Q1 medium
Reheat in Rankine cycle primarily reduces
A boiler fuel need to zero ✓ B moisture content at turbine exit and blade erosion ✓ C pump work to zero always ✓ D condenser size to zero always ✓ Show Answer 💡 Explanation: Reheat raises expansion path quality at LP turbine.
Q2 medium
Regeneration in Rankine cycle uses
A intercooling in Brayton compressor only as Rankine primary incorrectly ✓ B aftercooler in diesel only ✓ C feedwater heaters to preheat water using extracted steam ✓ D radiation shield only ✓ Show Answer 💡 Explanation: Bleed steam preheats feedwater reducing heat input.
Q3 Past Paper · PPSC/FPSC/NTS medium
Quality (dryness fraction) x in wet steam region is
A mass of vapor divided by total mass of mixture ✓ B volume of liquid only ✓ C pressure divided by temperature ✓ D enthalpy of liquid only ✓ Show Answer 💡 Explanation: x = m_v/(m_v + m_l).
Q4 hard
On T-s diagram, area under reversible process curve represents
A work always for any process without distinction incorrectly ✓ B heat transfer for that process ✓ C entropy generation equal to area always incorrectly ✓ D mass flow only ✓ Show Answer 💡 Explanation: Q = ∫T ds for reversible boundary heat transfer.
Q5 medium
On p-v diagram, area under process curve represents
A heat transfer always equal to area on pv incorrectly always ✓ B entropy change directly as area incorrectly ✓ C internal energy always ✓ D boundary work for quasi-static process ✓ Show Answer 💡 Explanation: W = ∫p dv for closed system work.
Q6 Past Paper · PPSC/FPSC/NTS medium
Critical point on steam property chart is where
A liquid and vapor phases become indistinguishable ✓ B boiling always occurs at 100°C regardless of pressure incorrectly ✓ C entropy is zero always ✓ D quality equals zero only always at critical incorrectly alone ✓ Show Answer 💡 Explanation: At critical T and p, phase boundary vanishes.
Q7 easy
Dry saturated steam has quality
A 1 ✓ B 0 ✓ C 0.5 always ✓ D undefined always incorrectly ✓ Show Answer 💡 Explanation: x = 1 means all vapor at saturation.
Q8 easy
Subcooled liquid has temperature
A above saturation always ✓ B equal to critical always ✓ C below saturation temperature at given pressure ✓ D zero always ✓ Show Answer 💡 Explanation: Compressed liquid region left of saturation curve.
Q9 Past Paper · PPSC/FPSC/NTS easy
Superheated steam has temperature
A below saturation ✓ B always at triple point ✓ C above saturation temperature at given pressure ✓ D equal to wet bulb always incorrectly ✓ Show Answer 💡 Explanation: Right of saturation dome on T-s diagram.
Q10 hard
Availability (exergy) of system interacting with environment at T0 is related to
A maximum useful work obtainable to dead state ✓ B total internal energy always equal to availability incorrectly ✓ C only kinetic energy ✓ D only gravitational PE always alone ✓ Show Answer 💡 Explanation: Exergy measures work potential relative to reference environment.
Q11 hard
Irreversibility I in process equals
A T0 times entropy generation ✓ B work output always ✓ C heat input only without T0 ✓ D enthalpy change only ✓ Show Answer 💡 Explanation: I = T0 S_gen for closed systems commonly.
Q12 Past Paper · PPSC/FPSC/NTS medium
Isothermal expansion of ideal gas in cylinder does
A no work ✓ B decrease U for ideal gas incorrectly in isothermal ✓ C positive work equal to heat added at constant T ✓ D zero heat transfer always ✓ Show Answer 💡 Explanation: ΔU=0 for ideal gas isothermal so Q=W.
Q13 medium
Polytropic process p v^n = C with n=0 gives
A constant volume ✓ B constant pressure process ✓ C isothermal always ✓ D isentropic always ✓ Show Answer 💡 Explanation: n=0 ⇒ p constant.
Q14 medium
Polytropic index n=1 corresponds to
A isochoric ✓ B isothermal process for ideal gas ✓ C isentropic ✓ D isobaric ✓ Show Answer 💡 Explanation: n=1: T constant for ideal gas.
Q15 Past Paper · PPSC/FPSC/NTS medium
Polytropic index n=γ corresponds to
A isentropic process for ideal gas ✓ B isothermal ✓ C isobaric ✓ D isochoric ✓ Show Answer 💡 Explanation: Reversible adiabatic ideal gas: n = γ.
Q16 hard
Helmholtz function A is
A H − T S incorrectly as Helmholtz ✓ B pV only ✓ C T S only ✓ D U − T S ✓ Show Answer 💡 Explanation: A = U − TS; natural variables T,V.
Q17 hard
Gibbs function G is
A U + pV only alone defines G incorrectly ✓ B T S − H ✓ C H − T S ✓ D pV − U ✓ Show Answer 💡 Explanation: G = H − TS; useful for constant T,p.
Q18 Past Paper · PPSC/FPSC/NTS hard
Maxwell relations connect
A only mechanical stress and strain ✓ B partial derivatives of thermodynamic potentials ✓ C only Fourier law ✓ D only Navier-Stokes ✓ Show Answer 💡 Explanation: Cross partial equality from exact differentials.
Q19 hard
Joule-Thomson coefficient μ_JT is
A (∂p/∂T)_v ✓ B (∂T/∂p)_h ✓ C (∂h/∂p)_T only alone without definition incorrectly ✓ D (∂s/∂v)_T only alone ✓ Show Answer 💡 Explanation: Throttling process is isenthalpic.
Q20 medium
Throttling of ideal gas causes
A no temperature change ✓ B always cooling regardless of gas incorrectly ✓ C always heating ✓ D phase change always ✓ Show Answer 💡 Explanation: Ideal gas enthalpy depends only on T; h constant ⇒ T constant.
Q21 Past Paper · PPSC/FPSC/NTS hard
Clausius inequality states for real cycles
A equal to zero always for irreversible too incorrectly ✓ B greater than zero always ✓ C independent of temperature ✓ D ∮ δQ/T ≤ 0 ✓ Show Answer 💡 Explanation: Equality for reversible; inequality for irreversible.
Q22 medium
Entropy generation is always
A negative in irreversible processes incorrectly ✓ B zero for all irreversible processes incorrectly ✓ C non-negative for real processes ✓ D undefined for heat transfer ✓ Show Answer 💡 Explanation: Second law: S_gen ≥ 0.
Q23 medium
Steady flow energy equation with negligible PE and KE changes reduces to
A Q − W_s = Δh for one stream ✓ B ΔU only always in SFEE incorrectly alone ✓ C ΔS only ✓ D p Δv only always alone sufficient ✓ Show Answer 💡 Explanation: SFEE: q + h1 + V1²/2 = h2 + V2²/2 + ws.
Q24 Past Paper · PPSC/FPSC/NTS medium
Nozzle converts
A KE into heat only always in nozzle incorrectly ✓ B pressure rise without velocity change always ✓ C enthalpy decrease into kinetic energy increase ✓ D mass into energy ✓ Show Answer 💡 Explanation: Adiabatic nozzle energy conversion.
Q25 medium
Diffuser converts
A pressure into vacuum always ✓ B heat into work only in closed piston always incorrectly as diffuser primary ✓ C entropy decrease without irreversibility always in real diffuser incorrectly stated as always ✓ D kinetic energy into pressure rise ✓ Show Answer 💡 Explanation: Deceleration increases static pressure.
Q26 Past Paper · PPSC/FPSC/NTS medium
Dew point temperature is temperature at which
A condensation begins for given moisture content at constant pressure ✓ B boiling occurs always at 100°C incorrectly regardless of pressure ✓ C entropy is maximum always ✓ D dry bulb equals wet bulb always for any air incorrectly ✓ Show Answer 💡 Explanation: Saturation partial pressure reached.
Q27 medium
Relative humidity is ratio of
A absolute humidity to density of dry air only without saturation reference incorrectly alone always as RH ✓ B wet bulb to dry bulb always as RH incorrectly ✓ C mass of liquid to mass of vapor in tank incorrectly as RH ✓ D partial pressure of vapor to saturation pressure at same temperature ✓ Show Answer 💡 Explanation: φ = p_v/p_g × 100%.
Q28 medium
Throttling valve process is approximated as
A isentropic always for irreversible throttle incorrectly always ✓ B isochoric always ✓ C isothermal always for real gas always incorrectly ✓ D isenthalpic with negligible heat and work transfer ✓ Show Answer 💡 Explanation: h1 ≈ h2 for throttling.
Q29 Past Paper · PPSC/FPSC/NTS easy
Mixture of ideal gases total pressure is
A product of partial pressures ✓ B average only without summing incorrectly ✓ C zero if mixed ✓ D sum of partial pressures (Dalton) ✓ Show Answer 💡 Explanation: P = Σ p_i.
Q30 medium
Partial pressure of component i is
A m_i P only mass fraction without conversion incorrectly always alone ✓ B P/n_i only incorrectly ✓ C y_i P where y_i is mole fraction ✓ D always equal for all components incorrectly ✓ Show Answer 💡 Explanation: p_i = x_i P for ideal gas mixture.
Q31 hard
Air-standard diesel cycle cut-off ratio affects
A only compression ratio of Otto only incorrectly alone always ✓ B only condenser pressure ✓ C heat addition at constant pressure and cycle efficiency ✓ D only pump work in Rankine incorrectly alone ✓ Show Answer 💡 Explanation: Higher cut-off adds heat at lower average T_add effect.
Q32 Past Paper · PPSC/FPSC/NTS hard
Mean effective pressure (MEP) in IC engines is
A peak pressure always equal to MEP incorrectly ✓ B work per cycle divided by displaced volume ✓ C fuel energy only without volume incorrectly ✓ D exhaust pressure only ✓ Show Answer 💡 Explanation: MEP compares engine output independent of size.
Q33 medium
Knocking in SI engine is caused by
A too much steam injection incorrectly ✓ B auto-ignition of end-gas ahead of flame front ✓ C low compression always beneficial incorrectly ✓ D rankine reheat only incorrectly ✓ Show Answer 💡 Explanation: Uncontrolled combustion raises pressure rapidly.
Q34 Past Paper · PPSC/FPSC/NTS easy
Octane number measures
A cetane for diesel incorrectly labeled as octane primary ✓ B viscosity only ✓ C calorific value only alone always as octane ✓ D knock resistance of fuel ✓ Show Answer 💡 Explanation: Higher octane resists knock.
Q35 easy
Cetane number measures
A knock resistance of gasoline incorrectly as cetane ✓ B ignition quality of diesel fuel ✓ C steam quality x incorrectly ✓ D hardness of coal incorrectly ✓ Show Answer 💡 Explanation: Higher cetane easier ignition delay shorter.
Q36 Past Paper · PPSC/FPSC/NTS medium
Turbocharger on engine uses
A exhaust energy to drive compressor for intake boost ✓ B crankshaft only without exhaust incorrectly alone always ✓ C electric battery only primary in standard turbo definition incorrectly alone ✓ D condenser cooling only ✓ Show Answer 💡 Explanation: Waste exhaust enthalpy increases intake density.
Q37 medium
Supercharger differs from turbocharger by being
A driven only by condenser pump incorrectly ✓ B only on steam turbine incorrectly ✓ C mechanically driven from crankshaft rather than exhaust turbine ✓ D only passive air filter incorrectly ✓ Show Answer 💡 Explanation: Supercharger draws crank work directly.
Q38 medium
Internal energy of ideal gas depends on
A pressure and volume independently always additionally incorrectly for ideal gas U ✓ B temperature only ✓ C only volume ✓ D only pressure ✓ Show Answer 💡 Explanation: u = u(T) for ideal gas.
Q39 Past Paper · PPSC/FPSC/NTS medium
Enthalpy of ideal gas depends on
A only pressure independently always for h incorrectly for ideal gas ✓ B only specific volume independently always for h incorrectly ✓ C temperature only ✓ D only quality x of steam always incorrectly for ideal gas air ✓ Show Answer 💡 Explanation: h = h(T) for ideal gas.
Q40 medium
Heat transfer at constant volume to ideal gas increases
A internal energy by Q since W=0 ✓ B enthalpy only without U change incorrectly always ✓ C nothing because W=0 incorrectly ✓ D only pressure without T change incorrectly always ✓ Show Answer 💡 Explanation: Isochoric: Q = ΔU.
Q41 medium
Heat transfer at constant pressure to ideal gas equals
A change in internal energy only always at cp incorrectly always alone ✓ B zero always ✓ C change in enthalpy ΔH ✓ D only entropy decrease incorrectly ✓ Show Answer 💡 Explanation: Isobaric: Q = ΔH.
Q42 Past Paper · PPSC/FPSC/NTS hard
Free expansion into vacuum (Joule expansion) of ideal gas has
A no temperature change and no work ✓ B large cooling always for ideal gas incorrectly always ✓ C large heating always ✓ D isentropic behavior always incorrectly ✓ Show Answer 💡 Explanation: Q=0,W=0, ideal gas ΔT=0.
Q43 hard
Gouy-Stodola theorem links
A stress and strain only ✓ B lost work to irreversibility and T0 ✓ C Fourier law only ✓ D Darcy law only ✓ Show Answer 💡 Explanation: W_lost = T0 S_gen.
Q44 medium
Saturated liquid and vapor lines meet at
A critical point on property diagrams ✓ B triple point only always incorrectly as meeting of sat lines termination on dome ✓ C origin always ✓ D absolute zero always ✓ Show Answer 💡 Explanation: Critical point top of vapor dome.
Q45 Past Paper · PPSC/FPSC/NTS medium
Isentropic expansion of steam in turbine ideally follows
A constant temperature line always incorrectly as isentropic always ✓ B constant h line on ph chart as isentropic primary incorrectly alone always ✓ C constant entropy line on T-s diagram ✓ D constant quality line in wet region always entire expansion incorrectly always ✓ Show Answer 💡 Explanation: Reversible adiabatic ⇒ Δs=0.
Q46 easy
Boiler in Rankine cycle operates ideally near
A constant volume heat addition as Otto primary incorrectly alone for boiler ✓ B constant pressure heat addition ✓ C isothermal heat rejection ✓ D adiabatic compression in boiler incorrectly ✓ Show Answer 💡 Explanation: Liquid heated and vaporized at approx constant p.
Q47 medium
Volumetric efficiency of engine is
A thermal efficiency incorrectly labeled ✓ B actual mass inducted divided by mass at STP filling displacement ✓ C mechanical efficiency incorrectly labeled ✓ D isentropic efficiency of turbine incorrectly labeled ✓ Show Answer 💡 Explanation: Breathing capacity measure.
Q48 Past Paper · PPSC/FPSC/NTS easy
Zeroth law of thermodynamics establishes
A conservation of energy ✓ B entropy always increases in isolated system ✓ C impossibility of perpetual motion only ✓ D transitivity of thermal equilibrium and temperature concept ✓ Show Answer 💡 Explanation: If A=B and B=C in thermal equilibrium then A=C.
Q49 Past Paper · PPSC/FPSC/NTS easy
First law of thermodynamics for closed system is
A ΔU = Q + W always regardless of convention ✓ B Q = 0 always ✓ C ΔU = Q − W with work done by system positive in common convention ✓ D W = m g h only ✓ Show Answer 💡 Explanation: Energy conservation: heat minus boundary work changes internal energy.
Q50 easy
Second law of thermodynamics implies
A energy is created ✓ B efficiency of heat engine can be 100% always ✓ C entropy always decreases in isolated system ✓ D natural processes tend toward increased total entropy of universe ✓ Show Answer 💡 Explanation: Clausius/Kelvin-Planck statements limit direction and efficiency.
Q51 medium
Third law of thermodynamics states entropy approaches
A infinity at 0 K always for all substances ✓ B constant at all temperatures ✓ C zero as absolute temperature approaches zero for perfect crystals ✓ D negative values only ✓ Show Answer 💡 Explanation: S → 0 at 0 K for perfect crystalline substances.
Q52 Past Paper · PPSC/FPSC/NTS easy
Enthalpy H is defined as
A U − pV ✓ B p/V ✓ C U/pV ✓ D U + pV ✓ Show Answer 💡 Explanation: H = U + pV convenient for constant pressure processes.
Q53 medium
Specific heat at constant pressure cp is
A (∂u/∂T)_v only always interchangeable without distinction ✓ B (∂p/∂T)_v ✓ C (∂v/∂T)_p only alone defines cp ✓ D (∂h/∂T)_p ✓ Show Answer 💡 Explanation: cp relates enthalpy change with temperature at constant p.
Q54 medium
Entropy change for ideal gas between two states
A depends on states not path when computed via reversible path ✓ B depends only on path always for all properties incorrectly ✓ C depends only on pressure always without temperature ✓ D depends only on volume always without mass ✓ Show Answer 💡 Explanation: Entropy is property; use reversible path between states.
Q55 Past Paper · PPSC/FPSC/NTS medium
Isentropic process for ideal gas satisfies
A p v = constant always for all processes ✓ B T/p = constant only always ✓ C p v^γ = constant ✓ D entropy increases always ✓ Show Answer 💡 Explanation: Reversible adiabatic: s = constant.
Q56 medium
Carnot cycle efficiency depends on
A temperatures of hot and cold reservoirs only ✓ B working substance only ✓ C pressure ratio only without temperatures ✓ D piston speed only ✓ Show Answer 💡 Explanation: η = 1 − T_L/T_H for reversible Carnot.
Q57 hard
Carnot theorem states no engine between two reservoirs exceeds efficiency of
A any irreversible engine always lower than Carnot incorrectly as theorem wording ✓ B 100% if operated quickly ✓ C engine independent of temperature difference ✓ D reversible Carnot engine between same reservoirs ✓ Show Answer 💡 Explanation: All reversible engines same η between given T_H and T_L.
Q58 Past Paper · PPSC/FPSC/NTS easy
Otto cycle is idealized model for
A steam turbine only ✓ B gas turbine Brayton only ✓ C vapor compression refrigerator only as power cycle ✓ D spark-ignition gasoline engine ✓ Show Answer 💡 Explanation: SI engines approximated by Otto cycle.
Q59 medium
Otto cycle efficiency increases with
A compression ratio ✓ B decreasing compression ratio ✓ C increasing exhaust back pressure only beneficial always ✓ D water injection alone as sole lever incorrectly ✓ Show Answer 💡 Explanation: η = 1 − 1/r^(γ−1) for air-standard Otto.
Q60 medium
Diesel cycle differs from Otto mainly by
A constant pressure heat addition ✓ B constant volume heat rejection only as sole difference incorrectly ✓ C no compression stroke ✓ D isothermal expansion only entire cycle ✓ Show Answer 💡 Explanation: Diesel: compression ignition with constant p heat addition.
Q61 Past Paper · PPSC/FPSC/NTS hard
Diesel cycle efficiency compared to Otto at same compression ratio is generally
A always lower without exception ✓ B always equal regardless of cut-off ✓ C zero ✓ D higher due to different heat addition path with cut-off ✓ Show Answer 💡 Explanation: Cut-off ratio affects η; comparison depends on parameters.
Q62 easy
Rankine cycle is used for
A steam power plants ✓ B gasoline engines only ✓ C vapor compression refrigeration only as primary power ✓ D hydraulic turbines only ✓ Show Answer 💡 Explanation: Boiler-turbine-condenser-pump steam cycle.
Q63 Past Paper · PPSC/FPSC/NTS medium
Rankine cycle efficiency improves by
A increasing average temperature of heat addition and lowering condenser pressure ✓ B lowering boiler temperature always beneficial incorrectly ✓ C increasing condenser pressure always beneficial incorrectly ✓ D removing pump always ✓ Show Answer 💡 Explanation: Higher T_avg add and lower T_cond increase η.
Q64 easy
Brayton cycle is model for
A gas turbine engine ✓ B Rankine steam plant only ✓ C Otto SI engine only ✓ D vapor compression refrigerator only as power primary incorrectly ✓ Show Answer 💡 Explanation: Compressor-combustor-turbine cycle.
Q65 Past Paper · PPSC/FPSC/NTS medium
Brayton cycle efficiency depends on
A only fuel type without cycle parameters incorrectly alone ✓ B only turbine blade count ✓ C pressure ratio and γ ✓ D only condenser temperature as Rankine incorrectly alone ✓ Show Answer 💡 Explanation: η = 1 − 1/rp^((γ−1)/γ) air-standard.
Q66 medium
Intercooling in multi-stage compression reduces
A turbine work always increases beneficially incorrectly as primary intercooling effect on turbine ✓ B compressor work and final temperature ✓ C mass flow to zero ✓ D cycle efficiency always decreases without exception incorrectly always ✓ Show Answer 💡 Explanation: Cooling between stages lowers work.
Q67 hard
Reheat in Brayton cycle can
A increase net work but may reduce efficiency if not optimized ✓ B always reduce work ✓ C eliminate combustion ✓ D make cycle identical to Rankine always ✓ Show Answer 💡 Explanation: Trade-off between work and heat input.
Q68 Past Paper · PPSC/FPSC/NTS medium
Combined cycle plant pairs
A only two Rankine cycles in parallel without GT incorrectly alone ✓ B Brayton topping with Rankine bottoming using exhaust heat ✓ C only refrigeration cycle ✓ D hydraulic and wind only ✓ Show Answer 💡 Explanation: Higher overall efficiency using waste heat.
Q69 medium
COP of heat pump is
A work divided by heat always inverted incorrectly ✓ B always less than 1 for heat pump incorrectly always ✓ C desired heat effect divided by work input ✓ D independent of temperatures ✓ Show Answer 💡 Explanation: COP_HP = Q_H/W.
Q70 Past Paper · PPSC/FPSC/NTS medium
COP of refrigerator is
A heating effect divided by work for refrigerator incorrectly ✓ B always greater than Carnot incorrectly always ✓ C Q_H/W for refrigerator incorrectly ✓ D cooling effect divided by work input ✓ Show Answer 💡 Explanation: COP_R = Q_L/W.
Q71 hard
Carnot COP for refrigerator between T_L and T_H is
A T_H/(T_H − T_L) which is heat pump COP incorrectly labeled as refrigerator ✓ B T_L/(T_H − T_L) ✓ C 1 − T_L/T_H ✓ D (T_H − T_L)/T_L inverted incorrectly ✓ Show Answer 💡 Explanation: Maximum COP from reversible Carnot refrigerator.
Q72 medium
Triple point of water occurs at approximately
A 100°C and 1 atm always incorrectly as triple point ✓ B 0 K ✓ C 374°C always ✓ D 0.01°C and 611 Pa ✓ Show Answer 💡 Explanation: Unique T,p where solid-liquid-vapor coexist.
Q73 Past Paper · PPSC/FPSC/NTS medium
Specific volume of saturated mixture equals
A v_g only always ✓ B v_f + x(v_g − v_f) ✓ C v_f only always regardless of x incorrectly ✓ D x v_f only incorrectly ✓ Show Answer 💡 Explanation: Linear interpolation by quality x.
Q74 medium
Enthalpy of wet steam mixture is
A h_g only always ✓ B h_f only always regardless of quality incorrectly always ✓ C h_f + x h_fg ✓ D x h_g only incorrectly alone always ✓ Show Answer 💡 Explanation: h = h_f + x h_fg.
Q75 hard
Entropy change of ideal gas between two states can be found using
A only density measurement ✓ B cp ln(T2/T1) − R ln(p2/p1) for ideal gas ✓ C only color of gas ✓ D only piston material ✓ Show Answer 💡 Explanation: Property change from state points.
Q76 Past Paper · PPSC/FPSC/NTS hard
Van der Waals equation accounts for
A only ideal gas behavior always perfectly ✓ B only radiation heat transfer ✓ C only turbulence ✓ D molecular volume and intermolecular attraction ✓ Show Answer 💡 Explanation: Real gas correction to pV = RT.
Q77 medium
Compressibility factor Z equals
A RT/(pv) inverted always as definition incorrectly ✓ B γ always ✓ C cp/cv always ✓ D pV/(mRT) or pv/(RT) ✓ Show Answer 💡 Explanation: Z deviates from 1 for real gases.
Q78 hard
Law of corresponding states suggests
A all gases identical at any T,p without reduced coordinates incorrectly ✓ B real gases at same reduced T and p have similar Z ✓ C only liquids follow ✓ D only solids follow ✓ Show Answer 💡 Explanation: Reduced properties Tr, pr used.
Q79 Past Paper · PPSC/FPSC/NTS medium
Mollier chart for steam plots
A stress versus strain ✓ B velocity versus time only ✓ C current versus voltage only ✓ D enthalpy versus entropy with pressure overlays ✓ Show Answer 💡 Explanation: h-s diagram used in steam plant analysis.
Q80 medium
Isentropic efficiency of turbine is ratio of
A isentropic to actual inverted incorrectly as definition ✓ B actual work to isentropic work for same inlet state and exit pressure ✓ C heat transfer to work incorrectly ✓ D inlet KE to exit KE only ✓ Show Answer 💡 Explanation: η_s = w_actual/w_isentropic.
Q81 medium
Isentropic efficiency of compressor is ratio of
A actual to isentropic inverted for compressor definition incorrectly ✓ B heat to entropy always ✓ C isentropic work to actual work for same states ✓ D mass flow ratios only ✓ Show Answer 💡 Explanation: η_c = w_isentropic/w_actual.
Q82 Past Paper · PPSC/FPSC/NTS easy
Heat engine thermal efficiency η is
A Q_L/Q_H incorrectly ✓ B Q_H/Q_L inverted beneficial incorrectly ✓ C work input divided by heat ✓ D net work output divided by heat input from hot reservoir ✓ Show Answer 💡 Explanation: η = W_net/Q_H.
Q83 medium
Kelvin-Planck statement of second law denies
A heat flow from cold to hot without work incorrectly as Kelvin-Planck primary alone ✓ B device producing work from single heat reservoir alone ✓ C conservation of mass ✓ D existence of temperature ✓ Show Answer 💡 Explanation: Need heat rejection to complete cycle.
Q84 medium
Clausius statement of second law denies
A work from single reservoir incorrectly as Clausius primary ✓ B increase of entropy in universe incorrectly stated as denied ✓ C spontaneous heat transfer from cold to hot body without external work ✓ D phase change existence ✓ Show Answer 💡 Explanation: Refrigerator requires work input.
Q85 Past Paper · PPSC/FPSC/NTS easy
Perpetual motion machine of first kind violates
A second law only ✓ B first law of thermodynamics ✓ C third law only ✓ D zeroth law only ✓ Show Answer 💡 Explanation: PMM1 creates energy.
Q86 easy
Perpetual motion machine of second kind violates
A second law of thermodynamics ✓ B first law only ✓ C Newton laws only ✓ D continuity equation only ✓ Show Answer 💡 Explanation: PMM2 converts all heat to work without rejection.
Q87 medium
Psychrometric chart displays
A properties of moist air including humidity ratio and enthalpy ✓ B only steam inside boiler without air ✓ C only metal stress ✓ D only beam shear flow ✓ Show Answer 💡 Explanation: HVAC analysis tool.
Q88 medium
Entropy of isolated system during real spontaneous process
A always decreases ✓ B always zero change ✓ C increases or remains constant never decreases ✓ D decreases if heat added incorrectly always ✓ Show Answer 💡 Explanation: Second law for isolated system.
Q89 Past Paper · PPSC/FPSC/NTS medium
Air-standard Otto cycle consists of
A constant pressure heat addition only as Otto incorrectly ✓ B isothermal compression only entire cycle ✓ C rankine pump stage included ✓ D isentropic compression, constant volume heat addition, isentropic expansion, constant volume heat rejection ✓ Show Answer 💡 Explanation: Four processes modeling SI engine.
Q90 hard
Steam calorimetry throttling calorimeter measures
A bolt preload only ✓ B quality of wet steam by isenthalpic expansion to superheat region ✓ C shaft diameter only ✓ D Reynolds number only ✓ Show Answer 💡 Explanation: Measure T after throttle to find x.