It’s the ratio of V_{LL} to V_{LN}
It’s the ratio of V_{LL} to V_{LN}
In case the OP is looking for a less technical answer, I submit this (please note, this is very basic and ignores power factor or other such things).
Let's assume you have a balanced 3-phase load of 3600 watts at 208V and a single phase load of 3600 watts at 240V. How much amperage is on each conductor.
Well, the single phase is easy, W=EI, so
3600 = 240*I
3600/240 = I
15 = I.
So, now how do you figure out the 3-phase since you have 208V in 3 different configurations (AB,BC,CA). Because of the relationships of the 3-phase voltages explained above, you have to do something to make the calculations correct. That "something" is using 1.732 as a multiplier to the voltage, then use the same formula you would for the single phase calculation above. The new formula for 3-phase is W=(E*1.732)*I.
So 3600=(208*1.732)*I
3600 = 360*I
10 = I
You would work in reverse if you know the amperage and need to know the watts.
W = (208*1.732)*10
W = (360)*10
W = 3600
Just as an aside, in single phase, if you add the line to neutral voltages for each phase you come up with 240 (120+120=240).
In 3-phase, if you add the line to neutral voltages for each phase you come up with 360 (120+120+120) or ... 208*1.732
For the OP, here is a picture and basic equations, might be easier to understand than just verbiage.
Note that sqrt 3 is valid (as already noted by multiple responders) ONLY for a balanced system
V = line to line voltage for unity phase voltage.
1st column is Pythagoras relationships of right triangle, 2nd column is trig.
Tan 60 = sqrt 3 = 1.73....
Exercise for the student: displace phases by a few degrees and amplitude by a few percent and calculate positive, negative, and zero sequence components. (good luck)
(note; positive, negative, and zero are math tools only, 'solution' not programmed into most scopemeters so not easy to 'measure' - only ones I've seen with the capability are the windows based 4 channel Tek bench scopes)
190620-0925 EDT
The invention of the concept of phasors to do AC calculations on steady state AC sine waves was by Steinmetz of General Electric late 1800s. See https://en.wikipedia.org/wiki/Phasor
You must be dealing with sine waves and linear loads.
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While this has probably been beat to death, I think I might be able to provide a simple perspective.
Putting the OP’s original question into its most basic form:
How do we calculate the power delivered to a load from a given power source by using physically accessible and measurable properties of that power source?
Let the power source be one or more transformer windings, each of which has its own RMS voltage Vphase across it, and RMS current Iphase through it.
The voltage and currents on the load are Vline and Iline.
Assume a unity power factor and balanced voltages and currents where appropriate..
With 2-wire single phase, the power into the load is simply Vphase x Iphase = Vline x Iline.
In the 4-wire wye case Vphase = Vline and Iphase = Iline. The total power is simply 3 Vphase x Iphase = 3 Vline x Iline because we have three windings each contributing power, and the individual phase voltages and phase currents are both directly accessible for measurement. Therefore no scaling by sqrt (3) is necessary.
In the 3-wire wye case the power available is the same as the 4-wire above, but because the neutral point is no longer accessible Vphase is not a measurable quantity. So we have to use the line-to-line voltage Vline, but this overstates the actual voltage from the source winding by sqrt (3). So as mentioned before the power is 3 Vphase x Iphase = (3 /sqrt(3)) Vline x Iphase = sqrt (3) x Vline x Iphase = sqrt (3) x Vline x Iline.
In the 3-wire delta case the voltage Vphase is measurable because it equals Vline, but the phase current Iphase in an individual winding is not. The measurable quantity Iline overstates the current Iphase in the source winding (which is not accessible) by sqrt(3), which can be seen from a phasor diagram of the currents. So in the delta case the power is 3 Vphase x Iphase = 3 Vphase x (1/sqrt(3)) x Iline = sqrt(3) x Vphase x Iline = sqrt(3) x Vline x Iline
So the reason sqrt(3) appears is because:
In the wye connection the line-to-line voltage is a vector addition of two phase voltages resulting in a factor of sqrt(3) more voltage than that on one phase.
In the delta connection each line output current is a vector addition of two phase currents resulting in a factor of sqrt(3) more current than that on one phase.
So in each case either the measurable and delivered voltage or current is sqrt(3) higher than that on a phase (the transformer winding delivering the power). Therefore it must be divided by sqrt(3) to calculate the power delivered from the source.
Last edited by synchro; 06-20-19 at 02:02 PM.
That's not even remotely close to what the unit horsepower means at all. It makes no mention of how high the horse lifts the load, and 746 is really the number for the conversion factor of HP to Watts.
What happened is, James Watt needed a performance metric to sell the steam engine, and compare it to something people already understood. He studied the performance of draft horses, and how much load they could lift by pulling a mill wheel. Based on the speed the horses walked, and the load the millwheel lifted, he determined that horses could pull with the power equivalent to lifting a 550 pound vertical load at a rate of 1 ft per second.
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