Single-Phase Motors--Polyphase Induction Motors (part 2)

4. Development of an Equivalent Circuit

When a balanced three-phase induction motor is excited by a balanced three-phase source, the currents in the phase windings must be equal in magnitude and 120 -degree electrical apart in phase. The same must be true for the currents in the rotor windings as the energy is transferred across the air-gap from the stator to the rotor by induction. However, the frequency of the induced emf in the rotor is proportional to its slip [Eq. (5)]. Since the stator and the rotor windings are coupled inductively, an induction motor resembles a three-phase transformer with a rotating secondary winding. The similarity becomes even more striking when the rotor is at rest (blocked-rotor condition, s = 1). Thus, a three-phase induction motor can be represented on a per-phase basis by an equivalent circuit at any slip s as depicted in Fgr. 1. In this figure, …

PI = applied voltage on a per-phase basis

R, = per-phase stator winding resistance

L, = per-phase stator winding leakage inductance

XI = 27rfL, = per-phase stator winding leakage reactance

R, = per-phase rotor winding resistance

L, = per-phase rotor winding leakage inductance

x, = 27rf

Lb = per-phase rotor winding leakage reactance under blocked

X, = 2rsf

L6 = sX, = per-phase rotor winding leakage reactance at slip, s.

X, = per-phase magnetization reactance R, = per-phase equivalent core-loss resistance N, = actual turns per phase of the stator winding N2 = actual turns per phase of the rotor winding k,, = winding factor for the stator winding kw2 = winding factor for the rotor winding, am = amplitude of the per-phase flux El = 4.44fNlk,@, = per-phase induced emf in the stator winding rotor condition (s = 1)

Fgr. 1 Per-phase equivalent circuit of a balanced three-phase induction motor.

Eb = 4.44fN2kw2@, = per-phase induced emf in the rotor winding under blocked-rotor condition (s = 1) E, = sEb = per-phase induced emf in the rotor winding at slip s I, = per-phase rotor winding current I, = per-phase current supplied by the source = I, + I, = per-phase excitation current I, = per-phase core-loss current I, = per-phase magnetization current.

From the per-phase equivalent circuit ( Fgr. 1), it’s evident that the current ...in the rotor circuit is ...

Fgr. 2 Modified equivalent circuit of a balanced three-phase motor on a per-phase basis.

Based upon the above equation, we can develop another circuit of an induction motor as given in Fgr. 2. In this circuit, the hypothetical resistance RJs in the rotor circuit is called the effective resistance. The effective resistance is the same as the actual rotor resistance when the rotor is at rest (standstill or blocked-rotor condition). On the other hand, when the slip approaches zero under no-load condition, the effective resistance is very high (RJs -+ m). By defining the ratio of transformation, the a-ratio, as ....we can represent the induction motor by its per-phase equivalent circuit as referred to the stator. Such an equivalent circuit is shown in Fgr. 3, where ....

The per-phase stator winding current and the applied voltage are ....

The equivalent circuit of the rotor in Fgr. 3 is in terms of the hypothetical resistance RJs. In this circuit, I:R,/s represents the per-phase power delivered to the rotor. However, the per-phase copper loss in the rotor must be 1:R2. Thus, the per-phase power developed by the motor is ....

The above equation establishes the fact that the hypothetical resistance R,/s can be divided into two components: the actual resistance of the rotor R, and an additional resistance R, [(l - s)/s]. The additional resistance is called the load resistance or the dynamic resistance. The load resistance depends upon the speed of the motor and is said to represent the load on the motor because the mechanical power developed by the motor is proportional to it. In other words, the load resistance is the electrical equivalent of a mechanical load on the motor.

An equivalent circuit of an induction motor in terms of the load resistance is given in Fgr. 4. This circuit is proclaimed as the exact equivalent circuit of a balanced three-phase induction motor on a per-phase basis.

Fgr. 3 Per-phase equivalent circuit of a balanced three-phase induction motor as referred to the stator side.

Fgr. 4 The equivalent circuit of Fgr. 3 modified to show the rotor and the load resistances.

Power Relations:

Since the load resistance varies with the slip and the slip adjusts itself to the mechanical load on the motor, the power delivered to the load resistance is equivalent to the power developed by the motor. Thus, the performance of the motor at any slip can be determined from its equivalent circuit, as given in Fgr. 4.

For a balanced three-phase induction motor ...

P, = 3VJ, cos 0

....where 6 is the phase difference between the applied voltage c1 and the stator winding current il. Since the power input is electrical in nature, we must account for the electrical losses first. The immediate electrical loss that must be taken into consideration is the stator copper loss.

The total stator copper loss is ....

If the core loss is modeled by an equivalent core-loss resistance, as shown in the figure, we must also take into account the total core loss (magnetic loss) as ...

The net power that is crossing the air-gap and is transported to the rotor by electromagnetic induction is called the air-gap power. In this case, the air-gap power is ...

The air-gap power must also equal the power delivered to the hypothetical resistance RJs. That is, ...

The electrical power loss in the rotor circuit is ...

Hence, the power developed by the motor is ....

...is the per-unit (normalized) speed of the motor.

Fgr. 5 Power-flow diagram when the core loss is (a) simulated by R,, and (b) treated as a part of the rotational loss.

The electromagnetic torque developed by the motor is ...

By subtracting the rotational loss from the power developed, we obtain the power output of the motor as ...

Since the core loss has .already been accounted for, the rotational loss includes the friction and windage loss Pfi and the stray-load loss Psi. The corresponding power-flow diagram is given in Fgr. 5a. When the core loss P, is also considered a part of the rotational loss, the core-loss resistance R, in Fgr. 4 must be omitted, then Eq. (13a) becomes Pas = Pin - Psc,. The power-flow diagram when the core loss is a part of the rotational loss is given in Figure 5b.

Speed-Torque Characteristic:

Equation (16) reveals that the torque developed by an induction motor is directly proportional to the square of the current in the rotor circuit and the equivalent hypothetical resistance of the rotor. However, the two quantities, the rotor current and the hypothetical rotor resistance, are inversely related to each other. For instance, if the rotor resistance is increased, we expect the torque developed by ...the motor to increase linearly. But any increase in the rotor resistance is accompanied by a decrease in the rotor current for the same induced emf in the rotor.

A decrease in the rotor current causes a reduction in the torque developed.

Whether the overall torque developed increases or decreases depends upon which parameter plays a dominant role.

Let us examine the entire speed-torque characteristic of the motor. At stand-still, the rotor slip is unity and the effective rotor resistance is R,. The magnitude of the rotor current, from Fgr. 3, is

Note that the rotor winding resistance R, is usually very small compared with its leakage reactance…

That is, R, << X,.

The starting torque developed by the motor is ....

As the rotor starts rotating, an increase in its speed is accompanied by a decrease in its slip. As s decreases, R,/s increases. As long as R,/s is smaller than X,, the reduction in the rotor current is minimal. Thus, in this speed range, the rotor current may be approximated as ...

Since the rotor current is almost constant, the torque developed by the motor increases with the increase in the effective resistance R,/s. Thus, the torque developed by the motor keeps increasing with the decrease in the slip as long as the rotor resistance has little influence on the rotor current.

When the slip falls below a certain value called the breakdown slip sb, the hypothetical resistance becomes the dominating factor. In this range, RJs >> X, and the rotor current can be approximated as ...

Since the rotor current is almost constant, the torque developed by the motor increases with the increase in the effective resistance R,/s. Thus, the torque developed by the motor keeps increasing with the decrease in the slip as long as the rotor resistance has little influence on the rotor current.

When the slip falls below a certain value called the breakdown slip sb, the hypothetical resistance becomes the dominating factor. In this range, RJs >> X, and the rotor current can be approximated as ....

Fgr. 6 Typical speed-torque characteristic of a three-phase induction motor.

EXAMPLE 2:

A 6-pole, 230-V, 60-H~~ Y-connected, three-phase induction motor has the following parameters on a per-phase basis: R, = 0.5 R, R, = 0.25 R, X, = 0.75 R, X, = 0.5 R, X, = 100 R, and R, = 500 R. The friction and windage loss is 150 W. Determine the efficiency of the motor at its rated slip of 2.5%.

SOLUTION:

The synchronous speed of the motor is ....

The per-phase applied voltage is ....

The effective rotor impedance as referred to the stator is ....

Hence, the total input impedance is ...

The stator current: I, =

The power factor: pf =

Power input: Pin = 3V

Stator copper loss: Psce

Core-loss current: fc =

Magnetization current

Excitation current: f+ =

Hence, the rotor current…

Core loss: P, =

Air-gap power: Pa8 =

Rotor copper loss: Prce

Power developed: Pd =

Power output Po =

Efficiency: q =

Shaft torque: T,

Exercises:

A 10-hp, 4-pole, 440-V, 60-Hz, Y-connected, three-phase induction motor runs at 1725 rpm on full load. The stator copper loss is 212 W, and the rotational loss is 340 W. Determine (a) the power developed, (b) the air-gap power, (c) the rotor copper loss, (d) the total power input, and (e) the efficiency of the motor. What is the shaft torque? A 2-hp, 120-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor operates at 1650 rpm on full load. The rotor impedance at standstill is 0.02 + j0.06 Rl_phase. Determine the rotor current if the rotational loss is 160 W. What is the magnitude of the induced emf in the rotor? A 208-V, 50-Hz, 12-pole, Y-connected, three-phase induction motor has a stator impedance of 0.1 + j0.3 W_phase and a rotor impedance of 0.06 + j0.8 R/phase at standstill. The core-loss resistance is 150 R/phase, and the magnetization reactance is 750 W_phase. The friction and windage loss is 2 kW. When the motor operates at its full-load slip of 5%, determine (a) the power input, (b) the stator copper loss, (c> the rotor copper loss, (d) the air-gap power, (e) the power developed, (f) the power output, (g) the efficiency, (h) the shaft torque, and (i) the horsepower rating of the motor.

Fgr. 7 An approximate equivalent circuit on a per-phase basis of a balanced three-phase induction motor.

5. An Approximate Equivalent Circuit

A well-designed three-phase induction motor usually meets most of the following guidelines:

1. The stator winding resistance is kept small in order to reduce the stator copper loss.

2. The stator winding leakage reactance is minimized by reducing the mean-turn length of each coil.

Thin laminations of low-loss steel are used to cut down the core loss. Thus, the equivalent core-loss resistance is usually high.

The permeability of steel selected for laminations is high, and the operating flux density in the motor is kept below the knee of the magnetization curve.

Thus, the magnetization reactance is usually high.

An induction motor conforming to the above stipulations can be represented by an approximate equivalent circuit, as shown in Fgr. 7. In this case, we have placed the parallel branch (the excitation circuit) across the power source. We admit that the analysis of an induction motor using the approximate equivalent circuit is somewhat inaccurate, but the inaccuracy is negligible for a well-designed motor. On the other hand, the approximate equivalent circuit not only simplifies the analysis but also aids in comprehending various characteristics of the motor.

For instance, we use the approximate equivalent circuit to determine the speed at which (a) the torque developed is maximum, (b) the power developed is maximum, and (c) the motor efficiency is maximum. Prior to proceeding further, let us examine the error introduced when an induction motor is analyzed using an approximate equivalent circuit.

EXAMPLE 3

Using the data of Example 9.2 and the approximate equivalent circuit, determine the efficiency of the motor at its rated slip.

Core-loss current: I, =

Magnetization current: I

The equivalent impedance of the series circuit is ...

Hence, the rotor current is ...

The per-phase current supplied by the source is ...

Power input: Pi,...

Stator copper loss: Psce....

Rotor copper loss: Prce...

Core loss: P,

Power output:

Finally, the efficiency:

Exercises:

Redo Exercise 6 using the approximate equivalent circuit.

The equivalent circuit parameters of a 208-V, ~0-HZ, 6-p0le, Y-connected, three-phase induction motor in ohms/phase are R, = 0.21, R, = 0.33, X, = 0.6, X, = 0.6, R, = 210, and X, = 450. When the motor runs at a slip of 5% on full load, determine the torque developed by the motor using the approximate equivalent circuit. What is the starting torque developed by the motor?

6. Maximum Power Criterion

From the equivalent circuit as given in Fgr. 7, the rotor current is [...]

The power developed by the three-phase induction motor, from Eq. (15), is [...]

From the above equation it’s evident that the power developed by a three-phase induction motor is a function of slip. Therefore, we can determine the slip sp at which the power developed by the motor is maximum by differentiating the above equation and setting the derivative equal to zero. After differentiating and canceling most of the terms, we obtain ...

...where Z, is the magnitude of the equivalent impedance of the stator and the rotor windings at rest. That is ...

Equation (24) states that the power developed by a three-phase induction motor is maximum when the equivalent load (dynamic) resistance is equal to the magnitude of the standstill impedance of the motor. This, of course, is the well-known result we obtained from the maximum power transfer theorem during the study of electrical circuit theory.

From Eq. (24) we obtain the slip at which the induction motor develops maximum power as ...

Substituting for the slip in Eq. (23) we obtain an expression for the maximum power developed by a three-phase induction motor as ....

The net power output, however, is less than the power developed by an amount equal to the rotational loss of the motor. ....

EXAMPLE 4

A 120-V, 60-Hz, 6-pole, A-connected, three-phase induction motor has a stator impedance of 0.1 + j0.15 R/phase and an equivalent rotor impedance of 0.2 + j0.25 fl/phase at standstill. Find the maximum power developed by the motor and the slip at which it occurs. What is the corresponding value of the torque developed by the motor?

SOLUTION:

From Eq. (26), the slip at which the motor develops maximum power is ...

The maximum power developed by the motor is ...

The synchronous speed of the motor is ...

Thus, the torque developed by the motor is ....

Exercises:

A MO-V, 50-Hz, %pole, three-phase, A-connected induction motor has a stator impedance of 0.05 + j0.25 Rj_phase and an equivalent rotor impedance of 0.15 + 10.35 Rj_phase at standstill. Determine (a) the slip at which the motor develops maximum power, (b) the maximum power developed by the motor, and (c) the corresponding value of the torque developed.

A 208-V, 60-Hz, 4-pole, three-phase, Y-connected induction motor has the following constants in ohms/phase: R, = 0.08, XI = 0.5, R, = 0.1, X, = 0.6. Determine the slip at which the motor develops maximum power.

What is the maximum power developed by the motor? What is the corresponding torque developed by the motor?

Fgr. 8 Effect of rotor resistance on the breakdown slip.

7. Maximum Torque Criterion

The torque developed by a three-phase induction motor, from Eq. (23), is ... where Re = R, + R, and X, = XI + X,. Differentiating the above equation with respect to s and setting it equal to ...zero, we obtain an expression for the breakdown slip sb at which the motor develops the maximum (breakdown) torque as ....

Note that the breakdown slip is directly proportional to the rotor resistance. Since the rotor resistance can be easily adjusted in a wound-rotor induction motor by means of an external resistor, we can obtain the maximum torque at any desired speed, including the zero speed (starting). Substituting the above expression for the breakdown slip in Eq. (28), we obtain an expression for the maximum torque developed by the motor as ...

Note that the maximum torque developed by the motor is independent of the rotor resistance. In other words, the motor develops the same maximum torque regardless of its rotor resistance. The rotor resistance affects only the breakdown slip (or breakdown speed) at which the torque is maximum, as illustrated in Fgr. 8.

EXAMPLE 5:

Using the data of Example 4, determine (a) the breakdown slip, (b) the breakdown torque, and (c) the corresponding power developed by the motor.

SOLUTION

The motor parameters given in Example 4 are ...

The breakdown slip from Eq. (29) is ....

From Eq. (30), the maximum (breakdown) torque developed by the motor is ...

The power developed by the motor at the breakdown slip is ...

When we compare the expressions for the breakdown slip, Eq. (29), with the slip at which the motor develops maximum power, Eq. (26), we find that the denominator in Eq. (26) is larger than that in Eq. (29). In other words, the motor develops maximum power at a slip lower than that at which it develops maximum torque (sp < sb).

Further Approximations :

When the stator impedance is so small that it can be neglected in comparison with the rotor impedance at standstill, we obtain a very useful expression for the breakdown slip, from Eq. (29), as ....

This equation states that the breakdown slip is simply a ratio of rotor resistance to rotor reactance. When the rotor resistance is made equal to the rotor reactance, the breakdown slip is unity. In this case, the motor develops the maximum torque at starting.

The approximate expression for the breakdown torque, Eq. (30), becomes ....

In order to use Eq. (32), the induction motor must operate at a very low slip.

In fact, most three-phase induction motors operate below a slip of 5%, and the approximate relationship can be used to estimate the maximum torque developed by the motor.

The rotor current at any speed when the stator impedance is neglected is ....

The torque developed by the motor at any slip s is ...

The ratio of the torque developed at any slip s to the breakdown torque is ....

This is a very useful, albeit approximate, relation that can be used to determine the torque developed at any slip s once the breakdown torque Tdm and the breakdown slip sb are known.

EXAMPLE 6:

A 208-V, 60-Hz, 8-pole, Y-connected, three-phase wound-rotor induction motor has negligible stator impedance and a rotor impedance of 0.02 + j0.08 n/phase at standstill. Determine the breakdown slip and the breakdown torque. What is the starting torque developed by the motor? If the starting torque of the motor has to be 80% of the maximum torque, determine the external resistance that must be added in series with the rotor.

SOLUTION:

From Eq. (9.35), the starting torque (s = 1) in terms of the maximum torque is ...

Since the starting torque is only 47% of the maximum torque, we must add some resistance to the rotor circuit. As the rotor resistance is increased, the breakdown slip also increases. If sbn is the new breakdown slip, then ...

The two roots of the above quadratic equation are 2 and 0.5. Hence, the new breakdown slip must be 0.5. The corresponding value of the rotor resistance, from Eq. (9.31), is ...

Thus, the external resistance that must be added in each phase in series with the rotor resistance is 0.02 R. ...

Exercises:

Compute the breakdown slip and the breakdown torque of the motor of Exercise 9.9. What is the corresponding power developed by the motor? Using the data of Exercise 9.10, calculate (a) the breakdown speed, (b) the breakdown torque, and (c) the power developed by the motor at its breakdown slip.

The rotor impedance at standstill of a three-phase, Y-connected, 208-Vj. 60-Hz, 8-pole, wound-rotor induction motor is 0.1 + j0.5 Q/phase. Determine the breakdown slip, the breakdown torque, and the power developed by the motor. What is the starting torque of this motor? Determine the resistance that must be inserted in series with the rotor circuit so that the starting torque is 50% of the maximum torque.

8. Maximum Efficiency Criterion

When the core loss is considered a part of the rotational loss the power input to the motor using the approximate equivalent circuit, is ...

...where 0 is the power-factor angle between the applied voltage v, and the rotor current ';. ...

The power output is ...

The motor efficiency is ...

Differentiating q with respect to I, and setting the derivative equal to zero, we obtain ...

as the criterion for the maximum efficiency of an induction motor. It simply states that the efficiency of an induction motor is maximum when the sum of the stator and the rotor copper losses is equal to the rotational loss.