Three-Phase Power and the Single-Line Diagram

Chapter 1 followed a single-phase load and found something uncomfortable. With \(v = \sqrt2\,V\cos\omega t\) across an impedance carrying \(i = \sqrt2\,I\cos(\omega t - \phi)\), the instantaneous power was

The first term is the useful average; the second is an oscillation at twice supply frequency whose amplitude \(VI\) is larger than the average \(VI\cos\phi\) whenever the power factor is less than unity. A single-phase machine therefore receives its energy in a series of 100 pushes per second on a 50 Hz system, with a backward pull in between whenever \(\phi \ne 0\). The shaft must carry that torque ripple, the foundations must absorb it, and the machine cannot start without some auxiliary trick to produce a rotating field, because one alternating winding produces only a pulsating one.

Now suppose that instead of one winding we place three identical windings on the stator, spaced \(120^\circ\) apart in space, and drive them with three voltages of equal magnitude spaced \(120^\circ\) apart in time. Two things happen at once. The three currents produce a magnetic field of constant magnitude that rotates at synchronous speed — the induction motor and the synchronous machine both exist because of this. And the three pulsating powers, each peaking at a different instant, add to a total that does not pulsate at all. That second claim is proved in Section 3-5; it is the reason the whole industry is three-phase.

There is a third reason, and it is about money. Chapter 2 compared systems by the volume of conductor metal they need, holding the delivered power, the transmission loss and the length fixed. Apply that same accounting to the number of phases, comparing on the basis of the same voltage between conductors — the quantity the insulation must withstand.

Three conductors carrying \(1/\sqrt3\) of the single-phase current need only three-quarters of the metal, because loss scales with the square of the current while the number of conductors grows only from two to three. A fourth wire is often run as a neutral, but Section 3-2 shows it carries no current in a balanced system, so it is made of half the cross-section of a phase conductor and adds little.

A three-phase source is balanced when its three phase voltages have equal magnitude and are displaced by exactly \(120^\circ\). Naming the phases \(a\), \(b\), \(c\) (equivalently R, Y, B) and taking \(V_{an}\) as the reference,

The order in which the three reach their positive maxima is the phase sequence. The set above is \(a\)–\(b\)–\(c\): \(a\) leads \(b\) by \(120^\circ\), \(b\) leads \(c\) by \(120^\circ\). Reversing any two of the three connections gives the \(a\)–\(c\)–\(b\) sequence, which reverses the direction of the rotating field and therefore the direction in which every motor on the system turns. Sequence is a physical property of the supply, not a labelling convention, and it survives into Chapter 22 as the positive and negative sequence components.

Repeated \(120^\circ\) rotations occur so often that they deserve a symbol. Define the operator \(a\) as the unit phasor that advances an angle by \(120^\circ\):

With it the balanced set is written \(\mathbf{V}_{an} = V\), \(\mathbf{V}_{bn} = a^2 V\), \(\mathbf{V}_{cn} = aV\). Adding the three complex numbers, the real parts give \(1 - \tfrac12 - \tfrac12 = 0\) and the imaginary parts give \(0 + \tfrac{\sqrt3}{2} - \tfrac{\sqrt3}{2} = 0\):

Balance is an assumption, and like every assumption in this book it has to be paid for eventually. It holds well on transmission, where lines are transposed and loads are large and aggregated; it holds poorly on a distribution feeder with unequal single-phase connections, and it fails completely during an unsymmetrical fault. Chapters 22 to 24 build the machinery that handles the failure. Until then, balance is what makes everything in this chapter simple.

Four quantities have to be kept apart, and most three-phase errors come from confusing two of them.

• Phase voltage \(V_{ph}\) — measured across one winding of the source or one branch of the load.
• Line voltage \(V_L\) — measured between any two line conductors.
• Phase current \(I_{ph}\) — flowing through one winding or one load branch.
• Line current \(I_L\) — flowing in a line conductor between source and load.

In the star (\(Y\)) connection the three windings are joined at a common point \(n\), the star point, and the free ends become the three lines. Each winding sits between a line and the star point, so the line current is the phase current. The line voltage, on the other hand, is a difference of two phase voltages, and that difference is where the \(\sqrt3\) comes from:

The same result follows from the triangle of the two phasors without any complex algebra. \(\mathbf{V}_{an}\) and \(-\mathbf{V}_{bn}\) have equal length \(V\) and include an angle of \(60^\circ\), so by the cosine rule

Repeating for the other two pairs gives \(\mathbf{V}_{bc} = \sqrt3\,V\angle{-90^\circ}\) and \(\mathbf{V}_{ca} = \sqrt3\,V\angle 150^\circ\): the three line voltages form a balanced set of their own, \(\sqrt3\) times as large as the phase voltages and leading them by \(30^\circ\). That \(30^\circ\) is not decoration. It reappears in the star–delta transformer, which shifts the phase of every voltage passing through it by \(30^\circ\), and in the two-wattmeter method of Section 3-7 where it is the entire origin of the \(\cos(30^\circ \pm \phi)\) readings.

In the delta (\(\Delta\), or mesh) connection the three windings are joined end to end in a closed loop and the lines are taken from the three junctions. Now each winding is connected directly between two lines, so the phase voltage is the line voltage. It is the current that must be resolved: the current leaving junction \(a\) is the difference of the two branch currents meeting there.

Before going further, one point about the closed loop. Three windings in series around a mesh look like an invitation to a large circulating current, and would be one if the three generated voltages did not sum to zero. They do, by the identity of Section 3-2, so the net driving voltage around the loop is zero and no circulating current flows — provided the set is balanced and free of triplen harmonics. Third-harmonic voltages are all in phase with one another and do not cancel around the mesh, which is exactly why a delta winding is deliberately provided on many transformers to give those harmonics a path and keep them out of the lines.

Applying KCL at junction \(a\), with branch currents \(\mathbf{I}_{ab} = I\angle 0^\circ\), \(\mathbf{I}_{bc} = a^2 I\) and \(\mathbf{I}_{ca} = aI\):

The structure of the two derivations is identical — a difference of two phasors \(120^\circ\) apart — which is why the same \(\sqrt3\) and the same \(30^\circ\) appear, with the sign of the shift reversed. In star the voltages combine and lead; in delta the currents combine and lag.

Start with the promise made in Section 3-1. Let each phase carry \(v_k\) and \(i_k\) as in Chapter 1, with the phases displaced by \(120^\circ\) and the same impedance angle \(\phi\) in every phase. Each phase delivers the single-phase result, shifted:

The \(120^\circ\) displacement of the phase quantities becomes a \(240^\circ\) displacement of the double-frequency terms, because the argument contains \(2\omega t\). But three cosines whose arguments differ by \(240^\circ\) are still three unit phasors spaced \(120^\circ\) apart on the double-frequency plane, and by the identity \(1 + a + a^2 = 0\) they sum to zero at every instant. The oscillating parts cancel and only the constant parts survive:

For steady-state work the phasor form is what gets used. Chapter 1 defined complex power per phase as \(\mathbf{S} = \mathbf{V}\mathbf{I}^{*}\). Adding three identical contributions,

Two habits follow from this. First, every three-phase rating quoted in this book — a 30 MVA generator, a 400 kV line, a 2 MVA transformer — is the three-phase total, and every voltage is the line-to-line value, unless it is explicitly labelled otherwise. Second, when a load is described as "50 kW at 0.8 power factor lagging" the reactive demand follows immediately as \(Q = P\tan\phi = 50 \times 0.75 = 37.5\) kvar, exactly as in Chapter 1; three phases changes the arithmetic of the currents, not the shape of the power triangle.

Everything so far points to one conclusion: in a balanced system the three phases are copies of one another, differing only by a known \(120^\circ\) rotation. Solving all three is therefore three times as much work as it needs to be. Per-phase analysis makes the reduction formal, and it rests on two steps.

Step one: connect the neutrals. Take a star-connected source feeding a star-connected balanced load through three identical line impedances. The star point of the source and the star point of the load are both at the potential of the average of the three phase voltages, which by \(1+a+a^2=0\) is the same at both ends. Two points at identical potential may be joined by a wire of zero impedance without changing anything in the circuit, and the wire will carry \(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0\). With that wire in place, phase \(a\) forms a complete loop on its own: source, line impedance, load impedance, return through the neutral. The three phases are decoupled.

Step two: convert every delta to its star equivalent. A delta load has no star point to connect, so it must first be replaced by the star that draws exactly the same line currents from the same line voltages. The general transformation, applied to the three branches of a mesh, is

The factor of three can also be read off the power. A delta branch sees the full line voltage \(V_L\) while a star branch sees \(V_L/\sqrt3\), so for the same impedance the delta branch dissipates three times as much; to draw the same power the delta impedance must be three times as large. This is the reason a star–delta starter reduces motor starting current to a third of its direct-on-line value: reconnecting the same windings in star triples their effective per-phase impedance as seen from the line.

This is the single most important simplification in power system analysis, and it is worth being explicit about what it costs. Per-phase analysis is exact for a balanced system and meaningless for an unbalanced one. Every load-flow calculation in Part 4, every symmetrical fault calculation in Chapter 21 and every stability study in Part 6 is a per-phase calculation. When the assumption breaks — a line-to-ground fault, one open conductor — Chapter 22 restores it by a change of variables that turns one unbalanced three-phase problem into three balanced ones.

A wattmeter reads the average of the product of the voltage across its pressure coil and the current through its current coil. To measure three-phase power the obvious approach is one meter per phase, but that requires access to the star point, which a delta load does not have and a sealed motor terminal box rarely offers. Blondel's theorem says fewer meters are needed anyway.

The total instantaneous power in a three-wire system is \(p = v_a i_a + v_b i_b + v_c i_c\), with all voltages referred to an arbitrary common point. With no neutral connection, \(i_a + i_b + i_c = 0\), so \(i_b = -(i_a + i_c)\). Substituting:

The total power is the sum of just two products. One wattmeter with its current coil in line \(a\) and its pressure coil across \(a\)–\(b\), a second with its current coil in line \(c\) and its pressure coil across \(c\)–\(b\), together measure the whole thing — for any load, balanced or not, star or delta. In general a system of \(n\) wires needs \(n-1\) wattmeters, which is Blondel's theorem.

Now specialise to a balanced load with lagging angle \(\phi\), abc sequence, \(V_{an}=V\angle0^\circ\). Then \(\mathbf{I}_a = I\angle{-\phi}\), \(\mathbf{I}_c = I\angle(120^\circ-\phi)\), \(\mathbf{V}_{ab} = \sqrt3 V\angle 30^\circ\) and \(\mathbf{V}_{cb} = -\mathbf{V}_{bc} = \sqrt3 V\angle 90^\circ\). Each meter reads the product of magnitudes times the cosine of the angle between its own voltage and its own current:

Adding and subtracting, with \(\cos X + \cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}\):

The readings themselves are diagnostic. At unity power factor \(W_1 = W_2 = V_LI_L\cos30^\circ\): the meters agree. At \(\phi = 60^\circ\) (power factor 0.5), \(\cos(30^\circ+60^\circ) = 0\) and one meter reads exactly zero while the other carries all the power. Beyond \(\phi = 60^\circ\) the cosine goes negative, the pointer tries to go backwards, and the meter must be reversed and its reading subtracted. A student who forgets that subtraction on a low-power-factor load will report roughly twice the true power.

Section 3-6 established that a balanced network is solved one phase at a time. Draw that per-phase circuit for a real system — a generator, its step-up transformer, a transmission line, a step-down transformer, a load — and it is a chain of impedances between a source and a return. Two observations then finish the job of simplification.

The first is that the neutral return carries no current and drops no voltage, so drawing it adds nothing; it is understood to exist and is left out. The second is that the remaining single conductor stands for all three phases at once, since the other two differ only by \(\pm120^\circ\). What is left is one line per circuit, with a standard symbol wherever a piece of plant sits. That is the single-line diagram, or one-line diagram: the working drawing of power engineering.

The three-line diagram — every conductor drawn — still has its uses in protection and metering work, where the phases genuinely differ: current transformer connections, relay wiring, the two-wattmeter arrangement of Section 3-7. For system studies it would be three times the ink for no extra information.

A single-line diagram is drawn at whatever level of detail the study demands. A planning diagram may show a whole regional grid with each substation as one bus. An operational diagram of a single substation shows every breaker, isolator, current transformer and earth switch. The same conventions apply at both extremes.

The symbols are largely standardised (IEC 60617 and IEEE 315 agree on the important ones). The table lists those used throughout this book, together with what has to be written beside each.

Three conventions matter more than the shapes. Every voltage written on a single-line diagram is a line-to-line value, and every MVA is a three-phase total, following Section 3-5. Every impedance is given as a percentage or a per-unit value on the plant's own rating — a transformer marked "20 MVA, 33/11 kV, 8%" has a reactance of 8% on a 20 MVA, 33 kV base, and Chapter 4 exists to convert that onto a common base before it can be added to anything else. The transformer connection and earthing are shown even though the neutral wire is not drawn, because they decide what happens in an unsymmetrical fault.

Read the diagram above from left to right and notice what it does not tell you. The generator reactance is 0.20 per unit on 30 MVA; the transformer reactance is 0.10 per unit on 35 MVA; the line reactance is 100 ohms at 132 kV. Three numbers on three different bases, one of them not even normalised. They cannot be added, cannot be put in series, cannot be used until they are expressed against a single common reference. Producing that common reference is what Chapter 4 does, and it is the last piece of preparation before the network itself can be analysed.

Assume balanced conditions, abc phase sequence and 50 Hz throughout. Quote every voltage as a line value and every power as a three-phase total unless the question says otherwise.