Fourier’s Law and
the Heat Equation
Chapter Two
The rate of heat conduction in a specified direction is proportional to the
temperature gradient, which is the rate of change in temperature with distance in
that direction. One dimensional steady state heat conduction through
homogenous material is given by Fourier Law of heat conduction:
When the proportionality constant is inserted,
where qx is the heat-transfer rate and ∂T/∂x is the temperature gradient in the
direction of the heat flow. The positive constant k is called the thermal
conductivity of the material, and the minus sign is inserted so that the second
principle of thermodynamics will be satisfied.
Recognizing that the heat flux is a vector quantity, we can write a more general
statement of the conduction rate equation (Fourier’s law ) as follows:
To treat more than one-dimensional heat flow, we need consider only the heat
conducted in and out of a unit volume in all three coordinate directions, as shown in
Figure below. The energy balance yields as follows.
The heat rate in x direction is,
and the heat flux is,
Or can be written as,
where the quantity α is called the thermal diffusivity of the material. where
is thermal conductivity (W/(m-K)
is density (kg/m³)
is specific heat capacity (J/(kg-K)
Significance of Thermal Diffusivity
The larger the value of α, the faster heat will diffuse through the material.
Thermal diffusivity of a material is the ratio of its thermal conductivity to the
thermal storage capacity . The storage capacity essentially represents thermal
capacitance or thermal inertia of the material.
It signifies the rate at which heat diffuses in to the medium during change in
temperature with time. Thus, the higher value of the thermal diffusivity gives the
idea of how fast the heat is conducting into the medium, whereas the low value
of the thermal diffusivity shown that the heat is mostly absorbed by the material
and comparatively less amount is transferred for the conduction.
In short it measures the ability of a material to conduct thermal energy relative to
its ability to store thermal energy.
Here read
T=t
Here read
T=t
Question
Develop the general heat conduction equation for a solid cylinder (with cylindrical
coordinates) assuming unsteady state condition with internal heat generation.
Solution:
 st   in   out   gen ... (A)
 gen  q  Volume
 gen  q   dr  dz  rd  
and
dT
 st    dr  rd  dz  Cp
dt
Putting values in energy balance equation (A),
dT
qr  q  qz  (qr  dr  q  d   qz  dz )  q  dr  rd  dz    Cp
(dr  rd   dz )
dt
dT
qr  q  qz  qr  dr  q  d   qz  dz  q  dr  rd   dz    Cp
(dr  rd   dz ) ...(B)
dt
qr
q
qz
qr  dr  qr 
dr , Similarly q  d   q 
d & qz  dz  qz 
dz
r

z
Putting these value in ( B ) and simplifying we get ,
qr
q
qz
dT
 dr 
d 
dz  q  dr  rd  dz    Cp
(dr  rd  dz )
r

z
dt
T
T
T
T
Since qr   kA
 k (rd  dz )
Similarly q  k ( dr  dz )
& qz  k ( dr  rd )
r
r
r
z

T

T

T
dT
(kr )dr  d  dz  (k
)dr  d  dz  (kr )dr  d  dz  q  dr  rd  dz    Cp (dr  rd  dz )
r
r
 r
z
z
dt
Divding by ( dr  rd  dz ) throughout the equation,
1 
T
1  T
 T
dT
(kr )  2 (k )  (k )  q   Cp
r r
r r   z z
dt
Exercise Problems
Problem 1: Assume steady-state, one-dimensional heat conduction through the
symmetric shape shown. Assuming that there is no internal heat generation,
derive an expression for the thermal conductivity k(x) for these conditions:
where A is in square meters, T in Kelvin's, and x in meters.
Solution:
Problem 1 (Continued):
Exercise Problems
Problem 2: Consider steady-state conditions for one-dimensional conduction in a
plane wall having a thermal conductivity k = 50 W/m-K and a thickness L = 0.25 m,
with no internal heat generation.
Determine the heat flux and the unknown quantity for each case and sketch the
temperature distribution, indicating the direction of the heat flux.
Solution:
Problem 2 (Continued):
Exercise Problems
Problem 3: Consider a plane wall 100 mm thick and of thermal conductivity 100
W/m-K. Steady-state conditions are known to exist with T1= 400 K and T2= 600 K.
..
Determine the heat flux qx and the temperature gradient dT/dx for the coordinate
systems shown.
Solution:
Problem 3 (Continued):
Exercise Problems
Problem 4: A cylinder of radius ro, length L, and thermal conductivity k is
immersed in a fluid of convection coefficient h and unknown temperature T∞. At a
certain instant the temperature distribution in the cylinder is ,
where a and b are constants. Obtain expressions for the heat transfer rate at ro
and the fluid temperature.
Solution:
..
Problem 4 (Continued):
Exercise Problems
Problem 5:
Solution:
Problem 5 (Continued):
Additional Practice problems
P 1: The inner and outer surface temperatures of a glass window 5 mm thick are 15 and 5C.
What is the heat loss through a window that is 1 m by 3 m on a side? The thermal conductivity
of glass is 1.4 W/m -K.
P 2: A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the
bottom to be perfectly insulated. What is the minimum thickness of Styrofoam insulation
(k= 0.030 W/m-K) that must be applied to the top and side walls to ensure a heat load of less
than 500 W, when the inner and outer surfaces are -10 and 35 0C?
P32: An inexpensive food and beverage container is fabricated from 25-mm-thick polystyrene
(k=0.023 W/m-K) and has interior dimensions of 0.8 m x 0.6 m x 0.6 m. Under conditions for
which an inner surface temperature of approximately 2 0C is maintained by an ice-water mixture
and an outer surface temperature of 20 0C is maintained by the ambient, what is the heat flux
through the container wall? Assuming negligible heat gain through the 0.8 m x 0.6 m base of the
cooler, what is the total heat load for the prescribed conditions?