Engineering contexts & challenges · Higher

Energy & efficiency

9 cards · 5 quiz · 0 exam
What you're learning in this topic

Learning intention WALT

We are learning about energy & efficiency, specifically to:

understand

  • energy conversions in a system and draw an energy audit
  • ways of reducing energy losses

be able to

  • calculate kinetic, potential, electrical and heat energy
  • calculate power, work done and efficiency
  • solve multi-stage energy problems involving efficiency and losses at each stage

Success criteria WILF

  • I can describe energy conversions in a system and draw an energy audit
  • I can calculate kinetic, potential, electrical and heat energy
  • I can calculate power, work done and efficiency
  • I can explain ways of reducing energy losses
  • I can solve multi-stage energy problems involving efficiency and losses at each stage

Conservation of energy

Energy can't be created or destroyed, only changed from one form to another. In real systems some energy is always "lost" as unwanted heat and sound, often through friction. An energy audit lists the energy in, the useful energy out and the losses.

Electrical energy in: 1000 JUseful kineticenergy: 750 JWasted heat andsound: 250 JEfficiency = useful out ÷ in = 750 ÷ 1000 = 0.75 (75%)
An energy audit: the energy in equals the useful energy out plus the energy wasted. Efficiency compares the useful output with the input.

Forms and sources of energy

Forms include kinetic, potential, heat (thermal), electrical, chemical, light, sound, nuclear and magnetic.

  • Fossil fuels: coal, oil and gas: burned, so they release CO₂ and will run out.
  • Nuclear: fission of uranium produces heat, which makes steam to turn a turbine and generator.
  • Renewable: wind, wave, tidal, hydro-electric and solar.

Energy is the capacity to do work, measured in joules (J). Power is how fast energy is transferred, measured in watts (W = J/s). Think of energy as the fuel in the tank and power as how quickly it's used.

An energy transfer moves energy (electricity along wires); a transformation changes its form (chemical → kinetic). Every transformation wastes some energy, usually as heat and sound.

Prefixes and significant figures

PrefixMeansExample
M (mega)× 1 000 0002.4 MJ = 2 400 000 J
k (kilo)× 10003.5 kW = 3500 W
m (milli)÷ 100030 mA = 0.03 A
μ (micro)÷ 1 000 00047 μF = 0.000 047 F

Convert to base units before substituting, and convert time to seconds (7 minutes = 420 s). Give the final answer to 2 or 3 significant figures with its unit: 0.0078 A, 54%, 24 m/s.

Energy formulae

QuantityFormulaUnits
Work doneEw = F dJ = N × m
Potential energyEp = m g h (g = 9.8 N/kg)J
Kinetic energyEk = ½ m v²J (v in m/s)
Electrical energyEe = V I tJ (t in seconds)
Heat energyEh = c m ΔTJ (c in J/kg °C)
PowerP = E / tW = J/s
Efficiencyη = Eout / Ein = Pout / Pinno units (or ×100 for %)

These are in the data booklet; you must be able to rearrange them and use the answer from one in the next.

Calculation technique

  1. Write the formula.
  2. Convert units first: hours to seconds (7.5 h = 27 000 s), kJ to J, mm to m where needed.
  3. Substitute, then rearrange (transpose).
  4. Give the answer to 2 significant figures with a unit.
  5. Don't round in the middle of a calculation; round only the final answer.

Each step usually carries a mark: substitution, transposition and answer with unit. Follow-on marks mean a wrong earlier value doesn't stop you being awarded the later marks.

Worked examples: energy changing form

Potential to kinetic. A 50 kg object falls freely from 30 m. Ep = mgh = 50 × 9.8 × 30 = 14 700 J. All of it becomes kinetic energy at impact: 14 700 = ½ × 50 × v², so v² = 588 and v = 24 m/s.

Electrical to heat. A kettle uses 24 kJ to heat 2 kg of water to 100 °C (no losses). 24 000 = 4180 × 2 × ΔT, so ΔT = 2.87 °C and the water started at 97 °C.

Efficiency in three steps. A coffee machine (230 V, 22 A) heats 0.15 kg of water from 20 °C to 85 °C in 15 s.

  1. Energy in: Ee = VIt = 230 × 22 × 15 = 75 900 J
  2. Useful energy out: Eh = cmΔT = 4180 × 0.15 × 65 = 40 755 J
  3. Efficiency: η = 40 755 ÷ 75 900 × 100 = 54%

The other 46% (35.1 kJ) is wasted, as heat to the casing and air, and sound. Efficiency can never be more than 100%.

Energy audits

An energy audit lists all the energy going into and coming out of a system, so you can see where energy is wasted. Total energy in = useful energy out + wasted energy. Show it as a diagram: the input on the left, the useful output on the right, and the waste (heat, sound) coming off the side, each labelled with its value.

Interactive simulation: try this on the revision site, where you can change the values and see the result.

Improving efficiency

Explain with a cause linked to the context and its effect: "Lubricate the gears (cause) to reduce friction, so less energy is lost as heat (effect)." Other examples: bearings, streamlining, insulation, reducing weight.

Design challenge: school lift energy audit

The brief. An Aberdeen school is replacing an old lift. The motor draws 15 A at 230 V for 20 s to lift 600 kg through 9 m.

  1. Calculate the electrical energy supplied.
  2. Calculate the potential energy gained.
  3. Calculate the efficiency of the lift.
  4. Explain two ways the new lift could be more efficient.
What a good answer includes
  • Ee = VIt = 230 × 15 × 20 = 69 000 J.
  • Ep = mgh = 600 × 9.8 × 9 = 52 920 J.
  • η = 52 920 ÷ 69 000 = 0.77, so 77%.
  • Lubricated bearings reduce friction, so less energy is lost as heat; a more efficient motor or regenerative braking recovers energy.

Higher energy problems HIGHER

Higher questions chain several formulae together. Example: a pump lifts 2500 litres (2500 kg) of water 2.9 m at 87% efficiency from a 230 V, 5 A supply. Ew = (2500 × 9.8) × 2.9 = 71 050 J; Ein = 71 050 ÷ 0.87 = 81 667 J; t = 81 667 ÷ (230 × 5) = 71 s.

Key terms

Work doneE = F × d
Potential energyE = m × g × h (g = 9.8)
Kinetic energyE = ½ m v²
Electrical energyE = V × I × t (t in seconds)
Heat energyE = c × m × ΔT
PowerP = E ÷ t
Efficiencyη = useful energy out ÷ energy in
Law of conservation of energyEnergy can't be created or destroyed, only changed from one form to another.
RoundingRound only the final answer, usually to 2 significant figures, with a unit.

Quick check

1. A 24 V, 0.42 A device runs for 7.5 hours. Which time value must be used?

  1. 27 000
  2. 450
  3. 7.5
  4. 0.125
Show answer
A: 27 000. 7.5 × 60 × 60 s.

2. Kinetic energy depends on…

  1. Height
  2. Mass and velocity
  3. Current and time
  4. Temperature
Show answer
B: Mass and velocity. ½ m v².

3. A motor takes in 2000 J and gives 1500 J of useful work. Its efficiency is…

  1. 3500
  2. 1.33
  3. 0.75
  4. 500
Show answer
C: 0.75. 1500 ÷ 2000.

4. Unwanted energy in a gearbox is mainly lost as…

  1. Light
  2. Chemical energy
  3. Potential energy
  4. Heat and sound
Show answer
D: Heat and sound. Through friction.

5. The unit of power is…

  1. Watt
  2. Newton
  3. Joule
  4. Volt
Show answer
A: Watt. J/s = W.

Printable worksheet

An I do, we do, you do worksheet for this topic, with a separate answer sheet.

Worksheet (PDF)Answers (PDF)