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	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.

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	303.611	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:

Related Tutorials

Tutorial: Thermal Analysis of a Differential Casing

Last updated: May 19th, 2021