Unit 42: Heat Transfer and
Combustion
Lesson 1: Fourier’s Law
Aim
NDGTA
• LO1: Understanding Heat Transfer Rates for
Composite Systems.
Conduction
NDGTA
• Thermal conduction in solids and liquids seems to
involve two processes. The first is concerned
with the movement of atoms and molecules.
The second with free electrons.
• Atoms at high temperature vibrate more
vigorously about their equilibrium positions in
the lattice than their cooler neighbours. Since
atoms and molecules are bonded to one another,
they pass on some of their vibration energy.
Conduction
NDGTA
• This energy transfer occurs from atoms of
high vibrational energy to those of lower
vibrational energy, without appreciable
displacement.
• This energy has a knock-on effect, since high
vibrational energy atoms increase the energy
in adjacent low vibrational atoms, which in
turn causes them to vibrate more
energetically resulting in thermal conduction.
Conduction
NDGTA
• The second process involves material with a
ready supply of free electrons. Since electrons
are considerably lighter than atoms, any gain
in energy by electrons results in an increase in
the electrons’ velocity and they are able to
pass this energy on quickly to cooler parts of
the material.
Conduction
NDGTA
• Thus heat transfer by conduction involves two
processes, one involving the energy transfer
from high-energy vibrating atoms or
molecules to those of lower energy and the
second by the rapid transfer of energy
between high-energy free electrons and their
lower energy neighbours
Conduction
NDGTA
• This is one of the reasons why electrical
conductors which have many free electrons
are also good thermal conductors
• The way in which heat may be transferred by
conduction may be modelled using Fourier’s
Law.
Fourier’s Law
NDGTA
• Whenever a temperature gradient exists in a
solid material heat will flow from high
temperature to the low temperature region.
• The rate at which heat is transferred by
conduction, dQ/dt is proportional to the
temperature gradient dT/dx and the area A
normal to the direction of heat transfer i.e.
dQ/dt ∞ AdT/dx
Fourier’s Law
NDGTA
• The constant of proportionality is called the thermal
conductivity (k) of the material subject to the heat –
Handout Table 17.1 p302 Eng Sci.
• Thus k is an inherent property of the material to
conduct heat
Y
Steady flow represented
by straight line
.
T1
Area ‘A’ normal
to heat flow
Q = dQ/dt
T2
Positive heat flow
direction
X
X
Fourier’s Law
.
NDGTA
• Thus Q = - kAdT/dx
• Note the (-) sign which results from the second
law of Thermodynamics which essentially notes
that heat flow from a higher temperature to a
lower temperature in order for this flow to be
positive (by convention) a minus sign is added
here.
• This relationship was first determined by a French
Mathematician Fourier and as it has never been
disproven its referred to as Fourier’s Law
Fourier’s Law
NDGTA
.
• Provided that the whole heat flow Q (J.s-1) is
steady and perpendicular to the surface
through which it travels and the surface
temperatures remain constant, then the
temperature gradient dT/dx may be expressed
in terms of the change in temperature
(T2 – T1) and the distance x across which the
temperature
change
takes
place,
then
.
Q = - kA(T2 – T1)/x
or = kA(T1 – T2)/x
Fourier’s Law
NDGTA
• Another useful variant is to express
conduction in terms of the rate of. heat
transfer per unit area (q) i.e. q = Q/A
• The quantity q has the units of W.m-2. and is
often referred to as the heat flux. Thus
q = k(T1 – T2)/x
Fourier’s Law
NDGTA
• The outer surface temperature of a pane of
glass is 15oC and its inner surface temperature
is 25oC. Calculate the rate of heat loss
through the pane of glass given that its
thermal conductivity is 0.8 W/m.K and it has
dimensions of 1.0 m high, 0.5m wide and
0.75cm thick.
Fourier’s Law
NDGTA
• Assume steady state conditions prevail
• Only consider conduction in the x-direction
through the glass
• The glass is homogeneous (i.e. constant
properties throughout)
.
Q = k(T1 – T2)/x = (0.8)(0.5)(25 – 15)/0.0075
= 533.33 W
• Note the heat flux q = 1066.67 W/m2.