Answer
$42.4\%$
Work Step by Step
$$
\begin{aligned}
& h_1=h_{f @2 \text {psia }}=94.02\ \mathrm{Btw} / \mathrm{lbm} \\
& v_1=v_{f @2 \text {psia }}=0.01623\ \mathrm{ft}^3 / \mathrm{lbm} \\
& w_{p_{, i \mathrm{i}}}=v_1\left(P_2-P_1\right) \\
& =\left(0.01623 \mathrm{f}^3 / \mathrm{lbm}\right)(1250-2 \mathrm{psia})\left(\frac{1 \mathrm{Btu}}{5.4039 \mathrm{psia}-\mathbb{f}^3}\right) \\
& =3.75\ \mathrm{Btu} / \mathrm{bm} \\
& h_2=h_1+w_{p \text { in }}=94.02+3.75=97.77\ \mathrm{Btu} / \mathrm{bm} \\
& h_4=h_f+x_4 h_{f 8}=94.02+(0.9)(1021.7)=1013.6\ \mathrm{Btu} / \mathrm{lbm} \\
& s_4=s_f+x_4 s_{f g}=0.17499+(0.9)(1.74444)=1.7450\ \mathrm{Btu} / \mathrm{bm} \cdot \mathbb{R} \\
& P_3=1250\ \mathrm{psia} \quad h_3=1693.4\ \mathrm{Btw} / \mathrm{bm} \\
& s_3=s_4 \quad\left\{T_3=1337^{\circ} \mathrm{F}\right. \\
&
\end{aligned}
$$ (b) $\quad Q_{\text {in }}=n\left(h_3-h_2\right)=(75 \mathrm{lbm} / \mathrm{s})(1693.4-97.77)=119,672\ \mathrm{Btu} / \mathrm{s}$
(c) $\dot{Q}_{\mathrm{att}}=\dot{m}\left(h_4-h_1\right)=(75 \mathrm{lbm} / \mathrm{s})(1013.6-94.02)=68,967\ \mathrm{Btw} / \mathrm{s}$
$$
\eta_{\text {ot }}=1-\frac{\dot{\underline{Q}}_{\text {out }}}{\dot{Q}_{\text {in }}}=1-\frac{68,967 \mathrm{Btu} / \mathrm{s}}{119,672 \mathrm{Btu} / \mathrm{s}}=\mathbf{4 2 . 4} \%
$$
