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9.9 Comparison Between Motor and Transformer

This section looks at how the voltage and the current interact in the stator and rotor windings. We make it by means of drawing comparisons between the functioning of a transformer and of an asynchronous motor.

Suppose we have an asynchronous motor that generates a rotating field at synchronous speed n₁ whose rotor spins at n₂ and a sinusoidal mutual flux (between stator and rotor) with a maximum value of Φ₀.

With this, the slip is:

s = (n₁ - n₂)/n₁, (9.6)

and slip frequency is then:

fs = p·ns/60 = p·s·n₁/60 = s·f₁ (9.7)

Similar to what happens in transformers, due to the resulting rotating field, an emf value also generates an emf wave.

E₁ = 4.44·N₁·Φ₀·f₁, where N₁ is the number of turns per phase in series in the stator, while f₁ corresponds to the frequency of the stator supply network.

The rotor also experiences a corresponding phase voltage in response to this same mutual interaction.

ES = 4.44·N₂·Φ₀·fS = 4.44·N₂·Φ₀·s·f₁ (9.8)

where N₁ represents the number of series loops per phase of the stator and f₁ denotes the frequency of the stator supply network. In the case of the rotor being still (as it happens at start-up), the rotor speed equals zero (n₂ = 0), the slip ratio equals unity (s = 1), and hence the rotor frequency is equivalent to the stator supply frequency (fS = f₁).

In this case, the induced electromotive forces per phase on the rotor hold a specific value of:

ES(s=1) = E₂ = 4.44·N₂·Φ₀·f₁ (9.9)

Suppose the rotor constant speed, then:

Question: What are the rotor speed n₂, the slip s, and the rotor frequency fs? The emf that appears on the rotor in this case is zero.

ES(s=1) = E₂ = 4.44·N₂·Φ₀·f₁ (9.10)

Therefore, in any rotor rotation scenario, it is true that:

ES = s·E₂

Getting an asynchronous motor started is the same that an unplanned short-circuits in the transformer, resulting in high circulating currents during this process. Motor operation under load is similar to transformer operation under load. During no-load operation (when no mechanical load connects to the shaft, and it only has to overcome its own frictional torque), the rotor speed is very near to the synchronous speed. Therefore, the motor's no-load operation (with no mechanical load connected to the shaft) corresponds to the transformer's no-load operation (with no electrical load connected to the secondary).

练习题

An asynchronous motor has a synchronous speed of $n_{1} = 1500$ rpm and the rotor spins at $n_{2} = 1425$ rpm. What is the slip $s$ ?

s = 0.05

s = 0.075

s = 0.95

s = 1.05

The slip frequency $f_{s}$ in an asynchronous motor is equal to the stator supply frequency $f_{1}$ multiplied by the slip $s$ .

The induced electromotive force per phase in the stator winding is given by the formula $E_{1} =$ ___ $\cdot N_{1} \cdot Φ_{0} \cdot f_{1}$ , where $N_{1}$ is the number of turns per phase and $f_{1}$ is the stator supply frequency.

If the rotor EMF at standstill is $E_{2} = 100$ V and the motor operates with a slip of $s = 0.04$ , what is the rotor EMF $E_{S}$ during operation?

E_{S} = 100

E_{S} = 96

E_{S} = 4

E_{S} = 2500

At motor start-up, when the rotor is stationary ( $n_{2} = 0$ ), the slip equals zero and the rotor frequency is much lower than the stator supply frequency.

Which of the following correctly compare asynchronous motor operation to transformer operation?

A. Motor starting is similar to an unplanned short-circuit in a transformer, causing high circulating currents.

B. Motor operation under load is similar to transformer operation under load.

C. Motor no-load operation corresponds to transformer no-load operation (no electrical load on secondary).

D. Motor starting is similar to transformer no-load operation.

In any rotor rotation scenario, the relationship between the rotor EMF $E_{S}$ and the rotor EMF at standstill $E_{2}$ is given by: $E_{S} =$ ___ $\cdot E_{2}$ .

Explain why the rotor speed during no-load operation of an asynchronous motor is very close to the synchronous speed, and what this implies for the slip and rotor EMF.

A three-phase induction motor is supplied at $f_{1} = 50$ Hz and operates with a slip of $s = 0.02$ . The rotor EMF at standstill is $E_{2} = 120$ V. What are the rotor frequency $f_{s}$ and the operating rotor EMF $E_{S}$ ?

f_{s} = 50

Hz,

E_{S} = 120

f_{s} = 1

Hz,

E_{S} = 2.4

f_{s} = 25

Hz,

E_{S} = 60

f_{s} = 100

Hz,

E_{S} = 240

The rotor of an asynchronous motor can never rotate at exactly the synchronous speed because Faraday's law requires relative motion between the magnetic field and the conductor to induce an EMF and produce torque.

According to Faraday's law, a relative speed between the magnetic field and conductor is required for EMF induction. If an induction motor's rotor were to rotate at exactly synchronous speed ( $n_{2} = n_{1}$ ), what would happen to the rotor induced EMF and the motor's ability to produce torque?

A. The rotor EMF would be maximum, and the motor would produce maximum torque

B. The rotor EMF would be zero, and the motor could not produce any torque

C. The rotor EMF would equal the stator EMF, and the motor would maintain constant torque

D. The rotor EMF would double, and the motor would accelerate beyond synchronous speed

A three-phase induction motor operates from a 50 Hz supply with a slip of 0.04. The rotor iron losses are relatively low compared to the stator iron losses. What is the frequency of the rotor currents, and why can lower quality sheet metal be used for the rotor?

A. 50 Hz; because the rotor experiences the same magnetic field frequency as the stator

B. 2 Hz; because the low slip frequency reduces rotor iron losses, making high-quality laminations unnecessary

C. 0.04 Hz; because the rotor frequency equals the slip value directly

D. 25 Hz; because the rotor frequency is always half the stator frequency

Which of the following statements correctly describe the conditions when a three-phase induction motor starts from rest (standstill)?

A. The slip equals 1 (100%)

B. The rotor frequency equals the stator supply frequency

C. The rotor EMF per phase is at its maximum value

E_{2} = 4.44 \cdot N_{2} \cdot Φ_{0} \cdot f_{1}

D. The starting condition is analogous to a transformer short-circuit, resulting in high circulating currents

E. The rotor speed equals the synchronous speed

During no-load operation of an induction motor, the rotor rotates at exactly the synchronous speed, resulting in zero slip and zero rotor current, which is directly analogous to a transformer operating with an open secondary winding.

The rotating stator field induces electromotive forces in the rotor windings according to Faraday's law. When the rotor is stationary at start-up, the rotor EMF reaches its maximum value $E_{2}$ . During normal operation, the rotor EMF is given by $E_{S} = \cdot E_{2}$ , which shows it is directly proportional to slip.

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9.9 Comparison Between Motor and Transformer