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$$
\begin{array}{l}{\text { Steam at } 4 \mathrm{MPa} \text { and } 350^{\circ} \mathrm{C} \text { is expanded in an adiabatic turbine to } 120 \mathrm{kPa} \text { . What is the isentropic }} \\ {\text { efficiency of this turbine if the steam is exhausted as a saturated vapor? }}\end{array}
$$

$$
\left.\begin{array}{l}{P_{1}=4 M Pa} \\ {T_1=350 c^{\circ}}\end{array}\right]\rightarrow \ \begin{array}{l}{h_{1}=3093.3 kJ/kg} \\ {S_1=6.5843 kJ/kg.k}\end{array}
$$

$$
\left.\begin{array}{l}{P_{2}=120 \mathrm{KPa}} \\ {\mathrm{X}_{2}=1}\end{array}\right]\rightarrow \ h_2=2683.1 kJ/kg
$$

$$
\left.\begin{array}{l}{P_{25}=120 \mathrm{KPa}} \\ {S_{25}=S_1}\end{array}\right] \rightarrow \ \begin{array}{l}{X_2x=0.8798} \\ {h_{25}=2413.4 kJ/kg}\end{array}
$$

$$
\dot{E}_{i n}-\dot{E}_{out}=\Delta \dot E_{sys} \ \longrightarrow \ \dot E_{in}=\dot E_{out}
$$

$$
\rightarrow \dot m h _{1}=\dot W_{a\ out}+\dot m h_2
$$

$$
\dot{w}_{a \ out}=\dot m(h_1-h_2)=> \quad \dot m_1=\dot m_2=\dot m
$$

$$
\eta_{T}=\frac{\dot w_{a\ out}}{\dot w_{s \ out }}=\frac{\dot m(h_1-h_2)}{\dot m(h_1-h_{25})}=\frac{h_1-h_2}{h_1-h_{25}}
$$

$$
\rightarrow=\frac{3093 \cdot 3-263.1}{3093.3-2413.4}=0.603
$$

$$60.3\%$$

$$
\text { Refrigerant- } 134 \text { a enters an adiabatic com-pressor }
$$$$
\text { as saturated vapor at } 100 \text { kPa at a rate of } 0.7 \text { tn }
$$

$$
\text { Vmin and exits at } 1 \text { -MPa pressure. If the isentropic efficiency of the compressor is } 87 \text { percent, }
$$

$$
\begin{array}{l}{\text { determine (a) the temperature of the refrigerant at the exit of the compressor and (b) the power }} \\ {\text { input, in } \mathrm{kW} \text { . Also, show the process on a } 1 \text { -s diagram with respect to saturation lines. }}\end{array}
$$

(1) $$
P=100 \ \mathrm{KPa}
$$

$$
\text { Sat. V }
$$

$$
\dot V=0.7 \mathrm{m}^{3} / \mathrm{min} \ \rightarrow \ \frac{0.7}{60} \quad m^3/s
$$

$$
\longrightarrow A-12
$$$$
\begin{array}{l}{\left[\begin{array}{l}{v_{1}=v_{g}=0.14254} \\ {h_{1}=h_{g}=234.44} \\ {S_{1}=S_{g}=0.95183}\end{array}\right.}\end{array}
$$

(2) $$
P_{2}=1000 \mathrm{kPa} \ ,h_2=289.7 \ , \ T_2 ??
$$

(2s) $$
\left.\begin{array}{l}{P_{2s}=P_{2}=1000 \mathrm{KPa}} \\ {S_{2s}=S_{1}=0.95183}\end{array}\right]\rightarrow A-13
$$

$$
h_{1}=271.71 \quad s_{1}=0.9179
$$

$$
h=?? \quad s=0.95183
$$

$$
h_{2}=282.74 \quad S_{2}=0.9525
$$

\(∴ h_{2s}=282.52 \)

$$
\eta_{c}=\frac{h_1-h_{2s}}{h_1-h_2} \ \rightarrow \ h_2=289.7
$$

$$
A-13 \longrightarrow \quad T_{1}=50 \quad h_{1}=282.74
$$

$$
T=?? \quad h=284.7
$$

$$
T_{2}=60 \quad h_{2}=293.38
$$

$$
T_{2}=56.54 \mathrm{c}^{\circ}
$$

$$
\dot Q-\dot W=\dot m(\Delta h)
$$

$$
\begin{array}{l}{\Delta k_e=0} \\ {\Delta P e=0}\end{array}
$$

$$
-\dot w=\dot m\left(h_{2}-h_{1}\right)
$$

$$
-(-\dot W)=\dot m(h_2-h_1)
$$

$$
\dot{m}=\frac{\dot \forall}{v_{1}}=0.0605 \mathrm{kg} / \mathrm{s}
$$

$$\rightarrow
\dot{w}=\dot{m}\left(h_{2}-h_{1}\right)=0.0605(289.7-234.44)
$$

$$
={3343 \mathrm{kw}}
$$

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