Current Calc: 2 Phase 3 Wire

[Smart $]

Nice.:roll: For some reason, your version used 30 degrees for the imaginary portion. I'm not sure where you got that from, but it is not the way I would have done it so I posted the way I would have done it.

OK, I now see my error, where I used 30?... so why didn't you just say so?!!! I still say you like to read your own typing... especially calcs.

As to my mistake, I was thinking ahead to I23 when I stuck it in there. But does it really matter in unity pf cases, as θ2? will always be the negated value of θ1? and add to 0?. Now if you want to show the difference in non-unity pf cases, by all means have at it... but please, don't just show the math. The math takes care of itself. What I'm asking you to present is premise and purpose along with the substantiating math.

I have tried but I have not been able to get you to understand. If you feel like disbelieving everything I say just to be argumentative, at least read some of the other posts where this was explained.I started out with a conjugate because I had a conjugate in part of the equation to start with. The end result was not a conjugate. I fail to see why this is so hard for you to understand.If you mean I don't understand you, you are correct. You are posting things without them being substantiated. Some of them are so strange, that I would say you either don't understand what you are doing or are purposely posting errors just to argue.

If it is the former, I will be glad to help you learn something. If you don't want to listen to me, there are plenty here that could teach you as well and I think you would be better for it.

If it is the latter, I'm really not interested in the entertainment value at this point and it really brings no value to this forum.

What you don't understand is that you are not teaching me anything, and too blind to jump over this trivial (confirmed in your reponse to skeshesh ) and meaningless stuff and go straight to the heart of it.

Let me sum it up in calc form:

VA/V@θ=I@-θ*
I@-θ*≈I@θ
VA/V@θ=I@θ​

...where "≈" is used for the equal and congruent symbol (three-line equal symbol)

 

[rattus]

Too much ado:

Too much ado:

So I get back from a little Friday night trip and decide to see what's up with the forum before heading out to a bar with buddies and what do I see? This thread's at it again! lol

I really have been puzzled by this discussion from the very start, this should be basic math to engineers. I suggest we start a side pot bet on how long this thread will run

Yup! A pencil and the back of an envelope should do. It is easy enough to find the load current phasors; then apply Kirchoff's Current Law to node 2. The sine and cosine of 60 degrees should be burned into our minds so all we need is a 4-function calculator if we are too lazy to do it by hand.

 

[mivey]

Possibly, but we have a good number of engineers on this forum; the fact that you chose to weigh in on this issue for so long does allow for the possibility that you enjoy the argument quite a bit as well.

I would say it was spawned more in defense than a penchant for arguing. Sometimes I get drawn into an argument when I don't feel like it. Sometimes I get tired of them and want to quit, and sometimes I continue if there is a possibility of extracting some value. Sometimes that value is educational, sometimes it is entertainment value, sometimes it might just be to scratch an itch. But whatever my reasons or whatever the cause, it is absolutely true that I am posting strictly because I choose to.

When the hog comes up and nudges you and wants to play, you have to decide if you want to wrestle with the hog because you are bound to get dirty. Sometimes I do, sometimes I don't.

I seem to recall one, namely you, going on about it, for more than 6 pages. I only pointed out that it should be elementary for all engineers, but that does not mean others (electricians included) cannot understand such this concept.

I am not taking issue with the concepts. I took issue with someone claiming I was using improper math and also claiming those common concepts were wrong or inferior somehow.

I am well aware that you do not have to be an engineer to understand anything, and have said the same many times in the past. I have said that there are many non-engineers that understand some concepts better than many engineers.

The engineer just has a piece of paper from his peer group that says he completed a set of courses and it is supposed to be an indicator of knowledge in certain areas. That is not always true, and neither does it mean that only those with the piece of paper possess all knowledge. I have learned a lot more about the engineering world after I graduated than I did in school. School was just laying some groundwork.

 

[mivey]

OK, I now see my error, where I used 30?... so why didn't you just say so?!!! I still say you like to read your own typing... especially calcs.

Oh if my fingers only had mirrors so they could watch themselves type. Maybe I could get some grill-work for my fingers.:grin:

As to my mistake, I was thinking ahead to I23 when I stuck it in there. But does it really matter in unity pf cases, as θ2? will always be the negated value of θ1? and add to 0?. Now if you want to show the difference in non-unity pf cases, by all means have at it... but please, don't just show the math. The math takes care of itself. What I'm asking you to present is premise and purpose along with the substantiating math.


What you don't understand is that you are not teaching me anything, and too blind to jump over this trivial (confirmed in your reponse to skeshesh ) and meaningless stuff and go straight to the heart of it.

Let me sum it up in calc form:

VA/V@θ=I@-θ*
I@-θ*≈I@θ
VA/V@θ=I@θ
​

...where "≈" is used for the equal and congruent symbol (three-line equal symbol)

OK. I'll take another stab at it.

The heart of it began when I was responding to jghist when he was wondering about the conjugate effecting the sign but did not feel like getting into the detail. He was, in fact, correct and I worked the problem out the long way to illustrate his point and explained the complex conjugate for those who might be following along and might not have been sure what jghist was referring to.

You chimed in with insulting comments and seemed to think I was taking some kind of shortcut and manipulating the signs. The calculations I made were done the long way on purpose to show how we could arrive at the end result without taking the shortcut.

Your "VA/V@θ=I@θ" is a shortcut and is not mathematically correct. You have taken the portion of the equation on the left and forced the desired solution on the right. For the shorthand version, something like the following would closer to mathematically correct: (VA /|V|)@θ=I@θ. That is a fairer representation of what the shortcut does: divides two scalar values then assigns an angle to the result.

When you have an angle in the denominator, that angle will change sign if it is in the numerator in the result. That is why your equation is incorrect. Correctly keeping up with the sign will produce the correct result when using the long method that involves the complex conjugate.

Jghist made an error to start with because he blended the long version and short version without properly accounting for the signs. He recognized that was where he probably went wrong but did not elaborate on what happened. I was merely trying to clarify the issue.

 

[Smart $]

OK. I'll take another stab at it.

...

You chimed in with insulting comments...

First, nothing I wrote was intended to be insulting, so I do apologize if that was the case. And I apologize in advance if you take this as offensive, insulting, or personally derogatory... but you still have your head stuck in the sand.

The heart of it began when I was responding to jghist when he was wondering about the conjugate effecting the sign but did not feel like getting into the detail. He was, in fact, correct and I worked the problem out the long way to illustrate his point and explained the complex conjugate for those who might be following along and might not have been sure what jghist was referring to.
... and seemed to think I was taking some kind of shortcut and manipulating the signs. The calculations I made were done the long way on purpose to show how we could arrive at the end result without taking the shortcut.

But you really didn't work the problem out the long way. You just "unconjugated" (or conjugated the conjugate) and proceeded from there. How is that any different than the so-called shortcut you say I am taking

Your "VA/V@θ=I@θ" is a shortcut and is not mathematically correct. You have taken the portion of the equation on the left and forced the desired solution on the right. For the shorthand version, something like the following would closer to mathematically correct: (VA /|V|)@θ=I@θ. That is a fairer representation of what the shortcut does: divides two scalar values then assigns an angle to the result.

I don't care if it is mathematically correct or not... it is empirically correct.

Take the following for example:

Using the empirical data we plot voltage and current waveforms. We can and do represent those waveforms as Vrms=1.414V@90? and Irms=1.061A@90?. We also know from a very basic electrical formula that V?I = P (or VA in this and other unity pf cases). So the empirical data confirms we have 3VA.

Therefore, it follows that P/V = I, or for this case 3VA ? 1.414V@90? = 1.061A@90?.

We have no need for the complex conjugate you have been so profusely adamant about being the correct way to do the math. This is not taking a shortcut. It is quite simply using math that concurs with the empirical data.

When you have an angle in the denominator, that angle will change sign if it is in the numerator in the result. That is why your equation is incorrect. Correctly keeping up with the sign will produce the correct result when using the long method that involves the complex conjugate.

Blah-blah-blah... get past it!

Jghist made an error to start with because he blended the long version and short version without properly accounting for the signs. He recognized that was where he probably went wrong but did not elaborate on what happened. I was merely trying to clarify the issue.

I would be suprised if it wasn't you that had put the idea in his head in the first place, from another thread

 

[Smart $]

... So the empirical data confirms we have 3VA.

...

Ummm... should have been 1.5VA

 

[mivey]

First, nothing I wrote was intended to be insulting, so I do apologize if that was the case.

Fair enough. And I apologize for any like response. Done.

As for the rest: You are still missing the points. At this time, I can't think of any additional ways to explain them that would help you.

 

[Smart $]

... You are still missing the points. At this time, I can't think of any additional ways to explain them that would help you.

You still don't get it, too.

I'm not missing any point(s). No matter how much you expand upon your point(s), the same conclusion (mine ) will always be the result. Quite simply, no matter how much math you throw at this, you cannot renounce the empirical data.

Take, for example, the following situation: We have a column of three items and a row of four items. How many total items do we have? There are actually two answers given the provided data. No matter which total of the two you say it is, I can say you are wrong until the empirical data is known by both of us.

Another might be where the solution involves a quadratic equation, where only one of the two possible mathematical answers can be possible in the "real" world.

As I see it, the flaw in your premise is, when dividing P/V, you are assuming P is a vector product. It is not. It is, at best, P = |V?I|. If you need proof, just use the vector multiplication formula you posted earlier using my example with 1.414V@90? and 1.061A@90? and you will discover the result does not match the empirical data.

 

[mivey]

you cannot renounce the empirical data.

I have never renounced the result or using the short method when applicable. Just another point you have missed.

Take, for example, the following situation: We have a column of three items and a row of four items. How many total items do we have? There are actually two answers given the provided data. No matter which total of the two you say it is, I can say you are wrong until the empirical data is known by both of us.

If there is not enough information given to find the answer, I would say so.

Another might be where the solution involves a quadratic equation, where only one of the two possible mathematical answers can be possible in the "real" world.

If I think there are two answers, I'll give both.

As I see it, the flaw in your premise is, when dividing P/V, you are assuming P is a vector product. It is not. It is, at best, P = |V?I|.

You do not understand my premise. That is why you keep thinking there is an error. BTW, the formula is P=VI*, which seems to be one of the points you are missing.

If you need proof, just use the vector multiplication formula you posted earlier using my example with 1.414V@90? and 1.061A@90? and you will discover the result does not match the empirical data.

First, there are cases when the answer is intuitive and does not require a detailed calculation. Second, why even make a calculation that you know will result in an error? I have never condoned such action. If you think I have, that is another point you have missed.

 

[mivey]

BTW, the formula is P=VI*

For the general case, the complex power is VI* or IV*
Apparent power = |V||I| = |VI*|
P = real power = Re[VI*]
Q = reactive power = Im[VI*]

For the case where the current and voltage is in phase, you have P = |V||I|. Then P / |V| only gives you the scalar result for the current. You would have to assign the angle for the current. This is the shortcut method and I have no problem with that as long as you know the correct angle to assign.

 

[mivey]

More math but don't blow a gasket:
p(t) = 2|V||I|cos(ωt+Θv)cos(ωt+Θi)

The real power is the average over one period and is
P = |V||I|cos(Θi - Θv)

The reactive power is
Q = |V||I|sin(Θv - Θi)

The complex conjugate is a shorthand version of these two formulas:
P = |V||I|cos(Θi - Θv)
Q = |V||I|sin(Θv - Θi)

and gives the complex power. This complex power formula gives the real and reactive power such that the sides of the power triangle have P is on the x-axis and Q on the y-axis (in other words we have P + jQ).

 

[LarryFine]

. . . I apologize in advance if you take this as offensive, insulting, or personally derogatory... but you still have your head stuck in the sand.

Gee whiz, how could anyone find that to be offensive?

 

[skeshesh]

Gee whiz, how could anyone find that to be offensive?

hehe, it's the same as the good ol "not to sound racist, but..." [fill in racist comment].

 

[Smart $]

More math...

Are you done yet? :roll:

 

[Smart $]

hehe, it's the same as the good ol "not to sound racist, but..." [fill in racist comment].

I didn't say it could not or should not be construed as offensive... thus I apologized beforehand.

 

[mivey]

Are you done yet? :roll:

I guess as long as I'm poking you with a stick, I might as well whomp you over the head with it a few times for good measure:
Given:
V = 1.414@90? volts
I = 1.061@90? amps
P = 1.5 watts

Intuitively, we know from these two formulas:
P = |V||I|cos(Θi-Θv)
Q = |V||I|sin(Θv-Θi)
that P=|V||I| and Q=0 because Θi & Θv are the same (i.e., voltage & current are in phase). The short calc is a simple scalar product for power. But we can't have much fun with that so:

Drawing back the stick:
Complex power = VI* = P + jQ = Re[VI*] +jIm[VI*]

Front-hand whomp:
P+jQ = VI* = (1.414@90?)(1.061@90?)* = (0 + j1.414)(0 + j1.061)* = (0 + j1.414)(0 - j1.061) = 1.5 + j0
so, P = 1.5 watts and Q = 0 var

Back-hand whomp:
P+jQ = VI* = IV* => I = (P+jQ) / (V*)
I = (P+jQ)/(V*) = (1.5+j0) / (1.414@90?)* = (1.5+j0) / (0 + j1.414)* = (1.5+j0) / (0 - j1.414) = 1.061@90?
so, I = 1.061@90? amps

So you see, as long as you multiply and divide like you are supposed to, you match the "empirical data" and do not have to force the sign to match the results like you do with your "VA/V@θ=I@θ" which is not mathematically correct.

WHOMP! (no math with that last hit as it was just for spite.) :grin:

 

[Smart $]

I guess as long as I'm poking you with a stick, I might as well whomp you over the head with it a few times for good measure...

So you see, as long as you multiply and divide like you are supposed to, you match the "empirical data" and do not have to force the sign to match the results like you do with your "VA/V@θ=I@θ" which is not mathematically correct.

WHOMP! (no math with that last hit as it was just for spite.) :grin:

Sorry, but feelin' no whompin'

I don't mind the math when it's done [substantially more] correct. That's because you finally started using moduli, absolute values, and the true vector product in your formulas. If you look back through my posts, I've been "hinting" just that for some time now.

So, since we are now at Post# 97, what took you so long?

But then again... whoop-te-do!!! You have shown through complex calculation nothing more than what I have said all along with much less calculation. Since you have used the word intuitively... all you have to do is take it up a few more levels (maybe many in your case ) and you'll be there

 

[mivey]

Spin it however you want. The posts are there for everyone to draw their own conclusions.