|MadSci Network: Physics|
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
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