The problem I have today is to determine the differential equation for a square fin protruding from a wall that experiences surface to ambient radiation and has an internal heat generation of $\dot q$ per length. The fin does not experience convection. I performed an energy balance on a differential element of the fin and reduced the differential equation to this: $$\frac{d}{dx}(kA_{c}\frac{dT}{dx})+\dot q=\frac{P \sigma}{kA_{c}}(T(x)^{4}-T_{sur}^{4})$$ Because of constant corss-sectional area and constant specific heat of the fin, the equation reduces to: $$\frac{d^{2}T}{dx^{2}}-\frac{P \sigma}{kA_{c}}T(x)^{4}+\frac{P \sigma}{kA_{c}}T_{sur}^{4}+\frac{\dot q}{kA_{c}}=0$$

When a fin is normally subjected to convection, we can say that $\theta(x)=T(x)-T_{\infty}$ and $\frac{d \theta}{dx}=\frac{dT}{dx}$ and $\frac{d^{2}\theta}{dx^{2}}=\frac{d^{2}T}{dx^{2}}$ . In that scenario, the $h(T(x)-T_{\infty})$ term reduces to $\theta(x)$ and the differential equation becomes second-order and linear.

In this case with radiation, the differential equation is second order and highly nonlinear. I saw that I could "linearize" the radiation heat loss term but it still looks like there is a $T^{2}$ dependence which suggests it is still nonlinear. I am confused on the next steps I should take. If the differential equation that I created is fine then that is also great. Any Help is appreciated.


1 Answer 1


Try using the method of undetermined coefficients to solve this - that's usually a good way to tackle an equation such as this one.

  • $\begingroup$ This method is for linear ODEs. $\endgroup$
    – jinawee
    Jan 19, 2014 at 14:12

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