# Single-Phase Motors--part 2

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Continued from: Single-Phase Motors--part 1

6. Testing Single-Phase Motors

The methods to determine the winding resistances of the main, R,, and the auxiliary winding, R,, are not discussed in this section because we can use any method that can accurately determine the resistances of these windings. The other equivalent circuit parameters of a single-phase induction motor can be determined by performing the blocked-rotor and the no-load tests.

Blocked-Rotor Test

The blocked-rotor test is performed with the rotor held at standstill by exciting one winding at a time while the other winding is left open. The test arrangement with the auxiliary winding open is shown in Fgr13. The test is performed by adjusting the applied voltage until the main winding carries the rated current.

Since the slip at standstill in either direction is unity, the rotor circuit impedance is usually much smaller than the magnetization reactance. Therefore, for the blocked-rotor test we can use the approximate equivalent circuit of the main winding without the magnetization reactance, as depicted in Fgr14.

Fgr13 Experimental setup for blocked-rotor test with auxiliary winding open.

Fgr14 Approximate equivalent circuit as viewed from the main winding under blocked-rotor condition.

(a) Auxiliary winding open

Let Vb,, Ibmr and Pb, be the measured values of the applied voltage, the main-winding current, and the power supplied to the motor under blocked-rotor condition. The magnitude of the input impedance is ...

The total resistance in the circuit is ...

Thus, the total reactance is ...

From the approximate equivalent circuit ( Fgr14), we have ....

Since the main-winding resistance is already known, the rotor resistance at standstill, from Eq. (10.53), is ...

To separate the leakage reactances of the main winding and the rotor, we once again make the same assumption that they are equal. That is, ...

(b) Main winding open

We can also perform the blocked-rotor test by exciting the auxiliary winding with the main winding open. Let P,, V,, and I, be the power input, the applied voltage, and the current in the auxiliary winding when ....the rotor is at standstill. The total resistance of the auxiliary winding can now be computed as ...

Since the rotor winding resistance is already known, we can compute the rotor resistance as referred to the auxiliary winding as ...

We can now determine the a-ratio, the ratio of effective turns in the auxiliary winding to the main winding, by the square root of the ratio of the rotor resistance as viewed from the auxiliary winding to the value of the rotor resistance as viewed from the main winding. Thus, ....

No-Load Test with Auxiliary Winding Open

In the three-phase induction motor operating under no load, we neglected the copper loss in the rotor circuit because it was assumed to be very small. In fact, we considered the rotor branch an open circuit because of very low slip at no load.

In a single-phase motor running on main winding only, the no-load slip is considerably higher than that for a three-phase motor. If we still assume that the slip under no load is almost zero and replace the rotor circuit of the forward branch with an open circuit under no load, the error introduced in the calculation of the motor parameters based upon this test is somewhat greater than that for the three-phase motor. Making such an assumption, however, does simplify the equivalent circuit of the main winding under no load with auxiliary winding open. Such an equivalent circuit is given in Fgr15.

Let VnL, In,, and P,, be the measured values of the rated applied voltage, the current, and the power intake by the motor under no-load condition. Then the no-load impedance is ...

The equivalent resistance under no load is ....

Fgr15 Approximate equivalent circuit as referred to the main winding under no-load condition (s = 0).

Hence, the no-lo A reactance is ...

However, from the equivalent circuit ( Fgr15), we have ...

Thus, from Eq. (10.62), the magnetization reactance is ...

Finally, the rotational loss is ...

All the parameters of a single-phase induction motor are now known. Using these parameters we can now compute the torque developed, the power input, the power output, the winding currents, the line current, and the efficiency of the motor at any slip. ....

EXAMPLE 4

A 115-V, 60432, single-phase, split-phase induction motor is tested to yield the following data.

Voltage (V) Current (A) Power (W)

With auxiliary winding open No-load test 115 3.2 55.17 Blocked-rotor test 25 3.72 86.23 Blocked-rotor test 121 1.21 145.35 With main winding open

The main-winding resistance is 2.5 R and the auxiliary-winding resistance is 100 0. Determine the equivalent circuit parameters of the motor.

SOLUTION

From the blocked-rotor test on the main winding with the auxiliary winding open, we obtain ...

From the no-load test data on the main winding with auxiliary winding open, we have ....

From the blocked-rotor test on the auxiliary winding with the main winding open, we get ....

Exercises:

7. A 230-V, 60432, single-phase, CSCR motor is tested to yield the following data.

Voltage (V) Current (A) Power (W)

With auxiliary winding open

No-load test 230 2.2 63.1 Blocked-rotor test 120 8.22 743.1 Blocked-rotor test 201 2.08 309.7 With main winding open

The main-winding resistance is 6 R and the auxiliary-winding resistance is 38 R. Determine the equivalent circuit parameters of the motor.

A 230-V, 60-Hz, reversible, garage door, PSC motor is tested as follows with the auxiliary winding open:

Blocked-rotor test: 230 V, 2.16 A, 406.84 W

No-load test: 230 V, 1.45 A, 138.52 W

Since the two windings are identical for a reversible motor, the a-ratio is unity and the resistance of each winding is 42.4 R. Determine the equivalent circuit parameters of the motor.

When the auxiliary winding of a single-phase induction motor is in the form of a copper ring, it’s called the shaded-pole motor. Most often, a shaded-pole motor has a salient-pole construction similar to the stator of a dc machine. The pole is, however, always laminated to minimize the core loss. The pole is physically divided into two sections. A heavy, short-circuited copper ring, called the shading coil, is placed around the smaller section, as shown in Fgr16. This section usually covers one-third the pole arc and is called the shaded part of the pole. The larger section is logically referred to as the unshaded part. The main winding surrounds the entire pole, as shown in the figure. The rotor is die-cast, just like the rotor of any other single-phase induction motor.

As you may have already construed, a shaded-pole motor is very simple in construction and is the least expensive for fractional horsepower applications.

Since it does not require a centrifugal switch, it’s not only rugged but also very reliable in its operation. To keep the cost low, a shaded-pole motor usually operates in the magnetic saturation region. This is one of the reasons why the efficiency of a shaded-pole motor is low compared with that of other types of induction motors. These motors develop low starting torque and are suitable for blower applications. Shaded-pole motors are usually built to satisfy the load requirements up to 1/3 horsepower.

When the flux is increasing from zero to maximum

When the flux is almost maximum

When the flux is decreasing from maximum to zero

Fgr17 Shading-pole action during the positive half cycle of a flux waveform.

a, wt < n/2: Almost all the flux passing through unshaded region; b, wt = n/2: No shading action, flux is uniformly distributed over the entire pole; c, wt > n/2: Most of the flux is passing through the shaded region.

Principle of Operation

Let us now examine how the shading coil helps a shaded-pole motor to set up a revolving field. To do so, we consider changes in the flux produced by the main winding at three time intervals:

Any change in the flux in each pole of the motor is responsible for an induced emf in the shading coil in accordance with Faraday's law of induction. Since the shading coil forms a closed loop having a very small resistance, a large current is induced in the shading coil. The direction of the current is such that it always creates a magnetic field that opposes the change in the flux in the shaded region of the pole. With this understanding, let us now analyze the effect of the shading coils during the time intervals mentioned above.

Interval a: During this time interval the flux in the pole is increasing and so is the current induced in the shading coil. The shading coil produces a flux that opposes the increase in the flux linking the coil. As a result, most of the flux flows through the unshaded part of the pole, as shown in Fgr17. The magnetic axis of the flux is then the center of the unshaded section of the pole.

Interval b: During this time interval the magnetic flux in the pole is near its maximum value. Therefore, the rate of change of flux is almost zero. Hence, the induced emf and the current in the shading coil are zero. As a result, the flux distributes itself uniformly through the entire pole. The magnetic axis, therefore, moves to the center of the pole. This shift in magnetic axis has the same effect as the physical motion (rotation) of the pole.

Interval c: During this time interval the magnetic flux produced by the main winding begins to decrease. The current induced in the shading coil reverses its direction in order to oppose the decrease in the flux. In other words, the shading coil produces the flux that tends to prevent a decrease in the flux produced by the main winding. As a result, most of the flux is confined in the shaded region of the pole. The magnetic axis of the flux has now moved to the center of the shaded region.

Note that without the shading coil, the center of the magnetic axis would always be at the center of the pole. The presence of the shading coil forces the flux to shift its magnetic axis from the unshaded region to the shaded region. The shift is gradual and has the effect of revolving magnetic poles. In other words, the magnetic field revolves from the unshaded part toward the shaded part of the motor. The revolving field, however, is neither continuous nor uniform. Consequently, the torque developed by the motor in not uniform but vanes from instant to instant.

Since the rotor follows the revolving field, the direction of rotation of a shaded-pole motor cannot be reversed once the motor is built. To have a reversible motor, we must place two shading coils on both sides of the pole and selectively short one of them.

To increase the starting torque, the leading edge of the shaded-pole motor may have a wider air-gap than the rest of the pole. It has been found that if a part of the pole face has a wider gap than the remainder of the pole, the motor develops some starting torque without the auxiliary winding. Such a motor is called a reluctance start motor. Adding the reluctance feature to a shaded-pole motor increases its starting torque. This feature is commonly employed in the design of a shaded-pole motor.

A typical speed-torque characteristic of a shaded-pole motor is given in Fgr18. Take another look at this curve. It clearly shows one of the important drawbacks of a shaded-pole motor; the third harmonic dip. To cancel some of the third-harmonic effect, we can use a relatively high-resistance rotor. However, any increase in the rotor resistance is accompanied not only by a decrease in the operating speed of the motor but also by a drop of motor efficiency.

The analysis of a shaded-pole motor is quite involved and is not within the scope of this book. However, for the benefit of those readers who are interested in exploring further, we refer them to an IEEE publication, "Revolving-Field Analysis of a Shaded-pole Motor" by Bhag S. Guru, IEEE Trans., Power Apparatus and Systems, Vol. 102, No. 4, pp. 918-927, April 1983. This publication provides (a) a complete theoretical development including the harmonics analysis and non-quadrature placement of the shading coil with respect to the main winding, (b) all the necessary design equations, (c) a step-by-step procedure to develop a computer program for analyzing/designing a shaded-pole motor, and (d) a long list of major publications on the subject.

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Fgr18 Speed-torque characteristic of a shaded-pole motor. Breakdown torque Starting torque , Load line point; Percent of speed

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Fgr19 Current and flux directions in a universal motor during (a) the positive and (b) the negative half cycles.

8 Universal Motor

A dc series motor specifically designed for ac operation is usually referred to as a universal motor. A universal motor is wound and connected just like a dc series motor. That is, the field winding is connected in series with the armature winding. However, some modifications are necessary to transform a dc series motor into a universal motor. We discuss these modifications later. Let us first explore how a dc series motor can operate on an ac source.

Principle of Operation

When a series motor is operated from a dc source, the current is unidirectional in both the field and the armature windings. Therefore, the flux produced by each pole and the direction of the current in the armature conductors under that pole remain in the same direction at all times. Hence, the torque developed by the motor is constant.

When a series motor is connected to an ac source, the current in the field and the armature windings reverses its direction every half cycle, as shown in Fgr19 for a two-pole series motor.

During the positive half cycle ( Fgr19a), the flux produced by the field winding is from right to left. For the marked direction of the current in the armature conductors, the motor develops a torque in the counterclockwise direction.

During the negative half cycle ( Fgr19b), the applied voltage has reversed its polarity. Consequently, the current has reversed its direction. As a result, the flux produced by the poles is now directed from left to right. Since the reversal in the current in the armature conductors is also accompanied by reversal in the direction of flux in the motor, the direction of the torque developed by the motor remains unchanged. Hence, the motor continues its rotation in the counterclockwise direction.

If K, is the machine constant, i, is the current through the field and the armature windings at any instant, and aP is the flux per pole at that instant, the instantaneous torque developed by the motor is K,i,Op. When the motor operates in the linear region (below the knee of the magnetization curve), the flux Op must be proportional to the field current i,. When the permeability of the magnetic core is relatively high, the presence of the air-gaps ensures that for all practical purposes the flux Op is in phase with the current i,. Thus, the instantaneous torque developed by the motor is proportional to the square of the armature current, as shown in Fgr20. In other words, the average value of the torque developed is proportional to the root-mean-square (rms) value of the current. It’s obvious from Fgr20 that the torque developed by the universal motor varies with twice the frequency of the ac source. Such pulsations in torque cause vibrations and make the motor noisy.

Fgr21 shows the equivalent circuit, the phasor diagram, and the speed-torque characteristics of a universal motor. The back emf E,, the winding current I,, and the flux per pole QP are in phase with each other as shown. R, and X, are the resistance and the reactance of the series field winding. R, and X, are the resistance and the reactance of the armature winding.

Fgr20 The current and the torque developed by a universal motor.

Fgr21 (a) Equivalent circuit of a universal motor, (b) phasor diagram, and (c) typical speed-torque characteristic.

• Design Considerations

1. When a series motor is to be designed as a universal motor, its poles and yoke must be laminated in order to minimize the core loss produced in them by the alternating flux. If a series motor with an unlaminated stator is connected to an ac supply, it quickly overheats owing to excessive core loss.

2. Under steady-state operation of a dc series motor, the inductances of the series field and armature windings have little effect on its performance. However, the motor exhibits reactive voltage drops across these inductances when connected to an ac source. The high reactive voltage drops have a two-fold effect: (a) reducing the current in the circuit for the same applied voltage, and (b) lowering the power factor of the motor. The reactive voltage drop across the series field winding is made small by using fewer series field turns.

3. The decrease in the number of turns in the series field winding reduces the flux in the motor. This loss in flux is compensated by an increase in the number of armature conductors.

4. The increase in the armature conductors results in an increase in the armature reaction. The armature reaction can, however, be reduced by adding compensating windings in the motor. A universal motor with a conductively coupled compensating winding is shown in Fgr22a. The corresponding phasor diagram ( Fgr22b) shows the improvement in the power factor.

Under ac operation, an emf is induced by transformer action in the coils undergoing commutation. This induced emf (a) causes extra sparking at the brushes, (b) reduces brush life, and (c) results in more wear and tear of the commutator. To reduce these harmful effects, the number of commutator segments is increased and high-resistance brushes are used in universal motors.

One may ask a very logical question: With all these drawbacks, why do we use a universal motor? Some of the reasons are given below:

1. A universal motor is needed when it’s required to operate with complete satisfaction on dc and ac supply.

2. The universal motor satisfies the requirements when we need a motor to operate on ac supply at a speed in excess of 3600 rpm (2-pole induction motor operating at 60 Hz). Since the power developed is proportional to the motor speed, a high-speed motor develops more power for the same size than a low-speed motor.

3. When we need a motor that automatically adjusts its speed under load, the universal motor is suitable for that purpose. Its speed is high when the load is light and low when the load is heavy.

Fgr22 (a) Equivalent circuit and (b) phasor diagram for a conductively compensated universal motor.

In light of the above reasons, a universal motor is used quite extensively in the fractional horsepower range. Some applications that require variation in speed with load are saws and routers, sewing machines, portable machine tools, and vacuum cleaners.

EXAMPLE 5

A 120-V, 60-Hz, 2-pole, universal motor operates at a speed of 8000 rpm on full load and draws a current of 17.58 A at a lagging power factor of 0.912. The impedance of the series field winding is 0.65 + j1.2 0. The impedance of the armature winding is 1.36 + j1.6 R. Determine (a) the induced emf in the armature, (b) the power output, (c) the shaft torque, and (d) the efficiency if the rotational loss is 80 W.

SOLUTION

(a) From the equivalent circuit of the motor ( Fgr21b), we have ...

As expected, the induced emf is in phase with the armature current.

The power developed by the motor is (b) ....

(c) From the rated speed of the motor, we obtain ....

(d) The power input is ....

Exercises:

A 240-V, ~O-HZ, 2-pole, universal motor operates at a speed of 15000 rpm on full load and draws a current of 6.5 A at 0.94 pf lagging. The motor parameters are R, = 6.15 Q, X, = 9.4 R, R, = 4.55 Q, and X, = 3.2 R. Calculate (a) the induced emf of the motor, (b) the shaft torque, and (c) the efficiency if the rotational loss is 65 W.

If the motor of Exercise 10.9 draws a current of 12.81 A at 0.74 pf lagging when its load is increased, determine the operating speed of the motor.

Assume that the motor is operating in the linear region.

SUMMARY

In this section we examined three different types of ac motors: the single-phase induction motor, the shaded-pole motor, and the universal motor.

We explained with the help of the double revolving-field theory that a single-phase induction motor continues rotating even when the flux is established by a single (main) winding. The split-phase motor and the capacitor-start motor operate on a single winding when the operating speed is usually above 75% of its synchronous speed. Based upon the double revolving-field theory, we were able to represent these motors by their equivalent circuits. The rotor circuit was resolved into two equivalent circuits: one for the forward revolving field and the other for the backward revolving field.

For a single-phase induction motor to develop starting torque, it requires an auxiliary winding. This winding is usually placed in space quadrature with respect to the main winding.

In a split-phase motor, the phase difference between the currents in the two windings and thereby between the fluxes produced by them is obtained by making the auxiliary winding highly resistive while the main winding is highly inductive.

A phase difference of as high as 60-degr electrical can be obtained for a well-designed split-phase motor.

In a capacitor motor, the phase difference is created by including a capacitor in series with the auxiliary winding. For a properly designed capacitor motor, the phase difference of nearly 90-degr electrical can be obtained at the time of starting.

Thus, a capacitor motor comes close to a two-phase motor.

The power developed and the efficiency of a capacitor motor can be improved by retaining the auxiliary winding in the circuit at all times. Motors that fall into this category are permanent split-capacitor (PSC) motors and capacitor-start capacitor-run (CSCR) motors.

A CSCR motor uses two capacitors: One capacitor is optimized to develop high starting torque while the other is picked on the basis of high efficiency. The selection of the capacitor for a PSC motor is based upon a compromise between the starting torque and its full-load efficiency.

The auxiliary winding of a shaded-pole motor consists of a shorted copper ring. Most often, the two windings are not in space quadrature. The currents in the two windings set up an unbalanced revolving field, and the motor develops some starting torque. In order to increase the starting torque, most shaded-pole motors have a graded air-gap region.

The universal motor is a series dc motor properly designed for ac operation.

The motor is capable of operating on both dc and ac supply. The performance, however, is better when the motor operates on the dc supply. Compensating winding may be used to improve the power factor.

The equivalent circuit parameters of an induction motor can be obtained by performing the blocked-rotor and no-load tests. When the blocked-rotor test is performed with the auxiliary winding open, we obtain the rotor resistance, the leakage reactance of the main winding, and the leakage reactance of the rotor. By performing the blocked-rotor test with the main winding open, we obtain the a-ratio. Performing the no-load test with the auxiliary winding open yields the magnetization reactance and the rotational loss.

QUIZ:

Describe the construction of the following motors: split-phase motor, capacitor motor, shaded-pole motor, and universal motor.

Explain the principle of operation of a split-phase motor, a capacitor motor, a shaded-pole motor, and a universal motor.

Why do we refer to the split-phase motor, capacitor motor, and shaded-pole motor as induction motors? Is a universal motor also an induction motor? Give reasons to justify your answer.

State some of the practical applications of universal motors other than those given in this text.

Is it always possible to replace a shaded-pole motor with other types of induction motors? Can the direction of rotation of a shaded-pole motor be reversed? How can the direction of rotation of a split-phase or a capacitor motor be reversed? Why is it necessary to "cut off" the auxiliary winding from the circuit when a split-phase motor is operating at full load? What would happen if the centrifugal switch failed? Why is a capacitor motor better than a split-phase motor? Why is a capacitor-start capacitor-run motor better than a permanent split-capacitor motor? What is the effect of armature reaction on the speed of a universal motor?

Why does a universal motor perform better on dc than ac supply? What happens to the power factor of a universal motor when the load is increased?

What happens to the speed of a universal motor when the load is increased ? Is it possible for a universal motor to self-destruct under no load? A 1/3-hp, 12O-V, 60-Hz, shaded-pole motor takes 830 W at its full-load slip of 8%. Calculate (a) the speed and (b) the efficiency of the motor.

Determine the number of poles, the forward slip, and the backward slip for the following motors: (a) 1140 rpm, 60 Hz, (b) 1440 rpm, 50 Hz, and (c) 3200 rpm, 60 Hz.

How can a universal motor be reversed? Make a sketch using a double-pole double-throw (DPDT) switch.

When the operating speed of a motor under all conditions is less than 3600 rpm, is it still better to use a universal motor than an induction motor?

Problems:

A 120-V, 60-Hz, 2-pole, single-phase induction motor operates at a slip of 4% and has a rotor resistance of 2.4 R at standstill. Determine (a) the speed of the motor, (b) the effective rotor resistance in the forward branch, and (c) the effective rotor resistance in the backward branch.

The effective rotor resistance of a 120-V, 50-Hz, 6-pole, single-phase induction motor in the forward branch at a slip of 5% is 120 a. What is the rotor resistance at standstill? What is the effective resistance in the backward branch? What is the operating speed of the motor? A 230-V, 60-Hz, 6-pole, single-phase induction motor has a stator impedance of 1.5 + j3 R and a rotor impedance of 2 + j3 R at standstill. The magnetization reactance is 50 il. If the rotational loss is 150 W at a slip of 5'%, determine the efficiency and the shaft torque of the motor.

A 208-V, 50-Hz, 4-pole, single-phase induction motor has R, = 2.5 0, XI = 2.9 (1, R, = 2.1 ll, X, = 2.6 a, and X,,, = 42 R. If the friction, windage, and core loss is 50 W at a slip of 4%, determine the efficiency and the torque of the motor.

A 115-V, 60-Hz, &pole, single-phase induction motor operates at a speed of 1050 rpm and has R, = 3.8 fl, XI = 5.9 fl, R, = 4.2 a, X, = 5.9 a, and X = 70.8 ll. Determine the torque and the efficiency of the motor if the rotational loss is 25 W. Plot the speed-torque curve, current versus speed, efficiency versus speed, and power output versus speed characteristics of the single-phase induction motor of Problem 5. Assume that the rotational loss is proportional to the motor speed.

7. A 230-V, 60-Hz, \$-pole, two-value capacitor motor is rated at 1710 rpm.

The motor parameters are R, = 30 R, X, = 36 R, R, = 24 a, X, = 30 a, R, = 120 R, X = 250 R, a = 1.75, start capacitor = 8 pF, and run capacitor = 4 pF. The rotational loss at full load is 25W. Determine the shaft-torque and the efficiency at full load. What is the starting torque developed by the motor?

8. Plot the speed-torque curve and speed-efficiency characteristic for the motor of Problem 7. Assume that (a) the switching action takes place at 75% of the synchronous speed and (b) the rotational loss is proportional to the speed.

9. A 115-V, 60-Hz, 6-pole, CSCR motor is rated at 1152 rpm. The motor parameters are R, = 20 R, X, = 32 0, R, = 22 0, X, = 32 R, R, = 55 SZ, X, = 210 R, a = 1.8, start capacitor = 8 pF, and run capacitor = 5 pF. The rotational loss is 10 W. Determine the shaft-torque and the efficiency at full load. What is the starting torque developed by the motor?

10. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 9. Assume that (a) the switching action takes place at 75% of the synchronous speed and (b) the rotational loss is proportional to the speed.

11. A llS-V, 60-Hz, 6-pole, CSCR motor is rated at 1120 rpm. The motor parameters are XI = 6 0, XI = 4.8 a, R, = 5 R, X, = 3.3 R, R, = 38 R, X,n = 51 a, a = 2.6, start capacitor = 20 pF, and run capacitor = 10 pF. The rotational loss is 28 W. Determine the shaft-torque and the efficiency at full load. What is the starting torque developed by the motor?

12. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 11. Assume that (a) the switching action takes place at 75% of the synchronous speed and (b) the rotational loss is proportional to the speed.

13. A 115-V, 60-Hz, 2-pole, CSCR motor is rated at 3325 rpm. The motor parameters are R, = 16 R, X, = 12 n, R, = 11 R, X, = 6.8 R, R, = 26 R, X, = 120 a, a = 1.2, start capacitor = 25 pF, and run capacitor = 10 pF. The rotational loss is 10 W. Determine the shaft-torque and the efficiency at full load. What is the starting torque developed by the motor?

14. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 13. Assume that (a) the switching action takes place at 75% of the synchronous speed and (b) the rotational loss is proportional to the speed.

15. A 220-V, 60-Hz, 4-pole, PSC motor is rated at 1710 rpm. The motor parameters are R, = 20 a, X, = 30 0, R, = 24 a, X, = 30 R, R, = 60 a, X,, = 200 0, Q = 1.5, and C = 4 pF. The rotational loss at full load is 20 W.

Determine the shaft torque and the efficiency at full load. What is the starting torque developed by the motor? 10.16. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 15. Assume that the rotational loss is proportional to the speed.

17. A 230-V, 60-Hz, 6-pole, PSC motor is rated at 1152 rpm. The motor parameters are R, = 15 R, XI = 42 R, R, = 22 0, X, = 42 R, R, = 25 R, X, = 180 R, u = 1.4, and C = 7.5 KF. The rotational loss is 10 W. Determine the shaft torque and the efficiency at full load. What is the starting torque developed by the motor? 10.18. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 17. Assume that the rotational loss is proportional to the speed.

19. A 120-V, 60-Hz, 6-pole, PSC motor is rated at 1120 rpm. The motor parameters are R, = 5 R, X, = 5.8 R, R, = 7 0, X, = 5.8 R, R, = 12 R, X, = 80 0, a = 2, and C = 10 pF. The rotational loss is 20 W. Determine the shaft torque and the efficiency at full load. What is the starting torque developed by the motor? 10.20. Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 19. Assume that the rotational loss is proportional to the speed.

21. A 115-V, 6O-Hz, 2-poie, reversible PSC motor is rated at 3325 rpm. The motor parameters are R, = 4 0, XI = 8 61, R, = 15 R, X, = 8 R, R, = 4 R, X, = 120 0, u = 1, and C = 10 pF. Neglect the rotational loss. Determine the shaft torque and the efficiency at full load. What is the starting torque developed by the motor? 10.22. A 208-V, 50-Hz, 4-pole, split-phase induction motor has X, = 12.5 dZ, XI = 25 fl, R, = 25 i2, X, = 25 fZ, R, = 280 0, a = 0.5, and X,, = 150 0. If the friction, windage, and core loss is 20 W at a speed of 1400 rpm, determine the efficiency and the torque of the motor. What is the starting torque developed by the motor? The switching speed is 75% of the synchronous speed.

23. Sketch the speed torque and speed-efficiency curves for the motor of Problem 22. Assume that the rotational loss is proportional to the speed.

24. A 120-V, 60-Hz, 4-pole, split-phase induction motor has R, = 6.5 dZ, X, = 12.5 62, R, = 15 0, X, = 12.5 R, R, = 120 R, u = 0.5, and X,, = 150 I]. If the friction, windage, and core loss is 25 W at a speed of 1650 rpm, determine the efficiency and the torque of the motor. What is the starting torque developed by the motor? The switching speed is 75% of the synchronous speed.

25. Sketch the speed-torque and speed-efficiency curves for the motor of Problem 24. Assume that the rotational loss is proportional to the speed.

A 120-V, 60-Hz, 4-pole, capacitor-start induction motor has R, = 5 R, X, = 12 R, R, = 20 a, X, = 12 R, R, = 20 R, a = 1.5, and X, = 200 R. The value of the start capacitor is 20 pF. If the friction, windage, and core loss is 20 W at a speed of 1650 rpm, determine the efficiency and the torque of the motor. What is the starting torque developed by the motor? The switching speed is 75% of the synchronous speed.

Sketch the speed-torque and the speed-efficiency curves for the motor of Problem 26. Assume that (a) the rotational loss is proportional to the speed and (b) the switching action takes place at 75% of the synchronous speed.

A 230-V, 60-Hz, 6-pole, capacitor-start motor is rated at 1152 rpm. The motor parameters are R, = 10 R, XI = 22 R, R, = 20 R, X, = 22 R, R, = 18 R, X, = 220 R, u = 1.8, and start capacitor = 8 pF. The rotational loss is 10 W. Determine the shaft torque and the efficiency at full load.

What is the starting torque developed by the motor? Plot the speed-torque curve and the speed-efficiency characteristic for the motor of Problem 28. Assume that (a) the switching action takes place at 75% of the synchronous speed and (b) the rotational loss is proportional to speed.

A 4-pole, 120-V, 60-Hz, capacitor-start motor is tested to yield the following data: No-load test with auxiliary open: 120 V, 2.7 A, 56 W Blocked-rotor test with auxiliary open: 120 V, 15 A, 1175 W Blocked-rotor test with main open: 120 V, 5.2 A, 503.4 W The main-winding resistance is 2.5 fl and the auxiliary-winding resistance is 12.5 R. Determine the equivalent circuit parameters of the motor.

A 208-V, \$-pole, 50-Hz, reversible, PSC motor, when tested with auxiliary winding open, gave the following results: No-load test: 208 V, 3.8 A, 96 W Blocked-rotor test: 80 V, 8 A, 420 W For a reversible motor to develop the same torque in either direction of rotation at full load, the main and the auxiliary windings are identical (a = 1). The resistance of each winding is 3.8 R. Determine the equivalent circuit parameters of the motor.

Calculate the torque developed and the efficiency of the motor of Problem 10.31 at a slip of 5%. What is the rated speed of the motor? What must be the voltage rating of the capacitor if the motor uses a 5-pF capacitor? The following test results were obtained from a 120-V, 60-Hz, split-phase motor: Main-winding resistance: 3.5 R Auxiliary-winding resistance: 140 R Blocked-rotor test with auxiliary open: 96.8 V. 10 A. 600 W J1

No-load test with auxiliary open: 120 V, 2.5 A, 45 W Blocked-rotor test with main open: 120 V, 0.85 A, 102.2 W Compute the equivalent circuit parameters of the motor.

34. A 120-V, 6-pole, 50-Hz, reversible, PSC motor, when tested with the auxiliary winding open, gave the following results: No-load test: 120 V, 1.2 A, 60 W Blocked-rotor test: 90 V, 5 A, 420 W For a reversible motor to develop the same torque in either direction of rotation at full load, the main and the auxiliary windings are identical (a = 1). The resistance of each winding is 15 (1. Determine the equivalent 'circuit parameters of the motor.

10.35. Calculate the torque developed and the efficiency of the motor of Problem 34 at a slip of 4%. What is the rated speed of the motor? What must be the voltage rating of the capacitor if the motor uses a 7.5-pF capacitor? 10.36. A 110-V, 60-Hz, 2-pole, universal motor operates at a speed of 12,500 rpm on full load and draws a current of 5 A at 0.74 pf lagging. The motor parameters are R, = 0.5 R, X, = 2.235 R, R, = 5.2 R, and X, = 12.563 R. Calculate (a) the induced emf in the armature, (b) the shaft torque, and (c) the efficiency of the motor.

37. If the motor of Problem 36 draws a current of 5.65 A at a power factor of 0.65 lagging when the load is increased, determine (a) the operating speed of the motor, (b) the power output, and (c) the shaft torque. Assume that the rotational loss is proportional to the motor speed.

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