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Chapter 1 · 4 hours

Introduction

Practice questions

Practice questions and answers

3 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 · 6 marks

Classify automobiles on the basis of (i) fuel used, (ii) number of wheels and axles, (iii) body type and (iv) drive and engine position. Briefly name the major systems of a modern automobile.

Answer

An automobile is a self-propelled road vehicle that carries passengers or goods. It is classified in several ways.

Classification

BasisTypesExamples
Fuel usedPetrol (SI), diesel (CI), CNG/LPG, electric, hybrid, fuel cellMaruti Alto, Tata Nexon EV
Wheels / axles2-wheeler, 3-wheeler, 4-wheeler, heavy vehicle with 6 or more wheelsMotorcycle, tempo, car, truck
Body typeSaloon (sedan), hatchback, station wagon, SUV, pick-up, bus, tipperSedan, mini-bus
Load carriedLight (LMV), medium, heavy (HMV) vehiclesCar, mini-truck, tanker
DriveFront-wheel drive (FWD), rear-wheel drive (RWD), all-wheel / four-wheel drive (4WD)FWD hatchback, 4WD jeep
Engine positionFront, rear, mid-engineFront engine: most cars; rear: Tata Nano
SteeringLeft-hand drive, right-hand driveRight-hand drive in Nepal
TransmissionManual, semi-automatic, automaticAMT, CVT, DCT

Major systems of a modern automobile

 Engine --> Clutch --> Gearbox --> Propeller shaft
                                        |
 Fuel, ignition, cooling,         Differential
 lubrication, exhaust                   |
                                  Rear/front axle --> Wheels
 Frame + Body + Suspension + Steering + Brakes
 Electrical system + Electronics (ECU, sensors)
  1. Power unit: engine with fuel, ignition, cooling, lubrication and exhaust systems.
  2. Transmission: clutch, gearbox, propeller shaft, differential and axles.
  3. Chassis: frame, suspension, steering, brakes, wheels and tyres.
  4. Body: passenger compartment, doors, seats.
  5. Electrical and electronic systems: battery, alternator, starter, lighting, ECU and sensors.
  • Practice · 4+4 marks

(a) Explain the working of a four-stroke petrol engine with the help of a sketch of the four strokes. (b) Differentiate between two-stroke and four-stroke engines.

Answer

(a) Four-stroke petrol engine

In a four-stroke engine one power cycle is completed in four piston strokes, i.e. two crankshaft revolutions (720 degrees). The cycle is the Otto cycle.

 SUCTION      COMPRESSION    POWER        EXHAUST
 IV open      both closed    both closed  EV open
 piston down  piston up      spark, piston piston up
              |   |          pushed down   |   |
   v   |       ^   |          v   |        ^   |
  1. Suction stroke: inlet valve open, exhaust valve closed. Piston moves from TDC to BDC and draws in air-fuel mixture.
  2. Compression stroke: both valves closed. Piston moves BDC to TDC and compresses the mixture to about 1/8 to 1/10 of its volume. Near TDC the spark plug fires.
  3. Power (expansion) stroke: burnt gases expand and push the piston from TDC to BDC. Work is delivered to the crankshaft.
  4. Exhaust stroke: exhaust valve open. Piston moves BDC to TDC and sweeps burnt gases out.

Only the power stroke produces work, so a flywheel stores energy to carry the piston through the other three strokes.

(b) Two-stroke vs four-stroke

PointTwo-strokeFour-stroke
Strokes per cycle2 (one revolution)4 (two revolutions)
Power strokeEvery revolutionEvery second revolution
ValvesPorts covered by pistonPoppet valves with cam and camshaft
Power for same sizeTheoretically twice, practically 1.3 to 1.5 timesLower
LubricationOil mixed with petrol or injectedSeparate sump, pressure lubrication
Thermal efficiencyLower (fresh charge loss through exhaust port)Higher
Fuel consumption and emissionHigher, more unburnt HCLower
Weight and costLight, simple, cheapHeavier, complex
FlywheelLightHeavier
UseSmall scooters, chain sawsCars, buses, most motorcycles
  • Practice · 8 marks

A four-cylinder, four-stroke petrol engine has a bore of 80 mm and a stroke of 90 mm. The clearance volume of each cylinder is 55 cc. At 4000 rpm the engine develops a brake torque of 120 N m. Calculate (i) the swept volume per cylinder and the total engine displacement, (ii) the compression ratio, (iii) the brake power, (iv) the brake mean effective pressure and (v) the mean piston speed.

Answer

Given: D=80D = 80 mm, L=90L = 90 mm, n=4n = 4 cylinders, Vc=55V_c = 55 cc, N=4000N = 4000 rpm, T=120T = 120 N m.

(i) Swept volume and displacement

Vs=π4D2L=π4(8)2(9) cm3=452.4 cm3Vd=nVs=4×452.4=1809.6 cm3≈1.81 L\begin{aligned} V_s &= \frac{\pi}{4} D^2 L = \frac{\pi}{4}(8)^2(9)\ \text{cm}^3 = 452.4\ \text{cm}^3 \\ V_d &= n V_s = 4 \times 452.4 = 1809.6\ \text{cm}^3 \approx 1.81\ \text{L} \end{aligned}

(ii) Compression ratio

r=Vs+VcVc=452.4+5555=9.23r = \frac{V_s + V_c}{V_c} = \frac{452.4 + 55}{55} = 9.23

(iii) Brake power

BP=2πNT60=2π×4000×12060=50 265 W≈50.27 kWBP = \frac{2\pi N T}{60} = \frac{2\pi \times 4000 \times 120}{60} = 50\,265\ \text{W} \approx 50.27\ \text{kW}

(iv) Brake mean effective pressure

For a four-stroke engine, BP=pmbVdN2×60BP = \dfrac{p_{mb} V_d N}{2 \times 60} (one power stroke per two revolutions).

pmb=BP×120VdN=50 265×1201.8096×10−3×4000=8.33×105 Pa=8.33 barp_{mb} = \frac{BP \times 120}{V_d N} = \frac{50\,265 \times 120}{1.8096\times10^{-3} \times 4000} = 8.33\times10^{5}\ \text{Pa} = 8.33\ \text{bar}

(v) Mean piston speed

Sˉp=2LN/60=2×0.09×400060=12 m/s\bar{S}_p = 2LN/60 = \frac{2 \times 0.09 \times 4000}{60} = 12\ \text{m/s}

Answer: Vs=452.4V_s = 452.4 cc, Vd=1809.6V_d = 1809.6 cc, r=9.23r = 9.23, BP=50.27BP = 50.27 kW, bmep =8.33= 8.33 bar, Sˉp=12\bar S_p = 12 m/s.

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

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