]> No Title

Z=S2ρkTa=(VoltK12(ohmm)(WmK)K=(J/coulK12(Jscoul2m(J/s)mK)K=J2J2frac1s(1/s)1K2K1coul21K1coul2=1 OK

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

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

η=(1-TLTH)1+0Ta-11+0Ta+TL/TH=(1-TLTH)1-11+TL/TH=(1-TLTH)01+TL/TH=0 OK

o If TL=TH, then there is no temperature difference across the thermoelectric material, and thus no power can be generated. In this case

η=(1-1)1+ZTa-11+ZTa+1=(0)1+ZTa-11+ZTa+1=0

OK

o Even the best possible material (ZTa) cannot produce an efficiency greater than the theoretically best possible efficiency (called the Carnot cyde efficiency, see page 88) =1- TL/TH, for the same temperature range. As ZTa,

η(1-TLTH)ZTa-1ZTa+TL/TH(1-TLTH)ZTaZTa=1-TLTH OK

Side note #1: a good thermoelectric material such as Bi2Te3 has ZTa=1 and works up to about

200C before it starts to melt, thus

η=(1-TLTH11+1-11+1+(25+273)/(200+273)=0.203(1-TLTH1=0.203ηCamot

=0.203(1-25+273200+273)=0.0750=7.50%

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) 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 S, so produces a large voltage for a small temperature change, a low p so that the resistance of the material to the flow of electric current is

17