes I could connect the star point to neutral.... but I would then not get a
    warning when an element blows....

    I know I could install current detection on the neutral line, but using the
    star imbalance is cheap.

    I just rewired the machine, and had to spend £ 143 on a reel of pyro cable,
    I had to re use the old elements, and I've no idea how long they will last.

    I can't help it if the factory manager does not want to spend a penny more
    than he absolutely has to.

  7. Don Kelly

    Don Kelly Guest

    I gave you a way to calculate the voltages - however I didn't discuss the
    problem with using a lamp to indicate trouble. One point that you were
    concerned about- too low a voltage to give a bright enough lamp. 46V on a
    115V lamp will be dull and a coloured lens might obscure it (is it on or
    not?). Another is that people tend to ignore lamps if they do not come on
    when they are not actually looking at it unless quite bright and obvious. A
    third is that when one element blows- the remaining element on that leg is
    stressed at higher than normal voltage (276V vs 230) so heating will be
    about 44% higher than normal in this element- accelerating failure. It could
    be that some voltage sensing relay (rather than current sensing) driving a
    lamp and light would actually save money by saving the second element. Such
    a device could operate if the voltage exceeds a set limit. If your
    objective is to shut the heaters down in case of a failure- then such a
    relay would be easier and cheaper than hoping an operator will notice.
    Just an opinion. "Penny wise, pound foolish". Work on the plant managers.
    Don Kelly

    remove the urine to answer
  8. daestrom

    daestrom Guest

    It seems to me (although I haven't sat down and done this calc), that you
    could convert the unbalanced wye resistances into an unbalanced delta
    configuration using the well known wye-delta conversions. Then you could
    figure the phase current in each delta leg (one would be significantly
    different than the others because of the unbalanced wye). Sum the phase
    currents into line currents (the line directly opposite from the 'odd' delta
    leg would be of interest). (each phase current is in-phase with its
    line-line applied voltage)

    With line currents known, apply them to the original wye configuration and
    figure the IR drop in each leg of the wye. From this, you can readily find
    the relation between lines and neutral (i.e. where the neutral has shifted).

    Oh, WTH. Here's a go at it. Suppose each resistor is 10 ohms and you have
    six of them arranged in a two-parallel, wye configuration. So the
    resistance in each leg is 5 ohms to start with. Then one resistor opens so
    that two legs are still 5 ohms, but the third leg is now 10 ohms, and
    line-line voltage is 100V. Using the wye-delta transformations (say the 'C'
    phase has the faulty resistor)....

    Zab = (Zoa*Zob + Zog*Zoc + Zoc*Zoa) / Zoc
    Zab = (5*5 + 5*10 + 10*5)/10 = 12.5 ohm
    Zbc = (5*5 + 5*10 + 10*5)/5 = 25 ohm
    Zca = (5*5 + 5*10 + 10*5)/5 = 25 ohm

    So the current through Zab = 100/12.5 = 8 amps, through Zbc = 4 amps and
    through Zca = 4 amps. Now, 'by inspection' since the currents through Zbc
    and Zca are equal and inphase with their respective line-line voltages, they
    are the easiest to sum. But the others can be properly summed as well, just
    be sure to consider the angles involved. I'll leave the vector addition as
    'an excercise', but I believe you will find each of the two delta legs
    attached to the C line (Zbc and Zca) will contribute cos(30)*4 amps to the
    line current so the total line current is cos(30)*8 = ~6.93 amps. (for
    those interested, sin(30)*4 = 2 amps of the current in Zbc will 'cancel'
    sin(30)*4 = 2 amps of the current in Zca. And notice that sqrt(3)*4 =

    Going back to the wye configuration, and knowing that Cline current is ~6.93
    amps, the voltage drop across the one remaining resistor in the C leg (R=
    10ohms) is ~69.3 volts. Before the fault, when it's balanced, the voltage
    drop was ~57.7, so the neutral has shifted 11.6 volts (~20%), directly
    toward the midpoint between the A and B lines.

  9. My thanks to the group.

    I have received answers to my problem of ; 57.8 V, 46 V, and 11.6 V.

    I think my favourite theory is the one giving 46 V :)

    I am about to log off this group ( hit and run ) but when an element blows I
    will take a measurement and report back.

    If I can return the favour I will. ( post to sci.engr.mech )


    Barnes's theorem; for every foolproof device
    there is a fool greater than the proof.

    To reply remove AT
  10. daestrom

    daestrom Guest

    Mind you, the 11.6 volts is for a the assumed 100V line-line voltage. So if
    you had 400 V line-line, you would have a corresponding shift of 46.4 volts.
    (other voltages are ratio-able from this example) The same ~20% increase.
    And since the remaining heater in that leg would have 120% of normal voltage
    applied to it, it would disipate (120%)^2 power or 1.44 times its normal

  11. daestrom

    daestrom Guest

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