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$$
\begin{array}{l}{\text { An adiabatic steam diffuser has steam entering at } 500 \mathrm{kPa}, 200^{\circ} \mathrm{C}, \text { and } 30 \mathrm{m} / \mathrm{s} \text { , and leaving as }} \\ {\text { a saturated vapor at } 200 \mathrm{kPa} \text { . Calculate the second-law efficiency of the diffuser. Take } \mathrm{T} 0-25^{\circ} \mathrm{C} \text { . }}\end{array}
$$

$$
\left.\begin{array}{l}{P_{1}=500 \mathrm{KPa}} \\ {T_{1}=200 {\mathrm{c^{\circ}}}}\end{array}\right\}\rightarrow
$$$$
\begin{array}{l}{h_{1}=2855.8 \mathrm{KJ/kg}} \\ {s_{1}=7.061 {\mathrm{kJ/kg.k}}}\end{array}
$$

$$
\left.\begin{array}{l}{P_{2}=200 {k Pa}} \\ {X_{2}=1}\end{array}\right] \rightarrow
$$$$
\begin{array}{l}{h_{2}=2706.3 \mathrm{kJ/kg}} \\ {s_{2}=7.1270 kJ/kg.k}\end{array}
$$

$$
\dot{E}_{\text {in }}-\dot{E}_{\text {out }}=\Delta \dot E_{system}
$$

$$
\dot{E}_{i n}=\dot{E}_{out}
$$

$$
\dot{m}\left(h_{1}+\frac{v_{1}^{2}}{2}\right)=\dot{m}\left(h_{2}+\frac{v_{2}^{2}}{2}\right)
$$

$$
\frac{v_{2}^{2}-v_{1}^{2}}{2}=h_{1}-h_{2}=\Delta K e_{\text {actual }}
$$

$$
\longrightarrow \Delta Ke_{\text {actual }}=h_{1}-h_{2}=2855.8-2706.3=149.5 kJ/kg
$$

$$
\dot{X}_{\text {in }}-\dot{X}_{\text {out }}-\dot{X}_{\text {destr }}=\Delta X_{\text {syst }}
$$

$$
\rightarrow \quad \dot{X}_{i n}=\dot{X}_{out}
$$

$$
\dot m \psi_{1}=\dot{m} \psi_{2}
$$

$$
h_{1}-h_{0}+\frac{V_{1}^{2}}{2}-T_{0}\left(s_{1}-S_{2}\right)=h_2-h_0+\frac{V_{2}^{2}}{2}-T_0 (s_2-s_0)
$$

$$
\frac{V_{2}^{2}-V_{1}^{2}}{2}=h_{1}-h_{2}-T_{0}\left(S_{1}-S_{2}\right)
$$

$$
\Delta k e_{rev}=h_{1}-h_{2}-T_0(S_1-S_2)
$$

$$
=2855. 8-2706.3-298(7.061.7.127)
$$

$$
=169.2 \mathrm{kJ} / \mathrm{Kg}
$$

$$
\eta_{II}=\frac{\Delta ke_{actual}}{\Delta ke_{rev}}
$$

$$
=\frac{149.5}{169 .2}=0.884=88.4 \%
$$

$$
\begin{array}{l}{\text { Refrigerant-134a enters an adiabatic compressor at - } 26^{\circ} \mathrm{C} \text { as a saturated vapor at a rate of } 0.45 \mathrm{m} 3 / \mathrm{m}} \\ {\text { in and leaves at } 800 \mathrm{kPa} \text { and } 50^{\circ} \mathrm{C} \text { . Dettermine }} \\ {\text { (a) the power input to the compressor, }}\end{array}
$$

$$
\begin{array}{l}{\text { (b) the isentropic efficiency of the compressor, and }} \\ {\text { (c) the rate of exergy destruction and the second-law }} \\ {\text { efficiency of the compressor. Take } 7^{\prime}\left(,=27^{\circ} \mathrm{C} \text { . }\right.}\end{array}
$$

$$
\left.\begin{array}{l}{T_{1}=-26 c^{\circ}} \\ {X_{1}=1}\end{array}\right\}\rightarrow
$$$$
\begin{array}{l}{h_{1}=234.68 \mathrm{kJ} / \mathrm{kg}} \\ {S_{1}=0.9514 \mathrm{kJ} / \mathrm{kg.k}} \\ {V_1=0.18946 m^{3}/kg}\end{array}
$$

$$
\left.\begin{array}{l}{P_{2}=800{k P{a}}} \\ {T_{2}=50 c^{\circ}}\end{array}\right\}\rightarrow
$$$$
\begin{array}{l}{h_{2}=286. 69 \mathrm{kJ} / \mathrm{kg}} \\ {S_{2}=0.9802 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{k}}\end{array}
$$

$$
\left.\begin{array}{l}{P_{2}=800 \mathrm{kPa}} \\ {S_{2}=S_{1}=0.9514}\end{array}\right\}\rightarrow
$$$$
h_{2 s}=277.53 \mathrm{kJ} \mathrm{/kg}
$$

$$
\dot{m}=\frac{\dot{V}_{1}}{V_{1}}=\frac{\frac{0.45}{60}}{0.18946}=0.03959 \mathrm{kg} / \mathrm{s}
$$

$$
\rightarrow \ \dot w_{out}=\dot m\left(h_{2}-h_{1}\right)=0.03959(286.69-234.68)=2.059 k \dot w
$$

$$
\eta_\text {com, isen }=\frac{\dot w_{isen}\rightarrow??}{\dot w_{act}}
$$

$$
\rightarrow \dot{w}_{i s e n}=\dot{m}(h_{2s}-h_1)=0.03959(277.53-234.68
$$

\(∴ \eta_{com,isen}=\frac{1.696}{2.054}=0.8238=82.38 \% \)

$$
\dot{X}_{d e s t}=\dot m \  T_{0}\left(s_{2}-s_{1}\right)=0.03959 * 300(0.9802-0.4514)
$$

$$
=0.3417 \mathrm{kw}
$$

$$
\dot w_{rev}=\dot w_{a c t}-\dot w_{d e s t}=2.059 - 0.3417=1.717 \mathrm{kw}
$$

$$
\eta_{com, II}=\frac{\dot w_{rev}}{\dot w_{act}}=\frac{1.717}{2.059}=0.8341=83.4\%
$$

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