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Chapter 2 · 6 hours

Fuel Supply Systems

Practice questions

Practice questions and answers

4 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 5 marks

List and explain the basic requirements of a good fuel for (i) a spark ignition engine and (ii) a compression ignition engine. Explain octane number and cetane number.

Answer

Requirements of a good engine fuel

A good fuel should burn smoothly, give high energy per kg, be easy to store and supply, and cause little pollution.

Spark ignition (petrol) fuel

  • High octane number (anti-knock quality), so a high compression ratio can be used.
  • Good volatility: easy cold starting, but low enough to avoid vapour lock.
  • High calorific value (about 44 MJ/kg) and low sulphur and gum content.
  • Non-corrosive, stable in storage, low emission.

Compression ignition (diesel) fuel

  • High cetane number for a short ignition delay (smooth, quiet running).
  • Proper viscosity so that the injector atomises it well and lubricates the pump.
  • Low sulphur, low ash and carbon residue (less deposits and smoke).
  • Suitable pour point and cloud point for cold weather.

Octane and cetane number

  • Octane number: percentage by volume of iso-octane in a blend of iso-octane (rating 100) and n-heptane (rating 0) that matches the knock behaviour of the fuel in a standard CFR engine. Higher value means better knock resistance. Petrol is typically 87 to 95 RON.
  • Cetane number: percentage by volume of cetane (n-hexadecane, rating 100) in a blend with alpha-methyl naphthalene (rating 0) that has the same ignition delay as the diesel fuel. Automotive diesel is typically 45 to 55.
PropertyPetrolDiesel
Wanted qualityHigh octane (resist self-ignition)High cetane (ignite readily)
Rating scaleOctane numberCetane number
  • Practice · 6 marks

Explain why an SI engine needs different air-fuel mixture strengths for cold starting, idling, cruising and full-load (acceleration). State the approximate air-fuel ratio in each case and how a carburettor or injection system provides it.

Answer

The stoichiometric air-fuel ratio (AFR) of petrol is about 14.7:1 by mass (λ=1\lambda = 1). Engine demand varies, so the mixture must be changed with operating condition.

Requirements

ConditionAFR (approx.)MixtureReason
Cold start2:1 to 5:1 (supplied)Very richLittle fuel vaporises on cold walls; only light ends burn
Idling10:1 to 12:1RichThrottle nearly closed, high residual gas fraction, poor mixing
Part-load cruising15:1 to 17:1LeanBest fuel economy, throttle partly open
Full load / maximum power12:1 to 13.5:1 (about 12.5:1)RichGives maximum power; excess fuel cools charge
AccelerationMomentary richRichThrottle opens suddenly, fuel lags behind air

Maximum economy occurs near 16:1 to 17:1 (lean), maximum power near 12.5:1 (rich), and the three-way catalyst needs 14.7:1 at cruise.

How the mixture is provided

Carburettor

  • Choke enriches mixture for starting.
  • Idle jet supplies rich mixture below the throttle plate.
  • Main jet with compensating jet gives a nearly constant lean mixture for cruising.
  • Power (economiser) valve enriches at full load.
  • Accelerator pump gives an extra squirt on sudden throttle opening.

Electronic fuel injection

  • The ECU reads coolant temperature, throttle position, mass airflow, and lambda sensor.
  • It lengthens injector pulse width for cold start, idle, and acceleration, and trims to 14.7:1 in closed loop at cruise.
  • At full load it goes open loop to a rich mixture.
  • Practice · 4+6 marks

(a) Differentiate between a carburettor and a fuel injection system for SI engines. (b) Describe the TBI, MPFI and GDI systems with simple block diagrams, stating one advantage and one limitation of each.

Answer

(a) Carburettor vs fuel injection

PointCarburettorFuel injection (EFI)
Fuel meteringVenturi vacuum draws fuel through jets (mechanical)ECU controls injector pulse width (electronic)
Accuracy of AFRPoor, varies with altitude and temperatureAccurate, closed loop with lambda sensor
Cylinder-to-cylinder distributionUneven (long manifold)Even in MPFI/GDI
Cold startChoke neededAutomatic enrichment
Fuel economy and emissionHigher consumption, high CO and HCLower, can meet BS-VI
PowerVenturi restricts airflowHigher volumetric efficiency
Cost and maintenanceCheap, simple, but needs tuningCostly, needs sensors and ECU

(b) EFI systems

1. Throttle body injection (TBI) - one or two injectors above the throttle valve spray fuel into the intake manifold, replacing the carburettor.

 Fuel tank -> pump -> filter -> [Injector] -> throttle
                                      ^ ECU
 Manifold distributes mixture to cylinders
  • Advantage: simple, cheap, low pressure (about 1 bar).
  • Limitation: unequal distribution, wall wetting, slow response.

2. Multi-point fuel injection (MPFI) - one injector per cylinder near the intake port, injecting into the port (about 3 to 4 bar).

 Tank -> pump -> filter -> rail -> Inj1..Inj4 -> intake ports
 Sensors (MAF, TPS, CTS, O2) -> ECU -> injector pulse
  • Advantage: equal distribution, quick response, better economy and emission.
  • Limitation: some fuel still wets the port; costlier than TBI.

3. Gasoline direct injection (GDI) - injector inside the combustion chamber injects high-pressure fuel (50 to 200 bar) directly, with stratified charge at part load.

 Tank -> low-pressure pump -> high-pressure pump -> rail
                                   -> injector in cylinder head
  • Advantage: lean burn, higher compression ratio, 10 to 15 % better fuel economy, more power.
  • Limitation: carbon deposits on intake valves, high cost, particulate (soot) emission needing a gasoline particulate filter.
  • Practice · 5 marks

A 1.6 litre, four-cylinder, four-stroke petrol engine runs at 3000 rpm with a volumetric efficiency of 85 %. The intake air density is 1.18 kg/m³ and the brake power is 35 kW. Calculate (i) the mass flow rate of air, (ii) the fuel flow rate for a stoichiometric AFR of 14.7, (iii) the brake specific fuel consumption, and (iv) the fuel flow at full load when the AFR is 12.5 and the equivalence ratio λ\lambda.

Answer

Given: Vd=1.6×10−3V_d = 1.6\times10^{-3} m³, N=3000N = 3000 rpm, ηv=0.85\eta_v = 0.85, ρa=1.18\rho_a = 1.18 kg/m³, BP=35BP = 35 kW.

(i) Air flow

A four-stroke engine inhales once per two revolutions.

m˙a=ηvρaVdN2×60=0.85×1.18×1.6×10−3×3000120=0.0401 kg/s\dot m_a = \eta_v \rho_a V_d \frac{N}{2\times 60} = 0.85 \times 1.18 \times 1.6\times10^{-3} \times \frac{3000}{120} = 0.0401\ \text{kg/s}

That is 0.0401×3600=144.40.0401 \times 3600 = 144.4 kg/h.

(ii) Fuel flow at AFR=14.7AFR = 14.7

m˙f=m˙a14.7=144.414.7=9.83 kg/h\dot m_f = \frac{\dot m_a}{14.7} = \frac{144.4}{14.7} = 9.83\ \text{kg/h}

For petrol density 740 kg/m³ this is 9.83/0.74=13.39.83/0.74 = 13.3 L/h.

(iii) Brake specific fuel consumption

bsfc=m˙fBP=9.83×100035=281 g/kWhbsfc = \frac{\dot m_f}{BP} = \frac{9.83 \times 1000}{35} = 281\ \text{g/kWh}

(iv) Full-load mixture

m˙f=144.412.5=11.55 kg/h,λ=AFRactualAFRstoich=12.514.7=0.85\dot m_f = \frac{144.4}{12.5} = 11.55\ \text{kg/h}, \qquad \lambda = \frac{AFR_{actual}}{AFR_{stoich}} = \frac{12.5}{14.7} = 0.85

The mixture is 15 % rich (ϕ=1/λ=1.18\phi = 1/\lambda = 1.18).

Answer: m˙a=0.0401\dot m_a = 0.0401 kg/s; m˙f=9.83\dot m_f = 9.83 kg/h at λ=1\lambda = 1; bsfc =281= 281 g/kWh; full-load fuel =11.55= 11.55 kg/h with λ=0.85\lambda = 0.85.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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