This is treated as a quasi-one dimensional system because, if we label the horizontal direction as $x$ and the vertical one as $y$, then it is assumed that there are no temperature gradients in the $y$ or $z$ direction, so:
$$\frac{\partial T}{\partial y}=0, \frac{\partial T}{\partial z}=0$$
So the temperature in the $y$ and $z$-directions is considered uniform. To put it even more clearly, temperature is dependent only on $x$. Not doing so would require solving the 3D Fourier PDE:
$$\frac{\partial T}{\partial t}=\kappa\Big(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\Big)+\frac{\dot{Q}(x,y)}{c_p \rho}$$
Which is mathematically far more challenging (for stationary state: $\frac{\partial T}{\partial t}=0$)
Using the quasi-one dimensional approach is a good approximation as long as the fin is thin (compared to its length). The DE becomes:
$$\frac{d^2T}{dx^2}-\frac{ph}{kA}(T-T_{\infty})=0$$
(Where $p$ is the perimeter of the fin and $A$ the cross-section)
Differential equations need boundary conditions and for this problem we have two choices:
- Assume the tip temperature is:
$$T(L)=T_{\infty}$$
This tends to be true for very long fins.
- Assume that due to the small area of the tip, no convection takes place there and of course no conduction can take place either (it's the end of the fin!), which mathematically means:
$$\Big(\frac{dT}{dx}\Big)_{x=L}=0$$
Neither assumptions are really 100 % correct but here your teacher chose the second one.
Here's my own derivation of the fin problem.
I just watched a video where he said he neglected the sides because they have negligible area...makes sense but I am not sure that this is 100% right because he also ignored the tip and we don't ignore it in our course.
There's not need to ignore the sides. Read my derivation.