Power in Watts to Temperature

Shashank Singh 2

Joined May 9, 2018
1
How much cooling would a Peltier module provide in terms of temperature if its power consumption is 5W?

Thats the basic question I need to answer.
I figured the answer to thermo electric cooling per watt usage is not straightforward. Or is it?

Janis59

Joined Aug 21, 2017
742
No straight answer, but may guess in this way:
If efficiency of Peltier is ca 1...2% then read cool power (thermal power) will be ca 5*0,01=50 miliWatts.
Then, one might have a task to organize the box stay less power loss than Peltier itself.
If Peltier self-losses becomes dominant at dT=40-70 C for cheap one-stage devices and 110-120 C for rich end three-stage devices, then guess probably 60 C for Your case.
Then box losses must be N [W]=A(max)*(1/R)*dT/h(min) or 50 [mW]=A [m2] / h [m]*0,03 [W/m2*m] (given for foam polystyrene)*60 [C] the permissible thickness and area of schaum-wall may be easily calculated.
Other story is box content warming (sorry, cooling) time evaluation. As it seems 50 mW is not much. Apply the Q [J]=N [W] / t [sec]=c [J/kg]*m [kg]* dT [C] or Your case 0,05/t=4190*1*60 (figures here for 1 kg of water) so the whole summer probably be enough to cool that all down before the Daddy-Cool comes with the better cold.

WBahn

Joined Mar 31, 2012
24,581
How much cooling would a Peltier module provide in terms of temperature if its power consumption is 5W?

Thats the basic question I need to answer.
I figured the answer to thermo electric cooling per watt usage is not straightforward. Or is it?
Whatever it is cooling has a particular thermal conductivity to its surroundings. So let's say that what you are trying to cool is located in a box and the temperature inside the box is T1 and the temperature outside the box is T2. If T2 is higher than T1, then heat will move from outside the box to inside the box. The greater the difference, the greater the amount of heat energy will flow in the same amount of time. For many purposes we can reasonably model this linearly and say that

P = Q/T = C(T2-T1)

Q is the heat energy that flows in time T and C is the thermal conductivity between the inside and outside of the box.

P has units of power (energy per time) and so can be expressed in watts.

Now you mount your Peltier cooler on the box. Peltiers are like kitchen refrigerators, only far less efficient. For instance (using very rough, made up numbers), if it consumes 5W of electrical power, it might be able to move a few percent of that from the inside to the outside. Let's call it 2%. So that means that you are pulling heat out of the box at a rate of 100 mW. Now you go back to the equation above and determine what the temperature differential is that results in 100 mW of heat power flowing into the box from the surroundings.

Now, there's a really big and often overlooked caveat here. Because Peltier coolers are so inefficient, you are dumping heat into the surroundings at the rate of 5.1 W in order to move heat out of the box at the rate of 0.1 W. Unless you take care, that's going to heat up the outside of the box (i.e., raise T2). It's entirely possible, if you don't take care, that the inside of the box will actually heat up because of the rise in T2.