Skip to content
Electrical Engineering

Three-phase systems: star, delta and power

The balanced set and its zero sum, star and delta connections, phase and line voltages, three-phase power, constant instantaneous power and the rotating field.

3 min read 0 comments

Electrical energy is produced, transmitted and distributed almost everywhere in three-phase form: three sinusoidal voltages of the same amplitude and frequency, 120° apart from each other. Household sockets are single phase, but they are just one of the three phases of the neighbourhood grid. In this lesson we see why three-phase pays off, star and delta connections, the relations between phase and line quantities, and power.

The balanced set

A three-phase generator (an alternator with three windings at 120° on the stator) produces three voltages:

    \[ e_1 = \sqrt2 E\cos\omega t,\qquad e_2 = \sqrt2 E\cos(\omega t - 120°),\qquad e_3 = \sqrt2 E\cos(\omega t - 240°) \]

With phasors, setting \alpha = e^{j120°}: \bar E_1 = E, \bar E_2 = \alpha^2 E, \bar E_3 = \alpha E.

Balanced three-phase set: three phasors at 120° and the corresponding three sinusoids
On the left the three voltage phasors, equal in magnitude and 120° apart. On the right the three sinusoids in time: at every instant their sum is zero.

Theorem. The sum of the three voltages of a balanced set is zero at every instant.

Proof. It is enough to show it for the phasors: \bar E_1 + \bar E_2 + \bar E_3 = E(1 + \alpha + \alpha^2). Since \alpha^3 = e^{j360°} = 1 and \alpha \ne 1,

    \[ 1 + \alpha + \alpha^2 = \frac{\alpha^3 - 1}{\alpha - 1} = 0 \]

Geometrically: three equal vectors at 120° form a closed equilateral triangle. ∎

The star connection

The three windings share one end, the star point (connected to the neutral), and the other three ends go to the line conductors. We distinguish:

  • the phase voltages, between each line and neutral: E;
  • the line voltages, between two lines: \bar V_{12} = \bar E_1 - \bar E_2.

Working it out: \bar E_1 - \bar E_2 = E\left(1 - \left(-\tfrac12 - j\tfrac{\sqrt3}{2}\right)\right) = E\left(\tfrac32 + j\tfrac{\sqrt3}{2}\right) = \sqrt3\,E\angle 30°. So

    \[ V_L = \sqrt3\,E \]

This is where the European grid’s pair of numbers comes from: 230 V between phase and neutral, \sqrt3\cdot 230 \approx 400 V between two phases. In star, the line current equals the phase current.

If the load is balanced (three equal impedances), the three currents also form a balanced set and their sum, that is the neutral current, is zero: the neutral could be removed. In low voltage distribution it is needed anyway, because the single-phase loads of homes are never perfectly balanced across the phases.

The delta connection

The three windings (or the three load impedances) are connected in a ring and the three corners go to the lines. Now each element sees the line voltage, and the line current is the difference of two phase currents: with the same calculation,

    \[ I_L = \sqrt3\,I_F \]

A balanced delta load of impedance \bar Z_\Delta is equivalent to a star load with

    \[ \bar Z_Y = \frac{\bar Z_\Delta}{3} \]

This gives star-delta starting of motors: started in star, each winding gets 1/\sqrt3 of the voltage and the line current drops to one third; once running, the motor is switched to delta.

Three-phase power

For a balanced load the power is three times that of one phase. In terms of line quantities, which are the measurable ones, it holds for both star and delta:

    \[ P = \sqrt3\,V_L I_L\cos\varphi,\qquad Q = \sqrt3\,V_L I_L\sin\varphi,\qquad S = \sqrt3\,V_L I_L \]

Example. A 15 kW motor at 400 V with \cos\varphi = 0.85 draws I_L = \frac{15\,000}{\sqrt3\cdot 400\cdot 0.85} \approx 25.5 A per line. Single phase at 230 V, the same power would need 76.7 A.

Why three-phase

Constant instantaneous power. In one phase the instantaneous power is VI\cos\varphi + VI\cos(2\omega t - \varphi - \theta_k), where \theta_k is 0°, 240° and 480° (that is 120°) for the three phases. The three oscillating terms are 120° apart: they form a balanced set, and by the theorem above their sum is zero. What remains is

    \[ p(t) = 3\,VI\cos\varphi \]

which is constant. A three-phase motor gets a constant torque and does not vibrate, unlike a single-phase one, whose power pulses at 100 Hz.

Less copper. For the same transmitted power, voltage between conductors and losses, a three-phase line without neutral uses 75% of the copper of a single-phase one.

Rotating field. Three coils at 120° fed by the balanced set produce a magnetic field of constant magnitude that rotates at grid frequency: it is the principle of the induction motor, the most widespread electric motor in the world, rugged and brushless. Swapping two phases makes the field rotate the other way and the motor reverses.

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x