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d) Calculate the maximum cycle pressure. Pmaz bar e) Calculate the maximum cycle temperature. Tmaz = K f) Calulate the temperature at point 5. K %3D

d) Calculate the maximum cycle pressure. Pmaz bar e) Calculate the maximum cycle temperature. Tmaz = K f) Calulate the temperature at point 5. K %3D

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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d) Calculate the maximum cycle pressure. bar mar e) Calculate the maximum cycle temperature. Tmaz K f) Calulate the temperature at point 5. T; = K
Transcribed Image Text:d) Calculate the maximum cycle pressure. bar mar e) Calculate the maximum cycle temperature. Tmaz K f) Calulate the temperature at point 5. T; = K
The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant Нeat IN Adiabatic expansion Heat OUT (Constant volume heat rejection) Constant volume heat addition 1 Adiabatic compression Volume (V) Using the Dual Air Standard Cycle and the following data: • Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK; y = 1.4 Cycle compression ratio = 11 (v1/v2) Air conditions prior to compression: Pressure = 1.6 bar; Temperature = 298K Heat added at constant volume = 290000 J/kg %3D Heat added at constant pressure = 579000 J/kg Pressure (P) 2]
Transcribed Image Text:The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 PV = constant Нeat IN Adiabatic expansion Heat OUT (Constant volume heat rejection) Constant volume heat addition 1 Adiabatic compression Volume (V) Using the Dual Air Standard Cycle and the following data: • Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK; y = 1.4 Cycle compression ratio = 11 (v1/v2) Air conditions prior to compression: Pressure = 1.6 bar; Temperature = 298K Heat added at constant volume = 290000 J/kg %3D Heat added at constant pressure = 579000 J/kg Pressure (P) 2]
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