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

The star configuration allows for the provision of both single-phase and three-phase power as needed. It also permits a neutral and earth point which enhances safety.

A wattmeter measures the electric power dissipated by a load at a given moment in watts, while a kilowatt-hour meter measures the total energy consumed over time in kilowatt-hours.

3.4.1 The National Grid transmits electrical power to the customer. 3.4.2 The National control centre monitors and controls the National Grid.

Label waveform 2 as $V_B$ (or $V_{L1}$) and waveform 3 as $V_C$ (or $V_{L2}$).

The phase displacement between the three waveforms is 120 degrees.

Input Power, $P_{in} = \frac{P_{out}}{\eta} = \frac{25000}{0.85} = 29411.76 \text{ W}$

Using the formula $I_L = \frac{\sqrt{3} \cdot V_L \cdot Cos \phi}{P_{in}}$, we have: $I_L = \frac{\sqrt{3} \cdot 380 \cdot 0.87}{29411.76} \implies I_L = 51.36 A.$

The phase current can be calculated as $I_{ph} = I_L = 51.36 A$.

The method used is the two wattmeter method.

Total Power, $P_T = P_1 + P_2 = 14 kW + 18 kW = 32 kW$.

The power factor can be calculated using: $tan \phi = \frac{P_1 - P_2}{P_1 + P_2} = \tan^{-1}\left(-0.22\right) \implies \phi \approx -12.41^{\circ}$, Thus, $PF = 0.98$.

Using the formula $S = \sqrt{3} \cdot I_L \, V_L$, we can rearrange to find: $I_L = \frac{S}{\sqrt{3} \cdot V_{PH}} = \frac{20000}{\sqrt{3} \cdot 220} = 52.5 A.$