Shown in the figure below is an electronic device that is mounted on an insulation board. The total heat generation rate inside the

device is $q_{\text{e}\text{l}\text{e}}.$ The device only loses heat to the surrounding moving air by convection $(T_{}=26C,h=8\frac{W}{m^2}K)$ from its top surface

$(A_s=0\mathrm{.}01(m)**0\mathrm{.}01(m)).$ Select the device as the control volume, and apply energy conservation to the c$.$v$.$ under steady state $($the

figure below$),$ one finds:

$E_{in}^{}+E_g^{}-E_{\text{o}\text{u}\text{t}}^{}=d\frac{E_{st}}{d}t$

where $E_g^{}=q_{\text{e}\text{l}\text{e}},E_{in}^{}=0,$ and $E_{\text{o}\text{u}\text{t}}^{*}=h{A}_s(T_s-T_c).$ After re$-$

arrangement, one obtains the following expression:

$T_s=T_{}+\frac{q_{\text{e}\text{l}\text{e}}}{h{A}_s}$

Here we test it using a CPU of a Pentium PC$.$ The heat generation of a CPU is about $4W.$ The surface temperature at the CPU

surface is calculated as:

$T_s=T_{}+\frac{q_{\text{e}\text{l}\text{e}}}{h{A}_s}=26+\frac{4}{8**(0\mathrm{.}01**0\mathrm{.}01)}=5026C$

You definitely know that most computers are not overheated, i$.$e$.,$ the temperature of those CPUs in industry is controlled

below a threshold that is much lower than $5026C.$

Please check online to find out the geometry and heat generation rate of a typical CPU in PCs$,$ then use the above

expression to calculate the steady state temperature of the CPU. Please be careful on the units.

What methods have been implemented in the industry of electronic cooling. Make sure you understand how those

methods are related to the formula given above. Figures should be included in your writing to illustrate the methods.

You need to have at least one page in writing and figures $(12$ font, single space$).$