Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length \(L=25.4 \mathrm{~mm}\), made of copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\), and the convection coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by 10 -cm section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

First, let's determine the temperature distribution along the fin (\(x\) direction) using the general heat conduction equation: $$\frac{d}{dx}\left(kA\frac{dT}{dx}\right) = q_{conv}$$ where \(A\) is the fin cross-sectional area, and \(q_{conv}\) is the heat transfer due to convection. In the steady state, the temperature distribution in the fin is linear, and we can write: $$kA\frac{dT}{dx} = hP(T - T_\infty)$$ where \(P\) is the perimeter of the fin. Substituting the given data, we obtain: $$400 * \pi (0.0005^2) * \frac{d\theta}{dx} = 100 * \pi(0.001) * (132 - \theta)$$ Solinding for \(\frac{d\theta}{dx}\): $$\frac{d\theta}{dx} = \frac{-625}{2}(132 - \theta)$$ Integrating to find the temperature function: $$\int_{132}^{\theta} \frac{1}{132 - y} dy = \int_0^x -\frac{625}{2} dx$$ $$-\ln(132 - \theta) = -\frac{625}{2}x + C_1$$ Now, we need to apply the boundary condition: at \(x = L\), \(T = T_{s2} = 0^{\circ} \mathrm{C}\): $$C_1 = \ln(132)$$ So, the temperature function \(\theta(x)\) can be expressed as: $$\ln(132 - \theta) = \frac{625}{2}x + \ln(132)$$ Now, let's find the temperature at \(x = \frac{L}{2} = 12.7 \mathrm{mm}\): $$\theta\left(\frac{L}{2}\right) = 132 - e^{\frac{625}{2} \cdot 0.0127+\ln(132)}$$ Calculating: $$\theta\left(\frac{L}{2}\right) \approx 66.4^{\circ} \mathrm{C}$$

The rate of heat transfer from the hot surface through each fin, \(q_f\), can be calculated by: $$q_f = kA \frac{dT}{dx} |_{x = 0}$$ $$q_f = 400 * \pi (0.0005^2)(-\frac{625}{2}(132 - \theta(0^{\circ} \mathrm{C})))$$ Calculating: $$q_f \approx 32.7 \mathrm{W}$$ Now, let's determine fin effectiveness. Fin effectiveness, \(\eta_f\), is the ratio of the actual heat transfer rate through the fin to the maximum possible heat transfer rate if the entire fin were at the base temperature: $$\eta_f = \frac{q_f}{hPL(T_{s1} - T_\infty)}$$ $$\eta_f = \frac{32.7}{100*\pi(0.001)*25.4*(132 - 0)}$$ Calculating: $$\eta_f \approx 0.327$$ The use of fins is justified if the fin effectiveness is greater than 0.1. In this case, η_f = 0.327, so the use of fins is indeed justified.

To calculate the total rate of heat transfer from a \(10 \times 10\) cm section of the wall with 625 uniformly distributed fins, we first determine the total heat transfer rate from the fins, and then add it to the heat transfer rate from the unfinned wall surface. Heat transfer rate from the fins: $$q_{f, total} = 625 \times q_f = 625 \times 32.7 = 20437.5 \mathrm{W}$$ Next, we need to calculate the heat transfer rate from the unfinned wall surface. The total area of the fins can be found by multiplying the number of fins by their base surface area: $$A_{fins} = 625 * \pi (0.0005^2) = 0.0491 \mathrm{m^2}$$ So, the unfinned wall surface area is: $$A_{wall} = 0.1^2 - A_{fins} = 0.0100 - 0.0491 = 0.095091 \mathrm{m^2}$$ Now, we can calculate the rate of heat transfer from the unfinned wall surface: $$q_{wall} = hA_{wall}(T_{s1} - T_\infty) = 100 * 0.095091 * (132 - 0) = 1257.2 \mathrm{W}$$ Finally, the total rate of heat transfer from the wall section is: $$q_{total} = q_{f, total} + q_{wall} = 20437.5 + 1257.2 = 21694.7 \mathrm{W}$$