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Phasors: Multiphase Power

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

The electrical service to your home is a two-phase service.1 This means that two 110 volt, 60 Hz lines, plus neutral, terminate in the panel. The lines are π radians (180) out of phase, so we can write them as

x 1 ( t ) = 110 c o s [ 2 π ( 60 ) t + φ ] = Re { 110 e j [ 2 π ( 60 ) t + φ ] } = Re { X 1 e j 2 π ( 60 ) t } X 1 = 110 e j φ x 2 ( t ) = 110 c o s [ 2 π ( 60 ) t + φ + π ] = Re { 110 e j [ 2 π ( 60 ) t + φ + π ] } = Re { X 2 e j 2 π ( 60 ) t } X 2 = 110 e j ( φ + π ) . x 1 ( t ) = 110 c o s [ 2 π ( 60 ) t + φ ] = Re { 110 e j [ 2 π ( 60 ) t + φ ] } = Re { X 1 e j 2 π ( 60 ) t } X 1 = 110 e j φ x 2 ( t ) = 110 c o s [ 2 π ( 60 ) t + φ + π ] = Re { 110 e j [ 2 π ( 60 ) t + φ + π ] } = Re { X 2 e j 2 π ( 60 ) t } X 2 = 110 e j ( φ + π ) .
(1)

These two voltages are illustrated as the phasors X1 and X2 in Figure 1.

Figure 1: Phasors in Two-Phase Power
Figure one is a cartesian graph with a line passing through the origin from the third to first quadrant with a shallow positive slope. The line on the left side is labeled X_2, and on the right side is labeled X_1. The angle from the positive horizontal axis to the portion of the line in the first quadrant is measured and labeled Φ.  The angle from the positive horizontal axis to the portion of the line in the third quadrant (angle measured in a counter-clockwise direction) is labeled Φ+π. Above the line in the first quadrant is the number 110.

You may use x1(t)x1(t) to drive your clock radio or your toaster and the difference between x1(t)x1(t) and x2(t)x2(t) to drive your range or dryer:

x 1 ( t ) - x 2 ( t ) = 220 cos [ 2 π ( 60 ) t + φ ] . x 1 ( t ) - x 2 ( t ) = 220 cos [ 2 π ( 60 ) t + φ ] .
(2)

The phasor representation of this difference is

X 1 - X 2 = 220 e j φ . X 1 - X 2 = 220 e j φ .
(3)

The breakers in a breaker box span the x1-tox1-to-neutral bus for 110 volts and the x1x1-to-x2x2 buses for 220 volts.

Exercise 1

Sketch the phasor X1-X2X1-X2 on Figure 1.

Most industrial installations use a three-phase service consisting of the signals x1(t),x2(t)x1(t),x2(t), and x3(t)x3(t):

x n ( t ) = 110 Re { e j [ ω 0 t + n ( 2 π / 3 ) ] } X n = 110 e j n ( 2 π / 3 ) , n = 1 , 2 , 3 . x n ( t ) = 110 Re { e j [ ω 0 t + n ( 2 π / 3 ) ] } X n = 110 e j n ( 2 π / 3 ) , n = 1 , 2 , 3 .
(4)

The phasors for three-phase power are illustrated in Figure 2.

Exercise 2

Sketch the phasor X2-X1X2-X1 corresponding to x2(t)-x1(t)x2(t)-x1(t) on Exercise 3. Compute the voltage you can get with x2(t)-x1(t)x2(t)-x1(t). This answer explains why you do not get 220 volts in three-phase circuits. What do you get?

Figure 2: Three-Phase Power
Figure two is a cartesian graph with three arrows pointing in various directions. The first is labeled X_1 and is drawn along the horizontal axis to the right. The second is labeled X_2 and is drawn with a sharp negative slope up into the second quadrant. The third is labeled X_3 and is drawn with sharp positive slope into the fourth quadrant.

Constant Power. Two- and three-phase power generalizes in an obvious way to N-phase power. In such a scheme, the NN signals xn(n=xn(n=0,1,...,N-10,1,...,N-1) are

x n ( t ) = A cos ( ω t + 2 π N n ) = Re [ A e j 2 π n / N e j ω t ] X n = A e j 2 π n / N . x n ( t ) = A cos ( ω t + 2 π N n ) = Re [ A e j 2 π n / N e j ω t ] X n = A e j 2 π n / N .
(5)

The phasors X n X n are Aej2π(n/N)Aej2π(n/N). The sum of all N N signals is zero:

n = 0 N - 1 x n ( t ) = Re { A n = 0 N - 1 e j 2 π n / N e j ω t } = Re { A 1 - e j 2 π 1 - e j 2 π / N e j ω t } = 0 . n = 0 N - 1 x n ( t ) = Re { A n = 0 N - 1 e j 2 π n / N e j ω t } = Re { A 1 - e j 2 π 1 - e j 2 π / N e j ω t } = 0 .
(6)

But what about the sum of the instantaneous powers? Define the instantaneous power of the nthnth signal to be

p n ( t ) = x n 2 ( t ) = A 2 cos 2 ( ω t + 2 π N n ) = A 2 2 + A 2 2 cos ( 2 ω t + 2 2 π N n ) = A 2 2 + Re { A 2 2 e j ( 2 π / N ) 2 n e j 2 ω t } . p n ( t ) = x n 2 ( t ) = A 2 cos 2 ( ω t + 2 π N n ) = A 2 2 + A 2 2 cos ( 2 ω t + 2 2 π N n ) = A 2 2 + Re { A 2 2 e j ( 2 π / N ) 2 n e j 2 ω t } .
(7)

The sum of all instantaneous powers is (see Exercise 3)

P = n = 0 N - 1 p n ( t ) = N A 2 2 , P = n = 0 N - 1 p n ( t ) = N A 2 2 ,
(8)

and this is independent of time!

Exercise 3

Carry out the computations of Equation 3.33 to prove that instantaneous power P P is constant in the N-phase power scheme.

Footnotes

  1. It really is, although it is said to be “single phase” because of the way it is picked off a single phase of a primary source. You will hear more about this in circuits and power courses.

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