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Unit 42: Heat Transfer and
Combustion
Lesson 5: Lagging
Aim
NDGTA
• LO1: Understanding Heat Transfer Rates for
Composite Systems.
Critical Thickness
of Insulation
NDGTA
• Let’s consider a layer of insulation which might be installed around
a circular pipe.
• If the pipe is metal then as has been shown the temperature on the
inner surface of the pipe varies little to that of the outer surface
and therefore for practical purposes it can be ignored
ri
h, T∞
ro
T∞
Ti
To
Ln(ro/ri)
2πkL
1
2πkL
Critical Thickness
of Insulation
NDGTA
• The inner temperature of the insulation is fixed at Ti
and the outer surface is exposed to a convection
environment at T∞.
• From the equivalent resistor network the heat
transfer is…
q = 2πL(Ti – T∞)
Ln(ro/ri) + 1
k
roh
Critical Thickness
of Insulation
NDGTA
• In interesting consideration is the amount of
insulation (lagging) required (i.e. radius ro) that will
maximise the heat transfer…
i.e.
dq = 2πL(Ti – T∞)[(1/kro) – (1/hro2)] = 0
dro
Ln(ro/ri) + 1 2
k
roh
Thus at a maximum ro = k/h ro being referred to as
the critical radius
Critical Thickness
of Insulation
q
NDGTA
This shows that lagging
(insulation) when placed
onto the outer surface of a
pipe will increase the heat
flow if it is less that the
critical radius ro. Greater
that ro, the heat flow will
decrease.
ro
r
Critical Thickness
of Insulation
NDGTA
• Thus if the outer radius is less than the value
given by ro = k/h, then the heat transfer will be
increased by adding further insulation.
• For radii greater than the critical value an
increase insulation thickness will cause a
decrease in heat transfer.
• The central concept is that for sufficiently small of
h the convection heat loss may actually increase
with the addition of insulation because of the
increased surface area.
Critical Thickness
of Insulation
NDGTA
• Calculate the critical radius of insulation for
asbestos (k=0.17 W/m.oC) surrounding a pipe
and exposed to room air at 20oC with h = 3.0
W/m2.oC. Calculate the heat loss from a
200oC, 5.0 cm diameter pipe when covered
with the critical radius of insulation and
without insulation.
Critical Thickness
of Insulation
NDGTA
ro = k/h = 0.17/3.0 = 0.0567 m = 5.67 cm
q=
2π(200 – 20)
= 105.7 W/m
L
Ln(5.66/2.5) +
1
0.17
(0.0567)(3.0)
Without insulation the convection from the outer surface of
the pipe is…
q = h(2πr)(Ti – To) = 3 x 2π x 0.025 x (200 -20) = 84.8 W/m
L
Critical Thickness
of Insulation
NDGTA
• So the addition of 3.17 cm (5.67 – 2.5) of
insulation actually increases the heat transfer
by 25%.
• As an alternative, fibreglass having a thermal
conductivity of 0.04 W/m.oC might be
employed as the insulation material. Then the
critical radius would be…
ro = k/h = 0.04/3.0 = 0.0133 = 1.33 cm
Critical Thickness
of Insulation
NDGTA
• Now the critical radius is less than the outside
radius of the pipe (2.5 cm) so addition of any
fibreglass insulation would cause a decrease in
the heat transfer.
• In a practical pipe insulation problem, the
total heat loss will also be influenced by
radiation as well as convection from the outer
surface of the insulation.