4.1 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 changing the connection of any two of the three supply phases. By interchanging any two phase connections, the magnetic field direction will also change, thus reversing the motor's rotation.

Connecting the stator in delta usually develops the greatest torque as it allows higher line current to flow through the windings, resulting in increased motor torque. The star connection, while suitable for starting, limits the current and thus reduces the initial torque produced.

One advantage of a three-phase induction motor over a single-phase induction motor is that it provides a smoother and more continuous torque, making it more efficient and suitable for industrial applications.

The readings indicate a possible fault such as a short circuit in one of the windings, particularly if the resistive reading between the end terminals is significantly lower than expected.

A resistive reading of 0 Ω between U2 and E indicates a short circuit, meaning there is a direct connection between the two points, which could lead to potential motor failure and safety hazards.

To carry out the insulation test, disconnect the power supply and use a megger to apply a high voltage (usually between 500V and 1000V) between each winding and ground. Measure the insulation resistance for each winding; it should typically be above 1 MΩ. Repeat this process between all pairs of windings.

The rotor speed can be calculated using the formula: $n_r = n_s (1 - S)$ where:

• $n_s = 1500$ rpm (synchronous speed)
• $S = 0.06$ (slip) Thus, $n_r = 1500 imes (1 - 0.06) = 1410 ext{ rpm}$

The frequency of the supply determines the synchronous speed of the motor. A change in frequency alters the speed at which the magnetic field rotates. If the frequency is reduced, the motor's speed decreases, which can lead to a reduced torque output and potentially affect the system's load negatively.

To calculate the apparent power ($S$): $S = \frac{P_{out}}{\eta \cdot cos(\phi)}$ Here, substituting the values:

• $P_{out} = 6800$ W
• $\eta = 0.95$
• $cos(\phi) = 0.8$ Thus, $S = \frac{6800}{0.95 \cdot 0.8} \approx 8947.36 ext{ VA}$

To find the reactive power $Q$, we use the formula: $Q = S \cdot \sin(\theta)$ where $\theta = \cos^{-1}(0.8) \approx 36.87°$. Using the apparent power calculated earlier: $Q = 8947.36 \cdot \sin(36.87°) \approx 5368.42 ext{ VAR}$

One practical situation is in a conveyor belt system where two motors are used to drive different sections of the conveyor.

If contact MC1 / NO2 were permanently closed, the timer contactor would be energized constantly. This could lead to both motors starting simultaneously without the proper control sequence, which might cause overloading or mechanical failure.

Under normal conditions, when the start button is pressed, MC1 is energized, starting motor M1. As this contact closes, it will maintain the circuit. Once motor M1 starts and after a predetermined time, contact MC2 will then energize, starting motor M2 while MC1 remains energized.

The control circuit can accommodate two motors rated differently by using timers and contactors strategically. While starting one motor, the circuit ensures that each motor is energized independently and will protect against overload conditions by managing the timing and sequencing of motor starts.