Your Answer : (Not Answered)

Correct Answer : C

The maximum electricity produced is 3.15 kW.

The engine is coupled directly to the generator, so the generator receives the full brake power of 3.5 kW. Its electrical output is that shaft input times the generator efficiency:

\[ \begin{aligned} P_{e} &= \eta_{g} \times P_{b} \\ &= 0.90 \times 3.5 = 3.15\ \text{kW} \end{aligned} \]

3.50 kW ignores the generator losses. 3.00 and 2.85 kW would correspond to generator efficiencies of about 86% and 81%, not the 90% given. The engine's thermal efficiency from Q48 plays no part here, because 3.5 kW is already the brake (shaft) output.

Your Answer : (Not Answered)

Correct Answer : D

The compression ratio of a diesel engine is of the order of 15 to 25.

A diesel engine ignites its fuel by the heat of compression alone, so the air must be compressed enough to raise its temperature above the self-ignition temperature of diesel. That needs compression ratios of roughly 15 to 25.

5 to 8 and 10 to 14 are typical of petrol (SI) engines, which would knock at higher ratios, and 26 to 30 would bring very high pressures and heavy engines for little gain.

Your Answer : (Not Answered)

Correct Answer : A

Theoretically the inlet valve opens at Top Dead Centre (TDC).

On the theoretical valve timing diagram of a four stroke engine, the suction stroke runs from top dead centre to bottom dead centre, so the inlet valve opens at TDC and closes at BDC. In actual engines it opens a little before TDC to make use of the inertia of the gases.

BDC is where the inlet valve closes theoretically, and 'side centre position' and 'bottom side position' are not standard piston positions.

Your Answer : (Not Answered)

Correct Answer : C

The thermal efficiency of a two stroke engine is less than that of a four stroke engine.

In a two stroke engine the exhaust and fresh charge overlap during scavenging, so some fresh charge escapes with the exhaust and some exhaust stays in the cylinder. The effective compression stroke is also shorter because the ports occupy part of it.

These losses lower its thermal efficiency, even though it gives more power for its size, so 'equal to', 'greater than' and 'three times more' are all wrong.

Your Answer : (Not Answered)

Correct Answer : C

The key's answer is steam generator.

A fuel injector sprays fuel into the cylinder or intake of an internal combustion engine. A diesel engine depends on it, and modern SI engines, two stroke as well as four stroke, also use port or direct fuel injection in place of a carburettor.

A steam generator is a boiler, an external combustion device in which fuel is burnt in a furnace to raise steam, so it has no fuel injector of the engine kind. Older carburetted SI engines also lack injectors, but the key treats the steam generator as the device to which injectors do not belong.

Your Answer : (Not Answered)

Correct Answer : B

The process is called carburation.

In an SI engine the carburettor atomises and vaporises petrol and mixes it with air in the correct proportion before the charge enters the cylinder. This preparation of the mixture is carburation.

Ignition is starting combustion with the spark, scavenging is clearing burnt gases from the cylinder, and 'firing of engine' is not the term for mixing.

Your Answer : (Not Answered)

Correct Answer : B

A carburetor supplies petrol and air.

The carburetor is fitted to petrol (spark ignition) engines. It meters petrol into the air stream passing through it and delivers a combustible petrol-air mixture to the inlet manifold.

Diesel engines draw in air alone and inject the fuel directly, so they have no carburetor, which rules out the options with diesel. Lubricating oil is not supplied by the carburetor either: in two-stroke petrol engines it is premixed with the petrol in the tank.

Your Answer : (Not Answered)

Correct Answer : C

The device is the carburetor.

In a petrol (spark ignition) engine the carburetor breaks the fuel into a fine spray as air rushes through its venturi, vaporises it and mixes it with the air in the right proportion before the charge enters the cylinder.

The choke is only a valve in the carburetor's air passage that enriches the mixture for a cold start; it does not atomise anything by itself. The piston compresses the charge, and the battery supplies the ignition and electrical system.

Your Answer : (Not Answered)

Correct Answer : C

The indicated mean effective pressure is 5.250 bar.

Mean effective pressure is the work per cycle divided by the swept volume. With compression ratio r, the total volume is \(rV_c\), so the swept volume is \((r - 1)V_c\):

\[ \begin{aligned} V_s &= (5.5 - 1)V_c = 4.5V_c \\ p_m &= \frac{23.625 \times 10^5\, V_c}{4.5\, V_c} \\ &= 5.25 \times 10^5\ \text{Pa} = 5.25\ \text{bar} \end{aligned} \]

Dividing by the total volume, \(5.5V_c\), gives the 4.295 bar option, and multiplying by 4.5 instead of dividing gives 106.3 bar. The clearance volume cancels, which is why the answer can be found without knowing it.

Your Answer : (Not Answered)

Correct Answer : A

The order is \(\eta_{otto} \gt \eta_{dual} \gt \eta_{diesel}\).

At the same compression ratio and the same heat rejection, the Otto cycle adds all its heat at constant volume, at the highest average temperature, while the Diesel cycle adds heat at constant pressure during expansion, at a lower average temperature. The dual cycle adds part of its heat each way, so its efficiency lies between them.

The higher the mean temperature of heat addition for the same heat rejected, the higher the efficiency, which gives Otto, then dual, then Diesel. Any order putting the Diesel or dual cycle first reverses this.

Your Answer : (Not Answered)

Correct Answer : B

The compression ratio is more than the expansion ratio.

In a diesel cycle, heat is added at constant pressure while the piston moves out, so combustion continues to the cut-off point. With compression ratio \(r = V_{1}/V_{2}\) and cut-off ratio \(\rho = V_{3}/V_{2} \gt 1\), the expansion ratio is

\[ r_{e} = \frac{V_{4}}{V_{3}} = \frac{r}{\rho} \lt r \]

So expansion starts from a larger volume and the expansion ratio is smaller. They are equal only in the Otto cycle, where heat is added at constant volume. 'Variable' is wrong because the inequality always holds.

Your Answer : (Not Answered)

Correct Answer : B

P is volatility and Q is cetane number.

An SI engine must vaporise enough fuel in a cold intake to form an ignitable mixture. Hard or cold starting is therefore governed by the volatility of the fuel, in particular its front end (the temperature at which the first 10% distils).

In a CI engine the fuel ignites on its own in hot compressed air, and the time between injection and the start of combustion is the ignition delay. A high cetane number means a short delay, so the cetane number is the governing property.

Octane number measures resistance to knock in SI engines, not starting. Calorific value sets the energy per kilogram, flash point is a handling and safety property, and viscosity affects atomisation, none of which is the predominant control here.

Your Answer : (Not Answered)

Correct Answer : A

The power developed in the cylinder is the indicated horsepower (IHP).

Indicated horsepower is the power developed by the burning gases on the piston, worked out from the indicator diagram (pressure against volume) of the cylinder, hence the name. It is the highest of the engine's power figures, since nothing has yet been lost to friction.

BHP is what remains at the flywheel after friction, FHP is the power lost to friction (IHP − BHP), and DBHP is the power left at the tractor's drawbar. Mechanical efficiency is BHP/IHP.

Your Answer : (Not Answered)

Correct Answer : C

The mechanical efficiency is 67.21%.

Brake power from the torque:

\[ \begin{aligned} BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 2200 \times 55}{60} \\ &\approx 12.67\ \text{kW} \end{aligned} \]

Indicated power from the mean effective pressure, with one power stroke every two revolutions in a 4-stroke engine:

\[ \begin{aligned} A &= \tfrac{\pi}{4}(0.073)^{2} = 4.185 \times 10^{-3}\ \text{m}^{2} \\ IP &= p_{i} L A \cdot \tfrac{N}{2 \times 60} \cdot n \\ &= 1050 \times 0.078 \times 4.185 \times 10^{-3} \\ &\quad \times 18.33 \times 3 \approx 18.85\ \text{kW} \end{aligned} \]

\[ \eta_{m} = \frac{12.67}{18.85} = 0.6721 = 67.21\% \]

Forgetting the factor 2 for a 4-stroke engine doubles the indicated power and halves the efficiency, and the other listed values do not follow from consistent working.

Your Answer : (Not Answered)

Correct Answer : B

The shaft is the cam shaft.

The camshaft carries a cam for each valve. As it turns, each cam lifts its follower at the right moment, opening the inlet or exhaust valve through the push rod and rocker arm, and the valve spring closes it again. In a four-stroke engine the camshaft turns at half the crankshaft speed, since each valve opens once every two revolutions.

The crankshaft drives the camshaft but does not act on the valves directly. A rocker shaft only carries the rocker arms as a pivot; it does not decide the timing.

Your Answer : (Not Answered)

Correct Answer : B

Both valves are closed for 40.97% of the cycle.

A four-stroke cycle takes 720 degrees of crank rotation. Find how long at least one valve is open:

\[ \begin{aligned} \text{inlet open} &= 10 + 180 + 40 = 230^{\circ} \\ \text{exhaust open} &= 25 + 180 + 15 = 220^{\circ} \\ \text{overlap at TDC} &= 10 + 15 = 25^{\circ} \end{aligned} \]

\[ \begin{aligned} \text{any valve open} &= 230 + 220 - 25 = 425^{\circ} \\ \text{both closed} &= 720 - 425 = 295^{\circ} \\ \frac{295}{720} &= 40.97\% \end{aligned} \]

Forgetting to subtract the 25-degree overlap gives 270 degrees, or 37.5%, which is not offered. The overlap counts once only, because during it both valves are open together.

Your Answer : (Not Answered)

Correct Answer : C

The most important property is the pour point.

The pour point is the lowest temperature at which a fuel still flows. In very cold climates, diesel's waxy components crystallise as the temperature drops, the fuel thickens, and filters and lines can block so that the engine is starved of fuel. A fuel for such regions must have a pour point (and cloud point) below the lowest expected temperature, which is why winter grades of diesel are blended.

Heating value governs energy content and fuel consumption, flash point is a safety property for storage and handling, and boiling point (volatility) affects vaporisation, but none of these stops the fuel flowing at low temperature.

Your Answer : (Not Answered)

Correct Answer : C

The fan efficiency is about 63%.

Brake power of the engine:

\[ \begin{aligned} P_{b} &= \frac{2\pi N T}{60} = \frac{2(3.14)(2000)(382)}{60} \\ &\approx 79\,965\ \text{W} \end{aligned} \]

The fan takes 5% of this, about 3998 W. The useful (air) power it delivers is the volume flow times the pressure rise:

\[ \begin{aligned} P_{air} &= Q\,\Delta p = 2.8 \times 900 = 2520\ \text{W} \\ \eta &= \frac{2520}{3998} \approx 0.63 \end{aligned} \]

So the efficiency is nearest to 63%. Rounding the brake power to 80 kW (fan input 4000 W) still gives 63%. The distractors would need different fan inputs: 65% about 3880 W, 60% 4200 W and 58% about 4350 W, none of which is 5% of this engine's brake power.

Your Answer : (Not Answered)

Correct Answer : B

The pump is a centrifugal pump.

In the forced (pump) circulation cooling system of a tractor engine, a small centrifugal impeller pump, usually driven by the same V-belt as the fan, circulates coolant from the bottom of the radiator through the water jackets of the cylinder block and head and back to the radiator. The job needs a large flow at a low head, and a centrifugal pump does that simply, with no valves and no risk of damage if the flow is throttled by the thermostat.

Piston, gear and vane pumps are positive displacement pumps. They suit high pressure and small flow, such as fuel injection or the lubrication and hydraulic systems, and would be needlessly complex and noisy for coolant circulation.

Your Answer : (Not Answered)

Correct Answer : B

Specific fuel consumption is inversely proportional to the thermal efficiency of an engine. Thermal efficiency is the work output divided by the energy of the fuel burnt, and specific fuel consumption is the fuel used per unit of power output, so when more fuel is needed for the same power the efficiency is lower. Mean piston speed, specific power output and mean effective pressure are not inversely proportional to thermal efficiency in this sense.