# Cooling the Processor

In this project, we will investigate some of the considerations that goes into designing a cooling system for a computer's processor by modeling the heat flow in a laptop's heat sink. This is a picture of my laptop, when I open the back and look at its mainboard. I have put a ruler on top of it so that you can estimate the lengths. The processor is right above the 6cm mark. It has a copper heat sink, covered with a black plastic material, that takes the heat from the processor to the the fan on the left. For simplicity, we assume that the fan on the right is responsible for cooling other parts of the board, and therefore we ignore it.

When all the four cores in the processor are reasonably busy, it dissipates around $$\dot{q}_p = 25 W$$ of heat, as indicated by the specifications sheet. This heat gets absorbed by the heat sink, travels towards the fan, and moves out of the system through the air flow induced by the fan. Let's assume the cooling power that the fan provides is proportional to the temperature difference between the surface of the copper directly below it and the ambient temperature $$u_0$$, as well as the frequency of rotation of its blades $$\nu$$ (in units of rotations per second or Hz). $$\dot{q}_{f} = \alpha \nu (u - u_0)$$

Your job is to find $$\alpha$$ such that if the fan is working at 2000 rotations per minute (RPM), the temperature of the processor never goes above $$80 ^{\circ}$$. Looking at the specifications sheet, the processor dies at $$T_{junction} = 100 ^{\circ}$$. We want to stay well below that temperature. In order to do that solve the heat equation numerically in 1 spatial dimension and 1 temporal dimension: $$\rho c_p \frac{\partial u}{\partial t} - \kappa \frac{\partial^2 u}{\partial x^2} = \dot{q}_{p} - \dot{q}_{f}$$

Consider the initial situation where everything is at the ambient temperature of $$u_0 = 20^{\circ}$$.

Properties of Copper
Density $$\rho$$$$8.96 \times 10^3 \frac{kg}{m^3}$$
Thermal Conductivity $$\kappa$$$$4.01 \times 10^2 \frac{W}{m.K}$$
Specific Heat $$c_p$$$$3.9 \times 10^{-1} \frac{kJ}{kg.K}$$