Answer
$r_{\mathrm{bw}}=0.8387$
Work Step by Step
For the compression process, $$
\begin{aligned}
T_{2 s}=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(288 \mathrm{~K})(12)^{0.4 / 1.4} & =585.8 \mathrm{~K} \\
\eta_C=\frac{h_{2 s}-h_1}{h_2-h_1}=\frac{c_p\left(T_{2 s}-T_1\right)}{c_p\left(T_2-T_1\right)} \longrightarrow T_2 & =T_1+\frac{T_{2 s}-T_1}{\eta_C} \\
& =288+\frac{585.8-288}{0.80} \\
& =660.2 \mathrm{~K}
\end{aligned}
$$ For the expansion process, $$
\begin{aligned}
T_{4 s}=T_3\left(\frac{P_4}{P_3}\right)^{(k-1) / k}=(873 \mathrm{~K})\left(\frac{1}{12}\right)^{0.4 / 1.4} & =429.2 \mathrm{~K} \\
\eta_T=\frac{h_3-h_4}{h_3-h_{4 s}}=\frac{c_p\left(T_3-T_4\right)}{c_p\left(T_3-T_{4 s}\right)} \longrightarrow T_4 & =T_3-\eta_T\left(T_3-T_{4 s}\right) \\
& =873-(0.80)(873-429.2) \\
& =518.0 \mathrm{~K}
\end{aligned}
$$ The isentropic and actual work of compressor and turbine are $$
\begin{aligned}
& W_{\text {Comp}, s}=c_p\left(T_{2 s}-T_1\right)=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(585.8-288) \mathrm{K}=299.3 \mathrm{~kJ} / \mathrm{kg} \\
& W_{\text {Comp }}=c_p\left(T_2-T_1\right)=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(660.2-288) \mathrm{K}=374.1 \mathrm{~kJ} / \mathrm{kg} \\
& W_{\text {Turb, } s}=c_p\left(T_3-T_{4 s}\right)=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(873-429.2) \mathrm{K}=446.0 \mathrm{~kJ} / \mathrm{kg} \\
& W_{\text {Turb }}=c_p\left(T_3-T_4\right)=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(873-518.0) \mathrm{K}=356.8 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The back work ratio for $90 \%$ efficient compressor and isentropic turbine case is
$$
r_{\text {bw }}=\frac{W_{\text {Comp }}}{W_{\text {Tub }, s}}=\frac{374.1 \mathrm{~kJ} / \mathrm{kg}}{446.0 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{0 . 8 3 8 7}
$$ The back work ratio for $90 \%$ efficient turbine and isentropic compressor case is $$
r_{\mathrm{bw}}=\frac{W_{\text {Comp }, s}}{W_{\text {Turb }}}=\frac{299.3 \mathrm{~kJ} / \mathrm{kg}}{356.8 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{0 . 8 3 8 7}
$$ The two results are identical.
