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overall heat transfer coefficient calculation, coated pipe temperature distribution

A very long cylindrical coated pipe (infinite length) is maintained at temperature *T*_{inner} along its internal surface and *T*_{outer} along its external surface.

**Problem Type**

Plane and axisymmetric problems of heat transfer.

**Geometry**

Thin layer of 3 mm coating is placed between steel pipe and insulation.

**Given**
*T*_{inner} = 85°C, *T*_{outer} = 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.

**Task**

Derive heat flux at the internal diameter and calculate the overall heat transfer coefficient (OHTC) of the system.

**Solution**

The OHTC of 3-layered pipe can be calculated as (*)

λ_{total} = 2πL / [ 1/λ_{1}·ln(*r2*/*r1*) + 1/λ_{2}·ln(*r3*/*r2*) + 1/λ_{3}·ln(*r4*/*r3*)],

where L is the length of the tube.

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.

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.

**Results**

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 |

* Reference: H.Y. Wong Heat transfer for Engineers, Table 2.2 with the heat transfer coefficient set at infinity.

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