Why three-phase electric power is awesome

Three-phase power distribution involves three AC voltage lines which are 120 degrees or $2 \pi / 3$ radians out of phase with each other. The mathematics of this gives rise to several desirable physical properties, like constant power transfer and rotating magnetic fields, because of the trigonometric identity

$\cos (\theta) + \cos (\theta - \frac{2 \pi}{3}) + \cos (\theta - \frac{4 \pi}{3}) = 0$

Cosine sum-to-product identity

To derive this, we'll need a couple of other common trigonometric identities. First recall that

$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Add these to get the product-to-sum identity

$\cos (\alpha + \beta) + \cos(\alpha - \beta) = 2 \cos \alpha \cos \beta$

Substitute $\theta = \alpha + \beta$ and $\phi = \alpha - \beta$ to finally get the sum-to-product identity

$\cos \theta + \cos \phi = 2 \cos (\frac{\theta + \phi}{2}) \cos (\frac{\theta-\phi}{2})$

Three-phase identity

With this sum-to-product identity we can now start the derivation proper:

$\cos (\theta) + \cos (\theta - \frac{2 \pi}{3}) + \cos (\theta - \frac{4 \pi}{3})$

$= 2 \cos \left( \frac{\theta + (\theta - \frac{2 \pi}{3})}{2} \right) \cos \left( \frac{\theta - (\theta - \frac{2 \pi}{3})}{2} \right) + \cos (\theta - \frac{4 \pi}{3})$

Simplifying,

$= 2 \cos (\theta - \frac{\pi}{3}) \cos ( \frac{\pi}{3} ) + cos(\theta - \frac{4 \pi}{3})$

$= 2 \cos (\theta - \frac{\pi}{3}) \cdot \frac{1}{2} + cos(\theta - \frac{4 \pi}{3})$

$= \cos (\theta - \frac{\pi}{3}) + \cos (\theta - \frac{4 \pi}{3})$

Using the sum-to-product identity again

$\cos (\theta - \frac{\pi}{3}) + \cos (\theta - \frac{4 \pi}{3})$

$= 2 \cos \left( \frac{(\theta - \frac{\pi}{3}) + (\theta - \frac{4 \pi}{3})}{2} \right) \cos \left( \frac{(\theta - \frac{\pi}{3}) - (\theta - \frac{4 \pi}{3})}{2} \right)$

Which simplifies to

$= 2 \cos (\theta - \frac{5 \pi}{6}) \cos ( \frac{\pi}{2} ) = 0$

as the cosine of pi/2 is zero. Q.E.D.

Constant Power Transfer

A three-phase voltage supply will deliver constant instantaneous power to a star- or delta-connect resistive load. The instantaneous power is the sum of power of each of the three phases

$P = P_A + P_B + P_C = V \cos (\omega t) \cdot I \cos (\omega t - \phi) + V \cos (\omega t - \frac{2 \pi}{3}) \cdot I \cos (\omega t - \frac{2 \pi}{3} - \phi) + V \cos (\omega t - \frac{4 \pi}{3}) \cdot I \cos (\omega t - \frac{4 \pi}{3} - \phi)$

$P = V I \left( \cos (\omega t) \cos (\omega t - \phi) + \cos (\omega t - \frac{2 \pi}{3}) \cos (\omega t - \frac{2 \pi}{3} - \phi) + \cos (\omega t - \frac{4 \pi}{3}) \cos (\omega t - \frac{4 \pi}{3} - \phi) \right)$

Using the product-to-sum identity for each phases term,

$P = V I \left( \frac{\cos ( 2 \omega t - \phi) + \cos (\phi)}{2} + \frac{\cos ( 2 \omega t - \frac{4 \pi}{3} - \phi) + \cos (\phi)}{2} + \frac{\cos ( 2 \omega t - \frac{8 \pi}{3} - \phi) + \cos (\phi)}{2} \right)$

$= V I \left(\frac{ 3 \cos \phi } {2} + \frac{\cos ( 2 \omega t - \phi) + \cos ( 2 \omega t - \phi - \frac{2 \pi}{3}) + \cos ( 2 \omega t - \phi - \frac{4 \pi}{2})}{2} \right)$

The second term in the above expression is the three-phase identity, which sums to zero. So we have

$P = \frac{3}{2} V I \cos \phi = 3 V_{rms} I_{rms} \cos \phi$

which is constant.

Rotating Magnetic Field

A three-phase supply hooked up to coils offset by 120 degrees will generate a rotating magnetic field, making induction motors possible.

When each phase is distributed in a sinusoidal-distributed winding with N turns, the turn 'density' distribution by angle (for phase A) is

$n_{sa} = \frac{N_s}{2} \sin (\theta)$

The magneto-motive force (m.m.f.) produced by this winding is

$\Im_a = \int_\theta^{\theta+\pi} n_{sa} i_a d (\theta) = N_s i_a \cos (\theta)$

Phases B and C are the same, but offset by 2pi/3 and 4pi/3 respectively.

When supplied from a balanced 3-phase supply, the current in phase A is $i_a = I \cos (\omega t)$. Again, phases B and C are the same but offset by 2pi/3 and 4pi/3 respectively. The m.m.f. becomes

$\Im_a = N_s \left( I \cos(\omega t) \right) \cos (\theta) = N_s I \cos (\omega t) \cos (\theta)$

The total m.m.f. of the three windings is thus

$\Im(\theta) = \Im_a + \Im_b + \Im_c = N_s I \left( \cos(\omega t) \cos (\theta) + \cos(\omega t - \frac{2 \pi}{3}) \cos (\theta - \frac{2 \pi}{3}) + \cos(\omega t - \frac{4 \pi}{3}) \cos (\theta - \frac{4 \pi}{3}) \right)$

Using the product-sum identity for each term, we get

$\Im(\theta) = N_s I \left( \frac{\cos ( \omega t - \theta) + \cos (\omega t + \theta)}{2} + \frac{\cos ( \omega t - \theta ) + \cos (\omega t + \theta - \frac{4 \pi}{3})}{2} + \frac{\cos ( \omega t - \theta ) + \cos (\omega t + \theta - \frac{8 \pi}{3})}{2} \right)$

$= N_s I \left(\frac{ 3 \cos (\omega t - \theta) } {2} + \frac{\cos (\omega t + \theta) + \cos (\omega t + \theta - \frac{2 \pi}{3}) + \cos (\omega t + \theta - \frac{4 \pi}{3})}{2} \right)$

The second term in the above expression is the three-phase identity, which sums to zero. So we have

$\Im(\theta) = \frac{3 N_s I}{2} \cos (\omega t - \theta)$

This represents a revolving magnetic field (m.m.f.) rotating around the air gap at an angular speed of $\omega$ rad/sec.