Question 3.9: 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
q_{0}(y,t)=q_{y0}\biggl({\frac{y}{W}}\biggr)^{p_{y}}+q_{t0}\biggl({\frac{t}{t_{r e f}}}\biggr)^{p_{t}}=\sigma_{q}y^{p_{y}}+s_{q}t^{p_{t}}~~~(0\le y\le W;~~t\ge0) (3.100b)
with p_y = 1 (linear-in-space increase), a slope of σ_q = q_0/W = 10^6 (W/m² )/m, p_t = 1 (linear-in-time increase) and a rate of s_q = q_0/t_{ref} = 10^4 W/m² – s. 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 at time 20 seconds by using Eq. (3.94) with a space step of 5 cm and a time step of 10 seconds.
T_{M}(x,y)=T_{i n}+\sum\limits_{j=1}^{N}\sum\limits_{i=1}^{M}q_{0,i,j}\left[\Delta_{M-i}(\lambda_{j,\mathrm{xy},M-i})-\Delta_{M-i}(\lambda_{j-1,\mathrm{xy},M-i})\right]
=\,T_{i n}+\,\sum\limits_{j=1}^{N}\sum\limits_{i=1}^{M}q_{0,i ,j}\left[\Delta_{j}\Delta_{M-i}(\lambda_{j,x y,M-i})\right] (3.94)
The temperature at the surface point of interest y_P= 10 cm and time t_P = 20 s is T(0, y_P , t_P). If the piecewise-uniform and -constant approximations of the applied surface heat flux are based on a space step Δy = 5 cm (i.e., N = 4) and a time step Δt = 10 s (hence, M = 2), by using Eq. (3.94) the temperature T(0, y_P , t_P = 2Δt) = T_2(0, y_P) is given by
T_{2}(0,y_{P})=T_{i n}+\sum\limits_{i=1}^{2}\sum\limits_{j=1}^{4}q_{0,i ,j}\left[\Delta_{j}\Delta_{2-i}(\lambda_{j,2-i}(0,\tilde{y}_{P}))\right] (3.101)
where the space and time heat flux component, q_{0,i,j}, may be taken as
{{{q}}}_{0,i, j}=\sigma_{q}(j-1/2)\Delta y+s_{q}(i-1/2)\Delta t\quad(i=1,2;j=1,2,3,4) (3.102)
Then, bearing in mind Eq. (3.95), for j = 1, 2, 3, 4 it results in
\Delta_{j}\Delta_{M-i}\left(\lambda_{j,\mathrm{xy,}M-i}\right)=\Delta_{M-i}\left(\lambda_{j,\mathrm{xy,}M-i}\right)-\Delta_{M-i}\left(\lambda_{j-1,\mathrm{xy,}M-i}\right)
=\left(\lambda_{j,xy,j,M-i+1}-\lambda_{j,x,M-i}\right)-\left(\lambda_{j-1,x,M-i+1}-\lambda_{j-1,xy,M-i}\right)
=\left(\lambda_{j,x y,M-i+1}-\lambda_{j-1,x y,M-i+1}\right)-\left(\lambda_{j,x y,M-i}-\lambda_{j-1,x y,M-i}\right)
=\Delta_{j}\bigl(\lambda_{j,{xy},M-i+1}\bigr)-\Delta_{j}\bigl(\lambda_{j,{xy},M-i}\bigr)=\Delta_{M-i}\Delta_{j}\bigl(\lambda_{j,{xy},M-i}\bigr) (3.95)
\Delta_{j}\Delta_{1}\left(\lambda_{j,1}(0,\;\tilde{y}_{P})\right)=\Delta_{1}\lambda_{j,1}(0,\tilde{y}_{P})-\Delta_{1}\lambda_{j-1,1}(0,\tilde{y}_{P}) (3.103a)
\Delta_{j}\Delta_{0}\bigl(\lambda_{j,0}(0,\;\tilde{y}_{P})\bigr)=\Delta_{0}\lambda_{j,0}(0,\tilde{y}_{P})-\Delta_{0}\lambda_{j-1,0}(0,\tilde{y}_{P}) (3.103b)
where
\Delta_{1}\lambda_{j,1}(0,\tilde{y}_{P})=\lambda_{j,2}(0,\tilde{y}_{P})-\lambda_{j,1}(0,\tilde{y}_{P}) (3.104a)
\Delta_{1}\lambda_{j-1,1}(0,\tilde{y}_{P})=\lambda_{j-1,2}(0,\tilde{y}_{P})-\lambda_{j-1,1}(0,\tilde{y}_{P}) (3.104b)
\Delta_{0}\lambda_{j,0}(0,\tilde{y}_{P})=\lambda_{j,1}(0,\tilde{y}_{P}) (3.104c)
\Delta_{0}\lambda_{j-1,0}(0,\tilde{y}_{P})=\lambda_{j-1,1}(0,\tilde{y}_{P}) (3.104d)
The \lambda_{j,2-i}(0,{\tilde{y}}_{P}) “building block” solutions in Eq. (3.101) may be computed by using Eqs. (3.88) and (3.91)
\frac{L}{k}\tilde{T}_{\mathrm{X22B(y1pt1)oY22B00T0}}(\tilde{x},\tilde{y},\tilde{t}_{M-i+1},\tilde{W}_{0}=j\Delta\tilde{y})=\lambda_{j}(\tilde{x},\tilde{y},\tilde{t}_{M-i+1})=\lambda_{j,\mathrm{xy,}M-i+1} (3.88)
\lambda_{j-1,xy,M-i}=\lambda_{j-1}(\tilde{x},\tilde{y},\tilde{t}_{M-i})={\frac{L}{k}}\tilde{T}_{\mathrm{X22B(y1pt1)OY22B00T0}}[\tilde{x},\tilde{y},\tilde{t}_{M-i},\tilde{W}_{0}=(j-1)\Delta\tilde{y}] (3.91)
with {\tilde{x}}_{P}=0, \tilde{y}_{P}=2\ \Delta\tilde{t}=0.04,\tilde{W}=4, and \tilde{W}_{0}=j\Delta\tilde{y}=j\ \mathrm{as\,}\Delta\tilde{y} = Δy L = 1 and Δy = W/N. The numerical values are shown in Table 3.9, where an accuracy of ten decimal-places (A = 10) has been considered for the building block solution.
Also, the various terms appearing in Eq. (3.101) are shown in Table 3.10. The temperature rise at the location and time of interest is of about 88 °C.
Table 3.10 also shows that \Delta_{1}\Delta_{1}(\lambda_{1,1})=\Delta_{4}\Delta_{1}(\lambda_{4,1}),\ \Delta_{1}\Delta_{0}(\lambda_{1,0})\;=\;\Delta_{4}\Delta_{0}(\lambda_{4.0}),\ \Delta_{2}\Delta_{1}(\lambda_{2.1})\,=\Delta_{3}\Delta_{1}(\lambda_{3.1}),\;\mathrm{and}\;\Delta_{2}\Delta_{0}(\lambda_{2.0})=\Delta_{3}\Delta_{0}(\lambda_{3,0}) according to the fact that the surface point of interest is located at the middle of the heated region and the surface heat flux varies linearly with space. Hence, there exists a thermal symmetry. This is an application of “numerical” intrinsic verification of Eq. (3.94)
=\,T_{i n}+\,\sum\limits_{j=1}^{N}\sum\limits_{i=1}^{M}q_{0,i ,j}\left[\Delta_{j}\Delta_{M-i}(\lambda_{j,x y,M-i})\right] (3.94)
and, hence, of the fdX22By_t_0Y22B00T0_puca.m function (Cole et al. 2014; D’Alessandro and de Monte 2018).
| Table 3.9 | Building block temperatures at the surface point y_P = 10 cm for a space step Δy = 5 cm and a time step Δt = 10 s. | ||||
| j | i | \begin{array}{l}{{\lambda_{j-1,2-i}(0,{\tilde{y}}_{P}),}}\\ {{(^{o}C-{\mathfrak{m}}^{2}/W)}}\end{array} | \begin{array}{l}{{\lambda_{j,2-i}(0,{\tilde{y}}_{P}),}}\\ {{(^{o}C-{\mathfrak{m}}^{2}/W)}}\end{array} | \begin{array}{l}{{\lambda_{j-1,2-i+1}(0,{\tilde{y}}_{P}),}}\\ {{(^{o}C-{\mathfrak{m}}^{2}/W)}}\end{array} | \begin{array}{l}{{\lambda_{j,2-i+1}(0,{\tilde{y}}_{P}),}}\\ {{(^{o}C-{\mathfrak{m}}^{2}/W)}}\end{array} |
| 1 | 1 | 0 | 0.0000000036 | 0 | 0.0000002617 |
| 2 | 0 | 0 | 0 | 0.0000000036 | |
| 2 | 1 | 0.0000000036 | 0.0001410474 | 0.0000002617 | 0.0001994712 |
| 2 | 0 | 0 | 0.0000000036 | 0.0001410474 | |
| 3 | 1 | 0.0001410474 | 0.0002820912 | 0.0001994712 | 0.0003986807 |
| 2 | 0 | 0 | 0.0001410474 | 0.0002820912 | |
| 4 | 1 | 0.0002820912 | 0.0002820948 | 0.0003986807 | 0.0003989424 |
| 2 | 0 | 0 | 0.0002820912 | 0.0002820948 | |
| Table 3.10 | Surface heat flux components and sensitivity coefficients at the surface point y_P = 10 cm for a space step Δy = 5 cm and a time step Δt = 10 s using the piecewise-uniform and -constant approximations for the surface heat flux. | |||
| j | i | \Delta_{j}\Delta_{2-i}\left(\lambda_{j,2-i}(0,\;\tilde{y}_{P})\right)=X_{2,i,j}\left(^{o}C-{\bf m}^{2}/W\right) | q_{0,i,j}-\mathrm{Eq.}\left(3.102\right)\left(W/\mathrm{m}^{2}\right) | q_{0,i j}X_{2,i, j}\left({}^{\circ}{C}\right) |
| 1 | 1 | \Delta_{1}\Delta_{1}(\lambda_{1,1})=0.000002581 | 0.75\times10^{5} | 0.0193575000 |
| 2 | \Delta_{1}\Delta_{0}(\lambda_{1,0})=0.00000000036 | 1.75\times10^{5} | 0.00063 | |
| 2 | 1 | \Delta_{2}\Delta_{1}(\lambda_{2,1})=0.0000581657 | 1.25\times10^{5} | 7.2707125 |
| 2 | \Delta_{2}\Delta_{0}(\lambda_{2,0})=0.0001410438 | 2.25\times10^{5} | 31.734855 | |
| 3 | 1 | \Delta_{3}\Delta_{1}(\lambda_{3,1})=0.0000581657 | 1.75\times10^{5} | 10.1789975 |
| 2 | \Delta_{3}\Delta_{0}(\lambda_{3,0})=0.0001410438 | 2.75\times10^{5} | 38.787045 | |
| 4 | 1 | \Delta_{4}\Delta_{1}(\lambda_{4,1})=0.0000002581 | 2.25\times10^{5} | 0.0580725 |
| 2 | \Delta_{4}\Delta_{0}(\lambda_{4,0})=0.0000000036 | 3.25\times10^{5} | 0.00117 | |
| Temperature rise: T_{2}(0,y_{P})-T_{i n}=\sum\limits_{i=1}^{2}\sum\limits_{j=1}^{4}q_{0,i, j}X_{2,i, j} | 88.05{}^{\circ}\mathrm{C} | |||