Question 3.7: A steel plate (α = 10^−5 m²/s, k = 40 W/m/ °C), initially at......
A steel plate (α = 10^{−5} m²/s, k = 40 W/m/ °C), initially at 30 °C, is subject at x = 0 to a surface heat flux as given by Eq. (3.80),
q_{0}(y)=q_{y0}{\Big(}{\frac{y}{W}}{\Big)}^{P}=\sigma_{q}y^{p}\quad(0\leq y\leq W) (3.80)
starting at y = 0, with p = 1 (linear-in-space variation), and ending at y = W, with a slope of σ_q = q_{y0}/ W = 10^7 (W/m²)/m. The plate is insulated on the back side x=L= 5 cm as well as on the lateral sides y = 0 and y=W= 20 cm. Calculate the temperature of the rectangular body at the middle point of its heated surface and at time 12.5 seconds by using Eq. (3.75) with a space step of 5 cm.
T(x,y,t)=T_{i n}+\sum\limits_{j=1}^{N}T_{j}(x,y,t)=T_{i n}+\sum\limits_{j=1}^{N}q_{0,j}\Delta_{j}\lambda_{j,x y t} (3.75)
The temperature at the heated surface point of interest y_P = 10 cm and time t_P = 12.5 s is T(0, y_P, t_P). If the piecewise-uniform approximation of the applied surface heat flux is based on a space step Δy = 5 cm (i.e., N = 4), by using Eq. (3.75) the temperature is given by
T(0,y_{P},t_{P})=T_{i n}+\sum\limits_{j=1}^{4}q_{0,j}\Delta_{j}\lambda_{j}(0,\tilde{y}_{P},\tilde{t}_{P})
=\,T_{i n}+q_{0,1}\lambda_{1}(0,\tilde{y}_{P},\tilde{t}_{P})
+\,q_{0,2}[\lambda_{2}(0,\;\tilde{y}_{P},\;\tilde{t}_{P})-\lambda_{1}(0,\tilde{y}_{P},\tilde{t}_{P})]
+\ {q_{0,{3}}}[{\lambda}_{3}(0,\tilde{y}_{P},\tilde{t}_{P})-{\lambda}_{2}(0,\tilde{y}_{P},\tilde{t}_{P})]
+\ {q_{0,{4}}}[{\lambda}_{4}(0,\tilde{y}_{P},\tilde{t}_{P})-{\lambda}_{3}(0,\tilde{y}_{P},\tilde{t}_{P})] (3.81)
where \lambda_{j}(0,{\tilde{y}}_{P},{\tilde{t}}_{P}) may be computed by using Eq. (3.73)
\frac{L}{k}\tilde{T}_{\mathrm{X22B}(\mathrm{y1pt1})\mathrm{0Y22B00T0}}(\tilde{x},\tilde{y},\tilde{t},\tilde{W}_{0}=j\Delta\tilde{y})=\lambda_{j}({\tilde{x}},{\tilde{y}},{\tilde{t}})=\lambda_{j,xy t} (3.73)
with \tilde{y}_{P}=2,\tilde{t}_{P}=0.05,\,\tilde{W}=4\,{\mathrm{and}}\,\tilde{W}_{0}=j\Delta\tilde{y}=j\ {\mathrm{as~}}\Delta\tilde{y}=\Delta y/L=1 and Δy = W/N.
Also, the various terms appearing in Eq. (3.81) are shown in Table 3.7 where, for the numerical values of the “building block” temperature, \lambda_{j}(0,\tilde{y}_{P},\tilde{t}_{P}), the first ten decimal-places have been considered (A = 10). The temperature rise at the location and time of interest is of about 315 °C.
Table 3.7 also shows that \Delta_{1}\lambda_{1}=\Delta_{4}\lambda_{4}\,\mathrm{and}\,\Delta_{2}\lambda_{2}=\Delta_{3}\lambda_{3}\, according to the fact that the surface point of interest is located at the middle of the heated region and, hence, there exists a thermal symmetry. This is an application of “numerical” intrinsic verification of Eq. (3.75) and, hence, of the fdX22By_t10Y22B00T0_pua.m function (Cole et al. 2014; D’Alessandro and de Monte 2018).
| Table 3.7 | Surface heat flux components and building block temperatures at the surface point y_p = 10 cm and time t_p = 12.5 seconds for a space step Δy = 5 cm using the piecewise-uniform approximation for the surface heat flux, that is,q_{0,j}=\sigma_{q}(j-1/2)\Delta y. | |||
| j | q_{0,j},{\mathrm{W/m}^{2}} | \begin{array}{c}{{\lambda_{j}(0,\tilde{y}_{p},\tilde{t}_{p}),}}^{\circ}C\ m^2/W\end{array} | \begin{array}{c}{{\Delta_{j}\lambda_{j}(0,\tilde{y}_{p},\tilde{t}_{p}),}}^{\circ}C\ m^2/W\end{array} | \begin{array}{c}{{q_{0,j}\Delta_{j}\lambda_{j}(0,\tilde{y}_{P},\tilde{t}_{P})}}^{\circ}C\end{array} |
| 1 | 0.25\times10^{6} | 1.84E-08 | 1.84E-08 | 0.004602796 |
| 2 | 0.75\times10^{6} | 0.000157696 | 0.000157677 | 118.25805 |
| 3 | 1.25\times10^{6} | 0.000315373 | 0.000157677 | 197.09675 |
| 4 | 1.75\times10^{6} | 0.000315392 | 1.84E-08 | 0.032219573 |
| Temperature rise: T(0,y_{P},t_{P})-T_{i n}=\sum\limits_{j=1}^{4}q_{0,j}\Delta_{j}\lambda_{j}(0,\tilde{y}_{P},\tilde{t}_{P}) | 315.4^{\circ}C | |||