Re: Maxwell's Daemon and Thermoelectric effect

First of all, let me congratulate you on your thought process.  Testing 
your understanding by trying to integrate different concepts (such as 
energy conservation and thermoelectric modules) is exactly how you expand 
your knowledge.  Good job!

There are several good resources for understanding how thermoelectric 
coolers work, which I include below.  It may be, however, that these won’t 
answer your question.

For a short answer, I suspect it would help the equations in your head to 
keep track not of the energies, but rather the powers in a thermoelectric 
cooler: the energy flowing per second from one place to another.  The 
total amount of energy handled by the device can be large.  The thermal 
energy transferred per second, multiplied by hours of operation, can be 
many many joules of energy, but what happens to that energy has very 
little to do with the cooler.  What matters in the cooler is how energy is 
flowing through it.

Explaining where the energy flows requires a longer answer, so I consider 
a cooler in operation.  It has a cold side at temperature ‘Tcold’ and a 
hot side at ‘Thot’, with a total temperature difference ‘dT = Thot – 
Tcold’.

A DC current ‘I’ flows through the device, which requires a certain 
voltage ‘V’ to drive.  The electrical power ‘P_e’ pulled from the wall 
plug and dissipated in the cooler is given by the product of the current 
and voltage, ‘P_e = I V’.  This is somewhat misleading, however, because 
the voltage is not independent of current.  Generally, the module will 
have a linear relationship between ‘V’ and ‘I’, given by Ohm’s resistance 
law.  A given cooler will have some electrical resistance ‘R’ so, 
finally, ‘P_e = R I-squared’.

As for thermal energy, the operation of the cooler removes a constant 
amount of thermal energy per second from the cold side and transports it 
to the hot side, where the energy is dissipated by some heat exchanger 
(such as a fan blowing air across radiating fins).  In other words, while 
the device is in operation it is shuttling a constant thermal power ‘Q’ 
from the cold side to the hot side.  The references, below, explain how it 
does this.

It is important to recognize that this thermal energy does not come from 
the electric current: it is being drawn from the environment surrounding 
the cold side of the module and deposited into the environment on the hot 
side.  Therefore, it is not necessary to have the electric power ‘P_e’ be 
some amount added to ‘Q’.  The electrical power required to transport the 
thermal power is typically less than the total amount of power being 
transported.

However, it clearly requires some work to move the thermal energy around!  
This is where the ‘P_e’ comes in, which results in yet more heat the 
cooler has to dissipate.  So, while only ‘Q’ is being drawn from the cold 
side of the device, you have to dissipate ‘Q + P_e’ on the hot side.  
(Because we’ve set up the problem so that ‘Q’ is the net thermal power 
being transported from cold to hot, we can say that all of the electrical 
energy is being deposited on the hot side.  You can claim that some 
of ‘P_e’, say ‘dP’, is dissipated on the cold side, which is true, but 
we’ve agreed the net power transferred is ‘Q’.  You would have to balance 
the excess ‘dP’ going into the cold side by an increased power ‘Q+dP’ 
being drawn out, to the hot side.  For the _total_ power transfer, you 
still wind up with the original ‘Q’ coming from the cold side and ‘Q+P_e’ 
being dissipated at the hot side.)

Note that everything depends on the amount of energy absorbed and 
dissipated at the boundaries of the cooler.  This is one of the reasons 
that cooler performance can be difficult to understand.  You can drive the 
cooler with the same current ‘I’, but if you turn the fan on the heat 
exchanger off, everything changes.

How?  A thermal gradient, by itself, creates a flow of heat from hot to 
cold, proportional to ‘dT’.  This is just a restatement of the behavior of 
thermal conduction.  The electric current in a cooler creates a flow of 
heat from cold to hot, proportional to ‘I’.  The net thermal transfer 
rate ‘Q’ is given by the rate due to the current minus the rate due to the 
temperature gradient.

If you turn the fan off, the temperature of the hot side will clearly go 
up, which is saying that the temperature gradient increases.  You haven’t 
changed the cooler current, so the net effect is a reduced thermal energy 
transfer.  You could reduce the electrical current in the device to bring 
the hot-side temperature back down to the original ‘Thot’, but then you’ve 
reduced the thermal carrying capacity, reducing ‘Q’ once again.

Another important consideration is that while the amount of thermal power 
transported by the device is proportional to the current, the amount of 
electrical heat the device has to dump in addition to ‘Q’ goes up as the 
current, squared.  Doubling the current may double the potential ‘Q’, but 
it quadruples the overhead ‘P_e’.  Eventually, there will be a point where 
increasing the current creates so much extra heat that the device can no 
longer shuttle ‘Q’ effectively through the system.

You’ll note I haven’t tried to give exact relationships, such as ‘Thot’ 
will go up by a factor of two when you turn the fan off, because behavior 
at heat exchange interfaces is extremely complex.  Plus, the coefficients 
which determine cooler performance change with temperature, complicating 
things still further.  But I hope that I have described the flow of energy 
through the system well-enough for your question.

-	Guy

Some of the Internet references I ran across while researching your 
problem:
 http://www.tetech.com/techinfo/
 http://www.tellurex.com/12most.h
tml
 http://www.its.caltech.edu/~jsnyder/thermoelectrics/history_page.htm
 http://www.tf.uni-
kiel.de/matwis/amat/elmat_en/kap_2/advanced/t2_3_2.html