Benchmark solution ===================== Calculation method used for the reference solution -------------------------------------------------------- To reduce the time required to set up thermal calculations in the linear domain, a new methodology is proposed. A :math:`\Omega` structure is considered. A thermal shock is imposed on a part of the structure :math:`\partial \Omega`. An example is shown in the following figure. .. image:: images/100002000000008100000080925C7F1100785B57.png :width: 1.339in :height: 1.3319in .. _RefImage_100002000000008100000080925C7F1100785B57.png: Thermal shock is defined as a significant and instantaneous temperature increment. Figure 2 shows an example of several consecutive thermal shocks. .. image:: images/1000020000000122000000AE9B5D7AAA335C641A.png :width: 3.0161in :height: 1.8091in .. _RefImage_1000020000000122000000AE9B5D7AAA335C641A.png: Fourrier's law describing heat exchanges applied to this structure subjected to thermal shock can be written in this form: :math:`T(x,t)=\{\begin{array}{c}T(x,{t}_{0})\text{pour t <}{t}_{c}\\ T(x,{t}_{0})+\underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}T(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t}\ge \text{}{t}_{c}\end{array}` Where :math:`T(x,{t}_{0})` is the initial state of the structure, known, and :math:`{t}_{c}` the instant of the thermal shock. For various thermal shocks, it is therefore necessary to solve this problem and thus obtain, for each of the shocks, solution :math:`T(x,t)`. Case of unit shock ~~~~~~~~~~~~~~~~~~~~~~ If a unit shock :math:`\mathrm{\Delta }{T}_{U}` (:math:`\mathrm{\Delta }T=1` degree shock) is imposed on :math:`\partial \Omega` at time :math:`t={t}_{c}`, the problem is written as: :math:`{T}_{U}(x,t)=\{\begin{array}{c}{T}_{U}(x,{t}_{0})\text{pour t <}{t}_{c}\\ {T}_{U}(x,{t}_{0})+\underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}{T}_{U}(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t}\ge \text{}{t}_{c}\end{array}` With :math:`{T}_{U}(x,{t}_{0})` the initial state of the structure before the unity shock. The solution to this problem can be determined numerically using the finite element method, in particular the Code_Aster operator THER_LINEAIRE if the material parameters are independent of temperature. Once the problem is resolved, solution :math:`{T}_{U}(x,t)` is known. As we are only interested in the temperature differential, a simplified notation is proposed: :math:`\stackrel{̃}{{T}_{U}}(x,t)={T}_{U}(x,t)-{T}_{U}(x,{t}_{0})` So we have :math:`\stackrel{̃}{{T}_{U}}(x,t)=\{\begin{array}{c}0\text{pour t <}{t}_{c}\\ \underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}\stackrel{̃}{{T}_{U}}(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t}\ge \text{}{t}_{c}\end{array}` Case of any shock ~~~~~~~~~~~~~~~~~~~~~~~~~~~ A thermal shock, still on :math:`\partial \Omega`, but of any intensity :math:`\beta` is this time imposed. This problem is written in the form: :math:`{T}_{\beta }(x,t)=\{\begin{array}{c}{T}_{\beta }(x,{T}_{0})\text{pour t <}{t}_{c}\\ {T}_{\beta }(x,{T}_{0})+\underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}{T}_{\beta }(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t}\ge \text{}{t}_{c}\end{array}` With :math:`{T}_{\beta }(x,{t}_{0})` the initial state of the structure before the thermal shock. Using the simplified notation, we get: :math:`\stackrel{̃}{{T}_{\beta }}(x,t)=\{\begin{array}{c}0\text{pour t <}{t}_{c}\\ \underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}{T}_{\beta }(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t}\ge \text{}{t}_{c}\end{array}` Compared to the previous unit problem, only the intensity of thermal loading has been changed: :math:`\Delta T=\beta \Delta {T}_{U}`. Indeed, the shock is still on :math:`\partial \Omega`, it is simply amplified by the scalar :math:`\beta`. Therefore: This last equation is only valid for a linear thermal problem and therefore :math:`\beta` independent of position. The solution for any thermal shock is therefore: :math:`\stackrel{̃}{{T}_{\beta }}(x,t)=\{\begin{array}{c}0\text{pour t <}{t}_{c}\\ \underset{{t}_{c}}{\overset{t}{\int }}a\beta \frac{{\partial }^{2}\stackrel{̃}{{T}_{U}}(x,t)}{\partial {x}^{2}}\mathit{dt}\text{pour t >}{t}_{c}\end{array}` In addition, considering that the scalar :math:`\beta` is independent of time, which is consistent with a thermal shock, we have: :math:`\underset{{t}_{c}}{\overset{t}{\int }}a\beta \frac{{\partial }^{2}\stackrel{̃}{{T}_{U}}(x,t)}{\partial {x}^{2}}\mathit{dt}=\beta \underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}\stackrel{̃}{{T}_{U}}(x,t)}{\partial {x}^{2}}\mathit{dt}` The integral present in this problem equation is in reality the solution obtained previously for the unit shock: :math:`\underset{{t}_{c}}{\overset{t}{\int }}a\frac{{\partial }^{2}\stackrel{̃}{{T}_{U}}(x,t)}{\partial {x}^{2}}\mathit{dt}=\stackrel{̃}{{T}_{U}}(x,t)\text{}\mathit{pour}t\ge {t}_{c}` The solution to the problem of heat shock of any kind on :math:`\partial \Omega` solution obtained for the unit heat shock: :math:`\stackrel{̃}{{T}_{\beta }}(x,t)=\beta \mathrm{.}\stackrel{̃}{{T}_{U}}(x,t)` It is considered here that the unitary thermal shock takes place at the same instant as the thermal shock :math:`\Delta {t}_{c}`. If the shock occurs at another moment, simply postpone the solution: :math:`\stackrel{̃}{{T}_{\beta }}(x,t)=\beta \mathrm{.}\stackrel{̃}{{T}_{U}}(x,t-({t}_{c}-{t}_{\mathit{c2}}))` Where :math:`{t}_{\mathit{c2}}` is the moment of unity shock. Therefore, by choosing the unit shock so that :math:`{t}_{\mathit{c2}}` is at time :math:`t=\mathrm{0s}`, the solution becomes: :math:`\stackrel{̃}{{T}_{\beta }}(x,t)=\beta \mathrm{.}\stackrel{̃}{{T}_{U}}(x,t-{t}_{c})` It is therefore possible to determine the temperature field of the structure by proceeding as follows: * Define a uniform initial temperature field on the structure :math:`{T}_{\beta }(x,{t}_{0})` * Set up a thermal analysis of a unit shock at time :math:`t=\mathrm{0s}` on the structure and obtain the solution of the unit problem and determine :math:`\stackrel{̃}{{T}_{U}}(x,t)`. It may be a good idea to set an initial temperature of zero so that :math:`\stackrel{̃}{{T}_{U}}(x,t)={T}_{U}(x,t)`. * Calculate the solution (in space and time) from a linear combination of the initial situation :math:`T(x,{t}_{0})`, the intensity of the shock :math:`\beta` as well as the solution of the unit shock problem :math:`\stackrel{̃}{{T}_{U}}(x,t)`. Benchmark results ---------------------- Thermal analyses ~~~~~~~~~~~~~~~~~~~~~ The thermal behavior is axisymmetric. Several points across the thickness and two moments are chosen to test the temperature response. The reference results were obtained by Code_Aster and with a linear axisymmetric mesh that is four times finer. The geometric coordinates of the reference points are as follows: .. csv-table:: "Points", "Radius (m)" "Intrados", "0.018" "A", "0.0185" "B", "0.019" "C", "0.020" "D", "0.022" "Extrados", "0.024" To find these reference results, simply change the value of the "Temp_Choc" variable in the test case command file when performing the unit calculations for each of the two models and to read the refined mesh as input.