Complex Power Question, when to use current conjugate

[stcalle]

I was hoping someone could help me out with explaining when you use these two particular forumulas.

Solving for 3 phase power:
1) S = sqrt 3 * V line to line * I line to line
2) S = 3 * V line to neutral * I' line to neutral
where I' is the conjugate of I


I've seen equations #1 and #2 but I have not seen the equation
3) S = sqrt 3 * V line to line * I' line to line

Does equation #3 exist or is it incorrect?

Thank in advance.

 

[StephenSDH]

S = sqrt 3 * V line to line * I line to line
S = sqrt 3 * V line to line * I' line to line

These Equations can only both be true if I == I'.

KVA(3phase)=V(p-p)*I*1.73=(V(p-n)*I)3
KVA(1phase)=V(p-n)*I

 

[charlie b]

The notions of "line to line current" and "line to neutral current" are meaningless. Are you sure you are using the correct terminology?

 

[stcalle]

I labeled current just to be consistent with the equations. The issue that I'm trying to figure out is when you're supposed to use the conjugate of the current to calculate apparent power.

S = sqrt 3 * V line to line * I line to line
S = sqrt 3 * V line to line * I' line to line

These Equations can only both be true if I == I'.

I understand what you're saying, but I do not get why you have this variation. The thing is those two equations would yield the same magnitude for apparent power but a different phase angle in which case you would get different real and reactive powers in each equation, so there must be a particular case when to use each equation.

KVA(3phase)=V(p-p)*I*1.73=(V(p-n)*I)3
KVA(1phase)=V(p-n)*I

These are the equations I am familiar with so following that, where do the equations with the current conjugate come into play?

S (1phase) = V (line to neutral) * I
S (1phase) = V (line to neutral) * I', where I' is the conjugate of the current

I am using the equation you noted for apparent power but I have also seen the same equation but using the conjugate of the current. When would you use one equation over the other?

 

[charlie b]

I looked back into my old (OLD) textbook (Stevenson, ?Elements of Power System Analysis?). After dusting off my memory banks, what I understand is that you don?t use the conjugate notation unless the entire expression is handled using phasors. Here is what Stevenson has to say:

If the voltage across and the current into a certain load of part of a circuit are expressed by V = |V| <a, and I = |I|<b, the product of voltage times the conjugate of the current is VI* = |V| |I| <a-b.

Stevenson goes on to say that VI* is the complex power, S.

My handwritten notes on that page remind me that the use of the conjugate was necessitated by the selected convention that ?positive Q exists for inductive loads.? That was an arbitrary choice.

I think that once you bring the square root of three into the mix, you are no longer dealing with phasor notation. I don?t think you should be seeing the sqrt 3 and the I? in the same expression.

 

[Mayimbe]

the only way that a number can be conjugate, is because is a complex number.

Solving for 3 phase power:
1) S = sqrt 3 * V line to line * I line to line (NO, I line or line current)
2) S = 3 * V line to neutral * I' line to neutral (NO, I line)
where I' is the conjugate of I

in #1 all numbers are real or not complex. Or even more, they are the absolut value of a complex number.

in #2 all the numbers are complex. for example I = 120+j*56.

the two equatons are right, but be carefull what kind of numbers are you dilling with complex or real. The conjugation is used for keeping the complex number S (aparent power) in the first and fourth quadrant of the phasorial diagram.

 

[charlie b]

S (1phase) = V (line to neutral) * I

I think it would help if you dropped the expression ?line-to-neutral current,? and just said ?current.? Since voltage is a measure of the difference in potential between two points, it is proper to distinguish such things as ?voltage line-to-line? from ?voltage line-to-neutral.? Current is not a measure of a difference between two points. Current flows in a line, and that is all there is.

Do you have a reference source that uses this ?line to neutral? expression for current? If so, I think it is not a good reference source.

 

[stcalle]

ok, thanks guys. From reading the responses I think I have kind of figured it out. I will explain how I understand things to work and I'll let you guys let me know if I'm on the right track or not.

So when talking about apparent power, you can either work with magnitudes or complex numbers.

when working with magnitudes you have the equations:
1) |S| (3phase) = sqrt 3 * |V| (line to line) * |I|
2) |S| (3phase) = 3 * |V| (line to neutral) * |I|
3) |S| (1phase) = |V| * |I|
you are able to work like this because if you are given the power factor, you can use that to solve for angle of the apparent power and thus converting from polar to rectangular solve for the real and reactive powers.

when working with phasors you have the equations:
4) S (3phase) = sqrt 3 * V (line to line) * I'
5) S (3phase) = 3 * V (line to neutral) * I'
6) S (1phase) = V * I'
where S, V, I are phasors and will have a corresponding angle.
where I' is the conjugate of the current.

you can work like this when you may not have the power factor given but have the phase angles for the current and voltage. You can then solve for the phase angle of the apparent power and this calculate the power factor.

 

[charlie b]

May I ask why you persist in using the phrase "line-to-line current"? May I ask what that phrase means to you? It means nothing to me. Nor is there any difference, in my mind, between "line-to-line current" and "line-to-neutral current." What is the difference, in your mind?

 

[Mayimbe]

"line to line current" = "line to neutral current"/ sqrt(3)

line to line current is the current that circulates in the delta of a transformer for example.

and line to neutral current is the one that flows in the transmission lines

I line to line < I line to neutral

the equation 4 is wrong.

 

[stcalle]

"line to line current" = "line to neutral current"/ sqrt(3)

line to line current is the current that circulates in the delta of a transformer for example.

and line to neutral current is the one that flows in the transmission lines

I line to line < I line to neutral

the equation 4 is wrong.

Why is equation 4 wrong?

Also, in my latest post #8 I have not used the line to line or line to neutral current naming.

 

[steve66]

If I remember this correctly, a negative phase angle means the wave is advanced in time.

I at -45 degrees leads I at 0 degrees.

But for power factor, a negative number means we have a lagging power factor. That means the current lags the voltage. So for a lagging power factor, the current would actually have a positive phase angle.

That's why we need to use the conjugate of the current. It gives the power factor the right sign.

In most simple problems, it may be easier to just keep track of if the power factor is leading or lagging, and ignore the conjugate business.

 

[Mayimbe]

Why is equation 4 wrong?

its right. sorry. my mistake

 

[charlie b]

"line to line current" = "line to neutral current"/ sqrt(3)

Not true, since neither phrase has any meaning. I?ll explain shortly.

line to line current is the current that circulates in the delta of a transformer for example.

That is properly called ?phase current.? There is phase current in the windings of a WYE transformer as well. The difference between the WYE and the delta is that (1) The phase current and line current are equal in the WYE, and (2) The line current in a delta exceeds the phase current by a factor of 1.732.

. . . and line to neutral current is the one that flows in the transmission lines

That is properly called just the ?line current.?

I line to line < I line to neutral

The word ?to? does not belong in this conversation. It is not a ?line to? anything. It is just I-line, or ?line current,? or simply ?current.? The current in any given wire does not have anything to do with that wire?s position with respect to any other wire.

 

[mivey]

Not true, since neither phrase has any meaning. I?ll explain shortly.
That is properly called ?phase current.? There is phase current in the windings of a WYE transformer as well. The difference between the WYE and the delta is that (1) The phase current and line current are equal in the WYE, and (2) The line current in a delta exceeds the phase current by a factor of 1.732.
That is properly called just the ?line current.? The word ?to? does not belong in this conversation. It is not a ?line to? anything. It is just I-line, or ?line current,? or simply ?current.? The current in any given wire does not have anything to do with that wire?s position with respect to any other wire.

Yeah, yeah, yeah. Coming from the Kelvin-Watt-Henry guy.:grin:

What was the other pet peeve? Was it you or Jim that took issue with "line" vs "phase"? I'm bad about these things, although I think I have quit using line-to-line current, etc.

 

[Besoeker]

May I ask why you persist in using the phrase "line-to-line current"? May I ask what that phrase means to you? It means nothing to me. Nor is there any difference, in my mind, between "line-to-line current" and "line-to-neutral current." What is the difference, in your mind?

Suppose you have a transformer with a star (wye) four-wire secondary - a common arrangement for the LV service in many countries.
A load connected between L1 and N would take line to neutral current. A load connected between L1 and L2 would take line to line current.

 

[dkarst]

Charlie B has cleaned up the loose terminology but we are still a bit unclear if earlier equations # 4 and #5 are correct, and if they are correct are they very useful.

From Matsch (Electromagnetic and Electromechanical Mach. ~ same vintage as Charlie's version of Prof. Stevenson's text):

S = 3 x Vph x Iph*, where he has clearly specified PHASE quantities and Iph* is the complex conjugate of the phase current Iph and used 3 vs. sqrt(3). The bold quantities have been identified as phasors although I'll use a little looser terminlogy from here on. So equation #5 stated earlier is correct. An example may illustrate:

Assume balanced Y-connected load of 3+j4 per phase and source of 120 /_ 0 phase a voltage (Van = 120 /_ 0).

Using typical method P= sqrt(3) x Vline x Iline x cos(theta) = sqrt(3) x 208 x 24 x0.6 = 5184 W

And Q (using sin-theta) = 6912 VAR
Giving a power triangle of with an apparent power of 8640 VA @ 53 degrees

Now let's try equation #5 (using /_ to indicate polar angle and remembering to change the sign of the angle to get Iph*)
S = 3 x Vph x Iph*, S = 3x 120/_ 0 x 24/_ +53 = 8640 /_ +53

and so in one step we have the solution.

Now what about pesky equation #4 with the sqrt (3) term ?

S = sqrt(3) x Vline x Iline*

Now the terminology gets a bit messy. If we define Vab as our line voltage it would be 208 /_ +30 for std. phase rotation.
But what about a corresponding line current, let alone its conjugate? In this case if we are using the line voltage a-b, the current a-b would flow through two legs of the delta and is clearly ambiguous. We could give up but another way is to transform the load to a delta load using Y-delta transformation. The load impedance would then be Z delta = 3 Z wye or = 9+j12 per phase. Now we can solve for the phase currrent in load leg ab, (~ 14 amps @ -23 degrees). Recognize that the line current for the delta load is sqrt(3) times that load current, keep track of our angles (don't forget to flip the sign of the angle to get the conjugate of line current), and wa-la...

S = sqrt(3) x 208 /_ +30 x 24 /_ +23 = 8648 /_ +53 degrees which is same as before.

I guess after all this, I think we have proven the math works but you need to be very careful with terminology. I do think that for a balanced system,

S = 3 x Vp x Iph* is a useful relationship for balanced systems if one remembers it is a per phase basis.

 

[charlie b]

A load connected between L1 and N would take line to neutral current.

No, it would have current flowing from the source on a "line" and returning on the neutral. But the current is not "line to neutral current."

A load connected between L1 and L2 would take line to line current.

And this would have current flowing from the source on one line, and returning on another line. This is not "line to line current."

As has been mentioned, I am only making a point about terminology. So let's not get too far off the original topic. But to make my point clearer, consider the task of measuring voltage at a 120/208V panel. You take one voltage probe and touch it to one of the main lugs. What do you expect the voltage reading to be? Well, you don't get a reading, because the second voltage probe has not been used. If you touch the second probe to another lug, you will get 208 volts, and if you touch it to the neutral you will get 120 volts. It is in this sense that we speak of "line to line voltage" and of "line to neutral voltage." You need two probes, and you need to know where they are touching. But you measure current by wrapping a CT around a single wire. Not two wires, one wire. What you get is the current flowing through that wire. That is why there is no concept of "line to line current" or "line to neutral current." When you measure current, the measurement does not tell you anything about where that current is going, nor on what wire it will return to the source.

 

[Mayimbe]

Not true, since neither phrase has any meaning. I?ll explain shortly.

Its true for me, at least it was true the last time I check.

The difference between the WYE and the delta is that (1) The phase current and line current are equal in the WYE, and (2) The line current in a delta exceeds the phase current by a factor of 1.732.

Not true for me. see the figure. kirchoffs law matter. currents in pu.

 

[dkarst]

I mis-typed in my earlier post, hopefully everyone read what intended to say..

Now the terminology gets a bit messy. If we define Vab as our line voltage it would be 208 /_ +30 for std. phase rotation.
But what about a corresponding line current, let alone its conjugate? In this case if we are using the line voltage a-b, the current a-b would flow through two legs of the Wye load NOT delta as I stated) and is clearly ambiguous... sorry for confusion.