= Peltier coefficient
{c}
{title2=$\Pi$}
At uniform <temperature>, define the material <Peltier coefficient> by transported <heat current> along positive <conventional current>:
$$
\dot Q_{\rm tr}=\Pi I.
$$
Its units are $\mathrm{W/A}=\mathrm{J/C}=\mathrm V$. In one dimension, $q=\Pi j-\kappa\,dT/dx$, with <electric current density> $j$ and <thermal conductivity> $\kappa$. Here <heat> excludes the <electrochemical potential> contribution to total <energy flux>.<Thermoelectric effects by Leon van Dommelen>[A.11.1; A.11.4, Eq. (A.34): heat-flux convention]{ref}
Now orient positive <conventional current> $I$ from material A into material B, at a common junction <temperature> $T$. Define $\dot Q_{\rm abs}>0$ as <heat> taken from the junction's thermal surroundings. The outgoing transported <heat> minus the incoming transported <heat> gives
$$
\Pi_{A\to B}^{\rm abs}=\Pi_B-\Pi_A,\qquad
\dot Q_{\rm abs}=(\Pi_B-\Pi_A)I.
$$
The local released <heat> is $\dot Q_{\rm rel}=-\dot Q_{\rm abs}$. Authors defining a junction coefficient by released <heat> therefore use the opposite sign. Exchanging A and B reverses the coefficient; reversing $I$ at fixed A and B reverses the junction's thermal response. An individual material coefficient and a junction difference are different quantities.
For an original stipulated example, choose $S_A=-100\,\mu\mathrm{V/K}$ and $S_B=+150\,\mu\mathrm{V/K}$ at $T=300\,\mathrm K$. The <Kelvin thermoelectric relation> gives $\Pi_A=-0.030\,\mathrm V$, $\Pi_B=+0.045\,\mathrm V$. A positive $2\,\mathrm A$ flowing A to B absorbs $(0.075\,\mathrm V)(2\,\mathrm A)=0.150\,\mathrm W$. A separate $0.010\,\Omega$ junction <electrical resistance> would dissipate $0.040\,\mathrm W$ for either direction. If all that dissipation heats the junction, its net absorbed <power> is $0.110\,\mathrm W$ forwards and $-0.190\,\mathrm W$ backwards. These numbers illustrate a chosen thermal accounting boundary, not measured material data or a universal allocation of contact heating.
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