problems on steam cycle
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A steam turbine operates between 20 bar steam inlet and 0.07 bar exhaust. (1) The heat rejected is calculated as 1938.52 kJ/kg using isentropic expansion and steam tables. (2) The work done by the turbine is 923.48 kJ/kg. (3) The net work done is 921.487 kJ/kg after accounting for pump work. (4) The heat supplied is calculated as 2860 kJ/kg using the net work and heat rejected. (5) The thermal efficiency is 32.21% and theoretical steam consumption is 3.91 kg/kWh.
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Recently uploaded (20)
Building Production Ready Search Pipelines with Spark and Milvus
Building Production Ready Search Pipelines with Spark and Milvus
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GraphSummit Singapore | The Art of the Possible with Graph - Q2 2024
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20240609 QFM020 Irresponsible AI Reading List May 2024
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Infrastructure Challenges in Scaling RAG with Custom AI models
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HCL Notes and Domino License Cost Reduction in the World of DLAU
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Observability Concepts EVERY Developer Should Know -- DeveloperWeek Europe.pdf
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National Security Agency - NSA mobile device best practices
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20240605 QFM017 Machine Intelligence Reading List May 2024
problems on steam cycle
- 1. PRESENTED BY, KARTHIK P -16M628 JEYANTH M-16M626 KOTA AKHILESH-16M630 LEO MICHAEL KEVIN-16M631 MAISH DASH- 16M632 MANOJ V-16M633 MARIKUMAR P-16M634 BATCH 4
- 2. PROBLEM 2: A steam turbine receives steam at pressure 20 bar and superheated to 88.6°C. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically Using steam table, calculate the following. (a) Heat rejected (b) Heat supplied, assuming that the feed pump supplies water to the boiler at 20 bar (c) Net work done (d) Work done by the turbine (e) Thermal efficiency (f) Theoretical steam consumption.
- 3. P-V AND T-S DIAGRAM
- 4. Given: Superheated steam supplied to a turbine at pressure, p1 = 20 bar Temperature of superheated steam = ts,1 + 88.6°C. From Steam table (Dry saturated steam) At pressure, p1 = 20 bar → ts,1 = 212.4°C Therefore, temperature of superheated steam, tsup,1 = 2l2.4 + 88.6 = 300°C From Steam table (superheated steam) At pressure, p1 = 20 bar and tsup,1 = 300°C → hsup,1 = 3025 kJ/kg, ssup,1 = 6.768 kJ/kg Given: Steam expanded isentropically to pressure, p2 = 0.07 bar. From Steam table (Dry saturated steam) At pressure, p2 = 0.07 bar → hf,3 = 163 kJ/kg, hfg = 2409.2 kJ/kg, s f,3 = 0.552 kJ/kgK, sfg = 7.722 kJ/kgK, v f,3 = 0.001 m3/kg From Steam table (Dry saturated steam) At pressure, p2 = 0.07 bar → hf,3 = 163 kJ/kg, hfg = 2409.2 kJ/kg, s f,3 = 0.552 kJ/kgK, sfg = 7.722 kJ/kgK, v f,3 = 0.001 m3/kg
- 5. a)Determine heat rejected, qR Formula: Heat rejected, qR = h2 – hf,3 Finding unknown, h2: Isentropically expansion, therefore s2 = ssup,1 = 6.768 Since for wet steam at point ‘2’, s2 = sf,3+ x2 sfg,(p=0.07 bar) Therefore, sf,3+ x2 sfg,(p=0.07 bar) = 6.768 0.552 + x2 (7.722) = 6.768 x2 = 0.8047 Enthalpy for wet steam at point ‘2’, h2 = h f,3 + x2hfg,(p=0.07 bar) h2 = 163.4 + 0.8047 (2409.2) = 2101.52 kJ/kg Answer: Heat rejected, qR = h2 – hf,3 = 2101.52 - 163 = 1938.52 kJ/kg
- 6. (d) Determined work done by the turbine, wt : Formula: Turbine work, wt = hsup,1 – h2 Answer: Turbine work, wt = hsup,1 – h2 = 3025 - 2101.52 = 923.48 kJ/kg (c) Determine net work done, w : Formula: Net work done during cycle, w = wt – wp Finding unknown, wp : Pump work, wp = vf,3 (p1 – p2) 102 = 0.001 (20 – 0.07) 102 = 1.993 kJ/kg Answer: Net work done during cycle, w = wt – wp = [923.48 – 1.993] = 921.487 kJ/kg
- 7. b) Determine heat supplied, qA : Formula: Net work done during cycle (w) = Heat supplied (qA) – heat rejected (qR) qA = w + qR Answer: qA = w + qR = 921.487 + 1938.52 = 2860.0 kJ/kg (e) Thermal efficiency, ηth: Answer: Thermal efficiency, ηth = 32.21% (f) Theoretical steam consumption: Answer: Theoretical steam consumption = = 3.91 kg/kWh