The thermal insulated pipe problem
In order to calculate heat flow passing through a pipe with or without insulation, it is necessary to
calculate the thermal resistance for each material layer. For cylinders this thermal resistance varies
not only with the insulation thickness and thermal conductivity, but also with the pipe outer
diameter and the average material temperature.
The thermal resistance of each (insulation) layer is calculated using the following equation:
ln (𝑟𝑒 ⁄𝑟𝑖 )
𝑅𝑙𝑎𝑦𝑒𝑟 =
[m.K/W]
2𝜋∙𝜆𝑘
Where:
re
= the outer radius of that layer [m]
ri
= the inner radius of that layer [m]
λk
= the thermal conductivity of that layer [W/m.K]
The total thermal resistance of a pipe, with or without insulation layer(s) and surface heat transfer
resistances to inner fluid and extern ‘air’, can be calculated with following equation:
𝑅𝑖
𝑅𝑙 𝑡𝑜𝑡 =
+ ∑𝑛−1
𝑘=1
2𝜋 𝑟1
Where:
ln (
𝑟(𝑘+1)
)
𝑟𝑘
2𝜋𝜆𝑘
+
𝑅𝑒
2𝜋 𝑟𝑛
[m.K/W]
Ri
= inner surface heat transfer resistance between fluid and material (1/hi) [m2.K/W]
Re
= outer surface heat transfer resistance between fluid and material (1/hi) [m2.K/W]
Suppose there is a metal pipe with thermal conductivity of 52 W/m.K and an insulated layer of 5 cm
thick with thermal conductivity of 0,03 W/m.K. The inner radius of the pipe is 1,60 cm and outdoor
radius is 2,13 cm. The inner surface heat transfer resistance coefficient to inner flowing liquid is
1/2000 m2.K/W and outer resistance coefficient to the air is 1/8 m2.K/W.
The thermal resistance over the pipe without insulation is:
𝑅𝑙 𝑡𝑜𝑡 =
1/2000
2𝜋∙0,016
+
0,0213
)
0,016
ln (
2𝜋∙52
1/8
+ 2𝜋∙0,0213 = 0,94 m.K/W
The total resistance with 5 cm thick insulation is:
𝑅𝑙 𝑡𝑜𝑡 =
1/2000
+
2𝜋∙0,016
0,0213
)
0,016
ln (
2𝜋∙52
0,0713
ln (
)
1/8
0,0213
+ 2𝜋∙0,03
+ 2𝜋∙0,0713 = 6,69 m.K/W
The heat flow passing through the pipe surface is calculated using the following equation:
𝜃 −𝜃𝑒
∙𝐿
[W]
𝛷𝑙 𝑡𝑜𝑡 =
60−20
= 42,6
[W/m]
𝛷𝑙 𝑡𝑜𝑡 =
60−20
= 6,0
𝑇 −𝑇
𝑟
𝛷𝑙 𝑡𝑜𝑡 = 𝑅𝑖
𝑙 𝑡𝑜𝑡
When the temperature of the inner liquid is 60⁰C and the extern air has a temperature of 20⁰C, the
heat flow through the pipe surface without insulation is:
0,94
When the pipe is insulated the heat flow passing through the pipe is:
6,77
[W/m]
When the temperatures of inner and outer surfaces are known, the temperature at any distance in
the material layer(s) can be calculated with the following equation:
𝑇𝑥 = 𝑇1 − 1 𝑟22 ∙ 𝑙𝑛 𝑟𝑥
ln
𝑟1
1
[K or ºC]
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Numerical simulation with Comsol Multiphysics
First we have calculated the heat loss through the pipe surface with or without insulation.
Now we simulate this heat flow problem by using the software Comsol Multiphysics v3.5.
In Fig. 1 the two-dimensional model of the pipe with (left) and without (right) insulation layer is
given, the colors in this figure indicate the thermal conductivity of the materials.
Fig. 1 The two-dimensional model of the pipe with and without diameter, indicated by their thermal conductivities
In this simulation model the same temperatures, conductivities and surface heat transfer
coefficients are used as in the calculations before. Under these stationary conditions the
temperature distribution though the pipe and insulation are shown in Fig. 2:
Fig. 2 Temperature distribution through the pipe an insulation material
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The heat flux through the material is given by the colors and arrows in Fig. 3:
Fig. 3 Heat flux to outside indicated by colors and arrow length
By multiplying the maximum heat flux, which accorded to the inner surface with the circumference
of the inner surface, we get the total lost of heat flux.
This is for the pipe without insulation:
𝛷𝑙 𝑡𝑜𝑡 = 423,6 ∙ (2 ∙ 0,016 ∙ 𝜋) = 42,6 [W/m]
The heat loss of the insulated is:
𝛷𝑙 𝑡𝑜𝑡 = 59,5 ∙ (2 ∙ 0,016 ∙ 𝜋) = 6,0 [W/m]
So the results out the numerical simulation are the same than the results we have obtained out of
the calculations.
Fig. 4 shows that the total thermal resistance is not proportional to the thickness of the insulation
layer. These graphs are based on the same material thermal conductivities and pipe dimensions as
before.
Fig. 4 Equivalent thermal resistance of an insulated pipe depending on the insulation thickness in meter.
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