Physics • Year 12 • Module 6 • Lesson 20
Power Transmission and Distribution
Apply Ploss = I²R to real transmission scenarios, interpret power-loss graphs, and evaluate engineering trade-offs in grid design.
1. Transmission loss comparison — Snowy Mountains to Sydney
Electricity generated at a Snowy Hydro station is to be transmitted 400 km to Sydney. The total resistance of the transmission lines is 12 Ω. The station generates 600 MW of power. The table below shows four possible transmission voltages.
10 marks
| Transmission voltage (kV) | Current I (A) | Ploss (MW) | % power lost |
|---|---|---|---|
| 50 | |||
| 100 | |||
| 300 | |||
| 500 |
1.1 Complete all blank cells in the table above, showing the current, power loss, and percentage loss for each transmission voltage. 8 marks (2 per row)
1.2 By what factor does the power loss decrease when transmission voltage is increased from 50 kV to 500 kV? Explain why this factor is larger than the factor by which voltage increased. 2 marks
2. Interpret graph — power loss vs transmission voltage
A grid engineer plotted the relationship between transmission voltage and percentage power loss for a line transmitting 400 MW with R = 10 Ω. 6 marks
Figure 2. % power loss vs transmission voltage for P = 400 MW, R = 10 Ω. Calculated data.
2.1 Describe the shape of the curve and explain the mathematical relationship it represents. 2 marks
2.2 Using the graph, estimate the transmission voltage at which power loss falls below 5 %. Verify your estimate by calculation. 2 marks
2.3 A grid engineer argues that “above 400 kV, further voltage increases give minimal benefit.” Use the graph to evaluate this claim and suggest one other engineering factor to consider. 2 marks
3. Compare AC and DC transmission
Complete the table comparing alternating current and direct current for long-distance power transmission. 8 marks (1 per cell)
| Feature | AC transmission | DC transmission (HVDC) |
|---|---|---|
| Voltage stepping | ||
| Power loss at very long distances | ||
| Transformer compatibility | ||
| Grid infrastructure complexity |
4. Predict and justify — an outback Queensland scenario
A remote mine in outback Queensland is 800 km from the nearest power station. The mine requires 80 MW. Two transmission options are proposed: Option A at 220 kV and Option B at 440 kV. The transmission line has total resistance 20 Ω. 6 marks
4.1 Calculate the power loss for each option and determine which option wastes less than 5 % of the generated power. Show all working. 4 marks
4.2 Explain why running the transmission line at 440 kV requires more expensive infrastructure, and justify whether this cost is worth paying for this application. 2 marks
Q1.1 — Transmission loss table (P = 600 MW, R = 12 Ω)
50 kV: I = 600×10&sup6;/50×10³ = 12 000 A; Ploss = (12 000)²×12 = 1 728 MW; % = 1728/600×100 = 288% (more than generated — 50 kV is entirely impractical).
100 kV: I = 6 000 A; Ploss = 6000²×12 = 432 MW; % = 72%.
300 kV: I = 2 000 A; Ploss = 2000²×12 = 48 MW; % = 8.0%.
500 kV: I = 1 200 A; Ploss = 1200²×12 = 17.3 MW; % = 2.9%.
Marking notes. 1 mark for correct I and 1 mark for correct Ploss and % per row. Accept minor rounding differences (±0.5 MW). 50 kV loss exceeds generated power: accept either the calculated number or the conclusion that this is non-viable; award marks for correct method.
Q1.2 — Factor of decrease and explanation
50 kV to 500 kV is a factor of 10 increase in voltage. Power loss ∝ I² ∝ 1/V², so a factor of 10 increase in voltage reduces loss by a factor of 10² = 100. (1728 MW ÷ 17.3 MW ≈ 100.) The factor is the square of the voltage ratio because power loss depends on I², and I is inversely proportional to V.
Q2.1 — Shape and relationship
The curve is an inverse-square (hyperbolic) relationship: as voltage increases, % loss decreases rapidly at first, then more gradually. Mathematically, % loss = (P·R/V²)×100, so loss ∝ 1/V² — doubling V quarters the loss.
Q2.2 — Voltage for <5% loss
From the graph, approximately 280–300 kV. Verification: % = (10×400×10&sup6;)/(V²)×100 = 5. Solving: V² = 10×400×10&sup6;×100/5 = 8×10¹0;; V = 283 kV. Both graph estimate and calculation should be within 10 % of this value.
Q2.3 — Evaluate the engineer’s claim
The claim is broadly supported by the graph: from 100 kV (25% loss) to 400 kV (1.6% loss) the improvement is dramatic, but from 400 kV (1.6%) to 500 kV (1.0%) the further reduction is only 0.6 percentage points. One additional engineering factor to consider: higher voltages require more expensive insulation, larger pylons, wider safety corridors, and more complex switchgear, so the marginal benefit may not justify the marginal cost above a certain threshold.
Q3 — AC vs DC comparison
Voltage stepping: AC — easily stepped up/down using standard transformers (transformers require AC because they rely on changing magnetic flux). DC — cannot be stepped up/down by a simple transformer; DC produces constant flux so no emf is induced in the secondary coil. Power loss at very long distances: AC — I2R losses are minimised by using transformers to raise voltage (reducing current); the same P = I2R principle applies. DC — can also achieve low I2R losses at high voltage, but requires expensive electronic converter equipment at each end instead of transformers. Transformer compatibility: AC — fully compatible; the entire Australian grid is built around AC transformers at power stations and substations. DC — incompatible with standard transformers; would require a separate conversion infrastructure. Grid infrastructure complexity: AC — simpler and more established; substations use straightforward step-up and step-down transformers. DC — requires additional conversion equipment at each end, increasing capital cost and complexity, though DC has some advantages for very long distances (noted in the lesson).
Q4.1 — Outback Queensland (P = 80 MW, R = 20 Ω)
Option A (220 kV): I = 80×10&sup6;/220×10³ = 364 A; Ploss = (364)²×20 = 2.65 MW; % = 2.65/80×100 = 3.3% < 5% — passes.
Option B (440 kV): I = 80×10&sup6;/440×10³ = 182 A; Ploss = (182)²×20 = 0.66 MW; % = 0.66/80×100 = 0.83% < 5% — also passes.
Both options satisfy the 5% criterion, but 440 kV wastes only 0.83% compared to 3.3% at 220 kV.
Marking notes: 1 mark per option for correct I, 1 mark per option for correct Ploss and %. Accept rounding within ±5%.
Q4.2 — Cost justification
Higher voltage transmission requires heavier insulation on towers, taller pylons with greater clearances, and more expensive switchgear and transformers at substations. For an 800 km remote line, the 440 kV option saves approximately 2 MW of power continuously (2.65 − 0.66 MW). Over years of operation this represents significant economic savings and reduces fuel/water costs at the generating station, likely justifying the higher capital cost of 440 kV infrastructure. The precise decision requires a cost-benefit analysis over the asset lifetime.