Question 1

Typical Colour of Diesel Engine Exhaust – Diagnostic Indication In routine operation, what exhaust colour is most commonly associated with a diesel engine running under heavy load with incomplete combustion (soot formation)?

Accepted Answer

Introduction / Context: Exhaust colour is a quick diagnostic for combustion quality and possible faults. Diesel engines, especially under transient or heavy load, can produce visible smoke whose colour points to root causes like over-fuelling, oil burning, or coolant ingress. Given Data / Assumptions: Compression-ignition diesel operating with varying loads. No after-treatment (or it is ineffective during the observation). Observation in normal ambient lighting. Concept / Approach: Black smoke indicates soot—carbonaceous particulate matter from incomplete combustion, often due to over-fuelling relative to available air, poor atomization, restricted air intake, or inadequate boost. Bluish smoke indicates oil burning (lubricating oil entering the combustion chamber), while persistent white smoke suggests unburned droplets (low temperature misfire) or coolant/steam in the exhaust. Violet is not a recognized diagnostic colour. Step-by-Step Solution: Link colour to cause: black → soot from rich or air-starved operation.Identify operating condition: heavy load or transients increase fuelling demand.Infer that air–fuel mixing/time may be insufficient → soot forms.Therefore, black is the typical colour under the stated condition. Verification / Alternative check: Opacity meters and particulate measurements correlate visible black smoke with high soot mass concentrations. Modern DPF systems capture soot, reducing visible black smoke under regulated operation. Why Other Options Are Wrong: White is more indicative of coolant ingress or cold misfire. Bluish signals lubricating oil burning. Violet is not standard. Colourless exhaust occurs when after-treatment is effective and combustion is complete, but the question asks the 'generally' seen colour linked with soot. Common Pitfalls: Assuming all diesel smoke is black; conditions and faults matter. Also, confusing condensation plume (white) on cold mornings with faults. Final Answer: black

Question 2

Spark-ignition (SI) engines — effect of advancing ignition timing on knock In a petrol/gasoline spark-ignition engine, if the spark timing is advanced (spark occurs earlier before top dead centre, BTDC), how does this change influence the detonation…

Accepted Answer

Introduction / Context: Knock (detonation) in spark-ignition engines is the spontaneous auto-ignition of the end-gas after normal flame initiation. Ignition timing is a first-order control variable that strongly affects peak pressure, end-gas temperature, and the time available for auto-ignition reactions. Understanding how advancing the spark changes knock tendency is core to calibration and engine diagnostics. Given Data / Assumptions: Conventional spark-ignition engine operating on gasoline-like fuel. Operating point (speed, load, coolant temp) otherwise unchanged. Advancing timing means firing the spark earlier BTDC. Concept / Approach: Advancing the spark increases the time for the flame front to propagate before TDC. This raises the cylinder pressure and temperature earlier in the cycle. The unburned end-gas is thus exposed to higher pressure/temperature for longer, which accelerates pre-flame reactions and reduces auto-ignition delay. Therefore, earlier timing generally increases knock tendency, while retarding timing tends to reduce it (at the expense of efficiency and exhaust temperature). Step-by-Step Solution: Advance timing → earlier heat release near TDC.Earlier heat release → higher peak pressure and temperature in the end-gas region.Higher T and p → shorter chemical ignition delay of end-gas.Shorter delay + more residence time → higher probability of knock. Verification / Alternative check: Engine maps show that the knock-limited spark advance (KLSA) is retarded under high load/higher intake temperatures to avoid detonation, confirming that advancing timing increases knock susceptibility. Why Other Options Are Wrong: reduce: opposite of observed behavior. not affect: ignores thermochemistry and pressure-temperature history. lead to complete elimination of knock at high load: advancing timing usually worsens knock at high load. Common Pitfalls: Confusing knock with pre-ignition; pre-ignition is an abnormal early ignition source, not end-gas auto-ignition. Also, small timing changes at light load may not trigger audible knock, but the tendency still moves in the same direction. Final Answer: increase

Question 3

Nuclear basics — identifying naturally occurring isotopes of uranium Which of the following are naturally occurring isotopes of uranium commonly referenced in nuclear engineering and fuel discussions?

Accepted Answer

Introduction / Context: Uranium in nature is a mixture of several isotopes, each with different neutron-interaction properties. Recognizing these isotopes is foundational for topics such as enrichment, reactor physics, and radiological characteristics. Given Data / Assumptions: Naturally occurring uranium in the Earth’s crust. Focus on the principal isotopes relevant to nuclear fuel. Concept / Approach: Natural uranium consists primarily of U-238 (~99.27%), with a smaller fraction of fissile U-235 (~0.72%), and a trace amount of U-234 (~0.005–0.01%) arising from decay chains. All three are naturally present and are therefore valid answers when asked to identify uranium isotopes found in nature. Step-by-Step Solution: List common natural isotopes: U-238, U-235, U-234.Confirm their presence in natural uranium ore and refined natural uranium.Therefore, the correct choice is the inclusive option listing all three. Verification / Alternative check: Fuel cycle references and reactor-grade material specifications routinely cite these exact isotopic compositions for natural uranium feedstock. Why Other Options Are Wrong: Choosing only one isotope ignores that natural uranium is a mixture; engineering practice must consider the presence and roles of all three. Common Pitfalls: Confusing U-233 (bred from Th-232) with natural uranium isotopes; U-233 is not naturally abundant and is produced in thorium fuel cycles. Final Answer: All of these

Question 4

Four-stroke diesel valve timing — exhaust event specifics For a typical four-stroke cycle diesel engine, which timing most closely represents the exhaust valve event relative to dead centres (DC)?

Accepted Answer

Introduction / Context: Valve timing in four-stroke engines is optimized for gas exchange and efficiency, not set strictly at dead centres. Understanding typical opening/closing angles helps in diagnosing performance and emissions behavior. Given Data / Assumptions: Conventional naturally aspirated or turbocharged diesel. Cam phasing aimed at effective blow-down and scavenging. Angles are representative values; exact numbers vary by design. Concept / Approach: Exhaust opens before BDC during the power stroke to start blow-down, reducing cylinder pressure before the piston rises on the exhaust stroke. Closing slightly after TDC ensures complete scavenging and overlap with intake valve opening for better cylinder emptying, especially at higher speeds. Step-by-Step Solution: Power stroke nearing BDC: open exhaust early (~30–60° BBDC) to release pressure.Piston rises on exhaust stroke: flow continues out.Close after TDC (~5–20° ATDC) to allow final scavenging and overlap with intake opening.Therefore, the typical pair 30° BBDC / 10° ATDC is the best fit among the choices. Verification / Alternative check: OEM timing charts commonly show early exhaust opening and slight after-TDC closing; exact degrees depend on speed/load targets and turbo matching. Why Other Options Are Wrong: Opening ABDC or closing BTDC (Option B) would impair blow-down. Exact BDC/TDC events (Option C) are unrealistic. Option D ignores the critical role of tuned overlap and phasing. Common Pitfalls: Assuming gasoline and diesel timings are identical; trends are similar but optimized differently for combustion and turbocharging. Final Answer: Opens at 30° before bottom dead centre (BBDC) and closes at 10° after top dead centre (ATDC)

Question 5

Engine governing — qualitative vs quantitative (concept check) Evaluate the statement about qualitative governing: “In qualitative governing, the quantity of fuel is varied to suit the load on the engine and the total charge of air is altered.” Is t…

Accepted Answer

Introduction / Context: Governing maintains engine speed under varying load. Two classic approaches are qualitative governing (altering mixture quality) and quantitative governing (altering total charge). Distinguishing these is essential in IC engine control theory and older carburetted systems. Given Data / Assumptions: SI engines with carburetors exemplify qualitative governing. Load changes require either mixture strength changes or inducted charge changes. Modern EFI can emulate both strategies electronically. Concept / Approach: Qualitative governing varies the fuel-air ratio (mixture strength) while keeping the total inducted air charge approximately constant. Quantitative governing varies the total mass of charge admitted (by throttling or fuel quantity in diesels) while keeping mixture quality more nearly constant (stoichiometric in SI, excess air in CI). The statement mixes these and is therefore incorrect. Step-by-Step Solution: Define qualitative governing: adjust mixture richness/leaness at near-constant air charge.Define quantitative governing: adjust total charge mass at near-constant mixture quality.Compare with statement: it claims varying fuel and altering total air charge simultaneously under “qualitative,” which is inconsistent.Therefore, the statement is incorrect. Verification / Alternative check: Classic texts on carburation differentiate: qualitative (needle/jet control of richness) vs quantitative (throttle to change quantity). Why Other Options Are Wrong: Choosing “Correct” perpetuates the mix-up between mixture quality and quantity control strategies. Common Pitfalls: Assuming diesels use the same terminology; diesels primarily use quantitative control by varying fuel quantity at essentially unthrottled air. Final Answer: Incorrect

Question 6

Nuclear weapons vs reactors — nature of chain reaction An atom bomb (fission weapon) utilises which form of nuclear chain reaction control?

Accepted Answer

Introduction / Context: Fission weapons and power reactors both rely on neutron-induced fission, but their control philosophies are opposite. This question checks understanding of chain-reaction control in weapons versus civilian reactors. Given Data / Assumptions: Fissionable material (e.g., highly enriched U-235 or Pu-239) in supercritical configuration. Presence or absence of engineered control systems and moderators. Concept / Approach: A fission weapon rapidly assembles a supercritical mass and ensures that neutron multiplication far exceeds losses, producing an exponential power rise in microseconds. There is no intent to stabilise the reaction. In contrast, nuclear power reactors maintain a controlled, near-critical state using control rods, coolant/moderator feedback, and engineered safety margins. Step-by-Step Solution: Define “uncontrolled”: prompt supercriticality with no regulating feedbacks engaged.Recognise weapon design goal: maximise energy release in an extremely short time.Therefore the correct description for an atom bomb is “uncontrolled nuclear fission chain reaction.” Verification / Alternative check: Historical accounts of fission weapons describe rapid assembly techniques (gun-type or implosion) to create prompt supercritical conditions. Why Other Options Are Wrong: Controlled fission (Option B) describes power reactors. Pure fusion without fission trigger (Option C) does not represent typical thermonuclear designs which use a fission primary. Option D is incorrect because chain reactions are fundamental to yield. Common Pitfalls: Equating “moderation” with “control”; moderation slows neutrons but control requires active reactivity management. Final Answer: Uncontrolled nuclear fission chain reaction

Question 7

Mechanical efficiency comparison — two-stroke vs four-stroke IC engines Compared at similar size and technology level, a two-stroke cycle engine typically exhibits __________ mechanical efficiency than a four-stroke cycle engine. Choose the most app…

Accepted Answer

Introduction / Context: Mechanical efficiency is the ratio of brake power to indicated power and reflects internal mechanical losses. Two-stroke engines often differ from four-strokes in scavenging, lubrication, and port timing, which influence friction and pumping losses. Given Data / Assumptions: Conventional crankcase-scavenged or loop-scavenged two-strokes. Comparable displacement and technology to a four-stroke peer. Typical operating speeds and loads. Concept / Approach: Two-strokes generally have higher specific friction and pumping work due to continuous scavenging, port timing that can increase short-circuit losses, and lubrication regimes that raise friction. Consequently, for similar conditions, their mechanical efficiency tends to be lower than that of well-designed four-strokes that separate intake and exhaust strokes and often benefit from pressure-charging only when desired. Step-by-Step Solution: Identify mechanical loss sources: bearings, piston/ring friction, pumping/scavenging work.Two-stroke continuous gas exchange increases pumping effort per cycle.Therefore brake power/indicated power ratio is typically lower in two-strokes. Verification / Alternative check: Empirical brake-specific fuel consumption and friction mean effective pressure comparisons show higher FMEP for two-strokes at similar speeds. Why Other Options Are Wrong: “higher” reverses observed trends. “approximately the same, always” ignores architecture differences. “indeterminate because friction is zero” is physically false. Common Pitfalls: Confusing thermal efficiency with mechanical efficiency; two metrics are distinct. Final Answer: lower

Question 8

Nuclear moderation — suitability of light elements Evaluate the statement: “Light-weight substances cannot be effective neutron moderators.” Is this statement valid for thermal reactors?

Accepted Answer

Introduction / Context: Moderators in thermal reactors slow down fast neutrons to thermal energies where fission cross-sections of fuels like U-235 are high. The suitability of a material as a moderator is largely determined by its atomic mass and neutron-absorption characteristics. Given Data / Assumptions: Thermal reactor context (e.g., PWR, BWR, CANDU, graphite-moderated). Goal: high scattering per collision, low absorption. Concept / Approach: Light nuclei (hydrogen in light water, deuterium in heavy water, carbon in graphite) are highly effective moderators. Their low mass enables large energy loss per elastic collision. Materials that are both light and have low neutron absorption make excellent moderators. Therefore, the statement that light substances cannot be effective moderators is false. Step-by-Step Solution: Identify candidates: H2O, D2O, graphite, Be.Note large fractional energy loss per collision for low-A nuclei.Confirm low absorption cross-section is desirable; D and C are particularly favorable.Conclude that light elements are among the best moderators. Verification / Alternative check: Most commercial thermal reactors use light water or heavy water as moderators, demonstrating practical effectiveness of light elements. Why Other Options Are Wrong: “True” contradicts theory and practice. Other choices limit the statement improperly. Common Pitfalls: Confusing moderator with reflector or coolant; some materials serve multiple roles but moderation quality hinges on scattering and absorption properties. Final Answer: False

Question 9

Engine testing — purpose of the Morse test The Morse test is commonly used to determine which performance parameter of an internal-combustion engine?

Accepted Answer

Introduction / Context: The Morse test is a practical workshop method to estimate indicated power of multi-cylinder engines when direct cylinder-pressure indication is not available. It relies on measuring changes in brake power as cylinders are disabled sequentially. Given Data / Assumptions: Engine with multiple cylinders and independent ignition/fuel cut-off. Brake dynamometer available to measure brake power. Speed held approximately constant during the test. Concept / Approach: Indicated power is not directly measured on the dynamometer. The Morse method disables one cylinder at a time, observing the drop in brake power. Assuming mechanical losses remain approximately constant at a given speed, the loss in brake power when a cylinder is cut corresponds to that cylinder’s indicated power contribution. Step-by-Step Solution: Measure brake power with all cylinders firing: BP_all.Disable cylinder i; measure BP_i_off.Indicated power of cylinder i ≈ BP_all − BP_i_off.Sum over cylinders to estimate total I.P.; mechanical efficiency = BP_all / IP_total. Verification / Alternative check: Results can be cross-checked with indicator diagrams where available or with cylinder pressure transducers on modern engines. Why Other Options Are Wrong: Single-cylinder (Option B) does not need Morse; direct methods are used. Brake power (Option C) is measured directly by dynamometer, not via Morse estimation. Compression ratio (Option D) is a geometric parameter, unrelated to Morse testing. Common Pitfalls: Allowing speed to vary significantly during disabling skews results because friction and pumping losses are speed-dependent. Final Answer: Indicated power (I.P.) of a multi-cylinder engine

Question 10

Nuclear energy units — MeV usage The unit “million electron volts” (MeV) is commonly used as a measure of nuclear-scale energies (for example, binding energies and reaction Q-values). Is this statement correct?

Accepted Answer

Introduction / Context: Energy scales in nuclear physics are orders of magnitude larger than typical chemical energies. Convenient units reflect this scale so that reaction energies, particle rest-mass energies, and binding energies can be expressed without awkward exponents. Given Data / Assumptions: 1 electron volt (eV) is the energy gained by an electron moving through a potential difference of 1 volt. 1 MeV = 10^6 eV. Conversion: 1 eV ≈ 1.602 × 10^−19 J. Concept / Approach: Nuclear binding energies per nucleon are typically on the order of a few MeV. Reaction Q-values and gamma energies are likewise expressed in keV or MeV. While SI units require Joules, MeV is the standard convenient unit in nuclear engineering and physics literature. Step-by-Step Solution: Recall 1 MeV in Joules: 1 MeV ≈ 1.602 × 10^−13 J.Compare to chemical bond energies (~eV scale), showing nuclear energies are ~10^6 times larger.Therefore, using MeV is customary and practical for nuclear energy scales. Verification / Alternative check: Mass-energy equivalence uses MeV/c^2 for particle masses (e.g., proton ~938 MeV/c^2), reinforcing the conventional usage. Why Other Options Are Wrong: “False” contradicts standard practice in nuclear data tables and reactor physics. Common Pitfalls: Confusing MeV (energy) with MeV/c^2 (mass units); they are related but not the same dimension. Final Answer: True

Question 11

Spark-ignition (petrol) engines vs. compression-ignition (diesel): Why is the compression ratio in petrol engines intentionally kept lower than in diesel engines? Choose whether you agree with the statement: a higher compression ratio in petrol engi…

Accepted Answer

Introduction / Context: This item tests core thermodynamics and combustion knowledge for internal combustion engines. Petrol (spark-ignition, SI) engines and diesel (compression-ignition, CI) engines operate with different ignition modes. Understanding why SI engines use lower compression ratios clarifies the link between compression, end-gas temperature, and abnormal combustion (pre-ignition/knock). Given Data / Assumptions: Petrol engines ignite the mixture by a timed spark. Diesel engines ignite fuel by air heating due to compression alone. Raising compression ratio increases end-gas temperature and pressure. Knock in SI is auto-ignition of the unburned end-gas ahead of the flame front. Concept / Approach: Compression ratio influences peak temperature prior to ignition. In SI engines, too high a compression ratio drives the unburned end-gas toward its auto-ignition limit, causing detonation/knock or even pre-ignition from hot spots. Diesel engines demand high compression to achieve auto-ignition of injected fuel; the fuels also differ in knock resistance (octane for SI, cetane for CI). Step-by-Step Solution: Recognize that higher compression ratio → higher end-gas temperature.In SI, ignition should occur due to the spark, not spontaneous auto-ignition.Therefore SI engines cap compression ratio to avoid knock; CI engines require the opposite (high compression) to ignite fuel. Verification / Alternative check: Practical SI engines use compression ratios roughly 8:1 to 12:1 (varies with fuel and technologies like EGR/direct injection), while diesels often range 15:1 to 20+:1, confirming the principle. Why Other Options Are Wrong: Disagree: Contradicts well-established knock behavior in SI engines. Agree only at high altitude / only with alcohol fuels: Altitude and fuel type shift limits, but the fundamental reason (avoid auto-ignition) universally applies. Common Pitfalls: Confusing SI knock with diesel knock; in SI, higher octane resists knock, whereas diesels need high cetane for short ignition delay. Also, modern knock sensors allow operation near limits but do not change the underlying constraint. Final Answer: Agree

Question 12

Identify the component that does NOT belong to a compression-ignition (diesel) engine fuel/air system.

Accepted Answer

Introduction / Context: This question checks recognition of key subsystems in compression-ignition (CI) engines versus spark-ignition (SI) engines. CI engines rely on fuel injection into hot compressed air, while SI engines premix fuel with air via a carburettor (traditional) or port/direct injection with throttling. Given Data / Assumptions: CI engines need high-pressure fuel delivery at precise timing. Air is drawn in separately; fuel is not premixed upstream in a carburettor. Speed/load are regulated by controlling injected fuel quantity. Concept / Approach: Components typical for CI engines include the fuel pump (high-pressure), injectors (atomization and spray), and a governor (controls fuel quantity for speed regulation). A carburettor, whose job is to meter petrol into the intake air stream by pressure drop at a venturi, is characteristic of old-style SI engines and is not used on diesels. Step-by-Step Solution: List CI essentials: injection pump + injectors + governor.Recognize that a carburettor premixes fuel with air before the cylinder, unsuitable for diesel auto-ignition strategy.Therefore, the item that does not belong to CI is the carburettor. Verification / Alternative check: Modern common-rail diesels still use high-pressure rails and electronic injectors, not carburettors, confirming the functional distinction. Why Other Options Are Wrong: Fuel pump: Essential to pressurize diesel fuel for injection. Fuel injector: Creates spray for mixing and ignition. Governor: Maintains speed by modulating fuel delivery in many diesel systems. Common Pitfalls: Confusing fuel system components across SI and CI engines; note that some small diesel generators still have mechanical governors and pumps but never a carburettor. Final Answer: Carburettor

Question 13

Compression-ignition engines: What is the effect of higher combustion-chamber wall temperature on knocking tendency?

Accepted Answer

Introduction / Context: Diesel (CI) knock is linked to ignition delay: longer delay allows more fuel to accumulate before rapid combustion, producing an abrupt pressure rise and characteristic knock. Chamber temperature influences chemical kinetics of auto-ignition and thus affects knock propensity. Given Data / Assumptions: Ignition in a CI engine occurs when injected fuel auto-ignites in hot compressed air. Ignition delay shortens as ambient (wall/charge) temperature increases. All else equal (fuel cetane, injection timing, air–fuel ratio), delay dominates knock intensity. Concept / Approach: Higher wall and charge temperatures promote faster pre-flame reactions, reducing ignition delay. With shorter delay, less fuel accumulates before combustion starts, softening the pressure rise rate and reducing knock severity. Conversely, cold starts lengthen delay and aggravate knock, which is why glow plugs are used in small diesels. Step-by-Step Solution: Relate temperature to chemical reaction rates (Arrhenius behavior).Higher wall/charge temperature → shorter ignition delay.Shorter delay → smaller premixed burn fraction → reduced knock tendency. Verification / Alternative check: Practical observations: winter operation yields harsher diesel combustion sounds; warming the chamber (glow plugs, block heaters) improves smoothness and reduces knock, supporting the theory. Why Other Options Are Wrong: Increase: Opposite of the established ignition-delay relationship. Not effect: Temperature is a primary variable in auto-ignition kinetics. Increase only at full load: Load modifies temperature and injection, but the base trend (higher temperature reduces knock) holds broadly. Common Pitfalls: Confusing SI knock (end-gas) with CI knock; mechanisms differ though both are forms of abnormal/undesirable combustion. Final Answer: reduce

Question 14

Two-stroke vs. four-stroke: Do two-stroke cycle engines generally require a lighter flywheel due to more uniform torque delivery?

Accepted Answer

Introduction / Context: Flywheel sizing depends on torque fluctuation within a cycle. Two-stroke engines deliver a power event every revolution, while four-stroke engines deliver power every second revolution. This directly affects the angular speed variation and the inertia needed to smooth it. Given Data / Assumptions: Two-stroke: one power stroke per crank revolution. Four-stroke: one power stroke per two revolutions. Other variables (number of cylinders, firing order) being comparable. Concept / Approach: More frequent power strokes reduce the interval between torque inputs, decreasing cyclic speed ripple. Hence, the required energy storage in the flywheel to keep speed within limits is lower in a two-stroke for the same cylinder and mean torque, implying a lighter flywheel suffices. Multi-cylinder four-strokes mitigate this with overlapping firing orders but the single-cylinder comparison reveals the principle clearly. Step-by-Step Solution: Compare frequency of power events: 2-stroke (every 360°) vs 4-stroke (every 720°).Torque fluctuation and speed ripple scale with interval between power pulses.Therefore, two-strokes typically need a lighter flywheel to achieve the same speed regulation. Verification / Alternative check: Classic motorcycle and small two-stroke engines exhibit small/light flywheels relative to similar displacement single-cylinder four-strokes, supporting the concept empirically. Why Other Options Are Wrong: Disagree: Ignores fundamental power-event frequency difference. Conditional answers: While application specifics matter, the baseline engineering rationale remains valid. Common Pitfalls: Overlooking the effect of cylinder count; a multi-cylinder four-stroke can have smooth torque and smaller flywheel per cylinder, but the question refers to the inherent cycle property. Final Answer: Agree

Question 15

Typical scavenging efficiency for a four-stroke diesel engine (naturally aspirated or mildly boosted) falls within which range?

Accepted Answer

Introduction / Context: Scavenging efficiency describes how effectively fresh charge replaces residual gases. In two-stroke engines this is a central design challenge; in four-stroke engines, separate exhaust and intake strokes generally allow near-complete clearing of exhaust products, especially with proper valve timing and moderate boost. Given Data / Assumptions: Four-stroke CI engine with distinct exhaust and intake strokes. Conventional valve overlap optimized for breathing. No extreme EGR intentionally retained within the cylinder for this conceptual question. Concept / Approach: Because exhaust and intake occur on separate strokes with piston-driven displacement, the cylinder has opportunity to expel most burned gases before fresh air induction. As a result, scavenging efficiency approaches unity (near 100%) under typical conditions. This is fundamentally higher than many uniflow or loop-scavenged two-strokes, where short-circuiting and mixing reduce efficiency. Step-by-Step Solution: Define scavenging efficiency as mass of fresh charge retained divided by mass that would fill the cylinder at intake conditions.Note that the dedicated exhaust stroke clears most residuals.Hence, efficiency is commonly in the high 90% range for four-strokes. Verification / Alternative check: Textbook values and engine simulation results typically place four-stroke scavenging efficiency very close to 1.0, albeit volumetric efficiency may vary. This supports the 95–100% selection. Why Other Options Are Wrong: Below 50% or 50–85%: Typical of poorly scavenged two-strokes, not healthy four-strokes. 85–95%: Possible under unusual constraints, but well-tuned four-strokes commonly exceed this. Common Pitfalls: Confusing scavenging efficiency with volumetric efficiency; the latter may be 70–110% depending on speed and boosting, while scavenging efficiency in a four-stroke is usually very high. Final Answer: between 95 and 100%

Question 16

Select the typical compression ratio range for a diesel (compression-ignition) engine.

Accepted Answer

Introduction / Context: Compression ratio is a defining difference between SI and CI engines. Diesels require high compression to heat air sufficiently for fuel auto-ignition, while petrol engines are limited by knock. Recognizing typical ranges is basic engine design literacy. Given Data / Assumptions: Conventional road and stationary diesel engines. Modern trends may slightly shift ranges but fundamentals hold. Compression ratio defined as V_max / V_min in the cylinder. Concept / Approach: Diesel engines commonly operate around 15:1 to 20:1; heavy-duty or cold-start-friendly designs may be higher, while some modern light-duty diesels use ratios a bit below 16:1 with advanced injection to meet emissions and NVH. Values far below 15:1 struggle with reliable CI ignition; values above ~25:1 are unusual due to peak pressure and stress limits. Step-by-Step Solution: Recall SI engines: typically 8–12:1 (varies with fuel and tech).CI engines: target air temperatures after compression high enough for auto-ignition → higher ratio.Thus the most correct band among choices is 15 to 25. Verification / Alternative check: Reviewing manufacturer data (trucks, generators) confirms ratios in the teens to low twenties, validating the selection. Why Other Options Are Wrong: 6–10 or 10–15: Typical for SI, not CI. 25–40: Extremely high; structural loads and emissions constraints make this rare. Common Pitfalls: Confusing geometric compression ratio with effective compression in turbocharged engines; boost changes effective compression temperature but not geometric ratio. Final Answer: 15 to 25

Question 17

Typical ignition voltage requirement in a spark-ignition system: what voltage is needed across the plug gap to reliably produce a spark?

Accepted Answer

Introduction / Context: Ignition systems in petrol engines transform low-voltage battery power into high voltage capable of breaking down the air-fuel mixture across the spark plug gap. Knowing the order of magnitude for breakdown voltage helps diagnose ignition faults and design insulation. Given Data / Assumptions: Spark gaps typically 0.6–1.0 mm for conventional systems. Mixture pressure near end of compression increases breakdown voltage. Classic coil/ distributor systems; modern systems can supply even higher voltage. Concept / Approach: Paschen-type behavior indicates required voltage increases with pressure and gap. Traditional automotive systems are designed to deliver roughly 10–12 kV minimum under cylinder pressure. Modern coil-on-plug or CDI systems may reach 20–40 kV for reliability, but the textbook baseline for minimum reliable spark is in the 10–12 kV bracket. Step-by-Step Solution: Consider typical plug gap and in-cylinder pressure at spark timing.Estimate breakdown voltage in the low tens of kilovolts.Select 10 000 to 12 000 volts as the representative requirement. Verification / Alternative check: Service manuals list secondary ignition peak capability often above 20 kV to ensure margin, aligning with the minimum threshold near 10–12 kV for dependable ignition in standard conditions. Why Other Options Are Wrong: 2–4 kV or 4–6 kV: Insufficient under compressed mixture conditions. 6–10 kV: May fire at low load but marginal at high pressure; not a robust design value. Common Pitfalls: Confusing no-pressure bench spark with in-cylinder pressure conditions; the latter demands much higher voltage. Final Answer: 10 000 to 12 000 volts

Question 18

Radiation hazards in nuclear reactors: which particles/photons require the most shielding attention from reactor structures?

Accepted Answer

Introduction / Context: Reactor shielding designs must attenuate harmful radiation to protect personnel and equipment. Different radiations have different penetration and interaction characteristics, requiring tailored shielding materials and thicknesses. Given Data / Assumptions: Common nuclear radiations: alpha (helium nuclei), beta (electrons/positrons), gamma (high-energy photons), and neutrons (uncharged particles). Goal: minimize dose rates outside containment to safe limits. Typical shielding materials: concrete, lead, water, borated compounds. Concept / Approach: Alpha particles are easily stopped by paper or skin; beta particles are limited by a few millimeters of metal or plastic. Gamma rays are highly penetrating and require dense materials (lead, heavy concrete). Neutrons, being uncharged, interact via scattering and absorption; they are penetrating and require hydrogenous materials for moderation plus absorbers (boron, cadmium) to capture them. Consequently, reactor shielding focuses primarily on neutrons and gamma rays. Step-by-Step Solution: Identify penetration ability: α < β < γ, with neutrons also deeply penetrating.Match shielding: hydrogen-rich for neutrons, high-Z dense materials for gammas.Therefore, the critical shielding targets are neutrons and gamma rays. Verification / Alternative check: Standard reactor biological shields consist of thick concrete (for gammas and moderated neutrons) and specialized absorbers for thermalized neutrons, demonstrating design priority on these two radiations. Why Other Options Are Wrong: Electrons: modest shielding needed compared with gammas/neutrons. “α, β and γ rays”: Alpha and most betas are not primary design drivers for thick biological shielding. None of these: Incorrect; shielding is essential. Common Pitfalls: Underestimating secondary gamma production from neutron capture; proper shields often layer moderators and absorbers to handle both particles and capture gammas. Final Answer: neutron and gamma rays

Question 19

Air–fuel preparation for spark-ignition engines: which device mixes air and petrol in the required proportion and supplies it during the suction stroke?

Accepted Answer

Introduction / Context: Traditional spark-ignition (petrol) engines used carburettors to meter and mix fuel with incoming air based on pressure drops at a venturi. Although modern engines use electronic fuel injection, the conceptual role of a carburettor remains fundamental in engine basics and legacy systems. Given Data / Assumptions: Petrol (SI) engine with intake depression during the suction stroke. Need to create a combustible mixture over a range of operating conditions. Mechanical metering without high-pressure fuel injection. Concept / Approach: A carburettor uses the venturi effect: as air speeds up through a constriction, static pressure drops, drawing fuel through a calibrated jet into the airstream. Auxiliary circuits handle idle, transition, and enrichment. By contrast, diesel engines employ direct injection of fuel into hot compressed air, not a carburettor. Step-by-Step Solution: Identify the need: proportion air and petrol before entry to the cylinder.Recognize venturi-driven fuel draw as the classical method.Select carburettor as the device fulfilling this function. Verification / Alternative check: Historic automotive designs and small engines (mowers, generators) widely used carburettors; the description aligns directly with their operation. Why Other Options Are Wrong: Fuel pump: Moves fuel from tank to carburettor; does not mix air and fuel. Injector: Used in EFI systems or diesel CI, not the classic device described. None of these: Carburettor clearly fits. Common Pitfalls: Assuming all modern engines have carburettors; most modern SI engines use injection, but the question explicitly references suction-stroke mixing typical of carburettors. Final Answer: carburettor

Question 20

Power comparison (idealized): For engines of equal size and speed, how does the theoretical power of a four-stroke cycle engine compare with that of a two-stroke cycle engine?

Accepted Answer

Introduction / Context: Cycle frequency directly influences indicated power in reciprocating engines. Two-stroke engines produce a power stroke every revolution, while four-stroke engines produce one every other revolution. The idealized comparison highlights why two-stroke engines can deliver more power density, all else being equal. Given Data / Assumptions: Same displacement, same mean effective pressure, same rotational speed. No pumping/scavenging losses or timing penalties considered (idealized). Equal mechanical efficiency assumed for comparison. Concept / Approach: Indicated power ∝ number of power strokes per unit time. At the same rpm, a two-stroke fires once per revolution; a four-stroke fires once per two revolutions. Therefore, in the simplest theoretical sense, the four-stroke develops half the power of the two-stroke under identical mean effective pressure and displacement assumptions. Step-by-Step Solution: Two-stroke: 1 power stroke per 360° → N power events per N revolutions.Four-stroke: 1 power stroke per 720° → N/2 power events per N revolutions.Thus idealized power_four-stroke = 0.5 * power_two-stroke. Verification / Alternative check: Real-world factors (scavenging losses in two-strokes, valve timing in four-strokes) modify outcomes, but the baseline theoretical result is standard in textbooks and explains design trade-offs for power density versus emissions/efficiency. Why Other Options Are Wrong: Same/Double/Four times: Contradict the direct proportionality to firing frequency given equal BMEP and displacement. Common Pitfalls: Ignoring that many high-output two-strokes suffer from scavenging and emissions issues, so real power advantage may be less than the ideal prediction; however, the theoretical comparison is correct. Final Answer: half

IC Engines and Nuclear Power Plants Questions