Mechanical Engineering 2024 Paper II 50 marks Solve

Paper II — Q8

(a) A cogenerating steam power plant operates with a boiler output of 25 kg/s steam at 7 MPa, 500°C. The condenser operates at…

(a)

A cogenerating steam power plant operates with a boiler output of 25 kg/s steam at 7 MPa, 500°C. The condenser operates at 7·5 kPa and the process heat is extracted at 5 kg/s from the turbine at 500 kPa and after use is returned as saturated liquid at 100 kPa. Assuming all components are ideal, find :

(i)

temperature of water leaving the condenser pump

(ii)

total turbine output

(iii)

total process heat transfer

At inlet to the turbine, assume h = 3410 kJ/kg and s = 6·802 kJ/kg K

Also, use data from Steam Tables given at the end. 20 marks

(b)

The fuel of an IC engine contains 85% carbon, 10% hydrogen, 3% oxygen and the remaining is nitrogen in composition by weight. Determine the chemically correct Air/Fuel ratio. If 30% excess air is supplied, find the percentage composition of dry products of combustion exhaust by weight and by volume. 20 marks

(c)

Explain in brief, how the molecular structure of the IC engine fuels affects the tendency to knock. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

एक सह-उत्पादक भाप शक्ति संयंत्र बॉयलर उत्पादन 25 kg/s भाप 7 MPa, 500°C पर देते हुए कार्यरत है। संघनित्र 7·5 kPa पर कार्यरत है तथा प्रक्रम की ऊष्मा 500 kPa पर, 5 kg/s की दर से टरबाइन से निकाली जाती है तथा प्रयोग के बाद 100 kPa पर संतृप्त द्रव के रूप में वापस की जाती है। सभी अवयवों को आदर्श मानते हुए, ज्ञात कीजिए :

(i)

संघनित्र पंप को छोड़ने वाले जल का तापमान

(ii)

कुल टरबाइन उत्पादन

(iii)

कुल प्रक्रम ऊष्मा अंतरण

टरबाइन के प्रवेश पर मानिए, h = 3410 kJ/kg एवं s = 6·802 kJ/kg K

अंत में संलग्न भाप सारणियों से आँकड़ों का भी प्रयोग कीजिए। (20 अंक)

(b)

एक अंतर्दहन (IC) इंजन के ईंधन में 85% कार्बन, 10% हाइड्रोजन, 3% ऑक्सीजन तथा शेष नाइट्रोजन, भार के अनुसार संघटन में पाया जाता है। रासायनिक रूप से सही वायु/ईंधन अनुपात का निर्धारण कीजिए। यदि 30% अतिरिक्त वायु की आपूर्ति की जाती है, तो शुष्क उत्पादों के दहन निकास का प्रतिशत संघटन, भार एवं आयतन के आधार पर ज्ञात कीजिए। (20 अंक)

(c)

एक अन्तर्दहन (IC) इंजन के ईंधन की आणविक संरचना उसकी अपरफोटन प्रवृत्ति को कैसे प्रभावित करती है ? संक्षेप में समझाइए। (10 अंक)

Q8 of the 2024 UPSC Mains Mechanical Engineering Paper II, as printed
The question as printed in the 2024 Mechanical Engineering paper

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a)(i) At 7.5 kPa, saturated liquid has T_sat = 40.3°C and v_f = 0.001008 m³/kg. The condenser pump raises the condensate to the process-return/mixing pressure of 100 kPa. For an ideal pump with incompressible liquid, w_p = v_f(100 − 7.5) kPa = 0.001008 × 92.5 = 0.0932 kJ/kg. Temperature rise = w_p/c_p = 0.0932/4.187 = 0.022 K. Therefore, T leaving condenser pump = 40.3 + 0.022 = 40.32°C ≈ 40.3°C. Condition: ideal pump, negligible kinetic and potential energy changes.

Diagram: Cogeneration plant flows

(a)(ii) At turbine inlet: h₁ = 3410 kJ/kg, s₁ = 6.802 kJ/kg K. At 500 kPa: s_f = 1.8607, s_g = 6.8213, h_f = 640.2, h_fg = 2108.5 kJ/kg. x₂ = (6.802 − 1.8607)/(6.8213 − 1.8607) = 0.9961. h₂ = 640.2 + 0.9961 × 2108.5 = 2740.5 kJ/kg.

At 7.5 kPa: s_f = 0.5764, s_g = 8.2515, h_f = 168.8, h_fg = 2405.8 kJ/kg. x₃ = (6.802 − 0.5764)/(8.2515 − 0.5764) = 0.8111. h₃ = 168.8 + 0.8111 × 2405.8 = 2120.3 kJ/kg.

Total turbine output: W_T = 25(3410 − 2740.5) + 20(2740.5 − 2120.3) = 25 × 669.5 + 20 × 620.2 = 16737.5 + 12404 = 29141.5 kW. W_T ≈ 29.14 MW.

(a)(iii) Process steam leaves as saturated liquid at 100 kPa, where h_f = 417.5 kJ/kg. Q_process = 5(2740.5 − 417.5) = 5 × 2323 = 11615 kW. Q_process ≈ 11.62 MW.

(b) Basis: 1 kg fuel. C = 0.85 kg → 0.85/12 = 0.07083 kmol C H = 0.10 kg → 0.10/1 = 0.10 kmol H atoms O = 0.03 kg → 0.03/32 = 0.0009375 kmol O₂ equivalent N = 0.02 kg → 0.02/28 = 0.000714 kmol N₂

Stoichiometric O₂ required: n_O₂,st = C + H/4 − O₂ equivalent = 0.07083 + 0.10/4 − 0.0009375 = 0.09490 kmol O₂/kg fuel.

Mass O₂ required = 0.09490 × 32 = 3.037 kg. Taking air as 21% O₂ and 79% N₂ by volume, air mass required = 3.037/0.233 = 13.03 kg/kg fuel. Chemically correct A/F ratio = 13.03 : 1.

With 30% excess air: n_O₂,supplied = 1.3 × 0.09490 = 0.12336 kmol. n_N₂ from air = 0.12336 × (79/21) = 0.46409 kmol. Total N₂ = 0.46409 + 0.000714 = 0.46480 kmol. CO₂ formed = 0.07083 kmol. Excess O₂ = 0.12336 − 0.09490 = 0.02847 kmol. H₂O formed = 0.10/2 = 0.05 kmol (not in dry products).

Dry product masses: m_CO₂ = 0.07083 × 44 = 3.117 kg m_O₂ = 0.02847 × 32 = 0.911 kg m_N₂ = 0.46480 × 28 = 13.014 kg Total dry mass = 17.042 kg.

By weight: CO₂ = 3.117/17.042 × 100 = 18.29% O₂ = 0.911/17.042 × 100 = 5.35% N₂ = 13.014/17.042 × 100 = 76.36%

Dry moles: n_CO₂ = 0.07083, n_O₂ = 0.02847, n_N₂ = 0.46480. Total = 0.56410 kmol.

By volume: CO₂ = 0.07083/0.56410 × 100 = 12.56% O₂ = 0.02847/0.56410 × 100 = 5.05% N₂ = 0.46480/0.56410 × 100 = 82.39%

(c) In SI engines, knock is autoignition of the unburnt end gas. Molecular structure controls self-ignition temperature and pre-flame reaction rate. Straight-chain paraffins form peroxides readily and autoignite early, so they knock strongly and have low octane number; knock increases with chain length. Branching and compact iso-paraffins hinder peroxide formation, giving high octane and low knock. Olefins are more knock-resistant than corresponding paraffins but less than aromatics. Aromatics have stable ring structures, high self-ignition temperatures and hence least knock; naphthenes are intermediate, better than straight-chain paraffins but worse than aromatics. In CI engines the requirement reverses: long straight-chain paraffins have high cetane number, ignite readily and reduce diesel knock, while aromatics have low cetane number and increase diesel knock.

How this answer will be evaluated

Framework: UPSC Engineering Services (Preliminary/Mains) Technical Rubric

What earns full marks: Complete working with correct values, clear diagrams, and physical interpretation of results.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Ideal cogeneration plant (25 kg/s at 7 MPa, 500 °C; 5 kg/s extracted at 500 kPa; condenser 7.5 kPa): condenser-pump exit temperature, turbine output, process heat. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Condensate at 7.5 kPa (40.3 °C) pumped to 100 kPa; w_p ≈ 0.09 kJ/kg ⇒ T ≈ 40.3 °C
    • Isentropic states: x₂ ≈ 0.996, h₂ ≈ 2740 kJ/kg at 500 kPa; x₃ ≈ 0.811, h₃ ≈ 2120 kJ/kg at 7.5 kPa
    • W_T = 25(h₁ − h₂) + 20(h₂ − h₃) ≈ 29.1 MW
    • Q_process = 5(h₂ − h_f,100 kPa) ≈ 11.6 MW

    Loses marks

    • Full 25 kg/s through LP turbine

    Earns more

    • Plant schematic with mixing chamber

    Extra mark

    • Utilisation factor
  2. (b) Stoichiometric A/F for fuel (85% C, 10% H, 3% O, 2% N) and dry exhaust composition with 30% excess air. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • O₂ required = 0.85 × 32/12 + 0.10 × 8 − 0.03 ≈ 3.04 kg/kg fuel
    • A/F = 3.04/0.23 ≈ 13.0–13.2
    • With 30% excess: CO₂ ≈ 3.12 kg, O₂ ≈ 0.91 kg, N₂ ≈ 13.0 kg per kg fuel
    • Dry products by mass ≈ CO₂ 18.3%, O₂ 5.3%, N₂ 76.4%; by volume ≈ CO₂ 12.6%, O₂ 5.0%, N₂ 82.4%

    Loses marks

    • Including H₂O in dry analysis

    Earns more

    • Fuel nitrogen included in products

    Extra mark

    • Wet products including H₂O
  3. (c) Effect of fuel molecular structure on knock tendency. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Knock in SI: end-gas autoignition; depends on pre-flame reactions
    • Straight-chain paraffins knock more; knock rises with chain length
    • Branched paraffins, olefins, naphthenes, aromatics in increasing knock resistance (octane)
    • CI engines: opposite trend (cetane number) — straight chains preferred

    Loses marks

    • Same ranking for SI and CI

    Earns more

    • Examples: n-heptane vs iso-octane

    Extra mark

    • Role of additives
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