Fins are a type of extended surface that can enhance heat transfer between a solid and
its adjoining fluid.
Conductive heat transfer occurs within the solid, and then you have conductive and convective
heat transfer from the extended surface.
In fins, we generally assume that radiation is negligible.
So let's start with a differential element from a fin, and we're going to do an energy
balance around this.
Here we have our dx, we have coming in our q_x, and coming out from the back, we have
our q at x+dx.
In addition, we have coming from here convective heat transfer that we'll call dq_convective.
So our energy balance is gong to simplify to in = out, so looking at this element we
find coming in we have q_x, coming out we have our q at x+dx plus our dq_conv.
According to Fourier's law, q_x is equal to -k times the cross sectional area which it's
going through, dT/dx.
We assume in using Fourier's law this way that k is constant.
Now we look at x+dx.
Our q at x+dx is equal to our q at x plus dq_x/dx plus our differential x.
So we'll call this equation 1, call this equation 2, and this equation 3.
Now let's substitute equation 2 into our equation 3.
We'll call this equation 4.
Now we're going to find our convective heat transfer.
So using Newton's law of cooling, this is going to equal h, the convective heat transfer
coefficient, which we're going to assume is constant over the surface of the fin, dA_s
where this is the surface area, T minus T infinity.
And now, let's put all of these expressions back into our energy balance.
And if you notice that q_x cancels out, we can rewrite this as d/dx(A_c dT/dx) times
dx, and I divided through by our k, so this is minus h divided by k, dA_s, T minus T infinity,
and this is going to equal 0.
So we use the chain rule and divide through by both A_c and dx, and we're left with d
squared T/dx squared, plus 1 over A_c, dA_c/dx, all multiplied by dT/dx, minus 1 over A_c,
and now here's our h over k, dA_s/dx, multiplied by T minus T infinity, and this entire expression
equals zero.
We're going to look at a fin.
We're going to say that it has a uniform cross sectional area, and it has a base that has
a constant temperature, T_b, and it extends into a fluid that has a temperature of T infinity
that's also constant.
Our surface area can be written as the perimeter of our element times x, so if we find d_As/dx,
this is going to equal P. In addition, we said that our cross sectional area was constant,
so our derivative of that with respect to x is going to equal zero.
So now we're left with d squared T/dx squared, minus h times P, which is that perimeter,
divided by that thermal conductivity, times the cross sectional area, all multiplied by
T minus T infinity equals zero.
Now we're going to define an excess temperature, and this is going to be theta, and that's
defined as the difference between a temperature at some x, minus T infinity.
In addition, we're going to define an m squared as this h times P divided by k times our cross
sectional area, and because our T infinity is constant, our d(theta)/dT is equal to our
dT(x)/dx.
So what we're left with is d squared theta/dx squared minus m squared times theta equals
zero.
This is a linear, homogeneous, second order differential equation with constant coefficients,
and the general solution has the form that theta(x) is equal to C1 e to the mx plus C2
e to the minus mx.
So now we're going to use the boundary conditions; we're going to say at x=0, our theta is going
to equal theta_b, which is going to be this constant temperature of the base minus T infinity.
Then, at x=L, where L is the tip of the fin, we can have four different cases.
So one of them is that we have at the tip convective heat transfer.
Another condition is that the tip is adiabatic.
A third case is that we have some prescribed constant temperature, T_L, and finally, we
have what's known as an infinite or a very long fin.
So using one of these four boundary conditions, we get a different solution, and the temperature
distribution as well as the heat transfer rate of the fin can be found in appropriate
tables.