Heat Transfer from Extended Surfaces (Fins)

There are two ways to increase the convection heat transfer rate:

1. Increasing the convection heat transfer coefficient (h) by enhancing the fluid flow or flow velocity over the body.
2. Increasing the surface area As.

Increasing h may require installation of a pump or a fan or replacing the existing one with a new one having higher capacity, the alternative is to increase the effective surface area by extended surfaces or fins.

Fins are extended surfaces from a surface or body and meant for increasing the heat transfer rate from any surface to the fluid by increasing surface area of heat transfer.

Types of Fin

• Longitudinal Fins

• Radial Fins

• Pins

Analysis of fins with uniform cross sectional area

Assumptions:

• Fin material is homogeneous and isotropic.
• The temperature at any cross section of the fin is uniform.
• No heat generation and steady state conduction
• 1D conduction along x only
• Uniform h and k

Consider a differential element of length dx. $q_x$ is the heat conducted in to the element along x- direction given by

\[q_x = -kA\frac{dT}{dx}\]

where, k : thermal conductivity of fin material 

A : Cross section area of fin

\[q_{x+dx} = q_x + \frac{\partial{q_x}}{\partial{x}}dx\]

Convective heat transfer from the surface of element to fluid is given by

\[q_{conv} = h(Pdx)(T - T_∞)\]

T : Temperature of element 

h : convective heat transfer coefficient 

P : perimeter of fin cross section

Energy Balance for the element:

\[q_x = q_{x+dx} + q_{conv}\]

we get,

\[\frac{d^2 T}{dx^2} - \Bigg(\frac{hP}{kA}\Bigg)(T - T_∞) = 0\]

Define θ = $T - T_∞$  and \(m^2 = \frac{hP}{kA}\)

\[\boxed{\frac{d^2 θ}{dx^2} - m^2 θ = 0}\]

General solution of above 2nd order differential equations is given as:

\[θ = C_1 e^{-mx} + C_2 e^{mx}\]

C1 and C2 are constant of integration and can be obtained from boundary conditions.

one common boundary condition is:

T = To at x = 0  $\implies θ = θ_o = T_o - T_∞$ at x =0.

Case I : Fin is infinitely long

When the fin is infinitely long then the temperature at the tip of the fin will be essentially that of the fluid. 

$ θ \to 0  \quad as \quad x \to ∞$

On applying boundary conditions we get,

\[\boxed{\frac{θ}{θ_o} = \frac{T - T_∞}{T_o - T_∞} = e^{-mx}}\]

The heat transfer through fin is:

\[q_{fin} = \sqrt{hPkA}(T_o - T_∞) = \sqrt{hPkA} θ_o\]

Case II : Fin tip is insulated

When fin tip is insulated then conduction heat transfer at x = L will be 0.

\[\Bigg(\frac{dT}{dx}\Bigg)_{x = L} = 0 \quad and \quad \Bigg(\frac{dθ}{dx}\Bigg)_{x = L} = 0\]

The Temperature distribution given by:

\[\frac{θ}{θ_o} = \frac{T - T_∞}{T_o - T_∞} = \frac{cosh[m(L - x)]}{cosh(mL)}\]

The heat transfer through fin is:

\[q_{fin} = \sqrt{hPkA} θ_o  tan(mL)\]

Case III : Fin is finite in length and also loses heat by convection from its tip

Conduction heat transfer rate at x = L is equal to convection heat transfer from tip 

\[\Bigg(-kA\frac{dT}{dx}\Bigg)_{x = L} = hA(T_{x = L} - T_∞)\]

Then the temperature distribution given by:

\[\frac{θ}{θ_o} = \frac{T - T_∞}{T_o - T_∞} = \frac{cosh[m(L_c - x)]}{cosh(mL_c)}\]

The heat transfer through fin is:

\[q_{fin} = \sqrt{hPkA} θ_o tan(mL_c)\]

where Lc is corrected length 

For longitudinal fin (rectangular), $Lc = L + \frac{t}{2}$

For pin fin (cylindrical), $Lc = L + \frac{d}{4}$

Fin Efficiency η 

Fin efficiency is defined as the ratio of actual heat transfer rate through the fin and maximum possible heat transfer rate through the fin i.e. when entire fin is at its base temperature.

\[η_{fin} = \frac{q_{actual}}{q_{max, possible}}\]

\[η_{long fin} = \frac{\sqrt{hPkA}(T - T_∞)}{h(PL)(T - T_∞)} = \frac{1}{L} \sqrt{\frac{kA}{hP}} = \frac{1}{mL}\]

\[η_{insulated tip} = \frac{\sqrt{hPkA}(T - T_∞) tanh(mL)}{h(PL)(T - T_∞)} = \frac{tanh (mL)}{mL}\]

Fin Effectiveness ε

It is defined as the ratio of heat transfer rate with fin and the heat transfer rate without fin. This parameter highlights the usefulness of fin.

\[ε_{long - fin} = \frac{q_{fin}}{q_{without fin}} = \frac{\sqrt{hPkA}(T - T_∞)}{h(A)(T - T_∞)} = \sqrt{\frac{kP}{hA}}\]

\[ε_{insulated - tip} = \frac{q_{fin}}{q_{without fin}} = \frac{\sqrt{hPkA}(T - T_∞) tanh(mL)}{h(PL)(T - T_∞)} = tanh (mL)\sqrt{\frac{kP}{hA}}\]

$ε \propto  \sqrt{k} \Big(\frac{1}{\sqrt{h}}\Big) \sqrt{\frac{P}{A}}$

NOTE:

• Fin material should made of highly thermal conductivity, (copper, aluminium, iron). Aluminum is preferred: low cost and weight, resistance to corrosion. 
• Convection heat transfer coefficient (h) value is low.
• Lateral surface area i.e. $\frac{P}{A}$ of the fin should be as high as possible (Thin plate fins and slender pin fins).
• Fins with triangular and parabolic profiles contain less materialand are more efficient requiring minimum weight.
• Use of fin is justified if ε > 2.