Farm Power Sources and I.C. Engines: Cycles, Components, Systems, Fuels, Lubricants and Performance
2. Thermodynamic principles and I.C. engine cycles
An engine converts the chemical energy of fuel into work by burning it inside the working gas. Its ideal is an air-standard cycle: air as a perfect gas of constant specific heats, heat added and rejected reversibly, compression and expansion isentropic. The Otto cycle (spark-ignition, SI) adds heat at constant volume and has η = 1 − 1/r^(γ−1), depending only on the compression ratio r. The Diesel cycle (compression-ignition, CI) adds heat at constant pressure, with cut-off ratio ρ = V₃/V₂, and η = 1 − (1/r^(γ−1))·(ρ^γ − 1)/(γ(ρ − 1)); the bracket exceeds 1, so for the same r the Otto cycle is more efficient. The dual cycle adds heat partly at constant volume and partly at constant pressure and lies between them. CI engines are nevertheless more efficient in practice because they run at compression ratios roughly twice those of SI engines — an SI engine is limited by knock, a CI engine needs the high compression to ignite its fuel.
Worked: at r = 8 and γ = 1.4 the Otto efficiency is 1 − 8^(−0.4) = 1 − 0.435 = 56.5 per cent. At r = 16 with cut-off ratio 2, the Diesel efficiency is 1 − (1/16^0.4)(2^1.4 − 1)/(1.4 × 1) = 1 − 0.3299 × 1.1707 = 61.4 per cent — higher, because of the doubled compression ratio.
A four-stroke engine completes suction, compression, power and exhaust in two crankshaft revolutions, so it has one power stroke per cylinder every two revolutions and its camshaft turns at half crankshaft speed. A two-stroke engine uses ports and crankcase (or blower) scavenging to complete the cycle in one revolution, giving one power stroke per revolution and a higher power-to-weight ratio at the cost of scavenging losses and fuel economy.
3. Engine components and systems; fuels, combustion and lubricants
The components are the cylinder block and head, piston with compression and oil rings, gudgeon pin, connecting rod, crankshaft with flywheel, camshaft, valves with their springs and rockers, and the timing gears. Around them sit the systems the syllabus names. Fuel: an SI engine mixes air and fuel in a carburettor or by port injection; a CI engine has a fuel tank, filters, a fuel-injection pump that meters and pressurises, and injectors that atomise fuel into the hot compressed air, with a governor controlling speed. Cooling: air cooling by fins, or water cooling with a pump, radiator, fan and a thermostat that keeps the engine at working temperature. Lubrication: splash, or forced-feed from a gear pump through filters to the bearings. Ignition: battery-coil or magneto for SI engines (a magneto generates its own current and needs no battery); CI engines need no ignition system. Electrical: battery, alternator or dynamo, starter motor, lights. Intake and exhaust: an air cleaner (oil-bath or dry paper element, vital in dusty field work), the manifolds, a silencer, and on many engines a turbocharger.
| Item | SI engine (petrol) | CI engine (diesel) |
|---|---|---|
| Quality rating | octane number — resistance to self-ignition | cetane number — readiness to self-ignite |
| Knock | end-gas auto-ignites ahead of the flame (detonation), late in combustion | too long an ignition delay, so accumulated fuel burns at once, early in combustion |
| Remedy | higher-octane fuel, lower compression, retard the spark | higher-cetane fuel, higher compression and temperature |
Lubricants separate moving surfaces, cool, seal and clean. Their properties: viscosity (resistance to flow) and viscosity index (how little the viscosity changes with temperature — high VI is better); flash point and fire point (safety); pour point (the lowest temperature at which the oil still flows); oiliness and detergency. SAE grades classify viscosity; a multigrade oil such as 20W-40 behaves like a thin SAE 20 oil when cold (the W is for winter) and like an SAE 40 oil when hot, thanks to viscosity-index improvers.
4. Power, efficiencies and their measurement; performance curves
Indicated power is the power developed by the gas on the pistons: IP = p_mi·L·A·n·k/60, with p_mi the indicated mean effective pressure (Pa), L the stroke (m), A the piston area (m²), k the number of cylinders, and n the number of power strokes per minute per cylinder — N/2 for a four-stroke engine and N for a two-stroke. Brake power, measured at the crankshaft by a dynamometer, is BP = 2πNT/60. Friction power FP = IP − BP is lost in rubbing, pumping and driving accessories, and mechanical efficiency is η_m = BP/IP. Friction power is found by the Morse test (for multi-cylinder engines, cutting out one cylinder at a time at constant speed and measuring the fall in BP), by Willan’s line (for CI engines, extrapolating fuel consumption against BP back to zero fuel), or by motoring the engine.
Brake thermal efficiency is η_bth = BP/(ṁ_f × CV), and brake specific fuel consumption is bsfc = ṁ_f/BP, usually in kg/kWh; the two are reciprocal apart from the calorific value, since η_bth = 3600/(bsfc × CV) with CV in kJ/kg. An engine burning 5 kg/h of diesel of CV 42 000 kJ/kg (58.3 kW of fuel energy) to give 20 kW has bsfc = 0.25 kg/kWh and η_bth = 34.3 per cent. The heat balance accounts for the rest of the fuel energy: heat to the brake power, to the cooling water, to the exhaust gases, and an unaccounted remainder (radiation and the like). Volumetric efficiency compares the air actually inducted with the swept volume at intake conditions.
Performance curves plot torque, brake power and bsfc against speed at full throttle (or full rack). Torque peaks at a lower speed than power, because power keeps rising with speed after torque has begun to fall; bsfc is lowest near the speed of maximum torque. The torque backup (torque reserve) = (T_max − T_rated)/T_rated × 100 measures how well an engine lugs through a sudden load without stalling — a tractor engine with a rated torque of 150 N·m and a peak of 180 N·m has a 20 per cent backup. A governor holds speed against load changes, which is why a tractor engine at part load runs along a steep governor line rather than the full-throttle curve.
5. Selection, operation, maintenance and repair of I.C. engines
An engine is selected for the load it drives: the power the job needs at the shaft, the speed range, the duty (continuous for a pump, fluctuating for a thresher), the fuel available, and the ambient conditions. Engines lose power at altitude and in heat because the air is thinner, so a rating is derated for site conditions. For a stationary pump set, a CI engine is chosen for its fuel economy and long life; for light portable jobs, a small SI engine for its weight and ease of starting.
- Daily: check fuel, lubricating oil and coolant levels; clean or check the air cleaner in dusty work; drain water from the fuel filter.
- Periodic: change engine oil and oil filter, replace fuel filters, check fan-belt tension, service the injectors and set valve clearances at the hours the manufacturer specifies.
- Symptoms and causes: black smoke — overload, a choked air cleaner or faulty injection; blue smoke — oil burning past worn rings or guides; white smoke — water in the fuel or a cold engine; overheating — low coolant, a slipping fan belt, a stuck thermostat or a scaled radiator.
- Overhaul: when compression, oil consumption and blow-by show worn rings, liners and bearings — reboring or relining, new rings and bearings, valve grinding.
Key takeaways
- Wind power is ½ρAV³ times a power coefficient that cannot exceed the Betz limit of 16/27.
- Otto η = 1 − 1/r^(γ−1); at equal r the Otto cycle beats the Diesel, but CI engines win in practice by running at far higher r.
- IP = p_mi·L·A·n·k/60 with n = N/2 for four-stroke; BP = 2πNT/60; η_m = BP/IP.
- η_bth = BP/(ṁ_f·CV) and bsfc = ṁ_f/BP; they are reciprocal through the calorific value.
- Octane rates resistance to self-ignition, cetane rates readiness for it; biodiesel comes from transesterification.
- Torque peaks below the speed of peak power; torque backup = (T_max − T_rated)/T_rated.
Practice questions (12)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the air-standard efficiency of an Otto cycle with a compression ratio of 8, taking γ = 1.4? Give the answer in per cent to one decimal place.
Numerical answer — type the value.
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Answer: 56.5
η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4 = 1 − 1/2.297 = 1 − 0.4353 = 0.5647, i.e. 56.5 per cent. Using γ in place of γ − 1 gives 1 − 8^(−1.4) = 94.6 per cent, which no engine cycle approaches.An air-standard Diesel cycle has a compression ratio of 16 and a cut-off ratio of 2. Taking γ = 1.4, what is its efficiency, in per cent to one decimal place?
Numerical answer — type the value.
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Answer: 61.4
16^0.4 = 3.031, so 1/r^(γ−1) = 0.3299. The bracket (ρ^γ − 1)/(γ(ρ − 1)) = (2^1.4 − 1)/1.4 = (2.639 − 1)/1.4 = 1.1707. η = 1 − 0.3299 × 1.1707 = 1 − 0.3862 = 61.4 per cent. Leaving out the bracket gives the Otto value at r = 16, 67.0 per cent, which overstates the Diesel cycle.An engine develops a torque of 200 N·m at 2000 rev/min. What is its brake power, in kW (to one decimal place)?
Numerical answer — type the value.
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Answer: 41.9
BP = 2πNT/60 = 2π × 2000 × 200/60 = 41 888 W = 41.9 kW. Omitting the 2π gives 6.67 kW, and treating N as rev/s (without the 60) gives 2513 kW.An engine produces 20 kW of brake power while consuming 5 kg/h of diesel with a calorific value of 42 000 kJ/kg. What is its brake thermal efficiency, in per cent to one decimal place?
Numerical answer — type the value.
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Answer: 34.3
Fuel power = 5 × 42 000/3600 = 58.33 kW, so η_bth = 20/58.33 = 0.3429 = 34.3 per cent. Forgetting to convert kg/h to kg/s gives an efficiency of 0.0095 per cent; the same data give bsfc = 5/20 = 0.25 kg/kWh.A single-cylinder four-stroke engine has a bore of 100 mm and a stroke of 120 mm, and runs at 1500 rev/min with an indicated mean effective pressure of 6 bar. What is its indicated power, in kW (to two decimal places)?
Numerical answer — type the value.
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Answer: 7.07
A = π/4 × 0.1² = 7.854 × 10⁻³ m²; n = N/2 = 750 power strokes/min. IP = p·L·A·n/60 = 6 × 10⁵ × 0.12 × 7.854 × 10⁻³ × 750/60 = 7069 W = 7.07 kW. Using N = 1500 in place of N/2 doubles it to 14.14 kW, the standard four-stroke slip.An engine has an indicated power of 25 kW and a brake power of 20 kW. What is its mechanical efficiency, in per cent?
Numerical answer — type the value.
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Answer: 80
η_m = BP/IP = 20/25 = 0.80 = 80 per cent, and the friction power is 25 − 20 = 5 kW. Dividing IP by BP gives 125 per cent, impossible for an efficiency.A wind turbine of rotor diameter 10 m operates in a wind of 8 m/s. Taking air density 1.2 kg/m³ and a power coefficient of 0.4, what power does it deliver, in kW (to two decimal places)?
Numerical answer — type the value.
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Answer: 9.65
A = π/4 × 10² = 78.54 m². Power in the wind = ½ × 1.2 × 78.54 × 8³ = 24 127 W; delivered = 0.4 × 24 127 = 9651 W = 9.65 kW. Using V² instead of V³ gives 1.21 kW, and using the diameter as the radius quadruples the answer.A tractor engine has a torque of 150 N·m at its rated speed and a maximum torque of 180 N·m at a lower speed. What is its torque backup, in per cent?
Numerical answer — type the value.
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Answer: 20
Torque backup = (T_max − T_rated)/T_rated × 100 = (180 − 150)/150 × 100 = 20 per cent. Dividing by the maximum torque instead gives 16.7 per cent, which is not the standard definition.The cetane number of a fuel is a measure of its:
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Answer: A — ignition quality in a compression-ignition engine
A high cetane number means a short ignition delay, so the fuel self-ignites readily in a CI engine and diesel knock is avoided. Resistance to knock in an SI engine is the octane number — the opposite property, which is why the two are so often swapped in options.Which of the following statements about I.C. engines are correct?
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Answer: A — A magneto ignition system does not need a battery to produce the spark; B — A compression-ignition engine has no spark plug; C — A two-stroke engine has one power stroke per crankshaft revolution
A magneto generates its own current from a rotating magnet; a CI engine ignites fuel by the heat of compression; a two-stroke engine completes its cycle in one revolution. Each valve of a four-stroke engine opens once every two revolutions, so the camshaft turns at half crankshaft speed — the fourth statement is false.Willan’s line method of finding the friction power of a CI engine consists of:
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Answer: A — plotting fuel consumption against brake power at constant speed and extrapolating to zero fuel
At constant speed, fuel consumption rises almost linearly with brake power; extending the line to zero fuel reaches a negative brake power whose magnitude is the friction power. Cutting out cylinders is the Morse test, and driving the engine is motoring — both valid methods, but not Willan’s.Biodiesel is produced from vegetable oils mainly by:
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Answer: A — transesterification with an alcohol in the presence of a catalyst
Transesterification swaps the glycerol backbone of the triglyceride for methanol, giving thinner methyl esters that a CI engine can burn, with glycerol as a by-product. Anaerobic digestion gives biogas, yeast fermentation gives ethanol, and gasification gives producer gas.