Mechanical Engineering 2021 Paper II 50 marks Derive

Q4

(a) Two walls A and B are maintained at temperatures T_A and T_B, respectively. One end of a metal rod of length l is embedded in the wall A, while the other end is fixed to wall B, the rod loses heat by convection to the environment at T_∞. Derive an expression to determine (i) the temperature distribution in the rod (ii) the total heat lost by the rod (iii) the heat transferred from the wall A (20 marks) (b) Air enters a constant-area duct at p_1 = 90 kPa, V_1 = 520 m/s and T_1 = 558°C. It is then cooled with negligible friction until it exists at p_2 = 160 kPa. Estimate : (i) V_2 (ii) T_2 and (iii) the total enthalpy of cooling in kJ/kg. Use attached chart. (20 marks) (c) Why is it more difficult to turbocharge spark ignition engines than compression ignition engines ? Under what circumstances might supercharger be more appropriate ? (10 marks)

हिंदी में प्रश्न पढ़ें

(a) दो दीवारों $A$ और $B$ को क्रमशः $T_A$ और $T_B$ तापमानों पर बनाए रखा जाता है। $l$ लंबाई वाली धातु की छड़ का एक सिरा दीवार $A$ में अंतःस्थापित है, जब कि दूसरा सिरा दीवार $B$ में अंतःस्थापित है। छड़ $T_\infty$ तापमान पर पर्यावरण में संवहन द्वारा ऊष्मा का ह्रास करता है। निम्नलिखित को निर्धारित करने के लिए एक व्यंजक की व्युत्पत्ति करें : (i) छड़ में तापमान वितरण (ii) छड़ द्वारा समग्र ऊष्मा ह्रास (iii) दीवार $A$ से स्थानांतरित ऊष्मा (20 अंक) (b) वायु p_1 = 90 kPa, V_1 = 520 m/s और T_1 = 558°C के एक नियत क्षेत्रफल वाली बाहिनी में प्रवेश करती है। तब इसे नगण्य घर्षण के साथ ठंडा किया जाता है, जब तक कि यह p_2 = 160 kPa पर निर्गत न हो जाए। आकलन करें : (i) V_2 (ii) T_2 (iii) शीतलन की समग्र पूर्ण-ऊष्मा (एन्थैल्पी) kJ/kg में। संलग्न तालिका का उपयोग करें। (20 अंक) (c) संपीड़न प्रज्वलन इंजनों की तुलना में स्पार्किंग प्रज्वलन इंजनों को टर्बोचार्ज करना अधिक कठिन क्यों है ? किन परिस्थितियों में उच्चदाबी निवेशक (सुपरचार्जर) अधिक उपयुक्त हो सकता है। (10 अंक)

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Directive word: Derive

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Approach

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.

Key points expected

  • 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)

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness25%12.5Correctly identifies fin with convection along length and prescribed end temperatures (not insulated tip); recognizes part (b) as Rayleigh flow with heat removal from supersonic inlet; distinguishes SI knock sensitivity from CI combustion control for part (c).Sets up fin equation correctly but applies wrong boundary conditions (e.g., insulated tip); treats part (b) as isentropic or Fanno flow; states turbocharging difficulty for SI but misses detonation mechanism.Uses 1D conduction without convection term; confuses subsonic/supersonic behavior in part (b); generic discussion of forced induction without SI/CI distinction.
Numerical accuracy20%10Part (b): M₁ = 1.073, uses Rayleigh relations correctly to find M₂ ≈ 0.7-0.8 range, V₂ ≈ 280-320 m/s, T₂ ≈ 650-700 K, cooling enthalpy ≈ 80-120 kJ/kg with consistent units; part (a) symbolic answers left in terms of given parameters.Correct M₁ calculation but errors in Rayleigh relation application or chart reading; order of magnitude correct for V₂, T₂ but specific values off by 10-20%; cooling enthalpy computed but with wrong stagnation temperature handling.M₁ calculation wrong (forgets °C to K conversion or uses V/a incorrectly); isentropic relations used instead of Rayleigh; numerical values without units or with orders of magnitude errors.
Diagram quality15%7.5Part (a): Clear rod schematic with T_A, T_B, T_∞ labeled, temperature distribution curve showing exponential decay/growth from hyperbolic solution; part (b): T-s diagram for Rayleigh flow showing cooling process, constant area duct, stagnation pressure drop; part (c): p-V diagram comparison SI vs CI with supercharger effect.Rod diagram present but missing T_∞ or convection arrows; part (b) has generic T-s diagram without specific Rayleigh line labeling; part (c) has engine schematic without thermodynamic cycle comparison.No diagrams despite problem demanding visual representation; or incorrect diagrams (e.g., linear temperature profile for part (a), isentropic process for part (b)).
Step-by-step derivation25%12.5Part (a): Full derivation from energy balance on differential element → ODE → general solution → application of two Dirichlet BCs → explicit expressions for θ(x), Q_conv, Q_A; part (b): Clear identification of governing equations (continuity, momentum, energy) for Rayleigh flow → use of attached chart or analytical solution.Jumps from ODE to solution without showing characteristic equation; states final formulas for heat transfer without integration steps; part (b) uses chart with minimal explanation of how M₂ is determined.No derivation shown—only final formulas stated; or fundamental errors in setting up differential element (e.g., missing convection term, wrong sign convention).
Practical interpretation15%7.5Part (a): Discusses when infinite fin approximation fails and finite fin with end convection matters; part (b): Comments on practical implications of cooling supersonic flow (diffuser design, heat exchanger applications); part (c): Specific Indian context—Mahindra Scorpio mHawk turbo-diesel vs petrol supercharged applications, or ISRO cryogenic engine turbopump considerations.Brief mention of practical relevance for each part without specific examples; generic statement about turbocharging in automotive industry.No interpretation provided; treats all parts as purely mathematical exercises without physical insight or engineering application.

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