Consider a volume element of a fin at location x having a length dx, C/S area Ac and a perimeter P as shown in fig under steady state condition,

Energy equation for above fig can be written as

$\begin{pmatrix}rate of heat\\ conduction into\\ the element at x\\\end{pmatrix}$=$\begin{pmatrix}rate of heat\\ conduction from the\\ element at x+dx\\\end{pmatrix}$+$\begin{pmatrix}rate of heat\\ convection from\\ the element\\\end{pmatrix}$

Fig 2.1: Volume element of a fin at location x having a length of dx.

$Q_x$=$Q_(x+dx)$+$Q_conv$

$\frac{Q_(x+dx)-Q_x}{dx+hp(t-t_a )}$=0………(i) as $Q_conv$=$h(pdx)(t-t_a)$

$\frac{dQ_x}{dx+hP(t-t_a )}$=0

$Q_cond$=$-kA_c$ $\frac{dt}{dx}$………(ii)

Substituting the value of $Q_c$ond from the above equation in eq (i), we get the differential equation governing heat transfer in fins:

$\frac{d}{dx}$ ($kA_c$ $\frac{dt}{dx}$)-hP($t-t_a$)=0………(iii)

$\frac{d^2 t}{dx^2 }$-$\frac{hP}{kA_c }(t-t_a )$=0

$\frac{d^2 t}{dx^2}$-$m^2 θ$=0………(iv)

Where m=$\sqrt{\frac{hP}{A_c k}}$……….(v)

θ=$t-t_a$

Solving equation (iv) we will get the solution as follows

θ=$C_1 e^mx$+$C_2 e^(-mx)$………(vi)

To solve above equation we will apply boundary condition. boundary conditions are heat dissipation from a fin losing heat at the tip.

Fig2.1: fin of finite length losing heat at the tip

Please note $T_L$=t, $T_∞=t_a$

Boundary condition can be obtained by energy balance at fin tip.

$-kA_c\frac{dt}{dx}_(x=L)$=$hA_s (t-t_a )$

$-k\frac{dθ}{dx}_(x=L)=hθ_(x=L)$ ………(vii)

Substituting above equation in (vi) we get,

-$km[-C_1 e^(-mL)+C_2 e^mL ]$=$h[C_1 e^(-mL)+C_2 e^mL ]$

Rearranging as,

$C_1 e^(-mL)-C_2 e^mL=\frac{h}{mk} [C_1 e^(-mL)+C_2 e^mL ]$

Solving above

$θ_0 e^(-mL)-C_2 e^(-mL)-C_2 e^mL=h/mk [θ_0 e^(-mL)-C_2 e^(-mL)+C_2 e^mL ]$ as$θ_0=C_1+C_2$ from eq (vi)

Solving above we will get,

$C_2=(θ_0 \frac{e^(-mL)-\frac{h}{mk} e^(-mL)}{e^mL+e^(-mL)+\frac{h}{mk} [e^mL-e^(-mL)}$

Similarly, we get

$C_1=θ_0 \frac{e^mL+\frac{h}{mk} e^mL }{e^mL+e^(-mL)+\frac{h}{mk} [e^mL-e^(-mL)]}$

Substituting $C_1$ and $C_2$ in equation (vi) we get,

$\frac{θ(x)}{θ_0}$ =$\frac{(e^m(L-x)+\frac{h}{mk} e^m(L-x)+e^(-m(L-x))-\frac{h}{mk} e^-m(L-x))}{(e^mL+e^(-mL)+\frac{h}{mk} [e^mL-e^(-mL) ] )}$

It may be rearranged and expressed in terms of hyperbolic functions as

$\frac{θ(x)}{θ_0}$ =$\frac{t(x)-t_a)}{(t-t_a )}$=$\frac{(cosh{m(L-x)}+(\frac{h}{mk})sinh{m(L-x)})}{(cosh⁡(mL)+(\frac{h}{mk})sinh⁡(mL))}$

The total heat transfer from fin

$Q_0=Q_fin$=$-kA_c (\frac{dθ}{dx})_(x=0)$=

$\sqrt{(hPkA_c )}(t-t_a)\frac{(sinh⁡(mL)+\frac{h}{mk} cosh⁡(mL))}{(cos⁡(mL)+\frac{h}{mk} sinh⁡(mL))}$