FIGURE 4.1 below shows a block diagram of the national power grid in South Africa - 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 that allows electricity to be generated, transmitted, and distributed across the country.

Electricity is transmitted at high voltages to decrease the current flowing in the transmission lines, which reduces the copper losses associated with power transmission.

An industrial consumer typically uses a three-phase voltage of 400 V, while domestic consumers use a single-phase voltage of 230 V.

1. A three-phase system is more economical and efficient for powering large machinery.
2. A three-phase system can supply both three-phase and single-phase installations, offering flexibility in usage.

The line current can be calculated using the formula:

$S = \sqrt{3} \cdot V_L \cdot I_L$

Thus,

$I_L = \frac{S}{\sqrt{3} \cdot V_L} = \frac{200,000}{\sqrt{3} \cdot 400} = 288.68 \, A$

The phase current can be calculated as follows:

$I_{PH} = \frac{I_L}{\sqrt{3}} = \frac{288.68}{\sqrt{3}} = 166.67 \, A$

The power factor can be calculated using:

$Cos \phi = \frac{P}{S} = \frac{180,000}{200,000} = 0.9$

To find the reactive power, we use:

$\theta = \cos^{-1}(0.9) = 25.84°$

Then,

$Q = \sqrt{S^2 - P^2} = \sqrt{(200,000)^2 - (180,000)^2} \approx 87.17 \, kVAr$

Power factor correcting devices may be installed at central substations or built into power consuming equipment.

One improvement could be the reduction of reactive power.

Power factor correction reduces the phase angle and apparent power, which decreases the total current drawn from the supply while keeping the load and supply voltage stable.

Wattmeters are used to measure instantaneous power in circuits, while energy meters measure the total electrical energy consumed over a specific period.

The reading of the second wattmeter can be calculated as:

$P_2 = P_{total} - P_1 = 3500 - 1300 = 2200 \, W$