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Question 11.7: A rod is connected at both ends to electrodes. The electric ...

A rod is connected at both ends to electrodes. The electric wires that connect the rod to each electrode are strong enough to carry an electric current flowing through the rod and yet thin enough for the heat transfer to be negligible. The contact resistance and the heat radiated from the rod are negligible. In these experimental conditions, an adiabatic measurement of the resistivity of the material can be performed. As Harman suggested in his seminal paper, experimental conditions can be found such that the Joule and Thomson heating have negligible effects. Use the empirical linear equations (11.92) to show that the adiabatic resistivity thus measured is given by,

\begin{cases}j_s= -\Bigl(\frac{\kappa }{T} +\sigma \varepsilon ^2\Bigr) \nabla T - \frac{\sigma \varepsilon }{q_e}\nabla \bar{\mu } _e \\j_e = -\frac{\sigma \varepsilon }{q_e}\nabla T - \frac{\sigma }{q^2_e} \nabla \bar{\mu }_e\end{cases} .        (11.92)

ρ_{ad}= ρ \Bigl(1+ \frac{ε^2 }{κρ} T\Bigr) .

where ρ = 1/σ is the isothermal resistivity, κ is the thermal conductivity and ε is the Seebeck coefficient of the rod material.

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In the experiment analysed here, the conductive electric current density is j_q = q_e j_e and there is no chemical effect, i.e. ∇μ_e = 0 , which implies that ∇\bar{μ}_e = q_e∇\varphi according to relation (11.93) since the electric charge q_e is a constant. Thus, the transport equations (11.95) are written as,

\bar{μ}_e =μ_e + q_e ∇ \varphi                     (11.93)

\begin{cases} j_ Q = – \kappa \nabla T + T\varepsilon j_q \\ j_ q = -\sigma \varepsilon \nabla T – \sigma \nabla \varphi \end{cases}                      (11.95)

\begin{cases} j_ q = -\sigma \varepsilon \nabla T – \sigma \nabla \varphi \\ j_ Q = – \kappa \nabla T + T\varepsilon j_q \end{cases}

An adiabatic resistivity is measured in the absence of a heat current density, i.e. j_Q = 0 (§ 11.4.9). Hence, the second transport equation implies that the temperature is given by,

∇ T = \frac {ε}{κ}T j_q

Thus, according to the first transport equation, the conductive electric current density is written as,

j_q = − σ ∇\varphi − σ \frac {ε²}{κ} T j_q

The isothermal conductivity σ is the inverse of the isothermal resistivity σ, i.e. σ = 1/ρ.
Therefore, in view of definition (11.81) of the adiabatic resistivity ρad, the electric potential gradient is given by,

∇ \varphi = −ρ_{ad}(s, n_A, q) · j_q        (11.81)

∇ \varphi = −ρ \Bigl(1+ \frac{\varepsilon ^2}{\rho \kappa }T\Bigr) j_q = −ρ_{ad} j_q .

which implies that the adiabatic resistivity ρ_{ad} is written in terms of the isothermal resistivity as,

ρ_{ad} = ρ \Bigl(1+ \frac{\varepsilon ^2}{\rho \kappa }T\Bigr) .

Harman suggests to determine the resistivity ρ in the same sample by using an alternative current of high enough frequency so that no temperature gradient has time to build up in each half period of the current. Then, the ratio (ε²/ρ κ) T can be deduced from the measurements of the resistivities ρ and ρ_{ad} . This ratio is called the ZT coefficient of the material. It is a figure of merit that characterises materials for thermoelectric power generation. As an alternative to a high frequency, a transient method was suggested, and corrections for non-adiabatic conditions in the steady state measurements were also analysed.

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