Chapter 8: Problem 15
The entropy production, \(\sigma\), in a thermocouple can be written $$ T \sigma=-J_{S} \cdot \nabla_{r} T-I \cdot\left(\nabla_{r} \frac{\mu_{\text {el }}}{\boldsymbol{F}}-\boldsymbol{F} \boldsymbol{E}\right) $$ where \(J_{S}\) is the entropy current, \(I\) is the current carried by the electrons, \(F\) is Faraday's constant, \(E\) is the electric field in the metal wires, and \(\mu_{\mathrm{el}}\) is the chemical potential of the electrons. The balance equation for the entropy/volume, \(s\), is \(\partial s / \partial t=-\nabla \cdot J_{S}+\sigma\). The generalized Ohm's laws can be written $$ \boldsymbol{J}_{S}=-\frac{\lambda}{T} \nabla, T+\frac{\Gamma}{T} I \text { and } \quad E-\nabla_{r} \frac{\mu_{e l}}{F}=-\zeta \Gamma_{r} T+R I $$ where \(\lambda\) is the coefficient of thermal conductivity at zero electrical current, \(R\) is the isothermal electrical resistance, \(\zeta\) is the differential thermoelectric power, and \(\Gamma / T\) is the entropy transported per unit electric current. (a) Show that the Onsager relation, \(L_{S E}=L_{E S}\), implies that \(\Gamma=T \zeta\), (b) Show that the entropy balance equation can be written \(\partial s / \partial t=\lambda / T \nabla_{r}^{2}(T)-\nabla_{r} \cdot \Gamma I / T+R I^{2} / T\), The first term on the right is the entropy production due to thermal conductivity. The third term on the right is the entropy production due to Joule heating. The second term on the right is the entropy production due to the Peltier and Thomson effects.
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