# phase power LL vs LN
# phase power LL vs LN
(OP)
I will begin with some facts and then ask the question.
Phase to Phase Voltages Lead Phase to Neutral voltages by 30 degrees.
Three Phase Apparent Power Equations:
If using Phase to Phase voltages:
S=1.732*V*I'
If using Phase to Neutral Voltages:
S=3*V*I'
Now when solving for the power and you know lets say the Phase to neutral voltage and its angle, why don't you multiply by 1.732 and add 30 degrees to it and stick that value into the Phase to Phase 3 phase apparent power formula?
For example:
V_L-n=120 @ an angle of 31 degrees
I*= 22 @ an angle of 20 degrees
V_L-L= (120*1.732) @ an angle of 61 degrees
Apparent Power Calc:
Using Phase to Neutral:
S=3*(120@31)*(22@20)= 7920@51degrees
Using Phase to Phase:
S=1.732*(208@61)*(22@20)= 7920@81degrees
As you can see the two apparent power calculations do not agree. The 7920@51degrees is in fact the correct answer. My question is, if Phase to Phase voltages lead Phase to neutral voltages, why don't you use this fact when solving three phase power with the Phase to Phase three phase formula? I mean you are supposed to put in the Phase to Phase voltage magnitude, why not the angle as well?
Phase to Phase Voltages Lead Phase to Neutral voltages by 30 degrees.
Three Phase Apparent Power Equations:
If using Phase to Phase voltages:
S=1.732*V*I'
If using Phase to Neutral Voltages:
S=3*V*I'
Now when solving for the power and you know lets say the Phase to neutral voltage and its angle, why don't you multiply by 1.732 and add 30 degrees to it and stick that value into the Phase to Phase 3 phase apparent power formula?
For example:
V_L-n=120 @ an angle of 31 degrees
I*= 22 @ an angle of 20 degrees
V_L-L= (120*1.732) @ an angle of 61 degrees
Apparent Power Calc:
Using Phase to Neutral:
S=3*(120@31)*(22@20)= 7920@51degrees
Using Phase to Phase:
S=1.732*(208@61)*(22@20)= 7920@81degrees
As you can see the two apparent power calculations do not agree. The 7920@51degrees is in fact the correct answer. My question is, if Phase to Phase voltages lead Phase to neutral voltages, why don't you use this fact when solving three phase power with the Phase to Phase three phase formula? I mean you are supposed to put in the Phase to Phase voltage magnitude, why not the angle as well?






RE: # phase power LL vs LN
Use phase to neutral when considering phase to neutral loads.
I don't know where you are getting your voltage angles. The three phase voltage angles are 0 deg. 120 deg and 240 deg. The current may lead or lag these angles.
The voltage angle between phase to phase and phase to neutral does not affect apparent power calculations.
Solve your line to neutral loads for kW and KVAR
Solve your line to line loads for kW and KVAR
Add kWs
Add KVARs
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: # phase power LL vs LN
A wye source connected to a wye load should be able to be solved per phase and use the Line to neutral quantities. But the Line to Line quantities should yield the same results. The values I used were just representative of the situation I want to portray.
A realistic version of this problem can be the following. A 3-Phase wye connected load is consuming 300KVA @ 0.7 PF. The voltage (L-L) at the load terminals 1500V. The transmission lines have a per phase impedance of 10+j5 ohms. What is the magnitude and angle of the voltage at the source.
To solve this you can easily find the current that the load is consuming, convert to a per phase circuit and get the voltage drop in the transmission lines. From there you add the voltage to the 1500V and you get the magnitude and angle of the voltage the source. At this point you have two choices:
1. Solve the problem using S=1.732*V_LL*I_LL'
2. Solve the problem using S=3*V*I'
Up until this point you would have been solving everything in per phase and would have the voltage in per phase, and can directly go with option 2, but if you wanted to you should be able to solve it with option 1 by converting the L-N voltage to a L-L voltage and solving the equation. The results will differ by 30 degrees when they shouldn't.
This is where the issue arises.
RE: # phase power LL vs LN
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: # phase power LL vs LN
By accepting those two facts and as they can be easily seen by simple subtractions, you should be able to use either three phase apparent power formula. My question is, what is the missing link that I am not seeing? What little caveat makes the two 3 phase power formulas equal?
RE: # phase power LL vs LN
1) S=Vp x I ¼∅
RE: # phase power LL vs LN
Any how;
S=VLN x IL ∠∅
The 3-phase apparent power is;
S3P=3 x VLN x IL ∠∅
As you stated we can substitute "3 x VLN" with "√3 x VLL" but we're only interested in the magnitude,
so the correct way to write it is:
S3P=√3 x |VLL| x |IL| ∠∅
Even if we're using line to line voltage, "∅" is still the angle between line to neutral voltage and line current.
Inside a delta connected load things are a bit different.
RE: # phase power LL vs LN
RE: # phase power LL vs LN
Line to line is taken from a to b.
Line to neutral voltage is taken from a to n.
However Van is at the same angle as Vab on the delta primary.
The 30 degree phase shift between Van (secondary) at the same angle as Vab on the delta primary and Vab on the secondary should be apparent.
Said another way:
Van secondary is at the same angle as Vab primary.
Vab secondary is rotated 30 degrees from Van secondary.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: # phase power LL vs LN
It can be easily said to only use Line to neutral phase angle, and I can accept it as it is just the way it is but it would be reassuring to see the theory behind it since it seems like it could be so easy to use the wrong angle for the voltage. Anyway thanks for the replies guys. I'll keep pondering it, and if I ever come across a proof in any direction, I'll post it back here.
RE: # phase power LL vs LN
You wrote that VLL = √3 x VLN. Well, there's you fundamental error.
The correct way of putting it should be:
|VLL| = √3 x |VLN|.
See what I'm aming at?
Example:
VLN= 230V∠0º
VLL= 400V∠30º
You stated that "VLL = √3 x VLN":
.BS
RE: # phase power LL vs LN
RE: # phase power LL vs LN
RE: # phase power LL vs LN
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: # phase power LL vs LN
The complex power in an impedance is equal to:
voltage over the impedance multiplied by the conjugate of the current through the impedance. Your equation for the power using phase-phase voltage is not correct. The problem starts here.
Power using the phase-neutral voltage and current:
S = 3 times [(120V @ 31 degrees) times (22A @ 20 degrees)*] = 7920 @ 11 degrees
Power using the phase-phase voltage and current:
S = 3 times [(207.85 @ 61 degrees) times (12.7 A @ 50 degrees)*] = 7920 @ 11 degrees.
In addition, the phase of the power is not correct in your calculations. Use the difference of the phases of voltage and current, not their sum.
RE: # phase power LL vs LN
RE: # phase power LL vs LN
RE: # phase power LL vs LN
This is why you have to use the conjugate of the current, but you always add the angles.
(22A @ 20 degrees)*=22A @-20 degrees
RE: # phase power LL vs LN
RE: # phase power LL vs LN
Start at the simplebeginning: At it's core, S = V I* applies to a single-phase circuit.
Then we have to figure out how to apply it to other situations. Application to per-phase analyis is straightforward.
As an aside, there is also a network theorem where we can sum power crossing a boundary S = Sum { V I*} if all the V are referenced to the same voltage.
But how do you propose to extend it to phase to phase measurements? What is phase to phase current? Phase to phase current is not defined to my knowledge. If you assume balanced scenario then you can use per-phase analysis to develop suitable equation which involves VLL. But the Phi is still the same Phi as we used in per-phase analysis. See Shooter's first post.
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(2B)+(2B)' ?
RE: # phase power LL vs LN
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(2B)+(2B)' ?
RE: # phase power LL vs LN
Start with the per-phase V-I pair (VLN, Iphase)
or
Start with the phase-to-phase V-I pair (VLL, Idelta).
The angle between VLN and Iphase would turn out to be the same as the angle between VLL and Idelta (call it Phi in either case).
I don't think I've added anything new here. just summarizing for myself.
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(2B)+(2B)' ?
RE: # phase power LL vs LN
RE: # phase power LL vs LN