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Transformer Impedance Discussion

  1. Aug 7, 2014 #1

    psparky

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    I was working on trying to find the short circuit current rating of a 120 volt panel fed from a 7.5KVA transformer.

    Using P=IV you get 7.5KVA=120 * I

    I = 62.5 Amps.

    Then I believe you divide that current by the impedance of the transformer. .03 is the impedance in this case in the transformer specs. So I do the math and I get about 2,000 amp short circuit rating.

    Great. My question is this. If impedance is R + JX, how does .03 relate to that?

    Is it possible the R+jx add up to be .03? Or is that nonsense? I believe they even state the .03 as 3% as well.
     
    Last edited: Aug 7, 2014
  2. jcsd
  3. Aug 7, 2014 #2
    Ok, I am not an expert in electronics but if you have a circuit with impedances then its automatically an alternating current circuit. In this case I am only able to use complex way to calculate. (Z_complex * I_complex = V_complex)
    You have apparent power and therefore effective/active power and reactive power.
    There has to be a cos(phi) in your calculation of power. active power should be defined as P=V*I*cos(phi)
    You can't easily devide P by U for the current.
    U is the source voltage, but there is also R, and X (capacity or inductivity)
    that means you have dropout voltages on R and X (Vr and Vx)
    Also impedances generally have units, it's Ohm
     
  4. Aug 7, 2014 #3

    jim hardy

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    Sparky - I believe you are getting to the right answer.

    I was taught a simpler method
    Your 3% impedance is probably per-unit

    so 1 per unit volts / 0.03 per unit impedance = 33.3 per unit amps
    and your per-unit amperage is 62.5 .

    62.5 amps X 33.3 = 2081 amps into a zero impedance fault, about what you got

    (That's why your household 15 amp breakers say something preposterous on the back like 10,000 amps - don't [STRIKE]use[/STRIKE] attempt to use one downstream from a source capable of more fault current than that .)

    i've not seen it broken down into R and X for small transformers.

    Clearly the impedance of interconnecting wires and of the circuit breaker itself aid in limiting fault current.

    old jim
     
  5. Aug 8, 2014 #4

    psparky

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    Interesting approach Jim.

    I'm still looking for the relation of R+ jx to the transformer impedence (.03)

    I will hit up my old electronics mentor from college. I have yet to stump him.
     
  6. Aug 9, 2014 #5

    Baluncore

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    120V, 7.5kVA; I = 62.5A
    R load = V / I = 1.92 ohm

    I believe the 3% is the maximum percentage of voltage drop to be lost in the transformer secondary and circuit under full load conditions.
    3% of 1.92 ohm = 0.0576 ohm transformer secondary resistance.
    Short circuit current will be V / R = 120 / 0.0576 = 2083.3A

    If you know the current rating of the panel, you do not need to know the line voltage.
    Short circuit current will be 63A * 100% / 3% = 2100.A
     
  7. Aug 9, 2014 #6
    If we’ll represent the current flowing through transformer in complex:
    I=I*[cos(fi)-j*sin(fi)] and [neglecting the magnetic circuit of Xm and Rfe] the transformer impedance will be: Z=R1+R’2+j*(X1+X’2) where R1= primary winding resistance R’2=secondary winding resistance referred to primary and X1=leakage magnetic flux reactance of primary winding X’2=secondary winding leakage reactance referred to primary.
    Then we may take R1+R’2=R and X1+X’2=X and Z=R+j*X.
    abs(I*Z)[complex]=abs(I*[cos(fi)-j*sin(fi)]*(R+j*X)
    I*Z=I*{[R*cos(fi)+X*sin(fi)]+j*[X*cos(fi)-R*sin(fi)]}
    abs(I*Z)=I*SQRT{[R*cos(fi)+X*sin(fi)]^2+[X*cos(fi)-R*sin(fi)]^2}=I*SQRT(R^2+X^2)
    abs(I*Z)[complex]=abs(I)*abs(Z)=V1-V’2 [Voltage dip-or voltage sag- through transformer.
    This voltage dip -if the current will be Irated- it is the short-circuit voltage[vk] and it is represented in p.u.-vk%=vk/Vrated*100.
    Z=vk%*Vrated/Irated for single-phase [or two-phase] system and Z=vk%*Vrated/sqrt(3)/Irated.
    By multiplying [and dividing] by Vrated Z=vk%*Vrated^2/Srated where S=SQRT(3)*Vrated*Irated [the apparent rated power of the transformer].
     
  8. Aug 9, 2014 #7
    Correction:
    Z=vk%/100*Vrated/Irated for single-phase [or two-phase] system and Z=vk%/100*Vrated/sqrt(3)/Irated for three-phase system.
    By multiplying [and dividing] by Vrated Z=vk%/100*Vrated^2/Srated.
    Ik1=Vrated/Z or Ik3=Vrated/Z/sqrt(3).By substituting Z=vk%/100*Vrated/Irated
    Ik1=Irated*100/vk% or Ik3=Irated*100/vk% as Baluncore already said.
     
  9. Aug 9, 2014 #8

    jim hardy

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    Secondary? or both windings, ie lumped ?

    One per unit load gives voltage drop numerically equal to per unit Z, without having to worry about turns ratios .
    Handy, eh? That's why there's 'per unit' .
     
  10. Aug 9, 2014 #9
    Of course, if you will consider a Zbase=Vrated^2/Srated then vk=Z/Zbase[p.u.] or vk%=Z/Zbase*100
     
  11. Aug 11, 2014 #10

    psparky

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    Interesting stuff. Wish I had the brains to understand it.
     
  12. Aug 11, 2014 #11

    psparky

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    Gotcha, thanks.
     
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