Question

0.1 kg of air as an ideal gas executes a Carnot power cycle in a piston-cylinder assembly. The cycle has a thermal efficiency of 50%. The heat transfer to the air during the isothermal expansion is 50 kJ. At the end of the isothermal expansion, the pressure is 574 kPa and the volume is 0.3 m^3. Determine:a. the maximum and minimum temperatures for the cycle (ans: TH = 600K, TC = 300 K).b. the pressure and volume at the beginning of the isothermal expansion (ans: V1 = 0.224 m3, p1 = 769 kPa).c. the work and heat transfer for each of the for processes (ans: W12 = 50 kJ, Q12 = 50 kJ, W23 = 221 kJ, Q23 = 0 kJ, W34 = -25 kJ, Q34 = -25 kJ, W41 = -221 kJ, Q41 = 0 kJ).d. Sketch the cycle on a p-v diagram.

Answer 1

Step-by-step explanation:

Check attachment for p-v Diagram

a.)

Calculate the maximum temperature of the cycle by using ideal gas equation:

`P_2V_2=mRT_2\\(574)(0.3)=(1)(0.287)T_2\\T_2=600K\\T_(max)=T_2=600K`

Calculate the minimum temperature of the cycle:

`\eta_(th)=1-(T_(min))/(T_(max))\\\\(50)/(100)=1-(T_(min))/(600)\\\\T_(min)=300K`

b.)

Calculate the volume at the beginning of the isothermal expansion by using the following expression:

`Q_(in)=mP_2V_2In((V_2)/(V_1))\\\\50=(1)(574)(0.3)In((0.3)/(V_1))\\\\V_1=0.224m^3`

Calculate the pressure at the beginning of the isothermal expansion by using the following expression:

`(P_1)/(P_2)=(V_2)/(V_1)\\\\(P_1)/(574)=(0.3)/(0.224)\\\\P_1=768.75KPa=769kPa`

c.)

Process 1-2:

Heat addition for the Process 1-2 is
`Q_(1-2)=Q_(in)=50KJ`

Calculate the work transfer for the process 1-2:

`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`

Process 2-3:It is an Adiabatic Process:

Heat transfer for the Process 2-3 is
`Q_(2-3)=0kJ`

Calculate the work transfer for the process 2-3:

`W_(2-3)=(mR(T_2-T_3))/(\gamma - 1)\\\\=((1)(0.287)(600-300))/(1.4-1)\\\\W_(2-3)=215.25KJ`

Process 3-4:

Calculate the heat transfer for the process 3-4:

`\eta_(th)=1-(Q_(3-4))/(Q_(in))\\\\0.5=1-(Q_(3-4))/(50)\\\\Q_(3-4)=25KJ`

Calculate the work transfer for the process 3-4:

`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`

Process 4-1: It is an Adiabatic Process:

Heat transfer for the process 4-1 is
`Q_(4-1)=0kJ`

Calculate the work transfer for the process 4-1:

`W_(4-1)=(mR(T_4-T_1))/(\gamma - 1)\\\\=((1)(0.287)(300-600))/(1.4-1)\\\\W_(4-1)=-215.25KJ`