Heat Conduction in a Rectangular Fin

Thin fins are often attached to a hot surface to help cool that surface. They are used in many places (for example - motorcycle engines, refrigerators, etc) to increase the effective surface area when the convective heat transfer coefficient (Equation 3.21)to that surface is small.

In this example, we shall consider the steady state temperature distribution and heat conduction in a thin rectangular plate (the fin in Figure 5) attached perpendicularly to a hot wall as shown in Figure 6.
Figure 5
Figure 6

 

The fin has length l, width w and thickness t. Heat flows from the hot wall into the fin by conduction through the area wt. In this problem, the area wt is constant. Let the temperature in the fin at a distance x from the wall be T(x). T(x) decreases with distance from the wall because the fin loses heat from both its top and bottom surfaces. We suppose that the heat loss from the thin edge (area wt) at x = L is negligible. This also justifies the assumption that T depends only on x.

Now let's see why fins attached to a wall increase the heat loss from the wall.

In Figure 6 there are 8 fins attached perpendicular to the wall. The wall area for heat transfer in the absence of fins in wh. The fin area for 8 fins losing heat from both sides is 16wL. Even though the temperature drops in the fin, the area enhancement of the system leads to greater heat loss from the wall than without the fins.

Let us now develop the differential equation which when solved for the temperature distribution will allow us to calculate the heat loss from the fin. That heat loss is proportional to the temperature difference between the fin and the ambient air so that
(4.1)
where
Q = heat loss from the fin, W
 
h = convective heat transfer coefficient from fin to ambient air, W/m2K
 
T(x) = temperature of the fin, K
 
Ta = ambient air temperature, K
 
w = width of fin, m

Clearly Q can be calculated once T(x) has been determined.

Consider a strip (or "element") of the fin of width dx at a distance x from the wall. In the steady state heat flows in at x and out at x+dx through the area wt. Heat is lost from the top and bottom faces of the element through areas w dx. Thus
(4.2)
where
qx(x) = heat flux in the x direction at x, W/m2
 
qx(x+dx) = heat flow in the x direction at x + dx, W/m2
 
qy(x) = heat flux in the y direction at x, W/m2

We relate qx at x+dx to qx at x using Taylor's theorem
 
(4.3)
Since by Fourier's law
(4.4)
 
(4.5)
(4.6)
The right hand side of equation (4.2) is evaluated using Newton's law of cooling so that
(4.7)
where h = convective heat transfer coefficient, W/m2K

Using equations (4.6) and (4.7) in equation (4.2), we obtain
 
which gives the differential equation for the temperature distribution in the fin as
(4.8)
with the boundary conditions
(4.9)
The simplification of T(x = L) = Ta says that the fin is long enough for the temperature of the fin to equal the temperature of the ambient air.

The solution to this problem is
(4.10)
and using equation (4.1)
(4.11)

Recall that we assumed in our derivation that there was y - direction heat conduction with losses governed by Newton's law of cooling
(4.7)
If we represent qy (x) using Fourier's law then
(4.12)
The problem here is that the temperature across the fin (in the y direction) was assumed to be constant. Of course, cannot be zero everywhere. At y = t/2 using equations (4.7) and (4.12).
(4.13)
In a thin fin, the distribution in the y direction is flat and at y = t/2, must equal


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