Determining behavior of heat exchange system

Question

I'm trying to develop a thermoelectric (Peltier tile-based) beverage cooling system. Ideally, I'd also like a device to solve the warm beer problem if the thermodynamics give me any confidence.

First, I'll detail my thought flow in the theory sense and then I'll get to my specific setup that I had in mind.

Obviously, if one immerses a 100 W, 10% efficient Peltier tile in a glass of water at 298 K (let's say, containing 10 mol exactly), if the container system is assumed to be adiabatic in nature, to drop the temperature to 274 K (roughly 35 °F) requires:

$$ Q = m \, c_p \, \Delta T = 180.2 \times 4.1813 \times 24 = 18083.3 \:\mathrm{J} $$

Then, since the efficiency gives the nominal "heat transfer" power to be 10 W:

$$ t = \frac{18083.3\:\mathrm{J}}{10\:\mathrm{J/s}} = 1808.33 \:\mathrm{s} $$

So, the cooling would take about 30 minutes to chill the glass down, much more when you consider that a glass of water doesn't adhere closely to adiabatic behavior.

The actual setup (poorly-drawn hand sketch follows):

Now, obviously there are four sets of values to consider here:

• The specific heat and thermal conductivity of the beverage itself
• The specific heat and thermal conductivity of the glass container holding the beverage, easily approximated by making the assumption that the glass can be characterized by fused silica
• The specific heat and thermal conductivity of the water transfer fluid, well-established
• The specific heat and thermal conductivity of the 6061 aluminum working chamber, which is again well-established

My problem is how to model the system and get model calculations for the transfers between each step, so, for example, solving the equation so I know if I use a given cumulative wattage of Peltier tile, it'll reduce the temperature of the system by a given amount in a certain amount of time.

As before, I'm willing to concede, for the purposes of discussion, that this system can be assumed to be adiabatic with respect to the external environment; that is, no heat is absorbed from the environment during the cooling process.

Answer 1

There are two wattage ratings for a peltier plates. One is the power consumed, the other is the heat transported across the chip when there is a temperature difference of zero. An important note is that the amount of heat transported across the plate is dependent on the temperature difference across the plate.

Image courtesy of: http://www.heatsink-guide.com/peltier.htm

Using the graph above as an example, if the hot side of the peltier plate is not attached to a heat sink then it will quickly heat up to 70 degrees hotter than the beverage unit and heat will stop flowing outward. At that point if this plate was 10% efficient it would still be producing out about 400W of heat. That heat would be produced on the hot side of the chip* but that means that without a heat sink of some sort the hot side would quickly become even hotter than 70 degrees above the beverage unit and heat would start to flow back into the beverage unit, heating it rather than cooling it.

Thus the heat sinks on your beverage "cooling" unit will be very, very important to how well your system cools (or heats) your beverage, and without any information on them my guess is that your system will heat your beverage.

*The heat is actually produced throughout the chip but it gets transferred out along with the other heat and the graph is compensated to account for this extra heat transfer, and only show the heat transferred away from the cool side.