Mechanical Engineering 2025 Paper II 50 marks Calculate

Q7

(a) A two-stage turbine receives steam at 50 bar and 350 °C. At 1·5 bar, the high-pressure stage exhausts and 12000 kg of steam per hour is taken at this stage for process heating purposes. The remainder steam is reheated to 250 °C at 1·5 bar and then expanded through the low-pressure turbine to a condensate pressure of 0·05 bar. The power output from the turbine unit is to be 3750 kW. The isentropic efficiency of high-pressure stage is 0·84 and that of the low-pressure stage is 0·81. Calculate the rate of steam being generated per hour by the boiler. Steam properties : 50 bar, 350 °C : h = 3068·4 kJ/kg, s = 6·4492 kJ/kg-K 1·5 bar, 250 °C : h = 2972·65 kJ/kg, s = 7·8709 kJ/kg-K Use Steam tables given at the end to get other properties. (20 marks) (b) (i) What is meant by effective temperature? List and explain briefly the factors which affect optimum effective temperature in air-conditioning. (10 marks) (ii) Explain psychrometric process of adiabatic mixing of two air streams. (10 marks) (c) Why is it necessary to cool IC engines? What are the disadvantages of overcooling an IC engine? (10 marks)

हिंदी में प्रश्न पढ़ें

(a) एक द्वि-चरणीय टरबाइन (टू-स्टेज टरबाइन) 50 bar और 350 °C पर भाप प्राप्त करता है। उच्च-दाब चरण (हाई-प्रेशर स्टेज) से 1·5 bar पर भाप निष्कासित होती है तथा इस चरण से प्रति घंटा 12000 kg भाप को प्रसंस्करण तापन (प्रोसेस हीटिंग) के लिये निकाला जाता है। शेष भाप को 1·5 bar पर 250 °C तक पुनः गर्म किया जाता है और फिर निम्न-दाब टरबाइन (लो-प्रेशर टरबाइन) से होकर 0·05 bar कंडेंसेट प्रेशर तक विस्तारित किया जाता है। टरबाइन इकाई से शक्ति उत्पादन 3750 kW है। उच्च-दाब चरण की आइसेन्ट्रॉपिक दक्षता 0·84 है तथा निम्न-दाब चरण की आइसेन्ट्रॉपिक दक्षता 0·81 है। प्रति घंटा बॉयलर द्वारा उत्पन्न भाप की दर की गणना कीजिये। भाप के गुणधर्म : 50 bar, 350 °C : h = 3068·4 kJ/kg, s = 6·4492 kJ/kg-K 1·5 bar, 250 °C : h = 2972·65 kJ/kg, s = 7·8709 kJ/kg-K अन्य अवस्थागत गुण प्राप्त करने हेतु अंत में दी गई भाप तालिकाओं का प्रयोग कीजिये। (20 अंक) (b) (i) प्रभावी तापमान से क्या तात्पर्य है? वातानुकूलन में आदर्श प्रभावी तापमान को प्रभावित करने वाले कारकों की सूची बनाइये और उन्हें संक्षेप में समझाइये। (10 अंक) (ii) दो वायु प्रवाहों के रूद्रोष्म (एडियाबैटिक) मिश्रण की आर्द्रतामितीय (साइक्रोमैट्रिक) प्रक्रिया की व्याख्या कीजिये। (10 अंक) (c) आइ० सी० इंजनों को ठंडा करना क्यों आवश्यक होता है? आइ० सी० इंजन के अत्यधिक शीतलन से क्या-क्या नुकसान होते हैं? (10 अंक)

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Approach

Calculate the steam generation rate for part (a) by applying steady-flow energy equations across both turbine stages, accounting for extraction, reheating, and isentropic efficiencies. For parts (b) and (c), explain effective temperature factors and adiabatic mixing with psychrometric charts, then discuss IC engine cooling necessities and overcooling disadvantages. Allocate approximately 40% time to (a), 30% to (b), and 30% to (c) based on mark distribution.

Key points expected

  • Part (a): Determine isentropic enthalpy at 1.5 bar from s1 = 6.4492 kJ/kg-K using steam tables; find actual h2 using η_HP = 0.84
  • Part (a): Calculate mass flow split: 12000 kg/h extracted, remainder (ṁ-12000) goes to reheating and LP turbine
  • Part (a): Apply energy balance: Power output = ṁ_HP×(h1-h2) + ṁ_LP×(h3-h4_actual) = 3750 kW; solve for total ṁ
  • Part (b)(i): Define effective temperature as index of thermal comfort combining dry-bulb, humidity, and air motion; list factors (clothing, activity, age, sex, climate acclimatization, economic status)
  • Part (b)(ii): Explain adiabatic mixing on psychrometric chart: two states mix along straight line; mass and energy balances give final state; show mixing ratio and enthalpy equations
  • Part (c): Necessity of cooling—prevent thermal stresses, maintain lubricant properties, avoid material failure, control NOx emissions; disadvantages of overcooling—reduced thermal efficiency, increased friction, poor combustion, corrosion, increased fuel consumption
  • Correct use of steam table interpolations for wet region at 0.05 bar and superheated region at 1.5 bar
  • Final answer for (a): steam generation rate in kg/h with proper unit conversion and significant figures

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10For (a), correctly applies isentropic efficiency definition η = (h1-h2)/(h1-h2s) for HP stage and similar for LP stage; for (b), accurately defines effective temperature as comfort index and correctly states psychrometric mixing rules; for (c), distinguishes between necessity of cooling and specific disadvantages of overcooling without conflation.Applies efficiency formulas correctly but confuses η definition direction; defines effective temperature vaguely; lists cooling necessities and overcooling disadvantages with some overlap or missing key points.Uses η = h2s/h2 or other incorrect efficiency formulation; conflates effective temperature with wet-bulb temperature; fails to distinguish between undercooling and overcooling problems.
Numerical accuracy20%10For (a), correctly interpolates steam tables to find h2s at 1.5 bar (sf + x·sfg = 6.4492), finds h4s at 0.05 bar from s3 = 7.8709, applies both efficiencies, sets up correct power equation with mass extraction, and obtains ṁ_boiler ≈ 16800-17200 kg/h with all intermediate values shown.Correct final answer but with minor interpolation errors or unit conversion slips (kW to kJ/s); correct method but arithmetic error in mass balance or power equation setup.Major errors: ignores extraction mass flow, treats both stages with same efficiency, uses h instead of s for isentropic determination, or final answer off by >20% due to systematic errors.
Diagram quality20%10For (a), draws clear T-s or h-s diagram showing: state 1 (50 bar, 350°C), isentropic drop to 2s, actual to 2, extraction point, reheating to 3, isentropic to 4s, actual to 4 at 0.05 bar; for (b)(ii), sketches psychrometric chart with mixing line and two inlet states; all states labelled with p, T, h values.Diagrams present but missing key labels or values; T-s diagram shows process direction but not actual vs isentropic paths; psychrometric chart drawn but mixing line not clearly explained.No diagrams drawn despite (b)(ii) requiring psychrometric illustration; or diagrams completely wrong (e.g., showing Rankine cycle without extraction, or p-v diagram instead of T-s/h-s).
Step-by-step derivation20%10For (a), shows: (1) h2s determination from s1 = s2s using steam table, (2) h2 from η_HP, (3) h4s from s3 = s4s at 0.05 bar, (4) h4 from η_LP, (5) power equation with ṁ_total and extraction, (6) algebraic solution for ṁ; for (b) and (c), structured explanations with clear headings and logical flow.Shows main steps but skips intermediate calculations (e.g., directly states h2s without showing interpolation); some logical gaps in explaining why certain equations are used; parts (b) and (c) less structured.Final answers stated without derivation; or jumbled working with no clear sequence; energy balance written incorrectly (e.g., ignores kinetic/potential energy terms without justification, or confuses mass and energy flows).
Practical interpretation20%10For (a), comments on cogeneration benefits (process heat + power) and why extraction improves overall efficiency vs condensing turbine; for (b), relates effective temperature to Indian climatic zones and HVAC design for comfort; for (c), connects overcooling to specific engine problems like cylinder wear in Indian ambient conditions and emission control trade-offs.Brief mention of cogeneration for (a); generic comfort comments for (b); standard overcooling disadvantages listed for (c) without contextual application.No practical interpretation; treats (a) as pure mathematics, (b) as textbook definition, (c) as isolated facts; or makes incorrect practical claims (e.g., extraction reduces efficiency, overcooling improves performance).

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