3.1 State the size of the angles between the phases of a balanced three-phase AC generated waveform - NSC Electrical Technology Power Systems - Question 3 - 2018 - Paper 1

Apparent power is defined as the product of the current and voltage in an AC circuit, not considering the effect of reactance. It is the total power drawn from the supply.

Power factor is the ratio of the active power (real power) consumed in an AC circuit to the apparent power. It indicates the efficiency with which the electrical power is used.

1. Improved power factor results in reduced energy losses in the distribution system.
2. Suppliers can avoid penalties that may be applied for low power factor.
3. Increased capacity of the supply system, allowing for additional loads without requiring upgrades.

1. For the same frame size, single-phase generators produce less power than three-phase generators.
2. Single-phase generation is relatively more expensive for the same amount of energy required as compared to three-phase generation.
3. Single-phase systems do not allow for effective load balancing, which can complicate management of power supply.

Connecting a three-phase alternator in star allows for the creation of a neutral point which can be grounded. This configuration also enables greater flexibility in the use of phase voltages, supporting various load types.

Copper losses in overhead transmission lines are reduced by increasing the transmission voltage, which in turn decreases the current in the lines. Since losses are proportional to the square of the current, reducing the current helps minimize these losses.

To calculate the phase voltage in a star connection, use the formula: $V_P = \frac{V_L}{\sqrt{3}}$ Substituting the given values: $V_P = \frac{380}{\sqrt{3}} = 219.39 \text{ V}$

The line current can be calculated using the formula: $I_L = \frac{P_{in}}{\sqrt{3} \cdot V_L \cdot \cos \theta}$ Substituting the given values: $I_L = \frac{18000}{\sqrt{3} \cdot 380 \cdot 0.8} = 34.19 \text{ A}$

The apparent power can be calculated using the formula: $S_{app} = \sqrt{3} \cdot V_L \cdot I_L$ Substituting the values: $S_{app} = \sqrt{3} \cdot 380 \cdot 34.19 = 22.50 \text{ kVA}$

The total input power to the motor can be calculated by summing the readings from the two wattmeters: $P_T = P_1 + P_2 = 1.2 \text{ kW} + 2.3 \text{ kW} = 3.5 \text{ kW}$

1. The two-wattmeter method is simpler and requires fewer measurements, making it easier to implement.
2. It provides a reliable measurement method without needing to include a neutral point.
3. It is less complicated and more cost-effective compared to the three-wattmeter method.