ME 3345 Heat Transfer

Objective:

Heat Transfer from Extended Surfaces.

( )conv surface sq A h T T

How to enhance heat transfer

(without increasing the temperature difference) ??

Fins - Extended Surfaces

( )conv surface sq A h T T

How to enhance heat transfer

(without increasing the temperature difference) ??

(1) Increase h by strong forced convection (use fan, use

water instead of air, spray or inject water, etc.

(2) Increase the surface area A. The second is often

achieved by using fins.

Fins - Extended Surfaces

Mobile Pentium Processors

Extruded Heat Sink

Automobile Radiator

Examples of Extended Surfaces

Radiator (household heating)

Simple Structures:

We will perform the analysis for simple cases and discuss

engineering methods to deal with complicated geometry.

How much performance increase

Space

Weight/ Material

Manufacturing process

Cost

(a) Rectangular fin. (b) Pin fin.

Fins of Uniform Cross Section

Analysis of Heat Transfer Enhancement

The application of extended surfaces for heat transfer

enhancement must be carefully considered. This processes

induces additional manufacturing costs and complexity.

Thus, we must find a way to quantify the added benefits

of using extended surfaces to justify their application.

A) Determine the rate of heat transfer from an extended

surface. Involves finding the temperature distribution

in the fin structure.

B) Define some measure of efficiency for extended

surfaces. Use this as a basis for determining when to use them.

P = Perimeter

Ac = Cross-sectional area

x dx

qx qx+dx

qconv

, T hThis is because we have

included the convection

boundary in the control volume.

x dx xq q

x c

dTq kA

dx

Pseudo - 1D, steady-state

x x dx c

d dTq q kA dx

dx dx

( )convq h T T Pdx

( )cd dT

kA hP T Tdx dx

1-D Temperature Distribution. Heat Diffusion Equation.

Constant k, uniform cross-section

2

2( ) 0

c

d T hPT T

kAdx

( )cd dT

kA hP T Tdx dx

2Let ( ) ( ) , then . Let ,d dT hP

x T x T mdx dx kA

22

2Then, 0

dm

dx

22

20

dm

dxSolution of

1 2( )mx mxx C e C e

Linear, homogeneous, second-order differential

equation with constant coefficients.

Need Boundary Conditions to solve for temperature

distribution.

1 2 ( )mx mxT x C e C e

1) At base of fin, Tb = T(0).

1 2(0) b C C

1 2( )mx mxx C e C e

2) At fin tip:

(A)Convection at the tip surface: h L = -kd /dx at x =L

(B) Adiabatic tip. d /dx = 0 at x =L

(C) Prescribed tip temperature L. = L at x = L

(D) Infinite fin (L ) L = 0 at x = L

Tb T

qf

( )b b bq T T

,

bf

t f

qR

qf

Tb T

, depends on B.C.'s.t fR

Rate of Heat Transfer from Fin

qf

Equivalent Thermal Circuit :

Equivalent Thermal Circuit :

Effect of Surface Contact Resistance:

Example: A rod of diameter D and thermal conductivity k

protrudes from a furnace wall that is at temperature Tw. The

initial length of the rod, Lins, is insulated while the remainder

is exposed to convective heat transfer. Assume a convective

heat transfer coefficient of h and temperature T . Find an

expression for the temperature of the rod at the insulation

surface and the rate of heat transfer for the fin.

h, T Tw

Tins

L0 Lins

Assume tip condition is

Adiabatic.

Use and electrical resistance analogy:

qf

Tw Tins T Rins Rfin

Rins = Lins/ k Ac

Rfin = b/ qf = co hPkAmLtanh

1

fin ww insins

fin ins fin fin ins

R T TT T T TT T

R R R R R

From conservation of energy:

And the rate of heat transfer for entire fin structure is given by:

ins

inswf

R

TTq