4.1.1 State the voltage generated at block A - NSC Electrical Technology Power Systems - Question 4 - 2024 - Paper 1

The standard international colour code for a three-phase system is as follows:

• Phase 1 - Red
• Phase 2 - Yellow
• Phase 3 - Blue

The national grid is a network of over 25,000 km of high voltage power lines for various bulk users. It facilitates the generation, transmission, and distribution of electricity across regions.

Electricity is transmitted at high voltages to reduce power losses in transmission lines. Higher voltage reduces the current flowing through the lines, minimizing heat losses due to resistance.

Industrial consumers typically utilize a three-phase voltage of 400 V, while domestic consumers receive a single-phase voltage of 230 V, as the industrial load requires higher power levels.

1. A three-phase system is more economical for supplying power, especially for larger loads.
2. It allows for efficient supply of both three-phase and single-phase installations, which aids in better load balancing.

The line current can be calculated using the formula: $S = \sqrt{3} V_L I_L$ Given: S = 200 kVA, $V_L$ = 400 V, we have: $I_L = \frac{S}{\sqrt{3} V_L} = \frac{200 \times 10^3}{\sqrt{3} \times 400} = 288.68 A$

The phase current can be calculated using: $I_{PH} = \frac{I_L}{\sqrt{3}} = \frac{288.68}{\sqrt{3}} = 166.67 A$

The power factor can be calculated using: $Cos \phi = \frac{P}{S} = \frac{180,000}{200,000} = 0.9$

Using the power factor calculated: \theta = \cos^{-1}(0.9) \approx 25.84^\circ \ The reactive power is: $Q = \sqrt{S^2 - P^2} = \sqrt{(200,000)^2 - (180,000)^2} = 87.17 kVAr$

Power factor correcting devices can be installed at a central substation.

One improvement is the reduced reactive power after power factor correction.

Power factor correction reduces the phase angle and the apparent power. It leads to a lower current drawn from the supply due to better energy efficiency.

Wattmeters measure the instantaneous power in watts, while energy meters measure the total energy consumed over time, usually in kilowatt-hours (kWh).

If the total power is 3.5 kW (3500 W) and meter 1 reads 300 W, then the reading of meter 2 is: $P_2 = P_{total} - P_1 = 3500 - 300 = 3200 W$