Chapter Two
Introduction to Conduction
2.1 The Conduction Rate Equation
2.2 The Thermal Properties of Matter
2.3 The Heat Diffusion Equation
2.4 Boundary and Initial Conditions
Fourier’s Law
Fourier’s Law
• A rate equation that allows determination of the conduction heat flux
from knowledge of the temperature distribution in a medium
• 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  k   q/  T 




– Direction of heat transfer is perpendicular to lines of constant
temperature (isotherms).
– Heat flux vector may be resolved into orthogonal components.
Heat Flux Components
• Cartesian Coordinates:
T  x, y , z 

T 
T 
T 
q    k
i k
jk
k
x
y
z
q x
q z
q y
• Cylindrical Coordinates:
(2.3)
T  r, , z 

T 
T 
T 
q    k
i k
jk
k
r
r 
z
qr
q
(2.24)
q z
T  r ,  , 


T 
T 
T
q    k
i k
jk
k
r
r 
r sin  
q
q
q 
• Spherical Coordinates:
r
(2.27)
Heat Flux Components (cont.)
• In angular coordinates  or  ,   , the temperature gradient is still
based on temperature change over a length scale and hence has
units of C/m and not C/deg.
• Heat rate for one-dimensional, radial conduction in a cylinder or sphere:
– Cylinder
qr  Ar qr  2 rLqr
[W]
or,
qr  Ar qr  2 rqr
[W/m]
– Sphere
qr  Ar qr  4 r 2 qr
[W]
Heat Diffusion Equation
The Heat Diffusion 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)
Thermal energy
generation
(2.19)
Change in thermal
energy storage
Heat Equation (Radial Systems)
• Cylindrical Coordinates:
1   T  1   T    T 
T
kr

k

k

q


c
p




r r  r  r 2     z  z 
t
(2.26)
• Spherical Coordinates:
1   2 T 
1
  T 
1
 
T 
T
kr

k

k
sin


q


c
p




r  r 2 sin 2      r 2 sin   
 
t
r 2 r 
(2.29)
Heat Equation (Special Case)
• One-Dimensional Conduction in a Planar Medium with Constant Properties
and No Generation
  T 
T
k


c
p


x  x 
t
becomes
 2T
x


2
1 T
 t
k
 thermal diffusivity of the medium  m 2 /s 


cp
Boundary Conditions
Boundary and Initial Conditions
• For transient conduction, heat equation is first order in time, requiring
specification of an initial temperature distribution: T  x,t t=0 = T  x,0 
• Since heat equation is second order in space, two boundary conditions
must be specified for each coordinate direction. Some common cases:
Constant Surface Temperature:
T  0 ,t  = Ts
Constant Heat Flux:
Applied Flux
Insulated Surface
-k
T
|x=0= qs
x
Convection:
-k
T
|x=0= h T - T  0 ,t  
x
T
|x=0 = 0
x
Properties
Thermophysical Properties
Thermal Conductivity: A measure of a material’s ability to transfer thermal
energy by conduction.
Thermal Diffusivity: A measure of a material’s ability to respond to changes
in its thermal environment.
Property Tables:
Solids: Tables A.1 – A.3
Gases: Table A.4
Liquids: Tables A.5 – A.7
Thermal Conductivity, k
• The thermal conductivity of a material is a
measure of the ability of the material to conduct
heat.
• High value for thermal conductivity
good heat conductor
• Low value
poor heat conductor or insulator
Thermal Conductivity, k
• The thermal conductivity of a material is a
measure of the ability of the material to conduct
heat.
• High value for thermal conductivity
good heat conductor
• Low value
poor heat conductor or insulator
Properties
Thermal Conductivity of Materials
• At room temperature, for
different materials the
conductivity range spans over
many orders of magnitude
• The thermal conductivities of
gases such as air vary by a factor
of 104 from those of pure metals
such as copper.
• Pure crystals and metals have
the highest thermal
conductivities, and gases and
insulating materials the lowest.
Thermal Conductivities and Temperature
• The thermal
conductivities of
materials vary with
temperature.
• The temperature
dependence of thermal
conductivity causes
considerable complexity
in conduction analysis.
• A material is normally
assumed to be isotropic.
Properties
Thermal Conductivity of Solids
• Tables A.1 – A.3
• In the modern view of materials, a solid may be comprised
of free electrons and atoms bound in a periodic
arrangement called the lattice
• Heat conduction in solids effected by waves propagating in
the lattice of atoms and by free electrons (electric
conductors such as metals) carrying kinetic energy to the
lower temperature zones of the solid.
• Conductivity of pure metals > Conductivity of alloys
• Conductivity of highly ordered structured solids (crystals) >
Conductivity of lower ordered amorphous solids (glass,
rubber)
• For most solid materials conductivity reduces when
temperature is increased
Thermal Conductivity of Fluids
• Gases - Table A.4
• Liquids Tables A5– A.7
• Larger intermolecular spaces make heat transport in all
gases and nonmetallic liquids less effective compared to
solids
• Conductivity of gases increases with temperature because
molecules have higher kinetic energy at higher temperature
and that promotes heat transfer
• Conductivity of liquids may increase or decrease with
increased temperature.
• Conductivity of fluids is independent of their pressure
Thermal diffusivity
Heat conducted
𝑘
𝛼=
=
Heat stored
𝜌𝑐𝑝
(𝑚2 𝑠)
• The thermal diffusivity represents how fast heat
diffuses through a material
• Appears in the transient heat conduction analysis
• A material that has a high thermal conductivity or a low
heat capacity will have a large thermal diffusivity
• The larger the thermal diffusivity, the faster the
propagation of heat into the medium
• At room temperature:
𝛼𝑐𝑜𝑝𝑝𝑒𝑟 = 113 ∙ 10−6 𝑚2 /s 𝛼𝑔𝑙𝑎𝑠𝑠 = 0.34 ∙ 10−6 𝑚2 /s
Properties (Nanoscale Effects)
Nanoscale Effects
• Conduction may be viewed as a consequence of energy carrier (electron or
phonon) motion.
• For the solid state:
1
k  Cc mfp
3
energy carrier
specific heat per
unit volume.
(2.7)
mean free path → average distance
traveled by an energy carrier before
a collision.
average energy carrier velocity, c <  .
• Energy carriers also collide with
physical boundaries, affecting
their propagation.
 External boundaries of a film of material.
thick film (left) and thin film (right).
Simple example
The cost of heating loss through a roof
Solution
Conduction Analysis
Typical Methodology of a Conduction Analysis
• Consider possible microscale or nanoscale effects in problems involving
small physical dimensions or rapid changes in heat or cooling rates.
• 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:
Chapter 3:
Chapter 4:
Chapter 5:
One-Dimensional, Steady-State Conduction
Two-Dimensional, Steady-State Conduction
Transient Conduction
Example Pb 2.3 page 67
Pb 2.43 Thermal response of a plane wall to convection heat transfer.
KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating.
FIND: (a) Differential equation and initial and boundary conditions which may be used to find the
temperature distribution, T(x,t); (b) Sketch T(x,t) for the following conditions: initial (t  0), steadystate (t  ), and two intermediate times; (c) Sketch heat fluxes as a function of time at the two
surfaces; (d) Expression for total energy transferred to wall per unit volume (J/m3).
Problem: Thermal Response (Cont).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal
heat generation.
ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has the
form,
 2T
 x

2
and the
conditions are:
1 T
  t
 Initial, t  0 : T  x,0   T
i


 Boundaries: x=0  T/  x 0  0

x=L
 k T/  x L = h T  L,t   T 


uniform temperature
adiabatic surface
surface convection
Problem: Thermal Response (Cont).
(b) The temperature distributions are shown on the sketch.
Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also
that the gradient at x = L decreases with time.
Problem: Thermal Response (Cont).
c) The heat flux, qx  x,t  , as a function of time, is shown on the sketch for the
surfaces x = 0 and x = L.
Problem: Thermal Response (Cont).
d) The total energy transferred to the wall may be expressed as
E in  

0
q conv A s dt

T  T  L,t  dt

0
E in  hA s 
Dividing both sides by AsL, the energy transferred per unit volume is
Ein h 
  T  T  L,t  dt
V L 0
J/m 3 


Pb 2.24 1D-Heat Diffusion Equation - Temperature Distributions
2.24 Temperature distributions
within a series of one-dimensional
plane walls at an initial time (t=0),
at steady state (t∞), and at several
intermediate times are as shown.
For each case, write the appropriate
form of the heat diffusion equation.
Also write the equations for the
initial condition and the boundary
conditions that are applied at x = 0
and x = L.
NOTE: If volumetric generation
occurs, it is uniform throughout the
wall. The properties are constant.
Problem 2.29: Non-uniform Generation due to Radiation Absorption
Problem: Non-uniform Generation due to Radiation Absorption
Problem 2.29 Surface heat fluxes, heat generation and total rate of radiation
absorption in an irradiated semi-transparent material with a
prescribed temperature distribution.
KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux.
FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) The heat generation rate
q  x  , and (c) Expression for absorbed radiation per unit surface area.
SCHEMATIC:
Problem : Non-uniform Generation (cont.)
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3)
Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internal
volumetric heat generation term q  x  .
ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using
Fourier’s law,
 dT 
A

qx = -k   = -k  e -ax + B 
 dx 
 ka

Front Surface, x=0:
A

A

qx  0  = -k  + B  = -  + kB 
 ka

a

<
Rear Surface, x=L:
A

A

qx  L  = -k  e -aL + B  = -  e -aL + kB 
 ka

a

<
(b) The heat diffusion equation for the medium is
d  dT  q

+ = 0
dx  dx  k
q  x  = -k
or
q = -k
d  dT 


dx  dx 
d  A -ax

+
e
+
B
= Ae -ax .


dx  ka

( c ) Performing an energy balance on the medium,
Ein - Eout + E g = 0
<
Problem : Non-uniform Generation (cont.)
On a unit area basis
 + Eout
 = -qx  0  + qx  L  = +
E g = -Ein


A
1 - e -aL .
a
<
Alternatively, evaluate E g by integration over the volume of the medium,
L
L
0
0
E g =  q  x dx =  Ae -ax dx = -


A  -ax  L A
e
=
1 - e -aL .


0
a
a