Answer
$\Delta$$z_{min}=3.29\ m$
Work Step by Step
First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 13-5. Noting that $\mathrm{mf}_s=$ 0.00078 and $\mathrm{mf}_{\mathrm{w}}=1-\mathrm{mf}_s=0.99922$, $$
\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{0.00078}{58.44}+\frac{0.99922}{18.0}}=18.01 \mathrm{~kg} / \mathrm{kmol} \\
& y_i=\mathrm{mf}_i \frac{M_m}{M_i} \rightarrow y_w=\mathrm{mf}_w \frac{M_m}{M_w}=(0.99922) \frac{18.01 \mathrm{~kg} / \mathrm{kmol}}{18.0 \mathrm{~kg} / \mathrm{kmol}}=0.99976
\end{aligned}
$$ The minimum work input required to produce $1 \mathrm{~kg}$ of freshwater from brackish water is
$$ w_{\min , \text { in }}=R_w T_0 \ln \left(1 / y_w\right)=(0.4615 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(291.15 \mathrm{~K}) \ln (1 / 0.99976)=0.03230 \mathrm{~kJ} / \mathrm{kg} \text { fresh wate r }
$$ Therefore, $0.03159 \mathrm{~kJ}$ of work is needed to produce $1 \mathrm{~kg}$ of fresh water is mixed with seawater reversibly. Therefore, the required power input to produce fresh water at the specified rate is
$$
\dot{W}_{\text {min, in }}=\rho \dot{U}_{\text {min, in }}=\left(1000 \mathrm{~kg} / \mathrm{m}^3\right)\left(0.175 \mathrm{~m}^3 / \mathrm{s}\right)(0.03230 \mathrm{~kJ} / \mathrm{kg})\left(\frac{1 \mathrm{~kW}}{1 \mathrm{~kJ} / \mathrm{s}}\right)=\mathbf{5 . 6 5} \mathbf{~ k W}
$$ The minimum height to which the brackish water must be pumped is
$$
\Delta z_{\min }=\frac{w_{\min , \text { in }}}{g}=\left(\frac{0.03230 \mathrm{~kJ} / \mathrm{kg}}{9.81 \mathrm{~m} / \mathrm{s}^2}\right)\left(\frac{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^2}{1 \mathrm{~N}}\right)\left(\frac{1000 \mathrm{~N} \cdot \mathrm{m}}{1 \mathrm{~kJ}}\right)=3.29 \mathrm{~m}
$$
