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Direct Energy, 2018a

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8 THERMOELECTRICS 189<br />

or mass involved starts and ends in the same state, so the processes can<br />

continue indenitelyas long as the input is continuallysupplied.<br />

How much energyis supplied in to the system from the heater? The<br />

amount of energyrequired to maintain the hot side at temperature T h is<br />

given by<br />

E in = k B T h . (8.37)<br />

The device is composed of atoms. Each of these atoms has some internal<br />

energy. A device at temperature T contains k B T joules of energywhere<br />

k B is the Boltzmann constant. <strong>Energy</strong>ows from the hot side to the cold<br />

side of the device. Above, we assumed that the device was in a room that<br />

was so large that the heat from the heater did not raise the temperature of<br />

the room. Thus, we must continuallysupplythis energyat a constant rate<br />

to keep the hot side of the device at temperature T h . While the cold side<br />

of the device is at a lower temperature T c , it maintains that temperature<br />

regardless of the fact that there is a heater in the room.<br />

How much energy is extracted out of the system as electrical energy?<br />

In the Seebeck device, the hot side is held xed at temperature T h , and<br />

because of the environment it is in, the cold side remains at temperature T c .<br />

<strong>Energy</strong>is conserved in this system. Thus, the electrical energyextracted<br />

from the device is given by<br />

E out = k B T h − k B T c . (8.38)<br />

What is the eciencyof this system? Above we assumed that no other<br />

energyconversion processes occur, so this is an idealized case. The resulting<br />

eciencythat we calculate represents the best possible eciencyof<br />

a thermoelectric device operating with sides at temperatures T h and T c .<br />

Eciencyis dened as<br />

η eff = E out<br />

. (8.39)<br />

E in<br />

Using Eqs. 8.37 and 8.38 and some algebra, we can simplifythe eciency<br />

expression.<br />

η eff = E out<br />

E in<br />

= k BT h − k B T c<br />

k B T h<br />

(8.40)<br />

η eff = T h − T c<br />

T h<br />

(8.41)<br />

η eff =1− T c<br />

T h<br />

(8.42)

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