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 (S) is defined as the product of the current (I) and the voltage (V) in an AC circuit, considering no reactance. It is measured in volt-amperes (VA) and represents the total power drawn from the supply.

Power factor (PF) is defined as the ratio of the active power (P) consumed in an AC circuit to the apparent power (S). It indicates the efficiency of the power usage in the circuit and is a dimensionless number between 0 and 1.

1. Improved power factor reduces the load on the power supply system, which means thinner supply cables are required, minimizing costs.
2. It minimizes the overall energy losses in the transmission lines, leading to more efficient power distribution.
3. Cost of maintenance and equipment can be lowered, as more efficient systems typically require less upkeep.

1. Single-phase systems produce less power compared to three-phase systems for the same capacity.
2. Single-phase generation is generally more expensive for the same amount of energy production compared to three-phase.
3. A single-phase supply cannot balance loads as effectively as three-phase systems, leading to potential overload situations.

Connecting a three-phase alternator in star creates a neutral point, which allows for different phase voltages to be available. This setup can accommodate varying loads and enhances the system reliability as the neutral point can be grounded.

Copper losses are reduced by increasing the transmission voltage, which in turn decreases the line current. Given that power losses in conductors are proportional to the square of the current, lower currents lead to significantly reduced losses.

The phase voltage (V_p) is calculated using the formula: $V_p = \frac{V_L}{\sqrt{3}}$ Using the given line voltage of 380 V: $V_p = \frac{380}{\sqrt{3}} \approx 219.39 V$

The line current (I_L) is calculated using the formula: $I_L = \frac{P_{in}}{\sqrt{3} \times V_L \times Cos \theta}$ For this scenario: $I_L = \frac{18000}{\sqrt{3} \times 380 \times 0.8} \approx 34.19 A$

The apparent power (S) can be calculated using: $S = \sqrt{3} \times V_L \times I_L$ Substituting the known values: $S = \sqrt{3} \times 380 \times 34.19 \approx 22.50 kVA$

The total input power (P_T) to the motor is the sum of the power readings from the wattmeters: $P_T = P_1 + P_2 = 1200 + 2300 = 3500 W = 3.5 kW$

1. The two-wattmeter method requires fewer instruments, making it more cost-effective and simpler to set up.
2. It can measure power accurately in both balanced and unbalanced three-phase systems.
3. It involves less complex circuitry compared to the three-wattmeter method, simplifying the measurement process.