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The question:

"A three-phase 230 V, 27 kVA, 0.9 PF (lagging) load is supplied by three 10 kVA, 1330/230 V, 60 Hz transformers connected in Y-Δ by means of a common three-phase feeder whose impedance is 0.003+j0.015 Ω per phase. The transformers are supplied from a three-phase source through a three-phase feeder whose impedance is 0.8+j5 Ω per phase. The equivalent impedance of one transformer referred to the low-voltage side is 0.12+ j0.25 Ω. Determine the required supply voltage if the load voltage is 230 V."

The way I did this was to first define the load as a delta load, so that each load takes 27000/sqrt(3) VA. Then, through KVL, the line currents can be found. From the line currents, it's just a matter of working backwards by multiplying the line current by two times the line impedance and one times the transformer impedance. Then, this voltage would be transformed to the primary side (along with the current) in order to find the input voltage (line-to-neutral); this would then be multiplied by sqrt(3) to get the line-to-line input voltage.

However, my answer is wrong, and I'm not sure why (I believe it has to do with the line currents)...

If somebody could help me work this out, it'd be mighty appreciated.

Regards

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  • \$\begingroup\$ I would calculate the required supply voltage per phase by referring the high-side feeder impedance to the low side. Then convert the voltage to the high level. It is not clear to me if that violates the "without converting the secondary side to 'Y'" constraint. Figuring out a more difficult way to work the problem is not something that I would volunteer to do.. \$\endgroup\$ – Charles Cowie Apr 21 '17 at 23:43
  • \$\begingroup\$ Thanks for the response! I know the much easier way to figure this out, but I think I may have some form of OCD - if I don't know why I'm wrong, it causes me to think about it far too long. \$\endgroup\$ – Lerbi Apr 22 '17 at 0:00
  • \$\begingroup\$ Now that I look at your description of what you did, I see "each load takes 27000/sqrt(3) VA." VA does not change by changing from Y to delta. The Load per phase is 9 kVA. \$\endgroup\$ – Charles Cowie Apr 22 '17 at 0:27
  • \$\begingroup\$ Interesting. But what about the equation S = sqrt(3)*V_L*I_L? \$\endgroup\$ – Lerbi Apr 22 '17 at 0:49
  • \$\begingroup\$ The formula for 3-phase VA is 3 X VA per phase. That is 3 X Phase current X phase voltage. 3 X V/sqrt(3) X I = sqrt(3) X V X I. You should be able to delete the answer that you posted when you should have posted a comment. \$\endgroup\$ – Charles Cowie Apr 22 '17 at 1:15

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