Arrow WIDTH is proportional to the amount of energy.
Input arrow on the left. Useful output arrow goes right. Wasted outputs go downward.
Reading a Sankey diagram:
Efficiency = width of useful output ÷ width of input arrow.
The input arrow = sum of ALL output arrows.
EXAMPLE — incandescent bulb (very inefficient):
Input: 100 J electrical
Useful light output: 10 J (thin right arrow)
Wasted heat: 90 J (wide downward arrow)
Efficiency = 10 ÷ 100 = 0.10 = 10%
EXAMPLE — LED bulb (efficient):
Input: 100 J
Useful light: 90 J
Wasted heat: 10 J
Efficiency = 90 ÷ 100 = 0.90 = 90%
Improving Efficiency
REDUCE FRICTION: lubricate moving parts → less thermal wasted.
REDUCE AIR RESISTANCE: streamlined shapes → less energy to air.
BETTER INSULATION: less thermal energy escapes hot devices.
BETTER COMPONENTS: LED lights instead of filament bulbs → more light, less heat.
REGENERATIVE BRAKING: hybrid/electric cars capture braking KE → stored in battery rather than wasted as thermal.
WHY EFFICIENCY MATTERS:
Higher efficiency → less fuel for same useful output → lower costs.
Less fuel → less CO₂ → lower environmental impact.
But: no device can exceed 100% efficiency.
⚠️ Common Mistake
Efficiency = useful output ÷ TOTAL INPUT — not useful ÷ wasted. Result must be ≤ 1 (≤ 100%). If your answer exceeds 1, you have divided the wrong way. Wasted energy = total input − useful output (not the denominator).
📐 Key Equations
efficiency = useful output energy ÷ total input energy
efficiency = useful power output ÷ total power input
efficiency (%) = (useful output ÷ total input) × 100
📌 Key Note
Efficiency = useful output ÷ total input (decimal, 0–1) or × 100 for %. Wasted = total − useful. Sankey diagrams: arrow width ∝ energy amount. Improve: reduce friction (lube), reduce air resistance (streamline), reduce heat loss (insulation), use LEDs. Max efficiency = 100% (never exceeded).
🎯 Matching Activity — Efficiency Values
Match each device to its efficiency. — drag the symbols on the right to match the component names on the left.
A car engine uses 20,000 J of chemical energy and produces 7,000 J of useful kinetic energy. Calculate its efficiency.
F
efficiency = useful output energy ÷ total input energy
I
efficiency = 7000 ÷ 20,000
F
efficiency = 0.35
A
efficiency = 0.35 (35%)
⭐ Higher Tier Only
Describe and evaluate specific methods to increase the efficiency of a given energy transfer — e.g. lubrication reduces friction losses, streamlining reduces air resistance, thermal insulation reduces heat loss, LED technology reduces thermal waste in lighting, regenerative braking recovers kinetic energy. Evaluate cost-effectiveness and practicality of each improvement.
🎯 Test Yourself
Question 1 of 2
1. A device has efficiency 0.6 and is supplied with 500 J. How much energy is wasted?
2. An LED uses 11 J/s total and produces 10 J/s of light. A filament bulb produces the same 10 J/s of light but uses 100 J/s. How much more energy per second does the bulb waste?
⭐ How Well Do You Understand This Topic?
Be honest with yourself — this helps you know what to revise!
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🤖 Ask Mr Badmus AI
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