UPSC Mechanical Engineering 2021 — Paper II

Begin with part (a) using 'discuss' to compare fan/blower/compressor pressure ratios and applications; for (b) 'prove' requires rigorous derivation using Fanno/Rayleigh flow or area-velocity relations showing shock requires supersonic upstream Mach number. Part (c) demands 'solve' for temperature distribution and heat transfer using thermal resistance network. Part (d) involves 'compare' to extract C and m through dimensional analysis matching Gr and Pr exponents. Part (e) requires 'calculate' percentage stroke using constant pressure heat addition with variable specific heat integration. Allocate time proportionally: ~15% each for (a), (b), (d); ~25% each for (c) and (e) due to numerical complexity.

• Part (a): Fan (<1.1 pressure ratio, low pressure rise), Blower (1.1-2.5 pressure ratio, medium flow), Compressor (>2.5 pressure ratio, high pressure rise); cite applications like HVAC fans, FD/ID blowers in Indian power plants, multi-stage compressors
• Part (b): Prove using dA/A = (M²-1)/M² * dV/V; for shock, entropy must increase requiring dS>0 which demands supersonic M>1 upstream; or use Prandtl relation M₁*M₂=1 showing M₁>1 needed for real M₂
• Part (c)(i): Derive T(x) = T₁ - (T₁-T₂)x/L with T₂ found from convection boundary; linear temperature profile in wall
• Part (c)(ii): Q = (T₁-Tₐ)/(L/kA + 1/hA) = 2934.8 W; thermal resistance network with conduction and convection resistances in series
• Part (d): Match exponents to find m=0.25 from (T_s-T_∞) exponent 0.33 and Gr∝ΔT; calculate C=0.525 using properties at 100°C with β=1/T_film, ν=μ/ρ
• Part (e): Integrate CᵥdT from 870K to T₃ using Q=m_f*CV*(A/F+1); find T₃=2154K; use V₃/V₂=T₃/T₂ for constant pressure; percentage stroke = (V₃-V₂)/(V₁-V₂)*100 = 23.8%

Calculate numerical solutions for all three sub-parts with systematic derivations. For (a)(i), apply polytropic relations for steady flow; for (a)(ii), distinguish flow work from shaft work using control volume analysis. For (b), integrate the sine velocity profile to obtain displacement and momentum thickness. For (c), apply Brayton cycle analysis with given pressure ratio and temperature limits. Allocate approximately 40% time to part (a) combined, 35% to part (b), and 25% to part (c) based on mark distribution.

• Part (a)(i): T2 = 40*(1000/100)^(0.25/1.25) = 40*10^0.2 = 63.4 K → T2 = 313.4 K; W_shaft = n/(n-1)*mR(T1-T2) = 5*3*0.287*(313-40) = -1176 kJ; Q = W*(γ-n)/(γ-1) = -1176*(1.4-1.25)/0.4 = -441 kJ; ΔS = m[Cp*ln(T2/T1) - R*ln(P2/P1)]
• Part (a)(ii): pdv work is boundary work for closed systems (quasi-static process); -vdp is flow work or shaft work in steady flow (Euler work equation); relate via steady flow energy equation
• Part (b): δ* = ∫[0,δ](1 - u/U0)dy = δ[1 - 2/π] = 0.363δ; θ = ∫[0,δ](u/U0)(1 - u/U0)dy = δ[2/π - 1/2] = 0.137δ; show integration steps with substitution
• Part (c): T2/T1 = (18)^(0.4/1.4) = 2.58; T2 = 293*2.58 = 756 K; T4 = T3/2.58 = 973/2.58 = 377 K; η = 1 - 1/(r_p)^((γ-1)/γ) = 1 - 1/2.58 = 61.2%; Q_add = Cp(T3-T2) = 1.0035*(973-756) = 217.8 kJ/kg
• All parts: State assumptions clearly, show unit consistency, and verify results against physical expectations

Derive the effective conductance expression for part (a)(i) using radiation network analysis with re-radiating surfaces; for (a)(ii) set up heat balance equations for two shields and solve the resulting system for steady-state temperatures. For part (b), derive the laminar sublayer thickness by equating shear stresses at the interface using the Blasius friction factor and the 1/7th power law velocity profile. For part (c), explain reheating effects on thermal efficiency, specific work output, and turbine blade cooling requirements, citing operational trade-offs in Indian power plants like NTPC gas turbine stations.

• Part (a)(i): Derivation using radiation network with two black surfaces and re-radiating walls; application of reciprocity and summation rules to obtain A₁F̄₁₂ = A₁(1+F₁₂)/2
• Part (a)(ii): Heat balance equations for two shields: q = σ(T₁⁴-Tₛ₁⁴)/(1/ε₁+1/εₛ₁-1) = σ(Tₛ₁⁴-Tₛ₂⁴)/(1/εₛ₁+1/εₛ₂-1) = σ(Tₛ₂⁴-T₂⁴)/(1/εₛ₂+1/ε₂-1); solution yields Tₛ₁ ≈ 467 K, Tₛ₂ ≈ 389 K
• Part (b): Blasius relation f = 0.0791/Re^0.25; wall shear stress τ_w = fρū²/8; velocity gradient in sublayer du/dy = τ_w/μ; equate with 1/7th law at edge of sublayer to obtain δ_s = 11.6ν/u* where u* = √(τ_w/ρ)
• Part (b): Integration of 1/7th power law to find average velocity ū = 49u_c/60; substitution to express δ_s in terms of Re and pipe diameter
• Part (c): Reheating increases specific work output (w_net ↑) but decreases thermal efficiency (η_th ↓) due to additional heat addition at lower average temperature; reduces compressor work ratio; mention blade cooling challenges in Indian tropical conditions

Derive the governing differential equation for extended surface heat transfer in part (a), applying appropriate boundary conditions for the convective tip case. For part (b), use compressible flow relations or the attached chart to solve for Rayleigh flow with heat removal, identifying whether flow is subsonic or supersonic at inlet. For part (c), explain the detonation limits in SI engines versus knock tolerance in CI engines, then justify supercharger selection for specific duty cycles. Allocate approximately 40% time to (a), 35% to (b), and 25% to (c) based on marks distribution.

• Part (a): Governing equation d²θ/dx² = m²θ where m² = hP/kA; general solution θ = C₁e^(mx) + C₂e^(-mx) with θ = T - T_∞
• Part (a): Boundary conditions θ(0) = θ_A = T_A - T_∞ and θ(l) = θ_B = T_B - T_∞; solve for C₁, C₂ to get temperature distribution
• Part (a): Heat lost by convection integral ∫₀ˡ hP(T-T_∞)dx and heat from wall A = -kA(dT/dx)|_{x=0}
• Part (b): Inlet Mach number calculation: a₁ = √(γRT₁) = 484.5 m/s, M₁ = 1.073 (supersonic); use Rayleigh flow relations or chart for cooling
• Part (b): For Rayleigh flow with heat removal from supersonic inlet: M₂ found from p₂/p₁ = 1.778 using chart; V₂ = M₂a₂, T₂ = T₀₂/(1+0.2M₂²)
• Part (b): Total enthalpy of cooling = c_p(T₀₁ - T₀₂) where stagnation temperatures relate through Rayleigh relation
• Part (c): SI engine limitation: narrow flammability range, knocking tendency with increased pressure/temperature; CI engine: wider ignition delay control, no knock issue
• Part (c): Supercharger preferred over turbocharger at low engine speeds, high altitude operations, or when instant response needed (e.g., mining vehicles, marine applications)

Begin with a concise definition of EGR for part (a), then explain the thermal and dilution mechanisms for NOₓ reduction. For part (b), set up the combustion equation using Orsat data and solve systematically for fuel composition, A/F ratios, and excess air—this numerical part demands ~35% of time. For (c), define nozzle efficiency and velocity coefficient with clear formulas. For (d), explain the balance point concept with a pressure-enthalpy diagram. For (e), outline the Equal Friction Method steps with a simple duct layout example. Conclude each descriptive part with practical implications for Indian conditions (e.g., BS-VI norms, tropical climate design).

• Part (a): EGR definition as Exhaust Gas Recirculation; thermal effect (reduced flame temperature below 1800K) and dilution effect (lower O₂ concentration) suppressing NOₓ formation via Zeldovich mechanism
• Part (b): Carbon balance to find x=7.3, hydrogen balance to find y=14.6 giving fuel formula C₇.₃H₁₄.₆ or approximated as C₇H₁₅; stoichiometric O₂ from (x+y/4); actual O₂ from N₂=82.3% using 79:21 ratio
• Part (b): A/F mole ratio ~18.5, mass ratio ~12.8; excess air ~35% using (actual-theoretical)/theoretical × 100
• Part (c): Nozzle efficiency ηₙ = (actual KE)/(isentropic KE) = (h₁-h₂)/(h₁-h₂s); Velocity coefficient Cᵥ = √(actual KE/isentropic KE) = √ηₙ; typical values 0.95-0.99
• Part (d): Balance point as intersection of compressor pumping curve and capillary tube flow characteristic; subcooling control; stable operation zone; effect of condensing pressure variation
• Part (e): Equal friction method maintains constant pressure drop per unit length (typically 0.5-1.0 Pa/m); sizing ducts using friction chart/Moody diagram; iterative velocity adjustment; suitable for Indian commercial buildings with uniform loading

Explain the fundamental combustion differences between CI and SI engines with clear physical reasoning for part (a), supported by well-labelled sketches of swirl chambers. For part (b), set up the regenerative cycle analysis methodically: identify states, apply energy balance at the feed heater, and compute cycle efficiency with proper steam table interpolation. For part (c), apply psychrometric relations precisely to find relative humidity. Allocate approximately 40% effort to part (a) given its 20 marks and diagrammatic demand, 40% to part (b) for its computational complexity, and 20% to part (c).

• Part (a)(i): CI engines use heterogeneous diffusion-controlled combustion with fuel injection late in compression; SI engines use homogeneous premixed flame propagation with spark ignition; contrast flame speed, ignition delay, and combustion duration characteristics
• Part (a)(ii): Combustion-induced swirl is organized air motion created by piston motion into specially shaped pre-chamber or main chamber; explain how it enhances air-fuel mixing and reduces ignition delay
• Part (a)(ii): Two designs—Swirl Chamber (Ricardo Comet V) with tangential throat creating vigorous swirl, and Pre-combustion Chamber with restricted passage creating pressure differential and swirl; both sketched with flow arrows
• Part (b): Identify turbine inlet (state 1: 30 bar, 400°C), extraction pressure (0.3 MPa ≈ 3 bar), and condenser pressure (0.1 bar); determine state 2s (isentropic expansion to 3 bar) using s1 = s2 and steam tables
• Part (b): Energy balance at direct-contact heater: y·h2 + (1-y)·h4 = h6 where y is extraction fraction, h4 = hf at 0.1 bar, h6 = hf at 3 bar; solve for y
• Part (b): Compute h3s (isentropic expansion to 0.1 bar), actual h3 if efficiency given (or assume ideal), then W_turbine and Q_boiler; efficiency η = (W_net)/Q_boiler with pump work neglected
• Part (c): Apply ω = 0.622·(φ·ps)/(p - φ·ps) or rearrange to find φ = ω·p/(0.622·ps + ω·ps); substitute ω = 0.0115 kg/kg, p = 90 kPa, ps = 4.24 kPa
• Part (c): Correct numerical substitution yielding φ ≈ 0.305 or 30.5%; show intermediate calculation of partial pressure of vapour pv = ω·p/(0.622 + ω)

Calculate numerical solutions for all six sub-parts systematically. For (a)(i), apply velocity triangle analysis for impulse turbine; for (a)(ii), explain super-saturation with T-s diagram. For (b), use psychrometric chart data to solve air-conditioning calculations with mixing of fresh and recirculated air. For (c), compute fuel injection parameters using discharge equations. Allocate ~35% time to (a) parts combined (40 marks), ~35% to (b) (20 marks), and ~30% to (c) (10 marks), ensuring all derivations are shown stepwise with proper units.

• (a)(i) Blade speed u = πDN/60 = 188.5 m/s; velocity triangles constructed with α₁=18°, ρ=0.42; β₁ and β₂ calculated using relative velocity ratio 0.9 and β₂ = β₁ - 3°
• (a)(i) Axial thrust = ṁ(V_{w1} - V_{w2}) or ṁ(V_{f1} - V_{f2}) depending on velocity components; power = ṁ(V_{w1} + V_{w2})u or ṁV_{w}u for impulse
• (a)(ii) Super-saturated flow: steam expands below saturation line without condensation, T-s diagram shows metastable region with Wilson line; mass flow rate increases due to higher density than equilibrium conditions
• (b) Psychrometric properties: room air ω₁, h₁ from chart; supply air at 7°C saturated gives ω₂, h₂; mass flow rate from sensible heat equation Q_s = ṁc_p(ΔT)
• (b) Latent heat gain from moisture difference and total heat balance; cooling load of air washer using energy balance on mixing (30% fresh, 70% recirculated) and conditioning process
• (c) Fuel flow rate from BSFC and power; average pressure difference = [(220-30)+(550-60)]/2 = 340 bar; injection time from 15° crank angle at 1500 rpm; nozzle area from ṁ_f = C_d A √(2ρΔp)

Solve the three sub-parts sequentially, allocating approximately 40% time to part (a) COP calculation using ammonia properties, 40% to part (b) cogeneration mass flow rate with energy balance, and 20% to part (c) T-s diagram comparison of throttling processes. For each numerical part, state assumptions clearly, show property extraction from tables, and present final answers with units. For part (c), draw two distinct T-s diagrams with clear labelling of states and entropy changes.

• Part (a): Identify T1 = 266.3 K, T2 = 299.7 K; use h_f, h_g, s_f, s_g at given temperatures from ammonia tables; calculate h1, h2, h4 with dry compression and isentropic expansion assumptions; COP = (h1-h4)/(h2-h1)
• Part (b): Apply steady flow energy equation to HP turbine, heat exchanger, and LP turbine; use given h at 1 MPa, 200°C; find h at 0.3 MPa (saturated liquid for bleed, isentropic expansion for remaining steam to 40 kPa); set up simultaneous equations for power (1 MW) and heating (500 kW) to solve for mass flow rates
• Part (c)(i): Throttling at turbine inlet for governing — high pressure steam, large pressure drop, significant temperature drop, entropy increase, moves toward saturation, used for load control
• Part (c)(ii): Throttling at closed feed heater trap exit — saturated liquid at lower pressure, smaller enthalpy drop, minimal temperature change, entropy increase but different magnitude, used for condensate return
• Part (c): Two T-s diagrams showing: (i) superheated steam throttling with large Δs, moving closer to saturation line; (ii) subcooled/saturated liquid throttling with smaller Δs, remaining in liquid region or entering wet region slightly
• Correct use of steam table data: at 0.3 MPa, identify h_f for bleed condensate; for LP turbine exit, find quality or h at 40 kPa using s_2 = s_3 (isentropic)
• Mass balance: m_total = m_bleed + m_LP; energy balances: W_turbine = m_total*(h1-h2) + m_LP*(h2-h3); Q_heater = m_bleed*(h2-h_f at 0.3 MPa)