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Author Topic: 3-phase demand question
Paul G. Thompson
The Weenie Man

Posts: 4718
From: Mount Vernon WA USA
Registered: Nov 2000


 - posted 10-20-2002 10:22 PM      Profile for Paul G. Thompson   Email Paul G. Thompson   Send New Private Message       Edit/Delete Post 
Steve said:
quote:
Next I'm waiting for one of you guys to go through the math of WHY when you have 208 VAC when you have each leg starting with 120VAC (and a 120 degree phase shift).

Well, Steve...Lets turn that around and let you be the teacher.

Rotating Vectors are not my bag.....


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Steve Kraus
Film God

Posts: 4094
From: Chicago, IL, USA
Registered: May 2000


 - posted 10-21-2002 12:23 AM      Profile for Steve Kraus     Send New Private Message       Edit/Delete Post 
Too late at night, Steve, for math I can't recall but not too late for purty pictures.

Everyone knows that the AC voltages we usually speak of are RMS average, right? That is, the voltage is varying continuously as a sine wave so we describe it as the average of the absolute value (a straight average would be zero since it's negative half the time). For a sine wave the relationship between the RMS average and the peak to peak voltage is the square root of 2 divided by 2 which is about .707. So your 120V USA household voltage is 120/.707 or about 170 Vpp.


With single phase, the two legs from opposite ends of a transformer can be thought of as being 180° apart. The yellow line represents the difference between the two legs and would be the voltage you get connecting across them. I'm too lazy to show a scale but if the red and the blue are 170Vpp then the yellow is about 339Vpp, (times .707 = 240Vrms).

Here is three phase. Red and blue represent any pair of phases, and are 120° apart. Once again yellow is the difference between the two legs--you'll notice it is crossing the zero line when red and blue intersect. The lesser phase difference means this voltage isn't going to have as high a peak--only about 294Vpp. Times .707 gives you about 208Vrms.


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William Bunch
Film Handler

Posts: 87
From: Misawa, Japan
Registered: Nov 2001


 - posted 10-21-2002 12:53 AM      Profile for William Bunch   Author's Homepage   Email William Bunch   Send New Private Message       Edit/Delete Post 
Where I am in northern Japan the low side is 115 while the high side is always 200 volts. And at 50 Hertz !! Does not matter single or three phase. Makes for some real challenges when ordering equipment parts from the USA. US manufacturers usually only make stuff for either North America or Europe. There is no mathematical match for here.

Southern Japan is better off. They have the same strange voltages but at least they are on a 60 Hz system. (less damaging heat)

Bill BuncH
Misawa, Japan

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Paul G. Thompson
The Weenie Man

Posts: 4718
From: Mount Vernon WA USA
Registered: Nov 2000


 - posted 10-21-2002 06:34 AM      Profile for Paul G. Thompson   Email Paul G. Thompson   Send New Private Message       Edit/Delete Post 
Sooner or later, someone is going to ask how the RMS value is determined. Here is the answer:


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Steve Kraus
Film God

Posts: 4094
From: Chicago, IL, USA
Registered: May 2000


 - posted 10-21-2002 07:51 AM      Profile for Steve Kraus     Send New Private Message       Edit/Delete Post 
A common first year computer science class programming problem is to find the area under a curve like this sine wave by using a series of rectangles or trapezoids, repeating the process each time halving their width for greater accuracy and comparing the result with the prior estimate, stopping when the difference is less than some amount or beyond the precision of the computer.

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Steve Guttag
We forgot the crackers Gromit!!!

Posts: 12814
From: Annapolis, MD
Registered: Dec 1999


 - posted 10-21-2002 09:01 AM      Profile for Steve Guttag   Email Steve Guttag   Send New Private Message       Edit/Delete Post 
Actually, First year calculus often does the approximation by means of rectangles, then trapizoids and finally arcs...at least, we did.
Visually seeing the graph, as Steve K did it is easier to see how the two sources add up. The math form actually takes more effort due to the phase shift. The periodic sine wave takes on the form of A*Sine(theda + phi)

Where A = the amplitude, Theda is the degree (in radians) of interest, and phi is the phase shift. If there is no phase shift, phi goes to 0 and it is a simple equation.

When I have the time, I'll go through the whole process.

Steve

------------------
"Old projectionists never die, they just changeover!"

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