ME 259
Heat Transfer
Lecture Slides II
Dr. Gregory A. Kallio
Dept. of Mechanical Engineering,
Mechatronic Engineering &
Manufacturing Technology
California State University, Chico
1/22/05
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Steady-State Conduction
Heat Transfer
 Incropera &
DeWitt coverage:
– Chapter 2: General Concepts of
Heat Conduction
– Chapter 3: One-Dimensional,
Steady-State Conduction
– Chapter 4: Two-Dimensional,
Steady-State Conduction
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General Concepts of Heat
Conduction
Reading: Incropera & DeWitt
Chapter 2
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Generalized Heat Conduction

Fourier’s law, 1-D form:
  k dT
qx
dx

Fourier’s law, general form:

q    kT
- q” is the heat flux vector, which has three
components; in Cartesian coordinates:

q  qxiˆ  qy ˆj  qzkˆ

2
2
2








q  q x  q y  qz
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(magnitude)
4
The Temperature Gradient

T is the temperature gradient, which is:
– a vector quantity that points in direction of
maximum temperature increase
– always perpendicular to constant
temperature surfaces, or isotherms
T ˆ T ˆ T ˆ
T 
i
j
k
x
y
z
(Cartesian)
T ˆ 1 T ˆ T ˆ
T 
i
j
k (Cylindrical)
r
r 
z
T ˆ 1 T ˆ
1 T ˆ
T 
i
j
k (Spherical)
r
r 
r sin  
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Thermal Conductivity

k is the thermal conductivity of the material
undergoing conduction, which is a tensor
quantity in the most general case:

k  k ( x, y , z, t , T )
– most materials are homogeneous, isotropic,
and their structure is time-independent;
hence:
k  k (T ),
which is a scalar and usually assumed to
be a constant if evaluated at the average
temperature of the material
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Total Heat Rate

Total heat rate (q) is found by integrating
the heat flux over the appropriate area:


q   q  dA
A

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k and  T must be known in order to
calculate q” from Fourier’s law
– k is usually obtained from material property
tables
– to find T, another equation is required;
this additional equation is derived by
applying the conservation of energy
principle to a differential control volume
undergoing conduction heat transfer; this
yields the general Heat Diffusion
(Conduction) Equation
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Heat Diffusion (Conduction)
Equation

For a homogeneous, isotropic solid material
undergoing heat conduction:
  T    T    T 
T
 k    k    k   q  c
x  x  y  y  z  z 
t


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Cylindrical and spherical coordinate system
forms given in text (p. 64-65)
This is a second-order, partial differential
equation (PDE); its solution yields the
temperature field, T(x,y,z,t), within a given
solid material
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Heat Diffusion (Conduction)
Equation

For constant thermal conductivity (k):
 2T  2T  2T q 1 T
 2  2  
,
2
x
y
z
k  t
where :


k
c
(thermal diffusivit y)
For k = constant, steady-state conditions,
and no internal heat generation ( q  0) :
 2T  2T  2T
 2  2  0,
2
x
y
z
or
 2T  0
– this is known as Laplace’s equation, which
appears in other branches of engineering
science (e.g., fluids, electrostatics, and
solid mechanics)
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Boundary Conditions and Initial
Condition




Boundary Conditions: known conditions at
solution domain boundaries
Initial Condition: known condition at t = 0
Number of boundary conditions required to
solve the heat diffusion equation is equal to the
number of spatial dimensions multiplied by two
There is only one initial condition, which takes
the form
T ( x, y, z,0)  Ti
– where Ti may be a constant or a function of
x,y, and z
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Types of Boundary Conditions for
Conduction Problems

Specified surface temperature, e.g.,
T (0, y, z, t )  T0

Specified surface heat flux, e.g.,
k

x 0
 hT  T (0, y, z, t )
x 0
Specified radiation (, Tsur given), e.g.,
k
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 q0
Specified convection (h, T given), e.g.,
T
k
x

T
x
T
x

4
  Tsur
 T 4 (0, y, z, t )
x 0
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
11
Solving the Heat Diffusion
Equation







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Choose a coordinate system that best fits
the problem geometry.
Identify the independent variables (x,y,z,t),
e,g, is it a S-S problem? Is conduction 1-D,
2-D, or 3-D? Justify assumptions.
Determine if k can be treated as constant
and if q  0.
Write the general heat conduction equation
using the chosen coordinates.
Reduce equation to simplest form based
upon assumptions.
Write boundary conditions and initial
condition (if applicable).
Obtain a general solution for T(x,y,z,t) by
some method; if impossible, resort to
numerical methods.
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Solving the Heat Diffusion
Equation, cont.




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Solve for the constants in the general
solution by applying the boundary
conditions and initial condition to obtain a
particular solution.
Check solution for correctness (e.g., at
boundaries or limits such as x = 0, t = 0,
t   , etc.)
Calculate heat flux or total heat rate using
Fourier’s law, if required.
Optional: rearrange solution into a
nondimensional form
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Example:


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GIVEN: Rectangular copper bar of dimensions
L x W x H is insulated on the bottom and
initially at Ti throughout . Suddenly, the ends
are subjected and maintained at temperatures
T1 and T2 , respectively, and the other three
sides are exposed to forced convection with
known h, T.
FIND: Governing heat equation, BCs, and
initial condition
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One-Dimensional, SteadyState Heat Conduction
Reading: Incropera & DeWitt,
Chapter 3
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1-D, S-S Conduction in Simple
Geometries w/o Heat Generation

Plane Wall
L
x
– if k = constant, general heat diffusion
equation reduces to
d 2T
0
2
dx
or
d  dT 

0
dx  dx 
– separating variables and integrating yields
dT
 C1
dx
and then
T ( x)  C1 x  C2
– where T(x) is the general solution; C1 and
C2 are integration constants that are
determined from boundary conditions
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1-D, S-S Conduction in Simple
Geometries w/o Heat Generation

Plane Wall, cont.
– suppose the boundary conditions are
T ( x  0)  Ts1
and
T ( x  L)  Ts 2
– integration constants are then found to be
Ts 2  Ts1
C1 
L
and
C2  Ts1
– the particular solution for the temperature
distribution in the plane wall is now
x
T ( x)  (Ts 2  Ts1 )  Ts1
L
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1-D, S-S Conduction in Simple
Geometries w/o Heat Generation

Plane wall, cont.
– The conduction heat rate is found from
Fourier’s law:
q  kA
dT
kA
Ts1  Ts 2 
 kAC1 
dx
L
– If k were not constant, e.g., k = k(T), the
analysis would yield
 k (T )dT  qx  C
» note that the temperature distribution would be
nonlinear, in general
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1-D, S-S Conduction in Simple
Geometries w/o Heat Generation
Electric Circuit Analogy

– heat rate in plane wall can be written as
(Ts1  Ts 2 ) temperature difference
q

L / kA
material constant
– in electrical circuits we have Ohm’s law:
– analogy:
V
i
R
q (heat rate)  i (current)
T (temperature)  V (voltage)
L
(thermal resistance)  R (electric resistance)
kA
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Thermal Circuits for Plane Walls

Series Systems

Parallel Systems
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Thermal Circuits for Plane Walls,
cont.

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Complex Systems
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Thermal Resistances for Other
Geometries Due to Conduction

Cylindrical Wall
r2
r1
l
ln( r2 / r1 )
Rt 
2k

Spherical Wall
r2
r1
1 / r1  1 / r2
Rt 
4k
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Convective & Radiative Thermal
Resistance
Convection

q  hA(Ts  T ) 
1
 Rt ,conv
hA
Ts  T
1 / hA
(convective thermal resistance)
Radiation

q  hr A(Ts  T ) 
Ts  T
1 / hr A
where hr   (Ts  T )(Ts2  T2 )
1
 Rt ,rad
hr A
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(radiative thermal resistance)
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Critical Radius Concept

Since the surface areas of cylinders and
spheres increase with r, there exist competing
heat transfer effects with the addition of
insulation under convective boundary
conditions (see Example 3.4)

A critical radius (rcr) exists for radial systems,
where:
– adding insulation up to this radius will
increase heat transfer
– adding insulation beyond this radius will
decrease heat transfer

For cylindrical systems, rcr = kins/h

For spherical systems, rcr = 2kins/h
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Thermal Contact Resistance

Thermal contact resistance exists at solid-solid
interfaces due to surface roughness, creating
gaps of air or other material:
A
B
Rt ,c 
q
TA  TB Rt,c

qAc
Ac
where Ac  apparent contact area
Rt,c  thermal resistance per unit area (m 2 K/W)
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Thermal Contact Resistance

R”t,c is usually experimentally measured and
depends upon
–
–
–
–
–

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thermal conductivity of solids A and B
surface finish & cleanliness
contact pressure
gap material
temperature at contact plane
See Tables 3.1, 3.2 for typical values
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EXAMPLE


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Given: two, 1cm thick plates of milled, coldrolled steel, 3.18m roughness, clean, in air
under 1 MPa contact pressure
Find: Thermal circuit and compare thermal
resistances
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1-D, S-S Conduction in Simple
Geometries with Heat Generation

Thermal energy can be generated within a
material due to conversion from some other
energy form:
– Electrical
– Nuclear
– Chemical

Governing heat diffusion equation if k =
constant:
 2T  q / k  0
d 2T
where  T  2
dx
for Cartesian systems
2
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S-S Heat Transfer from Extended
Surfaces (i.e., fins)

Consider plane wall exposed to convection
where Ts>T:

How could you enhance q ?
– increase h
– decrease T
– increase As (attach fins)
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Fin Nomenclature











x = longitudinal direction of fin
L = fin length (base to tip)
Lc = fin length corrected for tip area
W = fin width (parallel to base)
t = fin thickness at base
Af = fin surface area exposed to fluid
Ac = fin cross-sectional area, normal to heat flow
Ap = fin (side) profile area
P = fin perimeter that encompasses Ac
D = pin fin diameter
Tb = temperature at base of fin
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1-D Conduction Model for Thin Fins

If L >> t and k/L >> h, then the temperature
gradient in the longitudinal direction (x) is much
greater than that in the transverse direction (y);
therefore

q  qxiˆ

1/22/05
(1 - D conduction)
Another way of viewing fin heat transfer is to
imagine 1-D conduction with a negative heat
generation rate along its length due to
convection
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Fin Performance

Fin Effectiveness
f 


HT from single fin
HT from base area w/o fin
qf
hAc ,b (Tb  T )
Fin Efficiency
HT from single fin
f 
HT if entire fin were at Tb

qf
qmax

qf
hAf (Tb  T )
– for a straight fin of uniform cross-section:
f 
tanh( mLc )
mLc
– where Lc = L + t / 2 (corrected fin length)
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Calculating Single Fin Heat Rate
from Fin Efficiency


Calculate corrected fin length, Lc
Calculate profile area, Ap
Ap ,rec  Lc t ,

Ap ,tri  12 Lt ,
Ap , par  13 Lt ,
Evaluate parameter

L3c / 2 h / kAp  mLc / 2 for rectangular fins



Determine fin efficiency f from Figure 3.18,
3.19, or Table 3.5
Calculate maximum heat transfer rate from fin:
q f ,max  hA f (Tb  T )

Calculate actual heat rate:
q f   f q f ,max
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Maximum Heat Rate for Fins of
Given Volume
Analysis:

Set

dq f
dL
0
with Ap  constant
“Optimal” design results:
L3c/ 2 h / kAp  1.0035 for rectangular profile
 1.3094 for triangular profile


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1.7536 for concave parabolic profile
3
r2 / r1  2
for annular, rectangular profile
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Fin Thermal Resistance

Fin heat rate:
q f   f q f ,max   f hA f (Tb  T )


Define fin thermal resistance:
Rt , f 

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Tb  T
1 /  f hAf
1
 f hAf
Single fin thermal circuit:
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Analysis of Fin Arrays

Total heat transfer =
heat transfer from N fins +
heat transfer from exposed base
qt  Nq f  qb  N f hAf  b  hAb b
 h b N f Af  Ab 
Thermal circuit:
– where
Rt ,c 
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Rt",c
NAc ,b
, Rt , f
1
1

, Rb,conv 
N f hAf
hAb
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Analysis of Fin Arrays, cont.

Overall thermal resistance:
Rt ,o ( c ) 
1
o ( c ) hAt
where o ( c )  1 
NAf   f 
1  
At  C1 
C1  1   f hA f Rt,c / Ac ,b
At  NAf  Ab (total surface area of array)
then
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qt 
Tb  T
Rt ,o ( c )
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Example

Given: Annular array of 10 aluminum fins,
spaced 4mm apart C-C, with inner and outer
radii of 1.35 and 2.6 cm, and thickness of 1
mm. Temperature difference between base and
ambient air is 180°C with a convection
coefficient of 125 W/m2-K. Contact resistance
of 2.75x10-4 m2-K/W exists at base.

Find: a) Total heat rate w/o and with fins
b) Effect of R”t,c on heat rate
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Two-Dimensional, SteadyState Heat Conduction
Reading: Incropera & DeWitt
Chapter 4
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Governing Equation

Heat Diffusion Equation reduces to:
 2T  0
(Laplace' s equation)
or
 2T  2T
 2 0
2
x
y

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(2 - D, cartesian)
Solving the HDE for 2-D, S-S heat conduction
by exact analysis is impossible for all but the
most simple geometries with simple boundary
conditions.
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Solution Methods



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Analytical Methods
– Separation of variables (see section 4.2)
– Laplace transform
– Similarity technique
– Conformal mapping
Graphical Methods
– Plot isotherms & heat flux lines
Numerical Methods
– Finite-difference method (FDM)
– Finite-element method (FEM)
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Conduction Shape Factor

The heat rate in some 2-D geometries that
contain two isothermal boundaries (T1, T2) with
k = constant can be expressed as
q  Sk (T1  T2 )
– where S = conduction shape factor
(see Table 4.1)

Define 2-D thermal resistance:
Rt ,cond ( 2 D )
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1

Sk
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Conduction Shape Factor, cont.

Practical applications:
– Heat loss from underground spherical
tanks: Case 1
– Heat loss from underground pipes and
cables: Case 2, Case 4
– Heat loss from an edge or corner of an
object: Case 8, Case 9
– Heat loss from electronic components
mounted on a thick substrate: Case 10
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