Question

Heat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is \(15 \mathrm{~cm}\) long and \(15 \mathrm{~cm}\) wide, and the thicknesses of the copper and epoxy layers are \(0.1 \mathrm{~mm}\) and \(1.2 \mathrm{~mm}\), respectively. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the copper \((k=386 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and epoxy $(k=0.26 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ layers. Also determine the effective thermal conductivity of the board.

Short answer

Based on the provided step-by-step solution, the copper layer accounts for 99.26% of heat conduction and the epoxy layer accounts for 0.74% of heat conduction in the circuit board. The effective thermal conductivity of the entire board is found to be approximately 382.67 W/m·K.

Step-by-step solution

Calculate the cross-sectional area for both layers

First, we need to find the cross-sectional area for the copper and epoxy layers. The dimensions are given in cm and mm, but we'll convert them to meters to be consistent with the units of thermal conductivity provided. The common length and width for both are 15 cm × 15 cm which is equal to 0.15 m × 0.15 m. Area of copper layer: \(A_C = L × W × t_C = 0.15 \mathrm{~m} \times 0.15 \mathrm{~m} \times 0.0001 \mathrm{~m} = 2.25 \times 10^{-6} \mathrm{~m}^2\) Area of epoxy layer: \(A_E = L × W × t_E = 0.15 \mathrm{~m} \times 0.15 \mathrm{~m} \times 0.0012 \mathrm{~m} = 2.7 \times 10^{-5} \mathrm{~m}^2\)

Calculate thermal resistances of the copper and epoxy layers

Using the dimensions and thermal conductivities of each layer, we can find the thermal resistance for each layer using the formula: \(R = \frac{L}{kA}\) Thermal resistance of the copper layer: $R_C = \frac{0.0001 \mathrm{~m}}{(386 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(2.25 \times 10^{-6} \mathrm{~m}^2)} = 1.225 \times 10^{-4} \mathrm{~K} / \mathrm{W}$ Thermal resistance of the epoxy layer: $R_E = \frac{0.0012 \mathrm{~m}}{(0.26 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})(2.7 \times 10^{-5} \mathrm{~m}^2)} = 1.6923 \times 10^{-2} \mathrm{~K} / \mathrm{W}$

Calculate the fraction of total heat conduction in each layer

The total thermal resistance of the board is \(R_{total} = R_C + R_E\). We can now find the fraction of heat conduction in each layer their relative resistances: Fraction of heat conduction in the copper layer: \(\frac{R_E}{R_{total}} = \frac{1.6923 \times 10^{-2}}{{1.225\times 10^{-4}}+{1.6923\times 10^{-2}}}=0.9926\) Fraction of heat conduction in the epoxy layer: \(\frac{R_C}{R_{total}}=\frac{1.225 \times 10^{-4}}{{1.225\times 10^{-4}}+{1.6923\times 10^{-2}}}=0.0074\) Thus, the copper layer accounts for 99.26% of heat conduction and the epoxy layer accounts for 0.74% of heat conduction.

Calculate the effective thermal conductivity of the board

We can now determine the effective thermal conductivity of the entire board by finding the weighted sum of the thermal conductivities for each layer relative to their heat conduction fraction: \(K_{eff}= K_C \times \frac{Q_C}{Q_{total}} + K_E \times \frac{Q_E}{Q_{total}}\) Since the fractions of heat conduction are given by \(\frac{R_E}{R_{total}}\) and \(\frac{R_C}{R_{total}}\), and \(Q_C\) and \(Q_{total}\) are proportional, we can rewrite this as: \(K_{eff}=(K_C \times \frac{R_E}{R_{total}})+(K_E \times \frac{R_C}{R_{total}})\) \(K_{eff}=(386\mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) \times 0.9926+(\frac{0.26 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{1})(0.0074)=382.67 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) The effective thermal conductivity of the board is approximately 382.67 W/m·K.