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.
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.
...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.
BB+/BB=?
Well guys if this was a PE exam question, it would be pencils down and the clock would have run out by now.
"Just because you're paranoid, doesn't mean they're not out to get you"
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.
Gentlepeople,
My Bad
A much older Phil is still intact and almost completely back! And, I admit my declaring Load Z_{Δ} 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 V_{Δ} = Load Ph-Ph Voltage, given as 12.5 kVÐ0º. Thus,
o Vye = Load Ph-n Voltage = 12.5 kV/Ö3Ð-30º.
o Z_{L } = Line Impedance, given as (5.00 + j10.0), W.
o I_{L} = Line Current, given as 70.0Ð-20º, A
o Z_{Δ} = Load Ph Impedance, not given.
o I_{L }* Z_{L} = Line V_{DRP} is 0.77 kV.
o E_{YE } = Source Phase-N voltage..
o E_{Δ} = Sqrt(3) * E_{YE}. = 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.
Comments
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 H^{2}4F calculator is adequate.
Simplest Solution
Using the H^{2}4F calculator:
E_{Δ} = Sqrt(3) * [ (I_{L }* Z_{L }) / 1k + Vye ) ] = 13.6 kV.
General Solution
Because of the intervening impedance, Z_{L,} located between Source and Load, I normally recommend the Voltage-Divider method, where K_{VD }is the dimensionless ratio of just two variables, namely, Zye / (Zye + Z_{L}). Then,
o Given Source-E to find Load-V, then |V_{Δ}| = Sqrt(3) * |E_{YE}| * |K_{VD}| (most problems).
o Given Load-V to find Source-E, then |E_{Δ}| = Sqrt(3) * |Vye| / |K_{VD}| (this problem) = 13.61 kV !
A copy of the Excel Solution is available for anyone requesting it!
Regards, Phil Corso
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