3.1 List THREE advantages of power factor improvement for the consumer - NSC Electrical Technology Power Systems - Question 3 - 2019 - Paper 1

The star connection in power distribution systems allows for a neutral point, which can be used to provide single-phase power as needed to consumers. This configuration also helps in maintaining a balanced load across all three phases and provides a return path for unbalanced currents, ensuring stable operation.

A wattmeter measures the instantaneous power (in watts) being consumed by a load at any given moment, while a kilowatt-hour meter records the total energy consumed over time, measured in kilowatt-hours, which accounts for both the load and the duration of its operation.

3.4.1 National Power Grid: The National Grid transmits electrical power from generation sources to consumers efficiently and reliably.
3.4.2 Eskom's National Control Centres: These centres monitor and control the National Grid's operation, ensuring stability, managing supply and demand, and coordinating maintenance and outage schedules.

Using the reference waveform marked as Vr:
2: Voltage waveform of Phase 2 (Vb)
3: Voltage waveform of Phase 3 (Vc)

The phase displacement between the three waveforms is 120°.

To find the input power, use the formula:
$P_{in} = \frac{P_{out}}{\eta}$
Where:
$P_{out} = 25 \, kW, \quad \eta = 0.85$
Therefore:
$P_{in} = \frac{25 \, kW}{0.85} \approx 29.41 \, kW$

To find the line current, use the formula:
$I_L = \frac{P_{in}}{\sqrt{3} V_L \cos \phi}$
Where:
$P_{in} \approx 29.41 \, kW, \quad V_L = 380 \, V, \quad \cos \phi = 0.87$
Calculating gives:
$I_L = \frac{29.41 \times 10^3}{\sqrt{3} \times 380 \times 0.87} \approx 51.36 \, A$

The phase current can be calculated as:
$I_{PH} = I_L = 51.36 \, A$

The method used is the two wattmeter method.

To find the total power used by the load, we sum the individual powers:
$P_T = P_1 + P_2 = 14 \, kW + 18 \, kW = 32 \, kW$

To find the power factor, we use the formula:
$\tan\phi = \frac{P_1 - P_2}{P_1 + P_2} = \frac{14 - 18}{14 + 18} \Rightarrow \tan\phi \approx -0.22$
Thus,
$\phi \approx \tan^{-1}(-0.22) \approx -12.41°$
The power factor is then given by:
$PF = \cos(-12.41°) \approx 0.98$

To calculate the line current at full load, use:
$S = \sqrt{3} V_L I_L$
Given:
$S = 20 \, kVA, \quad V_{PH} = 220 \, V$
Then,
$I_L = \frac{20 \times 10^3}{\sqrt{3} \times 220} \approx 52.5 \, A$