4.1 Name ONE application of a three-phase motor - NSC Electrical Technology Power Systems - Question 4 - 2016 - Paper 1

1. For the same size, a three-phase motor delivers more power due to better efficiency and performance.
2. Three-phase motors are self-starting, which simplifies operations.
3. They are more robust and have a simpler construction, leading to lower maintenance.

It is necessary to reduce the starting current to prevent unnecessary tripping of the motor and to minimize wear and tear on components at startup. High starting currents can lead to overheating and potential damage.

The star-delta starter reduces the starting current by initially connecting the motor in star configuration, which reduces the voltage across each phase. Consequently, this lowers the current drawn by the motor at startup.

Three-phase induction motors operate based on the principle of generating a rotating magnetic field. The alternating currents supply induces a magnetic field at 120° intervals across the three phases. This rotating magnetic field passes through the rotor, inducing an electromotive force (EMF) that causes current to flow in the rotor conductors, leading to rotational motion.

Please refer to the terminal box diagram in FIGURE 4.1, ensuring to illustrate the connection of the motor coils as configured in a star formation.

The megger should show a high resistive value, more than 1 MΩ. This indicates no electrical connection between the windings, confirming good insulation and a sound motor.

If the reading on the megger across U1 and E shows low resistance, this indicates there is a potential electrical fault. Energising the motor could lead to electric shock hazards for operators.

Using the formula for pole pairs:

o = rac{60 imes f}{n_s}
where:
f = 50 Hz and n_s = 600 r/min,
we calculate the number of pole pairs as follows:

o = rac{60 imes 50}{600} = 5
Thus, the motor has 5 pole pairs.

Using the formula:
S = rac{ ext{√3} imes V_l imes I_l}{1000}
Substituting the values, we have:
S = rac{ ext{√3} imes 380 imes 16}{1000} = 10.53 ext{ kVA}

At 100% efficiency:
P = S = 10.53 ext{ kVA} ext{ or } 8.95 ext{ kW}.

The actual output power of the motor is calculated using:
P_o = P imes η
Substituting the values gives:
P_o = 8.95 ext{ kW} imes 0.90 = 8.06 ext{ kW}.