A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

The formula for calculating the heat transfer rate with the adiabatic fin tip assumption is: $$q = h * A_f * (T_b - T_\infty)$$ Where: - \(h\) is the heat transfer coefficient (12 W/m²·K) - \(A_f\) is the surface area of the fin. We can calculate it using the formula \(A_f = 2 * \pi * r * L\), where \(r\) is the radius (2 mm or 0.002 m) and \(L\) is the length (10 cm or 0.1 m) - \(T_b\) is the base temperature - \(T_\infty\) is the ambient temperature Notice that we don't know the values of \(T_b\) and \(T_\infty\), so we cannot determine the heat transfer rate directly. However, we can still solve the problem as we can compare the ratios of heat transfer rates obtained with different assumptions.

Using the concept of thermal resistance, we can write the heat transfer rate for the infinitely long fin as: $$q = \frac{(T_b - T_\infty)}{R_{total}} $$ Where: - \(R_{total} = R_{cond} + R_{conv}\), where R_cond is the conduction resistance and R_conv is the convection resistance We can write the respective resistances as: $$R_{cond} = \cfrac{t_cilinder}{k * A_cilinder}$$ And $$R_{conv} = \cfrac{1}{h * A_f}$$ where: - \(t_cilinder\) is the thickness of the aluminum fin (we assume it to be negligible) - \(A_cilinder\) is the cross-sectional area of the aluminum fin (\(\pi r^2\)) We can find the ratio of heat transfer rates as: $$\cfrac{q}{(T_b - T_\infty)} = \cfrac{1}{R_{total}} $$

Now we have expressions for ratios of heat transfer rates for both adiabatic fin tip assumption (\(\cfrac{q_{ad}}{(T_b - T_\infty)} = h * A_f\)) and infinitely long fin assumption (\(\cfrac{q_{inf}}{(T_b - T_\infty)} = \cfrac{1}{R_{total}}\)). Therefore, we can find the percent error as: $$\text{Percent Error} = \cfrac{\left|q_{ad} - q_{inf}\right|}{q_{ad}} \times 100$$ Since we don't know the values of \(T_b\) and \(T_\infty\), we will express the percent error in terms of the ratio of heat transfer rates: $$\text{Percent Error} = \cfrac{\left|\cfrac{q_{ad}}{(T_b - T_\infty)} - \cfrac{q_{inf}}{(T_b - T_\infty)}\right|}{\cfrac{q_{ad}}{(T_b - T_\infty)}} \times 100$$ Plugging in the expressions for \(q_{ad}\) and \(q_{inf}\), and simplifying: $$\text{Percent Error} = \cfrac{\left|h * A_f - \cfrac{1}{R_{total}}\right|}{h * A_f} \times 100$$

Now we have all the information needed to compute the percent error. We can calculate \(A_f\), \(A_cilinder\), \(R_{cond}\), \(R_{conv}\), and \(R_{total}\) using the given data, and finally substitute into the percent error formula to obtain the result: - \(A_f = 2 * \pi * 0.002 * 0.1 \approx 0.004\,\text{m}^2\) - \(A_cilinder = \pi * 0.002^2 \approx 1.257 \times 10^{-5}\,\text{m}^2 \) - \(R_{cond} \approx 0\) (since we assume \(t_cilinder\) to be negligible) - \(R_{conv} = \cfrac{1}{12 * 0.004} = 20.833\,\text{K/W}\) - \(R_{total} = R_{cond} + R_{conv} \approx 20.833\,\text{K/W}\) Now, we can substitute these values into the percent error formula: $$\text{Percent Error} = \cfrac{\left|12 * 0.004 - \cfrac{1}{20.833}\right|}{12 * 0.004} \times 100 \approx 18.75\%$$ Therefore, the percent error in the rate of heat transfer from the fin when using the infinitely long fin assumption instead of the adiabatic fin tip assumption is approximately 18.75%.