Wye-Delta transformer turns ratio question

[jd5191]

Sample Question from PPI's practic problems (Ch.42, # 9)

Simplified Question:

A 20MVA transformer has 12,500V(wye):4160V(delta) What is the rated primary and secondary current?

Book's Answer:

Ipri = S/(sqrt(3)*Vline) = (20x10^6)/(sqrt(3)x12,500) = 923.76A

Isec = Vpri*Ipri/Vsec = (12,500x923.76)/4160 = 2775.72A

For Isec, shouldn't the Voltages be phase voltages and the equation should be Isec = (Vpri/sqrt(3))*Ipri/Vsec = (12,500/sqrt(3))*923.76/4160 = 1602.5A?

I guess the bigger question is I've assumed A = Vp/Vs means those are phase voltages. Whenever a transformer voltage ratio is given, my understanding is they're usually given as Line-Line values therefore if a 12,500V:410V wye-delta transformer is shown, then the ratio a would not be a=12,500/4160 but should be a=(12,500/sqrt(3))/4160.

 

[jd5191]

****, I figured it out as I was hitting submit, they asked for line current. Sorry for the poast.

 

[akyip]

The best way I think of it for 3-phase transformer turns ratio, line-to-line voltages, and line currents is this:

The overall turns ratio of the 3-phase transformer as a whole is:

a 3-phase = V1 LL / V2 LL = I2 Line / I1 Line

But for the individual phase windings themselves (or for each 1-phase transformer of a 3-phase transformer bank):

a 1-phase = V1 phase / V2 phase = I2 phase / I1 phase

The phase voltages and currents depend on how the windings are connected (Y or delta).

For Y-connected windings:

V phase = V LN = V LL / sqrt(3)

I phase = I Line

For delta-connected windings:

V phase = V LL

I phase = I delta = I Line / sqrt(3)

I hope that this helps!