Sorry for the length of this. I couldn't find any place to stop or
trim.
Hi, Tim.
Actually, you can put the neutral anywhere you want; for convenience,
it is usually placed between the A and B phases, though other
connections are possible (and are used.)
OK, I'll see your ASCII art and raise you one :)
Actually, I think I'll copy and edit yours (quite good, it is.) Here
are the A and B phases relative to neutral, just like you had them.
(And I am aligning all my phases relative to the A phase just as it is
here, with the leading "> "; vertical calibration is approximately
57V/line; horizontal is 30 degrees per column.)
> ** ** ** **
> * * * * * * * *
> * * * * * * * *
> A ----------------------------------------------------
> * * * * * * *
> * * * * * * *
> ** ** ** *
>
> ** ** **
> * * * * * *
> * * * * * *
> B ----------------------------------------------------
> * * * * * * *
> * * * * * * *
> ** ** ** *
OK, now I'll add a new phase C to neutral 208vRMS (294 peak),
lagging A by 90 degrees:
+294 ** ** ** **
* * * * * * * *
* * * * * * * *
C ------------------------------------------------
* * * * * * * *
* * * * * * * *
-294 * ** ** ** *
So from A to B will look like:
+339 ** ** ** **
* * * * * * * *
* * * * * * * *
A-B ------------------------------------------------
* * * * * * * *
* * * * * * * *
-339 ** ** ** **
The resultant waveform for B-C:
+339 ** ** ** **
* * * * * * * *
* * * * * * * *
B-C ------------------------------------------------
* * * * * * * *
* * * * * * * *
-339 ** ** ** **
and for C-A:
+339 ** ** ** **
* * * * * * * *
* * * * * * * *
C-A ------------------------------------------------
* * * * * * * *
* * * * * * * *
-339 ** ** ** **
These look just like they should and much like you drew below.
mathematically (and somewhat more succinctly):
A-n = 170 sin(wt)
B-n = 170 sin(wt - 180) = -170 sin(wt)
C-n = 294 sin (wt - 90)
[I'll be using this trigonometric identity later]
sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
A-B = A-n - B-n = 340 sin (wt)
B-C = B-n - C-n
= -170 sin(wt) - 294 sin (wt - 90)
= -170 sin(wt) + 294 cos(wt)
This equation crosses zero volts going positive at wt = -120 degrees.
0 =? -170 sin(-120) + 294 cos(-120)
0 = 147 - 147
An equivalent equation, incorporating the 120 deg phase shift and the
trig identity above gives us our final answer:
B-C = 340 sin (wt + 120)
checking:
= 340 (sin(wt) cos(120) + cos(wt) sin(120))
= 340 (-0.5 sin(wt) + 0.866 cos(wt))
= -170 sin(wt) + 294 cos(wt)
which is what we determined B-C should be (simply the difference between
B-n and C-n.)
The solution for phase C-A can be found similarly, beginning with:
C-A = C-n - A-n
= 294 sin (wt - 90) - 170 sin(wt)
etc.
And if you've suffered with me this far, yes it is possible to put the
neutral in the middle of the whole thing (it would probably be called
240-wye-140, because there would be three phases, wye-connected, with
240 volts phase-to-phase, as well as three phase-to-neutral voltages
available at 140 volts each. I've only seen 208-wye-120 installed
this way, though. The 140 is very non-standard in the USA.)
> Date: Fri, 12 Oct 2001 14:10:52 -0400 (EDT)
> From: "Timothy R. Hoerning" <hoerni@cooper.edu>
>
> On Fri, 12 Oct 2001, Donald H Locker wrote:
>
> >
> > Hmmmm. I'll try some ASCII art.
> >
> > C
> > / \
> > / \
> > / \
> > / \
> > / \
> > / \
> > / \
> > / \
> > A--------n--------B
> >
> > Phases A and B are the 120 legs, n is neutral, C is the third phase.
>
> With you so far.
>
> > The angle between A and B is 180 degrees,
>
> Not sure I'm believing this one yet... I'll explain later
>
> > while the angles CAB, ABC, and BCA are each 120 degrees.
>
> Actually they are 60 degrees in you diagram. Now if you want to
> create another neural in the center of the triangle (called in N), then the
> angles ANB ANC and BNC are 120 degrees. This probably has something to do
> with my misunderstanding, but I'll elaborate further below. Unfortunately
> my power electronics textbook is at home.
>
> > A-n is 120 volts; B-n is 120; C-n is 208 (and at 90 degrees to A-n and B-n,
> > FWIW); A-B is 240; B-C is 240; > C-A is 240.
>
> Okay, time for my ASCII Art. Since I haven't thought about Power in a
> while, the phasor abstractions aren't working for me and I'm going to need
> to do this in the time domain.
>
> According to my understanding, for a typical house wiring system, you
> need two phases A & B - 180 degrees out of phase (120Vrms to ground). If you
> read accross them you would get 240Vrms because they are completely out of
> phase.
>
> ** ** ** **
> * * * * * * * *
> * * * * * * * *
> A ----------------------------------------------------
> * * * * * * *
> * * * * * * *
> ** ** ** *
>
> ** ** **
> * * * * * *
> * * * * * *
> B ----------------------------------------------------
> * * * * * * *
> * * * * * * *
> ** ** ** *
>
> Now from what I remember about three phase power, the lagging waveforms
> would need to each follow the previous by 120 degrees as in the following
>
> ** ** ** **
> * * * * * * * *
> * * * * * * * *
> A ----------------------------------------------------
> * * * * * * *
> * * * * * * *
> ** ** ** *
> ** ** ** **
> * * * * * * * *
> * * * * * * * *
> B ----------------------------------------------------
> * * * * * * * *
> * * * * * * * *
> ** ** ** ** *
> ** ** ** **
> * * * * * * * * *
> * * * * * * * * *
> C ----------------------------------------------------
> * * * * * * * * *
> * * * * * * * * *
> ** ** ** ** *
>
> Now if you were to measure the voltage between A and B you would get
> only 180 volts RMS (254V peak) because of the phase offsets.
>
> So I guess my question is how can the phases be 120 degrees apart for
> three phase and simultaneously 180 degree apart for house wiring? Or am I
> remembering something wrong
>
> Now I could see it working if there was a transformer in between the
> Delta and the house, but then it wouldn't be a simple case of adding a third
> conductor to get three phase power.
[snip]
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