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1. Junior Member
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Final answer I get is 12959 volts

I convert Delta load to wye and solve it for phase voltage

Vsource - VD across line reactance = Load voltage

This gives v source per phase multiplied by SQRT 3 to line line to line voltage of 12959 volts.

2. Originally Posted by roy167
Final answer I get is 12959 volts

I convert Delta load to wye and solve it for phase voltage

Vsource - VD across line reactance = Load voltage

This gives v source per phase multiplied by SQRT 3 to line line to line voltage of 12959 volts.
Well then your method and execution would evidently be correct since you got the correct answer.

3. Originally Posted by mivey
The something wrong is:
...and that is not to take away from David who first detailed Phil's error as shown below. But for some reason, after multiple members agreed with David's assessment, Phil could not seem to understand and suggested David should go back to school.

David's circuit analysis skills may have laid dormant but he certainly did not forget how to use them and evidently needed no refresher.

Originally Posted by david luchini
Should be easy enough to verify. With a Z of 290+j106 or 308.8<20deg, IAB would be VAB÷Z or 12500<0deg ÷308.8<20deg = 40.4<-20deg...
IBC = 40.4<-140deg, and
ICA=40.4<100deg.

IaA= IAB-ICA= 40.4<-20deg - 40.4<100deg = 70<-50deg.

Since the question tells us that IaA = 70<-20deg, it is clear that a load impedance of 290+j106 is not correct.

Try a load impedance of 304.6-j53.7...capacitive, not inductive. That would give an IAB= 40.4<10deg, IBC= 40.4<-110deg, and ICA=40.4<130deg.

So, IaA=IAB-ICA= 40.4<10deg - 40.4<130deg = 70<-20deg.
A capacitive load impedance of 304.6-j53.7 will give the proper line current.
Originally Posted by david luchini
Phil, here is your mistake. VAn will not equal VAB/Sqrt(3). VAn will equal (VAB<-30deg)/Sqrt(3).

Using the wrong VAn (7216.9<0deg V) will calculate the delta impedance as 290+j106, just as you mentioned in post #23.

Using the correct VAn (7216.9<-30deg V) will calculate the delta impedance as 304.6-j53.7, as I mentioned in post #24.

Hope this helps.
Originally Posted by david luchini
What we noticed is that your method produced the wrong load impedance, the wrong load current, and the wrong source voltage.

Your method is, quite frankly, irrelevant since it produces the wrong results.

4. Well guys if this was a PE exam question, it would be pencils down and the clock would have run out by now.

5. Originally Posted by kingpb
Well guys if this was a PE exam question, it would be pencils down and the clock would have run out by now.
There are problems just like this on the PE exam. You don't get to discuss with your fellow examinees though!

6. Junior Member
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Originally Posted by mivey
Well then your method and execution would evidently be correct since you got the correct answer.
I can't be sure of that. NCEES is known to include wrong answers in the options.

7. Originally Posted by roy167
I can't be sure of that. NCEES is known to include wrong answers in the options.
Yes they do. Common calculation and concept errors seem to be available answers. As a perfect example, an impedance problem would probably have both the correct answer and the answer Phil gave. Another answer might be one given by 12500@0d ÷ 70@-20d.

8. Junior Member
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When I solve it using the method above, you don't need to go into phasor etc for this kind of problem. It is quick and dirty and most importantly correct.

9. Senior Member
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Gentlepeople,

A much older Phil is still intact and almost completely back! And, I admit my declaring Load ZΔ as inductive was incorrect! It is capacitive, but it’s magnitude has little impact on the solution! Furthermore, I apologize to those taking umbrage to my comments! That said, following are the steps of my solution:

Old Stuff

o Sequence, given as A-B-C.
o VΔ = Load Ph-Ph Voltage, given as 12.5 kVÐ. Thus,
o Vye = Load Ph-n Voltage = 12.5 kV/Ö3Ð-30º.
o ZL = Line Impedance, given as (5.00 + j10.0), W.
o IL = Line Current, given as 70.0Ð-20º, A
o ZΔ = Load Ph Impedance, not given.
o IL * ZL = Line VDRP is 0.77 kV.
o EYE = Source Phase-N voltage..
o EΔ = Sqrt(3) * EYE. = 13.61 kV

New Stuff

I hope you all don’t mind, but I usually identify Source entities as E with upper-case subscripts, and Load entities as V with lower-case subscripts.

I use several methods, prioritized by problem complexity. For example, Florida’s PE Exam allows using the Casio fx-300MS PLUS calculator, which is great for complex numbers. However, when calculating using just entity magnitudes, then the H24F calculator is adequate.

Simplest Solution

Using the H24F calculator:
EΔ = Sqrt(3) * [ (IL * ZL ) / 1k + Vye ) ] = 13.6 kV.

General Solution

Because of the intervening impedance, ZL, located between Source and Load, I normally recommend the Voltage-Divider method, where KVD is the dimensionless ratio of just two variables, namely, Zye / (Zye + ZL). Then,
o Given Source-E to find Load-V, then |VΔ| = Sqrt(3) * |EYE| * |KVD| (most problems).
o Given Load-V to find Source-E, then |EΔ| = Sqrt(3) * |Vye| / |KVD| (this problem) = 13.61 kV !

A copy of the Excel Solution is available for anyone requesting it!

Regards, Phil Corso

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