Lecture Week 3 2
description
Transcript of Lecture Week 3 2
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ME 3345 Heat Transfer
Objective:
Heat Transfer from Extended Surfaces.
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( )conv surface sq A h T T
How to enhance heat transfer
(without increasing the temperature difference) ??
Fins - Extended Surfaces
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( )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
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Mobile Pentium Processors
Extruded Heat Sink
Automobile Radiator
Examples of Extended Surfaces
Radiator (household heating)
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Simple Structures:
We will perform the analysis for simple cases and discuss
engineering methods to deal with complicated geometry.
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How much performance increase
Space
Weight/ Material
Manufacturing process
Cost
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(a) Rectangular fin. (b) Pin fin.
Fins of Uniform Cross Section
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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.
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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.
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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
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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
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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
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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
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Equivalent Thermal Circuit :
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Equivalent Thermal Circuit :
Effect of Surface Contact Resistance:
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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.
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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