Chapter 2
Steady-state and onedimensional conduction
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2.1 Heat conduction equation
• A rate equation that allows determination of the conduction heat flux from knowledge
of the temperature distribution in a medium
• Fourier’s Law : its most general (vector) form for multidimensional conduction is:


q = −kT
Implications:
– Heat transfer is in the direction of decreasing temperature
(basis for minus sign).
– Fourier’s Law serves to define the thermal conductivity of the

medium
q
k =− 
T
– Direction of heat transfer is perpendicular to lines of constant
temperature (isotherms).
– Heat flux vector may be resolved into orthogonal components.
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2
2.1 Heat conduction equation
Thermophysical Properties
Thermal Conductivity: A measure of a material’s ability to transfer thermal
energy by conduction.
Thermal Diffusivity:𝛼 ≡
𝑘
𝜌𝑐𝑝
A measure of a material’s ability to respond to changes
in its thermal environment. (It measures the ability of a material to conduct thermal energy relative
to its ability to store thermal energy. Materials of large  will respond quickly to changes in their thermal
environment.)
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Property Tables:
Solids: Tables A.1 – A.3
Gases: Table A.4
Liquids: Tables A.5 – A.7
3
2.1 Heat conduction equation
Heat Flux
• Cartesian Coordinates: TT((xx, ,yy
, z, )z )

T 
T 
T 
q = −k
i −k
j −k
k
x
y
z
qx
qz
q y
qy
qx
• Cylindrical Coordinates:
qz
T (r ,  , z )

T 
T 
T 
q = −k
i −k
j −k
k
r
r
z
q
qr
• Spherical Coordinates:
qz
T (r ,  ,  )


T 
T 
T
q = −k
i −k
j −k
k
r
r
r sin 
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qr
q
q
4
2.1 Heat conduction equation
• A differential equation whose solution provides the temperature distribution in a
stationary medium.
• Based on applying conservation of energy to a differential control volume
through which energy transfer is exclusively by conduction.
• Cartesian Coordinates:
  T    T    T 
T
k  +
 k  +  k  + q = C P
x  x  y  y  z  z 
t
Net transfer of thermal energy into the
control volume (inflow-outflow)
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Thermal energy
generation
Change in thermal
energy storage
5
2.1 Heat conduction equation
• Cylindrical Coordinates :
1   T  1   T    T 
T
 kr  + 2
 k  +  k  + q = C P
r r  r  r     z  z 
t
• Spherical Coordinates :
 T  1
1   2 T  1

  T 
T

kr
+
k
+
k
+
q
=

C






P
r  r 2 sin 2     r 2 sin    
t
r 2 r 
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6
2.1 Heat conduction equation
Typical example
• One-Dimensional Conduction in a Planar Medium with Constant Properties
and No Generation
 2T 1 T
=
2
x
 t
k

→ Thermal diffusivity of the medium
c p
SCHEMATIC:
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7
2.1 Heat conduction equation
Boundary and Initial Conditions
• For transient conduction, heat equation is first order in time, requiring
specification of an initial temperature distribution: T ( x, t = 0) = To ( x)
• Since heat equation is second order in space, two boundary conditions
must be specified. Some common cases:
Constant Surface Temperature:
T ( x = 0, t ) = Ts
Constant Heat Flux :
Applied Heat Flux
Insulated Surface
T
−k
= qs
x x =0
T
=0
x x =0
Convection :
−k
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T
= hT − T (0, t )
x x =0
8
2.1 Heat conduction equation
Methodology of a Conduction Analysis
• Solve appropriate form of heat equation to obtain the temperature distribution.
• Knowing the temperature distribution, apply Fourier’s Law to obtain the heat flux at any time,
location and direction of interest.
• Applications:
1:
One-Dimensional, Steady-State Conduction
Common Geometries :
– The Plane Wall: Described in rectangular (x) coordinate.
Area perpendicular to direction of heat transfer is
constant (independent of x).
– The Tube Wall: Radial conduction through tube wall.
– The Spherical Shell: Radial conduction through shell wall.
2:
3:
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Two-Dimensional, Steady State Conduction (Chapter 3)
Transient Conduction (Chapter 4)
9
2.2 Electrical analogy-Thermal resistance
The Plane Wall
•
Consider a plane wall between two fluids of different temperature:
•
Heat Equation:
d  dT 
k
=0
dx  dx 
•
Implications:
Heat rate (qx) is independent of x
Heat flux (qx ) is independent of x.
•
Boundary Conditions : T (0) = Ts ,1
•
Temperature Distribution for Constant k :
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T ( L) = Ts , 2
T ( x) = Ts ,1 + (Ts , 2 − Ts ,1 )
x
L
10
The Plane Wall
•
2.2 Electrical analogy -Thermal resistance
Heat Flux and Heat Rate:
qx = −k
dT k
= (Ts ,1 − Ts , 2 )
dx L
dT kA
=
(Ts ,1 − Ts , 2 )
dx
L
Thermal Resistances  Rt = T 
q 

q x = −kA
•
Conduction in a plane wall:
and Thermal Circuits :
Rt ,cond =
L
kA
Convection:
Rt ,conv =
1
hA
Thermal circuit for plane wall with adjoining fluids:
1
L
1
Rtot =
+
+
h1 A kA h2 A
•
Thermal Resistance for Unit Surface Area :
Units:
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Rt  K/W
Rt,cond =
L
k
Rt,conv =
qx =
T ,1 − T ,2
Rtot
1
h
Rt  m 2 K/W
11
2.2 Electrical analogy -Thermal resistance
The Plane Wall
•
Composite Wall with negligible Contact Resistance:
Rtot =

1  1 L A LB LC 1  Rtot
+
+
+
+
=


A  h1 k A k B k C h4 
A
qx =
•
T ,1 − T , 2
Rtot
Overall Heat Transfer Coefficient (U) :
A modified form of Newton’s Law of Cooling to encompass multiple resistances
to heat transfer.
q x = UAToverall
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Rtot =
1
UA
12
2.2 Electrical analogy -Thermal resistance
The Plane Wall
•
Series – Parallel Composite Wall:
•
Note departure from one-dimensional conditions for k F  kG .
•
Circuits based on assumption of isothermal surfaces normal to x direction or
adiabatic surfaces parallel to x direction provide approximations for qx .
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13
2.2 Electrical analogy -Thermal resistance
The Tube Wall
•
Heat Equation:
𝑑
𝑑𝑟
𝑘𝑟
𝑑𝑇
𝑑𝑟
=0
What does the form of the heat equation tell us about the variation of qr with r
in the wall?
Is the foregoing conclusion consistent with the energy conservation requirement?
•
How does qr vary with r?
Temperature Distribution for constant k :
TS ,1 − TS ,2
 r
T (r ) =
ln  + TS ,2
ln(r1 / r2 )  r2 
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14
2.2 Electrical analogy -Thermal resistance
The Tube Wall
•
Heat Flux and Heat Rate:
qr = −k
dT
k
(Ts ,1 − Ts,2 )
=
dr r ln(r2 / r1 )
qr = 2rqr =
•
qr = 2rLqr =
2Lk
(Ts,1 − Ts,2 )
ln(r2 / r1 )
2k
(Ts,1 − Ts,2 )
ln(r2 / r1 )
Conduction Resistance :
Rt ,cond =
ln(r2 / r1 )
2Lk
Units  K/W
Rt,cond =
ln(r2 / r1 )
2k
Units  m.K/W
Why is it inappropriate to use R
a unit surface area) ?
t ,cond
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(to base the thermal resistance on
15
2.2 Electrical analogy -Thermal resistance
The Tube Wall
•
Composite Wall with negligible Contact Resistance
qr =
T,1 − T, 4
Rtot
= UA(T ,1 − T, 4 )
Note that
−1
UA = Rtot
is a constant independent of
radius. But, U itself is tied to
specification of an interface.
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16
2.2 Electrical analogy -Thermal resistance
The Spherical Shell
•
Heat Equation
𝑑
𝑑𝑟
𝑘𝑟 2
𝑑𝑇
𝑑𝑟
=0
What does the form of the heat equation tell us about the variation of
qr with r? Is this result consistent with conservation of energy?
How does qr vary with r?
•
Temperature Distribution for Constant k :
T (r ) = Ts ,1 − (Ts ,1 − Ts , 2 )
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1 − (r1 / r )
1 − (r1 / r2 )
17
2.2 Electrical analogy -Thermal resistance
The Spherical Shell
•
Heat flux, Heat Rate and Thermal Resistance:
qr = −k
dT
k
= 2
(Ts ,1 − Ts , 2 )
dr r (1 / r1 ) − (1 / r2 )
qr = 4r 2 qr =
4k
(Ts ,1 − Ts , 2 )
(1 / r1 ) − (1 / r2 )
Rt ,cond =
•
(1 / r1 ) − (1 / r2 )
4k
Composite Shell:
qr =
Toverall
= UAToverall
Rtot
−1
UA = Rtot
 Constant
U i = ( Ai Rtot )  Depends on Ai
−1
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2.3 Contact Resistance
Rt,c =
TA − TB
qx
Rt ,c =
Rt,c
Ac
Values depend on: Materials
A and B, surface finishes,
interstitial conditions, and
contact pressure (Tables 3.1
and 3.2)
The temperature drop across the interface between materials may be appreciable.
The existence of a finite contact resistance is due principally to surface roughness effects
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19
2.3 Contact Resistance
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20
2.4 Conduction with Thermal Energy Generation
Typical Thermal Energy Generation
• Involves a local (volumetric) source of thermal energy due to conversion
from another form of energy in a conducting medium.
• The source may be uniformly distributed, as in the conversion from
electrical to thermal energy (Ohmic heating):
E g
I 2 Re
q =
=


or it may be non-uniformly distributed, as in the absorption of radiation
passing through a semi-transparent medium.
q  exp(−x)
• Generation affects the temperature distribution in the medium and causes the heat
rate to vary with location, thereby precluding inclusion of the medium in a thermal
circuit.
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21
2.4 Conduction with Thermal Energy Generation
The Plan Wall
• Consider one-dimensional, steady-state conduction
in a plane wall of constant k, uniform generation and,
and asymmetric surface conditions :
• Heat Equation:
d  dT 
d 2T q
k
 + q = 0 → 2 + = 0
dx  dx 
dx
k
Is the heat flux q” independent of x?
• General Solution :
T ( x) = −(q / 2k )x 2 + C1 x + C2
What is the form of the temperature distribution for
q = 0 ?
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q  0 ?
q  0 ?
22
2.4 Conduction with Thermal Energy Generation
The Plan Wall
Symmetric Surface Conditions or One Surface Insulated :
• What is the temperature gradient at the centerline
or the insulated surface?
• Temperature Distribution :
qL2  x 2 
1 −  + Ts
T ( x) =
2k  L2 
• How do we determine Ts ?
Overall energy balance on the wall →
− E out + E g = 0
− hAs (Ts − T ) + qAs L = 0
Ts = T +
qL
h
• How do we determine the heat rate at x = L?
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23
2.4 Conduction with Thermal Energy Generation
Radial Systems
Cylindrical (Tube) Wall
Solid Cylinder (Circular Rod)
• Heat Equations:
Cylindrical
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1 d  dT 
 kr
 + q = 0
r dr  dr 
Spherical Wall (Shell)
Solid Sphere
Spherical
1 d  2 dT 
 kr
 + q = 0
2
r dr 
dr 
24
2.4 Conduction with Thermal Energy Generation
Radial Systems
• Solution for Uniform Generation in a Solid Sphere of Constant k
with Convection Cooling:
Temperature Distribution
dT
qr 3
kr
=−
+ C1
dr
3
2
T =−
qr 2 C1
− + C2
6k
r
dT
= 0 → C1 = 0
dr r =0
Surface Temperature
Overall energy balance:
•
q ro
qro
→ Ts = T T
+ =T


+
− Eout + Eg = 0 →
s 3h 
3h
Or from a surface energy balance:
 T 
qcond (ro ) = qconv  −4ro2 k 

 r  r = r
o
•
qr
→ Ts = T + qor
= 4ro2 h(Ts − T )  Ts = T +3h o
3h
qro2
T (ro ) = Ts → C2 = Ts +
6k
qro2  r 2 
1 −  + Ts
T (r ) =
6k  ro2 
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25
2.5 Fins and their Applications
Typical Applications
Mechanical systems
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26
2.5 Fins and their Applications
Typical Applications
Electronic systems
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27
2.5 Fins and their Applications
Description of a fin
A Fin (or an extended surface also known as a combined conduction-convection
system) is a solid within which heat transfer by conduction is assumed to be one
dimensional, while heat is also transferred by convection (and/or radiation) from the
surface in a direction transverse to that of conduction.
Extended surfaces may exist in many situations but are commonly used as fins to
enhance heat transfer by increasing the surface area available for convection
(and/or radiation). They are particularly beneficial when h is small, as for a gas and
natural convection.
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28
2.5 Fins and their Applications
Typical fin configurations
Some typical fin configurations:
Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annular
fin, and (d) pin fin of non-uniform cross section.
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29
2.5 Fins and their Applications
General equation of a fin
x
Assuming one-dimensional, steady-state conduction in an extended surface of
 = 0) ,
constant conductivity k, without heat generation (q = 0) and negligible radiation (qrad
the general fin equation is of the form:
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30
2.5 Fins and their Applications
Fins of uniform cross section
For a uniform cross-sectional area Ac , the fin equation is of the form:
d 2 T hP
(T − T ) = 0
2 −
kAc
dx
1
R
=
conv and the reduced temperature  = T − T ,
or, with m = hP /t ,kA

c
hA
2
d 2
2
2 − m  = 0
dx
Rt ,cond =
The general solution is given as follows :
L
kA
𝜃 𝑥 = 𝐶1 𝑒 𝑚𝑥 + 𝐶2 𝑒 −𝑚𝑥
To determine the constants C1 and C2, boundary conditions should be specified.
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31
2.5 Fins and their Applications
Fins of uniform cross section
Solutions :
Base (x = 0) condition
 (0) = Tb − T   b
Tip ( x = L) conditions
A. Convection :
−k
d
= h (L)
dx x = L
d
=0
dx x =L
B. Adiabatic :
C. Fixed temperature :
D. Infinite fin :
Fin Heat Rate:
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 ( L) =  L
(mL  2.65) :  ( L) = 0
q f = −k Ac
d
= h ( x)dAs
dx x =0 Af
32
2.5 Fins and their Applications
Fins of uniform cross-section
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33
2.5 Fins and their Applications
Fins of uniform cross-section
Fin performance parameters
• Fin Efficiency :
f 
qf
q f ,max
=
qf
hA f  b
How is the efficiency affected by the thermal conductivity of the fin?
• Fin Efficiency for case B :
f =
tanh mL
mL
• Fin Resistance :
Rt , f 
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b
qf
=
1
hA f  f
34
2.5 Fins and their Applications
Fins of nonuniform cross-section
In this case, the fin thermal analysis becomes more complex. The solutions of fin
equation are no longer in the form of simple exponential or hyperbolic functions.
As a typical case, let consider the annular fin shown in figure below. Although, the
thickness t of the fin is uniform, i.e. it is independent of radius r, we have :
and
Thus, the resulting fin equation is given by :
With
and
This is a modified Bessel equation of order zero and its general solution of the
following form:
where,
are the modified zero-order Bessel functions of the first and
second kinds respectively. The Bessel functions are tabulated in Appendix B4
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35
2.5 Fins and their Applications
Fins of nonuniform cross-section
For prescribed temperature at the base:
And adiabatic condition at the tip :
The constants C1 and C2 can be determined to yield to the temperature distribution :
where,
, which are
tabulated in Appendix B5.
Hence,
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36
2.5 Fins and their Applications
Concept of corrected fin length
The concept of corrected fin length is based on assuming equivalence between heat
transfer from actual fin (of length L) with tip convection and heat transfer from a
hypothetical fin (of length Lc) with an adiabatic tip.
Hence,
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37
2.5 Fins and their Applications
Concept of corrected fin length
The table below shows efficiency of typical fin shapes (from L. Bergman et
al. Fundamentals of Heat and Mass Transfer, John Wiley & Sons) .
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38
2.5 Fins and their Applications
Concept of corrected fin length
The table below shows efficiency of typical fin shapes (from L. Bergman et
al. Fundamentals of Heat and Mass Transfer, John Wiley & Sons) .
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39
2.5 Fins and their Applications
Concept of corrected fin length
The table below shows efficiency of typical fin shapes (from L. Bergman et
al. Fundamentals of Heat and Mass Transfer, John Wiley & Sons) .
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40
2.5 Fins and their Applications
Concept of corrected fin length
The figure below shows efficiency of typical fin shapes (from L. Bergman et
al. Fundamentals of Heat and Mass Transfer, John Wiley & Sons) .
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41
2.5 Fins and their Applications
Concept of corrected fin length
The figure below shows efficiency of typical fin shapes (from L. Bergman et
al. Fundamentals of Heat and Mass Transfer, John Wiley & Sons) .
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42
2.5 Fins and their Applications
Fin Arrays
• Representative arrays of
(a) rectangular and
(b) annular fins.
– Total surface area:
At = NA f + Ab
Number of fins
Area of exposed base (prime surface)
– Total heat rate:
q t = ( N f hA f + hAb )b
– Overall surface efficiency and resistance:
q
q
o  t = t
qmax hAt b
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o = 1 −
NA f
At
(1 −  f )
Rt ,o =
b
qt
=
1
 o hAt
43
Typical Problems
Problem 2.1
Consider the composite wall shown in figure below. The hot and cold fluids properties, as well
as thermal conductivities of the different material layers and their thickness are given as follows:
Hot fluid: T∞, 1 = 100°C, h1 = 50 W/m2.°C. Cold fluid: T∞, 4 = 100°C, h4 = 25 W/m2.°C
Composite wall : kA = 90 W/m.°C, LA = 3 cm, kB = 60 W/m.°C, LB = 3 cm,
kC = 20 W/m.°C, LC = 30cm.
Surface area of the wall A = 1 m2
1) Determine the resulting heat rate, q
2) Calculate the different surface temperatures
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44
Typical Problems
Problem 2.2
Assessment of thermal barrier coating (TBC) for protection
of turbine blades. Determine maximum blade temperature
with and without TBC.
Schematic:
P.S. La marque est utilisée comme préfixe pour environ 25 alliages, les plus couramment utilisés étant l'Inconel 600,
l'Inconel 625 (NiCr22Mo9Nb), et l'Inconel 718 (NiFe38Cr16Nb).
La zircone est le nom commun de l'oxyde de zirconium (ZrO2).
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COMMENTS: Since the
durability of the TBC decreases
with increasing temperature,
which increases with increasing
thickness, limits to its thickness
are associated with reliability
considerations.
45
Typical Problems
Problem 2.3
Under steady-state conditions, the temperature distribution within a semi-transparent
medium subjected to laser irradiation is given by
Determine: (a) Expressions for the heat flux at the front and rear surfaces,
(b) The heat generation rate,
(c) Expression for absorbed radiation per unit surface area.
Laser : Light Amplification by Stimulated Emission of Radiation
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46
Typical Problems
Problem 2.4
Suitability of a composite spherical shell for storing
radioactive wastes in oceanic waters.
SCHEMATIC:
Are there any problems associated with this proposal?
COMMENTS: In fabrication, attention should be given to maintaining a good thermal contact. A protective
outer coating should be applied to prevent long term corrosion of the stainless steel.
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47
Problem 2.5
Typical Problems
Assessment of cooling scheme for gas turbine blade. Determination of whether blade
temperatures are less than the maximum allowable value (1050 °C) for prescribed
operating conditions and evaluation of blade cooling rate.
FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant.
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k,
(3) Adiabatic blade tip, (4) Negligible radiation.
Schematic:
(a) Whether blade operating conditions are acceptable
(b) Heat transfer to blade coolant
qf = M tanh (mL) = −517Wx0.983 = − 508W
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48
Typical Problems
Problem 2.6
Determination of maximum allowable power for a 20mm x 20mm
electronic chip whose temperature is not to exceed Tc = 85°C when
the chip is attached to an air-cooled heat sink with N=11 fins of
prescribed dimensions.
Tc = 85oC
W = 20 mm
R”t,c= 2x10-6 m2-K/W
k = 180 W/m-K
L b= 3 mm
Lf = 15 mm
Air
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Too = 20oC S
h = 100 W/m2-K
t
Rt,b
Tc
qc
Rt,c
Too
Rt,o
 = 1.8 mm
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Typical Problems
Problem 2.7
Determine heat rate without the fins and with the fins in place (N = 5 fins)
qt = hAt𝜂𝑜 𝜃𝑏
o = 1 −
2025-02-12
NA f
At
(1 −  f )
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