3-phase complex power calculations

[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.

 

[mivey]

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.

 

[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.

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.

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.

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.

 

[kingpb]

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

 

[mivey]

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

There are problems just like this on the PE exam. You don't get to discuss with your fellow examinees though!

 

[roy167]

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. :roll:

 

[mivey]

I can't be sure of that. NCEES is known to include wrong answers in the options. :roll:

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.

 

[roy167]

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.

 

[Phil Corso]

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 H²4F calculator is adequate.

Simplest Solution
Using the H²4F 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

 

[mbrooke]

Bump

 

[david luchini]

o E_Δ = Sqrt(3) * E_YE. = 13.61 kV

The source voltage is 12.95kV, or most equal to answer C) 13.0kV.

 

[mbrooke]

Question- this is based on a transposed line?

 

[Phil Corso]

Lucini...

What is your V-Drp value across ZL?

 

[david luchini]

Lucini...

What is your V-Drp value across ZL?

The voltage drop across ZL is 782.6 volts

 

[junkhound]

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

+1

Yep thread whent way past PE exam extent. Back in #51 one can get the correct answer to the multiple choice even without calculator (or pencil/paper) if one can do simple equations in your head and remember sqrt 3 value (1.73; or approximate fraction of 7/4 which is easier without penci/paper or calculator)

Back in November, almost took the 10 minutes needed to transcribe to matlab or pspice for 10 digit accuracy, still to lazy to do so :roll:

 

[Phil Corso]

Correct Luchini...
Your number is in Volts, which in kV is 0.78!
Now, when 0.78 is added to Vye, which is also in kV, then, EYE is 7.86! And when multiplied by Sqrt(3), EYE is 13.6 kV!

Brilliant Roy & Junkhound...
Unfortunately neither of you are sitting near the guy taking the exam! And, you certainly aren't educating others on this forum!

Last effort, what did you get when you applied the Voltage-Divider technique?

Finally, I used the V-load values in a 3-ph, balanced D-D configuration, and got the same answer!

 

[david luchini]

Correct Luchini...
Your number is in Volts, which in kV is 0.78!
Now, when 0.78 is added to Vye, which is also in kV, then, EYE is 7.86! And when multiplied by Sqrt(3), EYE is 13.6 kV!

Yes, 0.78kV is correct. And when you figure Vab= VaA+VAB-VbB, you will get the correct source voltage of 12.95kV, or (C) 13.0kV.

You're certainly not educating anyone by posting incorrect solutions over and over and over again.

 

[mivey]

You're certainly not educating anyone but posting incorrect solutions over and over and over again.

Which at this point is interesting in itself. He went down the wrong path but finally, after much ado, figured out the capacitive vs. inductive error.

He then proceeds to make another error after what you would think would be a much closer look.

So the interesting question is there a fundamental lack of understanding that may have caused errors throughout his career or is this a recent development which shows a lack of concentration or other issue?

If the first then it is time to refresh/tighten up. If the second then a genuine concern is raised. Based on past posts I know Phil has skills so now I'm curious as to why this one circuit is causing him so much trouble.

Phil, are you doing okay?

 

[david luchini]

Correct Luchini...
Your number is in Volts, which in kV is 0.78!
Now, when 0.78 is added to Vye, which is also in kV, then, EYE is 7.86! And when multiplied by Sqrt(3), EYE is 13.6 kV!

When you add the Line voltage drop (VaA=782.6V<43.43 ) to the Vye (7216.9V<-30), you will get 7.47kV<-24.24 . And when multiplied by Sqrt(3)<30, Eye is 12.95kV<5.76...which is the correct source voltage.

 

[Phil Corso]

3-ph complex power caalc

3-ph complex power caalc

Gentlepeople…

I promise, nay, pledge to God, Country, MHF’s Moderators, PEs, Newbees, Members, and guests, that this is my final reply to this topic!!! It is last of 4 methods (3 of which were previously given) that prove the PE Exam answer is wrong!

Really simple:

o Used Exam’s Delta-Source voltage, E_Δ,

o Converted them to their 1-ph, Y-equivalent, E_YE,

o Subtracted Line’s V_DRP yields load’s Y-equivelent, Vye,

o Converted it to its 3-phase Delta value, V_Δ..

Results are tabulated below. Note: all values are given in kV!

|E_Δ| => |E_YE| => |V_DRP| => |Vye | => |V_Δ|

15.0 8.67 0.78 7.88 13.6 :happysad:
13.8 7.97 0.78 7.18 12.4 :happyyes:
13.0 5.92 0.78 6.72 11.6 :happysad:
11.8 6.81 0.78 6.01 10.5 :happysad:

The fact that I won’t respond to this thread again, will not prevent me from giving details to those interested! Contact me at Cepsicon@aol.com or via the forum’s communication system!

Regards, Phil Corso

Ps: Anyone know the mfg of the first H^2-4F calculator ?