3-phase complex power calculations

[Phil Corso]

DL...

Dividing a source who's angle is zero, by an impedance R+jX, who's angle is positive, produces a current who's angle is negative, then the impedance must part inductive, hence current is lagging, and 'Load' pf is positive! Q.E.D. !

I agree... you should go back to the books!

Phil

 

[david luchini]

DL...

Dividing a source who's angle is zero, by an impedance R+jX, who's angle is positive, produces a current who's angle is negative, then the impedance must part inductive, hence current is lagging, and 'Load' pf is positive! Q.E.D. !

Where in the original question do you see that the Load Impedance has a positive angle? The question says no such thing.

I agree... you should go back to the books!

Phil

I didn't say I needed to go back to the books, I said its been years since I've done this stuff.

I'm sure if you go back to your books, and solve the problem completely, I suspect you will find that the load current IAB leads the load voltage VAB, and that the load impedance is R-jX, who's angle is negative.
In other words, the load is capacitive.

 

[Phil Corso]

David,.

Using the procedure I presented earlier resulted in a Delta-impedance of 290+j106, thus inductive... not capacitive! What value did you calculate?

Phil

 

[david luchini]

David,.

Using the procedure I presented earlier resulted in a Delta-impedance of 290+j106, thus inductive... not capacitive! What value did you calculate?

Phil

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.

 

[dkarst]

David, I used a delta-wye transformation and agree with your solution ~ 306 - j54 ohms

 

[david luchini]

Whenever a balanced delta-load is to be analyzed, then. most of the time, the solution is easier if you convert to a balanced 3ph, 4-wire, wye-load circuit ! Now, the circuit can be reduced to single-phase: having one source-voltage, Van; one load-voltage, VAn = VAB/Sqrt(3)! and one line impedance, ZaA !

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.

 

[Phil Corso]

David…

You are gravely mistaken! You shouldn't arbitrarily(?) assign the -30º angle to VAn! Transformation of a balanced D-D to a balanced Y-Y circuit, does not alter the associated angle! The given VAB voltage VAB is 12.5-kV at 0º! Thus VAn = VAB/Sqrt(3) @ 0deg !

Phil

 

[david luchini]

David…

You are gravely mistaken! You shouldn't arbitrarily(?) assign the -30º angle to VAn! Transformation of a balanced D-D to a balanced Y-Y circuit, does not alter the associated angle! The given VAB voltage VAB is 12.5-kV at 0º! Thus VAn = VAB/Sqrt(3) @ 0deg !

Phil

And that's why you got the WRONG answer!

Would you care to explain why with a load impedance of 290+j106, IaA=IAB-ICA=70<-50deg A instead of the 70<-20deg A that was given in the original question?

 

[Phil Corso]

David...

It's not Load-Voltage, VAn, that needs an angle adjustment ! It's GIVEN ! Source Voltage, Van, should be adjusted! That's why the problem's answers IGNORED angles !

Further more if the Z Load was capacitive, as you strongly believe, then, source voltage would have gone DOWN !
Do you need a Jiffies solution ?

Phil

 

[dkarst]

Here's my solution... literally the back of an envelope. Hmm, guess I should have tried to shrink?

 

[Phil Corso]

DKarst...

Some kudos... but, since VAn's angle is zero, then Vab's must be +30 , because Sequence is A-B-C!

Phil

 

[mivey]

DKarst...

Some kudos... but, since VAn's angle is zero, then Vab's must be +30 , because Sequence is A-B-C!

Phil

And it follows that if Vab is 0d then Van is -30d. This is what he used so there is no "but".

 

[dkarst]

Phil, I guess I have to move on. I think three people have shown you are incorrect and I don't know how to simplify further.

If in a balanced 3-phase abc sequence system, VAB is given at an angle of 0, then by definition, the voltage VAN lags it by 30 degrees and is sqrt(3) less in magnitude. Therefore in this case, VAB at load is given as 12.5kV /_ 0 so you know angle of VAN is -30, not 0 as you are continually incorrectly stating.

Once you have that, the current is given and since you have the voltage across load, the impedance is straightforward.

I guess this is it for me, time to move on... best regards

 

[Phil Corso]

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

 

[mivey]

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

Your proof is missing an angle change. On road right now so typing is a pain but can type it in later if no one else gets to it first.

add: and Happy Thanksgiving!

 

[Phil Corso]

Mivey...

I purposely did not want to solve for Source ph-ph... Instead, I only wanted to prove load was NOT capacitive, by determinng delta-load parameters !

There was an error though: ZΔ should have been = 1,000*VΔ/IΔ

BTW, what was your value for ZΔ ?

Phil

 

[david luchini]

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

Phil, as Mivey pointed out, here is where you went wrong (again).


Idelta = (IL / sqrt(3)) * (1<30)

So IAB = 40.42<10

With a VAB of 12500<0, you can see current leads voltage. A leading power factor is a capacitive load.

QED.

 

[mivey]

Mivey...

I purposely did not want to solve for Source ph-ph... Instead, I only wanted to prove load was NOT capacitive, by determinng delta-load parameters !

There was an error though: ZΔ should have been = 1,000*VΔ/IΔ

BTW, what was your value for ZΔ ?

Phil

You forgot to shift current angle from line to phase in the delta.

add: you get the same result either way (using line or phase solution).

 

[mivey]

Phil, as Mivey pointed out, here is where you went wrong (again).


Idelta = (IL / sqrt(3)) * (1<30)

So Idelta = 40.42<10

With a Vdelta of 12500<0, you can see current leads voltage. A leading power factor is a capacitive load.

QED.

Precisely!

 

[Phil Corso]

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