Calculate the effects of cooling fans on a system?

Question

If I have a home-audio receiver that draws a given amount of power (lets say 200W), and its passively cooled reaching a temperature differential of 50F over ambient, how can I calculate how much cooler it should run given a modest airflow like 10CFM?

Answer 1

There is no easy way to calculate that. The "real" way is to generate a 3-D CAD model of your system, complete with accurate heat sources, and simulate it. That particular field is called "Computational Fluid Dynamics".

There are so many things that must be taken into account including the turbulence of the air, the color of the components, the type of alloy used for the heat sink, and of course the actual airflow.

The software that does this modeling is expensive. At least US$10,000. For the average person it is easier to just try the fan and see what happens. Get some good thermocouples or at least a non-contact IR thermometer.

Answer 2

Here are a couple of pointers:

1) CFM and air speed are proportional... as air passing through a channel of a given volume passes through more quickly, the CFM will also relate that change (IE: faster moving air moves more air through the same volume in the same time) That being said, how effective a system is at removing heat is another school all unto itself (as noted above, thermodynamics of fluids).

2) a system's efficiency at removing heat shouldn't change dramatically with changes in airflow rate (but how much heat is removed should) in an ideal world, but turbulence and the formation of insulative pockets can create somewhat unpredictable results at certain flow rates. (IE: if efficiency drops dramatically at 12.5 CFM, but is higher above and below 12.5 CFM... then just try to avoid that number and you should be okay.)

3) Given that the system's efficiency should remain relatively constant, a little experimentation can teach you a lot about that specific system's ability to remove heat. The specific heat of pure dry air is about 1005.4 kj/(kg * degree C) ftp://ftp1.esrl.noaa.gov/users/cfairall/wcrp_wgsf/flux_handbook/Constants_functions_ELA_05.pdf, which means it takes about a megajoule of energy to raise the temperature of a kg of air by one degree celsius at approximately room temp and standard barometric pressure. The volume of a kg of air is about 0.8562 cubic meters, or about 30.25 cubic feet. You chose 10 CFM, which is about a third of a kg of air per minute. So now we're down to about 334 kJ/degree C. The easiest way to figure out what's actually happening is to measure the temperature of the air going in, coming out, and the rate of flow. You'd have to experimentally determine a few numbers on your own, but afterwards you could predict (to some extent) the change in temperature afforded by a given rate of flow.
For example: If you did have 10 CFM rate of flow, the temperature of the inlet is 25 degrees Celsius (about room temp), and the outlet temp is 43 degrees Celsius (about 110 Fahrenheit); then you've raised the temp by 18 degrees Celsius, and exhausted (over the course of a minute) about 6012 kJ, (since 1 Watt for one second = 1 joule) and expend heat at a rate of about 100.2 Watts. This doesn't seem very realistic, given that the system is only a 200 Watt system. So it seems reasonably safe to assume that the change you should expect from a flow of 10 CFM should be less than 18 degrees celsius (or roughly 60 degrees F)
A lot of factors have to be considered, you say the system runs at "50 degrees above ambient", does that mean that the system's heat sinks are 50 degrees above ambient, or that the air inside is 50 degrees above ambient? The specific heat for most lighter metals is quite low (which is part of what makes them good at transmitting heat) so they will read much hotter than the things around them. It's also important to bear in mind that "temperature" is a measure of the intensity of heat... not the amount of heat. In cooling electrical systems, generally the amount of heat is more important (IE: can this cooling system adequately dissipate X watts of heat)

In review, if you had all the numbers, you'd be able to calculate your answer for the given system. Measuring it's efficiency (specifically energy in watts lost due to heat) and understanding the relationship between the amount of heat present and the energy in watts that corresponds to those changes in heat will help you have a better idea of how much heat is building up, and thus how much you could expect the temperature to drop with any given rate of flow.

Answer 3

Something else to watch out for: the circuit is designed to be optimal at a specific temperature. Cooling it may actually worsen performance. I am not an audio expert, but I have built audio amplifier kits with BJTs that don't operate correctly until they heat up.