CHAPTER 4 POWER FOR INDUCTION MOTORS

4.1       OUTPUT AND LOSSES

A motor is installed for the purpose of driving some item of plant such as a pump, compressor or fan, and it imparts mechanical power to its driven load.  This is true power, and the motor takes it in electrically in the form of active power (i.e. kilowatts) from the main electrical system.

Most of the input electrical power is passed on to the driven load, but some of it is lost within the motor itself.  These losses can be listed as follows:

·         Friction

·         Windage

·         Iron losses

·         Copper losses due to load

·         Copper losses independent of load.

Since the speed of the motor is virtually constant, the first two (friction and windage) may be considered constant at all loads.  ‘Iron losses’, which are due to the imperfect magnetisation of the iron, depend on the magnetising flux and so on the applied voltage; as this is constant, the iron losses may also be regarded as constant at all loads.

FIGURE 4.1

MOTOR POWER AND LOSSES

The principal current drawn by the machine is the active current which powers the load and which therefore varies as the load varies.  The heating loss due to any current is I²R, therefore the ‘copper losses’ due to the load current are variable and are roughly proportional to the square of the load.  These variable losses occur in both the stator windings and the rotor bars.

Finally, even when there is no mechanical load on the motor, it is still drawing some active power to supply the losses; it is also drawing magnetising (reactive) power. Both these result in an appreciable no-load current, which itself gives rise to some I²R losses.  These are the ‘copper losses independent of load’ and are also almost constant.

The losses for which a motor must draw active power, in addition to the power required to supply the load and how they vary with load, are shown in Figure 4.1.  There is a constant element which is always present, even at no-load, and a variable element which is roughly proportional to the square of the load.

Suppose the input power and losses (both in kW_e) are plotted to the same scale against the mechanical output load (in kW_m).  Draw the line representing the input power actually needed for the load above the zero line; it will be a straight line OP at 45° through O.  Draw the loss curves below the zero line.  They will consist of two parts: a constant part representing friction, windage, iron losses and copper losses independent of load; this is the horizontal line TQ.  The other part representing copper losses depending on load will be a ‘square law’ curve which, when added to the constant loss line TQ, gives the total loss curve TR.  The point S represents full-load mechanical output.

Then at full-load the part PS is the mechanical output power, SQ the constant losses, and QR the variable copper losses.  The line PR is then the total power required to meet the load and to supply all losses - that is, the total input power in kW_e.

                                              

                                                               

and this is clearly less than unity.

At some other load S’ the efficiency is , and it can be seen that this is less than the full-load efficiency .  This shows that the pattern of losses causes the efficiency to fall with reduction of load until, at no-load where the output is nil, the efficiency becomes zero.

All losses result ultimately in heating, and the motor must be designed to dissipate heat at the maximum rate at which it can be produced without overheating the machine.  For small motors their cases can usually do this, but larger motors normally need a fan.  The largest may require cooling through an air/water heat exchanger.  Cooling is further discussed in Chapter 3.

4.2       MAGNETISATION

As explained in Chapter 2, the operation of an induction motor requires a rotating magnetic field - that is to say, it needs to be magnetised.  In the manual ‘Fundamentals of Electricity 1’ it was stated that magnetisation causes inductance in a circuit, and in ‘Fundamentals of Electricity 2’ it was shown that inductance gives rise to ‘inductive reactance’ in an a.c. circuit. This causes the magnetising current to lag 90° on the applied voltage (neglecting resistance). 

Whereas active current creates a demand for active power (watts), reactive current calls for reactive power, or ‘vars’.

An induction motor therefore demands active power to meet its mechanical output load and to supply its losses, but it also demands reactive power to supply its magnetisation.  Whereas active power will vary with the loading, as shown in Figure 4.1, the magnetisation remains virtually constant and independent of the loading, since the rotating field is present whether the motor is working or running light.  Therefore the reactive (var) loading can be regarded as constant while the active (watt) loading varies with loading.

Figure 4.2 shows pictorially how the active and reactive power to a motor is generated, distributed and consumed.  Active, or true, power originates from the generator’s prime mover as mechanical output from the engine.  On the other hand reactive power emanates from the generator’s excitation system through its main field.

In Figure 4.2 the active power route is indicated in red, and the reactive power route in blue.  Both powers come from the generator itself through a common cable.  At the switchboard they give a common current indication on the generator ammeter, and both combine to give a power-factor indication.  Then they separate to give independent wattmeter and varmeter indications.  They recombine to feed the motor through a common cable.  At the motor the reactive power is used to magnetise the machine, and the active power supplies the (variable) mechanical load and also the losses.

FIGURE 4.2

ACTIVE AND REACTIVE MOTOR POWER

The following example shows how these two types of power combine at various loadings.

Example           Q        A 6.6kV, 400 hp motor has a full-load power factor of 0.8. What are its load current and power factor at full-load, half-load, quarter-load and no-load?  Assume a full-load efficiency of 90%.  Neglect no-load losses.

                         A        400 hp is equivalent to ¾ x 400 = 300kW_m

                                   If efficiency is 90%, fuII-Ioad input power is 

                                   Prepare a table as follows ready for insertion of values as the calculation proceeds:

	KWe	kvar	kVA	/  (amps)
Full ½ ¼ No-load	333 167 83 0	250 250 250 250	416 301 263 250	36 26 23 22	0.8 0.55 0.31 0

                                    (The two figures in ‘boxes’ were given in the question or deduced as above.  The remainder are calculated as follows.)

                          

                                   (Note: The zero power factor at no-load is theoretical, as no-load losses have been neglected.  In practice the losses at no-load are appreciable, and a typical no-load power factor is 0.2 to 0.3.)

All the above calculated figures have been inserted in the table.

Deductions from the Table

If the table is examined carefully, certain conclusions, some perhaps surprising, can be drawn.

(i)            As loading is reduced, both motor current and the kVA are also reduced, but by no means in proportion.  Due to the continued presence of the magnetising current, the current and kVA at no-load are still more than half their values at full-load (in the example taken).  This means that the reading of a local motor ammeter must be taken with reserve as an indication of motor loading.

(2)        The power factor also falls with reduction of load; theoretically to zero at no-load but in practice to a figure about 0.2 to 0.3 because of the continuing losses.  On an offshore or onshore installation where many motors may be running at the same time, it is unlikely that all the motors will be running at full-load simultaneously, and some may even be running light.  Therefore the overall power factor of the combined motor load is likely to be well below that given on the motors’ individual rating plates.  It is of course improved overall by the addition of high power factor loads such as lighting and heating.