# General Heat Transfer Guide

#### What Is Our Task?

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**What is our task?**

It is surprisingly simple to summarize what we must do. The computer takes power in the form of electricity and converts it to heat. We must take that heat and get it out of the computer case. If we do not remove power as quickly as the computer uses it, the temperature inside the case will increase until one of two things happens. Either we will reach a point of equilibrium where we remove heat as quickly as the computer generates it, or the computer will crash due to excess temperature.

Whether your PC is completely air-cooled, water-cooled, or uses a Peltier cooler, ultimately all the heat inside the PC must be transferred to the air in the room.

In order to understand things a little better, we’re going to need to know how much heat the computer generates, where that heat is generated, and how we go about removing that heat. We’ll also need to understand the basics of conduction, convection, and radiation.

**Understanding Conduction, Convection, and Radiation**

**Conduction:** As defined previously, conduction is the transfer of heat through a solid material. The simplest case is one-dimensional heat transfer. Consider a wall that has a uniform temperature “T1” on one side and “T2” on the other. The thickness of the wall is “L”. The wall is perfectly insulated at its edges. The wall is made of a single material that is uniform through the thickness of the wall. Here is a diagram of the wall.

The heat transfer through the wall follows a simple equation:

We can draw some interesting conclusions from this equation. First, heat transfer is proportional to the temperature difference on the object. If the temperature differential doubles, the heat transferred doubles. Second, the conduction coefficient "k" is proportional to heat transfer. If the conduction coefficient doubles, the heat transfer doubles. Alternatively, for the same differential temperature, twice as much heat may be transferred. The final observation is "L". As thickness increases, heat transfer decreases. Alternatively, to maintain the same heat transfer through a material twice as thick requires twice the temperature differential.

The same equation applies to three-dimensional heat transfer, but it becomes much more difficult to analytically determine heat transfer.

To put this in perspective, consider a stock 1.2GHz Athlon running at default core voltage. Such a chip produces 66 watts of power at full load. A typical heat sink base made of pure copper may be five millimeters thick. If the 66 watts had to flow through the heat sink as per the one-dimensional diagram above, what would the temperature differential be? To answer this question, we must solve the equation for the temperature differential as follows:

Copper’s conductivity at 300 K (80°F, 27°C) is 401 watts/ (meter*K). Five millimeters equals 0.005 meters.

The answer to this equation is 0.000823 K*meter^2. This tells us the temperature differential to flow through a square plate measuring one meter per side. The die area of the Athlon is 117 mm^2, which equals 0.000117 m^2. To determine the differential across the heat sink area of 117 mm^2, we must divide our initial answer by 0.000117 m^2. The final answer is about 7K = 7°C.

The true differential across the bottom heat sink plate will be lower because the heat transfer is not one-dimensional. Some heat spreads out over the entire area of the heat sink dropping the peak differential to a somewhat lower value.

The materials most often used in the construction of heat sinks are aluminum, copper, and silver. Note that silver or gold plating does not affect overall conduction due to the extreme thinness of the plating. For those wondering about the thermal conductivity of common materials used in heat sinks, here’s a graph. Note that copper and silver are superior to aluminum at all reasonable temperatures.

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