Specification: AQA A-level Physics (7408) 3.11.2.4 Engineering physics option | Paper: 3 Section B | Data and formulae booklet: given in the exam, including indicated power, P = Tω, friction power, input power = calorific value × fuel flow rate and the mechanical, thermal and overall efficiencies
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Indicated power: the net rate at which work is done by the gas on the pistons, found from the area of the indicator (p–V) diagram: indicated power = (area of p–V loop) × (number of cycles per second) × (number of cylinders).
Output (brake) power: the useful power delivered at the crankshaft, P = Tω. Friction power = indicated power − brake power: the power used to overcome friction in the engine.
Efficiencies: mechanical efficiency = brake power ÷ indicated power; thermal efficiency = indicated power ÷ input power; overall efficiency = brake power ÷ input power, where input power = calorific value × fuel flow rate.
The equation
indicated power = (area of p–V loop) × (number of cycles per second) × (number of cylinders) and output power = torque × angular velocity P = Tω and friction power = indicated power − output power
| Quantity | Symbol | Unit |
|---|---|---|
| area of p–V loop (work done per cycle) | W | joules (J) |
| number of cycles per second | – | s⁻¹ |
| indicated, output or friction power | P | watts (W) |
| torque at the crankshaft | T | newton metres (N m) |
| angular velocity of the crankshaft | ω | radians per second (rad s⁻¹) |
| calorific value of the fuel | – | joules per kilogram (J kg⁻¹) |
| fuel flow rate | – | kilograms per second (kg s⁻¹) |
Rearranged: T = P ÷ ω | area of loop = indicated power ÷ (cycles per second × cylinders) | ω = 2π × (rev min⁻¹) ÷ 60 | overall efficiency = mechanical efficiency × thermal efficiency | In a four-stroke engine each cylinder completes one cycle every two revolutions, so cycles per second = (rev min⁻¹ ÷ 60) ÷ 2.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) The area of the p–V loop for one cylinder of an engine represents 450 J. The engine has 4 cylinders and each completes 20 cycles per second. Calculate the indicated power, in kW.
Show solution
F: indicated power = (area of p–V loop) × (cycles per second) × (number of cylinders)
I: indicated power = 450 × 20 × 4
F: indicated power is already the subject
A: 36 000 W = 36 kW
Q2 (Core) An engine delivers a torque of 150 N m at its crankshaft, which rotates at 300 rad s⁻¹. Calculate the output power in kW.
Show solution
F: P = Tω
I: P = 150 × 300
F: P is already the subject
A: 45 000 W = 45 kW
Q3 (Core) A car engine produces an output torque of 220 N m at 3000 rev min⁻¹. Calculate the output power, to 2 significant figures.
Show solution
Convert first: ω = 3000 × 2π ÷ 60 = 314.1… rad s⁻¹
F: P = Tω
I: P = 220 × 314.1…
F: P is already the subject
A: 69 115… W = 69 kW (2 s.f.)
Watch out: rev min⁻¹ is not an angular velocity. Multiply by 2π to get radians and divide by 60 to get per second.
Q4 (Core) A single-cylinder four-stroke lawnmower engine has an indicated power of 3.0 kW when running at 3600 rev min⁻¹. Calculate the work done by the gas in each cycle.
Show solution
Convert first: 3.0 kW = 3000 W; 3600 rev min⁻¹ = 60 rev s⁻¹; four-stroke, so 60 ÷ 2 = 30 cycles per second
F: indicated power = (area of p–V loop) × (cycles per second) × (number of cylinders)
I: 3000 = area × 30 × 1
F: area = 3000 ÷ 30
A: 100 J per cycle
Q5 (Exam) A four-cylinder four-stroke car engine runs at 4800 rev min⁻¹. The area of the p–V loop for each cylinder represents 520 J. Calculate the indicated power, to 2 significant figures.
Show solution
Convert first: 4800 rev min⁻¹ = 80 rev s⁻¹; four-stroke, so 80 ÷ 2 = 40 cycles per second
F: indicated power = (area of p–V loop) × (cycles per second) × (number of cylinders)
I: indicated power = 520 × 40 × 4
F: indicated power is already the subject
A: 83 200 W = 83 kW (2 s.f.)
Watch out: forgetting to halve the revolutions per second for a four-stroke engine doubles the answer. Each cylinder has only one power stroke every two revolutions.
Q6 (Exam) An engine has an indicated power of 62 kW. Its brake torque is 160 N m at 3300 rev min⁻¹. Calculate (a) the brake power, to 3 significant figures, (b) the friction power, to 2 significant figures, (c) the mechanical efficiency, to 2 significant figures.
Show solution
Convert first: 62 kW = 62 000 W; ω = 3300 × 2π ÷ 60 = 345.5… rad s⁻¹
Step 1 – (a) brake power
F: P = Tω
I: P = 160 × 345.5…
A: 55 292… W = 55.3 kW (3 s.f.)
Step 2 – (b) friction power
friction power = indicated power − brake power = 62 000 − 55 292… = 6707.9… W = 6.7 kW (2 s.f.)
Step 3 – (c) mechanical efficiency
mechanical efficiency = brake power ÷ indicated power = 55 292… ÷ 62 000 = 0.891…
A: 89% (2 s.f.)
Q7 (Exam) An engine burns fuel of calorific value 4.5 × 107 J kg⁻¹ at a rate of 3.2 g s⁻¹. Its indicated power is 50 kW and its brake power is 42 kW. Calculate the thermal efficiency and the overall efficiency, to 2 significant figures.
Show solution
Convert first: 3.2 g s⁻¹ = 3.2 × 10−3 kg s⁻¹
Step 1 – input power
input power = calorific value × fuel flow rate = 4.5 × 107 × 3.2 × 10−3 = 144 000 W
Step 2 – thermal efficiency
thermal efficiency = indicated power ÷ input power = 50 000 ÷ 144 000 = 0.347…, so 35% (2 s.f.)
Step 3 – overall efficiency
overall efficiency = brake power ÷ input power = 42 000 ÷ 144 000 = 0.291…, so 29% (2 s.f.)
Check: mechanical efficiency × thermal efficiency = (42 ÷ 50) × 0.347… = 0.84 × 0.347… = 0.291…
Q8 (Exam) The indicator diagram for one cylinder of a two-cylinder four-stroke motorcycle engine is drawn on a grid. Each grid square represents 0.50 × 105 Pa on the pressure axis and 0.050 × 10−3 m³ on the volume axis. The loop encloses 96 squares. Calculate the indicated power when the engine runs at 6000 rev min⁻¹, to 2 significant figures.
Show solution
Step 1 – work represented by one square
0.50 × 105 × 0.050 × 10−3 = 2.5 J
Step 2 – area of the loop
96 × 2.5 = 240 J per cycle
Step 3 – indicated power
Convert first: 6000 rev min⁻¹ = 100 rev s⁻¹, so 100 ÷ 2 = 50 cycles per second
F: indicated power = (area of p–V loop) × (cycles per second) × (number of cylinders)
I: indicated power = 240 × 50 × 2
A: 24 000 W = 24 kW (2 s.f.)
Watch out: multiply the scale values of one square to get its energy, including both powers of ten, before counting squares.
Q9 (Stretch) Each cylinder of a four-cylinder four-stroke engine has an idealised indicator diagram that is a triangle with corners at (5.0 × 10−5 m³, 1.0 × 105 Pa), (5.0 × 10−5 m³, 25 × 105 Pa) and (5.0 × 10−4 m³, 1.0 × 105 Pa). The engine runs at 3600 rev min⁻¹ and its mechanical efficiency is 85%. Calculate (a) the indicated power, (b) the friction power, (c) the brake torque, to 3 significant figures.
Show solution
Step 1 – area of the loop
base = 5.0 × 10−4 − 5.0 × 10−5 = 4.5 × 10−4 m³; height = 25 × 105 − 1.0 × 105 = 24 × 105 Pa
area = ½ × 4.5 × 10−4 × 24 × 105 = 540 J per cycle
Step 2 – (a) indicated power
3600 rev min⁻¹ = 60 rev s⁻¹, so 30 cycles per second; indicated power = 540 × 30 × 4 = 64 800 W = 64.8 kW
Step 3 – (b) friction power
brake power = 0.85 × 64 800 = 55 080 W; friction power = 64 800 − 55 080 = 9720 W = 9.72 kW
Step 4 – (c) brake torque
ω = 3600 × 2π ÷ 60 = 376.9… rad s⁻¹
F: P = Tω
I: 55 080 = T × 376.9…
F: T = 55 080 ÷ 376.9…
A: 146.1… N m = 146 N m (3 s.f.)
Watch out: the brake torque comes from the brake power, not the indicated power. Using 64 800 W would overestimate the torque.
Q10 (Stretch) A car travels at a steady 25 m s⁻¹ on a level road, with its engine delivering a brake power of 20 kW. The engine’s thermal efficiency is 35% and its mechanical efficiency is 80%. The fuel has a calorific value of 4.6 × 107 J kg⁻¹ and a density of 740 kg m⁻³. Calculate (a) the indicated power and the friction power, (b) the fuel flow rate in g s⁻¹, to 3 significant figures, (c) the volume of fuel used in 100 km, in litres, to 2 significant figures.
Show solution
Step 1 – (a) indicated and friction power
indicated power = brake power ÷ mechanical efficiency = 20 000 ÷ 0.80 = 25 000 W = 25 kW
friction power = 25 000 − 20 000 = 5000 W = 5.0 kW
Step 2 – (b) fuel flow rate
input power = indicated power ÷ thermal efficiency = 25 000 ÷ 0.35 = 71 428.5… W (overall efficiency = 0.35 × 0.80 = 0.28)
fuel flow rate = input power ÷ calorific value = 71 428.5… ÷ 4.6 × 107 = 1.552… × 10−3 kg s⁻¹
A: 1.55 g s⁻¹ (3 s.f.)
Step 3 – (c) fuel for 100 km
time = 100 000 ÷ 25 = 4000 s; mass = 1.552… × 10−3 × 4000 = 6.211… kg
volume = mass ÷ density = 6.211… ÷ 740 = 8.393… × 10−3 m³ = 8.39… litres
A: 8.4 litres (2 s.f.)
Watch out: 1 litre = 10−3 m³. Only 28% of the fuel energy reaches the crankshaft.
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