Multilayer coated pipe - QuickField simulation example
A very long cylindrical coated pipe (infinite length) is maintained at temperature Tinner along its internal surface and Touter along its external surface.
How to find overall heat transfer coefficient of multilayer coated pipes?
Answer Typical applications Geometry
Thin layer of 3 mm coating is placed between steel pipe and insulation.
Given
Task
Solution
For very long pipe the heat transfer in axial direction may be neglected, thus the temperature distribution in any cross section will be the same. In this case the plane-parallel can be solved. In plane-parallel problems all integral values are calculated per 1 meter of depth.
Results
OHTC, W/K-m
* Reference: H.Y. Wong Heat transfer for Engineers, Table 2.2 with the heat transfer coefficient set at infinity.
Engineering question
Set up a plane-parallel QuickField Steady-state Heat Transfer problem for a multilayer coated pipe and evaluate overall heat transfer coefficient from computed field results.
multilayer coated pipelines, insulated pipe sections, composite pipe insulation systems
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Simulation problem
Problem Type
Plane and axisymmetric problems of heat transfer.
Tinner = 85°C, Touter = 4°C;
Thermal conductivity of steel λ=40 W/K-m,
Thermal conductivity of coating λ=0.4 W/K-m,
Thermal conductivity of insulation λ=0.15 W/K-m.
Derive heat flux at the internal diameter and calculate the overall heat transfer coefficient (OHTC) of the system.
The OHTC of 3-layered pipe can be calculated as (*)
λtotal = 2πL / [ 1/λ1·ln(r2/r1) + 1/λ2·ln(r3/r2) + 1/λ3·ln(r4/r3)],
where L is the length of the tube.
The axial length of the axisymmetric model does not affect to results, and was set to some small value (0.05 m) to reduce the mesh size. To convert heat flux (and other integral values) calculated in axisymmetric problem to flux calculated in plane-parallel problem the former should be multiplied by 20.
Temperature distribution in multilayer coated pipe:
QuickField
Reference*
plane
axisymmetric
Heat Flux, W
178.2
8.89·20 = 177.8
-
Temperature difference, K
81
81
81
2.200
2.195
2.211
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