To make the calculation of the motor input current simpler, it is customary to define the impedances Z_P and Z_R, where Z_P is a single impedance that is equivalent to all impedance elements of the progressive magnetic field and Z_R is a single impedance that is equivalent to all impedance elements of the retrograde magnetic field. This representation is shown in Figure 108-07a.

This way, simply by looking at the circuit shown in Figure 108-07a it is possible to determine the values ​​of Z_P and Z_R, as it is enough to verify that it is the calculation of the equivalent impedance represented by the elements which are connected in series and parallel. Thus, using eq. 108-05 and eq. 108-06 we will find the impedance values progressive and retrograde.

Note carefully that after the numerical calculation of the values ​​of Z_P and Z_R, looking at these equations in rectangular form, we verify that there will be a real value added to an imaginary value. In Z_P, the real part represents the progressive electrical resistance, R_P , while the imaginary part represents the progressive electrical reactance, X_P. Obviously, the same is true when we analyze the retrograde impedance. In Z_R, the real part represents the retrograde electrical resistance, R_R , while the imaginary part represents the retrograde electrical reactance, X_R. These values ​​represent the Thévenin equivalent of the circuit. So, it is possible to write:

In Figure 108-08 we can see how we transform the circuit to obtain R_P and X_P. Obviously, the same is true when we analyze the retrograde impedance.

To complete this analysis, using eq. 108-05 and carrying out the operations indicated in the equation, we arrive at a final result where we obtain the real part and the imaginary part of Z_P. Taking into account that in general, X_m >> X₂ and, therefore, we can write X_m + X₂ ≈ X_m getting the results shown in equations eq. 108-7a and eq. 108-7b.

These equations are valid for the progressive field. Doing the same for the retrograde field, starting from eq. 108-06 and carrying out the appropriate operations and simplifications, we arrive at the equations eq. 108-8a and eq. 108-8b.

"Note that when using these equations there is no need to work with complex numbers.   There is a pdf available showing how we arrived at these equations. In addition, the results obtained with these equations are compared to those results of the example 9-1 from the book by Chapman, page 595,   5th edition.   If the reader is interested in read the pdf,   enough   Click here"

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    4.4   Power in the Air Gap of a Single-Phase Motor

After understanding the theoretical part involved in a single-phase induction motor, we can prepare a study to determine the various engine parameters.

Let's start by determining the motor input current, represented by I₁. Note that eq. 108-09 is obtained from the circuit shown in Figure 108-07a.

Note that the denominator of eq. 108-09 is the total impedance of the motor. In this way, we can write the eq. 108-09a.

It is important to note that the progressive air gap power of a single-phase induction motor is the power consumed by R_P, given that the progressive resistor R_P is the only resistor present in the progressive impedance Z_P. Similarly, the engine's air gap retrograde power is the power consumed by R_R. Thus, the air gap power of the engine can be found by calculating the difference between the power consumed by R_P and the power consumed by R_R. To maintain a nomenclature compatible with what was studied in three-phase induction motors, we will call the air gap power P_gap. Thus, the progressive power in the air gap will be represented by P_Pgap and the regressive power in the air gap as P_Rgap. So, we can write:

This equation represents the value of the power developed by the rotor to produce a torque necessary to maintain the rotor rotation. This difference between the quantities P_Pgap and P_Rgap is due to the fact that the progressive and retrograde fields act in opposite directions and, as a result, they subtract each other. Thus, we can determine how much of this power will be transformed into mechanical power using eq. 108-27 (see item 4.7). Note carefully that P_gap   NO represents losses in the motor copper due to the progressive and retrograde field. These losses are provided by eq. 108-21 (see item 4.6).

And the progressive power in the air gap is given by:

And the retrograde power in the air gap is given by:

Replacing the respective values ​​in eq.108-10 with eq. 108-11 and eq. 108-12, we obtain:

"Attention must be paid to eq. 108-10, as this equation expresses the power that will be transformed into power "mechanics" for the rotor. That is, This power will be directly related to the torque induced in the engine rotor, a torque that is essential to keep the engine moving. It is worth noting that, in technical literature, the symbols used in equations are fundamental for the correct understanding of mathematical concepts. The choice between adding or subtracting quantities can significantly change the meaning and result of an equation. Like this, Many technical literatures present this equation, using the same symbology, however, adding the two quantities instead of subtracting them. In reality, this sum refers to the electrical losses in the copper winding of the rotor."

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    4.5  Calculation of Torque in a Single-Phase Motor

When we studied the three-phase induction motor we saw that the torque induced in the rotor was given by eq. 107-32, repeated below for greater clarity.

Note that we use the synchronous speed in rad/s according to the frequency of the network that powers the motor, as remember that the progressive and retrograde fields rotate at this speed. It is also worth remembering the relationship between the machine's synchronous speed, ω_sync, the number of machine poles, P, and the network frequency, f, given by eq 108-14.

Since the two fields, progressive and retrograde, are rotating in opposite directions, the torque produced by the two fields are opposite. Then, the resulting torque developed is the difference between them, according to eq. 108-15.

Using eq. 108-13, it is also possible to express torque as eq. 108-16.

All these equations make it possible to calculate the induced torque in the engine. When we need to calculate the torque on the shaft of the motor, that is, when the motor is under load, then we must use the eq. 108-17, below.

To calculate ω_r we must use eq. 108-22, below.

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    4.6   Rotor Copper Losses

The joule losses occur in the copper of the rotor both due to the progressive field and the retrograde field. Losses due to progressive field are given by:

Losses due to the retrograde field are given by:

Therefore, the rotor copper losses are given by adding the losses due to the forward and backward field. Then, the losses are:

Here it is important to point out that eq. 108-21 only calculates the copper losses in the rotor due to the progressive and retrograde fields. It does not express the total losses in the motor's copper. To do this, it is necessary to take into account the joule loss in the resistor R₁, which can be calculated by P_R1 = R₁ I₁². This loss expresses the power lost in the main winding resistance. So, the total losses in the copper of the motor, which we will represent by P_cu , can be calculated by eq. 108-21a below.

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    4.7   Mechanical Power in Single-Phase Motor

After deducting the losses inherent to the operation of the engine, such as copper losses due to electrical resistance, ventilation and friction losses, and supplementary and core losses, mechanical power represents the energy effectively available to perform work, that is, that power that can be converted from electrical to mechanical form. This power, known as mechanical power can be expressed in several ways. Initially, we will define the relationship between the synchronous angular velocity and the angular velocity of the shaft of the motor, ω_r, which are related by the slip , according to eq. 108-22.

Thus, making the necessary equivalences we can write the mechanical power as:

Substituting eq. 108-22 in eq. 108-23, we obtain:

Substituting eq. 108-16 in eq. 108-24, we obtain:

And, based on eq. 108-11 and eq. 108-12, it is also possible to write eq. 108-26, or:

Simplifying the above expression further, we can also write that

Note that depending on the data provided in the problem statement, we have several alternatives to calculate the mechanical power.

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    4.8   Nominal Power of a Single-Phase Motor

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   4.9  Input or Electrical Power of a Single-Phase

Note that the electrical power (measured in watts) absorbed from the power supply network by a single-phase induction motor is considered the engine input power and is given by:

In this equation, cos φ represents the power factor of the motor, that is, it is the phase difference between the voltage applied V and the input current I₁.

We must also be aware that the machine's input power must satisfy eq. 108-30.

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    4.10   Efficiency in a Single-Phase Motor

The efficiency, η, of a single-phase induction motor is given by the ratio between the nominal power, that is, the power that the motor available on its axis and the input power, both measured in watts or another equivalent unit. So, we can write:

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    4.11   Study of Single-Phase Induction Motor no

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When a motor operates at no load, that is, without load, it consumes a minimum electrical current necessary to overcome the internal losses, such as iron losses and friction and ventilation losses. The empty condition is important to determine engine behavior in situations close to ideal operation. During a no-load test, the rated voltage is applied and the electric current is measured by an ammeter connected in series with the motor, as well as the real power consumed, which is indicated by the wattmeter. These measurements allow calculate motor rotational losses, which are crucial for understanding motor efficiency and for designing systems that use electric motors more effectively and economically. Therefore, this test is a valuable tool in electrical engineering to evaluate the performance of motors under controlled conditions.

Due to the high reluctance in the air gap of an induction motor, the current circulating through the winding of the stator is very high. Thus, the magnetization reactance, X_m, will be much smaller than the resistances that are connected in parallel with this reactance. As a consequence, the factor motor power becomes very low, indicating that most of the current supplied to the motor is out of phase with tension, contributing little to the production of actual work. This delay current, predominantly inductive, causes a significant voltage drop across the inductive components of the circuit.

Therefore, when the engine is running without load, there are two relevant simplifications to be made. The first is in the case of the progressive rotating field, where the portion R₂ / 2 s tends to be a very large value due to the small slip value. So, in this case, eq. 108-33 is valid.

The second simplification is in the case of the retrograde rotating field, where the portion X₂ / 2 + R₂ / 2 (s - 2) is very small when compared to X_m / 2, as shown in eq. 108-34.

This allows us to assemble the equivalent circuit for the case of the engine operating empty, as shown a Figure 108-10.

It is observed that a resistance called R_rot was added to the circuit, whose purpose is to represent the rotational losses of the machine. Thus, the power consumed by this resistance represents rotational losses, including losses in the iron-silicon core of the motor stator.

The current in the rotor due to the progressive flux is very small, so copper losses are neglected. However, the current due to retrograde flow is significant, and the corresponding copper losses are represented by 0.5 R₂, as shown in the circuit in Figure 108-10. Therefore, the wattmeter reading corresponds to the powers consumed by R₁ + R_rot + 0.5 R₂. This series association of resistances can be represented by a single resistance, which we will call empty resistance, whose symbol representative is R_NL. See eq. 108-35.

In the same way as we did for the empty resistance, we will define a empty reactance, as we can deduce from the circuit shown in Figure 108-10. Represented by X_NL, this reactance is defined by eq. 108-36.

And, after defining the no-load resistance and the no-load reactance, we can define the no-load impedance. Representing by Z_NL and using the usual definition of impedance, the modulus of this impedance is defined by eq. 108-37.

And finally, the power read by the wattmeter will be represented by P_NL. This power is directly related to the current measured by the ammeter and with R_NL, according to eq. 108-38.

Remembering that eq. 108-39 is also valid.

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We must note that since the motor has a blocked rotor and, therefore, s = 1, the resistance R₂ / s reduces to R₂, which is a very small value. Therefore, as the values ​​of R₂ and X₂ are small, the motor input current, I_SC, will circulate almost entirely through these components, instead of circulating by the magnetization reactance, X_m, which has a much higher value. So, we can say that under these conditions the circuit can be considered as a series of components R₁, X₁, R₂ and X₂. Thus, from the circuit shown in Figure 108-11, we easily see that the power, P_SC, measured by the wattmeter is given by:

It can be seen that we can calculate the power factor directly, as we know the other variables. This data informs the phase angle between the applied voltage, V_SC, and the electric current, I_SC. From the circuit, it is also possible to conclude that:

Another information that can be removed from the circuit is the electric current, I_m, according to eq. 108-42.

Where the value of Z_m is given by eq. 108-42a.

Now, knowing the value of I_SC and I_m, we can easily determine the value of I_r using eq. 108-41. And knowing the value of I_r, we can determine the impedance formed through the circuit where the current I_r flows. Calling this impedance Z_sec (secondary) we can express it through eq. 108-43 in its Cartesian form.

It is important to solve many problems the relationship between the modulus of the impedance and its respective parts real and imaginary. Therefore, we must pay close attention to eq. 108-44.

Another important fact that we must highlight is that this method finds the total resistance value, that is, R₁ + R₂. Then, To find the value of R₂ we must know the value of R₁. To find the value of R₁ we can use the method described in item 4.11.3, below.

It is worth noting that, in general, it is a consensus to establish a relationship between X₁ and X₂ of shape quite practical, or X₁ = X₂.

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Another widely used method to determine the electrical resistance of the motor magnetization winding, in this case R₁, is with the use of a variable DC voltage source and a DC ammeter. The DC source voltage, V_DC, is adjusted until the ammeter reads a current, I_DC, equal to rated current of the motor. As a result, after some time, the winding will heat up and the measured resistance value will be very close to the real value when the motor is operating at full load (as we know that the electrical resistance of the winding is a function of temperature).

It is also known that the reactance of the motor is null when subjected to direct current, as well as there will be no voltage induced in the rotor. So, we can determine R₁, given that the only quantity that limits the current in the winding is the stator resistance. This value is given by eq. 108-45.

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5.   Types of Single-Phase Induction Motors

All of these starting techniques are methods in which one of the motor's two rotating magnetic fields is made stronger than the other. In this way, the rotor receives a starting torque in a certain direction.

We will study each type of engine separately.

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    5.1   Split Phase Windings

The split phase motor is a classic type of induction motor, widely used in applications that do not require a high starting torque. The 90° phase shift between the stator windings is essential to create a magnetic field rotary, which is what allows the engine to operate. The main winding is responsible for continuous operation, while the auxiliary winding, in conjunction with the centrifugal switch, is used only during starting to help overcome the initial moment of inertia. The centrifugal switch is connected in series with the auxiliary winding. The split phase motor is a classic type of induction motor, widely used in applications that do not require a high starting torque. The 90° phase shift between the stator windings is essential to create a magnetic field rotating, allowing the engine to operate. The main winding is responsible for continuous operation, while the auxiliary winding, in conjunction with the centrifugal switch, is used only during starting to help overcome the initial moment of inertia. The centrifugal switch is connected in series with the auxiliary winding.

An important fact in this type of motor is to design an auxiliary winding that has a ratio resistance/reactance higher than in the main winding. To achieve this objective, typically, a thinner gauge wire is used for the auxiliary winding. This is possible because electrical current only flows in the auxiliary winding during starting. This avoids continuous circulation current, which would cause the winding to overheat. Due to the higher ratio of resistance/reactance in auxiliary winding, the electric current in this winding will suffer a small delay in relation to the applied voltage. Already in the main winding this delay will be much greater. As a result, the current in the auxiliary winding will be advance in relation to the current in the main winding. In this way, the auxiliary winding causes one of the opposing rotating magnetic fields of the rotor to be greater than the other, producing a net starting torque for the engine. The phasor plot shown in Figure 108-14 enhances what was stated above.