3-phase complex power calculations

[david luchini]

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 ?

Seriously...

Phil, are you doing okay?

Are you OK? The correct answer has been demonstrated over and over and over again. You keep getting the wrong answer over and over and over again?!?!?!

The magnitude of the source voltage Vab is most nearly 13.0kV

12.95KV /5.76deg.

I get the same.

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

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

The correct answer is 12.95kV or "C"

Final answer I get is 12959 volts

 

[david luchini]

Really simple:

This is literally as simple as 2 steps:

Step one: Calculate the line voltage drop VaA by multiplying the Line Impedance 5+j10 (which is given) by the line current IaA 70<-20 (which is given): VaA=782.6V<43.43

Step two: Add/subtract three voltages

...............VaA 782.6V<43.43 (which you just calculated)
........plus VAB 12500V<0 ......(which is given)
.....minus VbB 782.6V<-76.57 (which is known once you calculated VaA in step one.)
Equals.... Vab 12952V<5.76

So the magnitude of the source voltage Vab is most equal to 13.0kV

 

[david luchini]

This is literally as simple as 2 steps:

Of course for a couple of extra steps, you could convert the delta to a wye as you suggested:

Step one: Calculate the line voltage drop VaA by multiplying the Line Impedance 5+j10 (which is given) by the line current IaA 70<-20 (which is given): VaA=782.6V<43.43

Step two: Convert to delta load voltage to an equivalent wye voltage VAN=7216.9V<-30

Step three: Add VaA to VAN to get VaN: 782.6V<43.43 + 7216.9V<-30 = 7477.7V<-24.24

Step four: Convert the magnitude of VaN back to a delta source voltage by multiplying by the sqrt(3): 7477.7V*Sqrt(3)=12952V, which is most equal to 13.0kV

Either method will get you the correct source voltage Vab of 12.95kV

 

[dkarst]

Gentlepeople,



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.



Regards, Phil Corso

I guess I'm still as astonished as I was a couple months ago that this many methods can lead one to an incorrect solution but I'm tired of fighting that. I am going to point out an error just in case a PE exam candidate reads this and get tossed from the exam for violating the calculator policy. The Casio referenced by Phil above is not allowed in any state as it violates NCEES policy. You can find the list here... https://ncees.org/exams/calculator/

 

[Phil Corso]

LUCHINI,

I WISH YOU HAD PRODUCED YOUR 'PROCEDURE' LIKE I DID, EARLIER! IT WOULD HAVE SAVED ME AND OTHERS ( I'M SURE) A LOT OF GRIEF !

BUT YOU, AND OTHERS, USING THE TECHNIQUE YOU JUST POSTED, ARE ALL IN NEED OF ADDITIONAL POWER SYSTEM EDUCATION !

ONE DOES NOT JUST ADD THE V-DROP CAUSED BY AN INTERVENING LINE-IMPEDANCE TO NEITHER THE SOURCE'S PHASE-PHASE VOLTAGE NOR THE LOAD'S PHASE-PHASE VOLTAGE!

I'M RENEGING ON MY PLEDGE UNTIL YOU OR SOMEONE ELSE, SHOWS ME YOUR 'PROCEDURE' TAKEN FROM A "POWER THEORY" BOOK!

FROM A VERY OLD, VERY FRUSTRATED, VERY DISAPPOINTED, VERY VERY EXPERIENCED, 'ELECTRICAL' POWER ENGINEER!

IN ALL SINCERITY,

PHIL CORSO

 

[david luchini]

LUCHINI,

I WISH YOU HAD PRODUCED YOUR 'PROCEDURE' LIKE I DID, EARLIER! IT WOULD HAVE SAVED ME AND OTHERS ( I'M SURE) A LOT OF GRIEF !

BUT YOU, AND OTHERS, USING THE TECHNIQUE YOU JUST POSTED, ARE ALL IN NEED OF ADDITIONAL POWER SYSTEM EDUCATION !

ONE DOES NOT JUST ADD THE V-DROP CAUSED BY AN INTERVENING LINE-IMPEDANCE TO NEITHER THE SOURCE'S PHASE-PHASE VOLTAGE NOR THE LOAD'S PHASE-PHASE VOLTAGE!

I'M RENEGING ON MY PLEDGE UNTIL YOU OR SOMEONE ELSE, SHOWS ME YOUR 'PROCEDURE' TAKEN FROM A "POWER THEORY" BOOK!

FROM A VERY OLD, VERY FRUSTRATED, VERY DISAPPOINTED, VERY VERY EXPERIENCED, 'ELECTRICAL' POWER ENGINEER!

IN ALL SINCERITY,

PHIL CORSO

Phil, give it a rest. You are a very old, very frustrated, very disappointed electrical power engineer who cannot find the proper solution to a simple problem.

For the life of me, I cannot figure out why you are having such trouble with something so elementary.

The voltage Vab = VaA + VAB - VbB. Even engineers who don't go into the electrical field learn this.

 

[Phil Corso]

David,

Lets get elemental: A 3-ph D-D network has 6-loops, 5-unknowns, 5-equations; convert to Y-Y and you have 3-loops w/neutral, 3-unknowns, 3-equations; remove neutral you still have 2-loops, 3-unknowns, and 3 equations, but with simple math trick the 3 are reduced to 2!

If the 3-ph load is balanced, regardless of configuration, than you move directly to a single-loop equivalent having 1-equation! Simple, but only if the right equations are in play ! Your's aren't/weren't ! The Ab, bc, aa is just a smoke screen... totally necessary!

So, prove otherwise ! I always offered detail in the past, and will in the future !

And, knock-off the negativity... I'm doing fine now! Even failed two Alzheimer's Tests! :lol:

Phil

 

[david luchini]

And, knock-off the negativity... I'm doing fine now!

You mean this type of negativity...

ARE ALL IN NEED OF ADDITIONAL POWER SYSTEM EDUCATION !

A half a dozen engineers all get the same answer. You keep getting different answers.

Which do you think would be more likely to be correct. The same answer by six different people or the myriad of changing answer by you?

 

[david luchini]

Lets get elemental: A 3-ph D-D network has 6-loops, 5-unknowns, 5-equations; convert to Y-Y and you have 3-loops w/neutral, 3-unknowns, 3-equations; remove neutral you still have 2-loops, 3-unknowns, and 3 equations, but with simple math trick the 3 are reduced to 2!

There is only one single loop that needs to be evaluated in order to find Vab. There are 3 known voltages in that loop (VaA, VAB, and VbB,) and one unknown (Vab.)

Perhaps you'd like to point us to the electrical reference that says you can subtract the magnitudes of 2 voltages with different phase angles to find the resultant voltage? Because that is exactly what you are doing here.

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

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

15.0 8.67 0.78 7.88 13.6 :happysad:

This solution from your calculator is clearly wrong.

Simplest Solution
Using the H²4F calculator:
E_Δ = Sqrt(3) * [ (I_L * Z_L ) / 1k + Vye ) ] = 13.6 kV.

Just do the math. You have told us that IL is 70<-20 and ZL is 5+j10. The resultant voltage is 782.6V<43.43

Divide that by 1k to get kV instead of volts. 782.6V<43.43/1000 = 0.783kV<43.43

Add that to the Vye, which you told us is 7.217kV<-30. 7.217kV<-30 + 0.783kV<43.43 = 7.478kV<-24.24

If you multiply the magnitude |7.478kV| times Sqrt(3), you will get a source voltage magnitude of 12.95kV, which corresponds to the answer that everybody but you has derived.

If you want to include the phase angle in the conversion back from Wye to Delta, then 7.478kV<-24.24 * 1.732<30 will equal 12.95kV<5.76, which also corresponds to correct answer that everyone else has derived.

 

[BatmanisWatching1987]

I just notice this post and maybe the solution to the Text will help



Also, is the angel of V_ab,Source be 5.76 degrees?

 

[mivey]

For the life of me, I cannot figure out why you are having such trouble with something so elementary.

That is a mystery. This is pretty basic stuff and not something a newbie EE should struggle with much less a seasoned professional.

Perhaps it is just a developed bad habit. In a long career, someone would have called him on it already. If he listened, that is.

 

[mivey]

I just notice this post and maybe the solution to the Text will help

View attachment 22426

Also, is the angel of V_ab,Source be 5.76 degrees?

See David's detailed solutions.

 

[mivey]

I'M RENEGING ON MY PLEDGE UNTIL YOU OR SOMEONE ELSE, SHOWS ME YOUR 'PROCEDURE' TAKEN FROM A "POWER THEORY" BOOK!

This type step-through analysis is covered in every circuit analysis book I have ever seen and I have seen quite a few. This type of analysis is fundamental in any EE circuits course. I am at a loss as to why you continue to struggle with this solution.

It seems to me you are have a fundamental misunderstanding of phase angles or some similar concept. Perhaps you have never had to solve a problem quite like this since your school days but I would bet that if you are a EE graduate you covered this in school. Things can be forgotten over time and I'm not faulting you for that. I am faulting you for failing to refresh your knowledge after so many erroneous posts on your part.

You can run the values in a circuit modeling program and it will verify that your solution is wrong. At the start of the thread, I did run this out of respect for what I assumed was your knowledge level because your insistence made me think that maybe there was something I was missing. I also make mistakes from time to time and was willing to double-check.

May I respectfully suggest you take a breath, step back, and re-think what you are doing because you are wrong. Take it from a fellow long-timer.

 

[Phil Corso]

Gentlepeople...

Remember, I said I would stop contributing to this thread when I saw a solution in print! Well, fortunately, Batman produced it! It is exactly as mine, but with one major difference... in his, line-impedance is entered as... 5+j10, in mine it's 10+j5 (as well, in the four different solutions) and I can't explain why!

So to Batman... thank you ! And, to all others I fervently apologize !

Sincerely,

Phil Corso

 

[Phil Corso]

Will produce answers shortly -Phil

 

[mbrooke]

That is a mystery. This is pretty basic stuff and not something a newbie EE should struggle with much less a seasoned professional.

Perhaps it is just a developed bad habit. In a long career, someone would have called him on it already. If he listened, that is.

Truth being few EEs are as gifted as you- same pertaining to many other exceptionally brilliant inviduals on this forum. Being the 1% of 1% makes it easy to loose visibility of typical skill sets, aptitudes, or normal forgetfulness that comes with not using a certain equations, theories or rules daily. Especially true when someone has ASPEN Oneliner, Siemens PSS-E, Power World, ect at their fingertips. Further some are students, self taught or haven't the privilege to go through a full spectrum education.



Good example are the code experts on this forum. They can recite entire entire NEC chapters word for word and find 50 code violations in a single picture that past inspection- yet the typical electrician has to consult a code book for something basic as conductor re-identification as evidenced by all the posts here. That and for many the NEC is just something that sits on the dash. Not meant to put anyone down, just the facts of human nature as applicable to all humans and I am no exception for from it.



Further I've noticed a recurrent pattern that when someone rightfully challenges incorrect information, it turns into personal critique instead of a constructive rebuttal.

Ie, axiom made:

https://forums.mikeholt.com/showthread.php?t=196760&page=3&p=1982838#post1982838

After axiom was respectfully questioned from a humble place:

https://forums.mikeholt.com/showthread.php?t=196760&page=4&p=1982845#post1982845

https://forums.mikeholt.com/showthread.php?t=196760&page=4&p=1982846#post1982846


Of course said axiom is no swimmer as I knew:

https://forums.mikeholt.com/showthread.php?t=197606&p=1991320#post1991320


With that said lets give others the benefit