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4.1 Name the processes at A, B and C - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

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4.1 Name the processes at A, B and C. 4.1.2 Draw the voltage waveforms generated at A. 4.1.3 Explain why Transformer 1 is a step-up transformer. 4.1.4 Determine t... show full transcript

Worked Solution & Example Answer:4.1 Name the processes at A, B and C - NSC Electrical Technology Power Systems - Question 4 - 2021 - Paper 1

Step 1

Name the processes at A, B and C.

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Answer

A - Generation B - Transmission C - Distribution

Step 2

Draw the voltage waveforms generated at A.

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Answer

The voltage waveforms can be represented as follows:

   +V   +V   +V
   |    |    |
 -V   -V   -V
   |    |    |
    -----------
   R   Y   B

The three waveforms are 120 degrees apart.

Step 3

Explain why Transformer 1 is a step-up transformer.

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Answer

Transformer 1 is a step-up transformer because it increases the voltage from the generation process to reduce the current in the transmission lines. This reduction in current leads to lower copper losses (I²R losses) in transmission, making it more efficient.

Step 4

Determine the type of connection used on the secondary windings of Transformer 3 and motivate your answer.

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Answer

The type of connection used on the secondary windings of Transformer 3 is a star connection. This configuration allows for both three-phase and single-phase equipment to be powered efficiently, which is essential for compatibility with various load types. Additionally, it facilitates the use of neutral for single-phase equipment.

Step 5

Calculate the phase current.

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Answer

To find the phase current ( I_{PH} ), we use the formula:

IPH=IL3=1538.66AI_{PH} = \frac{I_L}{\sqrt{3}} = \frac{15}{\sqrt{3}} \approx 8.66 \, A

Step 6

Calculate the apparent power.

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Answer

The apparent power (S) can be calculated using the formula:

S=3×VL×IL=3×380×159.87kVAS = \sqrt{3} \times V_L \times I_L = \sqrt{3} \times 380 \times 15 \approx 9.87 \, kVA

Step 7

Calculate the phase angle.

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Answer

The phase angle ( θ ) can be calculated using the power factor:

θ=cos1(0.9)25.84°θ = \cos^{-1}(0.9) \approx 25.84°

Step 8

Calculate the reactive power.

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Answer

The reactive power (Q) can be calculated using the formula:

Q=3×VL×IL×sinθ=3×380×15×sin(25.84°)4.3kVARQ = \sqrt{3} \times V_L \times I_L \times \sin θ = \sqrt{3} \times 380 \times 15 \times \sin(25.84°) \approx 4.3 \, kVAR

Step 9

State TWO advantages of power factor correction to the consumer.

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Answer

  1. Reduced current drawn from the supply, leading to a lower electricity bill.
  2. Improved efficiency and longevity of equipment, as reduced heat generation results in less wear and tear.

Step 10

Does the reading show a leading or lagging power factor?

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Answer

The meter reading indicates a lagging power factor.

Step 11

State the cause of this type of reading.

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Answer

The lagging power factor is caused by predominantly inductive loads connected to the system.

Step 12

State how the meter reading could be brought closer to unity.

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Answer

The reading can be brought closer to unity by connecting capacitors in parallel with the load or using synchronous motors.

Step 13

Calculate the total power of the load.

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Answer

The total power (P_T) is calculated by summing the individual powers:

PT=P1+P2=250+460=710WP_T = P_1 + P_2 = 250 + 460 = 710 \, W

Step 14

State TWO quantities, other than power, that can be determined by using the two-wattmeter method.

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Answer

  1. Reactive power (Q)
  2. Phase angle (θ)

Step 15

Explain why the two-wattmeter method is preferred over the three-wattmeter method.

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Answer

The two-wattmeter method is preferred as it can be easily connected to the lines without requiring additional equipment, making it more economical for load measurement.

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