Answer:0.646 KJ
Explanation:
Using First law for cycle
[tex]\sum Q=\sum W[/tex]
[tex]\sum Q=Q_{1-2}+Q_{3-4}[/tex]
For adiabatic process heat transfer is zero and for isothermal process
d(Q)=d(W)
[tex]Q_{1-2}=mRT_1\ln {\frac{P_1}{P_2}}[/tex]
Given [tex]P_1=2000KPa[/tex]
[tex]P_3=20KPa[/tex]
[tex]\left (\frac{T_2}{T_3}\right )^{\frac{\gamma }{\gamma -1}[/tex]=[tex]\left (\frac{P_2}{P_3}\right )}[/tex]
[tex]P_2=1089.06K[/tex]
[tex]Q_{1-2}=0.0058\dot 0.287\dot 940\ln \frac{2000}{1089.06}[/tex]=[tex]0.95KJ[/tex]
[tex]Q_{3-4}=mRT_2\ln {\frac{P_3}{P_4}}[/tex]
[tex]\left (\frac{T_1}{T_4}\right )^{\frac{\gamma }{\gamma -1}[/tex]=[tex]\left (\frac{P_1}{P_4}\right )}[/tex]
Now we have to find [tex]P_4=36.72KPa[/tex]
[tex]Q_{3-4}=0.0058\dot 0.287\dot 300\ln \frac{20}{36.72}[/tex]=[tex]-0.30341KJ[/tex]
[tex]Q_{net}=Q_{1-2}+Q_{3-4}[/tex]
[tex]Q_{net}=0.95-0.303=0.646KJ[/tex]
[tex]Q_{net}=W_{net}=0.646KJ[/tex]
