5.1 State THREE mechanical inspections to be conducted on a three-phase motor after installation, but before commissioning - NSC Electrical Technology Power Systems - Question 5 - 2020 - Paper 1

1. It is cheaper and more robust than other motor types.
2. It has slightly higher efficiency and power factor.
3. These motors are explosion-proof, reducing the risk of sparking due to the absence of slip rings and brushes.

The control circuit shown in FIGURE 5.3 is a sequential motor starter without a timer.

The function of the stop button is to disconnect the supply from the control circuit and to stop both motors.

The function of MC1 (N/O1) is to allow current to flow in the parallel circuit even after the start button is released; it is the hold in contact.

1. When the start button 1 is pressed, the current flows through the stop button and O/L1.
2. MC1 (Motor 1) will energize.
3. MC1 N/O1 and MC1 N/O2 will close, allowing Motor 1 to start running.
4. When start button 2 is pressed, MC2 (Motor 2) will energize and close holding in contact with MC2 N/O1, allowing Motor 2 to start.
5. The two motors will run respectively.

The synchronous speed can be calculated using the formula: $n_s = \frac{120 \times f}{p}$ Substituting the given values: $n_s = \frac{120 \times 50}{6} = 1000 \text{ rpm}$.

The rotor speed is calculated using the formula: $n_r = n_s(1-s)$ For a slip of 0.05: $n_r = 1000(1-0.05) = 950 \text{ rpm}.$

Using the formula for line current: $I_L = \frac{\sqrt{3} \times V_L \times \cos \theta}{1000}$ Substituting values: $I_L = \frac{\sqrt{3} \times 380 \times 0.8}{1000} = 0.34 \text{ kA}.$

Using the formula for apparent power: $S = \frac{\sqrt{3} \times V_L \times I_L}{1000}$ Substituting values: $S = \frac{\sqrt{3} \times 380 \times 0.34}{1000} = 22.5 \text{ kVA}.$