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# Air is expanded in an adiabatic turbine of 90 percent isentropic efficiency from an inlet state of 2800 kPa and $400^{\circ} \mathrm{C}$ to an outlet pressure of 150 kPa. Calculate the outlet temperature of air, the work produced by this turbine, and the entropy generation.

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First the final temperature for the isentropic process can be determined:

\begin{align*} T_{\text{2s}}&=T_{1}\Bigg(\dfrac{P_{\text{2s}}}{P_{1}}\Bigg)^{(k-1)/k}\\ &=673\cdot\Bigg(\dfrac{150}{2800}\Bigg)^{(1.4-1)/1.4}\\ &=292\:\text{K} \end{align*}

The actual final temperature can be determined from the efficiency:

\begin{align*} \eta=\dfrac{h_{1}-h_{\text{2a}}}{h_{1}-h_{\text{2s}}}=\dfrac{T_{1}-T_{\text{2a}}}{T_{1}-T_{\text{2s}}} \end{align*}

\begin{align*} T_{\text{2a}}&=T_{1}-\eta(T_{1}-T_{\text{2s}})\\ &=673\:\text{K}-0.9\cdot(673-292)\:\text{K}\\ &=\boxed{330\:\text{K}} \end{align*}

The work produced is determined from the energy balance:

\begin{align*} \dot mh_{1}&=\dot W+\dot mh_{\text{2a}}\\ h_{1}&=w+h_{\text{2a}} \end{align*}

\begin{align*} w&=h_{1}-h_{\text{2a}}\\ &=(684.34-330.34)\:\dfrac{\text{kJ}}{\text{kg}}\\ &=\boxed{354\:\dfrac{\text{kJ}}{\text{kg}}} \end{align*}

The entropy generation is:

\begin{align*} s_{\text{gen}}&=s_{2}-s_{1}\\ &=\ln\Bigg(\Bigg(\dfrac{T_{2}}{T_{1}}\Bigg)^{c_{p}}\Bigg(\dfrac{P_{1}}{P_{2}}\Bigg)^{R}\Bigg)\\ &=\ln\Bigg(\Bigg(\dfrac{330}{673}\Bigg)^{1.005}\Bigg(\dfrac{2800}{150}\Bigg)^{0.287}\Bigg)\:\dfrac{\text{kJ}}{\text{kg}\text{K}}\\ &=\boxed{0.124\:\dfrac{\text{kJ}}{\text{kg}\text{K}}} \end{align*}

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