Question 1

Polytropic efficiency in multistage compressors: is it constant across stages? Assess the statement: “For a well-designed multistage compressor operating at similar conditions in each stage, the polytropic efficiency can be taken as the same for all…

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Introduction / Context:Polytropic efficiency is a differential (small-step) measure that characterizes how closely a real compression process follows an isentropic path. In multistage machines designed with similar geometry, Reynolds/Mach numbers, and cooling arrangements, it is common engineering practice to assume the same polytropic efficiency in each stage for preliminary design and performance prediction.Given Data / Assumptions:Stages have comparable tip speeds, clearances, valve behavior (reciprocating) or blade aerodynamics (dynamic).Intercooling restores similar inlet temperatures for each stage (nearly perfect intercooling).Pressure ratios per stage are chosen to be approximately equal.Concept / Approach:Because polytropic efficiency is relatively insensitive to the absolute pressure level when flow regime and geometry are similar, designers often assign a single value across stages. Detailed tests may reveal small differences, but the equal-efficiency assumption is accurate enough for sizing and for estimating overall work and discharge temperatures.Step-by-Step Solution:Define polytropic efficiency as the ratio of ideal small-step work to actual small-step work.Note similarity: identical machine elements and operating regimes → similar losses per stage.Conclude: use a common polytropic efficiency for all stages in preliminary design.Verification / Alternative check:Factory test results frequently report stage efficiencies within a narrow band; averaging to a single representative polytropic efficiency yields accurate overall predictions.Why Other Options Are Wrong:Limiting the statement to single-acting machines or zero clearance is unnecessary; the assumption pertains to similarity, not acting mode.Declaring it false ignores widespread design practice and the basis of stage-stacking methods.Common Pitfalls:Confusing polytropic efficiency (a differential metric) with isentropic efficiency for the whole machine; the latter can vary more with pressure ratio distribution.Final Answer:True

Question 2

Axial-flow compressor design practice: typical degree of reaction In axial flow compressors used in gas turbines and process plants, designers generally aim for a balanced degree of reaction so that the static enthalpy rise is shared between rotor a…

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Introduction / Context:The degree of reaction is fundamental in axial-flow compressor design. It indicates how the static enthalpy (or static pressure) rise is split between the rotating blades (rotor) and the stationary blades (stator). Choosing an appropriate value affects efficiency, blade loading, and stall margin.Given Data / Assumptions:Axial-flow compressor with repeating stages.Conventional subsonic design intent for good efficiency and operability.Standard definition: degree of reaction R = static enthalpy rise in rotor / stage static enthalpy rise.Concept / Approach:A stage with R ≈ 0.5 shares the work nearly equally between rotor and stator. This symmetry yields well-proportioned velocity triangles, moderate diffusion factors, and a good compromise between efficiency and stall resistance. Many classical stage designs target R in the range 0.4–0.6, with 0.5 being the canonical textbook value.Step-by-Step Solution:Define R = (static rise in rotor) / (total stage static rise).For balanced velocity triangles and manageable blade loading, select R near 0.5.Such staging supports consistent flow turning and diffusion in both rotor and stator rows.Hence, the usual target is approximately 0.5.Verification / Alternative check:Stage loading and diffusion-factor limits commonly plotted against R show favorable regions around 0.5. Empirical compressor maps also reflect robust operation near this value.Why Other Options Are Wrong:0.2, 0.3: too rotor-light; places excessive rise in stator, often leading to high stator diffusion.0.4: feasible but slightly below the standard balanced point.0.65: workable for some high-loaded stages but not the usual general target.Common Pitfalls:Confusing degree of reaction with isentropic efficiency or with stage loading coefficient; they are related but distinct non-dimensional measures.Final Answer:0.5

Question 3

Compressor capacity definition (standard delivery basis) In industrial air systems, compressor capacity is typically specified as the actual volume flow delivered by the machine but referenced back to normal temperature and pressure (standard condit…

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Introduction / Context:Air is compressible, so the same mass flow can occupy very different volumes depending on temperature and pressure. To compare compressors fairly, the capacity is stated at a standard reference, often called Free Air Delivery (FAD) at normal temperature and pressure.Given Data / Assumptions:Reference conditions are fixed (e.g., 1.013 bar and 15°C or as per standard).Capacity is used for sizing tools, piping, and receivers.Delivery basis, not suction stroke displacement alone.Concept / Approach:Capacity equals the volumetric flow that the compressor delivers, converted back to a standard reference state using the ideal gas relation. This allows apples-to-apples comparison across sites and climates and decouples performance from ambient variations.Step-by-Step Solution:Measure actual mass flow delivered.Convert mass flow to volume at NTP: V_NTP = m * R * T_NTP / p_NTP.Report capacity as this standardized volume per unit time.Verification / Alternative check:Vendor datasheets use SCFM or m^3/h at standard conditions. Site readings are corrected back using temperature and pressure data.Why Other Options Are Wrong:Raw delivered or sucked volume (options b and c) depend on instantaneous conditions and are not comparable.“None of the above” and mass-only definitions omit the standardized volumetric basis used in practice.Common Pitfalls:Confusing displacement (geometric) with capacity (delivered); ignoring correction to reference conditions leads to sizing errors.Final Answer:actual volume of the air delivered by the compressor when reduced to normal temperature and pressure conditions

Question 4

Flow direction in a centrifugal (radial) compressor Considering the machine geometry and blade motion, the primary flow of air through a centrifugal compressor is oriented in which direction relative to the rotor axis?

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Introduction / Context:Centrifugal compressors, also known as radial-flow compressors, transfer energy to the fluid by whirling it radially outward in a rotating impeller and then diffusing it in a stationary diffuser.Given Data / Assumptions:Classic shrouded impeller and vaned or vaneless diffuser.Neglect minor inlet pre-swirl and exit blade lean effects.Axis refers to the shaft centerline.Concept / Approach:Fluid enters near the eye of the impeller and is flung outward by centrifugal action. Thus the main flow path is predominantly radial. A radial path is perpendicular to the machine axis, unlike axial compressors where flow is largely parallel to the axis.Step-by-Step Solution:Identify machine class: centrifugal (radial-flow).Track flow path: from impeller eye to impeller tip is radial.Relative to axis: radial = perpendicular to axis.Therefore, the correct orientation is perpendicular.Verification / Alternative check:Cross-sectional diagrams show streamlines moving outward from the centerline to the periphery before entering the diffuser.Why Other Options Are Wrong:Parallel/inclined: describe axial or mixed-flow machines.“None of these” and “reverse to axis” do not match standard geometry.Common Pitfalls:Confusing compressor classes; “centrifugal” implies radial motion by definition.Final Answer:perpendicular

Question 5

Pressure coefficient definition check for turbomachinery A statement claims: “The ratio of the isentropic work to the Euler work in a compressor is called the pressure coefficient.” Decide whether this terminology is correct.

Accepted Answer

Introduction / Context:Non-dimensional parameters help compare turbomachines. However, consistent terminology is important. The “pressure coefficient” commonly relates pressure rise to dynamic head or to blade speed squared, not to a ratio of two different works as defined here.Given Data / Assumptions:Isentropic work = ideal thermodynamic head rise for a given inlet-to-exit state change.Euler work = ideal mechanical energy transfer per unit mass, U * delta V_theta.No losses or only conceptual comparison of definitions.Concept / Approach:Standard dimensionless groups include head coefficient (psi = delta h / U^2), flow coefficient (phi = V_ax / U), and pressure coefficient (often delta p / (0.5 * rho * V^2)). Isentropic efficiency is defined as isentropic work divided by actual shaft work, not by Euler work. Therefore, naming the ratio (isentropic work)/(Euler work) as “pressure coefficient” is incorrect usage.Step-by-Step Solution:Identify accepted definitions: head coefficient and pressure coefficient relate pressure or head to velocity scales.Recognize efficiency definition: eta_isentropic = isentropic work / actual work input.Euler work is a kinematic ideal; its ratio with isentropic work is not termed “pressure coefficient”.Hence, the statement is not correct.Verification / Alternative check:Textbook glossaries consistently separate “coefficients” (psi, phi) from “efficiencies.”Why Other Options Are Wrong:Conditional answers (low Mach, solidity dependence) do not fix the terminology error.“Only for pumps” is still incorrect; pumps use similar non-dimensional groups.Common Pitfalls:Mixing up head/pressure coefficients with various efficiency definitions; Euler head is not the denominator in isentropic efficiency.Final Answer:No

Question 6

Double-acting compressor: location of compression For a double-acting reciprocating air compressor, where does compression take place with respect to the piston and cylinders?

Accepted Answer

Introduction / Context:Reciprocating compressors may be single-acting or double-acting. The distinction is whether one or both sides of the piston contribute to compression and delivery, which changes capacity and mechanical design details.Given Data / Assumptions:Classic crosshead-type construction allowing rod-side sealing.Valving on both head and crank ends.Single cylinder considered for the definition.Concept / Approach:In a double-acting machine, both the head end and the crank end of the same cylinder undertake compression alternately as the piston shuttles. This effectively doubles the number of compression events per revolution compared to a single-acting cylinder of the same size and speed.Step-by-Step Solution:Identify acting surfaces: two sides of the piston.Each stroke: one side compresses while the other side intakes or delivers.Therefore, compression occurs in a single cylinder on both sides of the piston.Verification / Alternative check:Indicator diagrams show two loops per revolution (one for each side) for a double-acting cylinder.Why Other Options Are Wrong:“In one cylinder” describes single-acting behavior.“Two cylinders” or “both sides in two cylinders” confuses multi-cylinder with double-acting architecture.“Multistage only” is unrelated to single vs. double acting.Common Pitfalls:Assuming double-acting implies two cylinders; the term refers to two working faces of one piston.Final Answer:in a single cylinder on both sides of the piston

Question 7

Compressor classification by flow direction If the airflow through a compressor is predominantly perpendicular to the machine axis, what type of compressor is it?

Accepted Answer

Introduction / Context:Compressors are classified by operating principle and by flow direction. The latter gives quick insight into geometry, blade shapes, and specific performance characteristics.Given Data / Assumptions:Flow direction specified as perpendicular to the axis.Continuous-flow turbomachine implied.No mixed-flow exceptions considered.Concept / Approach:Axial compressors carry flow roughly parallel to the shaft axis. Centrifugal (radial) compressors send flow radially outward, which is perpendicular to the axis. Reciprocating and scroll compressors are positive-displacement and are not classified by a single dominant flow direction through rotating stages.Step-by-Step Solution:Perpendicular to axis → radial motion dominates.Radial motion is characteristic of centrifugal compressors.Therefore, the correct identification is centrifugal compressor.Verification / Alternative check:Machine schematics show impeller eye to tip flow path orthogonal to the shaft centerline.Why Other Options Are Wrong:Reciprocating: positive-displacement piston machine.Axial flow: flow parallel to axis.“Turbo compressor” is a generic term that includes both axial and centrifugal types.Scroll: orbiting spiral PD device.Common Pitfalls:Mixing naming by principle (turbo vs. PD) with naming by flow direction (axial vs. radial).Final Answer:centrifugal compressor

Question 8

Minimum work in three-stage reciprocating compression Under optimum staging and perfect intercooling, how does the total minimum work required by a three-stage reciprocating air compressor compare to the work required by each individual stage?

Accepted Answer

Introduction / Context:Multi-stage compression with intercooling reduces work input. When pressure ratios are optimally split, each stage does the same amount of work, which leads to a simple relation between per-stage work and the total work.Given Data / Assumptions:Three stages with perfect intercooling between stages.Equal pressure ratio per stage for minimum work.Similar efficiencies and speeds across stages.Concept / Approach:For minimum work, the interstage pressures are chosen such that the per-stage work is equal. The total work is then the sum of the three equal contributions, hence three times the work of any single stage.Step-by-Step Solution:Choose interstage pressures so that p2 = (p1 * p3)^(1/2) and p3 = (p2 * p4)^(1/2), etc., equalising pressure ratio per stage.With perfect intercooling, inlet temperature to each stage returns to the initial value, equalising stage work.Total work = W1 + W2 + W3 = 3 * W_stage.Verification / Alternative check:Indicator diagrams and analytical formulas for polytropic or isentropic compression confirm equal areas (works) for equal pressure ratios with perfect intercooling.Why Other Options Are Wrong:Equal to or double: underestimate the sum of three equal stage works.Six times: overestimates and has no basis.Half: contradicts additive nature of work.Common Pitfalls:Confusing “minimum total work” with “minimum per-stage work”; both are satisfied simultaneously when work is equalised per stage.Final Answer:three times

Question 9

Condition for minimum work in multistage compression State whether the following is true: For minimum total work in multi-stage compression with intercooling, the work done on each stage should be equal.

Accepted Answer

Introduction / Context:Optimizing interstage pressures in multi-stage compressors is critical to reduce shaft power. A classic result shows that minimum total work occurs when every stage shares an equal pressure ratio and, consequently, equal work input (with perfect intercooling).Given Data / Assumptions:Multiple stages with intercoolers.Intercoolers bring the air back near the initial intake temperature.Similar mechanical and volumetric efficiencies stage-to-stage.Concept / Approach:Total work is the sum of stage works. By calculus of variations or by symmetry arguments using equal temperature resets, equalising the pressure ratio per stage minimises the sum for a fixed overall ratio. With equal ratios and equal inlet temperatures, the per-stage work becomes equal.Step-by-Step Solution:Let overall ratio be r_total; divide into k stages with r_stage = r_total^(1/k).Perfect intercooling → each stage starts at the same temperature.Work per stage becomes equal; total is k * W_stage, and minimised under the equal-split condition.Verification / Alternative check:Deriving with polytropic relations shows d(W_total)/d(p2) = 0 gives p2 such that ratios are equal. This generalises to any number of stages.Why Other Options Are Wrong:“No” contradicts the optimal-split principle.“Only two stages” is unnecessarily restrictive.Dependence on n = 1 is false; the result holds for practical polytropic indices with proper intercooling.Common Pitfalls:Ignoring pressure drops and intercooler effectiveness; while these shift exact values, the equal-work principle remains the guiding design rule.Final Answer:Yes

Question 10

Compression path in a closed-cycle gas turbine (theory vs. reality) In a closed Brayton-cycle gas turbine, how is the compression process most accurately described in practice?

Accepted Answer

Introduction / Context:Thermodynamic textbooks idealize compression as isentropic for efficiency calculations. Real compressors, however, exhibit losses and follow a polytropic path, which more accurately captures the small-step efficiency across many blade rows.Given Data / Assumptions:Closed Brayton cycle using a working gas (e.g., helium, nitrogen, or air in sealed loop).Finite mechanical and aerodynamic losses.Steady-state operation.Concept / Approach:Polytropic compression assumes p * v^n = constant with an exponent n between 1 (isothermal) and gamma (isentropic for ideal gas). This provides an incremental efficiency representation and aligns with measured compressor maps.Step-by-Step Solution:Ideal (reference) path: isentropic.Real path: polytropic with n slightly greater than isothermal and less than isentropic exponent-related equivalent.Therefore, “polytropic” best describes the practical compression process.Verification / Alternative check:Manufacturers quote polytropic efficiency to characterise multi-row compression more robustly than a single isentropic efficiency figure.Why Other Options Are Wrong:Isothermal: unattainable without ideal heat removal during compression.Isentropic: a useful ideal limit, not the real path.Isochoric: volume does not remain constant during compression.Common Pitfalls:Confusing “closed cycle” with “isothermal”; the cycle closure does not imply isothermal compression.Final Answer:polytropically

Question 11

Why volumetric efficiency of reciprocating compressors is below 100% Identify the principal reasons why the volumetric efficiency of a reciprocating air compressor is less than unity (less than 100%).

Accepted Answer

Introduction / Context:Volumetric efficiency measures how much fresh charge actually enters the cylinder compared with the swept volume. In reciprocating compressors, several real effects reduce the volumetric efficiency below 100%.Given Data / Assumptions:Conventional valve plates with finite pressure loss.Non-zero clearance volume to avoid piston-head contact.Some heating of incoming air and minor leakage paths.Concept / Approach:At the end of delivery, compressed air remains in the clearance volume at discharge pressure. During expansion at the start of the suction stroke, this residual air expands, occupying part of the swept volume and delaying fresh intake. Valve pressure drops and heating further reduce the mass taken in per cycle.Step-by-Step Solution:Clearance effect: residual high-pressure air expands during suction, reducing intake volume.Valve losses: pressure drop across suction valves lowers the cylinder pressure, diminishing mass flow.Superheating: higher inlet temperature increases specific volume, reducing mass per unit volume admitted.Leakages: blowby past rings/valves reduces net intake.Verification / Alternative check:Empirical correlations for volumetric efficiency include terms for clearance ratio, pressure ratio, temperature rise, and valve losses, matching observed performance.Why Other Options Are Wrong:Any single cause alone (a, b, or c) is incomplete; multiple mechanisms act together.“None of these” contradicts standard compressor theory and practice.Common Pitfalls:Ignoring the dominant role of clearance; underestimating temperature effects when suction piping is hot or poorly cooled.Final Answer:all of these

Question 12

Capacity rating terminology for compressors: On a nameplate, a “3 m³/min compressor” most commonly indicates that the machine handles which volumetric flow basis?

Accepted Answer

Introduction / Context:Compressor capacities are quoted in different “bases” such as free air, standard air, or compressed (actual) conditions. Knowing what a “3 m³/min compressor” means prevents sizing mistakes and ensures apples-to-apples comparison between vendors.Given Data / Assumptions:Nameplate says 3 m³/min, no further qualifiers.Industry convention: unless explicitly marked as “standard” or given discharge state, volumetric flow refers to free air basis.Free air means volume referenced to the site’s atmospheric pressure and temperature.Concept / Approach:Three common bases exist. Free Air Delivery (FAD) expresses the delivered mass flow as an equivalent volume at local ambient conditions. Standard air uses a fixed reference (e.g., 1.013 bar and 15–20°C). Compressed (actual) volume is the discharge volume at delivery pressure/temperature. Unqualified ratings in general practice imply FAD, not standard or compressed volume.Step-by-Step Solution:Identify unspecified rating → adopt industry default (FAD).Relate mass flow to FAD volume at site ambient.Conclude the nameplate indicates 3 m³/min of free air.Select the option that explicitly states “free air”.Verification / Alternative check:Vendor datasheets typically label exceptions explicitly (SCFM/Nm³/h or “at discharge state”). Absence of such labels supports interpreting the figure as FAD.Why Other Options Are Wrong:Standard air requires a declared standard; compressed-state volume depends on discharge conditions; stating “at delivery pressure and temperature” would need exact values, which are not given.Common Pitfalls:Confusing FAD with SCFM/Nm³/h; comparing different bases without conversion; overlooking ambient altitude/temperature impacts on FAD.Final Answer:compresses 3 m³/min of free air (at local atmospheric conditions)

Question 13

Understanding slip factor in centrifugal compressors: Does the slip factor quantify leakage from a storage vessel, or something else within the impeller exit flow?

Accepted Answer

Introduction / Context:“Slip factor” is a core performance parameter for centrifugal compressors. It is often misunderstood as a mechanical leakage measure, whereas it actually relates to aerodynamics at the impeller exit and the head predicted by Euler’s equation.Given Data / Assumptions:Centrifugal (radial) compressor with finite blade number.Flow leaving the impeller experiences deviation from ideal blade-to-flow guidance.No mechanical leakage from a separate storage vessel is implied by the term.Concept / Approach:Slip factor σ is typically defined as Vw2(actual)/Vw2(ideal) or as Vw2(actual)/u2 depending on convention. Because real blades cannot turn the flow perfectly, the actual tangential (whirl) component Vw2 is less than the ideal value, reducing the ideal Euler head and the pressure rise achievable per stage.Step-by-Step Solution:State Euler relation: ideal head ∝ u2*Vw2(ideal).Recognize finite blade number → flow slip → Vw2(actual) Define slip factor σ capturing this shortfall.Conclude it has nothing to do with vessel leakage.Verification / Alternative check:Empirical correlations (Stodola, Wiesner) estimate σ from geometry and blade count, validating the aerodynamic nature of the parameter.Why Other Options Are Wrong:Options suggesting storage or bearing leakage misinterpret an aerodynamic efficiency parameter as a mechanical loss metric.Common Pitfalls:Equating “slip” with “leak”; assuming σ reflects casing clearances rather than exit flow deviation.Final Answer:No, it measures the shortfall in tangential (whirl) velocity at impeller exit due to finite blade effects

Question 14

Compressor pressure ratio check: For any practical compressor, the compression ratio is always __________ unity.

Accepted Answer

Introduction / Context:Pressure ratio is the cornerstone metric for compressors, defined as discharge pressure divided by inlet (usually atmospheric) pressure. Its magnitude indicates the machine’s ability to raise pressure and drive downstream processes.Given Data / Assumptions:Pressure ratio r_p = p_out / p_in.Normal compressor operation (not a throttling valve).Positive-displacement or turbocompressors considered.Concept / Approach:By definition, a compressor increases pressure; therefore p_out > p_in and r_p > 1. An r_p of 1 would imply no compression, and less than 1 is a pressure reducer (throttle/expander), not a compressor. This simple inequality holds across single-stage and multi-stage machines when referenced at the same stations.Step-by-Step Solution:Define r_p = p2 / p1.Compressor function → p2 > p1.Hence r_p > 1 (strictly greater).Select “more than”.Verification / Alternative check:Datasheets list r_p values (e.g., 1.5–3 for blowers, 3–10+ for multistage air compressors), all exceeding unity, confirming the definition.Why Other Options Are Wrong:“Equal to” contradicts compression; “less than” corresponds to throttling/expansion devices, not compressors.Common Pitfalls:Mixing up absolute and gauge pressures; comparing pressures at different elevations or temperatures without proper station definition.Final Answer:more than

Question 15

Ways to temporarily increase jet-engine thrust: Which method reliably boosts thrust in a turbojet/turbofan power unit?

Accepted Answer

Introduction / Context:Aircraft engines sometimes require short-term thrust augmentation for takeoff or combat. Several techniques are discussed in propulsion courses, but not all are standard or broadly effective. This item separates proven practices from distractors.Given Data / Assumptions:Conventional turbojet/turbofan with turbine exit stream available for tailpipe reheat.No fundamental hardware redesign beyond accepted augmentation systems.Safety, operability, and certification considerations matter.Concept / Approach:Afterburning (reheat) injects fuel in the exhaust duct and ignites it, significantly increasing jet temperature and exit velocity for a marked thrust rise, albeit with high specific fuel consumption. Water or water–methanol injection at the compressor inlet has historically increased mass flow and limited compressor temperature for some engines, but it is not generally used on modern turbofans due to performance and maintenance penalties. Direct ammonia injection into the main combustor is not a standard certified thrust-boost practice for aviation gas turbines.Step-by-Step Solution:Identify proven augmentation → afterburning.Evaluate water injection → niche, historical; not a general, reliable thrust booster for modern fleets.Ammonia injection → nonstandard fuel; operationally unsuitable.Choose the single best correct method: option highlighting afterburning.Verification / Alternative check:Military jet specs and propulsion texts document afterburner thrust increases on the order of 50–70% or more, confirming its effectiveness.Why Other Options Are Wrong:Water injection is conditional and dated; ammonia injection is atypical and problematic for combustion and materials; “all of the above” incorrectly groups them as equally valid.Common Pitfalls:Assuming any injected fluid increases thrust safely; overlooking turbine temperature limits and compressor surge margins.Final Answer:Burning additional fuel downstream of the turbine (afterburning/ reheat in the tailpipe)

Question 16

Effect of pressure ratio on volumetric efficiency: As the overall compression ratio decreases in a reciprocating air compressor with clearance, the volumetric efficiency generally __________.

Accepted Answer

Introduction / Context:Volumetric efficiency measures how effectively a reciprocating compressor fills its cylinder with fresh charge. It depends on clearance and the re-expansion of trapped clearance gas, which is sensitive to the pressure ratio. Understanding this trend guides stage selection and intercooling strategy.Given Data / Assumptions:Nonzero clearance volume at top dead center.Ideal valve action; negligible pressure losses in suction.Comparison at the same speed and ambient conditions.Concept / Approach:During suction, part of the stroke is used to re-expand the compressed clearance gas from discharge pressure back toward suction pressure. A higher pressure ratio means a greater expansion requirement, reducing the net fresh air admitted and hence lowering volumetric efficiency. Conversely, lowering the pressure ratio reduces clearance gas expansion work and increases the effective filling of the cylinder.Step-by-Step Solution:Define η_v ≈ 1 + C − C * (p2/p1)^(1/n) for a polytropic re-expansion (illustrative form).As p2/p1 decreases, the term (p2/p1)^(1/n) decreases.Therefore η_v increases for a fixed clearance C.Conclude: decreasing compression ratio → volumetric efficiency increases.Verification / Alternative check:Indicator-diagram geometry shows longer effective suction for lower discharge pressure, matching the analytical trend.Why Other Options Are Wrong:“Does not change” ignores clearance gas effects; “decreases” is opposite of the real trend; “negative” is unphysical for normal operation.Common Pitfalls:Confusing volumetric efficiency with isentropic efficiency; forgetting that zero-clearance (ideal) would make η_v approach unity regardless of ratio.Final Answer:increases

Question 17

Overall (plant) isothermal efficiency of a compressor system: Choose the correct definition used in practice for overall isothermal efficiency.

Accepted Answer

Introduction / Context:Different efficiency definitions exist for compressors: isentropic, polytropic, isothermal, and volumetric. “Overall isothermal efficiency” is a plant-level metric that compares the ideal isothermal requirement with the actual shaft input to the compressor drive (motor or engine).Given Data / Assumptions:Continuous operation at steady conditions.Isothermal power computed for the same inlet and discharge states.Shaft power is the brake power delivered to the compressor input shaft.Concept / Approach:Overall isothermal efficiency, η_iso,overall = P_isothermal / P_shaft. It judges how close the whole system (including mechanical and internal losses) operates to the theoretical minimum work case (isothermal). It is distinct from cylinder/process isothermal efficiency (option a) and from isentropic efficiency (option d). Volumetric efficiency (option c) is capacity-related, not energy-related.Step-by-Step Solution:Compute ideal isothermal power for given mass flow and ratio.Measure actual shaft (brake) power to drive compressor.Form η = P_isothermal / P_shaft.Select the definition that states this ratio explicitly.Verification / Alternative check:Manufacturer test codes report both isentropic and isothermal efficiencies; the latter referenced to shaft input for overall comparisons.Why Other Options Are Wrong:Option (a) is a process efficiency inside the machine; (c) is volumetric efficiency; (d) defines isentropic efficiency, not isothermal.Common Pitfalls:Confusing “overall” with “internal”; mixing thermodynamic and volumetric metrics.Final Answer:Isothermal power divided by the shaft power (brake power) supplied to the compressor drive

Question 18

Minimum compression work (single-stage, no clearance): For a single-stage, single-acting reciprocating air compressor without clearance, which ideal process yields the minimum work input for a given pressure ratio?

Accepted Answer

Introduction / Context:For a fixed inlet/outlet pressure, different compression paths require different works. Comparing isothermal, polytropic, and isentropic processes clarifies why intercooling is beneficial in multistage compressors.Given Data / Assumptions:Ideal gas behavior with constant properties.No clearance; steady operation.Same pressure ratio across all compared processes.Concept / Approach:Work input is the integral of p dV. For a given pressure ratio, the area under the curve is smallest for the isothermal path (n = 1), larger for general polytropic (1 Step-by-Step Solution:Compare processes on p-V diagram.Recognize isothermal curve lies below polytropic and isentropic for same end points.Therefore work_isothermal Select isothermal as minimum-work process.Verification / Alternative check:Analytic expressions confirm: w_isothermal = RT1ln(p2/p1), while w_isentropic = (k/(k−1))RT1*[(p2/p1)^((k−1)/k) − 1], with the latter numerically larger for the same ratio.Why Other Options Are Wrong:Isentropic has the highest work for a given ratio; generic polytropic lies in between; “none” is incorrect because isothermal is the known minimum.Common Pitfalls:Assuming “adiabatic is efficient” in compressors; forgetting that heat removal during compression reduces required input work.Final Answer:Isothermal compression

Question 19

Definition check — volumetric efficiency: Is volumetric efficiency correctly defined as the ratio of free-air volume delivered per stroke to the piston’s swept volume?

Accepted Answer

Introduction / Context:Volumetric efficiency links what a reciprocating compressor geometrically displaces to how much fresh charge it truly admits and delivers, referred to a common basis (free air). The definition must be precise to compare cylinders with different clearances and valve characteristics.Given Data / Assumptions:Swept volume is the piston displacement per stroke.Delivered volume is converted to free-air basis (ambient pressure and temperature).Single-acting cylinder considered for simplicity.Concept / Approach:Volumetric efficiency, η_v, quantifies cylinder filling: η_v = (free-air volume delivered per stroke) / (swept volume). It accounts for clearance re-expansion and suction losses, allowing meaningful comparison across designs and operating ratios. Because it uses free-air reference, η_v connects directly to mass flow at given ambient conditions.Step-by-Step Solution:Measure delivered flow and refer it to free-air conditions.Divide by geometric swept volume per stroke.Interpret the result (typically 0.7–0.95 for healthy machines).Hence the provided definition is correct.Verification / Alternative check:Indicator diagrams and standard test codes adopt this definition; models express η_v in terms of clearance and pressure ratio, reaffirming its basis.Why Other Options Are Wrong:“No” would be valid only if delivery were not converted to free-air or if a different basis were specified; here, the statement is standard.Common Pitfalls:Using discharge-state volume instead of free-air; ignoring temperature differences; confusing volumetric with isentropic efficiency.Final Answer:Yes

Question 20

Cooling process in a closed-cycle gas turbine plant: In the external cooler of a closed Brayton (Joule) cycle, the working gas is cooled at which thermodynamic condition?

Accepted Answer

Introduction / Context:Closed-cycle gas turbines circulate a working gas (e.g., helium, nitrogen) through compressors, heaters, turbines, and coolers. Understanding which processes are at constant pressure versus constant volume clarifies how the Brayton cycle is realized in hardware.Given Data / Assumptions:Steady flow through a heat exchanger (cooler).Negligible pressure drop in ideal analysis.Standard Brayton sequence: compression → heating → expansion → cooling.Concept / Approach:In the ideal Brayton (Joule) cycle, both heat addition and heat rejection occur at constant pressure in flow heat exchangers. The cooler removes heat at essentially constant pressure as the gas flows through tubes/plates, reducing temperature from turbine exhaust value back toward compressor inlet temperature.Step-by-Step Solution:Identify component: external cooler (heat exchanger).Idealize pressure losses → p ≈ constant across the cooler.Therefore, cooling occurs at constant pressure with decreasing temperature.Select “constant pressure”.Verification / Alternative check:T-s and p-V diagrams of Brayton cycle show horizontal heat-rejection line at constant pressure; textbooks label both the heater and cooler as constant-pressure devices.Why Other Options Are Wrong:Constant volume would require a rigid, closed vessel without flow; constant temperature would be ideal only with variable pressure and is not how the Brayton cooler is defined; “none” contradicts standard cycle representation.Common Pitfalls:Confusing closed-cycle (recirculated gas) with constant-volume processes; assuming coolers behave like tanks rather than flow heat exchangers.Final Answer:Constant pressure

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