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$$
\text { Refrigerant- } 134 \mathrm{a} \text { enters a compressor at } 100 \mathrm{kPa} \text { and }-24^{\circ} \mathrm{C} \text { with a flow rate of } 1.35 \mathrm{mV} \text { min and leaves at } 800 \mathrm{kPa} \text { . }
$$

$$
\text { and } 60^{\circ} \mathrm{C} . \text { Determine the mass flow rate of } \mathrm{R}-134 \mathrm{a} \text { and the power input to the compressor. }
$$

$$
P_1=6 \mathrm{MPa}
$$

$$
T_{1}=400 c^{\circ}
$$

$$
V_{1}=80 \mathrm{m/s}
$$

$$
P_{1}=180 \mathrm{kPa}
$$

$$
\dot V_{1}=0.35 \mathrm{m}^{3} / \mathrm{min}
$$

Sat.vap

$$
P_{2}=700 \mathrm{kPa}
$$

$$
\left.\begin{array}{l}{P_{1}=100 \mathrm{kPa}} \\ {T_{1}=24 \mathrm{c}^{\circ}} \\ {V_1}=1.35m^3/min\end{array}\right\}\rightarrow \begin{array}{l}{h_{1}=236.33 \mathrm{kJ/kga}} \\ {V_{1}=0.1947 \mathrm{m^3/k}}\end{array}
$$

$$
\left.\begin{array}{l}{P_{2}=8001 \mathrm{cm}} \\ {T_{2}=60^{\circ} \mathrm{c}}\end{array}\right\}\rightarrow \mathrm{h}_{2}=296.81 kJ/kg
$$

$$
\dot E_{i n}-\dot E_{o u t}=\Delta E_{sys}=0
$$

$$
\dot E_{i n}=\dot{E}_{ou t}
$$

$$
\dot{W}_{i n}+\dot{m} h_{1}=\dot m h_{2}
$$

$$
\dot W_{i n}=\dot m\left(h_{2}-h_{1}\right)
$$

$$
\rightarrow \dot m=\frac{\dot{V}_{1}}{V_{1}}=\frac{1.35 / 60}{0.1947}=0.1155 \mathrm{kg} / \mathrm{s}
$$

$$
\dot w_{in }=0.1155(296.81-236.33)=6.99 \mathrm{KW}
$$

$$
\text { Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are } 6 \mathrm{MPa}, 400^{\circ} \mathrm{C}
$$

$$
\begin{array}{l}{\text { and } 80 \mathrm{m} / \mathrm{s} \text { , and the exit conditions are } 40 \mathrm{kPa}, 92 \text { percent quality, and } 50 \mathrm{m} / \mathrm{s} \text { . }} \\ {\text { The mass flow rate of the steam is } 20 \mathrm{kg} / \mathrm{s} \text { . Determine (a) the change in kinetic energy, }} \\ {\text { (b) the power output, and (c) the turbine inlet area. }}\end{array}
$$

$$
\left.\begin{array}{l}{P_{1}=6 M P{a}} \\ {T_{1}=400 c^{\circ}}\end{array}\right] \rightarrow A-4, A-5_{1} A-6
$$

$$
V_{1}=0.047420 \mathrm{m}^{3 }/ \mathrm{kg}
$$

$$
h_{1}=3178.3 \mathrm{kJ/kg}
$$

$$
\left.\begin{array}{l}{P_{2}=40 \mathrm{KPa}} \\ {X_{2}=0.92}\end{array}\right] \rightarrow A-4,5,6
$$

$$
h_{2}=h_{f}+x_{2} h _{fg}=317.62+0.92(2392.1)
$$

$$
h_{2}=2318.5 \mathrm{kJ} / \mathrm{kg}
$$

$$
\Delta k e=\frac{v_{2}^{2}-v_{1}^{2}}{2}=\frac{(50)^{2}-(80)^{2}}{2} * \frac{1}{1000}=-1.95 \mathrm{kJ} / \mathrm{kg}
$$

$$
\dot{m}_{1}=\dot{m}_{2}=\dot{m}, \quad \dot{E}_{i n}=\dot{E}_{out}
$$

$$
\rightarrow \dot m\left(h_{1}+\frac{v_{1}^{2}}{2}\right)=\dot w_{out}+\dot m\left(h_{2}+\frac{v_{2}^{2}}{2}\right) \rightarrow \dot w_{out}=-\dot m(h_2-h_1+\frac{V_{2}^{2}-V_{1}^{2}}{2})
$$

\(∴ \dot w_{out}=-20(2318.5-3178.3-1.95)=14590 kw= 14.6 MW \)

$$
\dot{m}=\frac{1}{v_{1}} A_{1} V_{1} \Rightarrow
$$$$
A_{1}=\frac{\dot m V_{1}}{V_{1}}=\frac{20 * 0.04742}{80}
$$

$$=0.0119 m^2$$

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