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This validation case belongs to heat transfer, with the case of a hollow sphere under release of power condition. The aim of this test case is to validate the following parameters:

- Steady state heat transfer
- Volumetric heat source

The simulation results of SimScale were compared to the numerical results presented in [TPLV06]\(^1\).

The geometry used for the case is as follows:

It represents a section of a hollow sphere with an internal radius of 1 \(m\) and an external radius of 2 \(m\). Face ABCD is the internal face and EFGH is the external face. Axis X passes through the centroid of both faces, making the volume symmetric across the XY and XZ planes.

**Tool Type**: Code_Aster

**Analysis Type**: Heat transfer, linear, steady state.

**Mesh and Element Types**:

Case | Mesh Type | Number of Nodes | Element Type |
---|---|---|---|

A | 1st order hexahedral | 125 | Standard |

B | 2nd order hexahedral | 425 | Standard |

C | 1st order tetrahedral | 2094 | Standard |

D | 2nd order tetrahedral | 14939 | Standard |

The tetrahedral meshes were computed using SimScale’s standard mesh algorithm and automatic sizing. The hexahedral meshes were computed locally and uploaded into the release of power simulation project.

**Material**:

- Density \( \rho = \) 1 \( kg/m^3 \)
- Thermal conductivity \( \kappa = \) 1 \( W/(m.K) \)
- Specific heat \( C_p = \) 1 \( J/(kg.K) \)

**Boundary Conditions**:

- Constraints:
- Fixed temperature of 20 \(°C\) on faces ABCD and EFGH (internal and external respectively)
- Volume heat source of 100 \(W/m^3\) on the whole volume.

The reference solution is of the analytical type, as presented in [TPLV06]\(^1\), originally from [VPCS]\(^2\):

$$ T = T_i + \frac{Q}{6\kappa} \Bigg[ \frac{ (R_e^2 – R_i^2 ) \Big[ \frac{1}{R_i} – \frac{1}{r} \Big] }{ \Big[ \frac{1}{R_i} – \frac{1}{R_e} \Big] } – ( r^2 – R_i^2 ) \Bigg] $$

$$ T_i = 20\ °C $$

$$ Q = 100\ W/m^3 $$

$$ \kappa = 1\ W/(m.K) $$

$$ R_i = 1.0\ m $$

$$ R_e = 2.0\ m $$

The reference solution will be taken at points \( r = \) 1.25, 1.5 and 1.75 \(m\):

$$ T(r = 1.25) = 30.625\ °C $$

$$ T(r = 1.5) = 32.500\ °C $$

$$ T(r = 1.75) = 28.482\ °C $$

Comparison of temperatures at radii \( R = \) 1.25, 1.5 and 1.75 \(m\) with the reference solution is presented:

CASE | R \([m]\) | COMPUTED \([K]\) | COMPUTED \([°C]\) | REFERENCE \([°C]\) | ERROR |
---|---|---|---|---|---|

A | 1.25 | 303.611 | 30.461 | 30.625 | -0.54 % |

1.5 | 305.484 | 32.334 | 32.500 | -0.51 % | |

1.75 | 301.531 | 28.381 | 28.482 | -0.35 % | |

B | 1.25 | 303.79 | 30.64 | 30.625 | 0.05 % |

1.5 | 305.649 | 32.499 | 32.500 | -0.00 % | |

1.75 | 301.621 | 28.471 | 28.482 | -0.04 % | |

C | 1.25 | 303.699 | 30.549 | 30.625 | -0.25 % |

1.5 | 305.576 | 32.426 | 32.500 | -0.23 % | |

1.75 | 301.597 | 28.447 | 28.482 | -0.12 % | |

D | 1.25 | 303.776 | 30.626 | 30.625 | 0.00 % |

1.5 | 305.65 | 32.5 | 32.500 | 0.00 % | |

1.75 | 301.632 | 28.482 | 28.482 | 0.00 % |

Illustration of the temperature distribution from the release of power simulation, case D:

Tutorial: Thermal Analysis of a Differential Casing

References

- TPLV06 – Dégagement de puissance dans une sphère creuse – Code Aster validation case
- Guide de validation des progiciels de calcul de structures. Société Française des Mécaniciens, AFNOR 1990 ISBN 2-12-486611-7

Note

If you still encounter problems validating you simulation, then please post the issue on our forum or contact us.

Last updated: May 19th, 2021

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