“Do few things if you would enjoy tranquility.” —Marcus Aurelius
I received an email from someone who is not from Texas looking for a formula to calculate the current in an unbalanced 3-phase power distribution system. I could tell he wasn’t from Texas because he was going out of his way to find a solution to a complex problem.
Gittin’ Her Done
It’s not that people in Texas are lazy or slow. On the contrary, there are many very ambitious and quick people in Texas. It’s just that most of them came from New York. There something about the heat in Texas that makes us just want to git ‘er done and get out of the sun.
So when Texas secedes from the Union we’re going to pass a law that all 3-phase power distribution systems have to be perfectly balanced so that we won’t have to do any complex calculations. And once all the loads are evenly distributed among the three phases it will be against the law to unbalance them, so you’ll have to dim in groups of three lights distributed evenly across the three phases.
Fog machines, of course, will have to be connected three at a time, one on each of the three phases, and when one is triggered then all three must be triggered. The same goes for single-phase chain motors. It’s bad enough that we have to use the 3-phase power formula for a balanced load without having to expend even more time and brainpower trying to figure out the current when the system is unbalanced.
I’m Workin’ Up a Sweat
Until then, I guess we’re going to have to work harder to figure it out. I couldn’t find a formula that the fellow in the e-mail was looking for, but that’s not surprising considering I grew up in Texas. So I drew a series of cartoons to help figure it out. The first one shows a delta-wye connected feeder transformer supplying three equal 208V loads, two of which are connected across phases A and B, and one of which is connected across phases B and C. (See Figure 1.)
If you look at the drawing you’ll see that the current flowing through phase A is twice the current in phase C. So the question is, how much current is flowing in phase B?
If the currents IA and IC were in phase with each other then it would be a simple addition problem and even Texans like me could solve it. But in a 3-phase system, each of the three phases are 120° out of phase with each other, so it becomes more challenging.
Once again I simplified the problem by drawing another picture. In this picture I drew some arrows that are proportional in length to the amount of current flowing and that point in the direction of the instantaneous phase angle. (See Figure 2).
{mosimage} These are actually vectors showing the magnitude and phase angle of the current in each of the two phases. By summing them graphically we can figure out how much current is flowing in phase B.
Doin’ the Math
The drawing shows the two vectors, one representing the current in phase A (IA) and the other the current in phase C (IC). It also shows the sum of (or the difference between) the two, which is the resultant AC. All we have to do to figure out the current flowing in phase B is to solve for the length of the vector AC.
We started out by saying that the current in phase A is twice the current in phase C, and we know the phase angle between the two is 120°. Since the vector drawing is an obtuse triangle (that’s a triangle in which one of the angles is greater than 90°), we can use the law of cosines to figure out the length of AC, which is the same as the current in phase B.
As high tech and sophisticated as this industry is, we still use the ancient art of trigonometry to help solve problems. This goes back to the time of Euclid, the Greek mathematician who lived in 300 BC, long before Texas was established. Euclid reasoned the solution using drawings of triangles and squares. The answer he came up with was:
AC2 = OA2 + OC2 – (2 × OA × OC × cos α)
Suppose, for example, that the current IA was 14 amps and the IC was 7 amps. Then we could plug some numbers into the equation:
AC2 = IA2 + IC2 – (2 × IA × IC × cos 120°)
AC2 = 142 + 72 – (2 × 14 × 7 × cos 120°) = 196 + 49 – (–98) = 343
AC = SQRT(343) = 18.52 amps
If you didn’t know how to do this before, now you know why we like to avoid it in Texas; it’s a formula that involves more than just pluses and minuses. But if some Greek guy from the third century B.C. can figure it out, surely a smart person like you can do it.
Before all of you Bubbas in Texas get out your crayons and start writing angry letters, let me explain. Every time some local yokel from Texas opens his mouth and says Texas should secede or something equally embarrassing to the rest of us, the outside world thinks we’re all cut from the same cloth. So just for fun I thought I’d make them all wonder how someone like me, who was born and raised in Texas, could figure this out. Trust me; this is going to drive some people crazy. So put down your shootin’ irons and pass me the barbeque sauce.
