thermal conductivity of a liquid cooler

Question

I'm a high school student in a class called EDD, where we design a product and produce a prototype of said product. I have one issue though: My team bit off more than we could chew, and we have no idea how to calculate the thermal efficiency of the system. We are currently looking through potential designs, but we can't actually make anything, before we prove that it will work.

So, given an R5 1600 as the processor, how would I calculate the heat transfer from the processor to a copper water block, then to a liquid (water), then to an aluminum radiator? The aluminum radiator is being recycled from a car to save money and hopefully increase efficiency. We were just thinking: more aluminum means less heat.

Also, if there is a thermal engineer that knows how to do these things and would be willing to partner with us to guide us through the process, that would be wonderful.

Answer 1

So, given an R5 1600 as the processor, how would I calculate the heat transfer from the processor to a copper water block, then to a liquid (water), then to an aluminum radiator?

Actually calculating these things would be extremely difficult, since it would require you to know things like the distribution of heat on the surface of the processor and the exact properties of things like the thermal interface material. Instead of calculation from theory, a more common approach is to simply measure the temperature on both sides of a component and the power through it, from which you can calculate the thermal resistance in units of degrees per watt.

Answer 2

First, realize that a car radiator is overkill. A very rough calculation shows that a car radiator is capable of dissipating 37 kW!

Calculating the heat flow and temperature drops will be very challenging, but here is a start.

Heat flow can be solved using equations analogous to Electrical circuit analysis (ohms law for simple heat flow). If you have multiple paths, you can use KVL/KCL (which you probably haven't learned yet), or a circuit simulator can solve it for you.

Heat Dissipated (Watts) is analogous to Current

Temperature (DegC) is analogous to Voltage

Thermal resistance (DegC/W) is analogous to resistance (ohms)

For simple geometries of solid objects, the thermal resistance is easy to calculate.

ThermalResistance = Thickness / (CrossSectionArea * ThermalConductivity)

Where the thickness is the path that the heat travels.

Example:

3 cm x 3 cm x 1 cm thick of aluminum (~240 W/mK)

Thermal Resistance = 0.01m / ( 0.03m * 0.03m * 240 W/mK ) = 0.046 K/W [or degC/W]

Thermal calculations for moving fluids are much more complicated.

https://en.wikipedia.org/wiki/Thermal_resistance

^{simulate this circuit – Schematic created using CircuitLab}