## Related questions with answers

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.

Solution

VerifiedFirst 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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