Mechanical Engineering 2022 Paper II 50 marks Calculate

Q8

(a) A VCR cycle refrigerator driven by a 60 kW compressor has a COP of 6·0. The enthalpies of saturated liquid and saturated vapor refrigerant at condenser temperature of 35 °C are 114·95 kJ/kg and 283·89 kJ/kg, respectively. The saturated refrigerant vapor leaving evaporator has an enthalpy of 275·76 kJ/kg. Find the temperature of refrigerant at the exit of compressor. The Cₚ of refrigerant is 0·62 kJ/kg-K. (20 marks) (b) In a combined gas turbine-steam turbine power plant, the exhaust gas from the open-cycle gas turbine is the supply gas to the steam generator of the steam cycle at which additional fuel is burnt in the gas. The pressure ratio for the gas turbine is 7·5, the air inlet temperature is 15 °C and the maximum temperature is 750 °C. Combustion of additional fuel raises the gas temperature to 750 °C and the gas leaves the steam generator at 100 °C. The steam is supplied to the steam turbine at 50 bar and 600 °C and the condenser pressure is 0·1 bar. The total power output of the plant is 200 MW. The calorific value of the fuel burnt is 43·3 MJ/kg. Neglecting the effect of the mass flow rate of fuel on the air flow, determine (i) the flow rate of air and steam required, (ii) the power outputs of the gas turbine and steam turbine, (iii) the thermal efficiency of the combined plant and (iv) the air-fuel ratio. Take Cₚ = 1·11 kJ/kg-K and γ = 1·33 for combustion gases; and Cₚ = 1·005 kJ/kg-K and γ = 1·4 for air. Neglect pump work. Condensate enthalpy at 0·1 bar = 192 kJ/kg. [Mollier diagram is attached in Page No. 14] (20 marks) (c) Two vapor power cycles are coupled in series where heat lost by one is absorbed by the other completely. If η₁ is the thermal efficiency of the topping cycle and η₂ is the thermal efficiency of the bottom cycle, determine the efficiency of the combined cycle in terms of these efficiencies. Assume cycles to be reversible. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) 60 kW के एक संपीडक से चलने वाले एक वाष्प संपीडन प्रशीतन (वी० सी० आर०) चक्र प्रशीतित्र का निष्पादन गुणांक (सी० ओ० पी०) 6·0 है। संघनित्र द्रव तथा संतृप्त प्रशीतक वाष्प की एन्थैल्पी, 35 °C के संपीडित तापमान पर क्रमशः: 114·95 kJ/kg तथा 283·89 kJ/kg है। वाष्पित्र से निकलने वाले संतृप्त प्रशीतक वाष्प की एन्थैल्पी 275·76 kJ/kg है। संपीडक के निर्गम पर प्रशीतक का तापमान ज्ञात कीजिये। प्रशीतक का Cₚ = 0·62 kJ/kg-K है। (20 अंक) (b) एक गैस टरबाइन-भाप टरबाइन संयुक्त शक्ति संयंत्र में, विवृत चक्र गैस टरबाइन की निकास गैस, भाप चक्र के भाप जनित्र को आपूर्ति की जाती है, जिसमें गैस में अतिरिक्त ईंधन जलाया जाता है। गैस टरबाइन का दाब अनुपात 7·5 है, वायु का अंतर्गम तापमान 15 °C तथा अधिकतम तापमान 750 °C है। अतिरिक्त ईंधन के दहन से गैस तापमान 750 °C तक बढ़ जाता है तथा भाप जनित्र से गैस 100 °C पर निकलती है। भाप टरबाइन में भाप की आपूर्ति 50 बार तथा 600 °C पर की जाती है और संघनित्र दाब 0·1 बार है। संयंत्र का कुल शक्ति उत्पादन 200 MW है। दहन किये गये ईंधन का कैलोरी मान 43·3 MJ/kg है। ईंधन की मात्रा प्रवाह दर के वायु प्रवाह पर पड़ने वाले प्रभाव को नगण्य मानते हुए, गणना कीजिये (i) वांछित वायु तथा भाप प्रवाह दर, (ii) गैस टरबाइन तथा भाप टरबाइन का शक्ति उत्पादन, (iii) संयुक्त शक्ति संयंत्र की तापीय दक्षता और (iv) वायु-ईंधन अनुपात। दहन गैसों के लिये Cₚ = 1·11 kJ/kg-K तथा γ = 1·33; वायु के लिये Cₚ = 1·005 kJ/kg-K तथा γ = 1·4 लीजिये। पम्प कार्य नगण्य है। संघनित्र की 0·1 बार पर एन्थैल्पी 192 kJ/kg लीजिये। [पृष्ठ संख्या 14 में मोलियर आरेख संलग्न है] (20 अंक) (c) दो वाष्प शक्ति चक्रों को श्रृंखला में जोड़ा गया है जहाँ कि एक की लुप्त ऊष्मा, दूसरे द्वारा पूर्ण रूप से अवशोषित की जाती है। यदि अधियोजी चक्र की तापीय दक्षता η₁ हो तथा अधस्तलन चक्र की तापीय दक्षता η₂ हो, तो युगल-चक्र की दक्षता इन दक्षताओं के रूप में निर्धारित कीजिए। चक्रों को उत्क्रमणीय मानिए। (10 अंक)

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Approach

Calculate the compressor exit temperature for part (a) using energy balance and isentropic relations; for part (b) systematically determine air/steam flow rates, power outputs, thermal efficiency and air-fuel ratio using gas turbine and steam cycle analysis with the given Mollier diagram data; for part (c) derive the combined cycle efficiency formula using reversible heat engine principles. Allocate approximately 35% time to part (a), 50% to part (b) due to its four sub-requirements, and 15% to part (c) for the derivation.

Key points expected

  • Part (a): Refrigeration effect = COP × Ẇ = 360 kW; mass flow rate ṁ = 360/(275.76-114.95) = 2.237 kg/s; compressor work per kg = 26.82 kJ/kg; T₂ = T₁ + (h₂-h₁)/Cp with h₂ from energy balance yielding T₂ ≈ 52-55°C
  • Part (b): Gas turbine: T₂ = T₁×(r_p)^((γ-1)/γ) = 288×2.143 = 617.3 K; T₃ = 1023 K; T₄ = T₃/(r_p)^((γ-1)/γ) = 477.4 K; heat added in combustor = Cp,g×(1023-477.4); additional heat in HRSG raises to 1023 K; steam cycle: h₁ from 50 bar/600°C using Mollier chart ≈ 3666 kJ/kg, h₂ from 0.1 bar ≈ 2260 kJ/kg (isentropic)
  • Part (b)(i): Energy balance on HRSG gives ṁ_air and ṁ_steam; ṁ_steam = 200 MW total / [(h₁-h₂) + (Ẇ_GT/ṁ_steam)] after iteration
  • Part (b)(ii): Ẇ_GT = ṁ_air×Cp,g×[(1023-617.3)+(1023-477.4)]; Ẇ_ST = ṁ_steam×(h₁-h₂)
  • Part (b)(iii): η_combined = (Ẇ_GT + Ẇ_ST)/(ṁ_fuel×CV) with total fuel in GT combustor and HRSG
  • Part (b)(iv): AFR = ṁ_air/ṁ_fuel from energy balance Q = ṁ_fuel×CV = ṁ_air×Cp,g×ΔT
  • Part (c): For reversible cycles, η_combined = η₁ + η₂ - η₁η₂ derived from Q₁→W₁+Q₂ with Q₂ = Q₁(1-η₁) and W₂ = η₂Q₂

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness22%11Correctly applies VCR cycle energy balance for part (a); identifies topping-bottoming arrangement with supplementary firing for part (b); uses appropriate γ values for air vs combustion gases; correctly models HRSG as heat exchanger with temperature pinch; derives η_combined = η₁ + η₂ - η₁η₂ for part (c) using reversible heat engine principles.Uses correct basic formulas but confuses Cp values between air and combustion gases; treats gas turbine exhaust temperature as fixed rather than calculating with supplementary firing; states correct final formula for part (c) but derivation lacks rigor.Applies wrong cycle (e.g., heat pump instead of refrigerator); ignores supplementary firing in combined cycle; adds efficiencies directly for part (c) giving η₁+η₂.
Numerical accuracy24%12Part (a): T₂ ≈ 52-55°C with clear interpolation; Part (b): ṁ_air ≈ 450-500 kg/s, ṁ_steam ≈ 60-70 kg/s, Ẇ_GT ≈ 80-90 MW, Ẇ_ST ≈ 110-120 MW, η_combined ≈ 52-58%, AFR ≈ 50-60; all values within 5% of expected with proper unit handling.Correct methodology but arithmetic errors in temperature ratios or enthalpy readings; final answers within 10-15% of expected; inconsistent units (kJ vs MJ).Order of magnitude errors in flow rates; ignores Cp differences leading to 20%+ errors; wrong steam enthalpy values from Mollier diagram; part (c) gives numerical answer instead of symbolic expression.
Diagram quality14%7T-s diagram for VCR showing compression 1-2, condensation 2-3, throttling 3-4, evaporation 4-1 with superheated compression end; combined cycle layout showing GT, combustor, HRSG with supplementary firing, ST, condenser; temperature-heat diagram for part (c) showing heat cascading between cycles.Basic T-s diagram for refrigeration cycle but missing superheat region; schematic of combined plant but unclear on supplementary firing location; no diagram for part (c).No diagrams provided; or incorrect cycles drawn (Rankine instead of VCR for part a); confused flow directions in combined cycle.
Step-by-step derivation22%11Explicit energy balances at each component; clear statement of assumptions (isentropic efficiencies=1, neglecting pump work, fuel mass negligible); systematic solution of simultaneous equations for part (b); rigorous derivation for part (c) starting from first law for reversible engines showing Q_H1 = W1 + Q_L1 = W1 + Q_H2.Some intermediate steps skipped but key equations present; jumps from given data to final answers without showing HRSG energy balance; part (c) derivation starts correctly but skips algebraic manipulation.Final answers stated without derivation; no clear identification of control volumes; part (c) gives answer without derivation or uses incorrect starting point.
Practical interpretation18%9Comments on superheated compression in part (a) preventing liquid damage; discusses why supplementary firing is used in part (b) vs heat recovery alone; notes typical Indian combined cycle efficiencies (NTPC plants ~55%); explains why η_combined > η₁ or η₂ individually in part (c) due to waste heat utilization; mentions environmental benefits of combined cycles.Brief mention of why combined cycles are efficient; notes superheat is desirable but no explanation; generic statement about fuel saving.No practical interpretation; treats as pure mathematical exercise; no connection to real power plant operation or Indian energy sector context.

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