Refer to FIGURE 4.1 below and answer the questions that follow - NSC Electrical Technology Electronics - Question 4 - 2017 - Paper 1

The direction of rotation of the motor can be reversed by swapping any two of the three phase connections to the motor. This change in phase sequence reverses the magnetic field direction, causing the rotor to turn in the opposite direction.

The delta connection would develop the greatest torque compared to the star connection. This is because a delta connection provides higher voltage and current to the stator windings, resulting in greater starting torque, which is beneficial for motors that require high initial torque.

One advantage of a three-phase induction motor is that it provides a smoother and more continuous torque, leading to higher efficiency and better performance compared to single-phase induction motors.

The fault can be identified if there are significantly lower resistance values between the windings, indicating a short circuit or an insulation failure.

A reading of 0 Ω between U2 and E indicates a short circuit situation, meaning that there is no insulation resistance and current can flow directly, potentially causing damage to the motor.

The insulation test should be carried out by disconnecting the motor from the power supply, using a megger to apply a suitable DC voltage (usually 500V or 1000V), and measuring the resistance between each winding and the earth ground, ensuring that the readings are above the acceptable threshold (typically above 1 MΩ).

The rotor speed can be calculated using the formula:

$n_R = n_S(1 - S)$

Where:

• $n_S = 1500$ r/min (synchronous speed)
• $S = 0.06$ (slip)

Substituting the values gives:

$n_R = 1500(1 - 0.06) = 1410 \, r/min$

The frequency of supply is critical as it dictates the synchronous speed of the motor. Any change in frequency can lead to a change in the rotor speed, which may affect the performance of the motor under load conditions, possibly causing it to stall or operate inefficiently.

To calculate the apparent power ($S$), we can use the formula:

$S = \frac{P_{out}}{\eta \cdot \cos \phi}$

Given:

• $P_{out} = 6.8 kW$
• $\eta = 0.95$
• $\cos \phi = 0.8$

Calculating:

$S = \frac{6800}{0.95 \cdot 0.8} = 8947.36 \, VA$ So, the apparent power is approximately 8.95 kVA.

The reactive power ($Q$) can be calculated using the formula:

$Q = S \cdot \sin(\theta)$

Where $\theta = \cos^{-1}(0.8)$. Calculating:

1. Calculate $\theta$:

$\theta = \cos^{-1}(0.8) \approx 36.87^{\circ}$

2. Then, $Q = 8947.36 \cdot \sin(36.87) \approx 5368.42 \, VAR$ So, the reactive power is approximately 5.37 kVAR.

One practical situation is in a conveyor belt system, where two motors may need to operate in harmony to move materials effectively.

If the contact MC1 / NO2 was faulty and permanently closed, the timer contactor would be energized, starting both motors immediately without a delay, which could cause excessive wear or damage due to simultaneous starting under load.

Under normal conditions, pressing the start button energizes MC1, which then closes NO1, allowing the circuit to hold itself once the start button is released. The timer contactor is then activated, starting motor M2 after a predefined time, simultaneously ensuring MC1 releases T and that both motors are safely started.

The control circuit can accommodate two motors rated differently by utilizing the timer which can stagger the start-up of each motor, ensuring that the one with lower ratings is not overloaded while the other starts. This separation is crucial to prevent torque discrepancies that may arise from different power ratings.