3.1 Explain the following terms: 3.1.1 Efficiency 3.1.2 Power factor correction 3.2 State THREE disadvantages of three-phase generation in comparison with single-phase generation - NSC Electrical Technology Power Systems - Question 3 - 2021 - Paper 1

Power factor correction involves adjusting the power factor of an electrical system to be closer to unity (1.0). This is achieved by adding power-factor correction capacitors that counteract the inductive effects of motors and transformers, leading to improved efficiency and reduced energy costs.

1. Complexity: Three-phase systems require more complicated circuitry and protection devices, increasing installation and maintenance costs.
2. Equipment Size: Three-phase generators and motors are generally larger and heavier than their single-phase counterparts, making them less portable.
3. Imbalance Issues: Unequal loading across phases can cause efficiency losses and overheating in equipment, posing operational challenges.

In a three-phase system, the relationship between the phase voltage ( V_{ph}) and the line voltage (V_L) is given by: $V_{ph} = \frac{V_L}{\sqrt{3}}$

The phasor diagram should illustrate three vectors representing the phase voltages, each separated by 120 degrees, with the line voltage represented across the points of the star connection.

The generated electricity is lower at the point of distribution due to losses in the transmission system, which include resistive losses due to the resistance of the conductors, as well as induction and capacitance effects in the transmission lines.

Power-factor correction capacitors offset the inductive effects of motors, which draw lagging current. By introducing capacitance, they reduce the overall current drawn from the supply, improving the overall power factor and leading to better energy efficiency.

1. Reduced Demand Charges: Suppliers can incur lower demand charges from utilities as improved power factors lead to lower apparent power requirements.
2. Enhanced System Capacity: With a better power factor, suppliers can transmit more real power without exceeding load limits, increasing overall system efficiency.

The phase voltage ( V_{ph}) can be calculated using: $V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{380}{\sqrt{3}} \approx 219.39 \, V$

The active power (P) is calculated as: $P = S \cdot \cos \theta = 250000 \cdot 0.9 = 225000 \, W = 225 \, kW$

The reactive power (Q) can be calculated as: $Q = S \cdot \sin \theta = 250000 \cdot \sin(25.84^\circ) \approx 109 \, kVA$

A kilowatt-hour meter is used to measure the total electrical energy consumed by a load over time. It is essential for billing purposes and assessing energy usage.