Tutorial 13 B - Power in 3-phase Circuits

Learning Objective

To consider power transfer in balanced loads;

To work out phase angle when there are reactive elements;

To use the power factor in 3-phase circuits.

Key Questions

What is the total power?

Does the circuit configuration matter?

How do reactive components affect the power?

How is power measured in three phase?

Power in 3-Phase Systems

In a balanced circuit, the total power is simply three times the power in each phase.

In each phase, we need to remember that there is reactance in a load as well as resistance.  This is because most 3-phase loads are motors, which have coils of wire, so quite considerable amounts of inductance.  Power in each phase is the product between the square of the current and the impedance of the coil. The impedance of the phase is given by: Remember this phase vector diagram: As this is a series circuit, the current I is the same in the reactance of the coil and the resistance of the coil.  Both are 90o out of phase.  Therefore the resultant current will have a phase angle of f with the voltage. This is the same as the phase angle f between the impedance and the resistance.

So we can write for each phase: The total power is three times this, so: If the load is purely resistive, cos f = 1 as the phase angle is 0.  This is because for a resistive circuit, voltage and current are in phase.

 Worked Example A 2.2.kW star connected heater is connected to a 415 V 3 phase supply.  What is the current in each phase of the heater? Answer Power in each phase = 2200 ÷ 3 = 733 W   Voltage = 415 ÷ Ö3 = 240 V   Current = 733 ÷ 240 = 3.06 A

In this case the phase angle is zero, as a heater is a resistive load.  So cos f = 1

 Question 9 Three resistive loads, each of 50 W are connected in a star to a 400 V 3-phase supply.  Calculate: a.       The phase voltage; b.      The phase current; c.       The line current; d.       The power.

Star Configuration

If we looked at the power in a star configuration in terms of the line voltage and the line current, we can write: And: So we can substitute into the power equation: And this simplifies to: Delta Configuration

Of course we could put the loads in a delta configuration.

For a delta connection: And: As in the star connection, Power = 3 × power in each phase.  Therefore: This simplifies to: Remember the relationships:

 Quantity Star Delta Voltage  Current  Power  Question 10 Three resistive loads, each of 50 W are connected in a delta to a 400 V 3-phase supply.  Calculate: a.      The phase voltage; b.      The phase current; c.      The line current; d.      The power.

If we look at the answers to the two questions, we see that the power in the delta configuration is three times the power in the star configuration.  So the delta configuration is very good for maximum power transfer.

### Reactive Components and 3-Phase Power Transfer

In the previous section, we mention cos f in the formula, but since we were dealing with resistive components only, we said that the phase angle was 0.  Therefore the cos f term or the power factor was 1.

When we introduce reactive components the cos f term is no longer 1, but a fraction of 1.  It can never be more than 1 for these reasons:

·         Cos f can never be greater than 1;

·         If the power factor were greater than 1, we would be creating energy against all the rules of Physics.

Most devices that use 3-phase power have either a motor or a transformer in them.  Both of these have lots of coils of wire and big chunks of magnetic material in them.  Therefore there are big magnetic fields, leading to measurable values of reactance.  So we need to know what the reactances are at 50 Hz (60 Hz in USA).  In theory it is possible to have other frequencies in a 3-phase supply, but these would only be found in a specialist application.

It is possible to have capacitors as well as inductors, but in supply systems, the vast majority of reactive elements are inductors.  So we will stick to motors for the time being.

We can model a motor as an inductor in series with a resistor.  We know that the reactance of the inductor and the resistance make a vector sum to produce the impedance. Notice that the inductor symbol shows that there is an iron core.  The iron core intensifies the magnetic field. Motors with air-cored windings would be pretty useless.

So, going back to stuff we did before, we can draw the phase vector diagram: And we can easily see that: And: So, in a three phase motor the true power (in watts) is given by: Since the impedance is greater than the resistance, the product between the line voltage and the line current is greater.  This is the apparent power. In this case the product between volts and amps is not watts.  The apparent power is measured in volt-amperes (VA) and has the physics code S.

The apparent power (in volt-amperes) is given by the equation: It doesn’t take a genius to see that: So the power factor is the ratio of P to S.

There is also the reactive power, Q, which is measured in reactive volt-amperes (var): Sin f is worked out by: 3-phase motors are designed to be balanced; in other words the windings are identical.  It is reasonable to assume that this is the case for any calculations you have to do.  If they are unbalanced, there is clearly something wrong.  In these notes, all situations will be balanced.

 Worked Example A 415 V 50 Hz 3-phase electric motor has coils that have a resistance of 20 ohms and inductance of 0.08 H.  It is wired in a star configuration.  Calculate: (a)    The impedance of each coil; (b)   The current; (c)    The power factor; (d)   The true power; (e)    The apparent power. Answer (a)   XL =  2 × p × 50 × 0.08 = 25.1 W        Z2 =25.12 + 202 = 1032 Þ Z = 32.1 W   (b)   IL = IP = VP/Z = 240 ÷ 32.1 = 7.47 A   (c)    cos f = R/Z = 20 ÷ 32.1 = 0.623   (d)   P = Ö3 × 415 × 7.47 × 0.623 = 3345 W   (e)    S = Ö3 × 415 × 7.47 = 5370 VA

Now try one for yourself.  Same motor, but wired in delta configuration.

 Question 11 A 415 V 50 Hz 3-phase electric motor has coils that have a resistance of 20 ohms and inductance of 0.08 H.  It is wired in a delta configuration.  Calculate: a.      The impedance of each coil; b.      The current; c.      The power factor; d.      The true power; e.      The apparent power.

### Measurement of Power

Measurement of power in three-phase systems can be carried out using watt-meters. These are combined voltmeters and ammeters.  The diagram shows the idea of how the meter works: The current flows through a coil which acts as an electromagnet.  The higher the current, the stronger the electromagnet.  The voltage is measured using the coil that is attached to the pointer.  The higher voltage the more the coil turns.  The instrument is calibrated using the voltage and the current, and the scale displays power in watts.  Note that this diagram is for a DC power meter.  For AC, a diode needs to be placed in the circuit.

Watt-meters can be used in different ways:

1.      A single watt-meter is placed in one of the lines: Suggest a formula to work out the power.  What do you need to assume?

2.      Two watt-meters are used as below: In this case the total power is the sum of the two meter readings. You can work out the phase angle by this formula: If the loads are balanced, what is f ?  What is the power factor?

3.      Three watt-meters are used. In this case the neutral line has to be used.  The system can be balanced or unbalanced.  The power is simply the sum of the three meters: A balanced delta connected load has a line voltage of 400 V and a line current of 8.0 A.  The power factor is 0.94.  What is the total power dissipated by the load?  What is the apparent power?

### Advantages of 3-phase systems

·         For a given amount of power, the three phase system needs lighter conductors.  This saves on the costs of expensive metals like copper.

·         Two voltages are available rather than one (e.g. 415 V and 240 V).

·         Three phase motors a very robust, inexpensive.  They are self-starting.  (A single phase AC motor may start going the wrong way round, or do nothing.  A mechanical system can be used, or, much more commonly a capacitor to alter the phase.  3-phase motors don’t need these.)

·         Three-phase motors are smaller, provide a steadier output, and require less maintenance.

·         Three phase induction motors are much quieter than universal motors, and are controllable.

The picture shows a three-phase traction motor used on an electric train.  It is simple and relatively small for its power. (But if you drop it on your foot, it will still bring tears to your eyes.) Image from

By comparison a DC traction motor is shown. You can see how heavy and bulky they are.

Photo from Wikimedia Commons  (PtrS)