Question

Circular fins of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), are attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fins are made of material with thermal conductivity of \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), they are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is $250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of \(250^{\circ} \mathrm{C}\) (d) Convection from the fin tip

Short answer

Question: Calculate the heat transfer rate and plot the temperature variation for a fin with the given dimensions and material properties for four different boundary conditions: (a) infinitely long fin, (b) adiabatic fin tip, (c) fin with tip temperature of 250°C, and (d) convection from the fin tip. Diameter of the fin, D = 10 mm; Length of the fin, L = 50 mm; Surface temperature of the wall, T_s = 350 °C; Ambient temperature, T_inf = 25 °C; Thermal conductivity, k = 240 W/(m.K); Convective heat transfer coefficient, h = 250 W/(m².K).

Step-by-step solution

General Information and Constants

Diameter of the fin, D = 10 mm = 0.01 m Length of the fin, L = 50 mm = 0.05 m Surface temperature of the wall, T_s = 350 °C Ambient temperature, T_inf = 25 °C Thermal conductivity, k = 240 W/(m.K) Convective heat transfer coefficient, h = 250 W/(m².K) Fin area, A_fin = pi * D * L = 0.01 * pi * 0.05 m² Perimeter of fin, P = pi * D = 0.01 * pi m Now let's analyze the heat transfer for each boundary condition.

(a) Infinitely Long Fin

An infinitely long fin implies that heat is only lost through convection on the surface and not at the fin tip. So, for the infinitely long fin, we have to find the temperature distribution and the heat transfer rate. 1. Calculate the fin parameter, m: \(m = \sqrt{\frac{h \cdot P}{k \cdot A_c}}\) where \(A_c\) is the cross-sectional area of the fin. \(A_c = \frac{\pi \cdot D^2}{4} = \frac{\pi \cdot (0.01)^2}{4} \mathrm{m}^2\) 2. Find the temperature distribution for an infinitely long fin, T(x): \(T(x) = T_s - (T_s - T_{\infty})(e^{-mx})\) 3. Plot T(x) vs x for the fin. 4. Determine the heat transfer rate for a single fin, q_fin: \(q_{\text{fin}} = h \cdot A_{\text{fin}} \cdot \Big( \frac{T_s - T_{\infty}}{\frac{k \cdot A_c}{m} + \frac{1}{h \cdot P}} \Big)\)

(b) Adiabatic Fin Tip

An adiabatic fin tip means that there is no heat loss at the fin tip (x = L). The fin equation remains the same, but we change the boundary condition for the fin tip temperature. Here, the boundary condition at x = L is that dT/dx = 0. 1. Solve the fin equation with the new boundary condition to find the temperature distribution, T(x). 2. Plot T(x) vs x for the fin. 3. Find the heat transfer rate for a single fin, q_fin, using the same equation as in (a).

(c) Fin with Tip Temperature of 250°C

For this case, the boundary condition at x = L is T(L) = 250°C. 1. Solve the fin equation with the new boundary condition to find the temperature distribution, T(x). 2. Plot T(x) vs x for the fin. 3. Find the heat transfer rate for a single fin, q_fin, using the same equation as in (a).

(d) Convection from the Fin Tip

In this case, heat is being lost through convection even at the fin tip. 1. Set up a new fin equation accounting for the convection from the fin tip. 2. Solve the fin equation to find the temperature distribution, T(x). 3. Plot T(x) vs x for the fin. 4. Find the heat transfer rate for a single fin, q_fin, using the same equation as in (a). By following these steps for each of the four boundary conditions, the heat transfer rate and the temperature distribution for each case can be found and plotted.