# phase power LL vs LN

[LowSpark]

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?

 

[waross]

Use phase to phase when considering phase to phase loads.
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

 

[LowSpark]

Suppose you have the following Line voltages 120@120 degrees, and 120@0 degrees, To solve for Line to Line voltage you do 120@120 - 120@0. You will see that the Line to Line voltage is 208@30 degrees. So the line to line voltage between the two line to neutral voltages just mentioned is 208@30 degrees leading of the first vector.

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.

 

[waross]

What course is this for?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LowSpark]

This isn't for a course. I came across this as I am reviewing the theory/problems for my licensing exam. All review books will tell you in a wye connected system the LL voltage is 1.732 times the L to N voltage and is 30 degrees leading, in a delta connected system the LL current is 1.732 times the L - N current and is lagging by 30 degrees.

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?

 

[ShooterBoter]

Ok, here it goes:

1) S=V[sub]p[/sub] x I ¼∅

 

[ShooterBoter]

Hmm, sorry for my previous reply, thick fingers on a small keyboard is what caused that...
Any how;

S=V[sub]LN[/sub] x I[sub]L[/sub] ∠∅

Where

V[sub]LN[/sub]=Line to neutral voltage​

I[sub]L[/sub]=Line current​

∅=Angle between line to neutral voltage and line current​

The 3-phase apparent power is;
S[sub]3P[/sub]=3 x V[sub]LN[/sub] x I[sub]L[/sub] ∠∅

As you stated we can substitute "3 x V[sub]LN[/sub]" with "√3 x V[sub]LL[/sub]" but we're only interested in the magnitude,
so the correct way to write it is:
S[sub]3P[/sub]=√3 x |V[sub]LL[/sub]| x |I[sub]L[/sub]| ∠∅

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.

 

[LowSpark]

Interesting reply. But I don't see why it is only the Line to Neutral Voltage angle that we are concerned about.

 

[waross]

Draw a wye. Label it a,b,c,n
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

 

[LowSpark]

I think you are referring to a delta wye transformer. I was just referring to a transmission system that has a wye source and the load is connected wye.

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.

 

[ShooterBoter]

You wrote that V[sub]LL[/sub] = √3 x V[sub]LN[/sub]. Well, there's you fundamental error.
The correct way of putting it should be:
|V[sub]LL[/sub]| = √3 x |V[sub]LN[/sub]|.

See what I'm aming at?

Example:
V[sub]LN[/sub]= 230V∠0º
V[sub]LL[/sub]= 400V∠30º

You stated that "V[sub]LL[/sub] = √3 x V[sub]LN[/sub]":

"400V∠30º" = √3 x 230V∠0º, as you can see this is not correct (because the angle is not affected by scalar multiplication),​

but this is: "√3 x |V[sub]LN[/sub]| = |V[sub]LL[/sub]|" ⇒ "√3 x |230V∠0º|=|400V∠30º|"​

.BS

 

[LowSpark]

Yes, at the end of the day the Line to Line voltage is square root three time the magnitude and 30 degrees leading. You will see this from subtracting the two vectors. But we can talk about semantics but what is important is the proof as to why you need the Line to Neutral Phase angle.

 

[ShooterBoter]

Ok, I gave it a try, anyone else?

 

[waross]

There is no 30 degree phase shift between the delta primary and the wye secondary of a transformer. Think of a bank of three single phase transformers. The secondary of each transformer is not phase shifted with respect to the primary winding. When the three individual transforer are connected in delta wye, there is no magical phase shift. The line to neutral voltages are not shifted from the original delta angles. The secondary line to line voltage of a wye secondary is derived by the combination of two windings displaced by 120 degrees. The result of vector addition of two equal vectors displaced by 120 degrees is 1.73 rather than 2 and an angle of 30 degrees from either of the original vectors.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ijl]

As waross wrote, either use phase-phase or phase-neutral variables, but not a mixture.

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.

 

[LowSpark]

Consider the wye case. In a wye ll current is the same as phase current. I found the solution everyone by the way. The solution is what shooter said earlier, you need to only use phase to neutral angle as is - just because, I found it in my text book in undergrad. They specially said "take note that in the LL formula you need to still use line to neutral phase angles."

 

[LowSpark]

Here is the picture from the book:

 

[ShooterBoter]

"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."

This is why you have to use the conjugate of the current, but you always add the angles.

(22A @ 20 degrees)*=22A @-20 degrees

 

[LowSpark]

You guys realize that only in delta systems does the ll current lag the LN by 30 degrees. In that calc you are throwing 30 degrees on the voltage and on the current. Either you are working with a wye system and the voltage is affected, or you are working with a delta system and your current is affected.

 

[electricpete]

I think Shooter's first post was spot on. I didn't study the rest.

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)' ?