5.1 Name parts A and B - NSC Electrical Technology Power Systems - Question 5 - 2021 - Paper 1

Using a rotor without brushes and slip rings requires less maintenance since it eliminates the need to replace brushes and avoids issues with sparking.

Skewing the rotor bars helps to reduce cogging, which minimizes the tendency of the rotor to lock into fixed positions, improving smooth operation.

Slip is defined as the difference between the synchronous speed (S) and the actual rotor speed (R) of an induction motor, expressed as a percentage: ext{Slip} = rac{S - R}{S} imes 100. It indicates how much slower the rotor is compared to the rotating magnetic field produced by the stator.

Commissioning refers to the process where the electric motor is connected to the power supply and load after all preparations, including electrical and mechanical inspections, have been completed. It ensures that the motor is ready for its operational environment.

Inspecting mounting bolts to ensure they are properly tightened is crucial to prevent mechanical failure during operation.

The pole pairs per phase can be calculated as follows: $ext{Pole pairs per phase} = \frac{12}{3} = 4$. Thus, there are 4 pole pairs.

The synchronous speed (n_s) is given by the formula: $n_s = \frac{f \times 60}{p} = \frac{50 \times 60}{12} = 250$ The synchronous speed of the motor is 1500 rpm.

The rotor speed (n_r) can be calculated as: $n_r = n_s \cdot \left(1 - \frac{3}{100}\right) = 1500 \cdot 0.97 = 1455 \, \text{rpm}$ The rotor speed is 1455 rpm.

The efficiency (η) can be calculated using the formula: $\eta = \frac{P_{in} - \text{losses}}{P_{in}} \times 100$ Plugging in the values: $\eta = \frac{25000 - 800}{25000} \times 100 = 96.8\%$ The efficiency of the motor is 96.8%.

The control circuit shown in FIGURE 5.5 is a forward-reverse control circuit.

The OLN/C component will open the moment the current surpasses the preset rated current, stopping the flow of current to the motor and ensuring safe operation.

The MC2/N/O component is the retain contact for the reverse contactor, ensuring that current continues to flow to MC2 after the start button is released.

The MC2/N/C contact is used as an interlocking contact. When MC1 (forward) is activated, the MC2/N/C contact opens, ensuring that current cannot flow to MC2, preventing accidental activation of both forward and reverse operations simultaneously.

Let $I_{FL}$ be the full-load current. Given the maximum starting line current is 7 times full-load current, we can express it as: $I_{max} = I_{FL} imes 7 \Rightarrow 100 = I_{FL} \times 7 \Rightarrow I_{FL} = \frac{100}{7} \approx 14.29 \, A$ The full-load current of the motor is approximately 14.29 A.