Mechanical Engineering 2024 Paper II 50 marks Solve

Q2

(a) Air flows through a 5 cm diameter pipe. Measurements indicate that at the inlet to the pipe the velocity is 70 m/s, the temperature is 80°C and the pressure 1 MPa. Find the temperature, the pressure, and the Mach Number at the exit of the pipe if the pipe is 25 m long. Assume that the flow is adiabatic and the mean friction factor is 0.005. Use Fanno table attached. (20 marks) (b) Saturated liquid refrigerant at – 7°C flows through a horizontal copper (k = 330 W/mK) tube of inside diameter 25 mm, thickness 2·5 mm and length 10 m. The tube is exposed to surrounding air at 20°C. Find the exit dryness fraction of the refrigerant from the tube if the flow rate is 0·0012 kg/s and latent heat of evaporation is 400 kJ/kg. Take the property values of air at 280 K as given below: ρ = 1·271 kg/m³, k = 0·0246 W/mK, γ = 1·4 × 10⁻⁵ m²/s, Pr = 0·717. For natural convection from a horizontal tube, the following correlation be used: Nūf = (0·48)[Gr . Pr]⁰·²⁵. Neglect the temperature difference between the copper tube and the refrigerant. Also neglect the thermal resistance of copper tube. (20 marks) (c) (i) Explain the procedure to arrive at Stefan-Boltzmann law from the Planck's law. Also find the total emissive power of a black sphere of 5 cm diameter maintained at 500 K. Take σ = 5·67 × 10⁻⁸ W/m²K⁴. (5 marks) (ii) Explain the procedure of arriving at Wien's displacement law from Planck's law. Also find the temperature of the sun if the wavelength at which maximum monochromatic emissive power is received is 0·55 μm. Take Wien's constant = 2·9 mm K. (5 marks)

हिंदी में प्रश्न पढ़ें

(a) वायु 5 cm व्यास के पाइप से बह रही है। मापन बताते हैं कि पाइप के अन्तर्गम पर वेग 70 m/s, तापमान 80°C व दाब 1 MPa है। यदि पाइप 25 m लम्बी है, तो पाइप के निर्गम पर तापमान, दाब एवं मैक संख्या ज्ञात कीजिए। मान लीजिए कि प्रवाह रूद्धोष्म है तथा माध्य घर्षण गुणांक 0·005 है। संलग्न फैनो (Fanno) तालिका का प्रयोग कीजिए। (20 marks) (b) एक क्षैतिज ताँबे (k = 330 W/mK) की नलिका, जिसका आन्तरिक व्यास 25 mm, मोटाई 2·5 mm व लम्बाई 10 m है, के द्वारा एक संतृप्त द्रव प्रशीतक – 7°C पर प्रवाहित होता है। नलिका 20°C पर परिवेश वायु से अनावृत है। यदि प्रवाह दर 0·0012 kg/s व वाष्पन की गुप्त ऊष्मा 400 kJ/kg है, तो नलिका से निर्गम पर प्रशीतक का शुष्कतांश ज्ञात कीजिए। 280 K पर वायु के गुणों के मान, निम्नवत हैं: ρ = 1·271 kg/m³, k = 0·0246 W/mK, γ = 1·4 × 10⁻⁵ m²/s, Pr = 0·717. एक क्षैतिज नलिका से प्राकृतिक संवहन के लिए निम्नलिखित सहसंबंध का प्रयोग कीजिए: Nūf = (0·48)[Gr . Pr]⁰·²⁵. ताँबे की नलिका व प्रशीतक के मध्य ताप-अन्तर नगण्य मानिए। ताँबे की नलिका का ऊष्मीय प्रतिरोध भी नगण्य मानिए। (20 marks) (c) (i) प्लांक के नियम से स्टेफान-बोल्ट्ज़मान नियम पर पहुँचने की विधि समझाइए। एक काले गोले की कुल उत्सर्जक शक्ति भी ज्ञात कीजिए जिसका व्यास 5 cm एवं जिसको 500 K पर अनुरक्षित किया गया है। σ = 5·67 × 10⁻⁸ W/m²K⁴ लीजिए। (5 marks) (ii) प्लांक के नियम से वीन के विस्थापन नियम तक पहुँचने की विधि समझाइए। यदि वह तरंगदैर्ध्य जिस पर अधिकतम एकवर्णी उत्सर्जक शक्ति पाई जाती है, 0·55 μm हो, तो सूर्य के तापमान को भी ज्ञात कीजिए। वीन स्थिरांक = 2·9 mm K लीजिए। (5 marks)

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Approach

Solve each sub-part systematically, allocating approximately 40% time to part (a) Fanno flow calculation, 35% to part (b) heat transfer with phase change, and 25% combined to parts (c)(i) and (c)(ii) for derivation and calculation of radiation laws. Begin with stating given data and assumptions, proceed with step-by-step calculations using appropriate formulae and tables, and conclude with clearly boxed final answers for each sub-part.

Key points expected

  • Part (a): Calculate inlet Mach number using Ma = V/√(γRT); determine fL*/D from Fanno table; compute actual fL/D = 0.005×25/0.05 = 2.5; find exit Mach number by interpolation from Fanno table; determine exit T and p using Fanno relations or table
  • Part (b): Calculate Grashof number using β = 1/T_film, ΔT = 27K, characteristic length L = D_o = 0.03m; compute Nu = 0.48(Gr·Pr)^0.25; find h = Nu·k/D_o; calculate Q = h·A_s·ΔT; apply energy balance Q = ṁ·h_fg·(x_e - x_i) to find exit dryness fraction
  • Part (c)(i): Integrate Planck's law E_bλ = C₁λ⁻⁵/[exp(C₂/λT)-1] over all wavelengths 0 to ∞ using substitution ξ = C₂/λT; show reduction to σT⁴; calculate E_b = σT⁴ = 5.67×10⁻⁸×(500)⁴ = 354.375 W/m²; total power = E_b×πD²
  • Part (c)(ii): Differentiate Planck's law with respect to λ and set dE_bλ/dλ = 0; show transcendental equation leading to λ_maxT = C₃ = 2.898×10⁻³ m·K; calculate T_sun = 2.9×10⁻³/(0.55×10⁻⁶) = 5273 K
  • Correct use of Fanno table interpolation for subsonic flow with friction
  • Proper handling of natural convection correlation for horizontal cylinder with film temperature concept

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies Fanno flow physics for part (a) with proper choking concept; recognizes combined natural convection-conduction resistance for part (b); accurately states Planck's law and integration limits for (c)(i), differentiation approach for (c)(ii); distinguishes between total emissive power and total powerUses Fanno flow concept correctly but confuses subsonic/supersonic branches; applies natural convection correlation with minor errors in characteristic length; states Stefan-Boltzmann and Wien's laws correctly but derivation approach vagueTreats (a) as isentropic flow or uses Bernoulli; uses forced convection for (b); confuses Planck's law with Wien's law; states laws without any derivation procedure
Numerical accuracy20%10Part (a): Ma₁ ≈ 0.2, fL*/D ≈ 14.5, Ma₂ ≈ 0.4-0.5, T₂ ≈ 340-350K, p₂ ≈ 0.6-0.7 MPa; Part (b): Gr ≈ 10⁷, Nu ≈ 25-30, h ≈ 20-25 W/m²K, x_exit ≈ 0.15-0.25; Part (c): E_b = 354.4 W/m², Power ≈ 2.78 W; T_sun ≈ 5270 K; all calculations with proper significant figuresCorrect order of magnitude for all answers; minor arithmetic errors in interpolation or exponent handling; one part completely correct, others with small errorsOrder-of-magnitude errors (e.g., Ma > 1 for subsonic Fanno flow, negative dryness fraction, T_sun in millions of K); missing units; no intermediate calculations shown
Diagram quality15%7.5Fanno curve sketch for part (a) showing Rayleigh line, Fanno line, and process direction on h-s or T-s diagram; thermal resistance network for part (b) showing convection resistance R_conv and conduction resistance (stated as negligible); spectral blackbody emissive power distribution sketch for parts (c)(i) and (c)(ii) showing λ_maxOne clear diagram (either Fanno curve or thermal resistance network); other parts described verbally; sketches lack proper labellingNo diagrams; or incorrect diagrams (e.g., isentropic process line for Fanno flow, wrong resistance network)
Step-by-step derivation25%12.5Part (a): Complete Fanno table usage with interpolation shown; Part (b): Explicit Gr, Nu, h calculation sequence; Part (c)(i): Full integration of Planck's law with substitution ξ = C₂/λT, reduction to standard integral ∫x³/(eˣ-1)dx = π⁴/15; Part (c)(ii): Complete differentiation dE_bλ/dλ = 0, transcendental equation (4-x)eˣ = 4 where x = C₂/λT, numerical solution leading to Wien's constantKey steps shown but skips some algebraic manipulation; states standard integral results without derivation; correct differentiation setup for Wien's law but skips to final resultFinal formulae quoted without derivation; no Fanno table interpolation shown; jumps from Q to dryness fraction without energy balance equation
Practical interpretation20%10Part (a): Comments on friction causing subsonic flow to approach Mach 1 (choking), pipe length limitation for given inlet conditions; Part (b): Discusses implications for refrigeration system design (evaporator sizing, need for superheat section); Part (c): Relates Stefan-Boltzmann to thermal radiation engineering (furnace design, solar collectors) and Wien's displacement to pyrometry/temperature measurement; notes sun's effective temperature relevance to solar energy in IndiaBrief comment on one or two parts (e.g., choking in pipes, importance of dryness fraction in refrigeration); generic statements about radiation lawsNo physical interpretation; purely mathematical treatment; no connection to engineering applications or Indian context (solar energy potential)

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