Calculate unbalance current, capacitor bank

Discussion in 'Homework Help' started by EirikO, Dec 21, 2015.

  1. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    Can anyone please help me calculate the unbalance current I0 of this curcuit? (See attachment, I0 is down, right)

    I have simulated it, but would like to confirm with calculation by hand..
     
  2. JohnInTX

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    Jun 26, 2012
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    Is this homework for school?
     
  3. WBahn

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    Mar 31, 2012
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    What did your simulation say the result was?

    So show your hand calculations and we will look them over.
     
  4. WBahn

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    Mar 31, 2012
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    MOD NOTE: Moved from General Electronics Chat to Homework Help.
     
  5. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    Simulation gave 0.2467 A...
    do not know where to start to calculate by hand..
     
  6. WBahn

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    Mar 31, 2012
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    If you know the currents (magnitude and phase) in the six ammeters in the individual legs, could you determine the current in the one connecting the two horizontal bus bars?

    If you knew the node voltage on the two horizontal bus bars (we are assuming that the ammeters are ideal, right), could you find the currents in the six ammeters in the individual legs?

    If none of the ammeters were there (i.e., replaced by shorts), could you find the voltage on the common node connecting all of the capacitors?
     
  7. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    That circuit ? You have three AC sources and zero AC returns.
     
  8. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Looks like a wye-configured AC generator to me. AC return not required, no?
     
  9. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    @joeyd999

    It looks Y to me also, but the loading has no ground.

    If it were delta, that would be a different story.
     
  10. WBahn

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    Mar 31, 2012
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    It's just a wye-connected source and two wye-connected loads that happen to have their "neutral" points tied to each other. The two loads are pretty close to being balanced, so I wouldn't expect much current to flow through the interconnect. This jives with the fact that the generators are 10 kV and the capacitive reactances are about 120 Ω yet the simulated interconnect current is only a quarter of an ampere.
     
  11. WBahn

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    From a circuit standpoint, the common point on a Y-connected load doesn't have to be grounded. It's still a valid circuit. But the three phases of the load interact if it isn't balanced.
     
  12. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Someone will have to show how the current flows from AC + to AC - ... I just don't see a common connection on the loads, at least not on the schematic as presented. To me, this is an open circuit, as presented.
     
  13. WBahn

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    Mar 31, 2012
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    3PhaseWyeLoad.png

    I've shown the three loops for one of the wye-connected loads.

    It might help to recall that, for a wye-connected generator and load, the neutral conductor going between the center points of point carries no current if the system is balanced, and hence can be removed without changing anything.
     
    EirikO likes this.
  14. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    Thank you all, so far.
    The amperemeters are just added to show in simulation, except the I0-amperemetere.

    The system is a high-voltage capacitor bank (this one at 11 kV (ph-ph)). I'm supposed to calculate I0 to confirm that the legs are balanced, hence no current SHOULD flow through I0... But when one of the six capacitors fail, the current in I0 will increase.
     
  15. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    Bump. Anyone having a clue to calculate the current?
     
  16. sailorjoe

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    Jun 4, 2013
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    All the AC voltage sources have the same amplitude and frequency, but each has a different phase. This is a lesson in the effects of phase on an otherwise balanced circuit. The capacitors aren't of equal value, and that adds a twist to the unbalancing as well.

    EirikO, I'm not sure exactly where you are in your studies, but if you have the math for this, consider that the AC sources are represented as v = A*sin(omega*t + phi). The phi is the fixed phase shift of each source.

    Now try something simple, with just a single source, two capacitors, and an ammeter connecting the two caps together. Take this a a single chunk out of your problem circuit.

    Can you write the equation for that?
     
  17. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    I'mm cant get rid og both U1 and U2... :(

    [​IMG]
     
  18. WBahn

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    Mar 31, 2012
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    You've been given several hints as to how to go about it. But we need to see YOUR effort to solve it before we can begin to discern where YOU are running into problems on YOUR homework.
     
  19. EirikO

    Thread Starter New Member

    Dec 21, 2015
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    I added my attempt for an easier circuit above... But run into the typical problem of having more unknown than equations..
    Most of the discussion so far is if this really is a circuit or not.. Which it is..

    The problem, as pointed out above, is to find the voltage in the common node. The ammeters are just added in simulation, so the values can not be used in calculations...
     
  20. WBahn

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    It's hard to follow your work because you give no indication of the reference polarities for any of your voltages, so we are left to guess what you meant and we don't know if we guessed correctly. Engineering is not about guessing.

    Your notation is also needlessly confusing. Why do you have the current going through the right-hand capacitor labeled as I1 but the voltage across it labeled as U2? And then you don't even bother to label which capacitor is Z1 and which is Z2, forcing us back into that guessing game again.

    Your work is wandering around without clear direction. Keep in mind form the start that the goal is to find I0 in terms of U0 and the two capacitances (which you don't label at all). You started off okay, namely by starting with the constraints imposed by KVL and KCL. You need to work to replace unknown quantities with known quantities -- you are replacing unknowns with other unknowns.
     
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