Answer
$\dot{m}=786.6\text{ kg/s}$
Work Step by Step
Using variable specific heats, $$
\begin{aligned}
T_1 & =310 \mathrm{~K} \longrightarrow \begin{array}{l}
h_1=310.24 \mathrm{~kJ} / \mathrm{kg} \\
P_{r_1}=1.5546
\end{array} \\
P_{r_2} & =\frac{P_2}{P_1} P_{r_1}=(8)(1.5546)=12.44 \longrightarrow h_{2 s}=562.26 \mathrm{~kJ} / \mathrm{kg} \\
T_3 & =900 \mathrm{~K} \longrightarrow \begin{array}{l}
h_3=932.93 \mathrm{~kJ} / \mathrm{kg} \\
P_{r_3}=75.29
\end{array} \\
P_{r_4} & =\frac{P_4}{P_3} P_{r_3}=\left(\frac{1}{8}\right)(75.29)=9.411 \longrightarrow h_{4 s}=519.32 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {net,out }} & =w_{\mathrm{T}, \text { out }}-w_{\mathrm{C}, \text { in }}=\eta_T\left(h_3-h_{4 s}\right)-\left(h_{2 s}-h_1\right) / \eta_C \\
& =(0.86)(932.93-519.32)-(562.26-310.24) /(0.80) \\
& =40.68 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ and $$
\dot{m}=\frac{\dot{W}_{\text {net,out }}}{w_{\text {net, out }}}=\frac{32,000 \mathrm{~kJ} / \mathrm{s}}{40.68 \mathrm{~kJ} / \mathrm{kg}}=786.6 \mathrm{~kg} / \mathrm{s}
$$
