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$$
\begin{array}{l}{\text { Saturated water vapor at } 150^{\circ} \mathrm{C} \text { is compressed in a reversible steady-flow device to } 1000 \mathrm{kPa}} \\ {\text { while its specific volume remains constant. Determine the work required, in } \mathrm{k} / \mathrm{kg} \text { . }}\end{array}
$$

$$
\left.\begin{array}{l}{T_{1}=150 c^{\circ}} \\ {X_{1}=1}\end{array}\right] \rightarrow A-4
$$

\(∴ P_{1}=476.16 \mathrm{kPa} \)

$$
v_{1}=0.39248 \mathrm{m}^{3} / \mathrm{Kg}
$$

$$
W_{i n}=\int_{1}^{2} v d P=V_{1}\left(P_{2}-P_{1}\right)
$$

$$
=0.39248(1000-476.16)
$$

$$
=205 .6 \mathrm{kJ} / \mathrm{kg}
$$

$$
\begin{array}{l}{\text { Calculate the work produced, in } \mathrm{kJ} / \mathrm{kg} \text { , for the reversible isothermal, steady-flow process } 1-3} \\ {\text { shown in Fig. } \mathrm{P} 7-110 \text { when the working fluid is an ideal gas. }}\end{array}
$$

$$
W_{12}=\int_{1}^{2} v d P
$$                                            $$
v= \frac{R T}{P}
$$

$$
=\int_{1}^{2} R \frac{ T}{p} d p
$$

$$
=-R T \int_{1}^{2} \frac{d P}{P}
$$

$$
=-R T \ln \frac{P_{2}}{P_{1}}=-P_{2} V_{2} \ln \frac{P_{2}}{P_{1}}
$$

$$
=-(600)(0.002) \ln \frac{600}{200}
$$

$$
=-1.32 \ \mathrm{kJ} / \mathrm{kg}
$$

 

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