Mechanical Engineering 2024 Paper II 50 marks Calculate

Q6

(a) A four-cylinder gasoline engine has a bore of 75 mm and a stroke length of 100 mm. It is operated at 3000 rpm and tested at this speed against a brake which has a torque arm of 40 cm. The net brake load is 150 N and the fuel consumption is observed as 7·8 l/h. A Morse test is carried out and the cylinders are cut-out in the order 1, 2, 3, 4 with the corresponding brake loads of 110 N, 108 N, 106 N and 104 N respectively. The specific gravity of the fuel may be taken as 0·79 and it has a calorific value of 44000 kJ/kg. Calculate the following: (i) brake power (ii) bmep (iii) bsfc (iv) indicated power (v) mechanical efficiency (vi) imep (b) A heat pump that operates on ideal vapour compression cycle with R-134a is used to heat a house and maintain it at 20°C, using underground water at 10°C as the heat source. The house is losing heat at a rate of 75 MJ/h. The evaporator and condenser pressure are 320 kPa and 800 kPa respectively. Determine the power input to the heat pump and the electric power saved by the heat pump instead of using a resistance heater. The enthalpy of superheated R-134a at 800 kPa at compressor outlet may be taken as 420 kJ/kg. Use the R-134a property table attached at the end. The inlet to the compressor may be taken as saturated vapour. 20 marks (c) Explain using neat sketches any two types of governing used in steam turbines. 10 marks

हिंदी में प्रश्न पढ़ें

(a) एक चार सिलिंडर गैसोलीन इंजन का बोर 75 mm तथा स्ट्रोक की लंबाई 100 mm है। इसको 3000 rpm पर चलाया जाता है तथा इस गति पर एक ब्रेक के विरुद्ध, जिसकी बल-आघूर्ण की भुजा 40 cm है, परीक्षण किया जाता है। नेट ब्रेक भार 150 N तथा ईंधन की खपत 7·8 l/h निरीक्षण के दौरान पाई गई। एक मोर्स टेस्ट किया गया तथा सिलिंडरों को 1, 2, 3, 4 के क्रम में तब कट-आउट किया गया है जबकि तदनुकूली ब्रेक भार क्रमशः: 110 N, 108 N, 106 N व 104 N है। ईंधन का विशिष्ट गुरुत्व 0·79 लिया जा सकता है तथा उसका कैलोरिफिक मान 44000 kJ/kg है। निम्नलिखित की गणना कीजिए: (i) ब्रेक शक्ति (ii) bmep (iii) bsfc (iv) सूचित शक्ति (v) यांत्रिक दक्षता (vi) imep (b) एक उष्मा पंप जो एक आदर्श वाष्प संपीडन चक्र पर R-134a के साथ काम करता है, उसका उपयोग करके एक घर को गर्म किया जाता है तथा घर को 20°C पर बनाए रखा जाता है जबकि 10°C के भू-जल को उष्मा-स्रोत के रूप में उपयोग में लाया जाता है। घर 75 MJ/h की दर से उष्मा खो रहा है। वाष्पित्र व संघनित्र दाब क्रमशः: 320 kPa व 800 kPa हैं। उष्मा पंप में शक्ति निवेश एवं प्रतिरोध तापक के प्रयोग के बिना उष्मा पंप द्वारा बचत की गई विद्युत शक्ति निर्धारित कीजिए। अतिरिक्त R-134a का 800 kPa पर संपीडक के निर्गम पर एन्थैल्पी 420 kJ/kg ली जा सकती है। R-134a की गुण तालिका, जो कि अंत में संलग्न है, का प्रयोग कीजिए। संपीडक के प्रवेश पर संतृप्त वाष्प ली जा सकती है। 20 अंक (c) भाप तरबाइनों में प्रयोग होने वाले किन्हीं दो प्रकार के अधिनियंत्रणों को स्वच्छ चित्रों के माध्यम से समझाइए। 10 अंक

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How this answer will be evaluated

Approach

Calculate all performance parameters for the IC engine in part (a) using Morse test principles, then solve the heat pump cycle in part (b) using R-134a property tables and COP analysis. For part (c), sketch and explain two steam turbine governing mechanisms. Allocate approximately 45% time to (a) due to extensive calculations, 35% to (b) for thermodynamic cycle analysis, and 20% to (c) for descriptive diagrams.

Key points expected

  • Part (a): BP = 2πNT/60 = 2π×3000×(150×0.4)/60000 = 18.85 kW; use Morse test to find IP per cylinder from load differences
  • Part (a): IP_total = Σ(IP_individual) = (150-110)+(150-108)+(150-106)+(150-104) × 0.4 × 2π × 3000 / 60000 = 23.56 kW; η_mech = BP/IP = 80%
  • Part (a): bmep = BP×60000/(L×A×n×N/2) = 18.85×60000/(0.1×π/4×0.075²×4×1500) = 640 kPa; imep = bmep/η_mech = 800 kPa
  • Part (a): Fuel mass flow = 7.8×0.79/3600 = 1.712×10⁻³ kg/s; bsfc = ṁ_f/BP = 0.091 kg/kWh
  • Part (b): From R-134a tables at 320 kPa: h₁ = 251.9 kJ/kg (sat vap), s₁ = 0.932 kJ/kgK; at 800 kPa: h₃ = h₄ = 93.4 kJ/kg (sat liq); h₂ = 420 kJ/kg (given)
  • Part (b): COP_HP = (h₂-h₃)/(h₂-h₁) = (420-93.4)/(420-251.9) = 1.94; Ẇ = Q_H/COP = 75/1.94 = 38.66 MJ/h = 10.74 kW; Power saved = 75/3.6 - 10.74 = 10.1 kW
  • Part (c): Throttle governing - sketch showing throttle valve before nozzles, pressure-velocity diagram, suitable for small turbines
  • Part (c): Nozzle control governing - sketch showing multiple nozzle groups with valves, maintains full admission pressure, better part-load efficiency

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%12Correctly applies Morse test principle that IP of cut cylinder = BP_all - BP_remaining; identifies heat pump as reverse refrigeration cycle with COP_HP = COP_R + 1; distinguishes throttle vs nozzle governing by admission control mechanism.Uses Morse test but confuses load difference with IP directly; calculates COP but treats as refrigerator; describes governing types but mixes up mechanisms.Treats Morse test as simply adding powers; confuses heat pump with heat engine; describes governing vaguely without distinguishing features.
Numerical accuracy20%12All six engine parameters correct within 2%: BP=18.85kW, bmep≈640kPa, bsfc≈0.091kg/kWh, IP=23.56kW, η_mech=80%, imep≈800kPa; heat pump power input ≈10.7kW, power saved ≈10kW.Most values correct but minor arithmetic errors in unit conversions (e.g., rpm to rps, l/h to kg/s); heat pump COP calculation correct but final power values slightly off.Major errors in torque calculation (forgets arm length), wrong stroke/cycle factor for 4-stroke engine, or uses wrong enthalpy values from tables.
Diagram quality20%12Two clear governing sketches: throttle governing showing single valve with p-V or h-s implications; nozzle control showing 3-4 nozzle groups with separate valves; both labelled with flow paths, pressure stages, and valve positions.Sketches present but missing key labels (valve positions, pressure values) or showing only one governing type clearly; heat pump or engine cycle diagrams absent or minimal.No sketches for part (c); or diagrams are generic turbine pictures without governing-specific features; messy or unlabelled drawings.
Step-by-step derivation20%12Shows complete derivation: torque = load × arm, BP formula with unit handling, Morse test IP calculation per cylinder with clear tabulation, bmep/imep from fundamental PV work, heat pump energy balance with state table (p, T, h, s for all 4 states).Jumps to final formulas with minimal working; shows some intermediate steps but skips state table or Morse test detail; heat pump states identified but not tabulated.Final answers only with no derivation; or incorrect formulas used throughout without justification; no systematic approach to multi-part problem.
Practical interpretation20%12Comments on engine: η_mech ≈ 80% is typical for petrol engines, bsfc competitive; notes heat pump advantage (COP>1) vs resistance heating for Indian residential heating; explains why nozzle governing preferred for large central Indian power station turbines (better efficiency at part load).Brief mention that heat pump saves energy or that mechanical efficiency is 'reasonable'; no specific context or comparison with Indian conditions.No interpretation; purely numerical exercise without physical insight or engineering relevance.

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