How does a fan keep a computer cool

[g.lemaitre]

A little while ago I asked: why does wind feel cold. The molecules are moving faster so it should feel hot. The answer I got was that it increases the evaporation rate of water from your skin. Less water makes you feel cool, I think. Maybe there were some other reasons which I've forgotten. Anyway, why should a fan keep a computer cool. The air is moving faster, increasing the speed of molecules so it should become hotter.

 

[Integral]

The speed due to movement of air from the fan is very small compared to the speed of the air molecules due to the ambient temperature. What is important is the difference of temperature between the air and the electronics. What you increase is the rate of heat flow from the much hotter electronics parts to the cooler air. If the air is not moved away it will rapidly come to the same temp as the electronics thus reducing the amount of heat energy removed from the electronics.

 

[phinds]

A little while ago I asked: why does wind feel cold. The molecules are moving faster so it should feel hot. The answer I got was that it increases the evaporation rate of water from your skin. Less water makes you feel cool, I think. Maybe there were some other reasons which I've forgotten. Anyway, why should a fan keep a computer cool. The air is moving faster, increasing the speed of molecules so it should become hotter.

No, the bolded part is not correct. You could be doused in boiling water and you most definitely would NOT feel cool. What is going on is that the evaporation of the moisture on your skin requires ENERGY and this energy is provided by your body in the form of heat and that is removed from your body and carried off by the vapor. THAT'S what makes you cooler.

 

[Naty1]

Heat is transferred three different ways: conduction, convection and radiation.

All heating cooling questions can be addressed by considering these three possibilities.

With conduction and convection, temperature difference is important...If 100 degree air passes over a 100 degree object, not much happens; If the air is,say, 50 degrees, it will cool very effectively.

 

[rcgldr]

why should a fan keep a computer cool.

What's mostly happening is that the cooler air from outside the case is being drawn through the case, blowing out the hot air from inside the case back outside the case. The internal flow also cause air to circulate around the hot components (convection), which would get hotter still if the air was not moving since it's not an efficient conductor of heat. Some of the hotter components will have good heat conductors to further transfer heat to the circulating air.

 

[Chestermiller]

When cool air is blown past a hotter object, heat flows from the hotter object to the cooler air. This cools off the object. The cooling mostly takes place in a thin "thermal boundary layer" adjacent to the surface of the object. Conceptually, outside the boundary layer, the air temperature is at its original cool temperature, but within the thermal boundary layer, the air temperature varies from the object surface temperature to the "free stream" temperature (the cooler bulk temperature of the air). This temperature gradient within the boundary layer results in heat conduction through the boundary layer. The heat flux (heat per unit area) is given by q = k (T_S - T_{bulk air} )/δ , where T_S is the surface temperature of the object, T_{bulk air} is the cooler air temperature outside the boundary layer, k is the thermal conductivity of air, and δ is the boundary layer thickness. The higher the air velocity past the object, the thinner the boundary layer δ. Fluid mechanicists have worked out how to predict quantitatively the boundary layer thickness as a function of the air velocity. But the thing to remember is that, as the velocity of the air increases, the boundary layer gets thinner and the heat transfer rate from the object to the air (cooling rate) increases. This type of heat transfer is often referred to as convective heat transfer, and is often described in terms of a heat transfer coefficient h, which is equal to the thermal conductivity k divided by the boundary layer thickness:

h = k/δ

Thus, in terms of the convective heat transfer coefficient h, the heat flux is given by:

q = h (T_S - T_{bulk air} )

where h increases with the velocity of the cooling air. I hope this helps.