Answer
a) $0.874$
b) $1080.7\text{ kJ/kg}$
c) $194.3\text{ kg/s}$
Work Step by Step
$$
\begin{aligned}
& h_1=h_{f @10 \mathrm{kPa}}=191.81 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 10 \mathrm{kPa}}=0.00101 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-R_1\right) / \eta_p \\
& =\left(0.00101 \mathrm{~m}^3 / \mathrm{kg}\right)(10,000-10 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa}-\mathrm{m}^3}\right)(0.85) \\
& =11.87 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p, \text { in }}=191.81+11.87=203.68 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=10 \mathrm{MPa} \quad h_3=3375.1 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_3=500^{\circ} \mathrm{C}\right\} s_3=6.5995 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_{4 x}=10 \mathrm{kPa} \\
s_{4 x}=s_3
\end{array}\right\} x_{4 \mathrm{~s}}=\frac{s_{4 x}-s_f}{s_{\text {fir }}}=\frac{6.5995-0.6492}{7.4996}=\mathbf{0 . 7 9 3 4} \\
& h_{4 \mathrm{~s}}=h_f+x_4 h_{/ \mathrm{B}}=191.81+(0.7934)(2392.1)=2089.7 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_T=\frac{h_3-h_4}{h_3-h_{4 x}} \longrightarrow h_4=h_3-\eta_T\left(h_3-h_{48}\right) \\
& =3375.1-(0.85)(3375.1-2089.7)=2282.5 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_4=10 \mathrm{kPa} \\
h_4=2282.5 \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} x_4=\frac{h_4-h_f}{h_{f g}}=\frac{2282.5-191.81}{23921}=0.874 \\
&
\end{aligned}
$$ (b) $$
\begin{aligned}
q_{\text {in }} & =h_3-h_2=3375.1-203.68=3171.4 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =h_4-h_1=2282.5-191.81=2090.7 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {act }} & =q_{\text {in }}-q_{\text {out }}=3171.4-2090.7=1080.7 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ and $$
\eta_{\text {th }}=\frac{w_{\text {act }}}{q_{\text {in }}}=\frac{1080.7 \mathrm{~kJ} / \mathrm{kg}}{3171.5 \mathrm{~kJ} / \mathrm{kg}}=34.1 \%
$$ (c) $$
\dot{m}=\frac{\dot{W}_{\text {ret }}}{w_{\text {rat }}}=\frac{210,000 \mathrm{~kJ} / \mathrm{s}}{1080.7 \mathrm{~kJ} / \mathrm{kg}}=194.3 \mathrm{~kg} / \mathrm{s}
$$
