User:Dbrogioli/Concentration for maximum power production

Once all other parameters are fixed, the maximum power is obtained when the "fresh water" is not actually fresh, but has a concentration c1 that is about 1/3 of the "salt water" c2. Obviously, this does not mean that we must operate at that regime: if we already have anough power, we can increase the energy efficiency by reducing c1. Anyhow, I think that this calculation is interesting because shows that a partial mixing of solutions, before using them, can be useful.

Evaluation of the concentration of solutions that gives the maximum power

From Gouy-Chapman-Stern theory, we know that the potential $\varphi$ of the electrode is:

$\varphi =2{\frac {k_{B}T}{e}}\mathrm {sinh} ^{-1}\left(A{\frac {\sigma }{2{\sqrt {C}}}}\right)$

where $C$ is the concentration, $\sigma$ the charge density and $A$ is a constant, once the temperature and the solvent are fixed.

We work at a voltage of the order of 500 mV. This means that </math>\varphi</math> is much more than the thermal energy $K_{B}T/e$=25 mV. This implies that the sinh</math>^{-1}</math> is much more than 1. In this condition the formula can be approximated with:

$\varphi =2{\frac {k_{B}T}{e}}\mathrm {log} \left(A{\frac {\sigma }{\sqrt {C}}}\right)$

The total energy extracted per cycle is $Q(\varphi _{1}-\varphi _{2})$:

$E=2{\frac {k_{B}T}{e}}Q\left[\mathrm {log} \left(A{\frac {\sigma }{\sqrt {C_{1}}}}\right)-\mathrm {log} \left(A{\frac {\sigma }{\sqrt {C_{2}}}}\right)\right]$

where $Q$ is the total charge in the capacitor, and $C_{1}$ and $C_{2}$ are the concentrations of the solutions; the subscript 1 is for the solution with lower concentration.

The rate at which we can perform the cycles is determined by the total resistance of the circuit, including the internal resistance of the capacitor, $R_{I}$, and the resistance of the load $R_{L}$. The maximum power transfer to the load is obtained when $R_{I}=R_{L}$ (traditional electronic knowledge). In turn, $R_{I}$ is inversely proportional to the concentration. So the delivered average power $P$ is proportional to the energy per cycle and to the concentration $C_{1}$:

$P\propto C_{1}\left[\mathrm {log} \left(A{\frac {\sigma }{\sqrt {C_{1}}}}\right)-\mathrm {log} \left(A{\frac {\sigma }{\sqrt {C_{2}}}}\right)\right]$

With trivial calculations:

$P\propto C_{1}\mathrm {log} \left({\frac {C_{2}}{C_{1}}}\right)$

The maximum of $P$ is found by deriving with respect to $C_{1}$ ($C_{2}$ is fixed at the maximum value, sea water). The maximum is at $C_{1}=C_{2}/e$.