Electric Motors and Drives - Third Edition

Solutions to Review Questions - Chapter 7

1) Primary and secondary voltages and turns are related by , hence

Similarly , hence

To check we calculate the primary VA ( = 240 2), and the secondary VA ( = 120 4) to confirm that they are the same.

2) If the impedance connected to the secondary is , then the secondary current is given by

The referred (or apparent) impedance looking in at the primary is given by the ratio

of primary voltage to primary current, i.e.

Using the relationships and , and then eliminating and between

the four equations yields the referred impedance as

In this case, is a 5 resistor, so the referred impedance is resistive and its magnitude is given

by .

3) From the voltage/turns relationship we deduce that the turns ratio is given by

Now we can use the same approach as in question 2 to find the apparent resistance looking in at

the primary, i.e.

The primary current is therefore given by

Viewed from the mains, the transformer and its secondary load of 30 look like a 4,320

resistor. The primary input power is therefore simply given by

4) Our instinctive reaction to the thought of taking a hacksaw and cutting through the magnetic circuit would surely be to recoil in horror and expect serious adverse consequences. In fact, as

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we will see, this is one of the few situations where things turn out much better than we might expect.

With the primary supplied at the rated voltage and frequency (assumed to be sinusoidal), the peak magnetic flux in the core is given by the basic design equation (7.5), i.e.

Strictly, this equation applies to the ideal transformer in which the resistance of the windings is zero, but as explained in chapter 7 the resistance of the real transformer winding is sufficiently small for us to apply the equation to the real transformer.

The flux equation shows that the flux is determined only by the voltage, the frequency and the number of turns. The other variables that we might intuitively expect to have a bearing on the flux (the reluctance of the core, and the current in the winding) are absent from the formula, which of course tells us that they are of little or no consequence. We therefore conclude that if we increase the reluctance of the magnetic circuit by making a cut, the flux will remain the same.

This remarkable result stems from the feedback or self-stabilising nature of the flux, which, if the primary resistance is zero, must induce in the primary winding an e.m.f. that is equal to the applied voltage. The induced e.m.f. is determined by the magnitude of the flux, its rate of change (i.e. frequency) and the number of turns, so these are the only parameters in the flux formula.

In a real transformer with finite reluctance a current is required to produce the MMF needed to drive the flux around the core: the magnitude of this magnetising current is directly proportional to the reluctance of the core, so the better the magnetic circuit, the smaller the magnetising current. So if the reluctance is increased by including an air-gap, the current will automatically increase to produce the extra MMF now needed to maintain the flux at the level specified by the voltage, turns and frequency.

We would therefore see an increase in the no-load (magnetising) current, but no change in the secondary voltage since the flux is maintained at its rated value.

The magnetising reactance expresses the ratio of applied voltage to magnetising current, and because the magnetising current has increased the reactance decreases in inverse proportion.

The on-load performance will be virtually unaffected, except that the increased magnetising current will result in increased copper loss and a decrease in power-factor. But overall we would rightly conclude that the saw-cut had remarkably little effect - certainly much less than we might have feared!

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5) The approximate equivalent circuit is shown in Figure 7A: the upper part shows the no-load condition (i.e. with the secondary open-circuited), and the lower part shows the situation at full-load.

Fig 7A here

Under no-load conditions there is no current through the leakage reactance and the resistance so there is no volt-drop between the primary voltage and the referred secondary voltage , so

the latter is given by

Using the primary voltage as the reference for phase-angles, at full-load, the referred secondary

current is given by , and hence the referred secondary voltage is given

by

The output voltage on load is therefore 1.1% less than at no-load, and it lags the input by 2.8°.

To find the full-load current we must add the magnetising current to the referred full-load current (given by the first equation we derived), i.e.

We can do a rough check on this by noting that at full-load the dominant impedance seen from the supply is the load resistance of 10 . If we ignore the other elements, the full-load current

would be simply This is very close in magnitude to the actual value (

), but because the approximation ignores the lagging magnetising current, the approximation yields an erroneous value for the phase-angle.

Turning now to the short-circuit condition, the referred secondary current is given by

To this we must add the

magnetising current to give the short-circuit current as

Hence

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Once again we can do a rough check by noting that the dominant impedance under short-circuit conditions is the leakage reactance of j0.5. If we ignore the resistance term, the short-circuit

current will be Again the magnitude is close to the actual value ( )

but the phase is wrong because we neglected the resistance.

The importance of leakage reactance in limiting the primary current to at least a manageable value in the event of a secondary short-circuit is very apparent form these calculations.

6) We can tackle this question by invoking one of the delightfully simple fundamental equations of the induction motor, viz:-

The question says ‘estimate’ which is reassuring and implies that we can make use of the approximate equivalent circuit, Fig 7.16.

Under locked-rotor conditions, the slip is 1, so the rotor branch resistance is simply . The

load branch then consists of and the leakage reactance. We know that all three

elements have much lower impedances than the impedance of the magnetising branch, so we can assume that all of the input current and power goes into the load branch, and is dissipated in

and . Because these resistances are equal, the power divides equally, so the rotor input

power is 12 kW.

The synchronous speed is given by

Hence the locked-rotor (starting) torque is given by

7) Like the previous question, this one does not give us the equivalent-circuit parameters, and it says estimate, rather than calculate, so we are expected to draw on our knowledge of the general properties of single-cage motors, and in particular equation 7.22, i.e.

the synchronous speed is 1000 rev/min, the full-load speed is 950 rev/min so the full-load slip is

50/1000 = 0.05. Hence the starting torque is given by

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The full-load torque is given by Hence using the result

above the starting torque is given by

8) As with the previous two questions, section 7.8 contains most of the material needed to tackle this one.

a) Equation 7.26 shows that the maximum or pull-out torque is inversely proportional to the leakage reactance, so if the reactance increases from to , the peak torque will be

reduced from to

b) The pull-out slip (i.e the slip at which maximum torque is developed) is given by equation

7.25, i.e. . So again, if the reactance increases by 10%, the pull out slip will reduce

from s to 0.91s, so the peak torque will occur at a higher speed.

c) Taking a ‘physical’ viewpoint first, in the majority of induction motors the slip and rotor frequency at full load are small, at most a few Hz, so the reactance associated with the leakage inductance is small in comparison with the rotor resistance. The rotor current, torque and power are therefore not much influenced by the rotor leakage inductance, so an increase of 10% would have very little effect.

Alternatively, from the equivalent circuit viewpoint (with everything happening at the supply frequency), we would say that under full-load conditions where the slip is much less than 1, the

referred rotor resistance ( ) will be the dominant impedance and will swamp the leakage

reactance, which will therefore have little influence on the rotor current, torque or output power.

d) Under locked-rotor conditions (s = 1), the dominant component of the impedance seen by the supply is usually the leakage reactance. (This point is well illustrated in question 5.) An increase of 10% in the leakage reactance will therefore reduce the starting current by a factor of 1/1.1, i.e. 0.91.

9) The torque is proportional to the total rotor power, which in the equivalent circuit is the power

in the referred rotor resistance, . If the torque is to remain the same, so must the rotor

power, and so therefore must the effective rotor resistance. Hence if the rotor resistance

increases by a factor of 1.2, the slip must increase by the same factor in order for to remain

the same. the new slip is therefore 1.2 3.5 = 4.2%.

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We could reach the same conclusion by arguing that for given torque we need to induce a given current in the rotor. The induced voltage in the rotor is directly proportional to the slip, and the rotor resistance is the dominant impedance, so a 20% increase in rotor resistance must be compensated by a 20% increase in the slip to maintain the same current.

10) In solving this question we will make extensive use of the sketches shown in Figure 7.B.

Fig 7B here

The diagram on the left represents conditions at full load (slip = s), while on the right the slip is half of the full-load value.

We are advised to use the approximate equivalent circuit, so at full load the right-hand branch consists of the total leakage reactance X and the effective resistance R2/s. The impedance triangle is shown at the top left of Figure 7B: we are told that at full load the load component of current lags the voltage by 15°, so we know that the impedance angle at full load is 15°.

Estimating the full-load current is simply a matter of adding the load and magnetising components. The in-phase component of current is while the lagging component is given by The total current is therefore given by and the power factor is 38.64/42.8 = 0.90.

When the slip is half of the full-load value, we first need to work out the rotor branch impedance, referring to the impedance triangle on the right in Figure 7B. First we note from

the left-hand triangle that and from the right-hand triangle

Combining these two equations gives

The rotor branch impedance is the hypotenuse of the triangle, so the full-load impedance is

given by while the impedance at half full-load slip is given by

Hence We know that when the impedance is , the load

current is 40 A, so when the impedance is , the load branch current will be 40/1.95 = 20.5 A.

As before , we now need to add the rotor branch current of 20.5 A, lagging by 7.6° to the magnetising current of 8 A lagging by 90°. The in-phase component is

while the quadrature component is The total current is therefore given by

and the power factor is 20.32/22.9 = 0.89.

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11) Fortunately we are told to assume that the voltage has been adjusted so that the air-gap flux remains at its rated value. Hence as far as the rotor is concerned, the factor that determines the torque is the speed of the rotor relative to the rotating flux wave.

At 50 Hz the synchronous speed is 3000 rev/min and so a slip of 0.04 corresponds to a slip speed of 120 rev/min.

a) At 25 Hz the synchronous speed is 1500 rev/min, so given that the magnitude of the flux wave is the same as when the motor was supplied at 50 Hz, the rotor will develop full torque when its speed is 120 rev/min less than 1500, i.e. 1380 rev/min. The slip is given by 120/1500 = 0.08.

b) Similarly when the supply is 3 Hz, the synchronous speed is 180 rev/min so full torque will be developed at a speed of 180 -120 = 60 rev/min. In this case the slip is given by 120/180 = 0.67.

In section 7.6.1 it is shown that the efficiency of the rotor is given by (1 - s) 100%, so in case (b) the efficiency of the rotor is 33%. The overall efficiency of the motor is clearly less than the efficiency of the rotor: in fact at this low frequency loss will be dominated by stator and rotor copper losses, which are typically of similar order, so we can expect the overall efficiency to be perhaps half the rotor efficiency, say around 15%.

The relevant part of the equivalent circuit for assessing the total current is shown in Figure 7C.

Fig 7C here

For the sake of simplicity the ‘core-loss’ resistance has been omitted, because we are not given any information to enable us to say how this would change with frequency, and in any case the corresponding current component will be very small in comparison with the full-load current.

The upper sketch shows 50 Hz operation, the total current being the sum of the magnetising

current and the rotor branch current . At 25 Hz, the reactances are halved as

compared with 50 Hz, and the supply voltage has been adjusted to keep the flux (and hence the

magnetising current) the same as at 50 Hz, i.e. the air-gap voltage is reduced from to .

Because the slip has doubled at 25 Hz, the rotor branch impedance sat 25 Hz is half of its 50 Hz value, so the rotor branch current remains the same and so therefore does the total current.

Exactly the same arguments can be advanced for the 6 Hz case, though the numbers are less convenient!

- End of Solutions Chapter 7 -

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