Unit 42: Heat Transfer and
Combustion
Lesson 2: Heat Transfer by
Conduction through Solid
Composite Walls
Aim
NDGTA
• LO1: Understanding Heat Transfer Rates for
Composite Systems.
Composite Walls
NDGTA
• Until now we have considered only plane walls,
which consist of one homogeneous material.
• There are, however, many cases in practice when
different materials are constructed in layers to
form composite a wall.
• For example, furnace walls have an inner lining of
refractory brick, followed by insulating material
and another layer of fire brick. There may also be
an external layer of plaster of similar finishing to
complete the composite wall.
Composite Walls
NDGTA
• The flow of heat through a composite wall and
the resistance to this heat flow by each of the
materials is similar to the resistance to current
flow set up in an electric circuit.
• Consider the following diagram…
Composite Walls
NDGTA
T0
T1
Q
T2
k1
k2
k3
x1
x2
x3
Q
T3
Q
R1
R2
R3
Composite Walls
NDGTA
• Note there are three different layers of
materials of thicknesses (x1, x2, x3) with
corresponding thermal conductivities of
(k1,k2,k3).
• The internal wall temperature is T0 and the
material interface temperatures are T1, T2 and
T3 respectively.
Composite Walls
NDGTA
• We can think of heat flow as being analogous
to the flow of an electric current.
• We know that the heat flow is caused by
temperature difference while current flow is
caused by potential (voltage) difference.
• It is thus possible to think of thermal
resistance in a similar manner to the way we
think of electrical resistance.
Thermal Resistance
NDGTA
• Thus from Ohm’s
Law,
I
=
V/R
.
• We know that Q = kA(T1 – T2)/dx
.
• So let the heat flow Q be equivalent to the
current flow I, and let the temperature
difference (T1 – T2) be equivalent to the
potential difference V (volts), then by
substitution I = kAV/x. Thus..
1/R = kA/x
Thermal Resistance
NDGTA
• So thermal resistance is given by…
R = x/kA
Kelvin/Watt (K/W)
• With regard to the previous diagram the total
thermal resistance is given by…
RT = R1 + R2 + R3
= x1/k1A + x2/k2A + x3/k3A
• Note as A is constant it is more common to
calculate the total thermal resistance for unit
surface area.
Thermal Resistance
NDGTA
• Thus RT = x1/k1 + x2/k2 + x3/k3
• Again with consideration to IT = V/RT,
we can consider the total (0verall) heat transfer
through the wall…
Q.T = (T0 – T3)/RT
• Thus a general equation for the heat flow
through the total width of a composite wall of
from
. surface A to B is given by…
QT = (TA – TB)/RT
Thermal Resistance
NDGTA
• The composite wall of a building consists of an
inner liner 10mm thick (k=0.04 W/m.K), bricks
25cm thick (k=0.5 W/m.K) and an outer
protective concrete-based coating 30 mm
thick (k=0.8 W/m.K). If the inner surface
temperature of the wall is 25oC and the outer
surface is 10oC, find the heat flow through the
composite wall.