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.
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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) |
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where |
Q = heat loss from the fin, W
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h = convective heat transfer coefficient
from fin to ambient air, W/m2K
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T(x) = temperature of the fin, K
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Ta = ambient
air temperature, K
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w = width of fin, m
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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) |
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where |
qx(x) = heat
flux in the x direction at x, W/m2
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qx(x+dx) = heat
flow in the x direction at x + dx, W/m2
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qy(x) = heat
flux in the y direction at x, W/m2
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(4.3) |
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(4.4) |
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(4.5) |
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(4.6) |
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(4.7) |
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Using equations (4.6) and (4.7) in equation (4.2), we obtain
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(4.8) |
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(4.9) |
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The solution to this problem is
(4.10) |
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(4.11) |
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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) |
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(4.12) |
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(4.13) |
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