Validation Case: Hollow Sphere, Release of Power

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]\(^1\).

Geometry

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.

Analysis Type and Mesh

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	245	Standard
B	2nd order hexahedral	515	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 generated in SimScale with extrusion refinements to obtain a hexahedral mesh.

Simulation Setup

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.

Reference Solution

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 \(m\):

$$ T(r = 1.25) = 30.625\ °C $$

$$ T(r = 1.5) = 32.500\ °C $$

$$ T(r = 1.75) = 28.482\ °C $$

Result Comparison

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	303.660	30.510	30.625	-0.38 %
	1.5	305.540	32.390	32.500	-0.34 %
	1.75	301.568	28.418	28.482	-0.23 %
B	1.25	303.774	30.624	30.625	-0.00 %
	1.5	305.649	32.499	32.500	-0.00 %
	1.75	301.632	28.482	28.482	0.00 %
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:

Related Tutorials

Tutorial: Thermal Analysis of a Differential Casing

Last updated: April 3rd, 2026