Three-phase power is a common method of electric power generation, transmission, and distribution with alternating current which has three phases of phase difference – 120 degree.

It is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power.

In order to understand the method of generating a three-phase supply, let us initially consider the generation of single phase electric power. Alternating voltage is provided by an a.c. generator, more commonly called an alternator.

We have written a separate article on Electromagnetic Induction to explain the basics of electric power generation. It is clearly explained that when a coil of wire, wound on to a rectangular former, is rotated in a magnetic field, an alternating (sinusoidal) voltage is induced into the coil.

You should also be aware that for electromagnetic induction to take place, it is the relative movement between conductor and magnetic flux that matters. Thus, it matters not whether the field is static and the conductor moves, or vice versa. I suggest reading our Electromagnetic Induction article for a basic understanding. Also, read how to save money on electricity.

To understand the basic operation of three phase power generation, consider an alternator. For a practical alternator, it is found to be more convenient to rotate the magnetic field, and to keep the conductors (coil or winding) stationery. In any rotating a.c. machine, the rotating part is called the rotor, and the stationary part is called the stator.

Thus, in an alternator, the field system is contained in the rotor. The winding in which the emf is generated is contained in the stator. Following are the reasons for stator and rotor arrangements,

- When large voltages are generated, heavy insulation is necessary. If this extra mass has to be rotated, the driving device has to develop extra power.

This will then reduce the overall efficiency of the machine. Incorporating the winding in the stator allows the insulation to be as heavy as necessary, without adversely affecting the efficiency. - The contact resistance between the brushes and slip-rings is very small. However, if the alternator provided high current output (in hundreds of ampere), the { I }^{ 2 }R power loss would be significant.

The d.c. current (excitation current) for the field system is normally only a few amps or tens of amps. Thus, supplying the field current via the slip-rings produces minimal power loss. The stator winding is simply connected to terminals on the outside of the stator casing. - For very small alternators, the rotor would contain permanent magnets to provide the rotating field system. This then altogether eliminates the need for any slip-rings. This arrangement is referred to as a brushless machine.

The basic construction for a single-phase alternator is illustrated in following Figure. The conductors of the stator winding are placed in slots around the inner periphery of the stator. The two ends of this winding are then led out to a terminal block on the casing.

The rotor winding is also mounted in slots, around the circumference of the rotor. This figure is used to illustrate the principle. A practical machine would have many more conductors and slots.

Since the conductors of the stator winding are spread around the whole of the slots, it is known as a distributed winding. As the rotor field sweeps past these conductors an emf is induced in each of them in turn. These individual emfs reach their maximum values only at the instant that the rotating field ‘cuts’ them at 90°.

Also, since the slots have an angular displacement between them, then the conductor emfs will be out of phase with each other by this same angle. In above figure, there are a total of twelve conductors, so this phase difference must be 30°.

The total stator winding emf will therefore not be the arithmetic sum of the conductor emfs, but will be the phasor sum, as shown in following figure.

The ratio of the phasor sum to the arithmetic sum is called the distribution factor. For the case shown (a fully distributed winding) the distribution factor is 0.644.

Now, if all of the stator conductors could be placed into a single pair of slots, opposite to each other, then the induced emfs would all be in phase. Hence the phasor and arithmetic sums would be the same, yielding a distribution factor of unity. This is not a practical solution.

However, if the conductors are concentrated so as to occupy only one third of the available stator slots, then the distribution factor becomes 0.966. In a practical single-phase alternator, the stator winding is distributed over two thirds of the slots.

Let us return to the option of using only one-third of the slots. We will now have the space to put two more identical windings into the stator. Each of the three windings could be kept electrically separate, with their own pairs of terminals. We would then have three separate single-phase alternators in the same space as the original.

Each of these would also have a good distribution factor of 0.966. The three winding emfs will of course be mutually out of phase with each other by 120°, since each whole winding will occupy 120° of stator space. What we now have is the basis of a three-phase alternator.

When we talk about three phase power generation, the term three-phase alternator is in some ways slightly misleading. What we have, in effect, are three identical single-phase alternators contained in the one three phase power generator. The three stator windings are brought out to their own separate pairs of terminals on the stator casing.

These stator windings are referred to as phase windings or phases. They are identified by the colors red, yellow and blue. Thus we have the **red, yellow and blue phases.** The circuit representation for the stator winding of a (machine) three phase power generator is shown in following figure.

In this figure, the three phase windings are shown connected, each one to its own separate load. This arrangement is known as a three-phase, six-wire system. However, three-phase alternators are rarely connected in this way.

we need to select a reference phasor. By convention, the reference is always taken to be the red phase voltage. The yellow phase lags the red by 120°, and the blue lags the red by 240° (or, if you prefer, leads the red by 120°).

The windings are arranged so that when the rotor is driven in the chosen direction, the phase sequence is red, yellow, blue. If, for any reason, the rotor was driven in the opposite direction, then the phase sequence would be reversed, i.e. red, blue, yellow. We shall assume that the normal sequence of R, Y, B applies at all times.

It may be seen from the waveform diagram that at any point along the horizontal axis, the sum of the three voltages is zero. This fact becomes even more apparent if the phasor diagram is redrawn as in following figure. In this diagram, the three phasors have been treated as any other vector quantity.

The sum of the vectors may be determined by drawing them to scale, as in the figure following , and the resultant found by measuring the distance and angle from the beginning point of the first vector to the arrowhead of the last one. If, as in following figure, the first and last vectors meet in a closed figure, the resultant must be zero.

It is not necessary to have six wires from the three phase windings to the three loads, provided there is a common ‘ return ’ line. Each winding will have a ‘ start ’ (S) and a ‘ finish ’ (F) end. The common connection mentioned above is achieved by connecting the corresponding ends of the three phases together.

For example, either the three ‘ F ’ ends or the three ‘ S ’ ends are made common. This form of connection is shown in following figure, and is known as a star or Y connection. With the resulting 4-wire system, the three loads also are connected in star configuration. The three outer wires are called the lines, and the common wire in the center is called the neutral.

If the three loads were identical in every way (same impedance and phase angle), then the currents flowing in the three lines would be identical. If the waveform and/or phasor diagrams for these currents were drawn, they would be identical. These three currents meet at the star point of the load.

The resultant current returning down the neutral wire would therefore be zero. The load in this case is known as a balanced load, and the neutral is not strictly necessary. However, it is difficult, in practice, to ensure that each of the three loads are exactly balanced. For this reason, the neutral is left in place.

Also, since it has to carry only the relatively small ‘ out-of-balance ’ current, it is made half the cross-sectional area of the lines. Let us now consider one of the advantages of this system compared with both a single-phase system and the three-phase 6-wire system.

Suppose that three identical loads are to be supplied with 200 A each. The two lines from a single-phase alternator would have to carry the total 600 A required. If a 3-phase, 6-wire system was used, then each line would have to carry only 200 A.

Thus, the conductor csa would only need to be 1/3 for the single-phase system, but of course, being six lines would entail using the same total amount of conductor material. If a 4-wire, 3-phase system is used there will be saving on conductor costs in the ratio of 3.5:6 (the 0.5 is due to the neutral).

If the power has to be sent over long transmission lines, such as the National Grid System, then the 3-phase, 4-wire system yields an enormous saving in cable costs. This is one of the reasons why the power generating companies use three-phase, star-connected generators to supply the grid system.

Consider the following figure, which represents the stator of a 3-phase alternator connected to a 3-phase balanced load. The voltage generated by each of the three phases is developed between the appropriate line and the neutral.

These are called the phase voltages and may be referred to in general terms as {V}_{ph } , or specifically as {V}_{RN} , {V}_{YN} and {V}_{BN} respectively. However, there will also be a difference in potential between any pair of lines. This is called a line voltage, which may be generally referred to as {V}_{L} , or specifi cally as {V}_{RY} , {V}_{YB} arid {V}_{BR} respectively.

A line voltage is the phasor difference between the appropriate pair of phase voltages. Thus, { V }_{ RY } is the phasor difference between { V }_{ RN } and { V }_{ YN }.

In terms of a phasor diagram, the simplest way to subtract one phasor from another is to reverse one of them, and then find the resulting phasor sum. This is, mathematically, the same process as saying that a-b=a+(-b). The corresponding phasor diagram is shown below.

Note: If {V}_{YN} reversed, it is denoted either as {-V}_{YN} or as {V}_{NY} . We shall use the first of these.

The phasor difference between {V}_{RN} and {V}_{YN} is simply the phasor sum of {V}_{RN} + ({-V}_{YN}). Geometrically this is obtained by completing the parallelogram as shown below.

This parallelogram consists of two isosceles triangles, such as OCA. Another property of a parallelogram is that its diagonals bisect each other at right angles. Thus, triangle OCA consists of two equal right-angled triangles, OAB and ABC. This is illustrated below. Since triangle OAB is a 30°, 60°, 90° triangle, then the ratios of its sides AB:OA:OB will be 1:2:\sqrt { 3 } respectively.

Hence, \frac { OB }{ OA } =\frac { \sqrt { 3 } }{ 2 }

So, OB=\frac { \sqrt { 3 } .OA }{ 2 }

But, OC=2\times OB=\sqrt { 3 } .OA

and since OC={V}_{RY}\quad and\quad OA={V}_{RN}

then, {V}_{RY}=\sqrt {3} {V}_{RN}

Using the same technique, it can be shown that:

{V}_{YB}=\sqrt {3} {V}_{YN} and {V}_{BR}=\sqrt {3} {V}_{BN}

Thus, in star configuration {V}_{L}=\sqrt {3} {V}_{ph}

Hence, in star configuration {I}_{L}=\sqrt {3} {I}_{ph}

We now have another advantage of a 3-phase system compared with single-phase. The star-connected system provides two alternative voltage outputs from a single machine. For this reason, the stators of all alternators used in electricity power stations are connected in star configuration. These machines normally generate a line voltage of about 25 kV.

By means of transformers, this voltage is stepped up to 400 kV for long distance transmission over the National Grid. For more localized distribution, transformers are used to step down the line voltage to 132 kV, 33 kV, 11 kV, and 415 V. The last three of these voltages are supplied to various industrial users.

The phase voltage derived from the 415 V lines is 240 V, and is used to supply both commercial premises and households.

EXAMPLE 1:

A 415 V, 50 Hz, 3-phase supply is connected to a star-connected balanced load. Each phase of the load consists of a resistance of 25 Ω and inductance 0.1 H, connected in series. Calculate,

(a) Phase voltage,

(b) The line current drawn from the supply, and

(c) The power dissipated.

(a) Whenever a three-phase supply is specified, the voltage quoted is always the line voltage. Also, since we are dealing with a balanced load, then it is necessary only to calculate values for one phase of the load. The figures for the other two phases and lines will be identical to these.

{V}_{L}={415}; f=50Hz; {R}_{ph}={25 Ω}; {L}_{ph}={0.1H}

(a). { V }_{ ph }=\frac { { V }_{ L } }{ \sqrt { 3 } } =\frac { 415 }{ \sqrt { 3 } }

So, { V }_{ ph }= {240V}

(b). Since it is possible to determine the impedance of a phase of the load, and we now know the phase voltage, then the phase current may be calculated:

{{ X }_{ L }=2πfL } Ohms = 2π × 50 × 0.1

hence, { X }_{ L }=31.42\Omega

{ Z }_{ ph }=\sqrt { { { R }_{ ph } }^{ 2 }+{ { X }_{ L } }^{ 2 } } =\sqrt { { 25 }^{ 2 }+{ 31.42 }^{ 2 } }

{{ Z }_{ ph }=40.15\Omega }

{ I }_{ ph }=\frac { { V }_{ ph } }{ { Z }_{ ph } } = \frac { 240 }{ 40.15 }

{ I }_{ ph }={5.98 A}

In a star-connected circuit, { I }_{ L }={ I }_{ ph }

therefore, { I }_{ ph }={5.98 A}

The power in one phase, { P }_{ ph }={ { I }_{ ph } }^{ 2 }{ R }_{ ph }

{ P }_{ ph }={ 5.98 }^{ 2 }×{ 25 }

{P}_{ph}=893.29 W

and Since, there are three phases, then the total power is:

P=3{{P}_{ph}} =3×893.29 watt

hence, P = 2.68kW

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