Answer
$\dot{m}_a=27,340\text{ lbm/h}$
$\dot{W}_{\text {net }}=1286\text{ kW}$
Work Step by Step
The mass flow rate of water is $$
\dot{m}_w=\frac{\dot{Q}_{\text {buildings }}}{h_4-h_1}=\frac{2 \times 10^6 \mathrm{Btw} / \mathrm{h}}{(962.8-161.25) \mathrm{Btw} / \mathrm{lbm}}=2495\ \mathrm{lbm} / \mathrm{h}
$$ The mass flow rate of air is then $$
\dot{m}_a=\frac{\dot{m}_w}{0.09126}=\frac{2495}{0.09126}=\mathbf{2 7 , 3 4 0\ l b m / h}
$$ The power outputs from each cycle are $$
\begin{aligned}
\dot{W}_{\text {net, gas cycle }} & =\dot{m}_a\left(w_{\mathrm{t}, \text { out }}-w_{\mathrm{C}, \text { in }}\right) \\
& =\dot{m}_a c_p\left(T_7-T_8\right)-\dot{m}_a c_p\left(T_6-T_5\right)\\
& =(27,340 \mathrm{lbm} / \mathrm{h})(0.240 \mathrm{Btw} / \mathrm{lbm} \cdot \mathrm{R})(2560-1449-1099+540) \mathrm{R}\left(\frac{1 \mathrm{~kW}}{3412.14 \mathrm{Btw} / \mathrm{h}}\right) \\
= & 1062 \mathrm{~kW} \\
\dot{W}_{\text {net, steam cycle }} & =\dot{m}_a\left(w_{\mathrm{T}, \mathrm{out}}-w_{\mathrm{P}, \mathrm{in}}\right) \\
& =\dot{m}_a\left(h_3-h_4-w_{\mathrm{P}, \mathrm{in}}\right) \\
& =(2495 \mathrm{lbm} / \mathrm{h})(1270.9-962.8-2.43)\left(\frac{1 \mathrm{~kW}}{3412.14 \mathrm{Btw} / \mathrm{h}}\right) \\
& =224 \mathrm{~kW}
\end{aligned}
$$ The net electricity production by this cycle is then $$
\dot{W}_{\text {net }}=1062+224=1286\text{ kW}
$$
