Question

A 0.2-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of $12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of $45 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$. (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a \(0.1\)-cm-thick, \(10-\mathrm{cm}\)-high, and 15 -cm-long aluminum plate $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( with \)200.2-\mathrm{cm}$-thick, 2 -cm-long, and 15 -cm-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}\) thick epoxy adhesive $(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. Determine the new temperatures on the two sides of the circuit board.

Short answer

#Short Answer# To calculate the surface temperatures on both sides of the circuit board, we first determine the initial thermal resistance based on the conduction resistance through the circuit. We then use the given heat transfer coefficient and generated heat to solve for the surface temperatures. After adding the aluminum plate and epoxy adhesive, we adjust the thermal resistance and re-calculate using the same equation to find the new temperatures.

Step-by-step solution

(Step 1: Calculate the Thermal Resistance)

First, we need to find the total thermal resistance which consists of the conduction resistance through the circuit board. The conduction resistance can be found using the equation: \(R_{cond} = \frac{L}{kA}\), where \(L\) is the thickness of the board, \(k\) is the thermal conductivity, and \(A\) is the surface area for heat transfer. For the circuit board: \(L = 0.2 \text{ cm}\), \(k = 12 \frac{\text{W}}{\text{m K}}\), and the surface area \(A = 10 \cdot 15 \cdot 0.0001 \text{ m}^2\) (converted from cm^2 to m^2). Conduction Resistance: \(R_{cond} = \frac{0.2 \cdot 10^{-2}}{12 \cdot 10 \cdot 15 \cdot 10^{-4}} \mathrm{K / W}\).

(Step 2: Calculate the Convection Heat Transfer Coefficient)

We are given the heat transfer coefficient (\(45 \frac{\text{W}}{\text{m}^2\text{K}}\)) for the circuit board's back side.

(Step 3: Calculate the Surface Temperatures)

The heat generated by the components, \(q_{gen} = 15 \text{ W}\), is equal to the heat transfer across the circuit board (\(q_{cond}\)) and dissipated from the back side (\(q_{conv}\)). We can write the equation as: \(q_{gen} = q_{cond} = q_{conv} = \frac{T_1 - T_2}{R_{cond}} = hA(T_2 - T_{\infty}) = 45 \cdot 10 \cdot 15 \cdot 10^{-4}(T_2 - 37)\), where \(T_{\infty} = 37^\circ\text{C}\) is the surrounding temperature, and \(T_1\) and \(T_2\) are the temperatures on both sides of the board. Now, we can solve this equation for both \(T_1\) and \(T_2\) temperatures. ## Part (b) - Determine the new temperatures on the two sides of the circuit board ##

(Step 1: Calculate the New Thermal Resistance)

We need to consider the additional thermal resistance introduced by the aluminum plate, fins, and epoxy adhesive. We'll find the thermal resistances for each of them and add them to the initial conduction resistance. Aluminum plate: \(L = 0.1 \text{ cm}\), \(k = 237 \frac{\text{W}}{\text{m K}}\), \(A = 10 \cdot 15 \cdot 0.0001 \text{ m}^2\) (same as before) Epoxy adhesive: \(L = 0.03 \text{ cm}\), \(k = 1.8 \frac{\text{W}}{\text{m K}}\), \(A = 10 \cdot 15 \cdot 0.0001 \text{ m}^2\) (same as before) Conduction Resistance: \(R'_{cond} = R_{cond} + \frac{L_{al}}{k_{al}A} + \frac{L_{ep}}{k_{ep}A}\)

(Step 2: Calculate the New Heat Transfer Coefficient)

The fins provide additional convection heat transfer, but we need to ensure all heat is transferred: \(q_{gen} = q_{cond} = q_{conv} = 45 \cdot 10 \cdot 15 \cdot 10^{-4}(T'_2 - 37)\), where \(T'_1\) and \(T'_2\) are the new temperatures on both sides of the board.

(Step 3: Calculate the New Surface Temperatures)

We can now solve for the new surface temperatures \(T'_1\) and \(T'_2\) using the new thermal resistance and heat transfer equation: \(15 = \frac{T'_1 - T'_2}{R'_{cond}} = 45 \cdot 10 \cdot 15 \cdot 10^{-4}(T'_2 - 37)\). By solving this equation, we can find the new temperatures on both sides of the circuit board.