Heat Transfer Control
We consider the transient heat transfer problem across a slab of thickness 1 and conductivity k=1 shown in Example 1 of pde solver function PDSOLVE. The slab is insulated at the right side, x=1, and is initially at 0 degrees. At time equals 0, the left side of the slab, x=0, is brought to 100 degrees. The simulation below shows that the right side of the slab reaches approximately 89.2 degrees after 1 second.
Objective
Our target temperature for the right side of the slab after 1 second is 75 degrees. In this exercise, we will compute the initial condition that will produce the target temperature.
We solve the heat equation for the following configuration
| Time period | t ∈ [0,1] |
|---|---|
| Spatial range | 0 ≤ x ≤ 1 |
| Parameter | k = 1 |
| Initial condition | |
| Left boundary condition | |
| Right boundary condition |
Solution
To be able to control the initial value, we introduce a variable p with initial value
of 100 and define a parametrized initial value formula for u, and left boundary condition formula as shown in Table 1.
We obtain the initial solution, shown in Table 2, be evaluating the array formula
=PDSOLVE(B8, B2:B6, C8,D8, {0,1}, {0,1,2}) in array G8:J30.
| A | B | C | D | |
| 1 | Variables with initial values | Parameters | ||
|---|---|---|---|---|
| 2 | t | k | 1 | |
| 3 | x | p | 100 | |
| 4 | u | =IF(x=0,p,0) | ||
| 5 | ux | |||
| 6 | uxx | |||
| 7 | Equations | Left Bc | Right Bc | |
| 8 | du/dt | =k*uxx | =u-p | =ux |
| G | H | I | J | |
| 8 | t | 0 | 0.5 | 1 |
| 9 | x | u | u | u |
| 10 | 0 | 100 | 100 | 100 |
| 11 | 0.05 | 0 | 97.09061 | 99.15264 |
| 12 | 0.1 | 0 | 94.19917 | 98.31051 |
| 13 | 0.15 | 0 | 91.34352 | 97.47881 |
| 14 | 0.2 | 0 | 88.54128 | 96.66268 |
| 15 | 0.25 | 0 | 85.80973 | 95.86714 |
| 16 | 0.3 | 0 | 83.16573 | 95.09712 |
| 17 | 0.35 | 0 | 80.62557 | 94.35736 |
| 18 | 0.4 | 0 | 78.20491 | 93.65242 |
| 19 | 0.45 | 0 | 75.91868 | 92.98665 |
| 20 | 0.5 | 0 | 73.78095 | 92.36414 |
| 21 | 0.55 | 0 | 71.8049 | 91.78874 |
| 22 | 0.6 | 0 | 70.00271 | 91.26397 |
| 23 | 0.65 | 0 | 68.38546 | 90.79306 |
| 24 | 0.7 | 0 | 66.96312 | 90.37892 |
| 25 | 0.75 | 0 | 65.74445 | 90.02409 |
| 26 | 0.8 | 0 | 64.73694 | 89.73074 |
| 27 | 0.85 | 0 | 63.94682 | 89.50069 |
| 28 | 0.9 | 0 | 63.37894 | 89.33534 |
| 29 | 0.95 | 0 | 63.03681 | 89.23573 |
| 30 | 1 | 0 | 62.92253 | 89.20245 |
Next, we define a simple constraint in C1 which calculates the difference between the value of J30 of Table 2 which corresponds to u(1,1)
and the target value of 75. To solve for the design variable p, we evaluate the formula =NLSOLVE(C1, p) in cell C2.
The result shown in Table 3 indicates that the initial left surface temperature should be at approximately 84.0785 in order for the right side temperature to reach 75 in one second.
| C | |
| 1 | =DYNVAL(J24)-75 |
| 2 | =NLSOLVE(C1,p) |
| C | |
| 1 | 14.20224037 |
| 2 | 84.07859699 |
Note that we use DYNVAL to select J24 from the solution array because we want to use its dynamic value during the optimization.
Please note. DYNVAL is a new function available in ExceLab 7.0 and later versions only. It replaces deprecated Criterion Functions used in earlier versions.
