(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4·2 J/g-K and that of ice to be half of this value and taking the latent heat of fusion of i

Question

(a) 10 g of water at 20 °C is converted into ice at –10 °C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4·2 J/g-K and that of ice to be half of this value and taking the latent heat of fusion of ice at 0 °C to be 335 J/g, calculate the total entropy change of the system. (20 marks)

(b) A shaft having diameter of 5 cm rotates in a bearing made of cast iron. The shaft rotates at 4500 r.p.m. The bearing is 15 cm long, 8 cm outer diameter and has thermal conductivity of 70 W/m-K. There is a uniform clearance between the shaft and the bearing of 0·6 mm. The clearance is filled with a lubricating oil having thermal conductivity of 0·14 W/m-K and dynamic viscosity of 0·03 N-s/m². The bearing is cooled externally by a liquid, and its outer surface is maintained at 40 °C. Disregarding the heat conduction through the shaft and assuming only one-dimensional heat transfer, determine (i) the rate of heat transfer to the coolant, (ii) the surface temperature of the shaft and (iii) the mechanical power wasted by the viscous dissipation in the lubricating oil. (20 marks)

(c) Air (C_p = 1·05 kJ/kg-K, γ = 1·38) at 3 bar pressure and T = 600 K is flowing with a velocity of 180 m/s inside a 20 cm diameter duct. Calculate the—

(i) mass flow rate;

(ii) stagnation temperature;

(iii) Mach number;

(iv) stagnation pressure assuming flow to be (1) compressible and (2) incompressible. (10 marks)

UPSC Answer Check · Accepted Answer

Calculate the required quantities for all three parts systematically. For part (a), compute entropy changes for cooling water, freezing, and cooling ice separately, then sum. For part (b), apply viscous dissipation in Couette flow with cylindrical coordinates, then use thermal resistance network for conduction through oil and bearing. For part (c), apply compressible flow relations for mass flow, stagnation properties, and Mach number, then compare compressible vs incompressible stagnation pressure results. Allocate approximately 35% time to (a), 40% to (b), and 25% to (c) based on complexity and marks distribution.
- Part (a): Three-stage entropy calculation — cooling water from 20°C to 0°C (ΔS₁ = mc_w ln(273/293)), freezing at 0°C (ΔS₂ = -mL_f/T_f), cooling ice from 0°C to -10°C (ΔS₃ = mc_ice ln(263/273)); total ΔS = ΔS₁ + ΔS₂ + ΔS₃ with c_ice = 2.1 J/g-K
- Part (b)(i)-(iii): Velocity gradient in annular gap τ = μ(du/dy) = μ(ωR_i)/c, viscous dissipation Φ = τ²/μ, heat generation Q = Φ×volume, thermal resistances R_oil = ln(R_o/R_i)/(2πk_oilL) and R_bearing = ln(R_b/R_o)/(2πk_bearingL), shaft temperature found from heat balance
- Part (c)(i): Mass flow rate using ṁ = ρAV = (P/RT)×(πD²/4)×V with R = 287 J/kg-K for air
- Part (c)(ii)-(iii): Stagnation temperature T₀ = T + V²/(2C_p), Mach number Ma = V/√(γRT), critical check whether flow is subsonic (Ma < 1)
- Part (c)(iv): Compressible stagnation pressure P₀ = P(T₀/T)^(γ/(γ-1)), incompressible P₀ = P + ½ρV² using ρ = P/RT; explicit comparison showing difference
- Unit consistency throughout: temperatures in Kelvin for thermodynamic calculations, SI units for all quantities, proper handling of kJ vs J conversion
- Physical interpretation: negative entropy change in (a) indicates irreversibility of freezing process; in (b) recognition that viscous dissipation dominates heat generation; in (c) quantification of compressibility effects at Ma ≈ 0.37

Dimension	Weight	Max marks	Excellent	Average	Poor
Concept correctness	20%	10	Correctly applies: for (a) entropy as state function with reversible path construction and proper sign conventions for heat removal; for (b) Couette flow viscous dissipation Φ = μ(du/dy)² and cylindrical thermal resistance network; for (c) isentropic stagnation relations and distinction between compressible (isentropic) and incompressible (Bernoulli) approaches. Recognizes that entropy change of surroundings must be considered for total entropy change in (a).	Uses correct basic formulas but makes errors in sign conventions for entropy (e.g., positive for all steps in (a)), or treats bearing as plane wall instead of cylinder, or applies incompressible stagnation pressure formula for both cases in (c).	Confuses entropy with enthalpy in (a), ignores viscous dissipation and treats problem as simple conduction in (b), or uses stagnation temperature formula for pressure in (c).
Numerical accuracy	25%	12.5	All numerical values accurate: (a) ΔS_total ≈ -12.8 J/K (or similar precise value with correct sign); (b) Q ≈ 83-85 W, T_shaft ≈ 52-55°C, power ≈ 83-85 W; (c) ṁ ≈ 1.17 kg/s, T₀ ≈ 615.4 K, Ma ≈ 0.367, P₀(compressible) ≈ 3.35 bar, P₀(incompressible) ≈ 3.32 bar. Temperatures consistently in Kelvin for thermodynamic calculations.	Correct methodology but arithmetic errors (e.g., factor of 10 in unit conversion, using Celsius in logarithms, or 5-10% deviation in final values due to calculation slips).	Major errors: uses grams instead of kg without conversion, applies wrong specific heat values, or order-of-magnitude errors in final answers (e.g., entropy change positive, heat transfer in kW instead of W).
Diagram quality	10%	5	Clear T-s diagram for part (a) showing three processes with areas representing heat transfer; thermal resistance network for part (b) showing R_oil and R_bearing in series with T_shaft and T_outer labelled; T-s or h-s sketch for part (c) showing stagnation state. All diagrams properly labelled with states, directions, and relevant values.	Diagrams present but incomplete labelling, or missing one part's diagram (typically (a) or (c)), or schematic sketches without proper thermodynamic representation.	No diagrams despite their utility for visualization, or incorrect diagrams (e.g., P-v instead of T-s for entropy problem, electrical circuit analogy without thermal context).
Step-by-step derivation	25%	12.5	Explicit stepwise working: (a) three separate ΔS calculations with formulas stated, then algebraic sum; (b) derivation of velocity profile, shear stress, viscous dissipation rate, volume integration for total heat, thermal resistance formulation, and simultaneous solution; (c) all formulas derived from first principles (stagnation enthalpy, speed of sound, Mach number definition) with substitution of values at each stage.	Uses standard formulas directly with substitution but skips some derivation steps; or presents final formulas without showing how constants (like R for air) are obtained.	Final answers stated without intermediate steps; or incorrect formula used throughout with no derivation to catch error; or jumps from given data to final answer.
Practical interpretation	20%	10	For (a): explains negative entropy change indicates irreversible process, relates to Second Law; for (b): discusses bearing design implications—need for adequate cooling, effect of clearance on temperature rise, material selection for bearing; for (c): comments on when incompressible assumption is valid (Ma < 0.3), relevance to aircraft engine intakes, industrial duct design. Cites Indian context: thermal power plant bearing cooling systems, ISRO rocket nozzle flows.	Brief comment on one part (typically that negative entropy in (a) is acceptable or that compressibility matters at high speeds) without connecting to engineering design.	No physical interpretation; treats as pure mathematics exercise; or incorrect interpretation (e.g., claims entropy decrease violates Second Law without mentioning surroundings).

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