TWO isentropic PROCESSES featured in the Carnot cycle are replaced by two constant-volume regeneration processes.Temperature at the beginning of isothermal compression is T2 = 127 C = (127 + 273) K = 400 KThe thermal efficiency of the Stirling cycle η = 1 \(- \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\) = 50% = 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, \(\frac{{400}}{{{{\rm{T}}_1}}}\) = 1 0.5 = 0.5∴ T1 = \(\frac{{400}}{{0.5}}\) = 800 KNow, HEAT transferred to the regenerator (Qreg) = Cv(T1-T2)(Qreg)= Cv (800-400) kJ/Kg∴ (Qr eg) = 400 Cv kJ/K

"> TWO isentropic PROCESSES featured in the Carnot cycle are replaced by two constant-volume regeneration processes.Temperature at the beginning of isothermal compression is T2 = 127 C = (127 + 273) K = 400 KThe thermal efficiency of the Stirling cycle η = 1 \(- \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\) = 50% = 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, \(\frac{{400}}{{{{\rm{T}}_1}}}\) = 1 0.5 = 0.5∴ T1 = \(\frac{{400}}{{0.5}}\) = 800 KNow, HEAT transferred to the regenerator (Qreg) = Cv(T1-T2)(Qreg)= Cv (800-400) kJ/Kg∴ (Qr eg) = 400 Cv kJ/K

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In an engine working on air standard Stirling cycle, the temperature at the beginning of isothermal compression is 127°C. The engine thermal efficiency is 50%. The specific heat of air at constant volume is Cv. The heat transferred to the regenerator is

Power Engineering Gas And Vapor Power Cycles in Power Engineering 8 months ago

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Explanation:Stirling cycle, which is an altered version of the Carnot cycle in which the TWO isentropic PROCESSES featured in the Carnot cycle are replaced by two constant-volume regeneration processes.Temperature at the beginning of isothermal compression is T2 = 127 C = (127 + 273) K = 400 KThe thermal efficiency of the Stirling cycle η = 1 \(- \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}\) = 50% = 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, 0.5 = 1 \(- \frac{{400}}{{{{\rm{T}}_1}}}\)Or, \(\frac{{400}}{{{{\rm{T}}_1}}}\) = 1 0.5 = 0.5∴ T1 = \(\frac{{400}}{{0.5}}\) = 800 KNow, HEAT transferred to the regenerator (Qreg) = Cv(T1-T2)(Qreg)= Cv (800-400) kJ/Kg∴ (Qr eg) = 400 Cv kJ/K

Posted on 11 Nov 2024, this text provides information on Power Engineering related to Gas And Vapor Power Cycles in Power Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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