Mechanical Engineering 2025 Paper II 50 marks Calculate

Q6

(a) A single-cylinder, two-stroke, high-speed diesel engine working on 'dual cycle' has a compression ratio of 15 : 1. The engine takes in atmospheric air at 1 bar and 27 °C. The maximum temperature in the cylinder is 1312 K. The cutoff ratio is 1·093. The engine has a bore of 200 mm and stroke length of 250 mm. The engine speed is 3000 r.p.m. Calculate the following: (i) The cycle efficiency (ii) The net work output per unit mass of air (iii) The power output (iv) The mean effective pressure (v) The ratio of heat added in the constant-pressure process to that in the constant-volume process Draw the P-V and T-s diagrams of the cycle. For air, γ = 1·4, R = 0·287 kJ/kg-K. (20 marks) (b) An ideal regenerative steam power cycle operates so that steam enters the turbine at 30 bar and 350 °C, and exhausts at 0·1 bar. A single open feedwater heater is employed which operates at 5 bar. Compute the thermal efficiency of the cycle. The steam properties at 30 bar and 350 °C are h = 3115·3 kJ/kg and s = 6·7427 kJ/kg-K. Use Steam tables given at the end to get other properties. (20 marks) (c) Why is vapour absorption refrigeration system considered to be a better option in comparison to vapour compression refrigeration system for large capacity refrigeration and air-conditioning plants? (10 marks)

हिंदी में प्रश्न पढ़ें

(a) एक एकल-सिलेंडर, द्वि-स्ट्रोक, उच्च गति वाला डीजल इंजन 'हैट चक्र (ड्युअल साइकिल)' पर कार्य करता है, जिसका संपीडन अनुपात (कम्प्रेशन रेशियो) 15 : 1 है। इंजन वातावरणीय वायु को 1 bar तथा 27 °C पर प्रदान करता है। सिलेंडर के भीतर अधिकतम तापमान 1312 K है। कट-ऑफ अनुपात 1·093 है। इंजन का बोर 200 mm तथा स्ट्रोक लम्बाई 250 mm है। इंजन की गति 3000 r.p.m. है। निम्नलिखित की गणना कीजिये: (i) चक्र की दक्षता (ii) वायु के प्रति इकाई द्रव्यमान पर शुद्ध कार्य-निष्पादन (iii) शक्ति उत्पादन (पावर आउटपुट) (iv) औसत प्रभावी दाब (मीन इफेक्टिव प्रेशर) (v) स्थिर-दाब प्रक्रिया पर जोड़ी गई ऊष्मा और स्थिर-आयतन प्रक्रिया पर जोड़ी गई ऊष्मा का अनुपात चक्र का P-V तथा T-s आरेख बनाइये। वायु के लिये γ = 1·4, R = 0·287 kJ/kg-K है। (20 अंक) (b) एक आदर्श पुनर्जीवी (रिजनरेटिव) भाप शक्ति चक्र इस प्रकार संचालित होता है कि भाप 30 bar और 350 °C पर टरबाइन में प्रवेश करती है तथा 0·1 bar पर निष्कासित होती है। इसमें एकल मुक्त प्रभरण जल तापक (ओपन फीडवाटर हीटर) प्रयुक्त होता है, जो 5 bar पर कार्य करता है। चक्र की तापीय दक्षता (थर्मल एफिशिएंसी) की गणना कीजिए। 30 bar और 350 °C पर भाप के गुणधर्म h = 3115·3 kJ/kg और s = 6·7427 kJ/kg-K हैं। अन्य अवस्थागत गुणधर्मों की प्राप्ति हेतु अंत में प्रदत्त भाप तालिकाओं का प्रयोग कीजिए। (20 अंक) (c) भाप अवशोषण शीतलन प्रणाली को बड़ी क्षमता वाले शीतलन एवं वातानुकूलन संयंत्रों के लिए भाप संपीडन शीतलन प्रणाली की तुलना में बेहतर विकल्प क्यों माना जाता है? (10 अंक)

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How this answer will be evaluated

Approach

Calculate all numerical sub-parts systematically: spend ~40% time on part (a) dual cycle (20 marks) covering all five parameters and diagrams; ~35% on part (b) regenerative cycle (20 marks) with steam table interpolation and feedwater heater analysis; ~25% on part (c) theory (10 marks) comparing absorption vs compression systems. Present derivations with clear state-point labelling on P-V and T-s diagrams for (a), and T-s diagram with extraction point for (b).

Key points expected

  • Part (a): Correct T2 = 300 × 15^0.4 = 886.3 K; T3 = 1312 K; ρ = 1.093 gives T4 = 1434 K; T5 from isentropic expansion; efficiency = 1 - (T5-T1)/[(T3-T2)+γ(T4-T3)]
  • Part (a): Net work = cv(T3-T2) + cp(T4-T3) - cv(T5-T1); power = W_net × ṁ × (N/60) for two-stroke; mep = W_net/(V1-V2)
  • Part (a): P-V diagram showing 1-2 isentropic compression, 2-3 constant volume heat addition, 3-4 constant pressure heat addition, 4-5 isentropic expansion, 5-1 constant volume heat rejection; T-s diagram with corresponding slopes
  • Part (b): State 1 at 30 bar, 350°C (h1=3115.3, s1=6.7427); s2=s1 at 5 bar gives x2 and h2; h3 at 5 bar saturated liquid; pump work to 30 bar; h6 from turbine at 0.1 bar with s6=s1
  • Part (b): Extraction fraction y = (h3-h5)/(h2-h5); thermal efficiency = [(h1-h2)+(1-y)(h2-h6)]/[(h1-h3)] or equivalent energy balance
  • Part (c): Absorption system uses low-grade heat (waste heat, solar, biomass) vs high-grade work; no CFC/HCFC refrigerants; silent operation; suitable for large tonnage (Indian railways, NHPC power station cooling)
  • Part (c): COP comparison: COP_absorption = COP_Carnot × η_heatengine vs COP_compression; practical limitations (crystallization, part-load efficiency) mentioned for balance
  • All parts: Proper use of steam table data, unit consistency (kJ/kg, kPa, K), and clear identification of state points with numerical values

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies dual cycle efficiency formula with proper heat addition split; identifies regenerative cycle improves efficiency by raising feedwater temperature; explains why absorption suits large Indian industrial plants using waste heat.Uses correct formulas but confuses cutoff ratio with pressure ratio; identifies regenerative purpose but makes enthalpy state errors; lists absorption advantages without linking to large-scale context.Treats dual cycle as Diesel or Otto cycle; confuses open vs closed feedwater heater; states absorption uses compressor or confuses with adsorption.
Numerical accuracy20%10Part (a): η ≈ 61-62%, W_net ≈ 580-600 kJ/kg, Power ≈ 450-470 kW, mep ≈ 6.5-6.8 bar, Qp/Qv ≈ 1.8-2.0; Part (b): η ≈ 42-45% with correct extraction fraction y ≈ 0.18-0.22; all values within 2% tolerance.Correct method but arithmetic errors in T5 or extraction fraction; final efficiency off by 3-5%; units mostly consistent.Order-of-magnitude errors (e.g., efficiency >100% or <10%); ignores two-stroke factor in power calculation; steam table values used without interpolation.
Diagram quality20%10P-V and T-s diagrams for dual cycle with all five processes labelled, correct slopes (isentropic steeper than isobaric on T-s), state points numbered; regenerative cycle T-s showing extraction point and feedwater heating line.Diagrams drawn but missing labels or incorrect slope representation; one diagram missing or hastily sketched.No diagrams; or diagrams with processes in wrong order (e.g., expansion before compression); omits constant volume lines on P-V.
Step-by-step derivation20%10Shows all state calculations: T2 from isentropic relation, T4 from cutoff ratio, T5 from isentropic expansion with v5/v4 = r/ρ; mass flow rate from displacement volume; complete energy balance for feedwater heater with y derivation.Skips intermediate steps (e.g., jumps to efficiency formula); shows main equations but not numerical substitution; feedwater heater balance stated but not solved.Final answers only with no working; or incorrect formula substitution without explanation of variables.
Practical interpretation20%10Relates dual cycle to high-speed diesel engines (Maruti, Tata motors); notes why two-stroke gives higher power density; cites Indian Railways or NTPC plants using regenerative cycles for efficiency; mentions LiBr-H2O or NH3-H2O systems for large AC in Indian climatic conditions.Generic statements about efficiency improvement; mentions large plants but no Indian examples; states absorption is 'better' without specifying capacity threshold.No practical context; treats all cycles as abstract thermodynamic exercises; confuses refrigeration applications.

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