Three-phase

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This article deals with the basic mathematics and principles of three-phase electricity. For information on where, how and why three-phase is used, see three-phase electric power.

In electrical engineering, three-phase electric power systems have at least three conductors carrying voltage waveforms that are 2π/3 radians (120°,1/3 of a cycle in-phase) offset in time. In this article angles will be measured in radians except where otherwise stated.

1 Variable setup and basic definitions
2 Balanced loads
- 2.1 Star connected systems with neutral
  - 2.1.1 Constant power transfer
  - 2.1.2 No neutral current
- 2.2 Star connected systems without neutral
3 Unbalanced systems
4 Revolving magnetic field
5 Conversion to other phase systems
6 References
7 See also

One voltage cycle of a three-phase system, labelled 0 to 360° (2 π radians) along the time axis. The plotted line represents the variation of instantaneous voltage (or current) with respect to time. This cycle will repeat 50 or 60 times per second, depending on the power system frequency. The colours of the lines represent the American color code for 120v three-phase. That is black=V_L1 red=V_L2 blue=V_L3

Let $x$ be the instantaneous phase of a signal of frequency $f$ at time $t$ :

$x=2\pi ft\,\!$

Using this, the waveforms for the three phases are

$V_{L1}=V_P\sin x\,\!$

$V_{L2}=V_P\sin \left(x-\frac{2}{3} \pi\right)$

$V_{L3}=V_P\sin \left(x-\frac{4}{3} \pi\right)$

where $V P$ is the peak voltage and the voltages on L1, L2 and L3 are measured relative to the neutral.

Generally, in electric power systems the loads are distributed as evenly as is practical between the phases. It is usual practice to discuss a balanced system first and then describe the effects of unbalanced systems as deviations from the elementary case.

This refers to a system with a resistive load $R$ between each phase and neutral.

An important property of three-phase power is that the power available to a resistive load, $P = V I = \frac{V^2}R$ , is constant at all times.

$P_{L1}=\frac{V_{L1}^{2}}{R}\,\!$

$P_{L2}=\frac{V_{L2}^{2}}{R}\,\!$

$P_{L3}=\frac{V_{L3}^{2}}{R}\,\!$

$P_{TOT}=P_{L1}+P_{L2}+P_{L3}\,\!$

To simplify the math, we define a nondimensionalized power for intermediate calculations, $p = \frac{P_{TOT} R}{V_P^2}$

$p=\sin^{2} x+\sin^{2} (x-\frac{2}{3} \pi)+\sin^{2} (x-\frac{4}{3} \pi)$

Using angle subtraction formulae

$p=\sin^{2} x+\left(\sin x\cos\left(\frac{2}{3} \pi\right)-\cos x\sin\left(\frac{2}{3} \pi\right)\right)^{2}+\left(\sin x\cos\left(\frac{4}{3} \pi\right)-\cos x\sin\left(\frac{4}{3} \pi\right)\right)^{2}$

$p=\sin^{2} x+\left(-\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\right)^{2}+\left(-\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)^{2}$

$p=\sin^{2} x+\frac{1}{4}\sin^{2} x+\frac{\sqrt{3}}{2}\sin x\cos x +\frac{3}{4}\cos^{2} x+\frac{1}{4}\sin^{2} x-\frac{\sqrt{3}}{2}\sin x\cos x +\frac{3}{4}\cos^{2} x$

$p=\frac{6}{4}\sin^{2} x+\frac{6}{4}\cos^{2} x$

$p=\frac{3}{2}(\sin^{2} x+\cos^{2} x)$

Using the Pythagorean trigonometric identity

$p=\frac{3}{2}$

Hence (substituting back): $P_{TOT}=\frac{3 V_P^2}{2R}$

since we have eliminated $x$ we can see that the total power does not vary with time. This is essential for keeping large generators and motors running smoothly.

For the case of equal loads on each of three phases, no net current flows in the neutral. The neutral current is the sum of the phase current.

$I_{L1}=\frac{V_{L1}}{R}\,\!$

$I_{L2}=\frac{V_{L2}}{R}\,\!$

$I_{L3}=\frac{V_{L3}}{R}\,\!$

$I_{N}=I_{L1}+I_{L2}+I_{L3}\,\!$

We define a nondimensionalized current, $i=\frac{I_{N}R}{V_P}$ .

$i=\sin x+\sin \left(x-\frac{2}{3} \pi\right)+\sin \left(x-\frac{4}{3} \pi\right)$

Using angle subtraction formulae

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

$i=\sin x-\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x$

$i=0\,\!$

Hence also $I_N=0\,\!$

Since we have shown that the neutral current is zero we can see that removing the neutral core will have no effect on the circuit, provided the system is balanced. In reality such connections are generally used only when the load on the three phases is part of the same piece of equipment (for example a three-phase motor), as otherwise switching loads and slight imbalances would cause large voltage fluctuations.

Practical systems rarely have perfectly balanced loads, currents, voltages or impedances in all three phases. The analysis of unbalanced cases is greatly simplified by the use of the techniques of symmetrical components. An unbalanced system is analyzed as the superposition of three balanced systems, each with the positive, negative or zero sequence of balanced voltages.

The rotating magnetic field of a three-phase motor.

Any polyphase system, by virtue of the time displacement of the currents in the phases, makes it possible to easily generate a magnetic field that revolves at the line frequency. Such a revolving magnetic field makes polyphase induction motors possible. Indeed, where induction motors must run on single-phase power (such as is usually distributed in homes), the motor must contain some mechanism to produce a revolving field, otherwise the motor cannot generate any stand-still torque and will not start. The field produced by a single-phase winding can provide energy to a motor already rotating, but without auxiliary mechanisms the motor will not accelerate from a stop when energized.

Provided two voltage waveforms have at least some relative displacement on the time axis, other than a multiple of a half-cycle, any other polyphase set of voltages can be obtained by an array of passive transformers. Such arrays will evenly balance the polyphase load between the phases of the source system. For example, balanced two-phase power can be obtained from a three-phase network by using two specially constructed transformers, with taps at 50% and 86.6% of the primary voltage. This Scott T connection produces a true two-phase system with 90° time difference between the phases. Another example is the generation of higher-phase-order systems for large rectifier systems, to produce a smoother DC output and to reduce the harmonic currents in the supply.

When three-phase is needed but only single-phase is readily available from the utility company a phase converter can be used to generate three-phase power from the single phase supply.

Stevenson, William D., Jr. (1975) Elements of Power Systems Analysis, McGraw-Hill electrical and electronic engineering series, 3^rd ed., New York: McGraw Hill, ISBN 0-07-061285-4

Retrieved from "http://en.wikipedia.org/wiki/Three-phase"

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