1. Reference problem#
1.1. Tank geometry#
The geometry studied is that of the core zone of a generic tank of the 900 MWe level, of which an azimuthal portion of 45° is shown schematically in FIG. 1.1-a.
Warning: odt2sphinx failed to: schema Frame2
Figure 1.1-a: View of a 45° section of the core zone of a generic 900 Mwe level tank
1.2. Characteristics of the defect under consideration#
In method \(K\beta\), the defect is not modeled in the mesh. The mesh makes it possible to calculate the stresses at the nodes. A first post-treatment is first applied to calculate the strength factor of the elastic stresses at the points of the defect from the stresses at the nodes (the method is detailed in [R7.02.10]). A second post-treatment is then applied to calculate the intensity factor of the elasto-plastic stresses using the so-called « correction b » method (see documentation R7.02.10]).
For this test, the under-coating defect in question is elliptical in shape, has a longitudinal orientation and is offset in the base metal. Its dimensions are as follows (see figure below):
Depth: \({\mathit{prof}}_{\text{def}}\mathrm{=}\mathrm{6mm}\)
Width: \(\mathrm{2b}\mathrm{=}\mathrm{60mm}\)
Several offsets in the base metal are considered:
\(\text{-}\mathit{decalage}\mathrm{=}0.0\mathit{mm}\)
\(\text{-}\mathit{decalage}\mathrm{=}2.5\mathit{mm}\)
\(\text{-}\mathit{decalage}\mathrm{=}5.0\mathit{mm}\)
\(\text{-}\mathit{decalage}\mathrm{=}7.5\mathit{mm}\)
\(\text{-}\mathit{decalage}\mathrm{=}10.0\mathit{mm}\)
Figure 1.1-b: Defect position
1.3. Material properties#
For thermal calculation:
Two properties are specified, they are:
LAMBDA: isotropic thermal conductivity as a function of temperature, expressed in \({\mathit{W.m}}^{\mathrm{-}1}\mathrm{.}{K}^{\mathrm{-}1}\),
BETA: volume enthalpy as a function of temperature, expressed in \({\mathit{J.m}}^{\mathrm{-}3}\).
For the coating:
Temperature ( \(°C\) ) |
LAMBDA |
0 |
14.7 |
20 |
14.7 |
50 |
15.2 |
100 |
15.8 |
150 |
16.7 |
200 |
17.2 |
250 |
18 |
300 |
18.6 |
350 |
19.3 |
Temperature ( \(°C\) ) |
BETA |
0 |
0.000000.E+00 |
50 |
1.102100.E+08 |
100 |
3.013300.E+08 |
150 |
5.014300.E+08 |
200 |
7.081300.E+08 |
250 |
9.188800.E+08 |
300 |
1.132910.E+09 |
350 |
1.348980.E+09 |
For the base metal:
Temperature ( \(°C\) ) |
LAMBDA |
0 |
37.7 |
20 |
37.7 |
50 |
38.6 |
100 |
39.9 |
150 |
40.5 |
200 |
40.5 |
250 |
40.2 |
300 |
39.5 |
350 |
38.7 |
Temperature ( \(°C\) ) |
BETA |
0 |
0.000000.E+00 |
50 |
1.061900.E+08 |
100 |
2.903300.E+08 |
150 |
4.829100.E+08 |
200 |
6.832800.E+08 |
250 |
8.921600.E+08 |
300 |
1.109440.E+09 |
350 |
1.335060.E+09 |
For mechanical calculation:
Four parameters are filled in, they are:
|
Young’s modulus, expressed in \(\mathit{Pa}\), |
|
Poisson’s ratio, |
|
isotropic thermal expansion coefficient, expressed in \(°C\), |
|
value of the temperature at which the values of the thermal expansion coefficient ALPHAont were determined, expressed in \(°C\). |
|
Reference temperature \({T}_{\mathit{Réf}}\), for which the thermal deformation is zero, expressed in \(°C\). |
For the coating:
Temperature ( \(°C\) ) |
\(E\) |
0 |
1.985E+11 |
20 |
1.97E+11 |
50 |
1.95E+11 |
100 |
1.915E+11 |
150 |
1.875E+11 |
200 |
1.84E+11 |
250 |
1.8E+11 |
300 |
1.765E+11 |
350 |
1.72E+11 |
Temperature ( \(°C\) ) |
ALPHA |
0 |
1.64E-05 |
20 |
1.64E-05 |
50 |
1.654E-05 |
100 |
1.68E-05 |
150 |
1.704E-05 |
200 |
1.72E-05 |
250 |
1.75E-05 |
300 |
1.777E-05 |
350 |
For the base metal:
Temperature ( \(°C\) ) |
\(E\) |
0 |
2.05E+11 |
20 |
2.04E+11 |
50 |
2.03E+11 |
100 |
2E+11 |
150 |
1.97E+11 |
200 |
1.93E+11 |
250 |
1.89E+11 |
300 |
1.85E+11 |
350 |
1.8E+11 |
Temperature ( \(°C\) ) |
ALPHA |
0 |
1.122E-05 |
20 |
1.122E-05 |
50 |
1.145E-05 |
100 |
1.179E-05 |
150 |
1.214E-05 |
200 |
1.247-05 |
250 |
1.278E-05 |
300 |
1.308E-05 |
1.4. Boundary conditions and loads#
1.4.1. Step 1: non-linear thermal calculation#
The limit conditions applied to the thermal calculation are summarized in Figure 1.3-a and are broken down as follows:
fluid temperature and exchange coefficient imposed on the inner wall,
thermal insulation on the external wall.
Figure 1.3-a: Thermal problem considered
1.4.2. Step 2: mechanical calculation in linear elasticity#
The boundary conditions of the mechanical problem are summarized in Figure 1.3-b below and are broken down as follows:
fluid pressure on the inner wall,
symmetry along the Oz axis on the lower segment,
shape effect as well as uniform movement along the Oz axis on the upper segment.
Figure 1.3-b: Mechanical problem considered