15 - Conduction - Fins

\[\theta= C_1 \cosh m(L-x) + C_2 \sinh m(L-x) \tag*{(1)}\] The heat transfer area at the fin tip is generally small compared with the lateral area of the fin for heat transfer. For such situations, the heat loss from the fin tip is negligible compared with that from lateral surfaces, and the boundary condition at the tip characterizing this situation is taken as \(d\theta/dx = 0\) at \(x=L\). Boundary Conditions (BC): \[\begin{aligned} {3} \theta &= \theta_o = (T_o-T_\infty) & \qquad \text{at} \quad x&=0 \\ \frac{d\theta}{dx} &= 0 \qquad & \text{at} \quad x&= L \end{aligned}\] Differentiating Eqn.(1), \[\frac{d\theta}{dx} = -mC_1\sinh m(L-x) - mC_2\cosh m(L-x)\]

Using B.C 2 on the above, \[\begin{aligned} 0 &= -mC_1\sin h m0 - mC_2\cosh m0 \\ 0 & 0 - C_2 \qquad (\because \sinh 0 = 0 \quad \text{and} \quad \cosh 0 = 1) \\ \Longrightarrow \quad C_2 &=0 \end{aligned}\] Therefore, Eqn.(1) becomes, \[\theta = C_1 \cosh m(L-x)\] From B.C 1, at \(x=0\), \(\theta=\theta_o\). Using this in above equation, \[\theta_o = C_1 \cosh mL \qquad \Longrightarrow \quad C_1 = \frac{\theta_o}{\cosh mL}\] Hence, we get \[\boxed{\frac{\theta}{\theta_o} = \frac{\cosh m(L-x)}{\cosh mL}}\]

Taking derivative of the temperature distribution equation, and at \(x=0\) we get \[\left.\frac{d\theta}{dx}\right|_{x=0} = -m\theta_o \frac{\sinh m(L-0)}{\cosh mL} = -\theta_o m\tanh mL\] Heat transfer through the fin is given by \[Q = -\left.kA\frac{d\theta}{dx}\right|_{x=0} = kA\theta_om\tanh mL = \theta_o\sqrt{PhkA}\tanh mL\]