UPSC Mechanical Engineering 2023 — Paper II

Solve each sub-part systematically with clear section headings. For (a), apply entropy balance for irreversible heat transfer to surroundings; for (b), derive the characteristic Mach number relation using normal shock relations; for (c), prove symmetrical velocity triangles using reaction definition; for (d), apply Planck's law and Wien's displacement law with proper unit conversions; for (e), state LMTD assumptions and compare method applicability. Allocate time proportionally: ~15% each to (a), (b), (c), (d) and ~20% to (e) combined.

• (a) Entropy change of iron: ΔS_iron = mc ln(T2/T1) = 50×0.45×ln(285/500) = -12.52 kJ/K; heat transfer Q = mcΔT = 4837.5 kJ; ΔS_atm = Q/T_atm = +16.97 kJ/K; S_gen = +4.45 kJ/K; Irreversibility I = T0×S_gen = 285×4.45 = 1268.25 kJ
• (b) Characteristic Mach number M* = V/a* where a* is critical speed of sound; using M*² = [(γ+1)M²]/[2+(γ-1)M²]; apply across normal shock using M_y² = [2+(γ-1)M_x²]/[2γM_x²-(γ-1)]; algebraic manipulation yields M*_x × M*_y = 1
• (c) 50% reaction R = (Δh_rotor)/(Δh_stage) = 0.5 implies equal enthalpy rise in rotor and stator; from velocity triangles, R = 0.5 gives V1 = V3 and V2 = V4 (mirror symmetry); blade shapes are mirror images → symmetrical blading
• (d) T = 2973 K; (i) E_bλ at 1.2 μm using Planck's law = 2.42×10¹⁴ W/m²·μm; (ii) λ_max = 2898/2973 = 0.974 μm (Wien's law); (iii) E_bλ,max using Planck's law at λ_max = 4.10×10¹⁴ W/m²·μm; (iv) E_b = σT⁴ = 3.72×10⁶ W/m²
• (e)(i) LMTD assumptions: steady-state, constant overall heat transfer coefficient U, constant specific heats, no phase change, negligible heat loss to surroundings, counter-flow or parallel-flow configuration
• (e)(ii) LMTD method: suitable when inlet/outlet temperatures are known, design problem; NTU method: suitable when effectiveness is required, rating problem, or when outlet temperatures are unknown

Solve all three parts systematically, allocating approximately 40% time to part (a) as it carries the highest marks (20), and 30% each to parts (b) and (c). For (a), establish equilibrium states of argon gas with varying water head; for (b), construct the complete T-s diagram for the modified Brayton cycle with regeneration, reheating and intercooling; for (c), use isentropic flow tables to determine choked conditions and exit properties. Present derivations clearly with state points labelled on all diagrams.

• Part (a): Initial argon pressure = P₀ + (m_piston·g)/A + (ρ_water·g·h_water)/A = 101 + 0.327 + 49.05 ≈ 150.4 kPa; mass of argon calculated using ideal gas law
• Part (a): Work done = ∫P_ext dV = area under P-V curve = P_avg·ΔV + mgh term for water lifting; final answer ~90-95 kJ
• Part (a): Heat transfer using First Law Q = ΔU + W = m·Cv·(T₂-T₁) + W; temperature rise from ideal gas law with variable pressure
• Part (b): T-s diagram shows 6 state points with two isentropic compressions (intercooled), two isentropic expansions (reheated), regenerator heat exchange, and heat addition/rejection
• Part (b): With pressure ratio 4 per stage, overall ratio = 16; regenerator effectiveness implied by 20°C temperature rise; net work and efficiency calculated from enthalpy changes
• Part (c): Check if nozzle choked: A_exit/A* = 2.0 > 1, so subsonic or supersonic solution possible; from gas tables at γ=1.4, A/A*=2.0 gives Mach ~0.3 (subsonic) or ~2.2 (supersonic)
• Part (c): Maximum flow occurs when throat is choked (M=1 at throat); use choked flow equation with stagnation conditions; mass flow rate ~0.45-0.50 kg/s
• Part (c): For supersonic exit solution: static pressure ~0.1 MPa, temperature ~175 K, velocity ~450 m/s; subsonic alternative also acceptable if justified

Calculate solutions for all three sub-parts systematically: spend ~40% time on part (a) given its 20 marks, ~35% on part (b) for 20 marks, and ~25% on part (c) for 10 marks. For (a), derive the temperature distribution from the heat conduction equation with internal generation; for (b), apply Carnot efficiency and COP relationships for the reversible heat engine-pump combination; for (c), use Euler's pump equation with slip factor and power input factor to determine impeller type. Present each part with clear headings, governing equations, substitutions, and final answers with units.

• Part (a): Governing equation d²T/dx² + q̇/k = 0 integrated to T(x) = -q̇x²/2k + C₁x + C₂; boundary conditions T(0)=200°C, T(L)=100°C applied to find C₁, C₂
• Part (a): Temperature distribution T(x) = -13.636x² - 954.55x + 200 (x in m, T in °C); maximum temperature location x = -0.035 m (outside plate, so max at left boundary) or recalculated correctly if interior
• Part (a): Heat fluxes q''_L = -k(dT/dx)|_{x=0} and q''_R = -k(dT/dx)|_{x=L} with directions stated (leftward at left face, rightward at right face)
• Part (b): Carnot efficiency η = 1 - T_L/T_H = 1 - 303/323 for engine; COP_HP = T_H/(T_H - T_L) = 423/(423-303) for heat pump; energy balance Q̇_W1 + Q̇_W2 = 5 MW
• Part (b): Work output from engine Ẇ = ηQ̇_W1 drives heat pump; simultaneous solution yields Q̇_H delivered at 150°C
• Part (c): Euler work W = σψu₂²/g where σ=1, ψ=1.04; compare with actual power to find u₂; then tanβ₂ = V_f2/(u₂ - V_w2) to determine blade angle and impeller type
• Part (c): Velocity triangles drawn with inlet radial flow (α₁=90°, V_w1=0) and outlet with calculated blade angle β₂

Derive the choking condition in part (a) by starting from isentropic relations and showing that mass flow rate maximizes at M=1. For part (b), calculate Reynolds number first, then apply the given Nusselt number correlation to find heat transfer coefficient, use the LMTD or energy balance to find tube length, then compute midpoint temperature and pressure drop. For part (c), apply unsteady filling analysis with adiabatic assumption first, then isochoric cooling to find final pressure. Allocate approximately 40% time to (a), 40% to (b), and 20% to (c) based on marks distribution.

• Part (a): Derivation showing d(m_dot)/dM = 0 leads to M=1; area-Mach relation A/A* = (1/M)[(2/(γ+1))(1+(γ-1)M²/2)]^((γ+1)/(2(γ-1)))
• Part (a): Physical explanation that sonic throat is the minimum area where upstream information cannot propagate
• Part (b): Re_D = VD/ν = 1×0.015/(0.497×10⁻⁶) ≈ 30181 (turbulent, valid for correlation)
• Part (b): Application of Gnielinski correlation (given) to find h, then energy balance Q = m_dot×Cp×(T_out-T_in) = h×A_s×LMTD to solve for length L
• Part (b): Midpoint temperature found from exponential temperature profile or iterative energy balance; pressure drop from ΔP = f(L/D)(ρV²/2)
• Part (c): Adiabatic filling gives T_tank = γT_line for ideal gas with constant specific heats; final pressure after cooling p_f = p_tank×(T_room/T_tank)
• Part (c): Final pressure calculation yielding approximately 4.5 MPa (or exact value based on γ=1.4)

Begin with part (a) presenting a clear table with seven variables and their effects on ignition delay, citing physical reasoning for each. For (b), enumerate 6-8 desirable characteristics of working fluids with brief justification. Part (c) requires defining reheat factor, then deriving the inequality RF > 1 using T-s diagram and isentropic efficiency concepts. Part (d) should present a structured comparison across 5-6 parameters (energy source, COP, components, applications). Part (e) demands systematic calculation: apply modified Apjohn equation, interpolate vapor pressures, then compute all four psychrometric properties with proper units. Allocate time proportionally: (e) ~25%, (a)-(d) ~18.75% each.

• Table in (a): Self-ignition temp ↑ → delay ↑; Cetane number ↑ → delay ↓; Compression ratio ↑ → delay ↓; Intake pressure ↑ → delay ↓; Intake temp ↑ → delay ↓; Air-fuel ratio ↑ → delay ↑; EGR ↑ → delay ↑
• Part (b): Ideal working fluid characteristics—high critical temp, low triple point, high latent heat, chemically stable, non-toxic, low cost, easy availability (water/steam, ammonia, refrigerants cited)
• Part (c): Reheat factor definition as ratio of cumulative enthalpy drop to single isentropic enthalpy drop; derivation showing RF = Σ(Δh_actual)/Δh_single > 1 due to reheat effect and increasing isentropic efficiency at lower pressures
• Part (d): Comparison table covering—energy input (work vs heat), COP range, moving parts, noise, maintenance, suitability for waste heat/remote areas (vapour absorption preferred for solar/EGR applications in Indian context)
• Part (e): Correct application of modified Apjohn equation with p = 1.0066 bar; interpolation for p'_v at 23°C (between 22°C and 24°C); calculation of p_v, then RH, ω, DPT by interpolation, and h = c_p*t + ω*h_g

Calculate all engine performance parameters in part (a) using systematic thermodynamic formulas, ensuring unit consistency throughout. For part (b), derive the optimum blade speed ratio and maximum efficiency expressions using velocity triangle analysis, with clear diagrams at ρ_optimum. For part (c), enumerate duct design methods with brief explanations. Allocate approximately 40% time to (a), 40% to (b), and 20% to (c) based on marks distribution.

• Part (a): Brake power P = WN/20000 = 67.5 kW; bmep = (60×P×1000)/(L×A×n×N/2) = 7.82 bar; bsfc = ṁ_f/P = 0.208 kg/kWh; brake thermal efficiency = P/(ṁ_f×CV) = 38.46%; volumetric efficiency = (ṁ_a×R×T_a)/(P_a×V_s×N/2×n) = 85.2%; stoichiometric air-fuel ratio from C/H=83/17 gives excess air ≈ 25%
• Part (b)(i): Velocity triangles showing inlet/outlet absolute velocities (V₁, V₂), relative velocities (V_{r1}, V_{r2}), blade velocities (u), and angles (α, β, φ, ψ) for 50% reaction stage
• Part (b)(ii): Derivation using symmetrical triangles (V_{r1}=V₂, V₁=V_{r2}), power output expression, and dP/du = 0 leading to ρ_opt = u/V₁ = cos α
• Part (b)(iii): Substitution of ρ_opt into efficiency expression η = 2ρ(cos α - ρ)/(1 - 2ρ cos α + ρ²) yielding η_max = 2cos²α/(1+cos²α)
• Part (b)(iv): Velocity triangles at ρ_opt showing α = β₂ and φ = ψ with V_{r2} perpendicular to blade or specific geometric relationships
• Part (c): Equal friction method, velocity reduction method, static regain method, and T-method (total pressure method) with brief principles of each

Calculate numerical solutions for all sub-parts with systematic working. For (a), apply isentropic flow relations for both equilibrium and supersaturated steam, determining exit area using continuity equation. For (b), use given R134a properties to compute cycle parameters and compare with reversed Carnot cycle. For (c), explain lubricant properties with clear IC engine operational implications. Allocate approximately 40% time to part (a) due to dual cases, 35% to part (b) for six numerical outputs, and 25% to part (c) for conceptual explanations.

• Part (a)(i): Determine exit enthalpy at 1 bar using s_1 = s_2 = 7.2709 kJ/kg-K, find quality x_2, then velocity from h_0 = h_2 + V_2^2/2000, and exit area A_2 = m_dot*v_2/V_2
• Part (a)(ii): Apply pv^1.3 = constant for supersaturated flow, calculate T_2 from isentropic relation, find v_2, h_2 using c_p for superheated steam, then velocity and area
• Part (b): Use 1 TR = 3.517 kW, extract h_f, h_g at -10°C and 44°C from tables, compute refrigerant effect, mass flow rate, compressor work using h_2 = h_g at 44°C + c_p(T_2' - T_2), COP = RE/W
• Part (b)(v)-(vi): Calculate Carnot COP and work, then find superheat horn loss (area under superheat on T-s diagram) and throttling loss (h_4 - h_4' on enthalpy basis)
• Part (c): Viscosity affects film strength and pumping losses; VI ensures performance across temperature; pour point determines cold-start capability; flash/fire points indicate safety against crankcase ignition
• Steam tables interpolation skills demonstrated where needed; all units consistent (kJ/kg, m/s, m², kg/s, kW, kJ/m³)
• Comparison between actual VCRS and reversed Carnot cycle clearly quantified with percentage or absolute differences

Calculate systematically across all three sub-parts: spend ~40% time on (a) Otto cycle (highest computational load with multiple salient points, efficiency, work, MEP and fuel consumption), ~35% on (b) psychrometric mixing and cooling coil analysis (requires chart reading and bypass factor application), and ~25% on (c) chimney draught and stack design (natural draught formula with temperature-dependent density). Present each part with clear headings, state assumptions, show all formulae with substitutions, and conclude with labelled diagrams as demanded.

• Part (a): Compression ratio r = (V_s + V_c)/V_c = 6.32; T2 = 562 K, p2 = 12.6 bar; T3 = 2233 K after heat addition; T4 = 1125 K, p4 = 3.97 bar; η_otto = 51.1%; W_net = 287 kJ/kg; MEP = 4.52 bar; fuel consumption = 0.202 kg/kWh
• Part (b): Mixing line on psychrometric chart yields mixture at ~37°C DBT, 50% RH; after cooling coil with BPF=0.2, outlet DBT = 15.4°C, RH ~95%; room SHF = 0.82; coil cooling capacity = 8.6 TR
• Part (c): Chimney height H = 42.3 m using draught equation Δp = gH(ρ_a - ρ_g); base diameter D = 1.85 m from continuity with mass flow; mass flow rate of flue gases = 14.4 kg/s
• P-v and T-s diagrams for Otto cycle with all four states labelled and heat/work arrows indicated; skeleton psychrometric chart showing mixing, cooling coil process, and room condition line
• Correct application of isentropic relations pV^γ = constant and T2/T1 = (V1/V2)^(γ-1) for Otto cycle; use of psychrometric relations ω = 0.622p_v/(p-p_v) and bypass factor definition
• Unit consistency throughout: pressures in bar or kPa, temperatures in K for calculations, specific volumes in m³/kg, mass flow rates in kg/s or kg/min as appropriate