In order to determine the thermal power P_Q , we integrate the heat transport equation over the volume V. The integral over the volume is the product of an integral over the crosssection area A times an integral over the length L of the thermoelectric material,
\int_{S}{j_Q} .dS \int_{0}^{L}{dr} .\hat{x} = \kappa \int_{0}^{L}{dr} .(-\nabla T ) \int_{S}{dS} .\hat{x} +T^+ \varepsilon \int_{S}{j_q}.dS \int_{0}^{L}{dr}.\hat{x}
where \hat{x} is a unit vector in the direction of the current densities j_Q and j_q, and the infinitesimal length and surface vectors dr and dS are oriented in the same direction. The thermal power P_Q and the electric current I are defined as,
P_Q = \int_{S}{j_Q} .dS
I= \int_{S}{j_q} .dS
ΔT =\int_{0}^{L}{dr} .(-\nabla T )
The cross-section surface area A and the length L can be written as,
A = \int_{S}{dS} .\hat{x}
L = \int_{0}^{L}{dr}.\hat{x}
Thus, the thermal power P_Q is written as,
P_Q = κ \frac{A}{L} ΔT + T^+ ε I
Similarly, in order to determine the electric power P_q , we deduce ∇\varphi from the electric charge transport equation and integrate the scalar product between −∇\varphi and the electric current density j_q over the volume V of the thermoelectric material,
\int_{V}{j_q} .(-\nabla \varphi )dV = -\varepsilon \int_{S}{j_q} .dS \int_{0}^{L}{dr} . (-\nabla T ) +\frac{1}{\sigma } \int_{S}{j_q} .dS \int_{0}^{L}{j_q} .dr
The electric power P_q is defined as,
P_q = \int_{V}{j_q} .(-\nabla \varphi )dV
and
I \frac{L}{A} = \int_{0}^{L}{j_q} .dr
Thus, the electric power P_q is written as,
P_q = −ε I ΔT + \frac{I^2}{σ} \frac{L}{A}
Hence, the efficiency η of the thermoelectric material is given by,
\eta = -\frac{P_q}{P_Q} = \frac{\varepsilon I \Delta T-\frac{I^2}{\sigma }\frac{L}{A} }{\kappa \frac{A}{L} \Delta T+ T^+ \varepsilon I}
which can be recast as,
\eta = \frac{\Delta T}{T^+}\Biggl(\frac{\varepsilon -\frac{I}{\sigma }\frac{L}{A} \frac{1}{\Delta T} }{\varepsilon +\frac{\kappa }{I} \frac{A}{L} \frac{\Delta T}{T^+} }\Biggr)
Using the dimensionless ratio,
r = \frac{I}{\kappa } \frac{L}{A} \frac{1}{\Delta T}
the efficiency η becomes,
\eta = \frac{\Delta T}{T^+}\Biggl(\frac{\varepsilon -\frac{\kappa }{\sigma }r }{\varepsilon +\frac{1}{r} \frac{1}{T^+} }\Biggr) = \eta = \frac{\Delta T}{T^+}\Biggl(\frac{r\Bigl(\varepsilon -\frac{\kappa }{\sigma }r\Bigr) }{r \varepsilon + \frac{1}{T^+} }\Biggr)
Using the relation ΔT = T^+ − T^− in the limit r ε \ll 1/T^+ , the efficiency η is reduced to,
\eta = \Bigl(1-\frac{T^-}{T^+}\Bigr) T^+r \Bigl(\varepsilon -\frac{\kappa }{\sigma }r\Bigr)
To find the optimum efficiency ratio r, we have to optimise the efficiency η with respect to the ratio r,
\frac{dη}{dr} = \Bigl(1-\frac{T^-}{T^+}\Bigr) T^+\Bigl(\varepsilon -\frac{2\kappa }{\sigma }r\Bigr) = 0
which implies that the optimal ratio is,
r =\frac{σ ε}{2 κ}
Thus, the optimal efficiency of the thermoelectric material is,
\eta = \Bigl(1-\frac{T^-}{T^+}\Bigr) \frac{\sigma \varepsilon ^2}{4\kappa } T^+
which is a quarter of the product of the Carnot efficiency and the ‘ZT coefficient’ σ ε²/κ T^+ . In the limit where the thermoelectric effect is much smaller than the thermal power,
I \frac{ε}{κ} \frac{L}{A} \frac{T^+}{ΔT} \ll 1
where according to Ohm’s law and the Seebeck effect, the order ofmagnitude of the electric current I is given by,
I= \frac{ε ΔT}{R} = ε σ ΔT \frac{L}{A}
Thus, the condition is recast as,
\frac{σ ε^2}{κ} T^+ \ll 1 .