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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer by Conduction

When discussing heat transfer, conduction refers to the process by which heat energy is transferred through direct contact between materials. In our exercise, this concept is applied to the interface between the electronic component and the heat sink. We use the formula: \[ Q = h_c \cdot A \cdot \Delta T_{cond} \]where

• \(Q\) is the rate of heat dissipation in watts (in this case, 45 W).
• \(h_c\) is the thermal contact conductance, given as 25,000 \( \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \).
• \(A\) is the interface area in m², converted from cm².

With these values, you solve for \(\Delta T_{cond}\), which indicates the temperature change solely due to conduction. This temperature difference reveals how effective the interface is in conducting heat away.

Thermal Resistance

Thermal resistance is the measure of an object’s resistance to the flow of heat. Think of it like how electrical resistance works, but for heat. The exercise provides the combined convection and radiation thermal resistance as \(1.3 \mathrm{K} / \mathrm{W}\). This value tells us how much the heat flow is resisted by those processes.To calculate the temperature difference from this thermal resistance, we use:\[ \Delta T_{conv+rad} = Q \cdot R_{conv+rad} \]where

• \(R_{conv+rad}\) is the thermal resistance for convection and radiation.
• \(Q\) is the heat in watts.

The calculated \(\Delta T\) helps us understand how much the electronic component's temperature decreases due to these heat dissipation processes.

Convection and Radiation Heat Transfer

Convection and radiation are two critical heat transfer modes alongside conduction. In our case, these two processes occur through the heat sink.

• **Convection** involves the transfer of heat by the movement of fluids or gases adjacent to a heat sink's surface. It’s like the breeze blowing away heat.
• **Radiation** is when heat is emitted as electromagnetic waves, like how the sun warms your skin.

The combined effect of these processes is characterized by the given thermal resistance \( R_{conv+rad} \), which impacts how efficiently heat is removed from the heat sink.

Surface Area Conversion

Converting units is a vital step in assuring accuracy in calculations. Our initial surface area was provided in \(\mathrm{cm}^2\), a common unit in engineering tasks. To use it in our calculations, it must be converted to square meters, \( \mathrm{m}^2 \), the standard SI unit for area.Use the conversion factor \(1 \mathrm{~m} = 100 \mathrm{~cm}\), and then: \[ \text{Area in } \mathrm{m}^2 = \text{Area in } \mathrm{cm}^2 \cdot \left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)^2 \]This step is essential for obtaining the accurate area of \(0.095 \mathrm{~m}^2\), used in subsequent calculations of heat transfer.

Temperature Difference Calculation

Understanding how temperature differences are calculated and utilized is crucial for managing electronics heat dissipation. We identify two key temperature differences: one from conduction and another from convection and radiation.First, find the conduction difference: \[\Delta T_{cond} = \frac{Q}{h_c \cdot A}\]where \(Q\), \(h_c\), and \(A\) are inputs from the exercise.Next, find the convection and radiation difference: \[\Delta T_{conv+rad} = Q \cdot R_{conv+rad}\]With both temperature differences, calculate the electronic component's temperature by summing these values and adding the surrounding temperature. The total temperature gives insight into how hot the component becomes under given conditions.