Q: Compressor definitions and ratios: Is the following statement correct? “Compressor efficiency is defined as the ratio of the volume of free air delivery per stroke to the piston’s swept volume.”

Introduction / Context: This question distinguishes between two commonly confused ideas in air-compressor terminology: volumetric efficiency and compressor (isentropic) efficiency. Understanding the precise definitions is essential for sizing compressors and interpreting test results. Given Data / Assumptions: Piston compressors considered; terms apply broadly across positive-displacement units. “Free air delivery” refers to flow rate referred to local atmospheric conditions. Swept volume is the geometric displacement per stroke. Concept / Approach: Volumetric efficiency, not compressor efficiency, is defined as the ratio of free air delivery to swept volume (adjusted to the same reference conditions). Compressor efficiency normally means isentropic (adiabatic) or polytropic efficiency, comparing ideal work to actual work, or ideal temperature rise to actual temperature rise. Step-by-Step Solution: Identify the ratio FAD per stroke / swept volume → this is volumetric efficiency.Recall that compressor (isentropic) efficiency = ideal work / actual work, or (T2s − T1) / (T2a − T1) at equal pressure ratio.Therefore the quoted definition mismatches the accepted term.Conclusion: the statement is incorrect. Verification / Alternative check: Textbook definitions consistently separate volumetric efficiency (capacity-related) from isentropic efficiency (thermodynamic performance). Manufacturers publish both independently. Why Other Options Are Wrong: “Only true for single-stage” and “only at zero clearance” still refer to volumetric efficiency, not compressor efficiency. “Correct only for isothermal” is also a definition mismatch. Common Pitfalls: Assuming any ratio involving delivered volume is an “efficiency” of all kinds; ignoring the thermodynamic nature of isentropic efficiency. Final Answer: Incorrect

Q: Application of Euler’s turbomachinery equation: Euler’s head/energy equation is directly applicable to which of the following machines?

Introduction / Context: Euler’s turbomachinery equation relates the change in angular momentum of a fluid to the shaft work exchanged in rotating machines. It underpins performance analysis for pumps, compressors, and turbines of various flow geometries. Given Data / Assumptions: Steady, one-dimensional flow at rotor inlet and exit. Neglecting losses to state the ideal relation. Applies to rotating blades/impellers where whirl components exist. Concept / Approach: The equation states: specific work = u2 * Vw2 − u1 * Vw1, where u is blade speed and Vw is the whirl (tangential) velocity component. This is valid for any turbomachine—compressor or pump, radial or axial—so long as rotor-relative momentum exchange occurs. Step-by-Step Solution: Recognize the presence of rotating blades → angular momentum change.Both centrifugal/axial compressors and pumps meet this criterion.Therefore Euler’s equation applies to all listed turbomachines.Select “All of these”. Verification / Alternative check: Derivations for compressor head rise and pump head both start from the same Euler relation, confirming universality across rotodynamic machines. Why Other Options Are Wrong: Restricting to only one type ignores the generality of the angular momentum principle. Reciprocating compressors are not turbomachines and are outside this framework. Common Pitfalls: Thinking the equation is exclusive to pumps or to compressors; overlooking that axial machines also have whirl at rotor inlet/outlet. Final Answer: All of these

Q: Condition at the start of compression (reciprocating air compressor): At the beginning of the compression stroke, the air temperature is typically __________ the local atmospheric temperature.

Introduction / Context: Before compression begins in a reciprocating air compressor, the cylinder has just completed the suction process. Clarifying the expected air temperature at this instant helps separate suction effects from compression effects. Given Data / Assumptions: Suction occurs from the ambient with small pressure losses. Cylinder walls near ambient; no significant preheating assumed. Clearance effects are neglected for this basic concept question. Concept / Approach: During suction, the intake valve opens and the cylinder fills with air drawn from the surroundings. Aside from minor heating due to throttling or frictional effects in the intake passage, the bulk air temperature is essentially the ambient (local atmospheric) temperature at the start of compression. Step-by-Step Solution: Suction draws air from atmosphere.Negligible heat transfer or compression during suction in ideal analysis.Thus T at start of compression ≈ atmospheric T.Choose “equal to”. Verification / Alternative check: Indicator diagrams and basic compressor models assume T1 equals ambient for isentropic-work calculations unless specific intake heating/cooling is provided. Why Other Options Are Wrong: “Less than” or “more than” would need pre-cooling or preheating hardware; “indeterminate” ignores the standard assumption; “much higher” contradicts the lack of compression before the stroke begins. Common Pitfalls: Confusing end-of-suction with mid-compression; overestimating throttling temperature rises in normal intakes. Final Answer: equal to

Q: Overall efficiency of a compressed-air system: How is the overall system efficiency defined when air is compressed, stored, and then expanded in an air motor to deliver shaft output?

Introduction / Context: Compressed-air systems are energy conversion chains: mechanical input drives a compressor, compressed air is stored/conditioned, and an air motor reconverts the pneumatic energy to mechanical output. A clear definition of overall efficiency is necessary for system-level economics and design. Given Data / Assumptions: Single compressor feeding an air motor (or pneumatic tools). Steady-state operation during the measurement interval. Losses in storage, piping, and conditioning implicitly included. Concept / Approach: Overall (system) efficiency equals useful shaft output divided by required shaft input across the entire chain. Therefore, η_overall = (air motor brake power) / (compressor shaft power). It aggregates all inefficiencies: compression losses, storage and distribution losses, and motor conversion losses. Step-by-Step Solution: Identify input: compressor shaft power.Identify useful output: air motor shaft power.Form the ratio output / input.Hence select option defining exactly that ratio. Verification / Alternative check: Energy balance over the combined equipment set confirms that any other formulation (e.g., inverse or product) lacks physical meaning for efficiency bounded between 0 and 1. Why Other Options Are Wrong: The inverse ratio exceeds 1 for typical conditions; the product has units of power squared and is not an efficiency; indicated-to-brake compares internal to shaft within only the air motor, not the system. Common Pitfalls: Confusing component efficiencies with overall system efficiency; neglecting throttling and leakage losses that the overall metric captures implicitly. Final Answer: Ratio of shaft output of the air motor to the shaft input to the compressor

Q: Centrifugal compressor slip factor: The slip factor is defined as the ratio of which velocity component to the blade tip speed (u2) at the impeller exit?

Introduction / Context: Due to finite blade number and flow separation, the actual tangential (whirl) component of absolute velocity at a centrifugal impeller exit is less than the ideal value. The slip factor quantifies this deficit and directly affects the achievable head rise from Euler’s equation. Given Data / Assumptions: Steady operation at the impeller exit plane. Blade tip speed u2 is known from rotor speed and radius. Whirl (tangential) component at outlet is Vw2 (actual). Concept / Approach: A common practical definition is σ = Vw2(actual) / u2 when referenced to blade speed, representing the fraction of ideal whirl achieved. More rigorously, some texts define σ = Vw2(actual) / Vw2(ideal). Both convey that finite blades cause “slip,” reducing tangential momentum transfer and head rise. Step-by-Step Solution: Identify relevant velocity: tangential (whirl) at outlet, Vw2.Relate to blade tip speed u2 → σ ≈ Vw2 / u2 (in this option set).Select the option stating “outlet whirl velocity to blade velocity”.Conclude that slip lowers Euler head from the ideal prediction. Verification / Alternative check: Performance correlations employ σ to correct the ideal head: Δh_actual ≈ σ * u2 * Vw2(ideal) / g, consistent with reduced Vw2. Why Other Options Are Wrong: Inlet quantities do not define exit slip. Flow (meridional) velocities govern continuity and diffusion but not the slip factor definition relative to whirl delivery. Common Pitfalls: Confusing slip factor with work or loading factor; mixing up whirl and meridional components; forgetting finite blade-number effects. Final Answer: Outlet whirl velocity to blade velocity

Q: Definition check: “Free air” means air stated at the atmospheric pressure and temperature of the specific location (reference ambient). Is this statement correct?

Introduction / Context: Flow ratings for compressors often use “free air delivery” (FAD). The meaning of “free air” must be clear to compare capacity across sites and conditions. Given Data / Assumptions: Local site atmospheric pressure and temperature define the reference state. Standard air is a different concept (fixed reference such as 1.013 bar and 15°C or 20°C). Humidity may be specified separately but is not implied to be zero. Concept / Approach: “Free air” refers to the actual ambient state where the compressor ingests air. Converting measured volumetric flow to this reference allows consistent volumetric comparisons. It differs from “standard air,” which uses a universal standard condition regardless of site altitude/temperature. Step-by-Step Solution: Identify reference: local atmospheric pressure and temperature.Express delivered volume at those ambient conditions → free air.Therefore the statement is correct.Note: standards may define exact reference for “standard air,” not to be confused with FAD. Verification / Alternative check: Industry datasheets routinely state “FAD at site conditions,” while “SCFM” or “Nm³/h” indicate standard conditions. The definitions co-exist for different purposes. Why Other Options Are Wrong: Sea-level or fixed-temperature caveats are unnecessary; humidity need not be zero for the definition, though moisture content can be specified. Common Pitfalls: Confusing free air with standard air, leading to erroneous capacity comparisons across elevations and climates. Final Answer: True

Q: Reheat to maximize work output: For two-stage expansion with ideal reheating in a gas turbine (or steam turbine), the reheating should bring the pressure to which intermediate value (p_i) between maximum p1 and minimum p2 for maximum total work?

Introduction / Context: In multi-stage expansion with reheating, choosing the optimal intermediate pressure maximizes total work and balances irreversibilities. This is a classic result in gas/steam turbine cycle analysis. Given Data / Assumptions: Two equal-efficiency stages with reheating to approximately the same turbine inlet temperature between stages. Idealized isentropic stage analysis for the optimum condition. Neglecting pressure losses in reheater and piping for the derivation. Concept / Approach: For maximum work, distribute the overall pressure ratio equally across stages. This occurs when the individual stage pressure ratios are equal, so that p1 / p_i = p_i / p2, leading to p_i^2 = p1 * p2 and hence p_i = √(p1 * p2). Equalizing stage ratios tends to equalize temperature drops, improving mean effective temperature of heat addition and reducing exergy destruction. Step-by-Step Solution: Let stage ratios be r1 = p1 / p_i and r2 = p_i / p2.Set r1 = r2 for optimum → p1 / p_i = p_i / p2.Rearrange: p_i^2 = p1 * p2.Therefore p_i = √(p1 * p2). Verification / Alternative check: Similar result appears for optimum intercooling pressure in multi-stage compression, reinforcing the symmetry between expansion with reheating and compression with intercooling. Why Other Options Are Wrong: Arithmetic means or ad hoc multiples do not equalize pressure ratios. The cube-root form pertains to three-stage problems, not two-stage. Common Pitfalls: Confusing arithmetic with geometric mean; ignoring stage equality assumptions and reheater losses when applying the ideal result to real plants. Final Answer: p_i = √(p1 * p2)

Q: Rocket engines — basic characteristics: Which statements correctly describe rocket combustion chambers and exhaust compared with air-breathing jet engines?

Introduction / Context: Rocket propulsion differs fundamentally from air-breathing jets because the oxidizer is carried onboard. Recognizing chamber conditions and exhaust characteristics is key to nozzle design and mission performance estimates. Given Data / Assumptions: Steady operation of a chemical rocket with a converging–diverging nozzle. Combustion completes in the chamber before the nozzle throat. Nozzle expands flow from high chamber stagnation conditions to near-ambient or vacuum pressure. Concept / Approach: The chamber behaves as a high-pressure plenum with near-stagnation velocity, analogous to a supersonic tunnel reservoir. The nozzle then accelerates the gases to very high supersonic speeds, typically exceeding air-breathing turbojet exhaust velocities because chamber temperatures and pressure ratios are much higher and no intake losses occur from atmospheric compression systems. Step-by-Step Solution: Identify chamber as stagnation region → yes.Relate to wind-tunnel reservoir analogy → valid.Compare exit speeds: rockets often achieve several km/s, higher than typical turbojets.Therefore, all listed statements are correct. Verification / Alternative check: Rocket nozzle theory shows exit Mach numbers and velocities determined by pressure ratio and chamber temperature; flight data confirm much higher exhaust speeds than jets which are limited by turbine inlet temperature and intake compression ratios. Why Other Options Are Wrong: Choosing any subset omits true statements; “None” contradicts standard gas-dynamics. Common Pitfalls: Confusing rocket nozzles with ram/scramjet intakes; overlooking that jets entrain ambient air and are constrained by Brayton cycle components. Final Answer: All of the above

Q: Work comparison for compression vs. constant-pressure heating: In a single-stage, single-acting reciprocating air compressor without clearance, the work done during isentropic compression from T1 to T2 is __________ the heat required to raise air fr…

Introduction / Context: Relating compression work to heating requirements helps build intuition for compressor energy demands and the role of specific heats in ideal analyses. Given Data / Assumptions: Isentropic (adiabatic, reversible) compression from temperature T1 to T2. No clearance volume; ideal gas with constant specific heats. Comparison is made to constant-pressure heating over the same temperature rise. Concept / Approach: For an isentropic compression of an ideal gas, the specific work input equals the change in specific enthalpy because there is no heat transfer and h = cp * T for an ideal gas with constant cp. Thus w_isentropic,in = h2 − h1 = cp * (T2 − T1). At constant pressure, the heat required for the same temperature rise is also q_cp = cp * (T2 − T1). Hence they are equal in magnitude. Step-by-Step Solution: For ideal gas with constant cp: h2 − h1 = cp * (T2 − T1).Isentropic compression work input equals enthalpy rise: w_in = h2 − h1.Constant-pressure heating requires q = cp * (T2 − T1).Therefore w_in = q, i.e., they are equal. Verification / Alternative check: Using the steady-flow energy equation for an adiabatic compressor (neglecting KE/PE changes) also yields w_in = Δh. The same Δh arises for constant-pressure heating by definition of cp. Why Other Options Are Wrong: “Less than” or “greater than” contradicts the enthalpy-based equality under the ideal assumptions; “indeterminate” is incorrect because Δh depends only on temperatures for an ideal gas. Common Pitfalls: Mixing isentropic with polytropic processes; assuming variable cp without stating it, which would slightly modify numerical equality but not the learning goal. Final Answer: equal to

Question 1

Temperature after air-motor expansion in compressed-air systems Considering a compressed-air system where an air motor expands the supplied compressed air to produce shaft work, how does the discharge air temperature compare to the original compress…

Accepted Answer

Introduction / Context: Air motors convert the pressure energy of compressed air into mechanical work by expanding it. Gas expansion is accompanied by a temperature drop, the magnitude of which depends on the expansion path and efficiency. Given Data / Assumptions: Air is supplied at a pressure higher than ambient. Expansion occurs in the motor to near-atmospheric exhaust pressure. Ambient compressor intake temperature is the reference for comparison. Concept / Approach: For an isentropic or near-isentropic expansion, T_out = T_in * (p_out/p_in)^((gamma-1)/gamma). Since p_out < p_in, the exponent is positive and the ratio is less than 1, leading to a lower discharge temperature. Even with non-idealities, the exhaust from an air motor is typically colder than ambient compressor intake temperature. Step-by-Step Solution: Recognise expansion process: pressure drop from supply to exhaust.Apply qualitative isentropic relation: temperature decreases with pressure in expansion.Conclude: discharge air temperature is less than the original intake temperature. Verification / Alternative check: Practical observations (frosting of outlets under heavy load) confirm significant cooling during expansion. Why Other Options Are Wrong: “More/greater” contradict expansion thermodynamics. “Same under all conditions” ignores process dependence. “Undefined” is incorrect; temperature change is fundamental to gas expansion. Common Pitfalls: Confusing compressor discharge temperature with intake temperature; aftercooling and piping losses do not negate the cooling effect of downstream expansion in the motor. Final Answer: less

Question 2

Compressor definitions and ratios: Is the following statement correct? “Compressor efficiency is defined as the ratio of the volume of free air delivery per stroke to the piston’s swept volume.”

Accepted Answer

Introduction / Context: This question distinguishes between two commonly confused ideas in air-compressor terminology: volumetric efficiency and compressor (isentropic) efficiency. Understanding the precise definitions is essential for sizing compressors and interpreting test results. Given Data / Assumptions: Piston compressors considered; terms apply broadly across positive-displacement units. “Free air delivery” refers to flow rate referred to local atmospheric conditions. Swept volume is the geometric displacement per stroke. Concept / Approach: Volumetric efficiency, not compressor efficiency, is defined as the ratio of free air delivery to swept volume (adjusted to the same reference conditions). Compressor efficiency normally means isentropic (adiabatic) or polytropic efficiency, comparing ideal work to actual work, or ideal temperature rise to actual temperature rise. Step-by-Step Solution: Identify the ratio FAD per stroke / swept volume → this is volumetric efficiency.Recall that compressor (isentropic) efficiency = ideal work / actual work, or (T2s − T1) / (T2a − T1) at equal pressure ratio.Therefore the quoted definition mismatches the accepted term.Conclusion: the statement is incorrect. Verification / Alternative check: Textbook definitions consistently separate volumetric efficiency (capacity-related) from isentropic efficiency (thermodynamic performance). Manufacturers publish both independently. Why Other Options Are Wrong: “Only true for single-stage” and “only at zero clearance” still refer to volumetric efficiency, not compressor efficiency. “Correct only for isothermal” is also a definition mismatch. Common Pitfalls: Assuming any ratio involving delivered volume is an “efficiency” of all kinds; ignoring the thermodynamic nature of isentropic efficiency. Final Answer: Incorrect

Question 3

Application of Euler’s turbomachinery equation: Euler’s head/energy equation is directly applicable to which of the following machines?

Accepted Answer

Introduction / Context: Euler’s turbomachinery equation relates the change in angular momentum of a fluid to the shaft work exchanged in rotating machines. It underpins performance analysis for pumps, compressors, and turbines of various flow geometries. Given Data / Assumptions: Steady, one-dimensional flow at rotor inlet and exit. Neglecting losses to state the ideal relation. Applies to rotating blades/impellers where whirl components exist. Concept / Approach: The equation states: specific work = u2 * Vw2 − u1 * Vw1, where u is blade speed and Vw is the whirl (tangential) velocity component. This is valid for any turbomachine—compressor or pump, radial or axial—so long as rotor-relative momentum exchange occurs. Step-by-Step Solution: Recognize the presence of rotating blades → angular momentum change.Both centrifugal/axial compressors and pumps meet this criterion.Therefore Euler’s equation applies to all listed turbomachines.Select “All of these”. Verification / Alternative check: Derivations for compressor head rise and pump head both start from the same Euler relation, confirming universality across rotodynamic machines. Why Other Options Are Wrong: Restricting to only one type ignores the generality of the angular momentum principle. Reciprocating compressors are not turbomachines and are outside this framework. Common Pitfalls: Thinking the equation is exclusive to pumps or to compressors; overlooking that axial machines also have whirl at rotor inlet/outlet. Final Answer: All of these

Question 4

Condition at the start of compression (reciprocating air compressor): At the beginning of the compression stroke, the air temperature is typically __________ the local atmospheric temperature.

Accepted Answer

Introduction / Context: Before compression begins in a reciprocating air compressor, the cylinder has just completed the suction process. Clarifying the expected air temperature at this instant helps separate suction effects from compression effects. Given Data / Assumptions: Suction occurs from the ambient with small pressure losses. Cylinder walls near ambient; no significant preheating assumed. Clearance effects are neglected for this basic concept question. Concept / Approach: During suction, the intake valve opens and the cylinder fills with air drawn from the surroundings. Aside from minor heating due to throttling or frictional effects in the intake passage, the bulk air temperature is essentially the ambient (local atmospheric) temperature at the start of compression. Step-by-Step Solution: Suction draws air from atmosphere.Negligible heat transfer or compression during suction in ideal analysis.Thus T at start of compression ≈ atmospheric T.Choose “equal to”. Verification / Alternative check: Indicator diagrams and basic compressor models assume T1 equals ambient for isentropic-work calculations unless specific intake heating/cooling is provided. Why Other Options Are Wrong: “Less than” or “more than” would need pre-cooling or preheating hardware; “indeterminate” ignores the standard assumption; “much higher” contradicts the lack of compression before the stroke begins. Common Pitfalls: Confusing end-of-suction with mid-compression; overestimating throttling temperature rises in normal intakes. Final Answer: equal to

Question 5

Overall efficiency of a compressed-air system: How is the overall system efficiency defined when air is compressed, stored, and then expanded in an air motor to deliver shaft output?

Accepted Answer

Introduction / Context: Compressed-air systems are energy conversion chains: mechanical input drives a compressor, compressed air is stored/conditioned, and an air motor reconverts the pneumatic energy to mechanical output. A clear definition of overall efficiency is necessary for system-level economics and design. Given Data / Assumptions: Single compressor feeding an air motor (or pneumatic tools). Steady-state operation during the measurement interval. Losses in storage, piping, and conditioning implicitly included. Concept / Approach: Overall (system) efficiency equals useful shaft output divided by required shaft input across the entire chain. Therefore, η_overall = (air motor brake power) / (compressor shaft power). It aggregates all inefficiencies: compression losses, storage and distribution losses, and motor conversion losses. Step-by-Step Solution: Identify input: compressor shaft power.Identify useful output: air motor shaft power.Form the ratio output / input.Hence select option defining exactly that ratio. Verification / Alternative check: Energy balance over the combined equipment set confirms that any other formulation (e.g., inverse or product) lacks physical meaning for efficiency bounded between 0 and 1. Why Other Options Are Wrong: The inverse ratio exceeds 1 for typical conditions; the product has units of power squared and is not an efficiency; indicated-to-brake compares internal to shaft within only the air motor, not the system. Common Pitfalls: Confusing component efficiencies with overall system efficiency; neglecting throttling and leakage losses that the overall metric captures implicitly. Final Answer: Ratio of shaft output of the air motor to the shaft input to the compressor

Question 6

Centrifugal compressor slip factor: The slip factor is defined as the ratio of which velocity component to the blade tip speed (u2) at the impeller exit?

Accepted Answer

Introduction / Context: Due to finite blade number and flow separation, the actual tangential (whirl) component of absolute velocity at a centrifugal impeller exit is less than the ideal value. The slip factor quantifies this deficit and directly affects the achievable head rise from Euler’s equation. Given Data / Assumptions: Steady operation at the impeller exit plane. Blade tip speed u2 is known from rotor speed and radius. Whirl (tangential) component at outlet is Vw2 (actual). Concept / Approach: A common practical definition is σ = Vw2(actual) / u2 when referenced to blade speed, representing the fraction of ideal whirl achieved. More rigorously, some texts define σ = Vw2(actual) / Vw2(ideal). Both convey that finite blades cause “slip,” reducing tangential momentum transfer and head rise. Step-by-Step Solution: Identify relevant velocity: tangential (whirl) at outlet, Vw2.Relate to blade tip speed u2 → σ ≈ Vw2 / u2 (in this option set).Select the option stating “outlet whirl velocity to blade velocity”.Conclude that slip lowers Euler head from the ideal prediction. Verification / Alternative check: Performance correlations employ σ to correct the ideal head: Δh_actual ≈ σ * u2 * Vw2(ideal) / g, consistent with reduced Vw2. Why Other Options Are Wrong: Inlet quantities do not define exit slip. Flow (meridional) velocities govern continuity and diffusion but not the slip factor definition relative to whirl delivery. Common Pitfalls: Confusing slip factor with work or loading factor; mixing up whirl and meridional components; forgetting finite blade-number effects. Final Answer: Outlet whirl velocity to blade velocity

Question 7

Definition check: “Free air” means air stated at the atmospheric pressure and temperature of the specific location (reference ambient). Is this statement correct?

Accepted Answer

Introduction / Context: Flow ratings for compressors often use “free air delivery” (FAD). The meaning of “free air” must be clear to compare capacity across sites and conditions. Given Data / Assumptions: Local site atmospheric pressure and temperature define the reference state. Standard air is a different concept (fixed reference such as 1.013 bar and 15°C or 20°C). Humidity may be specified separately but is not implied to be zero. Concept / Approach: “Free air” refers to the actual ambient state where the compressor ingests air. Converting measured volumetric flow to this reference allows consistent volumetric comparisons. It differs from “standard air,” which uses a universal standard condition regardless of site altitude/temperature. Step-by-Step Solution: Identify reference: local atmospheric pressure and temperature.Express delivered volume at those ambient conditions → free air.Therefore the statement is correct.Note: standards may define exact reference for “standard air,” not to be confused with FAD. Verification / Alternative check: Industry datasheets routinely state “FAD at site conditions,” while “SCFM” or “Nm³/h” indicate standard conditions. The definitions co-exist for different purposes. Why Other Options Are Wrong: Sea-level or fixed-temperature caveats are unnecessary; humidity need not be zero for the definition, though moisture content can be specified. Common Pitfalls: Confusing free air with standard air, leading to erroneous capacity comparisons across elevations and climates. Final Answer: True

Question 8

Reheat to maximize work output: For two-stage expansion with ideal reheating in a gas turbine (or steam turbine), the reheating should bring the pressure to which intermediate value (p_i) between maximum p1 and minimum p2 for maximum total work?

Accepted Answer

Introduction / Context: In multi-stage expansion with reheating, choosing the optimal intermediate pressure maximizes total work and balances irreversibilities. This is a classic result in gas/steam turbine cycle analysis. Given Data / Assumptions: Two equal-efficiency stages with reheating to approximately the same turbine inlet temperature between stages. Idealized isentropic stage analysis for the optimum condition. Neglecting pressure losses in reheater and piping for the derivation. Concept / Approach: For maximum work, distribute the overall pressure ratio equally across stages. This occurs when the individual stage pressure ratios are equal, so that p1 / p_i = p_i / p2, leading to p_i^2 = p1 * p2 and hence p_i = √(p1 * p2). Equalizing stage ratios tends to equalize temperature drops, improving mean effective temperature of heat addition and reducing exergy destruction. Step-by-Step Solution: Let stage ratios be r1 = p1 / p_i and r2 = p_i / p2.Set r1 = r2 for optimum → p1 / p_i = p_i / p2.Rearrange: p_i^2 = p1 * p2.Therefore p_i = √(p1 * p2). Verification / Alternative check: Similar result appears for optimum intercooling pressure in multi-stage compression, reinforcing the symmetry between expansion with reheating and compression with intercooling. Why Other Options Are Wrong: Arithmetic means or ad hoc multiples do not equalize pressure ratios. The cube-root form pertains to three-stage problems, not two-stage. Common Pitfalls: Confusing arithmetic with geometric mean; ignoring stage equality assumptions and reheater losses when applying the ideal result to real plants. Final Answer: p_i = √(p1 * p2)

Question 9

Rocket engines — basic characteristics: Which statements correctly describe rocket combustion chambers and exhaust compared with air-breathing jet engines?

Accepted Answer

Introduction / Context: Rocket propulsion differs fundamentally from air-breathing jets because the oxidizer is carried onboard. Recognizing chamber conditions and exhaust characteristics is key to nozzle design and mission performance estimates. Given Data / Assumptions: Steady operation of a chemical rocket with a converging–diverging nozzle. Combustion completes in the chamber before the nozzle throat. Nozzle expands flow from high chamber stagnation conditions to near-ambient or vacuum pressure. Concept / Approach: The chamber behaves as a high-pressure plenum with near-stagnation velocity, analogous to a supersonic tunnel reservoir. The nozzle then accelerates the gases to very high supersonic speeds, typically exceeding air-breathing turbojet exhaust velocities because chamber temperatures and pressure ratios are much higher and no intake losses occur from atmospheric compression systems. Step-by-Step Solution: Identify chamber as stagnation region → yes.Relate to wind-tunnel reservoir analogy → valid.Compare exit speeds: rockets often achieve several km/s, higher than typical turbojets.Therefore, all listed statements are correct. Verification / Alternative check: Rocket nozzle theory shows exit Mach numbers and velocities determined by pressure ratio and chamber temperature; flight data confirm much higher exhaust speeds than jets which are limited by turbine inlet temperature and intake compression ratios. Why Other Options Are Wrong: Choosing any subset omits true statements; “None” contradicts standard gas-dynamics. Common Pitfalls: Confusing rocket nozzles with ram/scramjet intakes; overlooking that jets entrain ambient air and are constrained by Brayton cycle components. Final Answer: All of the above

Question 10

Work comparison for compression vs. constant-pressure heating: In a single-stage, single-acting reciprocating air compressor without clearance, the work done during isentropic compression from T1 to T2 is __________ the heat required to raise air fr…

Accepted Answer

Introduction / Context: Relating compression work to heating requirements helps build intuition for compressor energy demands and the role of specific heats in ideal analyses. Given Data / Assumptions: Isentropic (adiabatic, reversible) compression from temperature T1 to T2. No clearance volume; ideal gas with constant specific heats. Comparison is made to constant-pressure heating over the same temperature rise. Concept / Approach: For an isentropic compression of an ideal gas, the specific work input equals the change in specific enthalpy because there is no heat transfer and h = cp * T for an ideal gas with constant cp. Thus w_isentropic,in = h2 − h1 = cp * (T2 − T1). At constant pressure, the heat required for the same temperature rise is also q_cp = cp * (T2 − T1). Hence they are equal in magnitude. Step-by-Step Solution: For ideal gas with constant cp: h2 − h1 = cp * (T2 − T1).Isentropic compression work input equals enthalpy rise: w_in = h2 − h1.Constant-pressure heating requires q = cp * (T2 − T1).Therefore w_in = q, i.e., they are equal. Verification / Alternative check: Using the steady-flow energy equation for an adiabatic compressor (neglecting KE/PE changes) also yields w_in = Δh. The same Δh arises for constant-pressure heating by definition of cp. Why Other Options Are Wrong: “Less than” or “greater than” contradicts the enthalpy-based equality under the ideal assumptions; “indeterminate” is incorrect because Δh depends only on temperatures for an ideal gas. Common Pitfalls: Mixing isentropic with polytropic processes; assuming variable cp without stating it, which would slightly modify numerical equality but not the learning goal. Final Answer: equal to

Question 11

Condition at the start of compression (pressure): At the beginning of the compression stroke in a reciprocating air compressor, the cylinder air pressure is typically __________ the local atmospheric pressure.

Accepted Answer

Introduction / Context: Correctly identifying the starting pressure for compression is foundational to calculating pressure ratios, work input, and volumetric performance in reciprocating compressors. Given Data / Assumptions: Air is drawn from the atmosphere during the suction stroke. Minor intake losses are neglected in the basic idealization. No pre-pressurization devices at the intake. Concept / Approach: In the ideal model, suction fills the cylinder with ambient air, so the pressure at the instant compression begins equals atmospheric. Real systems may have small pressure drops across filters/valves, but conceptually, p_start ≈ p_atm is used for introductory analysis and indicator-diagram interpretation. Step-by-Step Solution: Suction → cylinder communicates with the atmosphere.At the start of compression, the intake valve closes.Pressure inside equals atmospheric just before closure (ideal model).Therefore select “equal to”. Verification / Alternative check: Textbook examples take p1 as ambient in work and temperature ratio formulas, confirming the standard assumption for baseline calculations. Why Other Options Are Wrong: “Less than” or “more than” would require throttling losses or supercharging intake; “indeterminate” ignores the model assumptions; “twice” is physically unjustified. Common Pitfalls: Confusing start-of-compression with mid-stroke states; overemphasizing small intake pressure drops in conceptual questions. Final Answer: equal to

Question 12

Centrifugal compressor prewhirl — effect on inlet absolute velocity In centrifugal compressor practice, “prewhirl” refers to intentionally imparting swirl to the incoming air before it meets the impeller eye. The process that allows the air to enter…

Accepted Answer

Introduction / Context: Prewhirl is a widely used inlet treatment in centrifugal compressors. By giving the incoming air a controlled tangential component (usually in the direction of impeller rotation), the relative incidence at the impeller leading edge can be optimized. This reduces losses and can extend the stable operating range without altering the meridional (through-flow) velocity or the mass flow rate. Given Data / Assumptions: Flow approaching the impeller eye is given a finite tangential velocity by inlet guide vanes. Mass flow rate and meridional velocity component are maintained. Focus is on how the absolute inlet velocity magnitude seen by the impeller changes. Concept / Approach: Velocity triangles govern blade incidence. Adding prewhirl (in the direction of wheel rotation) reduces the relative inlet velocity and the work input required per unit mass. For constant meridional component, a favorable tangential component changes the absolute velocity vector so that its magnitude, relevant to incidence and shock formation at leading edges, is effectively reduced at entry. Step-by-Step Solution: Establish inlet velocity triangle with components: V_m (meridional), V_t (tangential due to prewhirl).With prewhirl in the direction of rotation, the blade-relative velocity W = V − U decreases at the leading edge.Maintaining mass flow and V_m while adding favorable V_t leads to a reduced effective absolute entry magnitude for the target incidence.Therefore, the description that fits prewhirl is that the air enters at a reduced absolute velocity. Verification / Alternative check: Design texts show prewhirl lowers inlet relative Mach number and mitigates leading-edge shock or separation, consistent with a reduced effective entry velocity for a given blade speed. Why Other Options Are Wrong: Increased velocity would worsen incidence. “Unchanged” ignores the vector change. “Reversed” is not used in practical prewhirl and would cause severe incidence. Common Pitfalls: Confusing absolute velocity reduction with meridional velocity; prewhirl targets the swirl component while preserving through-flow. Final Answer: reduced

Question 13

Why axial-flow compressors are used in aircraft gas turbines In aircraft gas-turbine engines, which primary advantage makes axial-flow compressors preferable over centrifugal compressors for high mass-flow propulsion applications?

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Introduction / Context: Compressor selection in aircraft propulsion is driven by aerodynamic packaging, efficiency, and achievable pressure ratios for the required mass flow. Axial-flow compressors dominate modern turbojets and turbofans because they can process very large airflows while keeping the engine diameter small, reducing drag and easing nacelle integration. Given Data / Assumptions: High mass-flow requirements typical of flight propulsion. Constraints on engine nacelle diameter (frontal area) due to drag and installation. Comparable technology levels and good design practice. Concept / Approach: For the same mass flow, a centrifugal stage tends to require a larger diameter (due to radial flow path and diffuser volute), increasing frontal area. Axial compressors build pressure gradually over many stages while maintaining a slender annulus, resulting in a small frontal area and favorable installation aerodynamics. Step-by-Step Solution: Identify the key aircraft need: high mass flow with minimal drag.Recognize axial-flow architecture allows compact diameter by stacking multiple stages axially.Conclude that low frontal area is the decisive advantage in aircraft gas turbines. Verification / Alternative check: Modern turbofan cores universally employ multistage axial compressors, validating the dominance of low frontal area for large mass flows. Why Other Options Are Wrong: “Higher thrust” is not an intrinsic compressor property; thrust depends on overall cycle and nozzle. Single-stage pressure rise is actually higher for centrifugal than for a single axial stage. Common Pitfalls: Confusing stage pressure rise with overall engine performance; an axial machine attains high overall pressure via many stages, not a single stage. Final Answer: Low frontal area for a given mass flow rate

Question 14

Centrifugal compressor basics — select the correct statement Which of the following statements correctly characterizes a centrifugal compressor in standard engineering practice?

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Introduction / Context: Centrifugal compressors add kinetic energy to air by high-speed impeller rotation and then convert part of that kinetic energy into static pressure in the diffuser. They are widely used in small turbochargers and industrial air systems where moderate pressure ratios and ruggedness are desired. Given Data / Assumptions: Conventional single- or multistage centrifugal compressor with diffuser and volute. Steady, subsonic to transonic relative speeds depending on design. Comparison against axial compressors on general traits. Concept / Approach: Energy transfer in turbomachinery follows Euler's pump/compressor equation. In a centrifugal compressor, high impeller tip speed imparts swirl and kinetic energy to the air, which the diffuser converts to static pressure. Therefore, the defining correct statement revolves around high rotational speeds of the impeller to add energy. Step-by-Step Solution: Check option (a): axial-parallel flow describes axial compressors, not centrifugal.Check option (b): static pressure generally rises from impeller exit into diffuser; the phrasing about ‘‘decreases’’ is incorrect.Check option (c): aligns with Euler's principle; correct.Check option (d): multistage axial compressors typically achieve higher peak efficiencies for large mass flows. Verification / Alternative check: Performance maps show typical impeller tip speeds several hundred m/s, confirming the need for high peripheral speed for effective compression. Why Other Options Are Wrong: (a) Confuses machine type; (b) misstates pressure behavior; (d) overclaims versus axial machines. Common Pitfalls: Equating single-stage pressure rise with overall efficiency; the efficiency comparison depends on application scale and staging. Final Answer: The impeller rotates at high peripheral speeds to add energy to the flow.

Question 15

Specific mass of gas turbine vs. reciprocating I.C. engine For approximately the same power output, the mass of a gas turbine plant per kW developed is about __________ compared with that of a reciprocating internal combustion engine.

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Introduction / Context: One advantage of gas turbines is their high power-to-weight ratio, which is critical in aviation and mobile applications. This question compares the specific mass (kg per kW) of gas turbines to reciprocating I.C. engines at similar output scales. Given Data / Assumptions: Comparable technology levels and materials. Continuous-flow gas turbine versus multi-cylinder reciprocating engine. Accessory mass differences are ignored for a first-order estimate. Concept / Approach: Gas turbines deliver smooth continuous torque with far fewer moving parts per unit power, allowing high rotational speeds and compact rotors. Reciprocating engines require heavy crankshafts, pistons, and blocks to handle cyclic loads, increasing mass for the same output. Step-by-Step Solution: Identify governing factor: power-to-weight ratio is higher for turbines due to continuous combustion and high rpm capability.Industry comparisons often cite gas turbines weighing a small fraction of comparable I.C. engines.Among the options, ‘‘one-sixth’’ best represents the typical order-of-magnitude advantage. Verification / Alternative check: Aero and industrial turbine datasheets show significantly lower specific mass than piston engines at similar outputs, validating the selection. Why Other Options Are Wrong: “The same” and “one-half” understate the advantage; “one-fourth” is closer but still typically conservative relative to many practical comparisons. Common Pitfalls: Ignoring application context; actual ratios vary with scale and design, but the turbine advantage remains pronounced. Final Answer: one-sixth

Question 16

Where expansion provides useful output in a turbojet After heat addition in the combustor of a turbojet (Brayton cycle), in which component does the working fluid expand essentially isentropically to deliver shaft work to drive the compressor?

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Introduction / Context: The ideal Brayton cycle features isentropic compression, constant-pressure heat addition, and isentropic expansion. In a turbojet, the turbine extracts enough work to drive the compressor, while the nozzle converts remaining enthalpy into jet kinetic energy for thrust. Given Data / Assumptions: Idealized processes with negligible losses (isentropic compressor and turbine in the ideal model). Combustion approximated as constant-pressure heat addition. Nozzle treated separately for thrust generation. Concept / Approach: Work transfer to the compressor comes from the turbine. The turbine expansion is modeled as isentropic in the ideal cycle. The exit nozzle is not constant-volume; it is an accelerating, near-isentropic flow passage that primarily produces thrust rather than shaft power. Step-by-Step Solution: Heat addition raises total temperature at essentially constant pressure.Turbine expands the hot gas isentropically, extracting shaft work for the compressor via the shaft.Remaining enthalpy is then converted in the nozzle into jet kinetic energy for propulsion.Thus, the component providing shaft work is the turbine, with essentially isentropic expansion. Verification / Alternative check: T-s diagrams of the Brayton cycle show the compressor-turbine pair as vertical (isentropic) lines in the ideal case, confirming the answer. Why Other Options Are Wrong: Exit nozzle produces thrust, not shaft work; there is no constant-volume process in turbine or nozzle in Brayton engines. Common Pitfalls: Confusing the roles of turbine (shaft work) and nozzle (thrust) in turbojets. Final Answer: Turbine blades, which is essentially an isentropic process

Question 17

Compressor performance at altitude Compared with operation at sea level, the performance (capacity and pressure ratio capability at a given speed) of an air compressor operating at high altitudes will be __________.

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Introduction / Context: Ambient conditions strongly influence compressor performance. At higher altitudes, air density and ambient pressure are reduced, affecting mass flow capacity, surge margin, and achievable pressure ratios at a given rotational speed. Given Data / Assumptions: Same mechanical speed and geometry at altitude as at sea level. Lower ambient static pressure and density at altitude. No variable geometry or control changes assumed for the comparison statement. Concept / Approach: Mass flow rate m_dot = rho * V_dot. For the same volumetric flow capability, lower ambient density rho directly reduces m_dot. The compressor map referenced to corrected conditions shows that at altitude the corrected speed and flow shift, reducing surge margin and practical pressure ratio at fixed speed. Step-by-Step Solution: Recognize density and inlet pressure drop with altitude.At fixed volumetric flow, mass flow decreases proportionally to density.Lower inlet pressure reduces achievable absolute delivery pressure for a given pressure ratio, impacting useful performance.Therefore, overall performance is lower compared to sea level. Verification / Alternative check: Engine and compressor manufacturers publish altitude derating curves showing reduced capacity and different surge/choke boundaries at higher altitudes. Why Other Options Are Wrong: “The same” ignores density effects. “Higher” contradicts basic continuity and compressor map behavior. Common Pitfalls: Confusing corrected (standardized) performance with actual performance; corrected data hides ambient effects by design. Final Answer: lower

Question 18

Two-stage reciprocating air compressor — optimum intercooler pressure For a two-stage reciprocating air compressor with perfect intercooling, the intermediate (intercooler) pressure that minimizes the total compression work is given by which relatio…

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Introduction / Context: Staging and intercooling in reciprocating compressors reduce compression work and discharge temperature. For two stages with perfect intercooling (the air is cooled back to the initial intake temperature between stages), there is an optimum intermediate pressure that minimizes the total work. Given Data / Assumptions: Two compression stages with equal polytropic/isentropic characteristics. Perfect intercooling to the intake temperature between stages. Negligible pressure losses in intercooler and piping (idealization). Concept / Approach: For minimum total work, the pressure ratio is split equally between stages. That means p2/p1 = p3/p2. Solving this yields p2^2 = p1 * p3, or p2 = (p1 * p3)^(1/2). This balances the work of both stages and achieves the lowest possible sum for a given overall ratio. Step-by-Step Solution: Set equal pressure ratios: r1 = p2/p1 and r2 = p3/p2.For minimum work: r1 = r2.Therefore, p2/p1 = p3/p2 → p2^2 = p1 * p3.Hence, p2 = (p1 * p3)^(1/2). Verification / Alternative check: Equal temperature rises per stage under perfect intercooling also result when pressure ratios are equal, consistent with the optimum condition. Why Other Options Are Wrong: (b) Arithmetic mean ignores thermodynamic work minimization; (c) and (d) have incorrect dimensional/ratio logic and do not balance stage works. Common Pitfalls: Forgetting to include the square root; writing p2 = p1 * p3 is a frequent mistake that overpressurizes the first stage. Final Answer: p2 = (p1 * p3)^(1/2)

Question 19

Flow directions in a centrifugal compressor Confirm the standard flow pattern: in a centrifugal compressor the air enters the impeller axially and leaves the vanes in a predominantly radial direction.

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Introduction / Context: Centrifugal compressors are defined by their radial flow path through the impeller. Understanding the basic flow directions helps with interpreting velocity triangles, diffuser design, and overall stage performance. Given Data / Assumptions: Standard single-stage centrifugal compressor with axial entry and radial exit. Negligible prewhirl for the basic statement (though prewhirl may be used in practice). Concept / Approach: To minimize inlet losses and maintain compact geometry, flow is brought axially into the eye of the impeller. The spinning impeller imparts tangential and radial components, throwing the air outward; the diffuser then recovers static pressure from the high-velocity radial flow. Step-by-Step Solution: Inlet guide ensures near-axial entry to the impeller eye.Rotating blades add swirl and kinetic energy; the flow path bends to radial through the impeller passages.At exit, the flow is predominantly radial before diffusion and volute collection.Therefore, the statement is correct. Verification / Alternative check: Standard compressor schematics and velocity triangles depict axial entry and radial discharge for centrifugal stages. Why Other Options Are Wrong: Answering ‘‘No’’ conflicts with the very definition and geometry of centrifugal compressors. Common Pitfalls: Confusing centrifugal compressors with axial compressors where both entry and exit are axial. Final Answer: Yes

Question 20

Effect of reheating on gas-turbine performance In a simple gas turbine with reheating (expansion in multiple turbine stages with heat addition between them), which statement best describes the impact on overall cycle performance?

Accepted Answer

Introduction / Context: Reheating inserts additional heat addition between turbine stages, allowing a higher average turbine inlet temperature during expansion. This boosts the specific turbine work but changes the average temperature at which heat is supplied, affecting overall efficiency. Given Data / Assumptions: Simple Brayton cycle baseline with fixed compressor pressure ratio. Reheat applied between turbine stages, with comparable total-to-total efficiencies. No change to compressor side or pressure ratio due to reheat alone. Concept / Approach: Thermal efficiency = net work output / heat supplied. Reheating increases turbine work but requires additional fuel at a comparatively higher average temperature drop sequence that, in simple cycles, leads to a larger increase in heat input than in net work gain, thus lowering thermal efficiency. However, reheating can be beneficial where higher specific power is desired, or where constraints exist on turbine metal temperatures per stage. Step-by-Step Solution: Add reheat → turbine-stage inlet temperature raised for the second stage.Turbine work increases due to higher average expansion temperature.But additional heat input required outweighs the extra net work → efficiency falls.Therefore, among the options, ‘‘decreases the thermal efficiency’’ is correct. Verification / Alternative check: Cycle analysis and T–s diagrams show increased area for heat addition without proportional net-work increase; practical data confirm lower simple-cycle efficiency with reheat. Why Other Options Are Wrong: (a) Opposite of the usual simple-cycle effect; (b) compressor work is largely unaffected; (c) true locally but not the best single statement about overall cycle performance when only one choice is allowed. Common Pitfalls: Assuming more turbine work automatically means higher efficiency; efficiency depends on both net work and total heat input. Final Answer: Reheating decreases the thermal efficiency

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