Paper II — 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…
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
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
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
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)
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)
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)
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) Given: W_c = 60 kW, COP = 6.0. Refrigeration effect: Q_L = COP × W_c = 6.0 × 60 = 360 kW.
In the VCR cycle, throttling gives h₄ = h₃ = 114.95 kJ/kg. Refrigeration effect per kg = h₁ − h₄ = 275.76 − 114.95 = 160.81 kJ/kg. Mass flow rate: m = Q_L / (h₁ − h₄) = 360 / 160.81 = 2.2387 kg/s.
Compressor work per kg: w_c = W_c / m = 60 / 2.2387 = 26.8017 kJ/kg. h₂ = h₁ + w_c = 275.76 + 26.8017 = 302.5617 kJ/kg.
At 35 °C, saturated vapour enthalpy h_g = 283.89 kJ/kg. For superheated vapour, taking Cₚ = 0.62 kJ/kg-K: h₂ − h_g = Cₚ(T₂ − 35). T₂ = 35 + (302.5617 − 283.89) / 0.62 T₂ = 35 + 30.12 = 65.12 °C.
Exit temperature of refrigerant ≈ 65.12 °C.
(b) Use isentropic relations for compressor and turbine. T₁ = 15 + 273 = 288 K. For air: γₐ = 1.4, Cₚₐ = 1.005 kJ/kg-K. T₂ = T₁ r_p^((γₐ−1)/γₐ) = 288 × 7.5^(2/7) = 512.16 K. w_c = 1.005(512.16 − 288) = 225.28 kJ/kg.
For gas turbine: T₃ = 750 + 273 = 1023 K, γ_g = 1.33, Cₚg = 1.11 kJ/kg-K. T₄ = T₃ / r_p^((γ_g−1)/γ_g) = 1023 / 7.5^(33/133) = 620.52 K. w_gt = 1.11(1023 − 620.52) = 446.75 kJ/kg.
Net gas-turbine work: w_net,GT = 446.75 − 225.28 = 221.47 kJ/kg air.
Steam properties from steam tables/Mollier chart: At 50 bar, 600 °C: h₁ = 3662.5 kJ/kg, s₁ = 7.2589 kJ/kg-K. At 0.1 bar: h_f = 192 kJ/kg, h_fg = 2392.7 kJ/kg, s_f = 0.6493 kJ/kg-K, s_fg = 7.5009 kJ/kg-K.
Isentropic expansion: s₂ = s₁. x₂ = (7.2589 − 0.6493) / 7.5009 = 0.8812. h₂ = 192 + 0.8812 × 2392.7 = 2300.5 kJ/kg. w_ST = h₁ − h₂ = 3662.5 − 2300.5 = 1362.0 kJ/kg steam.
Heat added to steam per kg: q_steam = h₁ − h_f = 3662.5 − 192 = 3470.5 kJ/kg.
HRSG gas temperatures: after supplementary firing T₅ = 1023 K; gas leaves at T₆ = 100 + 273 = 373 K. q_HRSG = Cₚg(T₅ − T₆) = 1.11(1023 − 373) = 721.5 kJ/kg air. Thus mₛ / mₐ = 721.5 / 3470.5 = 0.2079.
Total work per kg air: w_total = 221.47 + 0.2079 × 1362.0 = 504.62 kJ/kg air.
(i) Total power = 200 MW = 200000 kW. mₐ = 200000 / 504.62 = 396.3 kg/s. mₛ = 0.2079 × 396.3 = 82.4 kg/s.
Flow rate of air ≈ 396.3 kg/s; flow rate of steam ≈ 82.4 kg/s.
(ii) W_GT = 396.3 × 221.47 = 87770 kW = 87.8 MW. W_ST = 82.4 × 1362.0 = 112230 kW = 112.2 MW.
Gas turbine output ≈ 87.8 MW; steam turbine output ≈ 112.2 MW.
(iii) Heat supplied in main GT combustor: Q₁ = Cₚg T₃ − Cₚₐ T₂ = 1.11 × 1023 − 1.005 × 512.16 Q₁ = 1135.53 − 514.72 = 620.81 kJ/kg air. Heat supplied by supplementary fuel: Q₂ = 1.11(1023 − 620.52) = 446.75 kJ/kg air. Total heat input per kg air = 620.81 + 446.75 = 1067.56 kJ/kg air. η_combined = w_total / Q_in = 504.62 / 1067.56 = 0.4727.
Thermal efficiency of combined plant ≈ 47.3%.
(iv) Total fuel flow: m_f = mₐ Q_in / CV = 396.3 × 1067.56 / 43300 = 9.77 kg/s. Air-fuel ratio = mₐ / m_f = 396.3 / 9.77 = 40.6.
Overall air-fuel ratio ≈ 40.6 kg air/kg fuel.
(c) Let Q_H be heat supplied to the topping cycle. Topping work: W₁ = η₁ Q_H. Heat rejected by topping: Q_M = (1 − η₁)Q_H.
Since the bottom cycle completely absorbs this rejected heat: W₂ = η₂ Q_M = η₂(1 − η₁)Q_H.
Total combined work: W_c = W₁ + W₂ = η₁ Q_H + η₂(1 − η₁)Q_H.
Combined efficiency: η_c = W_c / Q_H = η₁ + η₂(1 − η₁) η_c = η₁ + η₂ − η₁η₂ η_c = 1 − (1 − η₁)(1 − η₂).
Efficiency of the combined reversible series cycle: η_c = η₁ + η₂ − η₁η₂ = 1 − (1 − η₁)(1 − η₂).
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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) derive: given > assumptions > stepwise derivation > result > check Full marks: All parts solved with correct method, clear steps, and physical interpretation.
Key points expected
- Calculate mass flow rate using COP and compressor power
- Apply energy balance to find enthalpy at compressor exit
- Use specific heat relation to find exit temperature
- State assumptions regarding superheat or state
- Calculate gas turbine power output and air flow rate
- Determine steam flow rate and steam turbine power
- Compute overall thermal efficiency of the plant
- Calculate air-fuel ratio using calorific value
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine the temperature of the refrigerant at the compressor exit. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate mass flow rate using COP and compressor power
- Apply energy balance to find enthalpy at compressor exit
- Use specific heat relation to find exit temperature
- State assumptions regarding superheat or state
Loses marks
- Incorrect mass flow rate calculation
- Missing energy balance equation
- Arithmetic errors in final temperature
Earns more
- Correct identification of VCR cycle states
- Consistent units throughout calculation
- Clear step-by-step substitution of values
Extra mark
- Sketch of T-s diagram marking states
- (b) Determine flow rates, power outputs, efficiency, and air-fuel ratio for the combined cycle. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate gas turbine power output and air flow rate
- Determine steam flow rate and steam turbine power
- Compute overall thermal efficiency of the plant
- Calculate air-fuel ratio using calorific value
Loses marks
- Incorrect pressure ratio application
- Neglecting heat recovery in steam generator
- Unit conversion errors in power calculations
Earns more
- Correct application of isentropic relations
- Accurate use of Mollier diagram for steam states
- Clear separation of gas and steam cycle calculations
Extra mark
- Schematic diagram of combined cycle plant
- (c) Derive the efficiency of the combined cycle in terms of η₁ and η₂. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Define thermal efficiency for each cycle
- Apply energy balance for heat transfer between cycles
- Derive combined efficiency formula algebraically
- State assumption of complete heat recovery
Loses marks
- Incorrect energy balance setup
- Algebraic errors in derivation
- Missing assumption of reversibility
Earns more
- Clear definition of topping and bottom cycles
- Logical step-by-step algebraic manipulation
- Physical interpretation of the result
Extra mark
- T-s diagram showing coupled cycles
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