A thin electronic component with a surface area of \(950 \mathrm{~cm}^{2}\) is cooled by having a heat sink attached on its top surface. The thermal contact conductance of the interface between the electronic component and the heat sink is \(25,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). According to the manufacturer, the heat sink has combined convection and radiation thermal resistance of \(1.3 \mathrm{~K} / \mathrm{W}\). If the electronic component dissipates \(45 \mathrm{~W}\) of heat through the heat sink in a surrounding temperature of \(30^{\circ} \mathrm{C}\), determine the temperature of the electronic component. Does the contact resistance at the interface of the electronic component and the heat sink play a significant role in the heat dissipation?

Step by step solution

01

Convert the area to SI units.

First, we need to convert the given surface area from cm² to m² by using the conversion factor (1 m = 100 cm): Area = \(950 \mathrm{~cm}^2 \cdot \left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)^2 = 0.095 \mathrm{~m}^2\).

02

Calculate the temperature difference due to conduction.

Using the equation for heat transfer by conduction, we can find the temperature difference between the surfaces of the electronic component and the heat sink: \(Q = h_c \cdot A \cdot \Delta T_{cond}\) Where \(Q = 45 \mathrm{~W}\) is the dissipated power, \(h_c = 25,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the thermal contact conductance, and \(A = 0.095 \mathrm{~m}^2\) is the surface area. Solving for the temperature difference, we get: \(\Delta T_{cond} = \frac{Q}{h_c \cdot A} = \frac{45 \mathrm{~W}}{25,000 \mathrm{W/m^2\cdot K} \cdot 0.095 \mathrm{m}^2} = 0.019 \mathrm{~K}\).

03

Calculate the temperature difference due to convection and radiation.

Using the equation for heat transfer by convection and radiation expressed in terms of the thermal resistance \(R_{conv+rad} = 1.3 \mathrm{~K} / \mathrm{W}\), we can find the temperature difference between the heat sink surface and the surrounding air: \(\Delta T_{conv+rad} = Q \cdot R_{conv+rad} = 45 \mathrm{~W} \cdot 1.3 \mathrm{~K} / \mathrm{W} = 58.5 \mathrm{~K}\).

04

Calculate the temperature of the electronic component.

Knowing the temperature differences due to conduction and convection, and the surrounding temperature, we can now find the temperature of the electronic component: \(T_{component} = T_{surrounding} + \Delta T_{cond} + \Delta T_{conv+rad} = 30^{\circ}\mathrm{C} + 0.019\mathrm{K} + 58.5\mathrm{K} = 88.519^{\circ}\mathrm{C}\).

05

Determine the contact resistance significance.

We can determine the significance of contact resistance by calculating the percentage of the total temperature difference caused by contact resistance at the interface: Percentage = \(\frac{\Delta T_{cond}}{\Delta T_{cond} + \Delta T_{conv+rad}} \times 100\% = \frac{0.019 \mathrm{K}}{0.019 \mathrm{K} + 58.5 \mathrm{K}} \times 100\% = 0.032\%\) The contact resistance at the interface of the electronic component and the heat sink contributes only 0.032% to the total temperature difference, which is negligible. Thus, the contact resistance does not play a significant role in the heat dissipation.

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