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2.1 State THREE advantages of a three-phase distribution system over a single-phase distribution system - NSC Electrical Technology Electronics - Question 2 - 2017 - Paper 1

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2.1 State THREE advantages of a three-phase distribution system over a single-phase distribution system. 2.2 Draw a fully labelled representation diagram of a three... show full transcript

Worked Solution & Example Answer:2.1 State THREE advantages of a three-phase distribution system over a single-phase distribution system - NSC Electrical Technology Electronics - Question 2 - 2017 - Paper 1

Step 1

State THREE advantages of a three-phase distribution system over a single-phase distribution system.

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Answer

  1. Load Distribution: Three-phase systems allow for more efficient load distribution compared to single-phase systems, which can lead to reduced current and losses in conductors.

  2. Voltage Stability: The presence of multiple phases provides greater voltage stability and reduces the likelihood of voltage drop.

  3. Higher Power Capacity: Three-phase systems can transmit more power over a given resistance when compared to single-phase systems, making them suitable for heavy industrial applications.

Step 2

Draw a fully labelled representation diagram of a three-phase generated voltage waveform in a three-phase system.

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Answer

The diagram should consist of three sine waves, each 120 degrees out of phase with one another. Label the waves as follows:

  • R Phase: Wave 1 (positive peak at 0°)
  • Y Phase: Wave 2 (positive peak at 120°)
  • B Phase: Wave 3 (positive peak at 240°)

Include a horizontal axis for time and indicate peaks above and troughs below the axis.

Step 3

State ONE disadvantage of using the two-wattmeter method to measure power in a three-phase system.

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Answer

One disadvantage of the two-wattmeter method is that it can be less accurate if the load is unbalanced or if the power factor is significantly leading or lagging, leading to erroneous power readings.

Step 4

Calculate the Line voltage

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Answer

Given:

  • S = 25 kVA
  • I_L = 38 A
  • p.f. = 0.9 lagging

The formula for calculating line voltage (V_L) is:

V_L = rac{S}{ rac{ ext{√3}}{3} imes I_L} Substituting in the values:

V_L = rac{25 imes 10^3}{ rac{ ext{√3}}{3} imes 38} = 379.84 V

Step 5

Calculate the Phase voltage

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Answer

To find the phase voltage (V_PH), use the formula:

V_{PH} = rac{V_L}{ ext{√3}} Substituting the value of V_L:

V_{PH} = rac{379.84}{ ext{√3}} = 219.31 V

Step 6

Calculate Impedance per phase

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Answer

The formula for calculating impedance per phase (Z_PH) is:

Z_{PH} = rac{V_{PH}}{I_{L}} Substituting the values:

Z_{PH} = rac{219.31}{38} = 5.77 Ω

Step 7

Describe how Eskom could benefit if consumers improved the power factor of their systems.

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Answer

Improving the power factor would lead to less wasted energy, thus making more power available for consumers. This increased efficiency in energy usage can help decrease the overall generation costs for Eskom, allowing them to allocate resources more effectively and potentially reduce electricity prices for consumers.

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