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#129279  12/30/04 07:24 AM
1.732 (square root of 3) where did this come from?

Member
Registered: 11/09/00
Posts: 261
Loc: La Junta, Co. USA

There are alot of 3phase formulas that involve multiplying the linetoline voltage by 1.732. I realize this is the square root of 3, but why is this step necessary. I think it is because of the 3 windings being out of phase angle, but I would like a layman's explaination, please.
TIA
Rick Miell

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#129280  12/30/04 10:06 AM
Re: 1.732 (square root of 3) where did this come from?

Member
Registered: 09/15/03
Posts: 652
Loc: boston, ma

Oooooh me! Oooooh me! I am intimately familiar with that particular issue, since I get to work with 'high phase order' equipment all the time. Once you've had to figure out the voltage issues for a 17 phase mesh connection, you know where root 3 comes from in 3 phase calculations.
Step 1 is to remember that voltage is _always_ measured between two points. So when you say that a particular circuit is 277/480V wye, you provide the following information: Measured phase to phase the voltage is 480V. Measured phase to neutral the voltage is 277V. The source is wye connected, eg. a transformer bank with 277V secondaries.
Step 2 remember that for AC circuits the voltage is a constantly changing value, and when we give a single number, it is a form of average voltage, called the RMS voltage. The RMS voltage of an AC waveform corresponds to the DC voltage which would deliver the same power to a resistive load. If you take AC at 480V RMS and apply it to a 480 ohm resistor, the average power delivered to that resistor would be 480W. If you take DC at 480V and apply it to a 480 ohm resistor, the power delivered to that resistor would be 480W.
So now we have the two important facts: voltage is measured between two points, and an AC voltage is an average over time.
The 'voltage is measured between two points' concept can be expanded with the idea of a 'zero reference'. If you pick one point on a system and _arbitrarily_ call it '0V', then you can reference every other point on that system to that zero reference. In electrical systems, it is most common to pick 'earth' as the zero reference. Once you have a zero reference, you can speak of any other single point as having a voltage. In reality we are still measuring voltage between two points, but one of those points is assumed, so we don't have to mention it.
Now for something nifty: the _instantaneous_ voltages between points must always add up. If I measure 5V (DC) from A to B, and 6V (DC) from B to C then I will get 11V from A to B. If I measure 5V from X to Y and 6V from X to Z, then I will measure 1V from Y to Z. Finally , if I measure 5V from A to B, I will measure 5V from B to A. But this is all DC, or instantaneous voltage.
To see how AC voltages add up, we have to remember that the AC voltage is a form of average, and we have to look at the instantaneous values, get an instantaneous sum, and take the average of that.
So now consider phase A in our 277/480V wye system. We can plot the voltage relative to our earth reference as a function of time, and get a graph, ideally a nice sine curve. At time zero the voltage will be zero. At 1/240 of a second, the voltage will be +392V (277V * 1.414, the square root of 2). Then at 2/240 second the voltage will again be zero. At 3/240 of a second the voltage will be 392V, and at 4/240 second (1/60 second) the voltage will again be zero. This cycle will repeat.
Now imagine another point in the system, also connected to phase A. If we measure the voltage at this point, we will get the exact same curve. Finally, try to measure the voltage between these two locations. If you look at each instant in time, the voltage _difference_ will be zero. The average of zero is still zero. Net result is that if you measure the voltage between two points, both phase A, you will get zero volts, as expected.
Now I'll set up a nifty little tool; a special transformer that lets me produce output of arbitrary phase angle. The output is still 277V RMS relative to our zero reference, but I can shift its phase relative to phase A. Lets call the output of this transformer 'phase W'. I adjust phase W so that it is very slightly delayed from phase A, and graph the two phases.
At time zero, phase A is zero, but phase W is negative, say by a couple of volts. Then at time 1/240 phase A is at its positive peak, but phase W is slightly below. A teensy bit later, phase A is on its way down, and phase W hits its peak. And so on through the cycle. Both phases have the same amplitude, but hit their peaks at different times. There is usually a slightly different voltage between the two, but since sometimes A is more positive and sometimes W is more positive, the two graphs _must_ cross.
Make a graph of the difference between A and W, and you will find out that it is a sine wave with low amplitude. In fact, it is a law of mathematics that the sum or difference of two sine waves of the same frequency (but possibly different amplitudes or phases) is another sine wave of the same frequency, again with different frequency or phase.
Now, imagine that I start delaying phase W more and more. As I do so, it gets more and more different from phase A. The sine wave that is the difference between phase A and phase W gets larger and larger. Eventually this resultant sine wave is actually _greater_ than phase A. This difference sine wave reaches its peak when phase W is exactly 1/2 cycle from phase A, meaning that the +peak of A corresponds to the peak of W. At this point, the amplitude of the _difference_ between the two phases is twice the amplitude of each phase alone. This is to say that if phase A is 277V relative to neutral, and phase W is 277V relative to neutral, and 180degrees out of phase with phase A, then the voltage A to W is 534V.
The last bit is to ask 'how do I calculate the voltage difference between two phases with some other phase angle?' Clearly this is some function of the amplitude of the phases, and also a function of the phase angle between them. Simplify the question by stating that both phase A and phase W have the same phase to neutral RMS voltage V. Call the phase angle between these two phases T. The voltage between the phases is then given by V * 2 * sin(T/2). So for the 180 degree phase difference we get V * 2 * sin(180/2) = V *2
And for a 120 degree phase difference we get V * 2 * sin(120/2). Quick high school trig question: what is sin(60) { or sin(pi/3) if you are into radians). Taaadaaa.
You can use this to figure voltages when you get strange phase angles, say in a 6 phase system for a rectifier, where you have two transformer secondaries wound to give different phase angles on the output. If you know the voltage and the phase angle, then you can calculate the voltage difference between any of the two phases.
Jon

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#129281  12/30/04 12:07 PM
Re: 1.732 (square root of 3) where did this come from?

Member
Registered: 12/29/04
Posts: 12
Loc: Houston, Texas, USA

excellent info on voltages, Winnnie!
Thank you

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#129282  12/30/04 05:40 PM
Re: 1.732 (square root of 3) where did this come from?

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Registered: 04/10/02
Posts: 2561
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Â One correlation may be termed the â€śMagic of Three Phaseâ€ťâ€”comparing a 1Ă¸ 2w circuit to a 3Ă¸ 3w circuit at the same current and voltage, with a 50% increase in copper [3 wires versus 2] 73% more power can be delivered, for a powertransfer increase of 15Â˝ percent. Â Â Â Â Square root of 3 shows up repeatedly in threephase systemsâ€”like the ratio of phasetophase and phasetoneutral voltages under balanced conditions, like 208Y/120, 480Y/277, 600Y/347 and 4160Y/2400. [Donâ€™t forget 127/220, 220/380, 230/400 and 400/690 in 50Hz regions.] Oops, almost forgot... the ultimate square root of 3 web page is www.gutenberg.org/dirs/etext96/3sqrt10.txt [This message has been edited by Bjarney (edited 12302004).]

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#129283  01/22/05 02:07 PM
Re: 1.732 (square root of 3) where did this come from?

Junior Member
Registered: 11/20/04
Posts: 3

First think about a 480 volt grounded wyefed system. There is 480 volts between each of the three phases.
Now what if I tried to draw three points on a line each 480 volts apart from the other two... I can't do it. But as I think you already are aware, ac voltages cannot be fully represented on a 1d number line (like dc voltages), they can be represented as an arrow on a 2d plot.
Now try to draw the three phasetophase voltages on a 2d plot. We have to pick three points and connect them together by three equallenght arrows. We have just drawn an equilateral triangle. Where does neutral voltage lie for a balanced system? Right in the middle of the equilateral triangle, equal distance from all three sides. It is a simple matter with geometry to determine that the distance from that center/neutral point to each corner is 1/(sqrt(3)) times the length of the sides of the triangle. Since the length of the sides of the triangle represent 480 volt, the length of the voltage to neutral is 480/sqrt(3)=277vac.
So now hook up a wyeconnected resistor load... each leg of the wye does not see 480vac, it sees only the voltage to neutral which is 480/sqrt(3). That is the beginning of many of the power formula's involving square root of 3.
Without all the math, we could just post a hypothetical experiment: if you went out and measured the voltages on a balanced 480 volt system you would see 480vac between phases and 480/sqrt(3) volts to neutral. And as you know of course this type of measurement should not be attempted for safey reasons unless absolutely necessary and then only with proper training and supervision.
[This message has been edited by electricpete (edited 01222005).]

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