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.

To calculate the area of the board, we have the length and the width. The area can be calculated using the formula: $$A = length \times width$$ Given length and width are both \(15 cm\), converting to meters: $$A = 0.15 \mathrm{~m} \times 0.15 \mathrm{~m} = 0.0225 \mathrm{~m}^2$$

We will use Fourier's Law of heat conduction to calculate the heat transfer rate for both the copper and epoxy layers. We will assume the same temperature gradient for both layers since we are disregarding heat transfer from side surfaces. For copper: $$q_{Cu} = -k_{Cu} \cdot A \frac{dT}{dx}$$ For epoxy: $$q_{Epoxy} = -k_{Epoxy} \cdot A \frac{dT}{dx}$$ Dividing \(q_{Cu}\) by \(q_{Epoxy}\): $$\frac{q_{Cu}}{q_{Epoxy}} = \frac{k_{Cu}}{k_{Epoxy}}$$ Given values of \(k_{Cu} = 386 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{Epoxy} = 0.26 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\): $$\frac{q_{Cu}}{q_{Epoxy}} = \frac{386}{0.26} = 1484.62$$ Since \(q_{Cu} + q_{Epoxy} = 1\), the fraction of heat conduction along copper and epoxy layers is given by: $$q_{Cu} = \frac{1484.62}{1484.62 + 1} \times 100\%$$ $$q_{Epoxy} = \frac{1}{1484.62 + 1} \times 100\%$$

With the calculated values for \(q_{Cu}\) and \(q_{Epoxy}\), we can find the percentages for heat conduction along the copper and epoxy layers: $$q_{Cu} \approx 99.93\%$$ $$q_{Epoxy} \approx 0.067\%$$

To find the effective thermal conductivity of the board, we use the formula for heat transfer rates: $$q_{eff} = -k_{eff} \cdot A \frac{dT}{dx}$$ Since \(q_{eff} = q_{Cu} + q_{Epoxy}\): $$q_{eff} = -k_{Cu} \cdot A \frac{dT}{dx} + -k_{Epoxy} \cdot A \frac{dT}{dx}$$ $$k_{eff} = \frac{q_{eff}}{A \frac{dT}{dx}}$$ Using the values found in the previous steps: $$k_{eff} \approx \frac{q_{Cu} \times k_{Cu} + q_{Epoxy} \times k_{Epoxy}}{A \frac{dT}{dx}}$$ However, the term \(A \frac{dT}{dx}\) is the same for both copper and epoxy layers, so it cancels out in the equation leaving: $$k_{eff} \approx q_{Cu} \times k_{Cu} + q_{Epoxy} \times k_{Epoxy}$$ Using the previously calculated values: $$k_{eff} \approx 0.9993 \times 386 + 0.00067 \times 0.26$$ $$k_{eff} \approx 385.7 \,\mathrm{W} / \mathrm{m} \cdot \mathrm{K}$$ Therefore, the effective thermal conductivity of the board is approximately \(385.7 \,\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\).