A 1D transient conduction problem in a 0.1 m steel bar (35 W/mK, 7200 kg/m³, 440 J/kgK), initially at 0 °C, with one end held at 0 °C and the other varied as 100 sin(πt/40) °C. The target is the temperature 0.08 m from the 0 °C end at 32 s; the reference is 36.60 °C.

How can the T3 answer be verified?

It can be solved exactly by an eigenfunction series that includes the boundary variation. That gives 36.63 °C, within 0.1% of the 36.60 °C reference. Finite differences also approach 36.6 °C as elements are refined and the time step reduced.

Does the time integration scheme matter?

Yes. With 40 elements, Crank–Nicolson gives 36.52 °C even with a 2 s step, but backward Euler gives 35.61 °C, about 1 °C low. Backward Euler is stable but first-order: even with 0.5 s it gives 36.33 °C. Checking time-step sensitivity is one purpose of this benchmark.

NAFEMS T3: 1D Transient Heat Conduction

Category: V&V – NAFEMS Benchmarks | Revised 2026-10-01

CAE visualization for nafems t3 theory - technical simulation diagram

NAFEMS T3: 2D Steady-State Heat Conduction

The problem: a bar heated by a sine wave

Overview

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NAFEMS T3 is a heat-conduction benchmark. What does such a simple 1D problem check?

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It's a transient problem where temperature changes with time, checking not only mesh fineness but whether time marching is right. One end's temperature rises and falls as a sine wave, and heat reaches the interior with a lag. The target is the value at 32 s, after the boundary has peaked and begun falling, so time-integration error shows up directly. Let's build the exact series solution and vary elements and time steps in finite differences to see how they approach the 36.60 °C reference.

Problem definition

$$ \rho c\,\frac{\partial T}{\partial t} = k\,\frac{\partial^2 T}{\partial x^2},\qquad T(x,0) = 0,\quad T_{hot}(t) = 100\sin\frac{\pi t}{40},\quad T_{cold} = 0 $$

Length 0.1 m, conductivity 35 W/mK, density 7200 kg/m³, specific heat 440 J/kgK (diffusivity 1.10×10⁻⁵ m²/s). Find the temperature at 32 s at 0.08 m from the 0 °C end — 0.02 m from the heated end. The reference is 36.60 °C.

Exact series solution

$$ T = T_{hot}(t)\left(1-\frac{\xi}{L}\right) + \sum_{n=1}^{\infty} u_n(t)\sin\frac{n\pi\xi}{L},\qquad \dot u_n + \alpha\left(\frac{n\pi}{L}\right)^2 u_n = -\frac{2}{n\pi}\,\dot T_{hot}(t) $$

$\xi$ is the distance from the heated end. Splitting off a linear part carrying the boundary temperature and expanding the rest in eigenfunctions turns each coefficient into a first-order ODE. With 4000 terms the value is 36.626 °C, unchanged at 8000.

Coffee Break Trivia

NAFEMS benchmarks

NAFEMS was founded in the UK in 1983 to improve confidence in analysis methods such as finite elements. From the late 1980s to around 1990 it compiled standard benchmarks for linear elasticity (LE), heat transfer (T), vibration (FV) and more, so any software could solve the same problems and compare results. Many commercial codes' verification manuals still list these problems and their references. T3 is the basic problem among them for checking time integration in transient conduction.

Worked example: convergence with finite differences

Elements, time step and time integration

Elements	CN, Δt 2 s	CN, Δt 0.5 s	Backward Euler, Δt 2 s	Backward Euler, Δt 0.5 s
5	34.83	34.89	34.22	34.73
10	35.90	35.96	35.09	35.75
20	36.38	36.44	35.49	36.20
40	36.52	36.58	35.61	36.33

Values in °C; CN is Crank–Nicolson. With 40 elements, CN and Δt = 0.05 s give 36.58 °C. The series solution is 36.63 °C and the reference 36.60 °C.

Temperature history

Time	Heated-end temperature	Target-point temperature
8 s	58.8 °C	2.8 °C
16 s	95.1 °C	14.9 °C
24 s	95.1 °C	28.8 °C
32 s	58.8 °C	36.6 °C
40 s	0.0 °C	33.9 °C

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The heated end peaks at 20 s, yet the target point is still rising.

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The time for heat to travel 0.02 m is roughly distance squared over diffusivity, about 36 s — comparable to the boundary period (80 s). So the target lags the boundary: at 32 s it's still rising, and at 40 s, when the boundary is back to 0 °C, it's still 34 °C. Capturing this lag needs time integration that doesn't distort the phase. Backward Euler has strong numerical damping that smooths the variation, coming out about 1 °C low at Δt = 2 s. Crank–Nicolson is second-order, differing by only 0.06 °C between Δt = 2 s and 0.5 s. Element count matters too — 5 elements are nearly 2 °C low — and only 40 elements with Δt = 0.5 s bring the error within 0.1% of the reference.

Using the benchmark

First read the definition precisely (target location, which end is heated).
Double the element count repeatedly and check the observed convergence order.
Vary the time step likewise, separating time-integration error from spatial error.
Check the software's default time integrator (backward Euler, CN, etc.).
Record results with element count and time step, not just the difference from the reference.

Coffee Break Trivia

“The location was read backwards”

Verifying one program on T3, the target temperature came out 0.09 °C — nowhere near the 36.60 °C reference. A software bug was suspected, but the cause was misreading the problem: '0.08 m' had been measured from the heated end. Re-evaluating at 0.08 m from the 0 °C end — 0.02 m from the heated end — gave 36.6 °C. The lesson: with benchmarks, read the definition precisely before computing.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Measuring from the wrong end	Wildly wrong value (e.g. 0.09 °C)	Check definition
Large time step	Peaks smeared	Vary Δt
Judging by backward Euler only	About 1 °C error	Try second-order schemes
Coarse mesh	Nearly 2 °C error	Refine by doubling
Wrong property units	Diffusivity off by orders	Compute α to check

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I'd like to learn related topics.

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Related pages include NAFEMS T2, NAFEMS LE1, NAFEMS LE10, solution verification and Richardson extrapolation.

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