Before knowing about fin efficiency, let’s have brief instructions about the fins.

## What is fin?

Fins are the extended surfaces attached to the object to increase the area exposed to the surrounding medium that helps to increase the rate of heat transfer between body and surrounding.

It helps to increase the rate of heat transfer in various applications like engine radiators heat exchangers etc.

## What is Fin efficiency?

Fin efficiency is the ratio of actual heat dissipated by the fin to the maximum heat transferred by the pin if the whole fin has a temperature equal to the base temperature of the fin.

Therefore,

## Fin efficiency equation for different fin conditions:

**A) For infinitely long fin:**

For infinitely long fin, the rate of heat transfer is given by,

`Q_{\text{actual}}=\sqrt{PhKA}(t_{0}-t_{a})`

While the maximum heat transfer through fin is given by,

`Q_{\text{max}}=hPl(t_{0}-t_{a})`

Where,

P = Perimeter of cross-section

h = Convective heat transfer coefficient

K = Thermal conductivity of fin material

A = Area of cross-section of fin

`t_{0}` = Base temperature of the fin

`t_{a}` = Surrounding temperature

l = Length of fin

Therefore fin efficiency is given by,

`\eta_{\text{fin}}` = `\frac{\sqrt{PhKA}(t_{0}-t_{a})}{hPl(t_{0}-t_{a})}`

The equation can be reduced as,

= `\frac{\sqrt{PhKA}}{hPl}`

= `\frac{\sqrt{KA}}{\sqrt{Ph}\times l}`

= `\frac{1}{\sqrt{\frac{hP}{KA}} \times l}`

But `\sqrt{\frac{hP}{KA}}` = m, Therefore, the fin efficiency equation will become,

`\eta_{\text{fin}} =\frac{1}{ml}`

**This is the fin efficiency equation for an infinitely long fin.**

**B) For the fin with insulated tip:**

The `Q_{\text{actual}}` for fin with insulated tip is given by,

`Q_{\text{actual}}`=`\sqrt{PhKA}(t_{0}-t_{a})tanh(ml)`

And the equation for maximum heat transfer is given by,

`Q_{\text{max}}=hPl(t_{0}-t_{a})`

Therefore fin efficiency is given by,

`\eta _{\text{fin}}` = `\frac{\sqrt{PhKA}\times (t_{0}-t_{a})tanh(ml)}{hPl(t_{0}-t_{a})}`

The equation can be reduced as,

`\eta _{\text{fin}}` = `\frac{\sqrt{PhKA}\times tanh(ml)}{hPl}`

= `\frac{\sqrt{KA}\times tanh(ml)}{\sqrt{hP}\times l}`

= `\frac{tanh(ml)}{\sqrt{\frac{hP}{KA}}\times l}`

But `\sqrt{\frac{hP}{KA}}` = m, Therefore, the fin efficiency equation will become,

`\eta _{\text{fin}}` = `\frac{tanh(ml)}{m\times l}`

**This is the fin efficiency equation for an fin with insulated tip.**

Read also:Fin effectiveness for the different fin conditions

Super website… It helps me a lot

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