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Homework Help: Electrical Engineering Question : 3 Phase AC - 4 wire star connected load with one phase blown

  1. Oct 4, 2014 #1
    Hello fellow engineers!
    I am a student doing a simple course in Electrical Engineering.

    I've got an enquiry regarding this question

    1. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.

    If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?

    We use a program to calculate the symmetric components. I believe I need to construct the resultant phasor of this system and derive the symmetric components

    This is my approach:

    1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V

    2) Calculate the balanced current that flows through each resistor I = V/R, I = 132.7906V/8 = 16.5988.

    3) If Phase C is eliminated from the fuse, I think that the line/phase voltages stay the same (as there is a neutral) and hence the current is the same.

    4) Therefore Phase A is 16.5988 amps and so is Phase B. Phase B is offsetted by 120 degrees taking A as the relative point.

    5) Phase C is 0 and it's on the 240 degree mark on the phasor diagram.

    I was wondering if this is all right in approach?
  2. jcsd
  3. Oct 4, 2014 #2
    Didn't you say phase to neutral voltage was 230 V?
  4. Oct 4, 2014 #3
    Yes, oh have I overlooked something?
  5. Oct 4, 2014 #4
    You should show by using symmetrical components what is new voltage phasor between phase B and star neutral after phase C is gone. Looks like a typical homework problem .
  6. Oct 4, 2014 #5
    Ah I ,maybe somehow I've interpreted "phase" as the "line", so the current is simply 230/8 = 28.75 amps.

    Indeed It's a typical homework problem, the problem is my course is an online one: they didn't go through the basics as through as I would've liked and we hardly do math problems, so I'm not as well practiced unfortunately.

    Zoki85, do you mean the new Phase A?
  7. Oct 4, 2014 #6
    I was under the impression I work backwards to find the symmetrical components, not the other way around?
  8. Oct 4, 2014 #7
    230 V is a common phase voltage (phase to neutral) in LV networks in Europe, so I guess it isn't mistake. This is simple situation and can be quickly solved even without method of symmetrical components. Phase to phase voltage is 230×√3 ≈ 400 V. Since phase C is open, current flowing through phases A and B is the same current that flows through series resistance 8+8=16 Ω. Current has magnitude 400/16=25 A. Note that in a phase diagram currents in phase A and phase B have same magnitude but are in phase oposition (angle difference between phasors is 1800 ).

    Last edited: Oct 4, 2014
  9. Oct 4, 2014 #8
    Hello Zoki85,

    Thanks for the reply. I was wondering if we should've considered current flowing to the neutral? would the phase C being broken, cause the load to be unbalanced and because it's (I assume a star connection): would cause current to flow into the neutral?

  10. Oct 4, 2014 #9
    Indeed, I forgot you were referring to the case with connection between neutral points of two stars. In that case you were right! Current through the phase A: IA= VA/R=230/8=28.75 A, and current through the phase B: IB=VB/R= 28.75 ∠1200 A.
    By Kirchoff's law, current through the neutral is I0=IA + IB. That would give I0= 28.75 ∠60° A (if I didn't make mistake without pencil&paper). For excersise, find from that, for both cases, direct, inverse, and zero sequence components of currents.
  11. Oct 5, 2014 #10
    I started my calculation when nobody was on the web. Now some of the correspondents already answered [partially in my opinion].

    230 V it is the voltage phase-to-neutral [your declaration].So phase-to-phase will be sqrt(3)*230=400 V [as usual].

    The current in phase A [let's take it as origin of the angles] will be IA= 230/8=28.75<0;

    IB=28.75<-120=28.75<240 and IC=0.Neutral current will be IN = - (IA+IB+IC) = -28.7*{[cos(0)+cos(240)]+j*sin(240)}

    cos(0)=1; cos(240)=cos(pi()/180*240)=-0.5; sin(240)=-0.866

  12. Oct 5, 2014 #11
    Hello Zoki and Babadag!, Thanks for the help!! This was vaguely what I thought I was supposed to do.

    I got Zoki's answer of 28.75 ∠60° by converting everything to the complex number and adding, but that's only because I used the convention that all the phase angles are referenced from Phase A and rotate counterclockwise. Babadag you're also right and I know that convention of everything anticlockwise of A is negative and clockwise is positive.

    You're both legends!
  13. Oct 5, 2014 #12
    The symmetric components are (if I take 28.75 arg(60)) for this question are:

    Zero =19.2 arg(60)
    Positive = 19.2 arg(-60)
    Negative = 9.58 arg(0)

    The symmetric components are (if I take 28.75 arg(240)) for this question are:

    Zero = 0
    Postive = 0
    Negative = 28.8

    Interesting differences.

  14. Oct 5, 2014 #13
    Just been told that there shouldn't be any zero sequences through the neutral, don't know why?
  15. Oct 5, 2014 #14
    Symmetrical components here you get from known phase currents IA,IB,IC .
    Zero sequence phasor definitelly isn't 0 due to fact sum of the currents in phases A and B isn't 0.
    I see that in my calcs I took oposite way of system rotation, and not the convential, counterclockwise one (like Baba did) .
    Better stick to the convention to avoid confusion with arguments
  16. Oct 6, 2014 #15
    Sorry I made a mistake in the inputting to calculate the symmetrical values.

    Start again: the resultant phasor values are (using Badabung's convention):

    Ia = 28.75 arg(0)
    Ib = 28.75 arg(240)
    Ic from Badabung =-14.38+24.9j = 28.75 arg(-60) = 28.75 arg(300) = this is the resultant neutral phasor

    The calculator for Symmetric components gave me

    I0 = 19.2 arg(-60)
    I1 = 9.60 arg(0)
    I2 =19.2 arg(60)
  17. Oct 6, 2014 #16
  18. Oct 6, 2014 #17
    Oh dear, Ic = 0. Phase C is dead broken (remember?) and no current can be flowing through it. You should really learn basics of this stuff...
  19. Oct 6, 2014 #18
    Sorry I've attached Ic as the resultant vector of A + B, it's a terminology error. I'd like to learn more but we don't get enough practice excersizes unfortunately:(
  20. Oct 6, 2014 #19
    I really do appreciate your help though Zoki!
  21. Oct 6, 2014 #20
    Hello again, I was wondering if anyone can confirm my answer?
  22. Oct 6, 2014 #21
    IA= 28.75 A
    IB= 28.75 ∠240° A
    IC= 0 A
    Use formulae:)
  23. Oct 6, 2014 #22
    Ah the take home lesson I guess is to "always allocate A B and C to their appropriate phases and you MUST do so" I made Ic = In subconsciously because I thought it would be needed in a Phasor diagram, but these Phasor Diagrams are strictly for Phase A, B and C - something that I wasn't so brought on about.

    Thank you heaps Zoki85, I've learn't more from you about these diagrams than reading them bit by bit in a year!
  24. Oct 7, 2014 #23
    Hello all:)
    Have to revive this thread just to clarify a new question:

    Now we must answer the same question, but with no Neutral, this was answered without intention in the beginning of the post

    I can do the maths involved, but I don't know how the phasors are 180 degrees apart and I don't know how to mathematically derive it. I'm under the impression that Phase A and B are still 120 degrees separated.

    I was wondering how A and B can be 180?
  25. Oct 7, 2014 #24


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    Staff: Mentor

    With no neutral, and phase C open, the line currents add at the "centre junction" to equal zero. ( Σ current in = Σ current out )

    So ##\mathbf{I_A = -I_B}## (equal and opposite means 180° apart)

    The voltage across one load is ##\mathbf{V_A - V_B}## and across the other load it's ##\mathbf{V_B - V_A}##

    Construct the phasor diagrams and you'll see these phase relationships.
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