UPSC Mechanical Engineering 2024

Solve each sub-part systematically, allocating approximately 12 minutes per 10-mark section. For (a), draw FBD and apply equilibrium with friction cone analysis; for (b), use Rayleigh's method or standard formula for simply supported beam with central mass; for (c), analyze bolt-tube assembly as statically indeterminate problem with compatibility condition; for (d), explain CCT diagram features and link to industrial hardening practice; for (e), apply balancing of reciprocating masses with analytical method. Present derivations first, then numerical substitution, with clear sub-headings for each part.

• Part (a): Correct FBD with normal reaction N, friction force μN, tension T=200N; moment equilibrium about ground contact gives tan(α) ≥ (W - Tsinα)/(Tcosα - μW); iterative or quadratic solution yields α_min ≈ 51.3°
• Part (b): Fundamental frequency f = (1/2π)√(k_eq/m_eq) where k = 48EI/L³ for simply supported beam with central load; I = πd⁴/64; calculate f ≈ 22.6 Hz
• Part (c): Compatibility condition δ_bolt + δ_tube = pitch/4 = 0.45mm; force equilibrium P_bolt = P_tube = P; solve for P using 1/k_bolt + 1/k_tube = 0.45/P; stresses σ_steel ≈ 127 MPa, σ_copper ≈ 60 MPa
• Part (d): Critical cooling rate defined as minimum cooling rate to avoid pearlite nose in CCT diagram; importance in preventing soft pearlite/ferrite formation, ensuring martensite for hardness in alloy steels like AISI 4340 used in automotive gears
• Part (e): Balancing mass m_b = [(2/3)m_rec × r + m_rev × r_rev]/r_b = [(40×0.15) + (50×0.15)]/0.4 = 33.75 kg; residual unbalance force at 60° using exact and approximate methods with ω = 26.18 rad/s

Solve all three sub-parts systematically, allocating approximately 40% time to part (b) as it carries the highest marks and involves complex flywheel design, 35% to part (a) for the pulley kinematics problem, and 25% to part (c) for the crystallography calculation. Begin each part with a clear free-body diagram or crystal structure sketch, show all derivations with proper units, and conclude with physically reasonable answers.

• Part (a): Correct FBD of pulley system with constraints; apply constraint equations relating accelerations of A, B, C; solve simultaneous equations for accelerations; integrate to find velocities at t=1s (v_A, v_B, v_C with proper signs/directions)
• Part (b): Calculate energy fluctuation from areas under turning moment diagram (convert mm² to N-m using scales); find ΔE = 16956 N-m; determine coefficient of fluctuation of speed C_s = 0.03; compute required moment of inertia I = ΔE/(C_s·ω²_mean); apply hoop stress formula σ = ρv² to find rim velocity and diameter; use I = m·k² with k≈D/2 for thin rim to find mass, then cross-section dimensions
• Part (c): Identify diamond cubic structure (8 atoms per unit cell: 8 corners×1/8 + 6 faces×1/2 + 4 internal); calculate atomic radius r = a√3/8; find packing efficiency = (√3π)/16 ≈ 34% or 0.34; compute packing density = mass of atoms in cell/volume of cell = 8×1.992×10⁻²⁶/(3.57×10⁻¹⁰)³ kg/m³
• Correct unit conversions throughout: mm² to N-m, degrees to radians, rpm to rad/s, N/mm² to Pa
• Physical reasonableness checks: flywheel diameter typically 0.5-2m for such engines; packing efficiency of diamond cubic < BCC < FCC as expected

Solve part (a) by reverse-engineering loads from given BMD shapes using d²M/dx² = -w and boundary conditions for simply supported beams. For part (b), apply stress concentration theory for elliptical holes under biaxial loading with shear, using Kirsch's solution and strain transformation to find deformed axes. For part (c), construct a comparative table covering gray, white, malleable, and spheroidal graphite cast irons. Allocate approximately 35% time to (a), 40% to (b), and 25% to (c) based on marks distribution.

• Part (a): Reverse BMD analysis—triangular BMD implies concentrated moment at midspan, parabolic BMD implies UDL, trapezoidal BMD implies combined loading; apply dV/dx = -w and dM/dx = V relationships
• Part (a): Correct identification of load cases—point loads, UDL, moments, or combinations that produce given BMD shapes with zero moments at supports A and B
• Part (b): Stress transformation to principal stresses: σ₁,₂ = (σₓ+σᵧ)/2 ± √[((σₓ-σᵧ)/2)² + τₓᵧ²] = 15 ± 9.01 = 24.01 and 5.99 MN/m²
• Part (b): Stress concentration at hole boundary using Kirsch solution; tangential stress σθ = σ₁(1+2cos2θ) + σ₂(1-2cos2θ) at r = a
• Part (b): Ellipse deformation calculation using strains ε₁ = (σ₁-νσ₂)/E and ε₂ = (σ₂-νσ₁)/E; major axis = d(1+ε₁), minor axis = d(1+ε₂)
• Part (c): Gray CI—flake graphite, ferrite/pearlite matrix, good damping, poor toughness; White CI—cementite, pearlite/ledeburite, hard, brittle; Malleable CI—temper carbon nodules, ferrite/pearlite, better toughness; SG/ductile iron—spheroidal graphite, ferrite/pearlite, best strength-toughness combination
• Part (c): Microstructural comparison with sketches of graphite morphology; mechanical properties table with tensile strength, % elongation, hardness values

Solve the three sub-parts sequentially, allocating approximately 40% time to part (a) due to its computational complexity, 40% to part (b) requiring double integration setup, and 20% to part (c) for enumeration. Begin with clear free-body diagrams for (a) and (b), show all equilibrium and integration steps, and conclude with practical applications for each mechanism.

• Part (a): Correct governor geometry with a/b = 125/65 = 1.923; sleeve lift h = 45 mm for r = 80 to 125 mm; spring stiffness s = 2S/h where S found from moment equilibrium at both speeds
• Part (a): Centrifugal forces F1 = mω1²r1 and F2 = mω2²r2; solving simultaneous moment equations yields spring stiffness ≈ 8.5-9.5 N/mm and equilibrium speed at r = 100 mm ≈ 308-312 rpm
• Part (b): Correct boundary conditions for double integration; Macaulay's method or direct integration with appropriate singularity functions for the given loading (UDL/point loads as per typical figure)
• Part (b): Deflection at x = 1 m from left end using y = ∫∫M/EI dx dx; integration constants from deflection = 0 at supports and continuity conditions; final answer typically 2-5 mm
• Part (c): Cams: radial/face/cylindrical/conical/spiral; Followers: knife-edge/roller/flat/mushroom; merits/demerits covering wear, cost, pressure angle, and manufacturing complexity
• Part (c): Specific applications: roller followers in IC engines (Maruti/Diesel locomotives), flat followers in textile machinery, knife-edge for precision instruments but high wear

Calculate numerical solutions for all five sub-parts with systematic working. For (a) apply Taylor's tool life equation VT^n=C; for (b) compute positioning accuracy from stepper motor resolution and lead screw pitch; for (c) determine cycle time and balance assembly line workstations; for (d) apply EOQ model for inventory control; for (e) perform linear regression analysis for demand forecasting. Present each part with formula, substitution, and final answer clearly labelled.

• (a) Tool life T = (C/V)^(1/n) = (300/40)^2 = 56.25 min; at 1.8× speed, new T = 17.36 min; percentage decrease = 69.1%
• (b) Pulses per revolution = 360°/3° = 120; linear displacement per pulse = 2.4mm/120 = 0.02 mm; positioning accuracy = ±0.02 mm
• (c) Available time = 2×8×3600 = 57600 sec; actual cycle time = 57600/28000 = 2.057 sec; theoretical stations = 10/2.057 ≈ 4.86; minimum stations needed = 5 (rounded up)
• (d) EOQ = √(2×2500×1200)/(0.20×1250) = 154.9 ≈ 155 boxes; total cost = purchase + ordering + carrying = ₹31,25,000 + ₹19,355 + ₹19,375 = ₹31,63,730
• (e) Regression: ΣX=1747, ΣY=1383, ΣXY=303,355, ΣX²=383,849, n=8; b=0.743, a=-2.86; Y = -2.86 + 0.743X; correlation coefficient r ≈ 0.67
• All five parts show correct formula application, unit consistency, and final answers rounded appropriately with % or ₹ symbols where relevant

Calculate the required parameters for part (a) using extrusion formulae for true strain, average strain, ideal force and work done. For part (b), construct a p-chart with given standard (2%), identify out-of-control points, revise limits after removing assignable causes, and interpret process capability. For part (c), enumerate the seven wastes of JIT and explain their connection to flow layout principles. Allocate approximately 40% time to (a), 40% to (b), and 20% to (c) based on marks distribution.

• Part (a): True strain ε = 2ln(D₀/Df) = 2ln(50/20) = 1.832; average true strain ε̄ = ε/2 = 0.916 for frictionless extrusion
• Part (a): Ideal extrusion force F = σ₀ × A₀ × ln(A₀/Af) = 300 × π×25² × 1.832 = 1.079 MN; work done W = F × L₀ × ln(A₀/Af) or σ₀ × V × ε = 300×10⁶ × π×0.025²×0.3 × 1.832 = 323.7 kJ
• Part (b): Calculate p̄ from data, establish p-chart with UCL/LCL using 2% standard; identify samples 12 (p=0.063) and 16 (p=0.013) as potentially out-of-control
• Part (b): Revise control limits after removing assignable causes; compare revised process average with 2% target; conclude on process capability and recommend actions
• Part (c): List seven JIT wastes: overproduction, waiting, transport, over-processing, inventory, motion, defects (TIMWOOD); explain how flow layout (cellular/line) minimizes transport, inventory and motion wastes
• Part (c): Connect flow layout to JIT principles: single-piece flow reduces WIP, U-shaped cells minimize walking, pull system eliminates overproduction waste

Calculate requires systematic numerical solutions for parts (a) and (b) with conceptual elaboration for part (c). Allocate approximately 40% time to part (a) for metal cutting calculations, 40% to part (b) for linear programming formulation and solution, and 20% to part (c) for quality costing elements. Begin each numerical part with clear identification of given data, apply correct formulae with derivations, and conclude with practical interpretations relevant to Indian manufacturing contexts.

• Part (a): Chip thickness ratio r = t/tc = 0.2/0.4 = 0.5; shear angle φ = arctan(r·cosα/(1-r·sinα)) = 28.18°; friction force F = Fc·sinα + Ft·cosα = 600sin5° + 300cos5° = 351.4 N; friction energy percentage = (F·Vc)/(Fc·V) × 100 where Vc = V·sinφ/cos(φ-α)
• Part (a): Normal force N = Fc·cosα - Ft·sinα = 571.6 N; coefficient of friction μ = F/N = 0.615; friction angle β = arctan(μ) = 31.6°; percentage energy for friction = (F·sinβ)/(Fc·cos(β-α)) × 100 or equivalent energy ratio approach yielding ~29-30%
• Part (b): Correct LP formulation: Maximize Z = 400000x₁ + 320000x₂ subject to 6x₁ + 4x₂ ≤ 24, x₁ + 2x₂ ≤ 6, x₂ - x₁ ≤ 1, x₂ ≤ 2, x₁, x₂ ≥ 0; corner points (0,0), (4,0), (3,1.5), (2,2), (0,2); optimal at (3, 1.5) with Z = ₹16,80,000
• Part (b): For modified objective Z = 480000x₁ + 320000x₂, re-evaluate corner points; new optimal at (4, 0) with Z = ₹19,20,000; comment on specialization in exterior paint and sensitivity to profit coefficient changes
• Part (c): Four elements of quality costing: prevention costs (training, process design), appraisal costs (inspection, testing), internal failure costs (scrap, rework), external failure costs (warranty, returns, reputation loss)
• Part (c): Benefits include: quantifying 'hidden' quality costs, prioritizing prevention over appraisal/failure, supporting ISO 9001 and TQM initiatives, enabling cost-benefit analysis of quality improvements, enhancing competitiveness of Indian MSMEs in global supply chains

Explain the EDM principle with a clear schematic diagram showing tool-workpiece gap, dielectric circulation, and pulse generator circuit, allocating ~40% time to part (a). For part (b), solve the break-even analysis stepwise using the given constraints, showing all formula derivations. For part (c), apply the centre of gravity formula with weighted coordinates, presenting calculations in tabular form. Conclude with practical implications for each part.

• Part (a): EDM principle based on controlled spark erosion between tool (cathode) and workpiece (anode) in dielectric fluid; diagram showing servo system, power supply, dielectric pump, and gap
• Part (a): Effect of parameters—pulse on-time (MRR↑, surface roughness↑), peak current (MRR↑, wear↑), gap voltage, flushing pressure; trade-off between MRR and surface finish
• Part (b): Current sales = 90,000 units; MOS = 50% of BEP → Sales = 1.5 × BEP → BEP = 60,000 units; Fixed cost = ₹6,00,000; Variable cost = ₹30/unit; MOS = 30,000 units
• Part (c): Centre of gravity coordinates X̄ = Σ(wi×xi)/Σwi = (200×125 + 450×350 + 175×450 + 350×700)/1175 = 407.45 units; Ȳ = (200×550 + 450×400 + 175×125 + 350×300)/1175 = 343.09 units
• Part (c): Location approximately (407, 343) on grid; sensitivity to weight distribution noted

Solve each sub-part sequentially with clear section headers. For (a), apply first law to each process and sum work terms; for (b), use continuity and Bernoulli to analyze diffuser condition; for (c), set up and solve the heat equation with given boundary conditions; for (d), derive the velocity ratio using stage work and Euler turbine equations; for (e), sketch three comparative boundary layer diagrams with Prandtl number reasoning. Allocate approximately 18-20 minutes per sub-part.

• (a) Process 1-2: T₂ = T₁(V₁/V₂)^(γ-1) = 423.15×(0.5)^0.4 = 323.6 K; W₁₂ = mR(T₁-T₂)/(γ-1); Process 2-3: W₂₃ = p₂(V₃-V₂) = mR(T₃-T₂); Process 3-1: W₃₁ = 0; Net work = W₁₂ + W₂₃
• (b) Nozzle can act as diffuser when A₂ > A₁ and pressure gradient reverses; continuity: ρA₁V₁ = ρA₂V₂; momentum: dp/dx = -ρV(dV/dx); energy: p₁/ρ + V₁²/2 = p₂/ρ + V₂²/2; deceleration occurs when subsonic flow in diverging section
• (c) Governing equation: d²T/dx² + q̇/k = 0; BCs: dT/dx = 0 at x = 0 (insulated), T = T_w at x = L; Solution: T(x) = T_w + (q̇/2k)(L² - x²); T_max = T_w + q̇L²/2k at x = 0; T_min = T_w at x = L; parabolic profile
• (d) Degree of reaction R = Δh_rotor/Δh_stage = 1/(k+1); Utilization factor ε = 2u(V₁cosα - u)/(V₁² + u² - 2uV₁cosα); For maximum utilization, dε/du = 0 yields u/V₁ = (k+1)cosα/(2k)
• (e) δ/δ_t = Pr^(1/3) for laminar flow; (i) Pr < 1: thermal boundary layer thicker than hydrodynamic (δ_t > δ), e.g., liquid metals; (ii) Pr = 1: δ_t = δ, gases; (iii) Pr > 1: δ_t < δ, e.g., water, oil; three clearly labelled comparative sketches

Solve each sub-part systematically, allocating approximately 40% time to part (a) Fanno flow calculation, 35% to part (b) heat transfer with phase change, and 25% combined to parts (c)(i) and (c)(ii) for derivation and calculation of radiation laws. Begin with stating given data and assumptions, proceed with step-by-step calculations using appropriate formulae and tables, and conclude with clearly boxed final answers for each sub-part.

• Part (a): Calculate inlet Mach number using Ma = V/√(γRT); determine fL*/D from Fanno table; compute actual fL/D = 0.005×25/0.05 = 2.5; find exit Mach number by interpolation from Fanno table; determine exit T and p using Fanno relations or table
• Part (b): Calculate Grashof number using β = 1/T_film, ΔT = 27K, characteristic length L = D_o = 0.03m; compute Nu = 0.48(Gr·Pr)^0.25; find h = Nu·k/D_o; calculate Q = h·A_s·ΔT; apply energy balance Q = ṁ·h_fg·(x_e - x_i) to find exit dryness fraction
• Part (c)(i): Integrate Planck's law E_bλ = C₁λ⁻⁵/[exp(C₂/λT)-1] over all wavelengths 0 to ∞ using substitution ξ = C₂/λT; show reduction to σT⁴; calculate E_b = σT⁴ = 5.67×10⁻⁸×(500)⁴ = 354.375 W/m²; total power = E_b×πD²
• Part (c)(ii): Differentiate Planck's law with respect to λ and set dE_bλ/dλ = 0; show transcendental equation leading to λ_maxT = C₃ = 2.898×10⁻³ m·K; calculate T_sun = 2.9×10⁻³/(0.55×10⁻⁶) = 5273 K
• Correct use of Fanno table interpolation for subsonic flow with friction
• Proper handling of natural convection correlation for horizontal cylinder with film temperature concept

Solve the numerical parts systematically: for (a) apply energy balance to find cooling water flow, then use continuity and LMTD to determine tube geometry; for (b) construct a comparative table highlighting operating principles, flow characteristics, and applications of both pump types, then explain slip with negative slip conditions; for (c) apply first and second law analysis using isentropic efficiency and exergy destruction concepts. Allocate approximately 40% time to part (a) given its 20 marks and four sub-parts, 35% to part (b) for detailed explanation, and 25% to part (c) for thermodynamic calculations.

• Part (a)(i): Energy balance Q = m_steam × h_fg = m_water × C_p × ΔT_water → m_water = 500 × 2600 / (4.2 × 10) = 30,952 kg/s
• Part (a)(ii): Continuity equation m_water = ρ × A × v × N → N = m_water / (ρ × πd²/4 × v) = 30,952 / (1000 × π × 0.05²/4 × 1) ≈ 15,746 tubes
• Part (a)(iii)-(iv): LMTD calculation for condenser (ΔT₁ = 60-35 = 25°C, ΔT₂ = 60-45 = 15°C), Q = U × A × LMTD → A = Q/(U×LMTD), then L = A/(N×πd)
• Part (b): Centrifugal vs reciprocating comparison—continuous vs intermittent flow, high vs low discharge, smooth vs pulsating delivery, suitable for high vs low viscosity fluids; slip = (Q_theoretical - Q_actual)/Q_theoretical; negative slip occurs with air vessel or when delivery valve opens before suction valve closes
• Part (c): Isentropic work from T₂s/T₁ = (P₂/P₁)^((γ-1)/γ), actual work = w_isen/0.90, actual T₂ from energy balance, irreversibility = T₀(s₂-s₁), second-law efficiency = w_isen/w_actual or exergy out/exergy in
• Proper unit conversions throughout: kJ/kg to J/kg, kPa to Pa where needed, temperature in Kelvin for thermodynamic calculations
• Physical interpretation: condenser design implications for NTPC plants, pump selection criteria for water supply vs high-pressure applications, compressor irreversibility sources

Solve this multi-part numerical problem by allocating approximately 40% time to part (a) given its 8 sub-parts and complexity in isentropic/shock table usage; 35% to part (b) involving iterative heat transfer coefficient determination; and 25% to part (c) requiring sketches and percentage calculation. Begin each part with stated assumptions, proceed with systematic calculations using provided tables/correlations, and conclude with clearly labelled final answers for each sub-part.

• Part (a): Correct use of isentropic relations/tables for M=2.5 to find exit pressure ratio (0.0585), temperature ratio (0.444), and area ratio (2.637); mass flow rate at throat using choked flow condition with gamma=1.4
• Part (a)(iii)-(viii): Design back-pressure (46.8 kPa), subsonic-only limit (528 kPa), normal shock at exit (136 kPa), no-shock limit (46.8 kPa), oblique shock regime (46.8-136 kPa), expansion wave regime (<46.8 kPa)
• Part (b): Energy balance q_bottom = q_top + q_conduction; iterative solution for h_top using given Nu correlation; verify T_max ≤ 400°C at bottom surface; minimum velocity ≈ 8-10 m/s
• Part (c)(i): Free convection on cooled vertical plate (T_w < T_a): velocity and thermal boundary layer profiles showing flow reversal, h_x ∝ x^(-1/4) decreasing along height
• Part (c)(ii): Nu ∝ Re^0.8 ∝ V^0.8, therefore doubling velocity gives 2^0.8 = 1.741 → 74.1% increase in h
• Appropriate air properties at film temperature: Pr ≈ 0.7, k ≈ 0.026 W/mK, ν ≈ 1.6×10^-5 m²/s for part (b)

Explain the theoretical concepts for parts (a), (b), (d), and (e) with clear diagrams, while solving part (c) with rigorous thermodynamic calculations. Allocate approximately 20% time each to (a), (b), (d), and (e) as they carry equal marks; dedicate 20% to (c) for the numerical solution. Structure as: (a) superheater types with T-s sketches, (b) cogeneration schematic with energy flow, (c) stepwise COP and power calculation, (d) SI/CI HC emission comparison with causes, (e) air washer diagram with process descriptions.

• (a) Three superheater types: radiant (wall-mounted, high temp), convective (flue gas, temp increases with load), combined; T-s diagram showing heat addition lines; desuperheater function for temperature control and turbine protection
• (b) Cogeneration principle: simultaneous electricity and useful heat production; schematic showing boiler → turbine → generator with extraction/condenser and process heat utilization; energy cascade and overall efficiency improvement
• (c) COP calculation using h_f and h_g from R-134a tables at given pressures; compressor work = h_2 - h_1; refrigeration effect = h_1 - h_4; electrical power = Q_L / COP; note: -20°C is specimen temp, not evaporator temp
• (d) SI engines emit higher UHC due to flame quenching in crevices, incomplete combustion during cold start, and fuel-rich mixtures; CI engines have lean operation but may have HC from over-mixing; causes: wall quenching, crevice volumes, oil film absorption, misfire, poor atomization
• (e) Air washer schematic: chamber with spray nozzles, eliminators, pump, recirculated water; processes: cooling and dehumidification, cooling and humidification, adiabatic saturation, heating and humidification

Calculate all performance parameters for the IC engine in part (a) using Morse test principles, then solve the heat pump cycle in part (b) using R-134a property tables and COP analysis. For part (c), sketch and explain two steam turbine governing mechanisms. Allocate approximately 45% time to (a) due to extensive calculations, 35% to (b) for thermodynamic cycle analysis, and 20% to (c) for descriptive diagrams.

• Part (a): BP = 2πNT/60 = 2π×3000×(150×0.4)/60000 = 18.85 kW; use Morse test to find IP per cylinder from load differences
• Part (a): IP_total = Σ(IP_individual) = (150-110)+(150-108)+(150-106)+(150-104) × 0.4 × 2π × 3000 / 60000 = 23.56 kW; η_mech = BP/IP = 80%
• Part (a): bmep = BP×60000/(L×A×n×N/2) = 18.85×60000/(0.1×π/4×0.075²×4×1500) = 640 kPa; imep = bmep/η_mech = 800 kPa
• Part (a): Fuel mass flow = 7.8×0.79/3600 = 1.712×10⁻³ kg/s; bsfc = ṁ_f/BP = 0.091 kg/kWh
• Part (b): From R-134a tables at 320 kPa: h₁ = 251.9 kJ/kg (sat vap), s₁ = 0.932 kJ/kgK; at 800 kPa: h₃ = h₄ = 93.4 kJ/kg (sat liq); h₂ = 420 kJ/kg (given)
• Part (b): COP_HP = (h₂-h₃)/(h₂-h₁) = (420-93.4)/(420-251.9) = 1.94; Ẇ = Q_H/COP = 75/1.94 = 38.66 MJ/h = 10.74 kW; Power saved = 75/3.6 - 10.74 = 10.1 kW
• Part (c): Throttle governing - sketch showing throttle valve before nozzles, pressure-velocity diagram, suitable for small turbines
• Part (c): Nozzle control governing - sketch showing multiple nozzle groups with valves, maintains full admission pressure, better part-load efficiency

Solve the multi-part thermodynamics problem by allocating approximately 40% time to part (a) cooling tower psychrometrics, 40% to part (b) regenerative Rankine cycle with feedwater heater, and 20% to part (c) IC engine cooling factors. Begin with clear state-point identification, apply mass and energy balances systematically, and conclude with practical implications for Indian thermal power plants.

• Part (a): Calculate specific humidity at inlet (ω₁ ≈ 0.0029 kg/kg) and exit (ω₂ ≈ 0.0286 kg/kg) using psychrometric relations or chart
• Part (a): Determine enthalpies h₁ ≈ 25.5 kJ/kg and h₂ ≈ 110 kJ/kg; total heat added = ṁₐ(h₂-h₁) ≈ 2.2×10⁶ kW
• Part (a): Evaporation loss = ṁₐ(ω₂-ω₁) ≈ 668 kg/s; WBT₁ ≈ 12°C, WBT₂ ≈ 31.5°C; volume flow change using specific volumes
• Part (a) II: Sketch psychrometric chart showing heating and humidification process (sensible heating + adiabatic saturation)
• Part (b): Apply mass balance at FWH: ṁₑₓₜ = ṁₛₜₑₐₘ × (h₂-h₃)/(hₑₓₜ-h₃) with h₃ = h_f at 600 kPa; find ṁₑₓₜ ≈ 4.2 kg/s
• Part (b): Condenser heat rejection Q̇_c = (ṁₛₜₑₐₘ-ṁₑₓₜ)(h₄-h₅); river water flow ṁ_river = Q̇_c/(c_p×ΔT_max) with ΔT = 5°C
• Part (c): Discuss factors—combustion temperature, engine speed, load, cooling system type, material conductivity, ambient conditions, lubrication

Solve this multi-part numerical problem by allocating approximately 40% time to part (a) cogeneration cycle analysis, 35% to part (b) combustion stoichiometry with excess air, and 25% to part (c) molecular structure-knock relationship. For (a), construct the T-s diagram showing extraction and mixing points; for (b), set up the combustion equation with 30% excess air; for (c), structure as brief comparative explanation with specific fuel examples.

• Part (a)(i): Condenser pump outlet temperature = saturation temperature at 7.5 kPa ≈ 40.3°C (from steam tables)
• Part (a)(ii): Total turbine output = ṁ_total×(h₁-h₂) + (ṁ_total-ṁ_extract)×(h₂-h₃) with proper extraction state determination at 500 kPa
• Part (a)(iii): Process heat transfer = ṁ_extract×(h_extract - h_return) where h_return is saturated liquid enthalpy at 100 kPa
• Part (b): Theoretical A/F = 14.57 kg air/kg fuel; with 30% excess air, actual A/F = 18.94 kg air/kg fuel
• Part (b): Dry products composition by weight: CO₂ ≈ 18.5%, N₂ ≈ 72.3%, O₂ ≈ 9.2%; by volume: CO₂ ≈ 12.1%, N₂ ≈ 75.4%, O₂ ≈ 12.5%
• Part (c): Straight-chain paraffins (n-heptane) knock more than branched iso-paraffins (iso-octane); aromatic rings and alcohols resist knock due to stable molecular structure
• Part (c): Octane rating scale reference (0-100) linking molecular structure to anti-knock quality; mention of Indian fuel standards (BS-VI)