Complex Imaginary Set 4 - Question 41: Autotransformer question

Hi guys,

I'm a bit confused about an autotransformer question from Complex Imaginary Set 4. This is question 41:

"A Y-Y 230 KV 3-phase transmission line is being stepped down to 200 KV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 KVA, and the turns ratio is 5:2 (common:series), what is the apparent power in the common coil?"

Normally, my take on this (and other similar 3-phase autotransformer questions I've seen from other practice exams) would be this.

Each autotransformer has the line to neutral voltages on its terminals. So:

Vp = 230 KV / √(3) = 132.791 KV

Vs = 200 KV / √(3) = 115.470 KV

These are the line-to-neutral terminals of each autotransformer.

S 1-ph = 500 MVA (as given in the problem)

Finding the primary and secondary currents:

Ip = S 1-ph / Vp = 500 MVA / 132.791 KV = 3765.315 A

Is = S 1-ph / Vs = 500 MVA / 115.470 KV = 4330.129 A

The current in the common coil of this step-down autotransformer is:

Ic = Is - Ip = 4330.129 - 3765.315 = 564.814 A

The complex power in the common coil is:

Sc = Vc x Ic = Vs x Ic = 115.470 KV x 564.814 A = 65.219 MVA

However, the solution says that the correct answer is 143 MVA. The explanation from the given solution is:

S/Sc = (Ns + Nc)/Ns = 1 + Nc/Ns

Sc = S / (1 + Nc/Ns) = 500 MVA / (1 + 5/2) = 143 MVA

I'm just trying to understand why my initial attempt at this is incorrect, and how the solution came up with this formula relating the autotransformer MVA rating, common coil complex power, and common/series turns ratio.

I appreciate any input or response on this! Thank you!

akyip said:

Finding the primary and secondary currents:

Ip = S 1-ph / Vp = 500 MVA / 132.791 KV = 3765.315 A

Is = S 1-ph / Vs = 500 MVA / 115.470 KV = 4330.129 A

The current in the common coil of this step-down autotransformer is:

Ic = Is - Ip = 4330.129 - 3765.315 = 564.814 A

Click to expand...

This is where things started to go south. The power rating, 500 MVA, is for the windings. Yet, the voltage you used (Vp = 132.8) is not the voltage across the winding. You have to use the voltage across the winding when calculating the current in the winding.

I'll be referencing the autotransformer diagram below. S will be the power rating of the windings.



To calculate I_C, you would do

To calculate I_SE, you would do

However, from your calculations you did

This is incorrect. Notice you cannot divide the power of the windings (S) by the high-side voltage (V_H). You must use the voltage across the winding (V_SE).

On to your question about the equation used. This is from Electric Machinery and Power System Fundamentals (Chapman).