The more practical use of the fin efficiency lies in the fact
that, once it is calculated, the resulting *dimensionless*
expression can be used to easily determine the actual heat flow
from any similar fin.

In other words, we can rearrange the fin efficiency definition to yield:

$\displaystyle{q_{fin} = \eta_fhA_f(T_{BASE} - T_{bulk})}$

so that simply "looking up" the efficiency value $\eta_f$ yields our solution.

OUTCOME:

Calculate the heat flow through a fin from the fin's efficiency

TEST YOURSELF

Let's look at an example...

You are evaluating the introduction of "pin"-type fins onto the
surface of your company's new heat exchanger. You suggest that, in
order to compare the two scenario's realistically, you need to
calculate the heat flow both with and without the fin. As long as
you get at least a 10% increase in heat flow, you will go ahead and
do it.

Should you do it? (Your proposal is to put 25 aluminum (k=200W/mK)
pins that are 3 cm long fins of 1 cm diameter on every 100
cm^{2} of surface. Your heat transfer coefficient is 10
W/m^{2}K.)

Since the heat flow off the unfinned surface and the fins
themselves are in **parallel**, we simply add them to get the
total heat flow, so

$\displaystyle{q = q_{unfinned}+q_{fin} = hA_o\Delta T +
\eta_fhA_f \Delta T = h(A_o+\eta_fA_f)\Delta T}$

where A_{o} is the area of the unfinned surface.

NOTE:

We can therefore think of the fin efficiency as a correction factor for the amount of the surface area of the fin that is actually "useful" (as useful as the unfinned surface, that is).

One final important issue to note is that the simple expression for fin heat flow,

$\displaystyle{q_{fin} = \eta_fhA_f \Delta T}$

lends itself quite easily to our resistor analysis, to yield

$\displaystyle{R_{fin} = \frac{1}{\eta_fhA_f}}$

NOTE:

Be careful to remember that the fin heat flow (and therefore the
fin resistance) accounts for **both** conduction and convection
through/off the fin.