Answer:
a) [tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex], b) [tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]
Explanation:
a) The process within the turbine is modelled after the First Law of Thermodynamics:
[tex]-q_{out} - w_{out} + h_{in}-h_{out} = 0[/tex]
[tex]w_{out} = h_{in} - h_{out}-q_{out}[/tex]
[tex]w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}[/tex]
[tex]w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}[/tex]
[tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex]
b) The entropy production is determined after the Second Law of Thermodynamics:
[tex]-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)[/tex]
[tex]s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)[/tex]
[tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]
