### Re: Maxwell's Daemon and Thermoelectric effect

Date: Tue Dec 23 10:59:43 2003
Posted By: Guy Beadie, Staff, Optical sciences, Naval Research Lab
Area of science: Physics
ID: 1067280977.Ph
Message:
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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

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

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

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-

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