Respuesta :
Answer:
a). [tex]\frac{\dot{W}}{m}= 311[/tex] kJ/kg
b). [tex]\frac{\dot{\sigma _{gen}}}{m}=0.9113[/tex] kJ/kg-K
Explanation:
a). The energy rate balance equation in the control volume is given by
[tex]\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0[/tex]
[tex]\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})[/tex]
[tex]\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})[/tex]
[tex]\frac{\dot{W}}{m}= -30+1.1(980-670)[/tex]
[tex]\frac{\dot{W}}{m}= 311[/tex] kJ/kg
b). Entropy produced from the entropy balance equation in a control volume is given by
[tex]\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0[/tex]
[tex]\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})[/tex]
[tex]\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}[/tex]
[tex]\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}[/tex]
[tex]\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978[/tex]
[tex]\frac{\dot{\sigma _{gen}}}{m}=0.9113[/tex] kJ/kg-K
