Dear Swami,
I work for a lighting company and we have a slew of power draw sheets that look nice on paper, but never seem to be entirely correct when I do a load test on a show site. I asked someone here if they knew why that is and they mentioned something about multiplying by the square root of three. What are they talking about?
—Confused Electrician
Dear Confused,
If you think three-phase power calculations are confusing, you should try figuring out the Bowl Championship Series. Texas beats OU, Tech beats Texas, and OU beats Tech; so who goes to the championship? Whoever the university presidents say will go.
{mosimage}Compared to the BCS, three-phase power is easy. What your friend is alluding to is the difference between single-phase power and three-phase power. In a single-phase system, the power in watts is the product of the voltage and the current (P = V × I). But in a three-phase system, you have to take into consideration that each of the three phases is 120° out of phase with the other. So when the current drawn by two different sources connected across two different phases shares a common conductor as shown in the illustration, then the currents are 120° out of phase with each other. (See Fig. 1).
If the two currents were in phase with each other, then they would simply add, as in I1 + I2. If, for example, L1 drew 100 amps and L2 drew 100 amps and they were in phase with each other, then the feeder cable connected to phase B would carry 200 amps because it is shared between the two loads.
On the other hand, if L1 drew 100 amps and L2 drew 100 amps and they were 180° out of phase with each other, then they would cancel each other out because they would exactly oppose each other. But in a three-phase system, phase A and phase B are 120 degrees out of phase with each other. So when the currents combine then the total is somewhere be-tween 0 amps and the sum of the two currents.
If the two loads draw the same amount of current, because they are 120° out of phase with each other, then the total current going through phase B is I1 times the square root of three (or I2 times the square root of three since I1 and I2 are the same). Why that number?
This is where you are going to wish you’d have paid attention in your high school trigonometry class. I know the waves were bitchin’ the week Mrs. Halfdeck taught sines and cosines, so you traded school for a few days at the beach. And you skipped school when Mr. Dimwatt taught vectors because your band was playing the Rumba Room and you had to do a sound check. But there’s no reason that you can’t figure out trig and vectors now. You’re a bright, curious professional now and there are lots of resources to help you fill in those gaps like putty on a putty knife.
So take some time to figure out vectors and trig, and then you will see that the total current in phase B is I1 (or I2) times twice the sine of 60°, which is 1.732 or the square root of 3.
Keep in mind that this is only true if I1 and I2 are the same (meaning L1 and L2 are drawing the same amount of current). If they are not the same, then you have to draw some vectors and work out the math to figure out the resulting current. But fret not because there is always more than one way to skin a cat.
A watt is a watt, regardless of how it’s connected to the grid. If you have a pile of 208V fixtures and a pile of 120V fixtures all connected to a three-phase system, you could calculate the three-phase load using the formula with the square root of three and the single-phase load without using the square root of three and add them together. Or you could simply total the watt-age and think of the total load as three 120V loads. The result will be the same.
To illustrate, suppose we get a call from Tammy Miami to light her Tiki Torch Tour 2009. We end up with 117 Force Four 120V 400-watt ERS fixtures and 24 Marlin Mako 2000 208V 1200-watt Hologram Projectors. The 208V fixtures will draw 79.9 amps three-phase [24 × 1200 watts = 28,800 watts; I = 28,800 ÷ (208 × 1.732) = 79.9]. The 120V fixtures will draw a total of 390 amps single-phase (117 × 400 watts = 46,800 watts; I = 46,800 ÷ 120 = 390). All together we have a draw of 629.7 amps, or 210 amps three-phase.
The reason we use the three-phase formula for the 208V fixtures is because we connect the loads evenly among the three phases to make sure it’s balanced evenly so the current will cancel in the neutral. And by doing so we end up sharing conductors between loads; L1 shares a conductor with L2, L2 shares with L3, and L3 shares with L1. It’s in these shared conductors that the phase interaction causes us to have to use the square root of 3 in the power formula.
The other way to do it is to simply add up the total wattage and treat is as a single-phase load. We have a total of 75,600 watts (28,800 watts + 46,800 watts = 75,600 watts), divided by 120 volts is 630 amps. Compare that to the results by using the other method and we find they are the same – 640 amps total (or 213.3 amps three-phase).
So the choice is yours. You can take the easy route or you can think like the president of a university or an athletic director and take the complicated route.
—Your friend, Swami Candela
