Current Calc: 2 Phase 3 Wire

[74one]

Using the direction conventions given in #18 (important to keep the signs consistent), and the method from #25:
I12* = 600/480@90? = 1.25@-90? = 0 - j1.25 => I12 = 0 + j1.25

I23* = 900/480@-30? = 1.875@30? = 1.6238 + j0.9375 => I23 = 1.6238 - j0.9375

I2 = I23 - I12 = (1.6238 - j0.9375) - (0 + j1.25) = (1.6238 - 0) + j(-0.9375 - 1.25) = 1.6238 - j2.1875 = 2.7243@-53.4132?

You can also solve for the impedances for another method:
R12 = V12^2 / P12 = 480^2 / 600 = 384 ohms
R23 = V23^2 / P23 = 480^2 / 900 = 256 ohms
|I12| = sqrt(P12 / R12) = 1.25 => I12 = 1.25@90?
|I23| = sqrt(P23 / R23) = 1.875 => I23 = 1.875@-30?
I2 = I23 - I12 = 2.7243@-53.4132?

Back to original question ?Calculate the current in the common leg (2)?
1.I believe we should not ignore V31 and I31 in the calculations ? the load is connected to the 3 phase power supply circuit.
2.And finally I see nothing wrong with the following formula for solving this problem: Leg#2=1500w/480v*1.73=1.8A

 

[Cold Fusion]

I haven't been checking mivey's math - the problem while tedious is pretty simple vector algebra. But it does appear he has kept the signs and angles correct.

...
1.I believe we should not ignore V31 and I31 in the calculations ? the load is connected to the 3 phase power supply circuit. ...

I don't think he did. There is no load connected across 1 and 3, so I31 = 0. And this gives any contribution to the power as V31I31* = 0. So they aren't really ignored, they just don't show up because the contribution = 0

...
2.And finally I see nothing wrong with the following formula for solving this problem: Leg#2=1500w/480v*1.73=1.8A

Following other posters:
If the load were spread evenly across all three phases, as in a 500W heater in each phase, you would be exactly correct. But for this example - they are not.

cf

 

[rattus]

Following other posters:
If the load were spread evenly across all three phases, as in a 500W heater in each phase, you would be exactly correct. But for this example - they are not.

cf[/QUOTE]

Actually,the 1.73 factor may be used to find the current in the line common to TWO identical loads. The third load has no effect.

 

[74one]

"... There is no load connected across 1 and 3, so I31 = 0. And this gives any contribution to the power as V31I31* = 0. So they aren't really ignored, they just don't show up because the contribution = 0..."
It is not accurate, the load connected between 1-2, 2-3 and 3-1 (thru leg #2),
so V31 will "see" the load.

 

[jghrist]

That's true, but I have done some of my best work on white boards - with a handfull of colors.

Well, except I am partial to the back of restaurant place mats - but that work is never in color

My favourite restaurants provide crayons with their placemats!:roll:

 

[jghrist]

"... There is no load connected across 1 and 3, so I31 = 0. And this gives any contribution to the power as V31I31* = 0. So they aren't really ignored, they just don't show up because the contribution = 0..."
It is not accurate, the load connected between 1-2, 2-3 and 3-1 (thru leg #2),
so V31 will "see" the load.

This doesn't make any difference unless you take voltage drops in the source into account. The problem statement doesn't have enough data to account for voltage drops and it would complicate things immensely. We have been assuming no voltage drops.

 

[Smart $]

... But it does appear he has kept the signs and angles correct.

...

So you condone his use of vectors...

I12* = 600/480@90? = 1.25@-90?...

I23* = 900/480@-30? = 1.875@30?...

 

[Cold Fusion]

"... There is no load connected across 1 and 3, so I31 = 0. And this gives any contribution to the power as V31I31* = 0. So they aren't really ignored, they just don't show up because the contribution = 0..."

It is not accurate, the load connected between 1-2, 2-3 and 3-1 (thru leg #2),
so V31 will "see" the load.

74 -
I guess I need a sketch showing me the current and voltage vectors to understand your model. The model I would use would not change any if I were to omit the source V31.

Hummm ... Actually rattus said it better, "The third load has no effect."

Current I2 is not changed by any load between 1 and 3. Current I2 is not changed by any currents from source V13 or even if source v13 is omitted.

cf

 

[Cold Fusion]

I can't see the picture you attached. it's my system. So if the picture is significant, you will need to tell me.

So you condone his use of vectors...

... But it does appear he has kept the signs and angles correct.

I12* = 600/480@90? = 1.25@-90?...

I23* = 900/480@-30? = 1.875@30?...

Recommend reading my whole post. I highlighted a part that perhaps you missed.

I haven't been checking mivey's math - the problem while tedious is pretty simple vector algebra. But it does appear he has kept the signs and angles correct ...

I neither condone nor condemn mivey's notation. I can follow his reasoning, I understand his model, so I don't care.

If your interest is a contest on who can fill a quart bottle first - I'm not a worthy opponent. You would really need to pick someone that thinks the topic is worth discussion.

If your interest is a discussion on math models incorporating pseudo vector systems. I'd listen but I likely wouldn't contribute.

cf

 

[74one]

...Current I2 is not changed by any load between 1 and 3. Current I2 is not changed by any currents from source V13 or even if source v13 is omitted.

cf

Yes you right, I forgot about fazes, V31 is irrelevant to calculations.

 

[jghrist]

So you condone his use of vectors...

The arrow you drew marked I23 is I23* in mivey's post:

I23* = 900/480@-30? = 1.875@30? = 1.6238 + j0.9375 => I23 = 1.6238 - j0.9375

I23* is the complex conjugate of I23. I23 has the same angle (-30?) as the voltage.

Similarly, you drew I12* instead of I12.

 

[Smart $]

The arrow you drew marked I23 is I23* in mivey's post:

I23* is the complex conjugate of I23. I23 has the same angle (-30?) as the voltage.

Similarly, you drew I12* instead of I12.

I understand what you are saying but it is not written exactly how you say. To be how you say, it should be written,for example:

I12* = 600/480@90? = 1.25@-90?*...
I23* = 900/480@-30? = 1.875@30?*...​

You can't change conjugation amid the equation without so notating it.

I know you brought up using the complex conjugate, but I've yet to fathom why anyone would use it. As CF wrote, "...the problem... is pretty simple vector algebra..."

 

[mivey]

I understand what you are saying but it is not written exactly how you say. To be how you say, it should be written,for example:

I12* = 600/480@90? = 1.25@-90?*...
I23* = 900/480@-30? = 1.875@30?*...
​

You can't change conjugation amid the equation without so notating it.

I know you brought up using the complex conjugate, but I've yet to fathom why anyone would use it. As CF wrote, "...the problem... is pretty simple vector algebra..."

I think I see where your issue is. It is a simple misunderstanding. The conjugation did not change until I noted it:

I12* = 600/480@90? = 1.25@-90? = 0 - j1.25 => I12 = 0 + j1.25

I'm now thinking you have misunderstood division using the polar coordinates system. No big deal as it can be easy to forget the sign change. For those who might not remember, this is a quick summary of the polar form of complex numbers and multiplication and division:

z = x + i*y = r[cos(θ?) + i*sin(θ?)] = r@θ?
where i = sqrt(-1)

with multiplication we have:
z1 * z2 = [r1*r2] * [cos(θ1?+θ2?) + i*sin(θ1?+θ2?)]
or (r1*r2)@(θ1?+θ2?)

with division we have:
z1 / z2 = [r1/r2] * [cos(θ1?-θ2?) + i*sin(θ1?-θ2?)]
or [r1/r2]@(θ1?-θ2?)

You can quickly check this using Euler's formula of e^(iθ) = cosθ + i*sinθ. With z = r*e^(iθ), then z1/z2 = { r1/r2 }*{ e^[i*(θ1-θ2)] }

Add: the "*" symbols in the new stuff are all multiplication symbols, not an indicator of a conjugate. This forum format is not great for math formulas.

 

[Smart $]

I think I see where your issue is. ...

Nope!

First let me start by stating vector multiplication (or division) is not defined. Wolfram's Mathworld agrees with me (or my assertion with it, if you prefer). I am not even going to go down that road, because the problem is more basal...

Put in simplest form, the math must match the physics. According to just one of your assertions...

with multiplication we have:
z1 * z2 = [r1*r2] * [cos(θ1?+θ2?) + i*sin(θ1?+θ2?)]
or (r1*r2)@(θ1?+θ2?)​

We have:

z1=480@90?, and
z2=1.25@-90?, therefore
z1?z2=(480?1.25)?{cos[90?+(-90?)]+i?sin[30?+(-30?)]}
xxxxx=600?(cos0?+i?sin0?)
xxxxx=600?(1+i?0)
xxxxx=600​

OK, so now you've established the result is 600... but 600 of what. Perhaps the modulus of power, but not power itself, for power does have a waveform and therefore would have a phasor (not to mention the period is 1/2 that of V and I).

IMO, you just showed us a bunch of manipulative math that proves nothing better than using the most basic of vector math.

 

[mivey]

Nope!

First let me start by stating vector multiplication (or division) is not defined. Wolfram's Mathworld agrees with me (or my assertion with it, if you prefer). I am not even going to go down that road, because the problem is more basal...

Your reference does not say vector multiplication is not defined. I suspect any other valid reference you find won't say it is not defined either. It says it is not uniquely defined. Two commonly used forms of vector multiplication are the dot product and cross product. I see no need to get into that for this thread.

Put in simplest form, the math must match the physics. According to just one of your assertions...

We have:

z1=480@90?, and
z2=1.25@-90?, therefore
z1?z2=(480?1.25)?{cos[90?+(-90?)]+i?sin[30?+(-30?)]}
xxxxx=600?(cos0?+i?sin0?)
xxxxx=600?(1+i?0)
xxxxx=600
​

OK, so now you've established the result is 600... but 600 of what. Perhaps the modulus of power, but not power itself, for power does have a waveform and therefore would have a phasor (not to mention the period is 1/2 that of V and I).

IMO, you just showed us a bunch of manipulative math that proves nothing better than using the most basic of vector math.

I would say:
Given z1=480@90?, and z2=1.25@-90?

So we have z1?z2=(480?1.25)?{cos[90?+(-90?)]+i?sin[90?+(-90?)]} = 600?(cos0?+i?sin0?) = 600?(1+i?0) = 600

Don't read too much into it as it was just a math example.

If division was not the issue, I am not understanding the bone you are trying to pick. I was not trying to prove anything other than finding the correct current for the problem presented. Like I said before, if there is something about my posts that you are not getting, ask some questions and I'll see if I can clarify them.

 

[Smart $]

Your reference does not say vector multiplication is not defined. I suspect any other valid reference you find won't say it is not defined either. It says it is not uniquely defined.

Well excuuuuussse me !!!

Two commonly used forms of vector multiplication are the dot product and cross product. I see no need to get into that for this thread.

You're right... there is no need because neither applies. You could discuss dot product if non-unity power factor were involved.

I would say:
Given z1=480@90?, and z2=1.25@-90?

So we have z1?z2=(480?1.25)?{cos[90?+(-90?)]+i?sin[90?+(-90?)]} = 600?(cos0?+i?sin0?) = 600?(1+i?0) = 600

OK, so you would say... but is there some reason you wrote out all the math again other than you like to read your own typing.

Don't read too much into it as it was just a math example.

Don't worry, I'm not...

If division was not the issue, I am not understanding the bone you are trying to pick.

Division by your method is not appropriate within the conjugate argument. If you want to use your method as a shortcut to arrive at the conjugate outside the argument, be my guest.

Personally, I would rather see it as follows:

I12 = 600VA/480V@90? = 1.25A@90?​

Forget about your sign changing for whatever reason you feel you have to. These values match the physics model.

I was not trying to prove anything other than finding the correct current for the problem presented.

But you started out using conjugates, and didn't finish with them. Was there a purpose? ...'cause I sure can't see one.

Like I said before, if there is something about my posts that you are not getting, ask some questions and I'll see if I can clarify them.

Oh, I got it... it's the other way around :roll:

 

[mivey]

Well excuuuuussse me !!!

Was that in your best Ace Ventura voice?:grin: You're excused.

OK, so you would say... but is there some reason you wrote out all the math again other than you like to read your own typing.

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.

Personally, I would rather see it as follows:

I12 = 600VA/480V@90? = 1.25A@90?
​

Forget about your sign changing for whatever reason you feel you have to. These values match the physics model.

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.

But you started out using conjugates, and didn't finish with them. Was there a purpose? ...'cause I sure can't see one.

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.

Oh, I got it... it's the other way around :roll:

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.

 

[skeshesh]

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

 

[mivey]

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

It is basic stuff for engineers. We are taught this stuff in school and have no problems understanding the concepts or the math. Like anybody else, we may make calc errors from time to time but at least recognize when corrections are made.

Here we have an electrician who does not seem to get the math or the concept of other trivial things like conjugates, nor does he seem willing to learn. I don't recall any of the engineers that chimed in taking issue with these common concepts.

 

[skeshesh]

Here we have an electrician who does not seem to get the math or the concept of other trivial things like conjugates, nor does he seem willing to learn.

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 don't recall any of the engineers that chimed in taking issue with these common concepts.

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.