Question

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e., \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is $200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(; and base temperature is \)40^{\circ} \mathrm{C}$. The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2}\). . Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.

Short answer

Answer: The estimated temperature at a distance of 5.0 cm from the base is approximately 38.1°C, and the rate of heat loss from the entire fin is approximately 0.1937 W.

Step-by-step solution

Calculate thermal resistance and heat transfer coefficient

First, convert the given dimensions into meters: Width: 5.0 cm = 0.05 m Thickness: 1.0 mm = 0.001 m Now, calculate the fin's cross-sectional area, perimeter, and m: Area, A = Width * Thickness = 0.05 m * 0.001 m = 0.00005 \(m^{2}\) Perimeter, P = 2 * (Width + Thickness) = 2 * (0.05 m + 0.001 m) = 0.102 m m = sqrt(\(\frac{hP}{kA}\)) = sqrt(\(\frac{20 W/m^2 * 0.102 m}{200 W/m\cdot K * 0.00005 m^2}\)) = 6.4 m^{-1}

Estimate the temperature at a distance of 5.0 cm from the base

We'll use the formula for the temperature of the fin at a given distance, x: \(\theta(x) = \frac{T_b - T_{\infty}}{T_0 - T_{\infty}} \cdot \frac{cosh(m(L-x))}{cosh(mL)}\) Where: \(\theta(x)\) is the temperature at distance x \(T_b\) is the base temperature \(T_{\infty}\) is the surrounding temperature (end of the fin) \(T_0\) is the initial temperature (base of the fin) L is the length of the fin x is the distance from the base Plug in the values: Distance, x = 5.0 cm = 0.05 m \(T_b = 40^{\circ} C\) \(T_{\infty} = 20^{\circ} C\) \(m = 6.4 m^{-1}\) \(L=0.05m\) \(\theta(0.05) = \frac{40 - 20}{40 - 20} \cdot \frac{cosh(6.4 * (0.05 - 0.05))}{cosh(6.4 * 0.05)}= \frac{cosh(0)}{cosh(0.32)} = \frac{1}{1.106} \approx 0.904\) Now, calculate the temperature at x = 0.05 m: \(T(0.05) = T_{\infty} + (\theta(0.05) \cdot (T_b - T_{\infty})) = 20^{\circ} C + (0.904 \cdot (40^{\circ} C - 20^{\circ} C)) \approx 38.1^{\circ} C\)

Determine the overall rate of heat loss from the fin

Now, let's calculate the overall rate of heat loss from the fin: Rate of heat loss = m * k * A * (Tb - T(0.05)) * tanh(mL) Plug in the values: Rate of heat loss = 6.4 * 200 * 0.00005 * (40 - 38.1) * tanh(6.4 * 0.05) = 0.1937 \(W\) The estimated temperature at a distance of 5.0 cm from the base is approximately 38.1\(^{\circ}\)C, and the rate of heat loss from the entire fin is approximately 0.1937 W.