Answer
$w_{\text {min, in }}=−0.191$ Btu/lbm brackish water
$w_{\text {min, in }}=0.0214$ Btu/lbm
Work Step by Step
Analysis $(a)$ First we determine the mole fraction of pure water in brackish water using Eqs. $13-4$ and $13-5$. Noting that mf $=0.0012$ and $\mathrm{mf}_{\mathrm{w}}=1-\mathrm{mf}_{\mathrm{s}}=0.9988$, $$
\begin{aligned}
& M_{\mathrm{m}}=\frac{1}{\sum \frac{\mathrm{mf}_i}{M_i}}=\frac{1}{\frac{\mathrm{mf}_s}{M_s}+\frac{\mathrm{mf}_w}{M_w}}=\frac{1}{\frac{0.0012}{58.44}+\frac{0.9988}{18.0}}=18.015 \mathrm{lbm} / \mathrm{lbmol} \\
& y_i=\mathrm{mf}_i \frac{M_m}{M_i} \rightarrow y_w=\mathrm{mf}_w \frac{M_m}{M_w}=(0.9988) \frac{18.015 \mathrm{lbm} / \mathrm{lbmol}}{18.01 \mathrm{bm} / \mathrm{lbmol}}=\mathbf{0 . 9 9 9 6 3} \\
& y_s=1-y_w=1-0.99963=\mathbf{0 . 0 0 0 3 7}
\end{aligned}
$$ (b) The minimum work input required to separate $1 \mathrm{lbmol}$ of brackish water is
$$
\begin{aligned}
& w_{\min , \text { in }}=-R_w T_0\left(y_w \ln y_w+y_s \ln y_s\right) \\
& =-(0.1102 \mathrm{Btu} / \mathrm{lbmol} \mathrm{R})(525 \mathrm{R})[0.99963 \ln (0.99963)+0.00037 \ln (0.00037)] \\
& =-0.191 \mathrm{Btu} / \mathrm{lbm} \text { brackish water } \\
\end{aligned}
$$ (c) The minimum work input required to produce $1 \mathrm{lbm}$ of freshwater from brackish water is $w_{\text {min, in }}=R_w T_0 \ln \left(1 / y_w\right)=(0.1102 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(525 \mathrm{R}) \ln (1 / 0.99963)=\mathbf{0 . 0 2 1 4 B t u} / \mathrm{lbm}$ fresh water
