Physics • Year 12 • Module 6 • Lesson 20
Power Transmission and Distribution
Build HSC Band 5–6 extended-response technique: multi-step quantitative analysis, data evaluation, and engineering justification for transmission voltage decisions.
1. Data + scenario: Eastern Australia interconnect upgrade (Band 5–6)
9 marks Band 5–6
Scenario. Engineers are planning to upgrade a 500 km transmission corridor that currently carries 1 200 MW at 330 kV. Line resistance is 15 Ω. Three upgrade options are under consideration:
- Option A: Keep 330 kV, replace wires with lower-resistance cable (R = 8 Ω).
- Option B: Increase transmission voltage to 500 kV, keep existing line (R = 15 Ω).
- Option C: Convert to HVDC at 800 kV with new cable (R = 6 Ω).
The table below shows available data for the current system and Option A.
| Parameter | Current (330 kV, R = 15 Ω) | Option A (330 kV, R = 8 Ω) | Option B (500 kV, R = 15 Ω) | Option C (HVDC 800 kV, R = 6 Ω) |
|---|---|---|---|---|
| Current I (A) | 3 636 | 3 636 | ||
| Ploss (MW) | 198 | 106 | ||
| % power lost | 16.5% | 8.8% |
P = 1 200 MW. Illustrative engineering data.
Q1. Analyse and evaluate the upgrade options to recommend the best choice for the corridor. In your response you must:
- Complete the missing cells for Options B and C using the appropriate formulas. (2 marks)
- Compare all three options quantitatively (current, power loss, % loss) to identify which minimises energy waste. (2 marks)
- Explain the physical reason why increasing voltage from 330 kV to 500 kV reduces power loss more than reducing R from 15 Ω to 8 Ω at the same voltage. (2 marks)
- Evaluate one advantage and one disadvantage of Option C (HVDC) that is not captured by the power-loss calculation alone. (2 marks)
- State your recommended option with a justified conclusion. (1 mark)
2. Engineering design — optimal voltage for a new renewable energy corridor (Band 5–6)
7 marks Band 5–6
Design brief. A new offshore wind farm off the South Australian coast will generate 900 MW. The electricity must be transmitted 600 km to Adelaide. Engineers must choose between 275 kV, 400 kV, and 550 kV AC transmission. The submarine cable has a resistance per kilometre of 0.03 Ω/km.
Constraints: The project requires total power loss to remain below 3 % of generated power. Capital cost increases approximately with the square of the voltage (higher insulation and infrastructure requirements).
Q2. Evaluate the three voltage options and recommend the optimal choice. Present your reasoning using the structure below.
- Calculate the total cable resistance, then compute current and % power loss for each of the three voltage options.
- Identify which options satisfy the 3 % constraint and which do not.
- Explain the engineering trade-off between choosing the lowest acceptable voltage and a higher voltage when both satisfy the constraint.
- State two additional real-world factors an engineer would consider beyond power loss, and explain how each affects the decision.
- State your recommended voltage with a justified conclusion.
Q1 — Sample Band 6 response (9 marks), annotated
Missing cells (2 marks):
Option B (500 kV): I = 1 200×10&sup6;/500×10³ = 2 400 A; Ploss = (2400)²×15 = 86.4 MW; % = 86.4/1200×100 = 7.2% [1 mark]. Option C (HVDC 800 kV): I = 1 200×10&sup6;/800×10³ = 1 500 A; Ploss = (1500)²×6 = 13.5 MW; % = 13.5/1200×100 = 1.1% [1 mark].
Quantitative comparison (2 marks): Current system: 198 MW (16.5%). Option A: 106 MW (8.8%). Option B: 86.4 MW (7.2%). Option C: 13.5 MW (1.1%). Option C achieves the lowest absolute and percentage power loss by a wide margin [1]. Option B reduces loss more than Option A despite using the same existing line resistance, because the voltage increase reduces current more powerfully [1].
Physical explanation (2 marks): Power loss Ploss = I²R. When voltage increases from 330 to 500 kV (ratio 1.52), current decreases by the same ratio, so I² decreases by (1.52)² = 2.3 — a 56% reduction in loss [1]. In contrast, reducing R from 15 to 8 Ω (ratio 0.53) reduces loss linearly by only 47%. Because loss depends on the square of current, voltage increases have a larger effect than linear resistance reductions [1].
HVDC advantage and disadvantage (2 marks): Advantage: HVDC has no reactive (inductive or capacitive) losses over very long distances, and power flow can be precisely controlled electronically, enabling better grid management and the ability to connect asynchronous AC networks [1]. Disadvantage: HVDC requires expensive and complex converter stations (inverters/rectifiers) at each end of the link, significantly increasing capital cost and maintenance complexity compared to AC infrastructure [1].
Recommendation (1 mark): Option C (HVDC 800 kV, R = 6 Ω) is recommended for a 500 km high-capacity corridor because it reduces power waste from 16.5% to 1.1%, saving approximately 184 MW continuously. Although converter station costs are high, the ongoing energy savings and improved grid control over the asset lifetime (30–50 years) strongly justify the investment. Accept Option B with equivalent justification [1].
Marking criteria summary: 1+1 = correct I and Ploss/% for B and C; 1+1 = comparison identifying Option C as minimum-loss and explaining B vs A; 1+1 = I² vs linear R argument with correct ratios; 1+1 = one valid HVDC advantage with explanation + one valid disadvantage with explanation; 1 = clear recommendation with specific supporting data.
Q2 — Sample Band 6 response (7 marks), annotated
Calculations (Rtotal = 0.03 × 600 = 18 Ω):
% loss = (P·R/V²)×100.
275 kV: I = 900×10&sup6;/275×10³ = 3 273 A; Ploss = (3273)²×18 = 193 MW; % = 21.4% — fails constraint [1].
400 kV: I = 2 250 A; Ploss = (2250)²×18 = 91.1 MW; % = 10.1% — fails constraint [1].
550 kV: I = 1 636 A; Ploss = (1636)²×18 = 48.1 MW; % = 5.3% — fails constraint [1].
None of the three AC options satisfies the 3% constraint on this 600 km submarine cable. To achieve 3%, the voltage would need to be V = √(900×10&sup6;×18/0.03) = √(5.4×10¹1;) ≈ 735 kV, which is above even the 550 kV option [1]. Award the mark for clear identification that all three fail and a quantitative or qualitative argument for what would be needed.
Trade-off (1 mark): Among options that satisfy the constraint (none in this case; or hypothetically if the constraint were 10%), choosing the lowest acceptable voltage reduces infrastructure cost (insulation, tower height, clearances) while the higher voltage wastes less power but incurs greater capital expenditure. The optimal choice minimises total lifetime cost (capital plus energy losses), not just one variable [1].
Additional factors (2 marks): (1) Submarine cable insulation: higher AC voltages dramatically increase capacitive reactive power losses in submarine cables, making HVDC the preferred choice for submarine links above ~50 km — this suggests HVDC should be evaluated for this offshore application. (2) Environmental impact: higher voltage pylons/cables require wider safety corridors, greater seabed disturbance, and regulatory approvals, adding time and cost to the project. Accept any two well-explained engineering factors [1+1].
Recommendation (1 mark): No AC option satisfies the 3% constraint; HVDC at an appropriate voltage should be investigated. If forced to choose from the three AC options, 550 kV minimises power loss (5.3%), but all three require engineering review of the constraint or consideration of HVDC alternatives [1]. Accept 550 kV with clear acknowledgement that it does not satisfy the stated constraint.
Marking criteria summary: 1+1+1 = correct calculation for each voltage (I, Ploss, %); 1 = identification of which options satisfy/fail the constraint with quantitative support; 1 = valid trade-off explanation (lowest-acceptable vs higher voltage); 1+1 = two additional factors explained; 1 = recommendation with justified conclusion referencing the calculated data.