Answer:
Explanation:
Check attachment for p-v Diagram
a.)
Calculate the maximum temperature of the cycle by using ideal gas equation:
[tex]P_2V_2=mRT_2\\(574)(0.3)=(1)(0.287)T_2\\T_2=600K\\T_{max}=T_2=600K[/tex]
Calculate the minimum temperature of the cycle:
[tex]\eta_{th}=1-\frac{T_{min}}{T_{max}}\\\\\frac{50}{100}=1-\frac{T_{min}}{600}\\\\T_{min}=300K[/tex]
b.)
Calculate the volume at the beginning of the isothermal expansion by using the following expression:
[tex]Q_{in}=mP_2V_2In(\frac{V_2}{V_1})\\\\50=(1)(574)(0.3)In(\frac{0.3}{V_1})\\\\V_1=0.224m^3[/tex]
Calculate the pressure at the beginning of the isothermal expansion by using the following expression:
[tex]\frac{P_1}{P_2}=\frac{V_2}{V_1}\\\\\frac{P_1}{574}=\frac{0.3}{0.224}\\\\P_1=768.75KPa=769kPa[/tex]
c.)
Process 1-2:
Heat addition for the Process 1-2 is [tex]Q_{1-2}=Q_{in}=50KJ[/tex]
Calculate the work transfer for the process 1-2:
[tex]Q_{1-2}=U_2-U_1+W_{1-2)\\50=mc_v(T_2-T_1)+W_{1-2}\\50=mc_v(T_1-T_1)+W_{1-2}\\W_{1-2}=50kJ[/tex]
Process 2-3:It is an Adiabatic Process:
Heat transfer for the Process 2-3 is [tex]Q_{2-3}=0kJ[/tex]
Calculate the work transfer for the process 2-3:
[tex]W_{2-3}=\frac{mR(T_2-T_3)}{\gamma - 1}\\\\=\frac{(1)(0.287)(600-300)}{1.4-1}\\\\W_{2-3}=215.25KJ[/tex]
Process 3-4:
Calculate the heat transfer for the process 3-4:
[tex]\eta_{th}=1-\frac{Q_{3-4}}{Q_{in}}\\\\0.5=1-\frac{Q_{3-4}}{50}\\\\Q_{3-4}=25KJ[/tex]
Calculate the work transfer for the process 3-4:
[tex]Q_{3-4}=U_4-U_3+W_{3-4)\\25=mc_v(T_4-T_3)+W_{3-4}\\25=mc_v(T_4-T_4)+W_{3-4}\\W_{3-4}=25kJ[/tex]
Process 4-1: It is an Adiabatic Process:
Heat transfer for the process 4-1 is [tex]Q_{4-1}=0kJ[/tex]
Calculate the work transfer for the process 4-1:
[tex]W_{4-1}=\frac{mR(T_4-T_1)}{\gamma - 1}\\\\=\frac{(1)(0.287)(300-600)}{1.4-1}\\\\W_{4-1}=-215.25KJ[/tex]
