3-phase complex power calculations

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Mivey... following are additional errors I uncovered...

When a complex-number is divided by a real-number, for example sqrt(3), only the magnitude changes... not the angle! Thus, the FWIW phase-current values you produced are dead wrong!

Secondly, sequence details the order of Voltages, that is vector rotation, A-B-C or A-C-B.! The A-B-C sequence was specified!

Phil
 
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Phil Corso said:
Mivey... following are additional errors I uncovered...

When a complex-number is divided by a real-number, for example sqrt(3), only the magnitude changes... not the angle! Thus, the FWIW phase-current values you produced are dead wrong!

Secondly, sequence details the order of Voltages, that is vector rotation, A-B-C or A-C-B.! The A-B-C sequence was specified!

Phil
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Nope.

This is a basic concept. The line current into the corner of a delta is the sum of the two phase currents out of the corner. That is why I listed the phase currents as leaving the corner. In other words you should recognize that

I_aA = I_AB + I_AC

70@-20d = 40.41@10d + 40.41@-50d



What is not true, and is one error you have repeated, is thinking delta phase current is equal to line current divided by the square root of three. Doing so neglects the phase shift common in any wye-delta conversion. In other words:

I_AB /= IaA ÷ sqrt(3)



What is true for this problem is that delta phase current is equal to line current divided by the square root of three at an angle of -30d. In other words:

I_AB = IaA ÷ sqrt(3)@-30d



I used A-B-C rotation in my calcs.
 
Phil Corso said:
Mivey, Dkarst, et al...

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

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

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

More later!
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I claimed I was done wasting time on this but since my name was quoted...

2. the load impedance in the delta has been shown without a doubt to be ~ 305 -j54 ohms. Most engineers would synthesize this with a resistor and capacitor in series and term the load capacitive.

3. I don't know what the term P.L.C. you are referring to is and don't care. If you can't calculate the correct load impedance, I can't imagine we can agree on any other term.

5. The problem in this book referenced by OP is only asking the voltage, not load but it has been vetting by literally thousands of examinees and subject matter experts over many years. There is 0% probability that "C" is incorrect.
 
Phil Corso said:
Mivey, Dkarst, et al...

1) Shouldn't all your 'FWIW' Load Phase angles be 120 deg apart? Luchini, what do you think?
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I think all his load phase angles are 120 deg apart.

Phil Corso said:
2) Exactly what given values led to the idea the load was capacitive?
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Exactly what given values led you to the idea that the load was inductive?

Phil Corso said:
4) I did exactly what you suggested! In a computer!
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Your computer seems to work as well as your slide rule. Or in other words...GIGO

Phil Corso said:
6) BTW, 7-decimal answers is not proper engineering!
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This is not proper engineering...

Phil Corso said:
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
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Phil Corso said:
Lucini...

Answer was shown in my Jiffies method!
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No it wasn't.

Phil Corso said:
Fellas...

Now, here’s my Gi.Fi.E.S. method (pronounced Jiffies) that proves the Delta-load is Inductive:

o Gi means ‘Given’
o Fi “ ‘Find’
o E “ ‘Equation’
o S “ ‘Solution’ The procedure follows:


Given: (System Parameters)
o VΔ = Load Phase Volts, given as 12.5kVÐ0º !
o IΔ = Load Phase Amperes, not given !
o ZΔ = Load Phase Impedance, not given !
o ZL = Line Impedance, given as (5 + j10) W !
o IL = Line Amperes, given as 70 @ -20º !
o Let (R+ jX) equal an Impedance that’s Inductive !
o Let (R - jX) equal an Impedance that’s Capacitive !

Find: Is Delta-Load Impedance Inductive of Capacitive ?

Equation(s):
o IΔ = IL / Sqrt(3)
o ZΔ = VΔ / IΔ

Solution:
o ZΔ = 309 @ 20º = 291 + j 106 (Q.E.D.)


Happy Post Thanksgiving, Phil
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Load currents are listed anywhere in that post.
 
Phil Corso said:
Fellas.

I offered my calcs!
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Post your work then. Please feel free to do so. Go right ahead.

Quit dancing around and show your work. Someone might take the time to mark it up with a red pen and correct your errors. However, you have thus far been unwilling to listen.
 
Phil,

You ignore a lot of discussion and seem to want to post bits and pieces without addressing the problem. Perhaps focusing on what appears to be the main problem will help.

So let's cut to the chase. Do you agree or disagree with the following:

I_AB = IaA ÷ sqrt(3)@-30d
 
Mivey...

I have never shirked from doing the math! And, NO, I don't agree with your approach! I did the math earlier! You don't arbitrarily affix a -30 deg! I said it before, but you won't listen! The correct procedure is to simply transpose the I-Line angle to I-ph! Your cohorts make the same mathematical error!

I'm sorry to say it, but we will never agree! But, know I now why!
 
Mivey...

I had a change of heart! So, i'll reveal what your error is! Affixing the-30 deg is done ONLY when using the "1-ph Equivalent For Balanced 3-ph Circuits"! It is not done for a 3-phase D-D problem!

Phil
 
Phil Corso said:
Mivey...

I have never shirked from doing the math! And, NO, I don't agree with your approach! I did the math earlier! You don't arbitrarily affix a -30 deg! I said it before, but you won't listen! The correct procedure is to simply transpose the I-Line angle to I-ph! Your cohorts make the same mathematical error!

I'm sorry to say it, but we will never agree! But, know I now why!
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Nothing arbitrary about it. It is determined by the given information. It is basic circuit analysis, not some new approach I came up with. It is taught at elementary levels as part of any electrical engineering curriculum.

It is common knowledge in any circuits course that you can change the MAGNITUDE from line to phase with a sqrt(3) operator but the VECTOR transformation requires a phase shift operator as well.

My cohorts do the same because that is the way it is done by the collective electrical community all around the world and has been done for more than a century.

I don't get why you struggle with this. Your knowledge usually seems much better. The best I can figure is that something is being lost in translation.
 
Phil Corso said:
Mivey...

I had a change of heart! So, i'll reveal what your error is! Affixing the-30 deg is done ONLY when using the "1-ph Equivalent For Balanced 3-ph Circuits"! It is not done for a 3-phase D-D problem!

Phil
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What?!

It does not matter if it is delta-delta or wye-delta. The line current is still the line current. We were not given delta currents at the source.

We do not know if the source is wye or delta and it does not matter because the solution is the same.

as for the problem answer: We know V_AB but want V_ab. With the IaA current and line impedance we can calculate VaA. With the rotation information and knowing the circuit is balanced we rotate IaA to get IbB and using the same line impedance (balanced circuit) we solve for V_Bb.

Now we have V_aA, V_AB, and V_Bb so we solve for V_ab using V_ab = V_aA + V_AB + V_Bb.

Basic circuit analysis.
 
Phil Corso said:
Mivey...

I have never shirked from doing the math! And, NO, I don't agree with your approach! I did the math earlier! You don't arbitrarily affix a -30 deg! I said it before, but you won't listen! The correct procedure is to simply transpose the I-Line angle to I-ph! Your cohorts make the same mathematical error!

I'm sorry to say it, but we will never agree! But, know I now why!
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You are at least correct that WE will never agree with your incorrect answer.

It's fairly simple math to calculate the load currents IAB, IBC and ICA from your proposed load impedance of 291 + j 106; given VAB=12500<0 V and given an ABC sequence. Yet, you won't do the math.

You have shirked from doing the math through this entire thread. For the simple reason that doing the math shows that your solution is wrong.
 
david luchini said:
You are at least correct that WE will never agree with your incorrect answer.

It's fairly simple math to calculate the load currents IAB, IBC and ICA from your proposed load impedance of 291 + j 106; given VAB=12500<0 V and given an ABC sequence. Yet, you won't do the math.

You have shirked from doing the math through this entire thread. For the simple reason that doing the math shows that your solution is wrong.
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I'll do it for him.

VAB = 12500@0d (given)
VBC = 12500@-120d (from given ABC rotation)
VCA = 12500@-240d (from given ABC rotation)


ZD = ZAB = ZBC = ZCA
ZD_Phil_Corso = (290.642 + j105.785)
ZD_Correct = (304.596 - j53.708)


IaA = 70@-20d (given) = IAB - ICA

IAB = VAB / ZD
ICA = VCA / ZD

IaA_Phil_Corso = IAB - ICA
= VAB / ZD - VCA / ZD
= 12500@0d / (290.642 + j105.785) - 12500@-240d / (290.642 + j105.785)
= 40.4145@-20d - 40.4145@100d
= 70.00@-50d

IaA_Correct = IAB - ICA
= VAB / ZD - VCA / ZD
= 12500@0d / (304.596 - j53.708) - 12500@-240d / (304.596 - j53.708)
= 40.4145@10d - 40.4145@130d
= 70.00@-20d

Pretty simple to prove that Phil's delta impedance is wrong.
 
Phil Corso said:
Mivey...

I did the math earlier! You don't arbitrarily affix a -30 deg! I said it before, but you won't listen! The correct procedure is to simply transpose the I-Line angle to I-ph! Your cohorts make the same mathematical error!

I'm sorry to say it, but we will never agree! But, know I now why!
Click to expand...

Let's just sum the currents at the upper node of delta. The given line current 70 <-20 is coming in and I- AB and I -AC are leaving. I assume (or at least hope we can all agree with KCL!!)

According to Mivey post #57, I-AB + I -AC = (40.4 < 10) + ( 40.4 <-50) = punch out on calculator = 70 <-20 which is line current in... whew!

According to Phil I think I-AB is in phase with line current in so = 40.4<-20. Now if we calculate current down right side of delta it is line current in minus I-AB which gives me (punch in calculator) ... whoa... 29.6 < -20 !!! The magnitude of current isn't correct for balanced load... something must be wrong here
 
dkarst said:
Let's just sum the currents at the upper node of delta. The given line current 70 <-20 is coming in and I- AB and I -AC are leaving. I assume (or at least hope we can all agree with KCL!!)

According to Mivey post #57, I-AB + I -AC = (40.4 < 10) + ( 40.4 <-50) = punch out on calculator = 70 <-20 which is line current in... whew!

According to Phil I think I-AB is in phase with line current in so = 40.4<-20. Now if we calculate current down right side of delta it is line current in minus I-AB which gives me (punch in calculator) ... whoa... 29.6 < -20 !!! The magnitude of current isn't correct for balanced load... something must be wrong here
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Another good proof that Phil's impedance is wrong.

dkarst said:
The magnitude of current isn't correct for balanced load... something must be wrong here
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The something wrong is:

mivey said:
[to Phil]... the E part of your GiFiES was wrong...
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