Mechanical Engineering 2023 Paper II 50 marks Calculate

Q3

(a) Heat is generated in a stainless steel plate (thermal conductivity = 22 W/m-K) of thickness 1 cm, at a uniform rate of 600 MW/m³. The left side of the plate is maintained at 200 °C and the right side is maintained at 100 °C. What will be the (i) temperature distribution across the plate, (ii) location and value of maximum temperature and (iii) heat flux from both sides of the plate and its direction? Assume one-dimensional, steady-state heat conduction. (20 marks) (b) A combination of a heat engine driving a heat pump (see the figure) takes waste energy at 50 °C as a source, Q̇W1, to the heat engine rejecting heat at 30 °C. The remainder, Q̇W2, goes into the heat pump that delivers Q̇H at 150 °C. If the total waste energy is 5 MW, find the rate of energy delivered at the higher temperature. Assume heat engine and heat pump as reversible : (20 marks) (c) A centrifugal compressor delivers 1·25 kg/s of air while running at 6000 r.p.m. The diameters at the inlet and outlet are 0·5 m and 1 m respectively. The power input factor is 1·04, while the slip factor is unity. The power consumed by the compressor is 50 kW. State the type of impeller used, whether forward, radial or backward curved. Draw velocity triangles. Assume no prewhirl at the inlet. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) 1 cm मोटाई की एक जंगरोधी इस्पात पट्टिका (ऊष्मा चालकता = 22 W/m-K) में 600 MW/m³ की एकसमान दर से ऊष्मा उत्पन्न होती है। पट्टिका के बायीं ओर का तापमान 200 °C पर स्थिर बनाए रखा जाता है और दाहिनी ओर को 100 °C पर स्थिर बनाए रखा जाता है। (i) पट्टिका में तापमान वितरण, (ii) अधिकतम तापमान का स्थान और मान क्या होगा तथा (iii) पट्टिका के दोनों ओर से ऊष्मा फ्लक्स और उसकी दिशा क्या होगी? एक-आयामी, स्थायी-दशा ऊष्मा चालन मान लीजिए। (20 अंक) (b) ऊष्मा पम्प चलाने वाले ऊष्मा इंजन का एक संयोजन (चित्र देखिए) 50 °C पर अपशिष्ट ऊर्जा को एक स्रोत, Q̇W1, के रूप में, 30 °C पर ऊष्मा परित्याग करने वाले ऊष्मा इंजन में ले जाता है। शेष, Q̇W2, ऊष्मा पम्प में चली जाती है जो 150 °C पर Q̇H प्रदान करता है। यदि कुल अपशिष्ट ऊर्जा 5 MW है, तो उच्च तापमान पर वितरित ऊर्जा की दर ज्ञात कीजिए। ऊष्मा इंजन और ऊष्मा पम्प को प्रतिक्रम्य मान लीजिए : (20 अंक) (c) एक अपकेन्द्री संपीडक 6000 r.p.m. पर चलते समय 1·25 kg/s वायु प्रदान करता है। अन्तर्गम तथा निर्गम पर व्यास क्रमशः: 0·5 m और 1 m हैं। शक्ति निवेश गुणक 1·04 है, जबकि सर्पण गुणक इकाई है। संपीडक द्वारा खपत की गई शक्ति 50 kW है। उपयोग किए गए प्रणोदक का प्रकार बताइए, चाहे वह अग्र, विप्र या पश्च वक्र है। वेग त्रिभुज बनाइए। मान लीजिए कि अन्तर्गम पर कोई पूर्व-आवर्त नहीं है। (10 अंक)

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Approach

Calculate solutions for all three sub-parts systematically: spend ~40% time on part (a) given its 20 marks, ~35% on part (b) for 20 marks, and ~25% on part (c) for 10 marks. For (a), derive the temperature distribution from the heat conduction equation with internal generation; for (b), apply Carnot efficiency and COP relationships for the reversible heat engine-pump combination; for (c), use Euler's pump equation with slip factor and power input factor to determine impeller type. Present each part with clear headings, governing equations, substitutions, and final answers with units.

Key points expected

  • Part (a): Governing equation d²T/dx² + q̇/k = 0 integrated to T(x) = -q̇x²/2k + C₁x + C₂; boundary conditions T(0)=200°C, T(L)=100°C applied to find C₁, C₂
  • Part (a): Temperature distribution T(x) = -13.636x² - 954.55x + 200 (x in m, T in °C); maximum temperature location x = -0.035 m (outside plate, so max at left boundary) or recalculated correctly if interior
  • Part (a): Heat fluxes q''_L = -k(dT/dx)|_{x=0} and q''_R = -k(dT/dx)|_{x=L} with directions stated (leftward at left face, rightward at right face)
  • Part (b): Carnot efficiency η = 1 - T_L/T_H = 1 - 303/323 for engine; COP_HP = T_H/(T_H - T_L) = 423/(423-303) for heat pump; energy balance Q̇_W1 + Q̇_W2 = 5 MW
  • Part (b): Work output from engine Ẇ = ηQ̇_W1 drives heat pump; simultaneous solution yields Q̇_H delivered at 150°C
  • Part (c): Euler work W = σψu₂²/g where σ=1, ψ=1.04; compare with actual power to find u₂; then tanβ₂ = V_f2/(u₂ - V_w2) to determine blade angle and impeller type
  • Part (c): Velocity triangles drawn with inlet radial flow (α₁=90°, V_w1=0) and outlet with calculated blade angle β₂

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies 1D steady conduction with heat generation for (a); recognizes reversible cycles with Carnot relations for (b); uses Euler pump equation with slip and power input factors for (c). Converts all temperatures to Kelvin for thermodynamic calculations.Uses correct governing equations but mixes Celsius/Kelvin in (b) or omits slip/power input factors in (c); minor conceptual gaps in heat engine-pump coupling.Treats (a) as no-generation conduction; uses wrong efficiency/COP formulas in (b); confuses compressor with turbine or ignores factors in (c).
Numerical accuracy20%10Precise values: (a) T_max location and value, heat fluxes with correct signs; (b) Q̇_H ≈ 6.25-6.5 MW range with clear working; (c) u₂ ≈ 314 m/s, blade angle determination leading to backward-curved identification. All unit conversions correct.Final answers approximately correct but arithmetic slips in intermediate steps; unit errors (e.g., cm vs m in thickness) partially compensated; one part significantly off.Major calculation errors: wrong integration constants, temperature in °C for Carnot, or power/energy confusion; answers without units or with wrong orders of magnitude.
Diagram quality20%10Clear schematic for (a) showing coordinate system and boundary conditions; T-x parabola sketched with T_max indicated; velocity triangles for (c) with all velocity components labelled (V₁, V₂, u₂, V_r1, V_r2, α₁, α₂, β₂) and flow angles marked.Diagrams present but missing labels or with incorrect angles; T-x curve shape wrong or velocity triangles incomplete; no schematic for heat engine-pump arrangement in (b).No diagrams despite explicit requirement in (c); or completely wrong sketches (e.g., temperature profile linear, velocity triangles for axial compressor).
Step-by-step derivation20%10Full derivation: (a) integration of Poisson equation, application of BCs, verification of T_max location; (b) explicit Carnot relations, energy balances, simultaneous equations solved; (c) Euler equation expansion, substitution of all terms, algebraic solution for blade angle.Key steps shown but skips some algebra (e.g., direct quote of integration results); jumps to final formulas without derivation; solution method unclear for coupled equations in (b).Final answers stated without derivation; no governing equations written; or incorrect derivation with missing terms (e.g., no q̇/k term in conduction equation).
Practical interpretation20%10Interprets (a) heat flux directions physically and comments on thermal stress implications for stainless steel; (b) discusses energy upgrading and practical irreversibilities; (c) justifies backward-curved impeller choice with reference to pressure ratio vs efficiency trade-offs in Indian HVAC/industrial applications.States impeller type with minimal justification; mentions heat flux directions without physical interpretation; no discussion of real-world irreversibilities.No interpretation; treats as pure mathematics; or gives physically absurd interpretations (e.g., heat flowing from cold to hot without work input).

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