Question 3.5: Calculate the temperature at 1 cm inside the steel plate of ......

The temperature at the interior point of interest x_P = 1 cm is T(x_P, t) and increases with time. If the piecewise-linear approximation of the applied surface heat flux is based on a time step of Δt = 1 s, by using Eq. (3.62) with q_{0,0} = 0 the temperatures at t = 1, 2, and 3 seconds are given, respectively, by

T_{1}(x_{P})=T_{i n}+q_{0,1}\gamma_{1}(\tilde{x}_{P})=30.187600^{\circ}\mathrm{C} (3.66a)

T_{2}(x_{P})=T_{i n}+q_{0,1}[\Delta(\Delta\gamma(\tilde{x}_{P}))_{1}]+q_{0,2}\gamma_{1}(\tilde{x}_{P})=34.074329^{\circ}{\mathrm{C}} (3.66b)

T_{3}(x_{P})=T_{i n}+q_{0,1}\left[\Delta(\Delta\gamma(\tilde{x}_{P}))_{2}\right]\,+\,q_{0,2}\left[\Delta(\Delta\gamma(\tilde{x}_{P})\right)_{1}]+\,q_{0,3}\gamma_{1}(\tilde{x}_{P})=46.725442\,^{\circ}\mathrm{C} (3.66c)

where \gamma_{M}(\tilde{x}_{P})=\tilde{T}_{\mathrm{X2LB20T0}}(\tilde{x}_{P},M\Delta\tilde{t},\,\tilde{t}_{r e f}=\Delta\tilde{t})\times L/k, with M = 1, 2 or 3. Also,\tilde{x}_{P}=x_{P}/L=0.2,\,\Delta\tilde{t}=\alpha\Delta t/L^{2}=0.004, and the first six decimal-places have been considered for the building block solution (A =6).

The exact temperatures can be obtained from the fdX22B20T0.m MATLAB function of Section 2.3.9 with an accuracy of one part in 10^{15} (A = 15) as the current problem is denoted by X22B20T1. They are 30.187600, 34.074329, and 46.725442 °C, respectively, as already given in Example 3.4. Note that the accuracy of A = 15 allows the computational analytical solution to be considered as exact.
Therefore, the approximate and exact temperatures are the same (as expected) as the applied surface heat-flux changes linearly with time and, hence, the piecewise-linear approximation cannot perform any approximation in the current case.
This is an application of “numerical” intrinsic verification of Eq. (3.62) and, hence, of the fdX22B_0T0_pla.m MATLAB function (D’Alessandro and de Monte 2018).