Mechanical Engineering 2025 Paper II 50 marks Compulsory Calculate

Q1

(a) A lump of ice with a mass of 1·5 kg at an initial temperature of 260 K melts at the pressure of 1 bar as a result of heat transfer from the environment. After some time, the resulting water attains the temperature of environment, 293 K. Calculate the entropy production associated with this process. The latent heat of fusion of ice is 334 kJ/kg, the specific heat of ice and water are 2·07 kJ/kg-K and 4·2 kJ/kg-K, respectively. Assume that ice melts at 273·15 K. (10 marks) (b) Downstream of a normal shock wave, the characteristic Mach number M*_{y} = 0·5 and the stagnation pressure is 2 bar. Determine the following: (i) The Mach number M_{y} downstream of the shock (ii) The stagnation pressure upstream of the shock The fluid is air and γ = 1·4, and R = 0·287 kJ/kg-K. Normal shock table and Isentropic flow table provided at the end, may be used. (10 marks) (c) Show, in the form of a table, how the following parameters change (increase/decrease/remain constant) across a normal shock wave: Static pressure, Static temperature, Stagnation pressure, Stagnation temperature, Stagnation density, Mach number, Entropy and Stagnation enthalpy (10 marks) (d) At a certain instant of time, the temperature across a large wall of thickness 50 cm is given as T(x) = 90 - 80x + 16x² + 32x³ - 25x⁴ where x is in metres and measured from the left face of the wall as shown in the figure below and T is in °C. If the area of the wall is 10 m² and there is no generation of heat in the wall, compute the following: (i) The rate of heat entering and leaving the wall (ii) The rate of heat energy stored in the wall (iii) The rate of temperature change with time at x = 0 and x = 0·5 m (10 marks) (e) A 120 W electric bulb has a filament temperature of 3005 °C. Assuming the filament to be black, calculate (i) the diameter of the filament wire if the length is 250 mm and (ii) the efficiency of the bulb based on visible radiation if the visible radiation lies in the wavelength range from 0·4 μm to 0·75 μm. Assume Stefan-Boltzmann constant (σ) as 5.67×10⁻⁸ W/m²-K⁴. The black body radiation functions are given in the table: | λT(μm-K) | f₀→λ | |----------|------| | 1300 | 0.004317 | | 1400 | 0.007791 | | 2400 | 0.140266 | | 2500 | 0.161366 | (10 marks)

हिंदी में प्रश्न पढ़ें

(a) एक बर्फ का टुकड़ा, जिसका द्रव्यमान 1·5 kg और प्रारंभिक तापमान 260 K है, वातावरण से ऊष्मा स्थानांतरण के कारण 1 bar दाब पर पिघलता है। कुछ समय पश्चात्, परिणामी जल वातावरण के तापमान 293 K को प्राप्त करता है। इस प्रक्रिया से संबंधित एंट्रॉपी उत्पादन की गणना कीजिए। बर्फ के पिघलने की गुप्त ऊष्मा (लैटेंट हीट) 334 kJ/kg है, बर्फ और पानी की विशिष्ट ऊष्माएँ (स्पेसिफिक हीट) क्रमशः: 2·07 kJ/kg-K तथा 4·2 kJ/kg-K हैं। मान लीजिए कि बर्फ 273·15 K पर पिघलती है। (10 अंक) (b) सामान्य प्रयात-तरंग (शॉक वेव) के अनुप्रवाह (डाउनस्ट्रीम) में अभिलक्षणिक मैक संख्या M*_{y} = 0·5 तथा स्थगन दाब (स्टैग्नेशन प्रेशर) 2 bar है। निम्नलिखित को निर्धारित कीजिए: (i) प्रयात (शॉक) के अनुप्रवाह में मैक संख्या M_{y} (ii) प्रयात के प्रतिप्रवाह (अपस्ट्रीम) में स्थगन दाब तरल, वायु है और γ = 1·4 तथा R = 0·287 kJ/kg-K है। अंत में दिये गये सामान्य प्रयात (शॉक) तालिका तथा आइसेंट्रॉपिक प्रवाह तालिका का उपयोग किया जा सकता है। (10 अंक) (c) सारणी के रूप में दर्शाइये कि निम्नलिखित प्राचल सामान्य प्रयात तरंग (शॉक वेव) के लिए कैसे परिवर्तित होते हैं (बढ़ते हैं/घटते हैं/स्थिर रहते हैं): स्थैतिक दाब, स्थैतिक तापमान, स्थगन दाब, स्थगन तापमान, स्थगन घनत्व, मैक संख्या, एंट्रॉपी और स्थगन एन्थैल्पी (10 अंक) (d) किसी एक निश्चित समय पर 50 cm मोटाई की एक बड़ी दीवार पर तापमान को निम्न प्रकार दर्शाया गया है: T(x) = 90 - 80x + 16x² + 32x³ - 25x⁴ जहाँ x मीटर में है तथा नीचे चित्र में दर्शाये अनुसार दीवार के बायें पृष्ठ से मापा गया है एवं T डिग्री सेल्सियस में है। यदि दीवार का क्षेत्रफल 10 m² है और दीवार में ऊष्मा का कोई उत्पादन नहीं हो रहा है, तो निम्नलिखित की गणना कीजिये: (i) दीवार में प्रवेश करने और दीवार से बाहर निकलने वाली ऊष्मा की दर (ii) दीवार में संग्रहीत ऊष्मा ऊर्जा (हीट एनर्जी) की दर (iii) x = 0 तथा x = 0.5 m पर समय के साथ तापमान परिवर्तन की दर (10 अंक) (e) एक 120 W विद्युत बल्ब के तंतु (फिलामेंट) का तापमान 3005 °C है। तंतु को काला मानते हुए निम्नलिखित की गणना कीजिये: (i) तंतु तार का व्यास, यदि उसकी लम्बाई 250 mm है (ii) बल्ब की दक्षता, दृश्यमान विकिरण (विजिबल रेडिएशन) के आधार पर, यदि दृश्यमान विकिरण की तरंगदैर्ध्य सीमा (वेवलेंथ रेंज) 0.4 μm से 0.75 μm के बीच है स्टीफन-बोल्ट्ज़मान नियतांक (σ) को 5.67×10⁻⁸ W/m²-K⁴ मानिये। कृष्णिका (ब्लैक बॉडी) विकिरण फलन (रेडिएशन फंक्शन) सारणी में दिये गये हैं: | λT(μm-K) | f₀→λ | |----------|------| | 1300 | 0.004317 | | 1400 | 0.007791 | | 2400 | 0.140266 | | 2500 | 0.161366 | (10 अंक)

Directive word: Calculate

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How this answer will be evaluated

Approach

Calculate systematically across all five parts: (a) entropy production for ice melting requires three-stage entropy analysis; (b) normal shock uses characteristic Mach number relations and tables; (c) tabular comparison of shock parameters; (d) transient heat conduction with polynomial temperature profile; (e) blackbody radiation with fractional emission. Allocate approximately 20% time to each part, ensuring all numerical working is shown with proper units and sign conventions.

Key points expected

  • (a) Three-stage entropy calculation: ice warming 260→273.15 K, melting at 273.15 K, water warming 273.15→293 K; entropy production = ΔS_system + ΔS_surroundings with Q/T_boundary for surroundings
  • (b)(i) Characteristic Mach number M* = √[(γ+1)/2 * M²/(1+(γ-1)M²/2)] inverted to find M_y = 0.534 downstream
  • (b)(ii) Stagnation pressure ratio across shock from normal shock tables: p_0x/p_0y = 1.113, giving p_0x = 2.226 bar upstream
  • (c) Tabular comparison: static pressure ↑, static temperature ↑, stagnation pressure ↓, stagnation temperature →, stagnation density ↓, Mach number ↓, entropy ↑, stagnation enthalpy →
  • (d) Heat rates using Fourier's law at x=0 and x=0.5 m: q = -kA(dT/dx); energy storage from ∂T/∂t = α(∂²T/∂x²) with no generation
  • (e)(i) Stefan-Boltzmann law Q = σAT⁴ with T = 3278 K to find filament diameter d = 0.457 mm
  • (e)(ii) Blackbody radiation function interpolation: f(0.4μm) and f(0.75μm) at λT values, visible efficiency η = (f_0.75 - f_0.4) × 100 = 6.53%
  • Consistent units throughout: kJ/kg-K for specific heats, bar for pressure, μm-K for λT products, W/m²-K⁴ for σ

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies: (a) entropy balance for irreversible heat transfer with environment at 293 K; (b) characteristic Mach number definition M*² = (γ+1)M²/(2+(γ-1)M²) and its inverse; (c) distinguishes between static and stagnation property changes across shock; (d) 1D transient conduction with polynomial profile using energy equation; (e) Planck's law integration via blackbody radiation functionsUses correct formulas but confuses: entropy of surroundings calculation, or M* vs M relation, or treats stagnation temperature as changing across shock, or misses that ∂T/∂t requires ∂²T/∂x² evaluationFundamental errors: uses ΔS = Q/T_system for entropy production, or treats shock as isentropic, or uses steady-state conduction for (d), or applies Wien's law instead of Stefan-Boltzmann
Numerical accuracy20%10(a) S_gen = 0.1543 kJ/K; (b)(i) M_y = 0.534, (ii) p_0x = 2.226 bar; (c) all 8 parameters correctly classified; (d) q_in = 6640 W, q_out = 1360 W, dE/dt = 5280 W, ∂T/∂t|_0 = 0.01024 K/s, ∂T/∂t|_0.5 = -0.1792 K/s; (e) d = 0.457 mm, η = 6.53%; all values within 2% of exactCorrect methodology but arithmetic slips: entropy value off by factor, M_y = 0.5 given as answer without calculation, shock pressure ratio inverted, heat flux signs reversed, radiation function interpolation error leading to η ≈ 5-8%Order-of-magnitude errors: entropy in J/K vs kJ/K confused, Mach number >1 downstream of shock, heat storage negative when energy entering exceeds leaving, diameter in meters instead of mm
Diagram quality15%7.5T-s diagram for (a) showing three process paths and entropy generation area; (c) clear table with ↑/↓/→ symbols and brief justification column; (d) wall sketch with coordinate system, heat flux arrows at boundaries, and temperature profile curvature indicated; shock wave schematic for (b) with stations x and y labelledTable for (c) present but without justifications; diagrams drawn but missing labels or arrows; T-s diagram shows processes but not the environment temperature lineNo diagrams despite (a) and (d) benefiting from visual representation; or incorrect diagrams like P-v for entropy problem, or shock shown as continuous expansion
Step-by-step derivation25%12.5Every formula stated before substitution: (a) ΔS = mc ln(T2/T1) for warming, ΔS = mL/T for melting, ΔS_surroundings = -Q/T_env; (b) explicit inversion of M*(M) relation; (c) brief physical reasoning for each table entry; (d) Fourier's law with dT/dx from polynomial, then ∂²T/∂x² for heat equation; (e) A = πdL solved before Stefan-Boltzmann, interpolation formula for f valuesSome steps skipped: jumps from given data to final entropy value, or uses M* = M without explanation, or states heat equation result without showing ∂²T/∂x² calculation, or interpolates f values without showing linear formulaBare answers with no working; or incorrect derivations like using C_p for ice warming, or treating shock as isentropic for stagnation pressure calculation
Practical interpretation15%7.5(a) Notes irreversibility quantified can guide insulation design; (b) Comments on total pressure loss implications for supersonic diffuser design (e.g., ramjet inlets); (c) Explains why stagnation pressure loss limits shock recovery; (d) Identifies when wall reaches steady state (∂T/∂t → 0); (e) Relates low visible efficiency to incandescent bulb replacement by LEDs in Indian energy conservation contextBrief comment on one or two parts: mentions irreversibility, or notes shock losses, or states that bulb is inefficient without quantitative contextNo interpretation; or incorrect physical statements like 'entropy production is undesirable so minimize heat transfer' missing that melting requires heat transfer

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