CHAPTER 6 MOTOR OPERATION AT REDUCED VOLTAGE

6.1          GENERAL CONSIDERATION

 

Due to system disturbances, especially when starting large drives on limited generating capacity, motors may at any time find themselves operating on a voltage lower than the nominal value.

FIGURE 6.1

CURRENT AND POWER IN A SIMPLE RESISTOR

If such a thing happened when voltage were applied to a simple resistor, as shown in Figure 6.1, the current, as determined by Ohm’s Law, would always remain proportional to the applied voltage.  For example, a 10% reduction of voltage (i.e. to 0.9 times nominal) would result also in a 10% reduction of current (i.e. to 0.9 times its previous value), and the power, which is the product of voltage and current, would be reduced to 0.9², or 0.81, times its previous value - that is, a power reduction of 19%.  For small voltage reductions the percentage power drop is approximately double the percentage voltage drop.

This is quite simple with a fixed resistance, but induction motors do not behave in this way, as will be explained below.  It is because they do not present a fixed impedance to the voltage, equivalent to the R of Figure 6.1.  In fact Ohm’s Law does not appear to apply.  The behaviour of motors when the applied voltage varies must be examined in more careful detail.

6.2       EFFECT ON TORQUE

It was explained in Chapter 2 that an induction motor operates by the interaction of the rotating stator field and the currents in the rotor bars.  The magnitude of the field depends directly on the applied voltage; also the emf induced in the rotor, and so the rotor bar current, depends on the strength of that field.  So the rotor current too depends on the voltage, and, since the developed torque is the product of both, that torque is proportional to the square of the applied voltage.  Expressed in symbols:

                                                                   T µ V²

It will be seen therefore that torque is very sensitive to voltage variation.  Taking the figures from the previous example, a 10% reduction of voltage (i.e. to 0.9 times nominal) reduces the torque to 0.9², or 0.81T.  For small voltage variations the percentage drop of torque, like that of the power, is also approximately double the percentage voltage drop.

For small voltage variations this might be acceptable, although it would increase the motor’s run-up time from starting and could so cause some overheating.  It can, however, pose a serious problem if the variation is large.  For example if, when starting a very large motor, the system voltage fell by 30% (i.e. to 0.7 times nominal - a situation not unknown), the torque would fall to 0.7², or 0.49T - that is, to about half its nominal value.  If this were the starting torque of a large motor with a high-friction drive, it might not be enough to overcome the initial break-away friction, or ‘stiction’, and the motor would then not start at all.

Consider now the effect of a running motor suffering a sudden reduction of applied voltage of 10% - that is, to 0.9 times nominal.  It has already been shown that the torque will be reduced by 19%, that is to 0.81T.

FIGURE 6.2

INDUCTION MOTOR OPERATING ON REDUCED VOLTAGE

The full lines of Figure 6.2 are reproductions, on a wider scale, of the right-hand parts of the torque/speed (black) and current/speed (blue) characteristics of Figure 5.1 in Chapter 5, for a motor operating at its rated, or nominal, voltage.  The dotted curves are the corresponding characteristics for the same motor operating at a 10% reduced voltage.  As already explained, the reduced-voltage torque/speed curve will lie at all points 19% below the nominal curve.

Point S is synchronous speed and point N the actual speed when a stated load torque is applied with the motor working at nominal voltage.  SN is then the slip, point P the working point for the given load torque, and point Q the working current point.

If now, while the motor is still loaded, the voltage is reduced by 10%, momentarily the external load torque and the speed remain unaltered.  All that happens is that the characteristics change suddenly from the full-line to the dotted-line curves, and the torque working point drops from P to P”.

The torque being delivered by the motor (NP”) is now less than the mechanical torque being imposed on it by the external load.  Since the latter has not changed, the motor and its driven load will begin to slow down, and the speed point will move to the left from position N.  The slip will increase, and, as it does so, the working point will move to the left from P” up the dotted characteristic.  As it does so the motor torque will increase.  When it reaches a new point P’, it is once again equal to the load torque; the deceleration ceases and the motor settles down to a new speed N’ slightly less than its previous full-voltage speed N, and with a slightly increased slip.  The progressive change in torque is indicated in Figure 6.2 in red.

6.3       EFFECT ON CURRENT

On the same Figure 6.2 the point Q represents the current with full voltage applied and with the given load.  After voltage reduction the new speed is at point N’, and the new current is given by point Q’ on the intersection of the vertical through N’ and the reduced-voltage dotted current curve.  This is higher than the current (Q) at full voltage - that is to say, the current rises as voltage is reduced, in direct contrast to Ohm’s Law.

This result should not be surprising if considered from the power point of view.  The mechanical power extracted from the motor is the product of torque and speed.  The speed decreases only slightly since there has been a slight increase in a small slip, and the load torque can be regarded as almost constant.  Consequently the mechanical power output is also nearly constant, demanding a near-constant electrical power input.  But electrical power (VI cos j) is proportional to the product of V and I, so, if V goes down, I must go up more or less in inverse proportion.

6.4       PULL OUT AND STALL

In Figure 6.2 the dotted torque curve was drawn for only a 10% voltage reduction, and the maximum available, or ‘pull-out’, torque shown there (point Z’) is greater than the load torque.  Therefore a point P’ will be established where the motor torque can rise to meet the demanded load torque, even with the 10% voltage reduction, and the motor settles down to a new steady but slightly lower speed.

But if the voltage reduction had been greater, the dotted reduced-voltage torque curve would be far lower (being proportional to V²), and its maximum, or pull-out, value at point Z” might then be lower than the demanded load torque, as indicated in part by the chain-dotted curve in Figure 6.2.  The point P’ could then never be found where the motor torque can rise to meet the load torque, and deceleration would continue until the motor stalled to a stop.  Its current would rise to its starting or ‘locked-rotor’ value at that voltage and would flow continuously, causing rapid overheating (see Chapter 10, Paragraph 10.3, ‘Overcurrent / Overloading’).  The motor would have to rely on its automatic protection to clear it off the line and prevent burnout.

Even without a voltage reduction, if the load torque should rise excessively - for example due to a malfunction of the driven machine, or to a bearing seizure - to a level exceeding the pull-out peak torque curve of Figure 5.1, the torque delivered by the motor cannot then rise to meet it, and the motor will decelerate into a stall.  The lower the voltage, the lower the pull-out torque.