Fin (extended surface)

Some finned elements

In the study of heat transfer, a fin is a surface that extends from an object to increase the rate of heat transfer to or from the environment by increasing convection. The amount of conduction, convection, or radiation of an object determines the amount of heat it transfers. Increasing the temperature difference between the object and the environment, increasing the convection heat transfer coefficient, or increasing the surface area of the object increases the heat transfer. Sometimes it is not economical or it is not feasible to change the first two options. Adding a fin to an object, however, increases the surface area and can sometimes be an economical solution to heat transfer problems.

Simplified case

To create a simplified equation for the heat transfer of a fin, many assumptions need to be made:

1. Steady state
2. Constant material properties (independent of temperature)
3. No internal heat generation
4. One-dimensional conduction
5. Uniform cross-sectional area
6. Uniform convection across the surface area

With these assumptions, the conservation of energy can be used to create an energy balance for a differential cross section of the fin.

$q_x=q_{x+dx}+dq_{conv}\,$

Fourier’s law states that

$q_x=-kA_c \left ( \frac{dT}{dx} \right )$,

where $A_c$ is the cross-sectional area of the differential element. Therefore the conduction rate at x+dx can be expressed as

$q_{x+dx}=q_x+\left ( \frac{dq_x}{dx} \right )dx$

Hence, it can also be expressed as

$q_{x+dx}=-kA_c\left ( \frac{dT}{dx} \right )-k\frac{d}{dx} \left ( A_c\frac{dT}{dx} \right )dx$.

Since the equation for heat flux is

$q''=h\left (T_s-T_\infty\right )$

then $dq_{conv}$ is equal to

$h*dA_s \left(T-T_\infty\right)$

where $A_s$ is the surface area of the differential element. By substitution it is found that

$\frac{d^2T}{dx^2} = - \left (\frac{1}{A_c}\frac{dA_c}{dx} \right )\frac{dT}{dx} + \left (\frac{1}{A_c} \frac{h}{k} \frac{dA_s}{dx} \right ) \left ( T - T_\infty \right )$

This is the general equation for convection from extended surfaces. Applying certain boundary conditions will allow this equation to simplify.

Convecting Tip

There is convection occurring all over a fin. The surface all along the length as well as the surface of the tip of the fin are undergoing convection. This calls for a correction of the area of the fin to adequately calculate the heat transfer of the fin. The area of the fin can be calculated by adding the surface area of the sides of the fin and the surface area of the tip.

$A=A_\text{lateral}+A_\text{tip}=pL+A_c$

The length of the fin can be defined by combining the two areas. The total surface area of the fin will become the product of the perimeter and a corrected length, $L_c$.

$A=p L_c$

Then, rearranging the expression to solve for $L_c$:

$L_c=L+A_c/p$

Expanding the equation using the equation for the area of a rectangle we find:

$L_c = L+\frac{wt}{2(w+t)} = L+\frac{t}{2}\frac{w}{w+t} \approx L+\frac{t}{2} \text{ if } t \ll w$

Where L is the length of the rectangle, w is its width, and t is its thickness. Expanding the equation using the equation for the area of a cylinder we find:

$L_c=L+\frac{\pi D^2/4}{\pi D}=L+\frac{D}{4}$

Where D is the diameter of the cylinder and L is its length. Now the length can be replaced in the insulated tip equation with the corrected length to make the equation more accurate for the fin with the convecting tip:

$\frac{T(x)-T_\infty}{T_b-T_\infty} = \frac{\cosh\sqrt{\frac{hp}{kA_c} }(L_c-x)}{\cosh\sqrt{\frac{hp}{kA_c} }L_c}$

$\dot Q_\text{convecting tip} = \sqrt{hpkA_c}(T_b-T_\infty)\tanh\left(\sqrt{\frac{hp}{kA_c} }L_c\right)$

Adiabatic Tip

When comparing the surface area of the fin to the surface area of the tip, it can be seen that the surface area of the tip is fairly negligible when calculating the heat transfer. In this instance, it is assumed that the tip is insulated. If the tip is completely insulated, there will be no heat loss from the tip. The boundary condition regarding the adiabatic tip can be expressed as:

$-kA_c\left(\frac{dT}{dx}\right)_{x=L}=-kA_c\left(\frac{d\theta}{dx}\right)_{x=L}=0$

From this, the general equation can be altered to:

$-kA_cmC_1e^{mL}-C_2e^{-mL}=0$

Then solving for $C_2$:

$C_2=C_1 e^{2mL}$

And substituting that value and solving for θ(x):

$C_1+C_1 e^{2mL}=\theta_b$

$C_1 = \frac{\theta_b}{1+e^{2mL} } = \frac{\theta_b}{e^{ml}\left(e^{-mL}+e^{mL}\right)} = \frac{\theta_b}{e^{-mL}+e^{mL} }e^{-mL}$

$C_2 = C_1 e^{2mL} = \frac{\theta_b}{e^{-mL}+e^{mL} }e^{-mL} e^{2mL} = \frac{\theta_b}{e^{-mL}+e^{mL} }e^{+mL}$

$\theta(x)=T-T_\infty=\theta_b\left[\frac{e^{-m(L-x)}+e^{+m(L-x)} }{e^{-mL}+e^{mL} }\right]=(T_b-T_\infty)\left[\frac{e^{-m(L-x)}+e^{+m(L-x)} }{e^{-mL}+e^{mL} }\right]$

Using the hyperbolic cosine function, the equation becomes:

$\frac{T(x)-T_\infty}{T_b-T_\infty}=\frac{\cosh\sqrt{\frac{hp}{kA_c} }(L-x)}{\cosh\sqrt{\frac{hp}{kA_c} }L}$

$\dot Q_\text{adiabatic tip}=\sqrt{hpkA_c}(T_b-T_\infty)\tanh\left(\sqrt{\frac{hp}{kA_c} }L\right)$

If the fin is very long, the hyperbolic tangent function will approach 1. This will cause the heat transfer equation to simplify again to:

$\dot Q_\text{adiabatic tip}=\sqrt{hpkA_c}(T_b-T_\infty)$

The heat transfer in this case is approximately the same as the calculations for the case of an infinitely long fin.

Infinitely Long Fin

As the tip of the fin is approached, the temperature of the tip can be seen approaching the ambient temperature of the air. In the case that the fin is assumed to be infinitely long, it can also be assumed that the temperature of the tip is equal to the temperature of the air.

$\theta(L) = T_\text{tip}-T_\infty=0 \text{ as } L \to \infty$

Due to this assumption, the equation can be simplified to:

$\theta(L) = 0 = C_1 e^\infty + C_2 e^{-\infty}$

In order to make this equation true, $C_1$ must equal zero and we find that $C_2$ is equal to $\theta_b$. Thus, the first term can be eliminated and the equation simplified once more to:

$\theta(x) = \theta_b e^{-x\sqrt{\frac{hp}{kA_c} } }$

The total heat transfer from the fin is equal to the heat which enters the fin. All of the heat that enters the fin comes from conduction from the heat source to the base, which eventually conducts from the base to the fin. All of this heat must leave the fin through convection, making the equation:

$\dot Q_\text{long fin} = \sqrt{hpkA_c}(T_b-T_\infty)$

Uniform cross-sectional area

For all four cases, the above equation will simplify because the area is constant and

$\frac {dA_s}{dx} = P$

where P is the perimeter of the cross-sectional area. Thus, the general equation for convection from extended surfaces with constant cross-sectional area simplifies to

$\frac{d^2T}{dx^2}=\frac{hP}{kA_c}\left(T-T_\infty\right)$.

The solution to the simplified equation is

$\theta(x)=C_1e^{mx}+C_2e^{-mx}$

where $m^2=\frac{hP}{kA_c}$

and $\theta_x=T(x)-T_\infty$

The constants $C_1$ and $C_2$ can be found by applying the proper boundary conditions. All four cases have the boundary condition $T(x=0)=T_b$ for the temperature at the base. The boundary condition at $x=L$, however, is different for all of them, where L is the length of the fin.

For the first case, the second boundary condition is that there is free convection at the tip. Therefore,

$hA_c\left(T(L)-T_\infty\right)=-kA_c\left.\left(\frac{dT}{dx}\right)\right\vert_{x=L}$

which simplifies to

$h\theta(L)=-k\left.\frac{d\theta}{dx}\right\vert_{x=L}$

Knowing that

$\theta_b=T_{base}-T_\infty$,

the equations can be combined to produce

$h\left(C_1e^{mL}+C_2e^{-mL}\right)=km\left(C_2e^{-mL}-C_1e^{mL}\right)$

$C_1$ and $C_2$ can be solved to produce the temperature distribution, which is in the table below. Then applying Fourier’s law at the base of the fin, the heat transfer rate can be found.

Similar mathematical methods can be used to find the temperature distributions and heat transfer rates for other cases. For the second case, the tip is assumed to be adiabatic or completely insulated. Therefore at x=L,

$\frac{d\theta}{dx}=0$

because heat flux is 0 at an adiabatic tip. For the third case, the temperature at the tip is held constant. Therefore the boundary condition is:

$\theta(L)=\theta_L$

For the fourth and final case, the fin is assumed to be infinitely long. Therefore the boundary condition is:

$\lim_{L\rightarrow \infty} \theta_L=0\,$

The temperature distributions and heat transfer rates can then be found for each case.

Temperature distribution and heat transfer rate for fins of uniform cross sectional area

Case	Tip condition (x=L)	Temperature distribution	Fin heat transfer rate
A	Convection heat transfer	$\frac{\theta}{\theta_b}=\frac{\cosh{m(L-x)}+\left(\frac{h}{mk}\right)\sinh {m(L-x)}}{\cosh{mL}+\left(\frac{h}{mk}\right)\sinh{mL}}$	$\sqrt{hPkA_c}\theta_b\frac{\sinh {mL} + (h/mk) \cosh {mL}}{\cosh {mL} + (h/mk) \sinh {mL}}$
B	Adiabatic	$\frac{\theta}{\theta_b}=\frac{\cosh {m(L-x)}}{\cosh {mL}}$	$\sqrt{hPkA_c}\theta_b\tanh {mL}$
C	Constant Temperature	$\frac{\theta}{\theta_b}=\frac{\frac{\theta_L}{\theta_b}\sinh {mx} + \sinh {m(L-x)}}{\sinh {mL}}$	$\sqrt{hPkA_c}\theta_b\frac{ \cosh {mL}-\frac{\theta_L}{\theta_b}}{\sinh {mL}}$
D	Infinite Fin Length	$\frac{\theta}{\theta_b}=e^{-mx}$	$\sqrt{hPkA_c}\theta_b$

Fin performance

Fin performance can be described in three different ways. The first is fin effectiveness. It is the ratio of the fin heat transfer rate to the heat transfer rate of the object if it had no fin. The formula for this is

$\epsilon_f=\frac{q_f}{hA_{c,b}\theta_b}$,

where $A_{c,b}$ is the fin cross-sectional area at the base. Fin performance can also be characterized by fin efficiency. This is the ratio of the fin heat transfer rate to the heat transfer rate of the fin if the entire fin were at the base temperature.

The fin efficiency is defined as $\eta_f=\frac{q_f} {h A_f \theta_b}$

$A_f$ in this equation is equal to the surface area of the fin. Fin efficiency will always be less than one. This is because assuming the temperature throughout the fin is at the base temperature would increase the heat transfer rate.

The third way fin performance can be described is with overall surface efficiency.

$\eta_o=\frac{q_t}{hA_t\theta_b}$,

where $A_t$ is the total area and $q_t$ is the sum of the heat transfer rates of all the fins. This is the efficiency for an array of fins.

• 

  Aluminium heat sink with low efficiency cooling fins

• 

  Aluminium heat sink with high efficiency cooling fins.

Fin uses

Fins are most commonly used in heat exchanging devices such as radiators in cars and heat exchangers in power plants. They are also used in newer technology such as hydrogen fuel cells. Nature has also taken advantage of the phenomena of fins. The ears of jackrabbits and Fennec Foxes act as fins to release heat from the blood that flows through them.

References

• Incropera, Frank; DeWitt, David P., Bergman, Theodore L., Lavine, Adrienne S. (2007). Fundamentals of Heat and Mass Transfer (6 ed.). New York: John Wiley & Sons. pp. 2–168. ISBN 0-471-45728-0.