Answer
$ W_{\text {net,in }}=14.2\text{ kJ}$
$Q_{\text {net,out }}=14.2\text{ kJ}$
Work Step by Step
Analysis The properties of the steam at various states are (Tables A-4 through A-6)
$
\begin{aligned}
& \left.\begin{array}{l}
P_1=400 \mathrm{kPa} \\
T_1=350^{\circ} \mathrm{C}
\end{array}\right\} \begin{array}{l}
u_1=2884.5 \mathrm{~kJ} / \mathrm{kg} \\
v_1=0.71396 \mathrm{~m}^3 / \mathrm{kg} \\
s_1=7.7399 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array} \\
& \left.\begin{array}{l}
P_2=150 \mathrm{kPa} \\
T_2=350^{\circ} \mathrm{C}
\end{array}\right\} \begin{array}{l}
u_2=2888.0 \mathrm{~kJ} / \mathrm{kg} \\
s_2=8.1983 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array} \\
& \left.\begin{array}{l}
P_3=400 \mathrm{kPa} \\
s_3=s_2=8.1983 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array}\right\} \begin{array}{l}
u_3=3132.9 \mathrm{~kJ} / \mathrm{kg} \\
v_3=0.89148 \mathrm{~m}^3 / \mathrm{kg}
\end{array}
\end{aligned}
$
The mass of the steam in the cylinder and the volume at state 3 are
$
\begin{aligned}
m & =\frac{V_1}{V_1}=\frac{0.3 \mathrm{~m}^3}{0.71396 \mathrm{~m}^3 / \mathrm{kg}}=0.4202 \mathrm{~kg} \\
V_3 & =m V_3=(0.4202 \mathrm{~kg})\left(0.89148 \mathrm{~m}^3 / \mathrm{kg}\right)=0.3746 \mathrm{~m}^3
\end{aligned}
$
Process 1-2: Isothermal expansion $\left(T_2=T_1\right)$
$
\begin{aligned}
& \Delta S_{\mathrm{l}-2}=m\left(s_2-s_1\right)=(0.4202 \mathrm{~kg})(8.1983-7.7399) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}=0.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& Q_{\mathrm{m}, 1-2}=T_1 \Delta S_{1-2}=(350+273 \mathrm{~K})(0.1926 \mathrm{~kJ} / \mathrm{K})=120 \mathrm{~kJ} \\
& W_{\text {out }, 1-2}=Q_{\mathrm{in}, 1-2}-m\left(u_2-u_1\right)=120 \mathrm{~kJ}-(0.4202 \mathrm{~kg})(2888.0-2884.5) \mathrm{kJ} / \mathrm{kg}=118.5 \mathrm{~kJ}
\end{aligned}
$
Process 2-3: Isentropic (reversible-adiabatic) compression $\left(s_3=s_2\right)$
$
\begin{aligned}
& W_{\mathrm{in}, 2-3}=m\left(u_3-u_2\right)=(0.4202 \mathrm{~kg})(3132.9-2888.0) \mathrm{kJ} / \mathrm{kg}=102.9 \mathrm{~kJ} \\
& Q_{2.3}=0 \mathrm{~kJ}
\end{aligned}
$
Process 3-1: Constant pressure compression process $\left(P_1=P_3\right)$
$
\begin{aligned}
& W_{\mathrm{in,3-1}}=P_3\left(V_3-V_1\right)=(400 \mathrm{kPa})(0.3746-0.3)=29.8 \mathrm{~kJ} \\
& Q_{\text {out } 3-1-1}=W_{\mathrm{in,3-1}}-m\left(u_1-u_3\right)=29.8 \mathrm{~kJ}-(0.4202 \mathrm{~kg})(2884.5-3132.9)=134.2 \mathrm{~kJ}
\end{aligned}
$
The net work and net heat transfer are
$
\begin{aligned}
& W_{\text {net,in }}=W_{\mathrm{in}, 3-1}+W_{\mathrm{in}, 2-3}-W_{\text {out }, 1-2}=29.8+102.9-118.5=14.2 \mathrm{~kJ} \\
& Q_{\text {net,in }}=Q_{\text {in, } 1-2}-Q_{\text {oat }, 3-1}=120-134.2=-14.2 \mathrm{~kJ} \longrightarrow Q_{\text {net, out }}=\mathbf{1 4 . 2}\ \mathbf{k J} \\
&
\end{aligned}
$
