UPSC Mechanical Engineering 2026 — Paper II

(a) A three-state closed-system problem on argon: 'Find the index of expansion n' requires the candidate to fix state 1 from the ideal-gas equation, get state 2 from a constant-volume energy balance using the electrical work minus 20% heat loss, and then apply the polytropic P-T relation between states 2 and 3 to extract n. Only n is asked, but P1, T2 and P2 must be computed on the way. (b) 'Find the maximum work done per kg of air and the corresponding cycle efficiency' for an ideal Brayton cycle between fixed Tmin and Tmax: the candidate must recognise that maximum specific net work occurs at the optimum pressure ratio where T2 = T4 = sqrt(Tmin x Tmax), then compute w_net,max and the efficiency at that condition (not the Carnot efficiency). (c) Three separate values are demanded for an air jet whose static temperature is 425 K at sonic condition (M = 1): (i) the speed of sound at the stagnation state, (ii) the maximum (zero-temperature) velocity attainable by isentropic expansion, and (iii) the Crocco number. The key recognition is that 'has sonic velocity' fixes M = 1 so T = T* = 425 K and T0 = 1.2 T. (d) 'Determine the heat transfer across the composite wall shown in the figure' requires building the thermal-resistance network for the five-slab wall (A in series, B and C in parallel, then D, then E in series) and evaluating Q = (T1 - T2)/R_total for the 600 degC overall temperature difference. (e) Three linked demands: (1) write the governing differential equation and both boundary conditions for steady 1-D conduction in a plane wall with constant k and no generation, (2) solve it to obtain T(x), and (3) evaluate the right-face temperature. The left boundary is a specified-flux (Neumann) condition of 750 W/m^2 and the left surface temperature 90 degC is the second condition.

• P₁ = mRT₁/V₁ = 124.9 kPa
• Net energy 17.6 kJ from 22 kJ electrical input less 20%, giving T₂ = 328.2 K and P₂ = 136.6 kPa
• State 3: T₃ = 313.15 K and P₃ = 0.5P₂
• n from T₃/T₂ = (P₃/P₂)^((n−1)/n), giving n ≈ 1.07
• Optimum condition T₂ = T₄ = √(T_min T_max) = 600 K
• Pressure ratio (T₂/T₁)^(γ/(γ−1)) ≈ 11.3
• Maximum work = 301.5 kJ/kg
• Efficiency 1 − T₁/T₂ = 50%
• M = 1 at 425 K so T* = 425 K and T₀ = 1.2T* = 510 K
• a₀ = 452.7 m/s
• V_max = √(2C_pT₀) ≈ 1012 m/s
• Crocco number = V/V_max ≈ 0.408
• Series-parallel resistance network with B and C in parallel
• Individual resistances R = L/kA with the correct areas
• R_BC = 0.01111 and R_total = 0.01911 K/W
• Q = 600/0.01911 ≈ 31.4 kW
• d²T/dx² = 0 for steady one-dimensional conduction with constant k and no generation
• Boundary conditions −k dT/dx = 750 W/m² and T(0) = 90 °C
• T(x) = 90 − 250x
• Right face T(0.25) = 27.5 °C

(a(i)) 'Show that' is a proof directive: starting from dS = m C_avg ln(T2/T1) for an incompressible substance, the candidate must argue that a thermal energy reservoir undergoes an infinitesimally small temperature change (T2 -> T1 = T) because its mass/heat capacity is very large, expand the logarithm, and recover dS = Q/T. (a(ii)) A refrigeration-load problem with two explicit sub-answers: (1) the total rate of heat removal from bread in kJ/hr, which must include both sensible cooling from 25 degC to -10 degC AND the latent heat of freezing, and (2) the volume flow rate of air in m^3/hr from the air-side energy balance with a 5 degC rise, converted using the ideal-gas specific volume at -25 degC and 101.3 kPa. (b(i)) 'Show that' the entropy rise across a normal shock in adiabatic duct flow equals -R ln(P0y/P0x): the candidate must start from the Tds/Gibbs entropy-change relation for a perfect gas written between the two stagnation states, invoke the adiabatic energy equation to prove T0y = T0x, and conclude that only the stagnation-pressure loss survives. (b(ii)) 'Differentiate between Fanno and Rayleigh flows' demands a point-by-point comparison of the two classical one-dimensional compressible-flow models: the physical assumptions, the governing conserved quantities, the shape and limit point of each curve on the h-s (or T-s) plane, and the direction of Mach-number change in each. (c) 'Determine the rate of heat loss from the wall by convection' for external parallel flow over a flat plate: convert the wind speed to m/s, form the Reynolds number on the 12 m flow length, apply the supplied combined laminar-turbulent Nusselt correlation, obtain h, and multiply by the 60 m^2 area and the 6 degC temperature difference.

• Reservoir has very large heat capacity so its temperature change is tiny, T₂ = T₁ + ΔT with ΔT ≪ T₁
• Expansion ln(1 + x) ≈ x with x = ΔT/T₁
• Q = mC(T₂ − T₁) substituted to give ΔS = Q/T₁
• Conclusion ΔS = Q/T with T ≈ T₁ ≈ T₂
• Bread mass rate 160 kg/h
• Sensible 101.5 kJ/kg plus latent 109.15 kJ/kg gives Q̇ = 33 704 kJ/h
• Air mass flow 6740.8 kg/h from ṁ c_p ΔT = Q̇
• Specific volume RT/P = 0.703 m³/kg at −25 °C and V̇ ≈ 4739 m³/h
• Gibbs relation ds = c_p dT/T − R dP/P for a perfect gas
• Stagnation states connected to the flow states isentropically
• Energy equation h₀x = h₀y gives T₀y = T₀x, so the temperature term vanishes
• Result ΔS = −R ln(P₀y/P₀x) with stagnation pressure loss
• Physical effect: friction in adiabatic flow (Fanno) versus heat transfer in frictionless flow (Rayleigh)
• Conserved quantities: mass and h₀ for Fanno, mass and impulse for Rayleigh
• Curves on the h–s plane with maximum entropy at M = 1
• Mach-number direction and choking limits of each
• Velocity 13.89 m/s and Re_L = 1.18 × 10⁷
• Nu = (0.037Re^0.8 − 871)Pr^0.33 ≈ 14 300
• h = Nu k/L ≈ 29.6 W/m²·°C with L = 12 m
• Q = hAΔT ≈ 10.6 kW with A = 60 m²

(a) Two explicit demands: 'calculate the average heat transfer coefficient between the water and the tubes' - obtained from an energy balance on the water together with the LMTD for the nearly isothermal 40 degC tube wall (the paper deliberately withholds k, nu and Pr) - and 'find the number of tubes needed to achieve the indicated heat transfer rate', which follows from the condenser duty 0.15 x 2406 kW divided by the duty carried by one tube. (b) An axial-flow compressor stage with four demands: draw the inlet and outlet velocity triangles, then determine (i) the blade angles at inlet and outlet, (ii) the power (work) input and (iii) the degree of reaction. The route is blade speed from pi D N/60, actual temperature rise from the isentropic rise divided by stage efficiency, work from Cp dT, whirl change from Euler's equation, and the angles from the constant axial velocity of 220 m/s. (c) 'Determine the air velocity above which the natural convection effects are less than 12%' is a mixed-convection criterion problem: the candidate must use the standard test Gr/Re^2 (the Richardson number), set it equal to 0.12, compute Gr at the film temperature for the 2.5 m plate, back out Re and hence the minimum forced velocity.

• Condenser duty Q = 0.15 × 2406 = 360.9 kW
• Water flow per tube 0.429 kg/s and heat per tube 14.36 kW
• LMTD = 8/ln(28/20) = 23.78 °C and h = Q_tube/(πDL × LMTD) ≈ 2564 W/m²·°C
• Number of tubes 25.1, rounded up to 26
• Blade speed U = 392.7 m/s and actual ΔT = 41.87 K from ΔT_s/η
• Work 42.08 kJ/kg and outlet whirl V_w2 = 187.2 m/s from Euler's equation
• Inlet and outlet velocity triangles with β₁ ≈ 54.9° and β₂ ≈ 43.1°
• Degree of reaction ≈ 0.66
• Criterion Gr/Re² = 0.12
• β = 1/T_film with T_film = 150 °C = 423 K
• Gr_L ≈ 1.15 × 10¹¹
• Re_L ≈ 9.8 × 10⁵ and V ≈ 11.2 m/s

(a(i)) Two connected conceptual demands: explain what NTU (= UA/C_min) physically represents in a heat exchanger, and then answer the judgement question 'is a very large NTU necessarily a good one?' - which requires the diminishing-returns argument on the effectiveness-NTU curve plus the cost/size/pressure-drop penalty. (a(ii)) Two demands: 'determine the effectiveness' and 'determine the flow configuration' of a double-pipe oil-to-oil exchanger whose arrangement is unknown. The configuration follows from the temperature cross (cold outlet 50 degC exceeds hot outlet 40 degC, impossible in parallel flow), and the effectiveness follows from eps = (dT of the C_min stream)/(Th,in - Tc,in). (b) A transient charging (filling) problem on an insulated rigid volume with four numbered answers: (i) final mass, (ii) final temperature, (iii) total entropy production and (iv) irreversibility. The candidate must recognise that the piston stays on its stops (so V = 0.25 m^3 throughout and no boundary work is done), apply the unsteady open-system energy balance m2 u2 - m1 u1 = m_in h_in, and then the entropy balance with T0 = 300 K. (c) Three linked demands for two arbitrarily oriented black surfaces: define the radiation shape (view/configuration) factor, derive a general expression for the net radiant energy exchange when they are at T1 and T2, and show the reciprocity relation A1 F12 = A2 F21 that follows when both are at the same temperature.

• NTU = UA/C_min as a dimensionless measure of the thermal size of the exchanger
• Effectiveness increases with NTU but with diminishing returns, with a ceiling
• Large NTU means large area, higher cost, weight and pressure drop
• Conclusion: not necessarily good, and the design is a cost–effectiveness trade-off
• Counterflow identified because T_c,out = 50 °C exceeds T_h,out = 40 °C
• Hot stream has the larger temperature change, so C_min = C_h
• Q_max = C_min(T_h,in − T_c,in) = 57 C_h
• ε = 35/57 ≈ 0.614
• Final pressure P₂ = F/A = 1000 kPa with the piston on the stops, V = 0.25 m³ throughout
• Initial mass 0.0645 kg and unsteady energy balance C_v(m₂T₂ − m₁T₁) = (m₂ − m₁)C_pT_in
• m₂ ≈ 0.1866 kg and T₂ ≈ 645 K
• S_gen ≈ 0.210 kJ/K and irreversibility T₀S_gen ≈ 63.1 kJ
• Shape factor as the fraction of radiation leaving one surface that reaches the other, depending only on geometry
• Double-integral expression with cosθ₁ cosθ₂/(πr²)
• Net exchange A₁F₁₂σ(T₁⁴ − T₂⁴)
• Reciprocity A₁F₁₂ = A₂F₂₁ from zero net exchange at equal temperatures

(a) 'Write the air-fuel ratio requirement with reasons' for three named SI-engine operating regimes - (i) idling/no load, (ii) cruising/part load and (iii) maximum power/full load. Each sub-part needs a numerical A/F range AND the combustion reason behind it; all three must be covered. (b) Two distinct lists are demanded: the FUNCTIONS performed by lubricating oil in an IC engine, and the essential PROPERTIES such an oil must have. Marks are split between the two halves, so both must be covered. (c) 'Describe the behaviour of steam flow in a CONVERGENT nozzle' under three named back-pressure regimes: (i) at the design (critical) pressure ratio, (ii) at a pressure ratio higher than design and (iii) at a pressure ratio lower than design. The candidate must relate each case to choking, exit Mach number and mass flow rate, and note that a convergent nozzle cannot expand supersonically. (d) 'List the deviations that occur in the COMPRESSION PROCESS of an actual vapour compression refrigeration cycle compared with the simple saturated cycle, and specify the reasons.' The scope is restricted to the compression leg - suction state, the compression path itself and the discharge - and each deviation must be paired with its cause. (e) Two demands: distinguish room ADP from coil ADP on the psychrometric chart (which line is drawn, from which state point, and what each physically represents), and state the condition under which the two become numerically equal.

• Idling: rich mixture about 11:1–13:1 because of low charge and high residual gas
• Cruising: lean mixture about 15:1–17:1 for economy and complete combustion
• Full load: rich mixture about 12:1–13:1 for maximum power and charge cooling against knock
• Reference to the stoichiometric ratio about 14.7:1
• Functions: lubrication, cooling, sealing, cleaning and corrosion protection
• Additional functions such as shock absorption
• Properties: viscosity, viscosity index, oiliness, flash point and pour point
• Properties: oxidation stability, detergency and dispersancy, low volatility and anti-foaming
• Critical pressure ratio and choking at the throat (0.546 for superheated steam)
• At the design ratio: exit at critical pressure, M = 1 and maximum mass flow
• Above the design ratio (higher back pressure): subsonic, exit pressure equal to back pressure, lower mass flow
• Below the design ratio: choked with exit pressure above the back pressure, expansion outside, mass flow unchanged
• Superheated suction vapour, to avoid liquid entering the compressor
• Suction-line and valve pressure drop
• Non-isentropic (polytropic) compression with friction and heat exchange
• Higher discharge pressure from valve and line losses, plus heat loss and volumetric and mechanical losses
• Coil ADP from the coil process line (entering to leaving state) extended to the saturation curve, the effective coil surface temperature
• Room ADP from the RSHF line through the room state extended to the saturation curve
• GSHF against RSHF distinction with the ventilation load
• Equal when RSHF = GSHF, that is 100% recirculation with no outside-air load into the coil

(a) An air-standard Otto cycle with four required answers: the theoretical thermal efficiency (from the compression pressure ratio), then (i) heat supplied per kg, (ii) net work per kg and (iii) the pressure at the end of expansion. The compression ratio must first be extracted from P2/P1 = r^gamma. (b) A single-stage impulse turbine with the condition 'exit absolute velocity is a minimum', which means axial discharge (Vw2 = 0). The candidate must find the whirl change from the given power, hence the nozzle angle, the inlet and outlet blade angles and the exit air angle, compute the diagram (blading) efficiency, and draw the combined (compound) velocity triangles. (c) 'Is this statement correct thermodynamically? Justify' - a critical-evaluation question. The candidate must recognise that the two COPs are defined on different energy inputs (high-grade shaft work versus low-grade heat), so a bare numerical comparison is not a fair thermodynamic comparison, and must supply the correct basis (ideal/Carnot COP of a heat-operated refrigerator, or second-law efficiency).

• T₁ = 320 K and T₂ = T₁(p₂/p₁)^((γ−1)/γ) ≈ 693.7 K, compression ratio r ≈ 6.92
• Efficiency η = 1 − 1/r^(γ−1) ≈ 53.9%
• (i) Q₁ = C_v(T₃ − T₂) ≈ 1008 kJ/kg; (ii) W = ηQ₁ ≈ 543 kJ/kg (T₄ ≈ 968.7 K)
• (iii) p₄ = p₁T₄/T₁ ≈ 3.03 bar
• ΔV_w = P/(ṁu) = 375 m/s; minimum exit velocity means axial exit, V_w2 = 0, V_w1 = 375 m/s
• Nozzle angle α₁ = cos⁻¹(0.75) ≈ 41.4°; V_f1 ≈ 330.7 m/s; V_r1 ≈ 374.2 m/s; β₁ ≈ 62.1°
• V_r2 = 0.92 V_r1 ≈ 344.2 m/s; cos β₂ = u/V_r2, β₂ ≈ 54.5°; V₂ = V_f2 ≈ 280.2 m/s at 90°
• Diagram efficiency 2uΔV_w/V₁² = 60%, and a labelled combined triangle with shared base u
• Numerically correct: VAR COP about 0.5–0.8 against VCR about 3–5, also for ideal cycles
• Inputs differ: VCR uses work (high grade) while VAR uses heat (low grade), so COP_VA(ideal) = (1 − T₀/T_g) × COP_Carnot
• Fair comparison needs second-law (exergy) efficiency or primary energy, with waste-heat use noted
• Verdict that lower COP is not thermodynamic inferiority

(a) A cogeneration plant with extraction: draw the schematic and T-s diagram, then determine (i) the boiler steam generation capacity in kg/s and (ii) the extraction mass flow rate to the process heater. The route is isentropic expansion from 40 bar/500 degC (s = 7.09) to 2 bar and to 0.1 bar using the printed saturated-water pressure table, then a process-heater energy balance and a net-power balance. (b) Two complete cycle calculations on ammonia (R717) plus a P-h diagram: (a) the simple saturation cycle between -16 degC and 40 degC - determine COP and power per TR; (b) the same cycle with a liquid-vapour regenerative heat exchanger raising the suction to 15 degC - determine the new COP and power per TR and state the effect. The superheated discharge enthalpy must be found from the entropy balance using the given vapour specific heats. (c) 'How are the ratings of SI and CI engine fuels done? Explain in brief' requires both rating systems: the octane number for SI fuels and the cetane number for CI fuels, each with its reference fuels, the standard test engine and procedure, and typical values. Both halves carry marks.

• Schematic and T–s diagram with the extraction point, process heater, both pumps and condenser
• Isentropic expansion at s = 7.09: x₂ = 0.9934, h₂ ≈ 2692 kJ/kg and x₃ = 0.8587, h₃ ≈ 2246 kJ/kg from the tables
• (ii) ṁ_e = 1500/(h₂ − h_f2) ≈ 0.686 kg/s
• (i) Net power balance giving ṁ ≈ 5.28 kg/s (5.26 kg/s without pump work)
• Simple cycle: T₂ ≈ 390.7 K from s₂ = s₁ with c_p = 3.1, h₂ ≈ 1730.9; COP ≈ 3.65 and about 0.964 kW/TR
• Heat exchanger: h₁′ = 1516.4, s₁′ ≈ 6.113, h₄′ ≈ 316.8, T₂′ ≈ 435.6 K (c_p = 2.9), h₂′ ≈ 1845.5
• COP ≈ 3.42 and about 1.028 kW/TR; refrigerating effect 1125.8 against work 329.1 kJ/kg
• Effect stated: COP falls and power per TR rises, and a correct P–h diagram of both cycles
• Octane number: iso-octane 100 and n-heptane 0, CFR variable compression ratio engine, knock intensity matching
• RON and MON test conditions (600 and 900 rpm) with typical values
• Cetane number: n-cetane 100 and α-methylnaphthalene 0 (or heptamethylnonane 15), ignition-delay matching in a CFR diesel engine
• Typical cetane values 40–55 and the opposite trend of the two scales

(a) A summer air-conditioning problem with an air washer and 25% fresh air. Three numbered answers are required - (i) the volume flow rate of supply air, (ii) the room sensible heat factor and (iii) the cooling load of the air washer in TR - together with a schematic diagram and the processes marked on a skeleton psychrometric chart. (b(i)) 'Describe the method of finding indicated power of an IC engine using Morse test' - a short procedural note requiring the test conditions, the cut-out sequence, the working formula and the restriction to multi-cylinder engines. (b(ii)) A four-cylinder four-stroke engine test with four numbered answers: (1) brake thermal efficiency, (2) indicated thermal efficiency, (3) indicated mean effective pressure and (4) brake specific fuel consumption. Brake power comes from the rope-brake data (net tension x drum circumference x speed), and imep must use the four-stroke, four-cylinder form of the IP equation. (c) A natural-draught chimney problem with four numbered answers: (i) draught in mm of water, (ii) equivalent draught as a column of hot gas in metres, (iii) chimney efficiency given an artificial-draught flue-gas temperature of 120 degC, and (iv) the percentage of the fuel's calorific value spent in producing the natural draught.

• Supply mass flow 1.60 kg/s from 24.5/(c_pm × 15) and volume flow about 1.30 m³/s (v_s ≈ 0.81 m³/kg)
• RSHF ≈ 0.73 from total room load 33.6 kW (latent about 9.1 kW)
• Mixed state M from 75% room and 25% outside air: W_m ≈ 0.0117, h_m ≈ 58.8 kJ/kg (T_m ≈ 28.8 °C)
• Washer load ṁ_a(h_m − h_s) ≈ 47 kW ≈ 13.4 TR, with a schematic and chart showing O, R, M, S and the room ratio line
• Constant speed and throttle setting, measure brake power B with all cylinders firing
• Cut out one cylinder (spark shorted for SI, fuel cut for CI) and restore the speed by reducing the load
• I_j = B − B_j for each cylinder, and total IP = nB − ΣB_j
• Assumptions and limits: friction power constant, multi-cylinder engines only
• Brake power = 450 × 3 × 25 = 33.75 kW and fuel energy 146.67 kW
• (1) Brake thermal efficiency ≈ 23.0%
• (2) IP = 42.19 kW and indicated thermal efficiency ≈ 28.8%
• (3) imep ≈ 8.59 bar with four cylinders and N/2 power strokes; (4) bsfc ≈ 0.356 kg/kWh
• Densities ρ_a ≈ 1.185 and ρ_g ≈ 0.655 kg/m³ with 17 kg flue gas per kg fuel
• (i) h_w ≈ 18.55 mm of water
• (ii) h_g = h_w/ρ_g ≈ 28.3 m of hot gas
• (iii) η ≈ 0.154% and (iv) heat spent ≈ 9.3% of 33 000 kJ/kg