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 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 :
temperature of water leaving the condenser pump
total turbine output
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
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
Explain in brief, how the molecular structure of the IC engine fuels affects the tendency to knock. 10 marks
हिंदी में प्रश्न पढ़ें
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 :
संघनित्र पंप को छोड़ने वाले जल का तापमान
कुल टरबाइन उत्पादन
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)
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)
एक अन्तर्दहन (IC) इंजन के ईंधन की आणविक संरचना उसकी अपरफोटन प्रवृत्ति को कैसे प्रभावित करती है ? संक्षेप में समझाइए। (10 अंक)
Model answer
Written 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.
(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.
(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 and naphthenes have stable ring structures, high self-ignition temperatures and hence least knock. 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.
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
How this answer will be evaluated
Approach
(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) explain: definition/context > points in order > small example > short close Full marks: Complete working with diagrams, correct units, and clear physical interpretation.
Key points expected
- Schematic diagram with numbered states
- Steam table data for h and s at each state
- Energy balance for turbine and process heater
- Pump work calculation using v_f * (P2 - P1)
- Stoichiometric A/F calculation based on C, H, O content
- Combustion equation with 30% excess air
- Mass of each dry product species (CO2, N2, O2)
- Conversion of mass fractions to volume fractions
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine pump outlet temperature, turbine work, and process heat for a cogeneration cycle. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Schematic diagram with numbered states
- Steam table data for h and s at each state
- Energy balance for turbine and process heater
- Pump work calculation using v_f * (P2 - P1)
Loses marks
- Missing state points on diagram
- Using h_f for saturated liquid instead of compressed liquid
- Ignoring pump work in energy balance
Earns more
- T-s diagram of the cycle
- Explicit mass flow rate split (20 kg/s vs 5 kg/s)
- Interpolation of steam table values
Extra mark
- Calculation of cycle thermal efficiency
- (b) Determine stoichiometric A/F ratio and dry exhaust composition by weight and volume. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Stoichiometric A/F calculation based on C, H, O content
- Combustion equation with 30% excess air
- Mass of each dry product species (CO2, N2, O2)
- Conversion of mass fractions to volume fractions
Loses marks
- Ignoring oxygen content in fuel for A/F calc
- Confusing mass and volume percentages
- Incorrect stoichiometric coefficients
Earns more
- Explicit molar mass calculations
- Balanced chemical equation for the fuel
Extra mark
- Calculation of wet exhaust composition
- (c) Explain the relationship between fuel molecular structure and knock tendency. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Definition of knock (auto-ignition)
- Effect of carbon chain length (straight vs branched)
- Effect of aromatic and olefinic structures
- Link to Octane Number (RON/MON)
Loses marks
- Confusing knock with pre-ignition
- Vague statements without structural examples
Earns more
- Mention of specific fuel types (e.g., iso-octane vs n-heptane)
- Brief mention of cetane number for diesel
Extra mark
- Mention of lead tetra-ethyl as historical anti-knock agent
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