3-phase complex power calculations

[david luchini]

Line-current, IaA, was equal to 17.2 A<20º !

And yet another mistake. :slaphead:

Seriously, how difficult is it to say "sorry, I made a mistake and came up with the wrong answer"?

 

[Phil Corso]

D.Lucini

I'm sorry you are displeased! But, I'm at the age where mistakes are a way of life! Hopefully, that's all you will have to face when you're my age!

And, yes, it should have been 72.1 !

However, to prove my method's correct, I'm willing to give my program to anyone asking for it... at no cost !

Phil Corso

 

[david luchini]

D.Lucini

I'm sorry you are displeased! But, I'm at the age where mistakes are a way of life! Hopefully, that's all you will have to face when you're my age!

I'm sure when I'm your age I will have no problem admitting when I am wrong.

And, yes, it should have been 72.1 !

And no, 72.1 A <20° is no more correct than 17.2 A<20°

 

[mivey]

Sorry fellas...:happysad:


I used my load values in a 3-ph Load-Flow calculator! Line-current, IaA, was equal to 17.2 A<20º ! Referenced load VAB, was equal to 12.5 kV<0º ! Calculated source, Vab, was 13.8 kV<54º ! Furthermore, Other phases were appropriately displaced,i.e, +,- 120deg

Depending on reaction I may show you how it's done using Parameter magnitudes! Even with 2-phases!
And I use a slide-rule ! It as served me well 65 years !:slaphead:

Best of luck !!!

Phil

The 20 degrees is negative, not positive. As for the solution posted, the E part of your GiFiES was wrong.

Once you correct your setup, you can share your corrected results. No need to change methods until you fix the first one.

add: I don't think your method is flawed, just the setup.

 

[Phil Corso]

Gentlemen ?

Here are my final two words on the topic… No, they aren’t Mea Culpa! Instead, they are Astonishment and Surprise!

First, I am astonished that no-one, repeat no-one, noticed that the solution I provided was a D-D analysis, not a D-Y transformation, nor a 1-ph equivalent method, not even the 2-phase approach!

Secondly, I am surprised no-one asked for a copy of my method !

Wait… I erred once again! There is a 3^rd comment… I fondly wish that the remainder of your
lives are Healthy, Joyous, and Prosperous !

Phil Corso

 

[david luchini]

Gentlemen ?

Here are my final two words on the topic… No, they aren’t Mea Culpa! Instead, they are Astonishment and Surprise!

First, I am astonished that no-one, repeat no-one, noticed that the solution I provided was a D-D analysis, not a D-Y transformation, nor a 1-ph equivalent method, not even the 2-phase approach!

Secondly, I am surprised no-one asked for a copy of my method !

Wait… I erred once again! There is a 3^rd comment… I fondly wish that the remainder of your
lives are Healthy, Joyous, and Prosperous !

Phil Corso

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.

 

[dkarst]

Gentlemen ?

Here are my final two words on the topic… No, they aren’t Mea Culpa! Instead, they are Astonishment and Surprise!

Phil Corso

I'm in the astonished camp as well since I posted the correct solution in post #30 which confirms David's response.

This whole string reminds me of the proud parents in the stands and as the marching band comes by "Look honey, everyone is out of step except for our son..."

 

[mivey]

Gentlemen ?

Here are my final two words on the topic… No, they aren’t Mea Culpa! Instead, they are Astonishment and Surprise!

First, I am astonished that no-one, repeat no-one, noticed that the solution I provided was a D-D analysis, not a D-Y transformation, nor a 1-ph equivalent method, not even the 2-phase approach!

Secondly, I am surprised no-one asked for a copy of my method !

Wait… I erred once again! There is a 3^rd comment… I fondly wish that the remainder of your
lives are Healthy, Joyous, and Prosperous !

Phil Corso

The solution you provided had a formula error. Specifically the "IΔ = IL / Sqrt(3)" is missing a phase shift. Line current into the corner of a delta is not in phase with the delta (phase) current. You know that of course. Otherwise, nothing wrong with your method in general.

It is common analysis and you made a mistake. There is no escape. No biggie so don't make it a biggie.

Who wants a copy of a method from someone unwilling to fix an error? How many errors would be in your other submittal? Peer review is a normal process, not something to be feared and railed against. Listen to what your peers are trying to tell you.

 

[mivey]

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.

The method in general is not the problem. The execution is the problem.

 

[mivey]

This whole string reminds me of the proud parents in the stands and as the marching band comes by "Look honey, everyone is out of step except for our son..."

 

[junkhound]

Scanned the thread, in service of the OP and just 'simply' answering the test question vs. long accurate analysis:

D. is obviously wrong, voltage will not drop

For rough estimate, Z=sqrt 125, so drop across line is 11.1 *70 = 777, ignore the angle for rough estimate.
777/(sqrt3) = about 450V, mostly inductive drop

12,500 +450 (ignore phase lag for first cut) = about 12,950, hence answer is 13 kV as it is the 'nearest'

If the question wanted an exact answer, then angle would have to be accounted for, but for multiple choice like the OP posted, no need.

 

[david luchini]

The method in general is not the problem. The execution is the problem.

I'm not so sure about that. An execution mistake (missing an angle change) led to an incorrect load impedance of 309<20.

However, even using the incorrect load impedance of 309<20, it would still not lead to these results...

Line-current, IaA, was equal to 72.1 A<20º ! Calculated source, Vab, was 13.8 kV<54º !

Unfortunately, I'm thinking that the "method" being employed is "trolling."

 

[mivey]

I'm not so sure about that. An execution mistake (missing an angle change) led to an incorrect load impedance of 309<20.

However, even using the incorrect load impedance of 309<20, it would still not lead to these results...



Unfortunately, I'm thinking that the "method" being employed is "trolling."

Not sure what is up. So many twists and turns. Where is Robert Stack when you need him?

 

[Phil Corso]

Junkhouse...

You're getting closer ! Except that your calculated 777 V-Drop isn't added to the Ph-Ph source magnitude (Vab)! Instead it should be added to the Ph-neutral value (Van) ! Then, Van is multiplied by sqrt(3) and the 30 deg angle applied !

Your method negatively impacts the P.L.C. !

Phil

 

[david luchini]

Junkhouse...

You're getting closer ! Except that your calculated 777 V-Drop isn't added to the Ph-Ph source magnitude (Vab)! Instead it should be added to the Ph-neutral value (Van) ! Then, Van is multiplied by sqrt(3) and the 30 deg angle applied !

Your method negatively impacts the P.L.C. !

Phil

Do you ever get tired of being wrong? Junkhound has the correct answer.

 

[Phil Corso]

Sorry Luchini,

You and other responders to the OP's question have put me in a precarious legal position !

And knock of the name calling! I never shown such disrespect to anyone on this forum !

 

[mivey]

Sorry Luchini,

You and other responders to the OP's question have put me in a precarious legal position !

What does that even mean? What does have to do with the circuit analysis and responses?

Are you ready to concede that the load impedance is capacitive?

It is really a simple thing to prove. Hook up a circuit (in a simulator) with:

1) a delta load with impedances of (304.595904 -j53.708476) ohms
2) line impedances of (5 +j10) ohms
3) a wye source of 7.4777 kV at -24.2425826 degrees with positive sequence (12.9518056 kV line-line)

and this will give you 70 amps at -20 degrees from a to A and will also give 12.50 kV at 0 degrees from A to B.

FWIW:
I_AB = 40.41 amps at 10.00 degrees
I_AC = 40.41 amps at -50.00 degrees
I_CB = 40.41 amps at 70.00 degrees

So what say you?

 

[dkarst]

Gosh, it's been so long the OP has probably moved on after shaking his head for a bit. They were really looking for a reference to help with understanding and not the solution to the problem as the problem/figure is out NCEES practice problems book and the worked out solutions are in the back of the book. The correct answer is 12.95kV or "C"

Back to the OP's request, the reference I would recommend is T. Wildi "Electrical Machines, Drives, and Power Systems"

 

[Phil Corso]

Mivey, Dkarst, et al...

1) Shouldn't all your 'FWIW' Load Phase angles be 120 deg apart? Luchini, what do you think?

2) Exactly what given values led to the idea the load was capacitive?

3) Again I ask, what was is P.L.C. Factor?

4) I did exactly what you suggested! In a computer!

5) What makes you so sure the Identified answer to the Test question is correct?

6) BTW, 7-decimal answers is not proper engineering!

7) If you don't understand my legal dilemma, talk to a lawyer!

More later!

 

[mivey]

Mivey, Dkarst, et al...

1) Shouldn't all your 'FWIW' Load Phase angles be 120 deg apart? Luchini, what do you think?

2) Exactly what given values led to the idea the load was capacitive?

3) Again I ask, what was is P.L.C. Factor?

4) I did exactly what you suggested! In a computer!

5) What makes you so sure the Identified answer to the Test question is correct?

6) BTW, 7-decimal answers is not proper engineering!

7) If you don't understand my legal dilemma, talk to a lawyer!

More later!

1)No. To get 120 degree differences you need to take the currents in a symmetric direction. I took them in the direction I used at the time when I summed current.

2) V_AB, IaA, Z_line, phase rotation.

3) Define P.L.C.

4) So did I. Your results and comments suggest you used the wrong settings. My computer results agree with my other computer results and with my hand calculations and with my other hand calculations and the results of every other poster but you.

5) Because I checked it several different ways and get the same result.

6) An opinion. Significant digits are used to indicate precision of measure. In a made-up problem the numbers are as precise as you want because there is no real measurement.

It is proper if you want to account for rounding and match a number. Normally the final result is rounded for presentation but with computing power you don't round during the calculation.

Engineers (or anybody for that matter) can set the precision they want based on the circumstance.

To return the nit-pick, it was not a 7 decimal answer.

7)
me: "Hey lawyer, what is Phil's legal dilemma?"

lawyer: "About what exactly and who is Phil?"

me: "I have no idea, some guy on the internet said that forum member's answers to a question from a different member caused him a legal dilemma."

lawyer: "what does their answers have to do with him?"

me: "he said to ask you."

lawyer: "How could I possibly know the answer without information about the relationship between answers from other members and Phil and how they are legally relevant or even what one has to do with the other?"