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 a three-phase induction motor can be reversed by swapping any two of the three supply phases. This change alters the magnetic field's direction, causing the rotor to rotate in the opposite direction.

In a star connection, each winding receives a lower voltage, but it provides a more stable start-up with less current drawn from the supply. In contrast, a delta connection allows for higher line voltage across the windings, leading to greater torque production. Hence, the delta connection typically develops the greatest torque due to the higher voltage applied.

One advantage of a three-phase induction motor over a single-phase induction motor is that it provides a smoother power output, which results in better starting torque and overall efficiency.

Upon reviewing the readings from FIGURE 4.3, if one or more of the windings shows a significantly lower reading compared to others, it indicates a potential fault such as a short circuit or winding failure. The precise nature of the fault would depend on the specific readings observed.

A resistive reading of 0 Ω between U2 and E indicates a short circuit between those two points, suggesting that there is a direct connection, causing a possible failure in the insulation of the winding or a fault in the motor. This condition could lead to equipment damage or electrical hazards.

To carry out the insulation test, follow these steps:

1. Disconnect power supply to the motor and ensure it is safely isolated.
2. Connect the megger leads to the windings you intend to test.
3. Apply a voltage (typically 500V DC) to the windings and measure the insulation resistance.
4. A reading of above 1 MΩ is generally acceptable, indicating good insulation. Record results and compare against standards.

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

• $n_s$ = synchronous speed, which can be calculated as: n_s = rac{120 imes f}{P} Given:
• Supply frequency, $f = 50 Hz$
• Number of poles, $P = 2$ (for a 1500 r/min synchronous speed, implies 2 poles) Thus, n_s = rac{120 imes 50}{2} = 3000 ext{ RPM} Substituting values: $n_r = 3000(1 - 0.06) = 2820 ext{ RPM}$

The frequency of the supply is critical in determining the synchronous speed of the motor. A higher frequency results in a higher synchronous speed, which affects how the motor performs under load. If the frequency fluctuates, it may lead to improper operation of the motor, inefficient power usage, or damage to connected loads. Consistent frequency ensures stable operation and optimal load performance.

The apparent power can be calculated with: S = rac{P_{out}}{ ext{Efficiency}} Given:

• Output power, $P_{out} = 6800 W$
• Efficiency, $ext{efficiency} = 95 ext{ } ( ext{or }0.95)$ Thus: S = rac{6800}{0.95} = 7157.89 VA = 7.16 kVA

To calculate the reactive power, we first find the power factor: $ext{Power Factor (p.f.)} = 0.8$ Using: $Q = S imes ext{sin}( heta)$ Where: $heta = ext{cos}^{-1}(0.8)$ Calculating:

1. Find $heta$ using a calculator or cosine table.
2. Substitute into the reactive power formula: $Q = S imes ext{sin}( heta)$ Using the previously calculated apparent power to find $Q$.

A practical situation where two motors may be started using the method in FIGURE 4.6 could be in a conveyor belt system where two different motors operate conveyor segments.

If contact MC1 / NO2 was faulty and permanently closed, it would cause the timer contactor to be energized constantly. This would result in both motors starting simultaneously whenever the start button is pressed, potentially leading to overload, equipment damage, or unsafe operational conditions.

In normal conditions, upon pressing the start button:

1. MC1 is energized, which starts motor M1.
2. Contact MC1 / NO1 closes to maintain the circuit for motor M1 once the start button is released.
3. The timer contactor energizes after a scheduled time, initiating the start of motor M2.
4. The process concludes with MC2 controlling motor M2, ensuring both motors operate safely and efficiently.