]> No Title

$\mathit{Z}\mathrm{=}\frac{{\mathit{S}}^{\mathrm{2}}}{\mathit{\rho }\mathit{k}}{\mathit{T}}_{\mathit{a}}\mathrm{=}\frac{\mathrm{\left(}\frac{\mathit{V}\mathit{o}\mathit{l}\mathit{t}}{\mathit{K}}{\mathrm{1}}^{\mathrm{2}}}{\mathrm{\left(}\mathit{o}\mathit{h}\mathit{m}\mathit{m}\mathrm{\right)}\mathrm{\left(}\frac{\mathit{W}}{\mathit{m}\mathit{K}}\mathrm{\right)}}\mathit{K}\mathrm{=}\frac{\mathrm{\left(}\frac{\mathit{J}\mathrm{/}\mathit{c}\mathit{o}\mathit{u}\mathit{l}}{\mathit{K}}{\mathrm{1}}^{\mathrm{2}}}{\mathrm{\left(}\frac{\mathit{J}\mathit{s}}{\mathit{c}\mathit{o}\mathit{u}{\mathit{l}}^{\mathrm{2}}}\mathit{m}\mathrm{‖}\frac{\mathrm{\left(}\mathit{J}\mathrm{/}\mathit{s}\mathrm{\right)}}{\mathit{m}\mathit{K}}\mathrm{\right)}}\mathit{K}\mathrm{=}\frac{{\mathit{J}}^{\mathrm{2}}}{{\mathit{J}}^{\mathrm{2}}}\mathrm{frac}\frac{\mathrm{1}}{\mathit{s}\mathrm{\left(}\mathrm{1}\mathrm{/}\mathit{s}\mathrm{\right)}}\frac{\frac{\mathrm{1}}{{\mathit{K}}^{\mathrm{2}}}\mathit{K}}{}\frac{\mathrm{1}}{\mathit{c}\mathit{o}\mathit{u}{\mathit{l}}^{\mathrm{2}}}\frac{\mathrm{1}}{\mathit{K}}\frac{\mathrm{1}}{\mathit{c}\mathit{o}\mathit{u}{\mathit{l}}^{\mathrm{2}}}\mathrm{=}\mathrm{1}$ OK

(b) show that the equation makes physical sense (passes function test)

$\mathrm{o}$ If the material $\mathrm{Z}\mathrm{=}\mathrm{0}$, it produces no electrical power thus the efficiency should be zero. If $\mathrm{Z}$ $\mathrm{=}\mathrm{0}$ then

$\mathit{\eta }\mathrm{=}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{\right)}\frac{\sqrt{\mathrm{1}\mathrm{+}\mathrm{0}{\mathit{T}}_{\mathit{a}}}\mathrm{-}\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{+}\mathrm{0}{\mathit{T}}_{\mathit{a}}}\mathrm{+}{\mathit{T}}_{\mathit{L}}\mathrm{/}{\mathit{T}}_{\mathit{H}}}\mathrm{=}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{\right)}\frac{\sqrt{\mathrm{1}}\mathrm{-}\mathrm{1}}{\sqrt{\mathrm{1}}\mathrm{+}{\mathit{T}}_{\mathit{L}}\mathrm{/}{\mathit{T}}_{\mathit{H}}}\mathrm{=}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{\right)}\frac{\mathrm{0}}{\mathrm{1}\mathrm{+}{\mathit{T}}_{\mathit{L}}\mathrm{/}{\mathit{T}}_{\mathit{H}}}\mathrm{=}\mathrm{0}$ OK

$\mathrm{o}$ If ${\mathrm{T}}_{\mathrm{L}}\mathrm{=}{\mathrm{T}}_{\mathrm{H}}$, then there is no temperature difference across the thermoelectric material, and thus no power can be generated. In this case

$\mathit{\eta }\mathrm{=}\mathrm{\left(}\mathrm{1}\mathrm{-}\mathrm{1}\mathrm{\right)}\frac{\sqrt{\mathrm{1}\mathrm{+}\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{-}\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{+}\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{+}\mathrm{1}}\mathrm{=}\mathrm{\left(}\mathrm{0}\mathrm{\right)}\frac{\sqrt{\mathrm{1}\mathrm{+}\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{-}\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{+}\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{+}\mathrm{1}}\mathrm{=}\mathrm{0}$

OK

$\mathrm{o}$ Even the best possible material $\mathrm{\left(}\mathrm{Z}{\mathrm{T}}_{\mathrm{a}}\mathrm{\to }\mathrm{\infty }\mathrm{\right)}$ cannot produce an efficiency greater than the theoretically best possible efficiency (called the Carnot $\mathrm{\setminus }\mathit{c}\mathit{y}\mathrm{d}\mathit{e}$ efficiency, see page $\mathrm{8}\mathrm{8}$) $\mathrm{=}\mathrm{1}\mathrm{-}$ ${\mathrm{T}}_{\mathrm{L}}\mathrm{/}{\mathrm{T}}_{\mathrm{H}}$, for the same temperature range. As $\mathrm{Z}{\mathrm{T}}_{\mathrm{a}}\mathrm{\to }\mathrm{\infty }\mathrm{,}$

$\mathit{\eta }\mathrm{\approx }\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{\right)}\frac{\sqrt{\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{-}\mathrm{1}}{\sqrt{\mathit{Z}{\mathit{T}}_{\mathit{a}}}\mathrm{+}{\mathit{T}}_{\mathit{L}}\mathrm{/}{\mathit{T}}_{\mathit{H}}}\mathrm{\approx }\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{\right)}\frac{\sqrt{\mathit{Z}{\mathit{T}}_{\mathit{a}}}}{\sqrt{\mathit{Z}{\mathit{T}}_{\mathit{a}}}}\mathrm{=}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}$ OK

Side note #1: a good thermoelectric material such as $\mathrm{B}{\mathrm{i}}_{\mathrm{2}}\mathrm{T}{\mathrm{e}}_{\mathrm{3}}$ has $\mathrm{Z}{\mathrm{T}}_{\mathrm{a}}\mathrm{=}\mathrm{1}$ and works up to about

$\mathrm{2}\mathrm{0}\mathrm{0}\mathrm{C}$ before it starts to melt, thus

$\mathit{\eta }\mathrm{=}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{1}\frac{\sqrt{\mathrm{1}\mathrm{+}\mathrm{1}}\mathrm{-}\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{+}\mathrm{1}}\mathrm{+}\mathrm{\left(}\mathrm{2}\mathrm{5}\mathrm{+}\mathrm{2}\mathrm{7}\mathrm{3}\mathrm{\right)}\mathrm{/}\mathrm{\left(}\mathrm{2}\mathrm{0}\mathrm{0}\mathrm{+}\mathrm{2}\mathrm{7}\mathrm{3}\mathrm{\right)}}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{2}\mathrm{0}\mathrm{3}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{{\mathit{T}}_{\mathit{L}}}{{\mathit{T}}_{\mathit{H}}}\mathrm{1}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{2}\mathrm{0}\mathrm{3}{\mathit{\eta }}_{\mathit{C}\mathit{a}\mathit{m}\mathit{o}\mathit{t}}$

$\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{2}\mathrm{0}\mathrm{3}\mathrm{\left(}\mathrm{1}\mathrm{-}\frac{\mathrm{2}\mathrm{5}\mathrm{+}\mathrm{2}\mathrm{7}\mathrm{3}}{\mathrm{2}\mathrm{0}\mathrm{0}\mathrm{+}\mathrm{2}\mathrm{7}\mathrm{3}}\mathrm{\right)}\mathrm{=}\mathrm{0}\mathrm{.}\mathrm{0}\mathrm{7}\mathrm{5}\mathrm{0}\mathrm{=}\mathrm{7}\mathrm{.}\mathrm{5}\mathrm{0}\mathrm{%}$

By comparison, your car engine has an efficiency of about 25%. So practical thermoelectric materials are, in general, not very good sources of electrical power, but are extremely useful in some niche apphcations, particularly when either (1) it is essential to have a device with no moving parts or (2) $\mathrm{a}$ ""free“ source of thermal energy at relatively low temperature is available, e.g. the exhaust of an internal combustion engine.

Side note #2: a good thermoelectric material has a high $\mathrm{S}$, so produces a large voltage for a small temperature change, a low $\mathrm{p}$ so that the resistance of the material to the flow of electric current is

17