Thermal Insulation
“R-Value”
as used in Commercial Practice
Heat
conduction in a composite wall
“R-value” as used in classifying performance of insulation in commercial
practice is
The
units of R-value are m2˚C /W.
Note: “R-value” differs from the
thermal resistance. In R-value we use heat flow per unit area.
Furthermore,
“R-value” is always expressed for a certain
thickness of the insulating layer. For example: 3.8 cm thickness of extruded
polystyrene has an “R-value” of 1.3.
Examples
of “R-values”
|
Insulation |
Thickness |
“R-Value” (m2 ˚C/W) |
|
Extruded Polystyrene |
3.8 cm |
1.3 |
|
Expanded Polyurethane |
1.3 cm |
0.6 |
|
Expanded Polyurethane |
3.2 cm |
1.6 |
|
Fiberglass batt |
8.9 cm |
2.1 |
Mean
Critical Thickness of Insulation
Heat loss from a
pipe:
• If A, is increased, q will
increase. When insulation is added to a
pipe, the outside surface area of the pipe will increase. This would indicate an increased rate of
heat transfer
• The insulation material
has a low thermal conductivity,
· it reduces the conductive
heat transfer
· lowers the temperature
difference between the outer surface temperature of the insulation and the
surrounding bulk fluid temperature.
• This contradiction
indicates that there must be a critical thickness of insulation.
· The thickness of insulation
must be greater than the critical thickness, so that the rate of heat loss is
reduced as desired.
· As the outside radius, ro, increases, then in the
denominator, the first term increases but the second term decreases.
· Thus, there must be a
critical radius, rc , that
will allow maximum rate of heat transfer, q.
· The critical radius, rc, can be obtained by
differentiating and setting the resulting equation equal to zero.
Ti,Tb, k, L, ro, ri are
constant terms, therefore: 
When outside radius becomes
equal to critical radius, or ro
= rc, we get,
When no insulation is
provided then for a metal pipe with an outside radius of r2,
The rate of heat transfer
from an insulated pipe, where the annular insulating shell has an inside radius
of r2 and an outer radius of r3,
Then,
Radius A B (k/h)A >(k/h)B q_insulated/q_bare
