Question 1

Use of ordinary (light) water as a reactor moderator with enriched uranium fuel: evaluate the statement.  Statement: “Ordinary water is sometimes used as a moderator when enriched uranium is used as fuel in nuclear power plants.” Choose whether the …

Accepted Answer

Concept/Approach Light-water reactors (PWR/BWR) use ordinary water as moderator and coolant but require enriched uranium fuel (typically 3–5% U-235) to maintain criticality. Natural uranium needs better moderators like heavy water or graphite. Final Answer Agree.

Question 2

Purpose of supercharging an engine: select the most complete objective.  Context: Supercharging increases intake air density for higher specific output. Choose the correct statement describing why engines are supercharged.

Accepted Answer

Concept/Approach By boosting intake density, supercharging raises the power produced by a given displacement. This enables downsizing (lower mass and space for a given power) and provides higher power on demand. Final Answer All of the above.

Question 3

Key requirement of a good I.C. engine combustion chamber: choose the best criterion.  Select the most appropriate characteristic of a well-designed combustion chamber.

Accepted Answer

Concept/Approach An effective combustion chamber promotes rapid, stable combustion (adequate turbulence and mixture motion), good volumetric efficiency, and supports higher compression ratios—leading to high efficiency and power density. Why others are wrong “Minimum turbulence” harms combustion speed; “low compression ratio” and “low volumetric efficiency” reduce performance. Final Answer High thermal efficiency and power output.

Question 4

Diesel ignition delay terminology: identify the correct name for the interval between injection and ignition.  Definition prompt: In a CI (diesel) engine, the time between start of injection and actual start of combustion is called what?

Accepted Answer

Concept/Approach After fuel injection begins, atomization, vaporization, and mixing with hot compressed air take finite time before auto-ignition occurs. This interval is the ignition delay period. Final Answer Delay period.

Question 5

Typical diesel injection pressure: choose the closest standard value.  Select the approximate fuel injection pressure used in conventional diesel engines.

Accepted Answer

Concept/Approach Traditional mechanically-controlled DI diesels commonly use injection pressures in the ~150–250 bar range. Modern common-rail systems can exceed 1000 bar, but for textbook basics, ≈150 bar is the representative value among the given choices. Elimination 10 bar and 100 bar are too low; 500 bar is high for older conventional systems. Final Answer 150 bar.

Question 6

Secondary (bred) nuclear fuels used in power reactors: select the correct pair.  Definition: “Secondary fuel” refers to fissile material bred from fertile isotopes inside reactors. Choose the correct pair commonly identified as secondary fuels.

Accepted Answer

Concept/Approach Fertile isotopes Th-232 and U-238 breed fissile U-233 and Pu-239 respectively under neutron capture and subsequent decay. These bred fissile nuclides are termed secondary fuels. Final Answer U-233 and Pu-239.

Question 7

Effect of compression ratio on knock in petrol (SI) engines: evaluate the statement.  Statement: “Detonating (knocking) tendency in petrol engines increases as compression ratio increases.” Choose whether the statement is correct.

Accepted Answer

Concept/Approach Higher compression ratio raises end-gas temperature and pressure, reducing the ignition delay of the end-gas and making auto-ignition (knock) more likely for a given fuel octane. Therefore, knock tendency increases with CR unless higher-octane fuel or mitigating design measures are used. Final Answer Agree.

Question 8

Nuclear marine/space propulsion: fuel form for compact propulsion reactors  Reactors designed for propulsion applications (e.g., naval or space power units) require which fuel form to achieve a compact, long-life core at practical power density? Cho…

Accepted Answer

Given data Application: compact propulsion reactors (e.g., submarines, icebreakers). Concept/Approach To achieve high power density in a small core with long refueling intervals, the fuel must have a high fraction of fissile nuclei. Enriched uranium (typically enriched in U-235) provides sufficient reactivity in a compact geometry, unlike natural uranium which generally requires heavy moderation and large cores. Why others are not best “Any form” is too broad and incorrect; natural uranium would require impractically large cores for most propulsion uses; plutonium fuels pose different handling and design issues and are not the standard baseline choice. Final Answer Enriched uranium

Question 9

Diesel engines on weak (lean) mixtures: effect on thermal efficiency  For compression-ignition (diesel) engines operating with weak or lean mixtures (excess air), how does the thermal efficiency change? Choose the correct option.

Accepted Answer

Given data Engine type: compression-ignition (diesel), typically unthrottled and operating lean with excess air. Concept/Approach Diesel engines admit air first and inject fuel later. Running lean reduces pumping losses and often reduces heat transfer per unit fuel, bringing operation closer to the idealized cycle. Hence, thermal efficiency generally increases on leaner operation compared with richer diesel combustion. Notes Real engines have multiple influences (injection timing, combustion phasing, EGR), but foundational MCQ theory states that leaner (weak) mixtures in diesels indicate higher efficiency. Final Answer Higher

Question 10

Diesel ignition principle: fuel spray ignition due to hot compressed air  Statement: In a diesel (compression-ignition) engine, fuel is injected as a fine spray into the cylinder and it ignites because the compressed air temperature is high. Decide …

Accepted Answer

Given data Engine type: compression-ignition diesel. Concept/Approach In CI engines, air is compressed to a high pressure and temperature. Near the end of compression, finely atomized fuel is injected; the high air temperature causes auto-ignition without a spark. Final Answer True

Question 11

Spark-ignition engines – stoichiometric mixture strength What is the approximate stoichiometric air–fuel ratio by mass for petrol (gasoline) used in spark-ignition engines?

Accepted Answer

Introduction / Context: The stoichiometric (chemically correct) mixture provides exactly enough oxygen to fully combust the fuel without excess oxygen or fuel. For petrol (gasoline), this ratio is a cornerstone of engine calibration, emissions control, and lambda-sensor feedback strategies. Given Data / Assumptions: Air is treated as 21% O2 and 79% N2 by volume (approximately). Gasoline is approximated by a representative hydrocarbon such as iso-octane, C8H18, for stoichiometric calculation. Complete combustion to CO2 and H2O is assumed. Concept / Approach: Balancing the reaction for a surrogate like C8H18 leads to a required oxygen amount that, when converted to air, yields a mass ratio near 14.7:1. Variations in actual gasoline composition and measurement conventions produce values around 14.5–15.0. Hence, 14.6:1 is a standard rounded figure used in many texts and engine maps. Step-by-Step Solution: Write ideal reaction: C8H18 + 12.5 O2 → 8 CO2 + 9 H2O.Convert O2 to air using air/O2 ≈ 100/21 by volume and corresponding mass ratio.Compute mass of air required per unit mass of fuel → ≈ 14.7:1; practical rounded value ≈ 14.6:1. Verification / Alternative check: Engine controllers target λ = 1 (stoichiometric) for three-way catalyst operation, corresponding to ≈ 14.7:1 AFR; textbooks and datasheets commonly quote 14.6–14.7:1. Why Other Options Are Wrong: 18.5:1, 20.4:1, and 22.6:1 are lean mixtures (excess air), not stoichiometric. 10.0:1 is a rich mixture used for high load or cold start, not chemically correct. Common Pitfalls: Confusing volume ratios with mass ratios; for control and energy calculations, mass-based AFR is standard. Final Answer: 14.6:1

Question 12

Internal Combustion Engines – Suction Stroke in a Spark-Ignition (Petrol) Engine During the suction (intake) stroke of a conventional carburetted petrol engine, what working charge is drawn into the cylinder through the intake valve?

Accepted Answer

Introduction / Context: This question targets the basics of the spark-ignition (SI) petrol engine induction process. Understanding what enters the cylinder during the suction stroke is essential for concepts like air–fuel ratio, throttling, volumetric efficiency, and combustion quality in automotive and small power-plant applications. Given Data / Assumptions: Conventional carburetted petrol (gasoline) engine or port fuel–injected SI engine. Throttle controls the intake manifold pressure and the air–fuel mixture quantity. Standard four-stroke cycle: suction (intake), compression, power (expansion), exhaust. Concept / Approach: In an SI engine, fuel is premixed with air before ignition by the spark plug. Traditionally, a carburettor atomizes fuel into the incoming airstream, producing a combustible air–fuel mixture. In modern systems, port fuel injection meters fuel into the intake port where it mixes with air before entering the cylinder. Thus, during the suction stroke with the intake valve open, the piston’s downward motion draws an air–fuel mixture into the cylinder. Step-by-Step Solution: Identify engine type: spark-ignition petrol engine.Recognize induction method: carburetion or port fuel injection creates a premixed charge.During suction, pressure in the cylinder falls below manifold pressure.Mixture flows through the open intake valve into the cylinder. Verification / Alternative check: Contrast with compression-ignition (diesel) engines, which draw air only on the intake and inject fuel directly near the end of compression. Also contrast with direct-injection SI engines (GDI), where some fuel may be introduced directly into the cylinder; yet even then, the inducted working fluid is still primarily air, and for most conventional petrol engines the inducted charge is air premixed with fuel. Why Other Options Are Wrong: 'Air only' describes diesel operation. 'Petrol only' and 'pure fuel vapour' are impossible because combustion requires oxygen. 'EGR only' is not a working charge; EGR, if used, is mixed with fresh charge and never alone. Common Pitfalls: Confusing SI and CI engine cycles; overlooking that carburettors/prefueling systems create a combustible mixture before intake valve closure. Final Answer: a mixture of petrol and air

Question 13

Nuclear Reactor Fundamentals – What Does a Moderator Do? In a thermal nuclear reactor, a 'moderator' primarily performs which role with respect to fast neutrons produced by fission?

Accepted Answer

Introduction / Context: Safe and sustained fission in most power reactors requires an abundance of thermal (slow) neutrons. The component responsible for turning fast neutrons into thermal neutrons, without excessively removing them from the system, is the moderator. Recognizing this function is foundational for reactor physics and engineering design choices (e.g., water, heavy water, graphite). Given Data / Assumptions: Thermal reactor using a moderator such as light water, heavy water, or graphite. Fast neutrons are born at a few MeV of energy from fission events. Goal is to increase probability of further fissions in fissile isotopes (e.g., U-235) which have higher cross-sections at thermal energies. Concept / Approach: A moderator reduces neutron energy primarily by elastic scattering, transferring kinetic energy from neutrons to moderator nuclei. Good moderators have low neutron absorption cross-sections and atomic masses conducive to effective energy loss per collision (e.g., hydrogen in H2O, deuterium in D2O, carbon in graphite). By slowing neutrons to thermal energies while avoiding significant capture, the moderator raises the likelihood that they cause further fissions, sustaining the chain reaction. Step-by-Step Solution: Identify desired outcome: convert fast neutrons to thermal neutrons.Match mechanism: multiple elastic collisions in the moderator.Qualify materials: low absorption, good slowing-down power.Select the option describing 'slows down with minimal absorption'. Verification / Alternative check: Reactor design texts quantify 'moderating ratio' = slowing-down power / absorption cross-section; high values (e.g., D2O, graphite) indicate good moderators, confirming the role stated. Why Other Options Are Wrong: Absorbing neutrons suppresses reactivity; saying 'does not absorb and has no interaction' ignores scattering; acceleration of neutrons does not occur in moderators; conversion to protons via beta decay is not a moderation process. Common Pitfalls: Confusing moderators with control rods (which deliberately absorb neutrons) or reflectors (which return neutrons to the core). Final Answer: slows down (thermalizes) fast neutrons with minimal absorption

Question 14

Turbojet Architecture – Does the Compressor Need a Turbine? Evaluate the statement: 'A turbo-jet does not require a turbine to drive the compressor.' State whether it is correct for practical turbojet engines.

Accepted Answer

Introduction / Context: Understanding the flow of energy within air-breathing propulsion systems distinguishes ramjets, turbojets, turbofans, and turboshafts. The key question is whether the compressor has a mechanical power source and what component provides it in a turbojet. Given Data / Assumptions: A conventional turbojet comprises an inlet, compressor, combustor, turbine, and exhaust nozzle. Steady operation at design conditions is considered. Ramjets (no compressor) are a different class of engine. Concept / Approach: In a turbojet, the turbine extracts part of the hot gas enthalpy downstream of the combustor. The extracted shaft power drives the compressor via a concentric shaft; the remaining exhaust kinetic energy produces thrust through the nozzle. Without the turbine, the compressor would have no mechanical drive and could not maintain the required pressure ratio, so sustained turbojet operation would be impossible. By contrast, ramjets rely on vehicle forward speed and inlet compression, not on a rotating compressor; therefore, they have no turbine either. But the statement as written about turbojets is incorrect. Step-by-Step Solution: Identify component roles: turbine → shaft power; compressor → pressure rise.Energy balance: turbine work equals compressor work plus mechanical losses.Conclude: turbojets necessarily include a turbine to drive the compressor.Therefore, the statement is false. Verification / Alternative check: Any standard turbojet schematic shows a mechanically linked compressor–turbine spool. Engine performance maps further imply turbine work closely matches compressor work at steady operation. Why Other Options Are Wrong: Afterburners do not drive compressors. Supersonic flight does not eliminate compressor power needs. The note about ramjets is true for ramjets, but irrelevant to turbojets. Common Pitfalls: Confusing turbojets with ramjets/scramjets; assuming inlet ram compression suffices to replace a compressor. Final Answer: False

Question 15

Comparing Exhaust Temperatures – Petrol (SI) vs. Diesel (CI) For the same maximum pressure and the same heat input, how does the exhaust gas temperature of a petrol (spark-ignition) engine compare with that of a diesel (compression-ignition) engine?

Accepted Answer

Introduction / Context: Exhaust temperature influences turbine matching, catalyst durability, and emission control strategies. Knowing which prime mover tends to have hotter exhaust for comparable limits helps in thermal management and material selection. Given Data / Assumptions: Same maximum cylinder pressure and identical heat input conditions are assumed. Typical operating strategies: petrol engines usually run near stoichiometric; diesels operate lean with higher expansion ratios. Friction and heat losses are of comparable order for the comparison. Concept / Approach: Diesel engines generally have higher compression and effective expansion ratios, extracting more work from the hot gases before exhaust valve opening. They also operate with excess air (lean mixtures), which lowers combustion temperatures and increases specific heat capacity of the working charge. Conversely, petrol engines often run near stoichiometric and with lower expansion ratios, leaving more sensible energy in the exhaust. Therefore, for the same maximum pressure and heat input, petrol engines tend to have higher exhaust temperatures than diesels. Step-by-Step Solution: Assume equal peak pressure constraint.Note CI engines’ lean operation and larger expansion reduce exhaust enthalpy.SI engines’ near-stoichiometric burn and smaller expansion retain more exhaust sensible heat.Hence, petrol exhaust temperature is higher → choose 'more'. Verification / Alternative check: Engine test data routinely show lower exhaust gas temperatures for naturally aspirated diesels than for comparable SI engines at similar load fractions, supporting the conceptual reasoning. Why Other Options Are Wrong: 'Less' contradicts typical thermodynamic outcomes; 'equal' is unlikely across wide operating ranges; altitude affects both similarly but does not reverse the trend under stated constraints. Common Pitfalls: Assuming turbocharging alone dictates EGT; while boosting changes details, the SI vs. CI trend under the given premise remains. Final Answer: more

Question 16

Effect of Supercharging – Comparison with a Naturally Aspirated Engine (for the Same Power Output) Which statement best describes a supercharged engine compared to an ordinary naturally aspirated engine delivering the same brake power?

Accepted Answer

Introduction / Context: Supercharging (or turbocharging) raises the intake density, permitting higher trapped mass per cycle. For a given required brake power, this allows a smaller displacement engine (and often smaller physical package) to achieve the target output, which affects weight, mounting, and installation design in vehicles and stationary plants. Given Data / Assumptions: Comparison at identical brake power and similar technology levels. Properly matched supercharger/turbo with intercooling as needed. No abnormal derates or emissions constraints altering the baseline comparison. Concept / Approach: Because the boosted engine ingests denser charge, it can be designed with smaller swept volume and fewer cylinders for the same power. The result is lower engine block mass and smaller mounting footprint for equal output. While the supercharger/turbo adds components, the reduced core engine size typically more than offsets that mass, yielding a net lighter package per unit power. Lubricating oil consumption is not inherently reduced by boosting and may increase at high specific loads; thus the blanket claim of 'consumes less lubricating oil' is not generally valid. Step-by-Step Solution: Fix required brake power (target).Design alternatives: naturally aspirated large-displacement vs. smaller boosted engine.Account for added compressor/turbine mass but reduced block size.Net outcome: lighter engine and smaller foundation for the same output. Verification / Alternative check: Specific power (kW per litre) increases strongly with boosting. Industry sizing rules for gensets and marine engines reflect smaller skid sizes for boosted units at given ratings. Why Other Options Are Wrong: 'Consumes significantly less oil' is not generally true; 'must be heavier' ignores the displacement reduction; 'cannot deliver same power at lower displacement' contradicts the purpose of supercharging; 'always higher vibration' overgeneralizes—balance, mounts, and cylinder pressures govern vibration. Common Pitfalls: Assuming hardware mass alone decides weight; overlooking the large reduction in engine size when boosted. Final Answer: It is lighter and requires a smaller foundation for the same power output.

Question 17

Engine Efficiencies – Define Mechanical Efficiency What do we call the ratio of brake power (delivered at the shaft) to indicated power (developed in the cylinder) for a reciprocating engine?

Accepted Answer

Introduction / Context: An engine’s performance is commonly broken down into indicated power (in-cylinder) and brake power (usable at the shaft). The ratio between them quantifies mechanical losses in bearings, valve gear, piston friction, and auxiliary drives. This ratio is fundamental when diagnosing shortfalls between combustion performance and delivered output. Given Data / Assumptions: B.P. = brake power measured at the shaft (e.g., dynamometer). I.P. = indicated power computed from indicator diagrams or cylinder pressure data. Frictional power F.P. = I.P. − B.P. Concept / Approach: Mechanical efficiency measures how effectively mechanical components transmit the power produced by combustion to the crankshaft. It is defined as eta_m = B.P. / I.P. Values less than 1 highlight losses due to mechanical friction and pumping. Improving lubrication, reducing accessory loads, or optimizing mechanical design can raise eta_m. Step-by-Step Solution: Start with power balance: I.P. = B.P. + F.P.Rearrange for mechanical efficiency: eta_m = B.P. / I.P.Interpretation: the closer eta_m is to 1, the lower the mechanical losses.Cross-check with frictional power trends vs. speed and load. Verification / Alternative check: Experimental motoring tests (no fuel, external drive) determine F.P., confirming the relationship and enabling calculation of eta_m from measured B.P. and inferred I.P. Why Other Options Are Wrong: Overall efficiency includes fuel energy and is eta_o = B.P./(mf * LCV). Indicated thermal efficiency relates I.P. to fuel energy. Volumetric efficiency concerns air-handling capability. 'Relative efficiency' usually compares thermal efficiency to an ideal cycle. Common Pitfalls: Confusing mechanical with overall efficiency; forgetting that F.P. rises sharply with speed because friction is largely speed-dependent. Final Answer: mechanical efficiency

Question 18

Engine Power Balance – Frictional Power Expression Select the correct relationship for frictional power (F.P.) in terms of indicated power (I.P.) and brake power (B.P.) for a reciprocating engine.

Accepted Answer

Introduction / Context: Separating indicated, brake, and frictional power helps quantify where losses occur in an engine. This is central to design optimization and to interpreting dynamometer and indicator measurements. Given Data / Assumptions: I.P. is gross in-cylinder power calculated from pressure–volume data. B.P. is net shaft output measured externally. F.P. represents mechanical and pumping losses between cylinder and shaft. Concept / Approach: The power balance is I.P. = B.P. + F.P. This states that the power developed in the cylinder must either appear at the shaft or be dissipated in friction and pumping. Rearranging gives F.P. = I.P. − B.P., a straightforward expression used in motoring tests and Willans line analysis. Step-by-Step Solution: Write balance: I.P. = B.P. + F.P.Rearrange: F.P. = I.P. − B.P.Check units consistency: all in kW (or hp).Interpret physically: if B.P. increases at constant I.P., F.P. decreases only if losses fall (rare); typically losses rise with speed/load. Verification / Alternative check: Motoring test yields F.P. directly (B.P. = 0 while turning the engine), then I.P. under firing conditions is inferred to verify the balance. Why Other Options Are Wrong: 'B.P. − I.P.' would make F.P. negative. Ratios do not represent power differences; 'sum' violates conservation since it would exceed I.P. Common Pitfalls: Mistaking frictional mean effective pressure (FMEP) trends and assuming F.P. is constant across speeds—it typically rises with RPM. Final Answer: F.P. = I.P. - B.P.

Question 19

Rocket Propulsion – Propulsive Efficiency Formula If V1 denotes the jet (exhaust) velocity relative to the vehicle and V0 is the vehicle flight velocity, which expression gives the propulsive efficiency of an ideal rocket?

Accepted Answer

Introduction / Context: Propulsive efficiency measures how effectively a propulsion device converts the power in its jet into useful thrust power on the vehicle. For rockets (self-contained propellant), the classical one-dimensional relation links the vehicle speed V0 and jet speed V1. Knowing this formula helps reason about why very high jet speeds are inefficient at low flight speeds. Given Data / Assumptions: Steady, 1-D ideal rocket with fully expanded exhaust. Neglect pressure thrust and nozzle inefficiencies for simplicity. Jet speed V1 is relative to the rocket; flight speed is V0. Concept / Approach: Propulsive efficiency is defined as useful power divided by rate of energy output in the jet. Useful power = T * V0, where thrust T ≈ m_dot * (V1 − V0) in the ideal model (neglecting pressure terms). Jet power available = 0.5 * m_dot * (V1^2 − V0^2) when referenced to the vehicle frame. Taking the ratio and simplifying yields eta_p = 2 * V0 / (V1 + V0), a compact expression commonly cited for rockets. Step-by-Step Solution: Write thrust: T = m_dot * (V1 − V0).Useful power: P_useful = T * V0 = m_dot * (V1 − V0) * V0.Jet power change: P_jet = (1/2) * m_dot * (V1^2 − V0^2) = (1/2) * m_dot * (V1 − V0) * (V1 + V0).Form efficiency: eta_p = P_useful / P_jet = [m_dot*(V1 − V0)V0] / [(1/2)m_dot(V1 − V0)(V1 + V0)] = 2*V0 / (V1 + V0). Verification / Alternative check: Limits check: as V0 → V1, eta_p → 1 (ideal matching). As V0 → 0 (static firing), eta_p → 0, meaning no useful propulsive power though thrust exists—consistent with intuition. Why Other Options Are Wrong: Other forms either invert the ratio or omit the factor of 2, giving incorrect limiting behavior or values outside [0, 1]. Common Pitfalls: Confusing propulsive efficiency with overall efficiency (which also includes conversion of chemical energy to jet kinetic energy), or using air-breathing formulas directly for rockets. Final Answer: eta_p = 2 * V0 / (V1 + V0)

Question 20

Air–Fuel Ratio for Petrol – Stoichiometric (Chemically Correct) Mixture Select the approximate stoichiometric air–fuel ratio by mass for a conventional petrol (gasoline) engine mixture used for complete combustion.

Accepted Answer

Introduction / Context: The air–fuel ratio (AFR) at which exactly enough oxygen is supplied to burn all the fuel without excess air or fuel is called stoichiometric. For petrol engines, this AFR is a key target for three-way catalysts and closed-loop lambda control to minimize emissions while maintaining good drivability. Given Data / Assumptions: Petrol approximated as a representative hydrocarbon (e.g., iso-octane C8H18). Complete combustion to CO2 and H2O assumed. Mass-based AFR values are considered. Concept / Approach: For iso-octane as a surrogate, the balanced oxidation is C8H18 + 12.5 O2 → 8 CO2 + 9 H2O. Air contains about 21% O2 by volume (≈23% by mass), so the mass of air needed is roughly 14.7 times the mass of fuel. Rounded to a practical figure used in calibration and textbooks, the stoichiometric AFR for petrol is about 15:1 by mass. Step-by-Step Solution: Start with idealized fuel: C8H18.Compute O2 moles: 12.5 per mole of fuel for complete oxidation.Convert O2 to air mass using O2 mass fraction in air.Obtain AFR ≈ 14.7:1 → commonly rounded to 15:1. Verification / Alternative check: Engine calibrations use lambda = 1 at AFR ≈ 14.7:1. Emissions control and stoichiometric operation reinforce this canonical value. Why Other Options Are Wrong: 10:1 is rich (used for power enrichment). 20:1 and 25:1 are lean burn values associated with fuel economy but higher NOx without special strategies. 12.5:1 is a common rich power AFR for knock avoidance and cooling but not stoichiometric. Common Pitfalls: Mixing up volume-based and mass-based ratios; assuming one exact value for all petrol blends—the 14.7–15:1 figure is an accepted practical average. Final Answer: 15:1

IC Engines and Nuclear Power Plants Questions