6. Assessing a thermal calculation

A plausible temperature is a starting point

You have a warmed center, a cold exterior, and a smooth transient. What question does each comparison answer? This chapter turns the previous results into a method for investigating numerical error and physical assumptions.

Begin with the observable that matters. A maximum temperature, an average stored energy, and power passing through a surface are different quantities. A calculation can approximate one well while approximating another poorly.

Keep three exact answers separate

For the square investigation, we have derived:

1. The steady four-cell Q1 center rise $F/K=3/32$ K.
2. The continuous-time rise for that spatial system, $(3/32)(1-e^{-24t})$ K.
3. The Backward Euler rise at the original step, $(3/32)(1-2^{-n})$ K.

The first tests the algebraic spatial system. The second lets us isolate time integration error. The third checks the chosen discrete time update. None is the exact temperature field of the continuous two-dimensional heated square. The exact slab parabola answers yet another boundary-value problem.

This separation makes a discrepancy useful. If the run matches the recurrence but differs from the exponential, the time step explains the difference. If a refined mesh no longer gives 300.09375 K at the center, the old one-coefficient calculation no longer describes that mesh.

Check the weighted balance at each step

For the free center row, compute

$$R_n=M\frac{\theta_n-\theta_{n-1}}{\Delta t}+K\theta_n-F.$$

$R_n$ has units W/m. Compare it with the load scale rather than quoting an unqualified small number. For example, $|R_n|/|F|$ is dimensionless when $F\ne0$. Check all boundary values separately: a correct free-row residual does not excuse an incorrectly imposed boundary.

The physical steady account is instead

$$\int_{\partial\Omega}\boldsymbol q\cdot\boldsymbol n\,\mathrm ds =\int_\Omega s\,\mathrm dA=1\,\mathrm{W/m}.$$

The free test function is not the constant function one; it vanishes at the boundary. Consequently its weighted source is $F=1/4$ W/m. Confusing the two integrals creates a spurious factor of four.

Why a raw boundary gradient can mislead

On a lower cell of the four-cell mesh, the center hat has the form $\phi=\xi\eta/h^2$ with its center corner at $(h,h)$. At the outer face $\eta=0$, its normal derivative has magnitude $\xi/h^2$. Its integral along that face is $1/2$. There are eight such exterior cell faces, giving

$$P'_{\mathrm{raw}}=4k\theta.$$

At the steady Q1 coefficient $\theta=3/32$ K and $k=1$ W/(m K), this gives $P'_{\mathrm{raw}}=3/8$ W/m, while generation is 1 W/m. The free equation $K\theta=F$ is still satisfied. A raw derivative of the coarse interpolated field is a different flux estimate from the boundary reactions of the finite element equations.

The practical lesson is to identify your flux observable and how it is constructed. A small temperature residual does not make every derivative or surface integral equally accurate. If heat rejection is the design quantity, it deserves its own comparison and refinement study.

Separate space and time errors

First hold the four-cell mesh fixed. Choose a common final time $t_*$ and run with $\Delta t$, $\Delta t/2$, and $\Delta t/4$, adjusting the number of steps to keep $t_*$ fixed. Compare the center rise with the exact exponential for this spatial system. A simple absolute temporal error is

$$E_t=\left|\theta_N-\frac FK(1-e^{-Kt_*/M})\right|.$$

Avoid dividing by the full absolute temperature of roughly 300 K: that can hide a large relative error in a small heating-induced rise. Normalize by a relevant rise such as $F/K$ when a relative error is useful.

Next study space while making the time error small enough for the intended comparison, or use the steady problem. Change the runner’s Cartesian cell counts, hold the physical square and all material data fixed, and compare a chosen observable across meshes. A sequence of similar values supports a resolution judgment for that observable; it does not prove that the physical assumptions fit a real device.

The numerical simulation book develops norms, convergence rates, and comparisons between spatial methods. Carry this heated body with you as a concrete example of why those tools matter.

A diagnosis table

Observation	A discriminating next question
Interior becomes colder under positive generation and a cool fixed exterior	Did the source or flux sign change?
Steady rise changes when only capacity changes	Is the run actually steady, or did another input change?
Doubling heating does not double steady rise	Are properties and boundary data still linear and unchanged?
Halving the step changes a transient temperature substantially	Does the error against the fixed-mesh exponential decrease?
Mesh refinement barely changes temperature but flux changes	Are you comparing a derivative-sensitive observable?
An insulated heated steady problem fails	Is the integrated energy account solvable?

Each question targets a specific premise. Re-running a larger calculation without deciding which premise is in doubt can obscure the diagnosis.

Capstone: a prediction sheet

Use the maintained heated-body runner and prepare a table before each run. Record the change, your predicted steady rise, your predicted warming direction, and the observable you will compare.

1. Double heating alone.
2. Double conductivity alone.
3. Double capacity alone.
4. Halve the time step at fixed final time.
5. Refine the steady mesh while preserving the square geometry.

For the first three experiments, derive $F/K$ and $M/K$ anew. In the fourth, keep the spatial system fixed. In the fifth, stop using the four-cell coefficient as an exact expectation and examine how the chosen observable changes.

Finish with a short physical interpretation: which change cools the eventual center, which only delays its warming, and which merely improves the approximation? The useful result is that explanation, supported by the numbers.

Exercises

1. The center value agrees to six digits after mesh refinement. Does that establish equally accurate boundary heat flux? Explain.
2. Two transient runs use different time steps but the same step count. What must change before their final values measure temporal error?
3. Derive the raw boundary-flux integral $4k\theta$ on the four-cell square.
4. The heat source is zero. Propose a residual normalization without division by zero.
5. Your simulated component stays cool because the boundary is fixed at 300 K. What additional physical information would you need to model a finite cooler?

Solution hints

1. No. Flux depends on gradients and the chosen reconstruction, with its own numerical error.
2. Compare the same physical time and initial condition.
3. Integrate the one-sided normal derivative on each of eight outer cell faces.
4. Use a declared characteristic power or a combination of nonzero storage and conduction scales; also report an absolute residual in physical units.
5. An exchange relation or cooler model, its surroundings, geometry, material data, and the operating regime that justifies that relation.

References

• Klaus-Jürgen Bathe, Finite Element Analysis of Solids and Fluids I, MIT 2.092 (2009), Lecture 20, pp. 3–4: finite element heat equations and time-step accuracy.
• The independent integrals in Chapters 4–6 derive the temperature and raw-flux comparisons for this particular mesh. The example explanation connects those coefficients to the model.

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