Question 14.7: Desalination through Reverse Osmosis (a) Determine the minim...

Use Eq. (14.37) to obtain the energy required for reverse osmosis. For distillation use an energy balance in an open-steady device.

\dot{W}_{ rev }=-\dot{m}_{A+B} \int_{i}^{e} v_{A+B} d p=-\frac{\dot{m}_{A+B} \Delta p}{\rho_{A+B}}=-\frac{\dot{m}_{A+B} \bar{R} T}{\bar{M}_{A}} \ln \frac{1}{y_{A}^{\prime}} ; \quad[ kW ]                             (14.37)

Assumptions
Ocean water behaves as an ideal solution. Vapor pressure of the salt or any of its ions is zero. There is no recovery of energy during distillation. Enthalpy of ocean water can be approximated by that of pure water.

Analysis
The mole fraction of pure water in the ocean can be obtained from Raoult’s law, which relates the vapor pressure on the ocean water’s surface with the mole fraction of water in the saline water of the ocean:

\begin{gathered}p_{v, \text { ocean }}=p_{v, \text { fresh }} y_{ H _{2} O }^{\prime}=p_{\text {sat@10°C }}y_{ H _{2} O }^{\prime} \\\Rightarrow \quad y_{ H _{2} O }^{\prime}=\frac{p_{v, \text { ocean }}}{p_{\text {sat@10°C }}}=\frac{1.207}{1.228}=0.983\end{gathered}

The reversible work, the minimum useful work required, per unit mass of ocean water is now obtained from Eq. (14.37):

\frac{\dot{W}_{\text {rev }}}{\dot{m}_{\text {ocean }}}=-\frac{\bar{R} T}{\bar{M}_{A}} \ln \frac{1}{y_{A}^{\prime}}=-\frac{(8.314)(283)}{(18)} \ln \frac{1}{0.983}=-2.25 \frac{ kJ }{ kg }

To calculate the energy required to distill ocean water, a steady state energy balance produces:

\frac{\dot{Q}-\dot{W}_{\text {ext }}}{\dot{m}_{\text {ocean }}}=j_{e}-j_{i} \cong h_{g @ 100 kPa }-h_{f @ 10^{\circ} C }=2675-42=2633 \frac{ kJ }{ kg }

The osmotic pressure can be calculated from Eq. (14.36):

\Delta p \cong \frac{\Delta z \rho_{A+B} g}{(1000 N / kN )}=\frac{\rho_{A+B} \bar{R} T}{\bar{M}_{A}} \ln \frac{1}{y_{A}^{\prime}} \quad [kPa]             (14.36)

\begin{gathered}\Delta p \cong \frac{\rho_{A+B} \bar{R} T}{\bar{M}_{A}} \ln \frac{1}{y_{A}^{\prime}}=\frac{\rho_{A+B} \dot{W}_{ rev }}{\dot{m}_{ sal.water }}=\frac{\rho_{A+B} g \Delta z}{(1000 Pa / kPa )}[ kPa ] \\\Rightarrow \quad \Delta z=\frac{\Delta p}{\rho_{A+B} g / 1000}=\frac{1000}{g} \frac{\dot{W}_{ rev }}{\dot{m}_{ sal.water }}=\frac{1000}{9.81}(2.25)=229 m\end{gathered}

Discussion
The comparison presented above, is somewhat flawed due to the fact that the best case scenario for reverse osmosis has been compared with the worst case scenario for distillation. The energy required for distillation can be supplied from a heat source at a low temperature (just above 100°C) to improve the exergetic efficiency and much of the waste heat can be recovered through regeneration. The work required for the reverse osmosis, on the other hand, can be significantly higher by using a realistic value for exergetic efficiency. Despite these practical considerations, desalination by reverse osmosis is far more energy efficient than distillation.