UPSC Mechanical Engineering 2021

Calculate requires systematic problem-solving across all five sub-parts. For (a), draw the FBD and apply equilibrium equations to find reactions at A. For (b), use kinematic equations to find final velocity and vector dot product for the angle. For (c), define CE formula and explain its effect on hardenability and cracking susceptibility. For (d), apply the tabular method or relative velocity method for epicyclic trains. For (e), account for both external load and self-weight elongation. Allocate approximately 15% time to (a), 15% to (b), 20% to (c), 30% to (d), and 20% to (e) based on computational complexity.

• (a) Correct FBD with all applied loads shown; equilibrium equations ΣFx=0, ΣFy=0, ΣMA=0 yielding reaction forces and moment at A
• (b) Velocity at t=3.67 s: v = 6.20î + 3.36ĵ m/s; angle θ = cos⁻¹[(a·v)/(|a||v|)] ≈ 15.3° or equivalent correct calculation
• (c) CE formula: CE = %C + %Mn/6 + (%Cr+%Mo+%V)/5 + (%Ni+%Cu)/15; higher CE increases hardenability but raises risk of quench cracking and distortion
• (d) Tabular method with speed ratios: y = -28 rpm (arm fixed), x + y = 60 for gear A; solving gives arm speed ≈ 4.5 rpm in same direction as A
• (e) Total elongation δ = δ_load + δ_self = WL/AE + ρgL²/2E = 0.947 mm + 0.373 mm ≈ 1.32 mm with proper unit conversions

Calculate systematically across all three parts: for (a) locate I-centres using Kennedy's theorem and apply velocity analysis (spend ~30% time, 15 marks); for (b) solve the free vibration problem using energy methods or direct integration (spend ~30% time, 15 marks); for (c) derive SF and BM equations for all beam sections and construct diagrams (spend ~40% time, 20 marks). Begin each part with clear free-body diagrams, show all derivations, and conclude with properly labelled diagrams and physical interpretations.

• Part (a): Locate all 6 I-centres (I12, I13, I14, I23, I24, I34) using Kennedy's theorem; I13 found via auxiliary points or direct construction
• Part (a): Velocity of slider = ω₂ × (I12I24) × (I13I34)/(I12I13) or equivalent; angular velocity of connecting rod ω₃ = ω₂ × (I12I13)/(I13I34)
• Part (b)(i): Static deflection δ_st = W/k = 100/4000 = 0.025 m = 25 mm
• Part (b)(ii): Natural frequency ω_n = √(k/m) = √(4000×9.81/100) = 19.81 rad/s or f_n = 3.15 Hz
• Part (b)(iii): x(t) = (v₀/ω_n)sin(ω_nt) = 0.0505 sin(19.81t) m, since x₀=0, v₀=1 m/s downward
• Part (b)(iv): Maximum acceleration a_max = ω_n² × X_max = (19.81)² × 0.0505 = 19.81 m/s²
• Part (c): Identify loading pattern on beam (UDL, point loads, moments as shown in figure); determine reactions at supports A and B
• Part (c): Derive piecewise V(x) and M(x) equations for each region; locate points of zero shear and maximum moment

Solve all three sub-parts systematically, allocating time proportional to marks: spend ~30% on (a) for conceptual explanation with diagrams, ~30% on (b) for shaft diameter calculation using combined loading theories, and ~40% on (c) for flywheel fluctuation analysis with energy and speed variation calculations. Begin each part with stated assumptions and end with clearly boxed final answers.

• Part (a): Purpose of tempering (reduce brittleness, relieve internal stresses, improve toughness); tempering temperature ranges (150-250°C low, 350-500°C medium, 500-650°C high) with corresponding microstructures (martensite → tempered martensite/ferrite + cementite)
• Part (a): Schematic showing hardness vs tempering temperature curve, and microstructural transformation diagram with time-temperature parameters
• Part (b): Application of maximum shear stress theory or distortion energy theory for combined bending and torsion; equivalent torque Te = √(M² + T²) or equivalent moment Me = ½[M + √(M² + T²)]
• Part (b): Correct shaft diameter calculation: d = [16Te/(πτ_allowable)]^(1/3) or using allowable bending stress; τ_allowable = 500/6 = 83.33 MPa, σ_allowable = 700/6 = 116.67 MPa
• Part (c): Work done per cycle = 20,000 × (60/125) = 9600 J; expansion work = 3× compression work, so W_exp = 7200 J, W_comp = 2400 J
• Part (c): Maximum fluctuation of energy ΔE = ½ × base × height of triangle = ½ × (π/2) × (T_max - T_mean) × (π/2) for triangular TM diagram; or ΔE = ½ × W_exp = 3600 J
• Part (c): Coefficient of fluctuation of speed Cs = ΔE/(Iω²) = ΔE/(mk²ω²) where ω = 2π×250/60 = 26.18 rad/s; I = 1500 × (0.6)² = 540 kg.m²

Calculate the three distinct mechanical quantities in sequence: (a) FCC atomic packing factor and Ir atomic radius using density relations—allocate ~30% time; (b) area moment of inertia for composite triangular lamina with circular hole using parallel axis theorem—allocate ~30% time; (c) complete balancing of four rotating masses using force and couple polygon methods—allocate ~40% time as it carries highest marks. Present each part with clear headings, state assumptions, show unit conversions, and conclude with physical significance.

• Part (a): APF = 0.74 (or 74%) derived from 4 atoms per unit cell and face-diagonal relation 4R = √2a; Ir atomic radius r ≈ 1.36 × 10⁻¹⁰ m or 0.136 nm using ρ = 4M/(Nₐa³) and a = 2√2r
• Part (b): Moment of inertia of solid equilateral triangle about base = √3a⁴/32; subtract moment of inscribed circle (radius r = a/2√3) using parallel axis theorem; final I = √3a⁴/32 - πa⁴/288 = (9√3 - π)a⁴/288
• Part (c): Angular positions: B at 0°, C at 90°, D at 240°, A at 270° (or -90°); masses satisfy m_A = 3 kg, m_C = 3 kg, m_D = 4 kg from ΣF = 0 and ΣM = 0 conditions
• Force polygon closure: horizontal components m_B + m_C cos90° + m_D cos240° + m_A cosθ_A = 0; vertical components similarly
• Couple polygon closure (equal radii and spacing): moments balance about reference plane
• Unit consistency throughout: g/cm³ to kg/m³, cm to m, degrees to radians where needed; final answers with appropriate significant figures

Calculate numerical solutions for all five parts systematically, spending approximately 20% time each on (a) welding energy, (b) Taylor's tool life, (c) inventory reasons, (d) forecasting methods, and (e) gauge design. For numerical parts, show all formulas, substitutions, and unit conversions; for part (c), enumerate 6-8 well-explained reasons; for part (d), compute all three forecasts and critically evaluate method appropriateness based on trend visibility.

• Part (a): Arc power = IV = 2400 W; melting power = 0.70 × 2400 = 1680 W; volume rate = 1680/20 = 84 mm³/s
• Part (b): Using Taylor's VT^n = C, find C = 60×(180)^0.27 ≈ 228.6; then T at 80 m/min = (228.6/80)^(1/0.27) ≈ 0.89 hr; V for T=2 hr = 228.6/(120)^0.27 ≈ 69.4 m/min
• Part (c): Reasons for inventory: demand variability, supply lead time, quantity discounts, production smoothing, safety stock, work-in-progress, seasonal demand, and speculative buying
• Part (d)(i): Last value = 39; Averaging = (5+17+29+41+39)/5 = 26.2; 3-month MA = (29+41+39)/3 = 36.33
• Part (d)(ii): Last value and simple averaging inappropriate due to clear upward trend; 3-month MA better but still lags; trend projection most suitable
• Part (e): Work tolerance = 0.06 mm; gauge tolerance = wear allowance = 0.006 mm; GO gauge: 30.000 to 30.006 mm; NO-GO gauge: 30.054 to 30.060 mm (unilateral system)
• Part (e) alternative: GO gauge = 30.006⁺⁰·⁰⁰⁶ mm; NO-GO gauge = 30.060⁻⁰·⁰⁰⁶ mm with wear allowance on GO only

Calculate the required parameters for all three sub-parts systematically. For (a), apply SPT rule to sequence jobs B-A-C-D-E, then compute flow times, tardiness and performance metrics. For (b), calculate total cost and profit for each city at 40,000 units, then derive break-even volumes using FC/(Revenue-VC). For (c), use Merchant's circle and shear plane relationships with given rake angle and chip thickness ratio to find shear angle, forces and friction angle. Allocate time proportionally: ~25% for (a), ~25% for (b), ~50% for (c) given its higher marks and complexity.

• Part (a): SPT sequence B-A-C-D-E; completion times 4,10,15,23,30; total completion time = 30 days; average flow time = 82/5 = 16.4 days; average jobs in system = 82/30 = 2.73; average tardiness = 16/5 = 3.2 days
• Part (b): Total cost Chennai = 20,00,000, Delhi = 22,00,000, Kolkata = 25,00,000; Profit Chennai = 4,00,000, Delhi = 2,00,000, Kolkata = -1,00,000; Chennai most attractive; BEP Chennai = 26,667, Delhi = 30,000, Kolkata = 33,333 units
• Part (c): Shear plane angle φ = arctan(0.2×cos10°/(1-0.2×sin10°)) = 12.13°; Friction force F = F_H×sinα + F_V×cosα = 2140.8 N; Normal force N = F_H×cosα - F_V×sinα = 1030.9 N; Friction angle β = arctan(F/N) = 64.3°
• Correct formulas: SPT sequencing by ascending processing time; Total cost = FC + VC×Q; BEP = FC/(R-VC); Shear angle from r = sinφ/(cos(φ-α)); Merchant's circle relationships
• Tabular presentation for (a) showing job sequence, processing time, completion time, flow time, due date and tardiness
• Cost-volume-profit analysis table for (b) comparing all three locations with clear recommendation
• Force diagram or Merchant's circle sketch for (c) showing F_H, F_V, F, N, R and angles

Calculate the SQC control limits for part (a) using n=4, A2=0.729, D3=0, D4=2.282; derive the forging force expression in part (c) starting from equilibrium and yield criterion; explain lean systems conceptually in part (b). Allocate approximately 40% time to (a) due to three numerical sub-parts, 30% each to (b) and (c), ensuring all five sub-parts are addressed with clear headings.

• Part (a)(i): Grand mean X̿ = 1000/25 = 40 min, R̄ = 250/25 = 10 min; X̄ chart: UCL = 40 + 0.729×10 = 47.29, LCL = 40 - 0.729×10 = 32.71; R chart: UCL = 2.282×10 = 22.82, LCL = 0×10 = 0
• Part (a)(ii): σ̂ = R̄/d2 = 10/2.059 = 4.857 min; Z = (50-40)/4.857 = 2.06; P(X≤50) = Φ(2.06) ≈ 98.03% or ~98% customers
• Part (a)(iii): 2σ X̄ limits: 40 ± 2×(4.857/√4) = 40 ± 4.857 → 35.14 to 44.86 min; or using A2 factor approximation
• Part (b): Lean systems function via JIT, jidoka, kaizen, pull production, waste elimination (muda); characteristics: zero inventory, small lot sizes, quick changeovers, multifunctional workers, cellular layouts; benefits: cost reduction, quality improvement, flexibility; risks: supply chain vulnerability, worker stress, disruption sensitivity (e.g., Maruti Suzuki lean implementation challenges)
• Part (c): Forging force per unit length F/L = 2k·a·exp(2μx/h) integrated or F/L = σ̄·h·(exp(2μa/h)-1)/(μ/h) for sticking friction; assumptions: plane strain, rigid-plastic material, constant friction (Coulomb or sticking), uniform deformation, isothermal, parallel flat dies, no barreling

Calculate the ECM material removal rate using Faraday's laws with efficiency correction for part (i), then determine working gap using Ohm's law and resistivity relationship for part (ii). For part (b), compute total material handling cost by summing individual load-distance products multiplied by unit cost. For part (c), explain coordinate systems (Cartesian, polar, cylindrical) and motion control types (point-to-point, continuous path) with NC machine context. Allocate approximately 35% time to numerical parts (a), 25% to cost calculation (b), and 40% to descriptive explanation (c).

• Part (i): Apply MRR = η × (C × I × t) with C = 3.42×10⁻² mm³/A-s, I = 1800 A, convert seconds to minutes; correct answer ≈ 3.32×10⁶ mm³/min
• Part (ii): Use V = I×R = I×(ρ×y/A) to solve for gap y = V×A/(I×ρ); correct answer ≈ 0.095 mm or 95 μm
• Part (b): Calculate cost = Σ(loads × distance × ₹1/m); incoming flows to area 6 total 4400 loads × 50m, outgoing 4400 × 100m; total cost = ₹6,60,000
• Part (c): Explain machine coordinate system (absolute vs. incremental), work coordinate system, and motion control types (PTP, continuous path, contouring) with NC applications
• Part (c): Describe interpolators (linear, circular) and control loops (open, closed, semi-closed) in CNC machines
• Correct unit conversions throughout (mm³/A-s to mm³/min, ohm-mm to consistent units)
• Recognition that total loads entering area 6 (4400) equal loads leaving to area 1 (4400), indicating system balance

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)