Question 7.6: The performance of the fin in Figure 7.12 is now examined; i...
The performance of the fin in Figure 7.12 is now examined; it depends on combined conduction, convection, and radiation. A gas at T_e is flowing over the fin and removing energy by convection. The environment to which the fin radiates is also assumed to be at T_e. The fin cross section has area A and perimeter P. The fin is nongray with total absorptivity α for radiation incident from the environment.
From the control volume approach, an energy balance on a fin element of length dx yields
kA\frac{\partial^2 T}{\partial x^2}dx=\sigma [\epsilon T^4(x)-\alpha T_e^4]Pdx+hPdx[T(x)-T_e] (7.16)
The term on the left is the net conduction into the element, and on the right are the radiative and convective losses. The radiative exchange between the fin and its base is neglected here but has been included in more detailed analyses. This equation is to be solved for T(x), which can then be used to obtain the energy dissipation. Multiply by [1/(kA dx)] dT/dx, and integrate once to yield.
\frac{1}{2}\left\lgroup\frac{dT}{dx} \right\rgroup^2=\frac{\epsilon \sigma P}{KA} \left\lgroup\frac{T^5}{5}-\frac{\alpha }{\epsilon }TT_e^4 \right\rgroup+\frac{hP}{KA} \left\lgroup\frac{T^2}{2}-TT_e \right\rgroup +C (7.17)
where C is a constant of integration.
For convection and radiation from the end surface of the fin, or for the end of the fin assumed insulated, the dT/dx in Equation 7.17 is expressed in terms of the appropriate end conditions. For a case that leads to an analytical solution let T_e ≈ 0 and let the fin be long. For large x, T(x) \longrightarrow 0 and dT/dx \longrightarrow 0 , and from Equation 7.17 C = 0. Solving for dT/dx gives
\frac{dT}{dx}=-\left\lgroup\frac{2}{5}\frac{P\epsilon \sigma }{KA} T^5+\frac{hP}{KA}T^2 \right\rgroup^{1/2} (7.18)
The minus sign is used for the square root since T decreases as x increases. The variables in Equation 7.18 are separated and the equation integrated with the condition that T(0) = T_h,
\int_{0}^{x}{dx}=-\int_{T_b}^{T}{\frac{dT}{T\left[\frac{2}{5}(P\epsilon \sigma /KA)T^3+hP/KA \right]^{1/2} } }
x=\frac{1}{3}M^{-1/2}\left[\ln\frac{(GT_b^3+M)^{1/2}-M^{1/2}}{(GT_b^3+M)^{1/2}+M^{1/2}} -\ln\frac{(GT^3+M)^{1/2}-M^{1/2}}{(GT^3+M)^{1/2}+M^{1/2}} \right] (7.19)
where G = 2/3P\epsilonσ/kA and M = hP/kA. For this simplified limiting case an analytical relation for T(x) is obtained. The solution can be carried somewhat further, as considered in Homework Problem 7.3. A detailed treatment of this type of fin is in Shouman (1965, 1968).