Mechanical Engineering 2025 Paper II 50 marks Calculate

Q2

(a) (i) A reversible power cycle engine is used to drive a reversible heat pump. The power cycle takes in Q₁ heat units at temperature T₁ and rejects heat Q₂ at temperature T₂. The heat pump abstracts heat Q₄ from the sink at temperature T₄ and discharges heat Q₃ at temperature T₃. Develop an expression for the ratio Q₄/Q₁ in terms of the four temperatures T₁, T₂, T₃ and T₄. (10 marks) (ii) Argon gas expands adiabatically in a turbine from 2 MPa, 1000 °C to 350 kPa. The mass flow rate of argon is 0·5 kg/s and the turbine develops power at the rate of 120 kW. Determine the following: (1) The temperature of argon at the turbine exit (2) The irreversibility rate (3) The second law efficiency Neglect kinetic and potential energy effects and take T₀ = 20 °C and P₀ = 1 atm. Take molecular weight of argon as 40 kg/kmol and γ = 1·67. (T₀ and P₀ are the environment temperature and pressure, respectively) (10 marks) (b) Compressed air is transported in an industrial pipeline of 50 mm internal diameter. The stagnation conditions at the inlet are P₀ = 10 bar and T₀ = 400 K. The average Fanning friction factor f̄ = 0·002. If the Mach number changes from 3 at the entry to 1 at the exit, determine the following: (i) The length of the pipe (ii) The velocity at the exit (iii) The change in the stagnation temperature (iv) The change in the stagnation pressure (v) The change in the entropy (vi) The mass flow rate Assume the flow to be adiabatic. For air, take γ = 1·4 and R = 287 J/kg-K. Isentropic flow table and Fanno flow table attached at the end, may be used. (20 marks) (c) A thin flat plate, that is 0·2 m×0·2 m on a side, is oriented parallel to an atmospheric air stream having a velocity of 40 m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120 °C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0·075 N. What is the rate of heat transfer from both sides of the plate to the air? (Assume ρair = 0·995 kg/m³, νair = 20·92×10⁻⁶ m²/s, Prair = 0·7, kair = 30×10⁻³ W/m-K) (10 marks)

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(a) (i) एक प्रतिवर्ती (रिवर्सिबल) शक्ति चक्र इंजन का उपयोग एक प्रतिवर्ती ताप पंप (हीट पंप) को चलाने के लिये किया जाता है। शक्ति चक्र तापमान T₁ पर Q₁ मात्रक ऊष्मा को ग्रहण करता है और तापमान T₂ पर Q₂ ऊष्मा का परित्याग करता है। ताप पंप तापमान T₄ पर स्थित सिंक से Q₄ ऊष्मा को अवशोषित करता है और तापमान T₃ पर Q₃ ऊष्मा को निष्कासित करता है। चारों तापमान T₁, T₂, T₃ एवं T₄ के संदर्भ में Q₄/Q₁ के अनुपात के लिये एक व्यंजक विकसित कीजिये। (10 अंक) (ii) आर्गन गैस एक टरबाइन में 2 MPa, 1000 °C से 350 kPa तक रूद्धोष्म (एडियाबैटिक) रूप से प्रसारित होती है। आर्गन की द्रव्यमान प्रवाह दर 0·5 kg/s है तथा टरबाइन 120 kW की दर से शक्ति उत्पन्न करती है। निम्नलिखित को निर्धारित कीजिए: (1) टरबाइन के निकास पर आर्गन का तापमान (2) अप्रतिक्रमता (इरिवर्सिबिलिटी) दर (3) द्वितीय नियम दक्षता गतिज और स्थितिज ऊर्जा प्रभावों की अनदेखी कीजिए तथा T₀ = 20 °C और P₀ = 1 atm लीजिए। आर्गन का आण्विक भार 40 kg/kmol और γ = 1·67 लीजिए। (T₀ और P₀ क्रमशः वातावरण के तापमान और दाब हैं) (10 अंक) (b) संपीडित वायु (कम्प्रेस्ड एयर) 50 mm आंतरिक व्यास वाली एक औद्योगिक पाइपलाइन में प्रवाहित की जाती है। अंतर्गम (इनलेट) पर ठहराव की स्थितियाँ (स्टेशन कंडीशन) P₀ = 10 bar और T₀ = 400 K हैं। औसत फैनिंग घर्षण गुणांक f̄ = 0·002 है। यदि मैक संख्या अंतर्गम पर 3 से बदलकर निर्गम पर 1 हो जाती है, तो निम्नलिखित को निर्धारित कीजिए: (i) पाइप की लंबाई (ii) निर्गम पर वेग (iii) ठहराव (स्टैग्नेशन) तापमान में परिवर्तन (iv) ठहराव दाब में परिवर्तन (v) एंट्रॉपी में परिवर्तन (vi) द्रव्यमान प्रवाह दर मान लीजिए कि प्रवाह रूद्धोष्म (एडियाबैटिक) है। वायु के लिये γ = 1·4 तथा R = 287 J/kg-K लीजिए। आइसेंट्रॉपिक प्रवाह तालिका तथा फैनो प्रवाह तालिका अंत में संलग्न हैं, जिनका उपयोग किया जा सकता है। (20 अंक) (c) 0·2 m×0·2 m आकार की एक पतली समतल पट्टी को वायुमंडलीय वायु प्रवाह के समानांतर रखा गया है, जिसका वेग 40 m/s है। वायु का तापमान T∞ = 20 °C है, जबकि पट्टी को Ts = 120 °C पर बनाये रखा गया है। वायु, पट्टी की ऊपरी और निचली सतहों पर प्रवाहित होती है तथा विकर्ष बल (ड्रैग फोर्स) के मापन से इसका मान 0·075 N प्राप्त होता है। पट्टी की दोनों सतहों से वायु की ओर उष्मा अंतरण दर (हीट ट्रांसफर रेट) क्या है? (मानिये ρवायु = 0·995 kg/m³, νवायु = 20·92×10⁻⁶ m²/s, Prवायु = 0·7, kवायु = 30×10⁻³ W/m-K) (10 अंक)

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Approach

Calculate and derive expressions for all nine sub-parts systematically, allocating approximately 20% time to (a)(i) reversible cycle derivation, 20% to (a)(ii) argon turbine analysis, 45% to (b) Fanno flow calculations using attached tables, and 15% to (c) Reynolds analogy application. Begin each part with governing equations, substitute given data with proper unit conversions, and conclude with numerical answers and physical interpretation.

Key points expected

  • (a)(i) Derivation: COP_heat_pump = Q₃/W = T₃/(T₃-T₄), η_power_cycle = W/Q₁ = 1-T₂/T₁, combine to get Q₄/Q₁ = (T₁-T₂)(T₃-T₄)/(T₁(T₃-T₄)) with algebraic manipulation for Q₄ = Q₃ - W
  • (a)(ii) Argon turbine: T₂ = T₁(P₂/P₁)^((γ-1)/γ) = 1273×(0.35/2)^0.402 = 773 K; actual work from power = 120/0.5 = 240 kJ/kg; actual T₂ = 1273 - 240/0.312 = 503 K; irreversibility = T₀(s₂-s₁) - q = 293×0.312×ln(773/503) = 38.6 kJ/kg; η_II = (h₁-h₂_actual)/(h₁-h₂_isentropic)
  • (b) Fanno flow: Use tables at M₁=3, M₂=1 to find 4fL*/D = 0.522; L = 0.522×0.05/(4×0.002) = 3.26 m; V_exit = a* = √(γRT*) = 347 m/s; T₀ constant (adiabatic) so ΔT₀=0; P₀₂/P₀₁ from tables = 0.408; Δs = Rln(P₀₁/P₀₂) = 287×ln(2.45) = 258 J/kg-K; ṁ = ρ*A*a*
  • (c) Reynolds analogy: St = Cf/2 = h/(ρCpU); from drag F_D = 2×(1/2)ρU²×Cf×A, find Cf = 0.075/(0.995×1600×0.08) = 0.000588; h = St×ρCpU = (Cf/2)×ρCpU = 0.000294×0.995×1005×40 = 11.7 W/m²K; Q = 2×h×A×(T_s-T_∞) = 2×11.7×0.04×100 = 93.6 W
  • Unit consistency: Convert all temperatures to Kelvin, pressures to consistent units (Pa or kPa), and verify dimensional homogeneity in each equation
  • Table usage in (b): Explicitly state Fanno flow parameter values extracted from tables (e.g., P/P*, T/T*, P₀/P₀* at M=3 and M=1) and show interpolation if needed
  • Second law analysis: Clearly distinguish between isentropic, actual, and dead states for exergy and irreversibility calculations in (a)(ii)
  • Physical verification: Check that results satisfy thermodynamic constraints (e.g., entropy generation ≥ 0, Mach number trends in Fanno flow, heat transfer direction in (c))

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies reversible cycle thermodynamics for (a)(i), recognizes argon as monatomic ideal gas with Cp = γR/(γ-1) in (a)(ii), identifies choked Fanno flow with proper table usage in (b), and applies Reynolds-Colburn analogy appropriately for turbulent flow in (c); distinguishes between static and stagnation properties throughout.Uses correct major formulas but makes minor errors in gas constant calculation or confuses static/stagnation properties in some parts; recognizes Fanno flow but table lookup has small errors.Applies wrong cycle concepts (e.g., treats heat pump as refrigerator), uses diatomic gas properties for argon, or applies Bernoulli instead of Fanno flow equations; fundamental misunderstanding of Reynolds analogy.
Numerical accuracy25%12.5All nine numerical answers within 2% of expected values: Q₄/Q₁ expression algebraically correct; T₂_isentropic ≈ 773 K, T₂_actual ≈ 503 K, I ≈ 38.6 kW, η_II ≈ 0.66; L ≈ 3.26 m, V_exit ≈ 347 m/s, ΔP₀/P₀₁ ≈ 0.592, Δs ≈ 258 J/kg-K, ṁ ≈ 0.42 kg/s; Q_total ≈ 94 W; all unit conversions correct.Most answers correct but 1-2 significant calculation errors (e.g., wrong exponent in isentropic relation, arithmetic slip in table lookup, or factor of 2 error in drag/heat area); final answers within 10%.Multiple numerical errors exceeding 15% deviation; consistent failure to convert °C to K; order-of-magnitude errors in mass flow or heat transfer rates.
Diagram quality10%5Clear T-s diagram for (a)(i) showing reversible engine and heat pump cycles with four temperature reservoirs; h-s or T-s diagram for (a)(ii) turbine showing actual vs isentropic expansion; Fanno flow diagram for (b) illustrating duct with Mach number variation; control volume sketches with all energy/entropy flows labelled.At least two correct diagrams present but missing some labels or temperature/pressure state points; sketches recognizable but not to scale.No diagrams or completely incorrect representations (e.g., P-v instead of T-s for cycles, no indication of irreversibility in turbine expansion).
Step-by-step derivation25%12.5Complete derivation for (a)(i) from first principles showing W = Q₁(1-T₂/T₁) and Q₃ = W×T₃/(T₃-T₄) leading to final ratio; explicit entropy balance for irreversibility in (a)(ii); clear Fanno flow equation integration or table interpolation shown in (b); Reynolds analogy derivation from momentum-heat transfer similarity in (c).Key steps shown but some algebraic manipulation skipped or assumed; uses given tables without showing how values are extracted; jumps from drag coefficient to heat transfer without showing Stanton number relation.Final formulas stated without derivation; no working shown for numerical substitutions; appears to use memorized results without understanding origin.
Practical interpretation20%10Interprets (a)(i) result for combined cycle efficiency optimization; discusses turbine blade cooling needs in (a)(ii); explains industrial implications of choked flow and pipe length limits in (b) with reference to ISRO propulsion or gas pipeline design; relates drag measurement to heat transfer in (c) for electronic cooling applications in Indian climate.Brief physical interpretation for 2-3 parts but superficial; mentions practical relevance without specific engineering context.Purely mathematical treatment with no physical insight; no discussion of why results matter for engineering design or operation.

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