Question 1

Why only rocket engines can operate in space  Select the key reason rockets are the only engines that can propel vehicles in outer space where there is no atmosphere. Choose the correct option.

Accepted Answer

Given data Space environment lacks atmospheric oxygen. Concept/ApproachRocket engines carry both fuel and oxidizer and therefore do not rely on atmospheric air. Air-breathing engines (turbojets, turbofans, ramjets) cannot operate in vacuum due to lack of oxygen mass flow. Final AnswerThey are not air-breathing engines

Question 2

Roots blower pressure rise mechanism  Statement: In a roots blower, the increase in delivery pressure is entirely due to backflow from the high-pressure side when trapped air pockets open to the outlet. Decide whether the statement is correct.

Accepted Answer

Given data Machine: roots blower (positive-displacement rotary blower). Concept/ApproachA roots blower does not compress internally while carrying air from inlet to outlet. When the trapped volume first communicates with the high-pressure outlet, backflow of air raises the trapped air pressure to delivery pressure, causing the observed pressure rise. Final AnswerTrue

Question 3

Propulsive efficiency condition for a rocket  A rocket achieves maximum overall propulsive efficiency when the aircraft (vehicle) flight speed is how related to the jet exhaust velocity? Choose the correct option.

Accepted Answer

Concept/ApproachPropulsive efficiency depends on the velocity match between the vehicle and the exhaust. For momentum devices (including rockets), efficiency is maximized when flight speed ≈ jet speed, minimizing kinetic energy wasted in the exhaust slipstream. Final AnswerEqual to the jet velocity

Question 4

Degree of reaction in an axial-flow compressor: precise definition  How is the degree of reaction (R) defined for an axial-flow compressor stage? Choose the correct option.

Accepted Answer

Concept/ApproachFor an axial-flow compressor, the degree of reaction is defined as the ratio of static enthalpy rise in the rotor to the static enthalpy rise in the entire stage (rotor + stator). Symbolically, R = (Δhstatic in rotor)/(Δhstatic in stage). Final AnswerStatic enthalpy rise in rotor to static enthalpy rise in the stage

Question 5

Typical overall compression ratio in simple gas-turbine cycles  Select a representative simple-cycle gas turbine compressor pressure ratio used in basic textbook problems and older industrial sets. Choose the correct option.

Accepted Answer

Given data Context: simple (single-spool) gas turbines in classical MCQ settings. Concept/ApproachEarly and small industrial simple-cycle gas turbines commonly used modest compressor pressure ratios (≈ 3–6). Modern aero/industrial engines may be far higher, but the textbook representative choice in such MCQs is often about 4. EliminationA ratio of 1 implies no compression; 9 or 12 are possible in advanced designs but exceed the typical simple-cycle baseline emphasized in entry-level questions. Final Answer4

Question 6

Compressor performance terminology: delivered air volume  The term used for the volume of air delivered by a compressor (referred to standard free-air conditions) is: Choose the correct option.

Accepted Answer

Concept/ApproachFree Air Delivery (FAD) denotes the delivered air volume referred back to standard free-air conditions. “Compressor capacity” is a general term and may be quoted as FAD, but the precise technical term for the delivered volume is FAD. “Swept volume” is the geometric displacement, not the delivered volume. Final AnswerFree air delivery

Question 7

Applications of compressed air: identify the correct set  Which of the following are common uses of compressed air in mechanical/thermal systems? Choose the correct option.

Accepted Answer

Given data Uses of compressed air across industries and power plants. Concept/ApproachCompressed air is widely used for pneumatic tools (e.g., drills), for starting and supercharging I.C. engines, and in gas turbine plant services (e.g., starting systems, purge, cooling and controls). Final AnswerAll of the above

Question 8

Compressor terminology: identify the parameter that equals work per cycle divided by stroke volume.  Definition focus: In reciprocating compressors, the ratio (workdone per cycle)/(stroke volume) is called what? Choose the correct term.

Accepted Answer

Given data / DefinitionQuantity of interest = (Workdone per cycle) ÷ (Stroke volume). Concept/ApproachMean Effective Pressure (MEP) is defined so that if it acted uniformly on the piston through the stroke volume, it would produce the same work as the actual cycle. Step-by-stepWcycle = MEP × VstrokeMEP = Wcycle / Vstroke Why others are wrongCompressor capacity is generally volumetric flow per unit time; compression ratio is p2/p1; compressor efficiency compares ideal vs actual work. Final AnswerMean effective pressure.

Question 9

Compressor fundamentals: select the correct statement among the following.  Evaluate each statement about definitions and work in air compression, and choose the only correct one.

Accepted Answer

Concept/ApproachCheck each statement against standard definitions and thermodynamic results for gas compression. Analysis of optionsA: The ratio p2/p1 is the compression ratio, not efficiency → false.B: For a compressor, discharge pressure p2 > inlet pressure p1, so p2/p1 > 1 → true.C: Work per cycle divided by stroke volume defines mean effective pressure, not capacity → false.D: For a given pressure ratio, isothermal compression requires the minimum work; polytropic/adiabatic require more → false. Final AnswerThe compression ratio for a compressor is always greater than unity.

Question 10

Clearance volume and volumetric efficiency in compressors: validate the statement for design practice.  Statement to evaluate: Clearance volume is kept as small as possible because it adversely affects volumetric efficiency. Choose whether the state…

Accepted Answer

Given / ContextClearance volume traps residual compressed air which re-expands during suction. Concept/ApproachResidual air reduces the fresh charge admitted each stroke, lowering volumetric efficiency (ηv). ReasoningLarger clearance → more re-expansion → less fresh intake → lower ηv. Hence designers minimize clearance consistent with mechanical needs. Final AnswerAgree.

Question 11

Perfect (complete) intercooling criterion in two-stage air compression: identify the temperature condition.  Condition to classify perfect intercooling: Temperature of air leaving the intercooler is ________ the original atmospheric (inlet) air temp…

Accepted Answer

Concept/ApproachPerfect (complete) intercooling brings the air temperature between stages back to the initial inlet (atmospheric) temperature, minimizing total compression work for given pressure ratio. ResultTafter intercooler = Tatmospheric inlet. Final AnswerEqual to.

Question 12

Pressure rise mechanism in a sliding vane blower: evaluate the two-stage explanation.  Statement to evaluate: In a vane blower, the pressure increases first due to internal compression (decreasing cell volume) and then due to some backflow from the …

Accepted Answer

Given / DeviceSliding vane rotary machines trap air between vanes and the casing; as the trapped pocket moves, its volume reduces, compressing the air. Concept/ApproachIn a vane compressor, internal compression occurs before the trapped air reaches the discharge port. When the pocket opens to the outlet, any remaining pressure difference is equalized by a brief backflow from the delivery line. ContrastA Roots blower has virtually no internal compression; pressure rise is primarily due to backflow/mixing at the outlet. Final AnswerAgree.

Question 13

Purpose of intercooling in multi-stage air compressors: choose the primary objective.  Select the main reason intercoolers are used between compression stages.

Accepted Answer

Concept/ApproachFor a given overall pressure ratio, the compression work W ∝ ∫ v dp. Reducing the specific volume by cooling between stages lowers the required work. Key resultPerfect intercooling (returning to the initial temperature) with equal pressure ratios per stage minimizes total work. Final AnswerTo minimise the work of compression.

Question 14

Where does stagnation pressure rise occur in a centrifugal compressor?  Identify the component(s) responsible for the increase in total (stagnation) pressure.

Accepted Answer

Concept/ApproachThe impeller does shaft work on the fluid, increasing total enthalpy and thus stagnation pressure (neglecting losses). The diffuser primarily converts kinetic energy to static pressure; ideally, total pressure would remain constant there (real diffusers have losses that can decrease stagnation pressure). ConclusionThe rise in stagnation pressure is produced in the impeller. Final AnswerIn the impeller only.

Question 15

Rotary compressors used in aircraft: select the common type for high-throughput, high-speed applications.  Choose the type of rotary compressor predominantly used in aeroplanes.

Accepted Answer

Concept/ApproachAircraft engines require very high mass flow rates with compact frontal area and favorable efficiency at high pressure ratios. Axial-flow compressors meet these requirements better than centrifugal units for large engines. Final AnswerAxial flow type.

Question 16

Effect of speed on centrifugal compressor performance at a given pressure ratio: choose the correct trend.  For a centrifugal compressor operating at a fixed pressure ratio, what is the typical effect of increasing rotational speed?

Accepted Answer

Concept/ApproachRaising speed generally shifts the characteristic to higher mass flow and head. At a fixed pressure ratio away from the new design point, the operating point tends to a higher flow with some off-design penalties, often reducing efficiency. ResultFlow increases; efficiency typically drops when operating off optimum speed-line at the same pressure ratio. Final AnswerIncrease in flow and decrease in efficiency.

Question 17

Heat rejection accounting in a multi-stage reciprocating air compressor with intercooling.  Validate the energy balance statement: The total heat rejected equals (heat rejected during polytropic compression per kg of air) + (heat rejected in the int…

Accepted Answer

Concept/ApproachApply the first law to each stage and the intercooler. Heat removed from the gas occurs during the compression process (if cooled) and additionally in the intercooler between stages. Energy balanceQtotal out = Σ Qpolytropic (per kg) + Σ Qintercooler (per kg)Work input is reduced by intercooling since the specific volume at the start of the next stage is lower. Final AnswerAgree.

Question 18

Gas turbines: effect of a high air–fuel ratio on performance and temperatures  Consider a gas turbine operating with a high (lean) air–fuel ratio relative to stoichiometric conditions. Choose the correct outcome that most directly follows from this …

Accepted Answer

Given data Operating condition: high air–fuel ratio (lean mixture) in a gas turbine. Concept/ApproachFor a fixed fuel flow, adding more air (i.e., making the mixture leaner) lowers the adiabatic flame temperature and consequently the turbine inlet and exhaust temperatures. Lower temperatures also reduce blade thermal stress, which is beneficial for durability. Step-by-step reasoning1) High A/F ⇒ excess air relative to stoichiometric requirement.2) Excess air absorbs heat ⇒ lower gas temperature at turbine inlet and outlet.3) Power output primarily follows turbine inlet temperature; with leaner mixture at the same fuel flow, power typically decreases rather than increases.4) Thermal efficiency has an optimum; too lean can lower efficiency due to reduced mean temperature levels and increased compressor work fraction. Common pitfallsAssuming that adding more air always increases power is incorrect unless matched with additional fuel to hold turbine inlet temperature constant. Final AnswerReduces exhaust temperature

Question 19

Two-stage reciprocating air compressor with complete intercooling: diameter ratio relation  Let p₁ be the suction pressure to the L.P. cylinder and p₂ the pressure leaving the L.P. cylinder (intercooler pressure). For a single-acting, two-stage reci…

Accepted Answer

Given data Single-acting, two-stage reciprocating air compressor. Complete intercooling (air cooled back to initial suction temperature between stages). D₁ = diameter of L.P. cylinder; D₂ = diameter of H.P. cylinder; p₁ = suction pressure to L.P.; p₂ = delivery pressure from L.P. (and intercooler pressure). Concept/ApproachWith complete intercooling to the initial temperature, the specific volume at each stage inlet is inversely proportional to the stage inlet pressure. For equal piston speeds, volumetric flow rate ∝ cylinder area, so the area (and hence diameter) ratio follows the square root of the pressure ratio seen across the stages. Key relationA₁/A₂ = V̇₁/V̇₂ = (p₂/p₁) ⇒ (D₁/D₂)² = p₂/p₁ ⇒ D₁/D₂ = √(p₂/p₁). Why given options are incorrectOptions (a), (b), and (c) omit the required square root; therefore the correct choice is None of these. Final AnswerNone of these (correct relation is D₁/D₂ = √(p₂/p₁)).

Question 20

Four-stage compressor with equal pressure ratio per stage: determine fourth-stage delivery pressure  In a four-stage compressor operating with equal pressure ratio per stage, the pressures at the first and third stage are given as 1 bar and 16 bar, …

Accepted Answer

Given data Number of stages = 4; equal pressure ratio per stage. Stated pressures: first stage = 1 bar; third stage = 16 bar. Concept/ApproachFor equal pressure ratio r per stage, overall pressure after n stages is p_out = p_in × rⁿ. Standard MCQ convention often uses the given intermediate pressure (16 bar) as the pressure after the second stage, leading to a clean power-of-four result. Step-by-step calculation (assumption noted)Assumption: The 16 bar value corresponds to the end of the second stage (a common textbook variant); then r² = 16 ⇒ r = 4.After four stages: p₄ = 1 × r⁴ = 4⁴ = 256 bar. Verification/AlternativeIf 16 bar were strictly after the third stage with equal r, r³ = 16 ⇒ r ≈ 2.52 and p₄ ≈ 40 bar, which does not match the provided options. Hence the usual intended reading yields 256 bar. Final Answer256 bar

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