When can a body be treated as a semi-infinite solid?

As long as the temperature change entering from the surface has not yet reached the opposite face. If the penetration depth 4√(αt) is smaller than the thickness — a Fourier number αt/L² below about 0.06 — a plate of finite thickness can be treated as semi-infinite.

Why does metal feel colder than wood at the same temperature?

When two semi-infinite bodies touch, the interface temperature is the average weighted by thermal effusivity √(kρc). With 34 °C skin touching a 20 °C object, the interface is about 21 °C for steel but about 31 °C for wood, and that difference is what you feel.

How should the mesh near the surface be chosen in FEM?

Refine near the surface so that several elements fit inside the penetration depth 4√(αt₁) at the first time t₁ you care about. With second-order elements, too small a time step can cause oscillations; the Abaqus documentation gives Δt > ρcΔℓ²/(6k) as a guideline.

Transient Conduction in Semi-Infinite Solids

Category: Thermal Analysis | Revised 2026-09-28

Transient conduction in semi-infinite solids: erfc-type temperature distribution and penetration depth during surface temperature transient

Theory: a body that extends forever on one side

Overview

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A “semi-infinite solid” — nothing real is infinitely thick. What is it good for?

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Heat spreads at a finite rate, so right after a surface is suddenly heated or cooled, only a thin layer near the surface has noticed. During that time it makes no difference where the back face is. That is why short-time events — the first seconds of quenching, frost in soil, the moment your hand touches something, a short laser pulse — can almost all be estimated with the semi-infinite solution. It is also the most common reference solution for checking FEM results.

Governing equation and three boundary conditions

Consider 1D transient conduction into a uniform body at initial temperature $T_i$, with heat entering from the surface $x=0$ into $x>0$:

$$ \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}, \qquad \alpha = \frac{k}{\rho c}, \qquad T(x,0)=T_i, \quad T(\infty,t)=T_i $$

(1) Surface temperature suddenly changed to $T_s$: the solution is an error function, and the surface heat flux decays as $1/\sqrt{t}$.

$$ \frac{T(x,t)-T_s}{T_i-T_s} = \operatorname{erf}\!\left(\frac{x}{2\sqrt{\alpha t}}\right), \qquad q_s''(t) = \frac{k\,(T_s-T_i)}{\sqrt{\pi \alpha t}} $$

(2) Constant surface heat flux $q_0''$: a model for laser heating or frictional heating. The surface temperature rises in proportion to $\sqrt{t}$.

$$ T(x,t)-T_i = \frac{2 q_0''\sqrt{\alpha t/\pi}}{k}\exp\!\left(-\frac{x^2}{4\alpha t}\right) - \frac{q_0'' x}{k}\operatorname{erfc}\!\left(\frac{x}{2\sqrt{\alpha t}}\right) $$

(3) Convection to a fluid at $T_\infty$ with heat-transfer coefficient $h$: reduces to (1) as $h \to \infty$.

$$ \frac{T(x,t)-T_i}{T_\infty-T_i} = \operatorname{erfc}(\eta) - \exp\!\left(\frac{hx}{k}+\frac{h^2\alpha t}{k^2}\right)\operatorname{erfc}\!\left(\eta+\frac{h\sqrt{\alpha t}}{k}\right), \qquad \eta=\frac{x}{2\sqrt{\alpha t}} $$

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They all contain the combination $x/(2\sqrt{\alpha t})$.

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That is the key feature. With no characteristic length, the only length scale is $\sqrt{\alpha t}$. So the same temperature change takes four times as long to reach twice the depth. Keep that square-root law in mind and you can sanity-check test data or FEM results at a glance.

Penetration depth and when “semi-infinite” holds

$$ \delta \approx 4\sqrt{\alpha t} \quad (\operatorname{erf}(2) = 0.9953), \qquad Fo = \frac{\alpha t}{L^2} \lesssim 0.06 $$

At $x = 4\sqrt{\alpha t}$ the temperature change is below 0.5% of the surface change. A plate of thickness $L$ behaves as semi-infinite until this depth reaches $L$, i.e. for Fourier numbers below about 0.06. Penetration depths for typical materials:

Material (diffusivity)	1 s	1 min	1 h
Aluminium (9.8×10⁻⁵ m²/s)	40 mm	306 mm	2.4 m
Carbon steel (1.4×10⁻⁵ m²/s)	15 mm	115 mm	0.89 m
Concrete (6.9×10⁻⁷ m²/s)	3.3 mm	26 mm	0.20 m

Coffee Break Trivia

Estimating the age of the Earth from heat conduction

In the 19th century Lord Kelvin treated the Earth as a semi-infinite solid that started uniformly hot and cooled from the surface, and worked back from the temperature gradient measured near the ground to the time the cooling had taken. His answer — some tens of millions up to about a hundred million years — was far shorter than geologists and evolutionists believed. Radioactive heating and mantle convection, discovered later, showed the estimate was wrong. Yet the idea of using the square-root-of-time behaviour of the erf solution to infer time from measurements is still used today, for example in non-destructive testing.

Worked examples

Example 1: frost depth in soil

Soil initially at 20 °C ($\alpha = 1.38\times10^{-7}$ m²/s) has its surface suddenly held at −15 °C. The depth of the 0 °C front follows from solution (1): $\operatorname{erf}(\eta) = 15/35 = 0.429$ gives $\eta \approx 0.400$, and $x = 2\eta\sqrt{\alpha t}$.

Days	Depth of 0 °C
30	0.48 m
60	0.68 m
90	0.83 m

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Tripling the duration increases the depth only by √3 ≈ 1.7. That is why water mains and foundations in cold regions only need to be buried a few tens of centimetres to about a metre. Snow cover, vegetation and the latent heat of freezing water make the real front shallower, so this is a conservative estimate.

Example 2: why metal feels cold (contact temperature)

When two semi-infinite solids at different temperatures touch, the interface temperature $T_c$ is constant in time and equals the average weighted by thermal effusivity $e$:

$$ T_c = \frac{e_A T_A + e_B T_B}{e_A + e_B}, \qquad e = \sqrt{k\rho c} $$

Taking skin as $k=0.37$ W/(m·K), $\rho=1000$ kg/m³, $c=3500$ J/(kg·K) ($e \approx 1140$) at 34 °C, the interface temperature on touching a 20 °C object is:

Material	Effusivity e [W·s^1/2/(m²·K)]	Interface temperature
Aluminium	24,000	20.6 ℃
Steel	13,400	21.1 ℃
Glass	1,620	25.8 ℃
Wood (softwood)	310	31.0 ℃
Expanded polystyrene	30	33.6 ℃

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Same 20 °C on a thermometer, yet the fingertip sees a 10 °C difference.

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Because we sense our skin's surface temperature, not the object's. The same formula is handy in product design: how hot or cold an appliance casing or door handle feels, burn-risk checks, the initial interface temperature between a mould and a polymer melt. Get the ballpark from this formula first, then refine with FEM.

Example 3: surface temperature rise under constant flux

Applying 1 MW/m² to steel ($k=50$ W/(m·K), $\alpha = 1.38\times10^{-5}$ m²/s), solution (2) at $x=0$ gives $\Delta T_s = 2q_0''\sqrt{\alpha t/\pi}/k$: about 27 K after 0.1 s and about 84 K after 1 s. Ten times the duration gives only $\sqrt{10}\approx3.2$ times the rise. This is why short, intense laser pulses limit the effect on the base material.

Solving it with FEM and checking it

Mesh near the surface

Refine the surface so that at least several elements lie within the penetration depth $4\sqrt{\alpha t_1}$ at the first time $t_1$ of interest. For steel at $t_1 = 0.1$ s that depth is only about 5 mm.
Elements may grow with depth, but avoid abrupt size changes between neighbours — grow by about 1.2–1.5× at a time.
Make the model thicker than the penetration depth at the final time analysed. Otherwise the back-face boundary condition leaks into the result and it will no longer match the semi-infinite solution.

Time step and early oscillations

Problems with a sudden surface change are hardest in the first few steps. Implicit time integration does not diverge with large steps, but coarse steps cannot capture the steep gradient near the surface. Conversely, with second-order elements an overly small step can make the temperature oscillate inside the surface element and produce unphysical values, such as temperatures below the initial value during heating. The Abaqus documentation gives $\Delta t > \rho c\,\Delta\ell^2/(6k)$ ($\Delta\ell$ = surface element size) as a guideline for transient heat transfer with second-order elements.

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So how do I avoid the oscillation?

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The basic fix is to refine the surface elements and start with a step that satisfies the guideline. Switching to first-order elements or ramping the surface temperature over a very short time also helps. Either way, compare a few points near the surface against the erf solution before moving to the production model.

Verifying against the erf solution

Build a 1D model (a long bar or thin slice) with the real material and temperature difference, thicker than the penetration depth at the final time.
Hold the surface temperature fixed, solve, and compare temperature histories at several depths and the surface heat flux with solution (1).
The surface flux varies as $1/\sqrt{t}$, so errors are largest early on. Adjust mesh and step until the error in the first few steps is acceptable.
Carry the same mesh and step settings over to the surface region of the real-geometry model.

Coffee Break Trivia

Why the first second of a quench simulation goes wrong

In quenching, the surface heat-transfer coefficient changes drastically just after a hot part enters water or oil. While a vapour film covers the surface heat flows poorly; when the film collapses and boiling starts, cooling suddenly accelerates. The near-surface temperature reacts at once, so a model with a constant coefficient often misses the cooling curve over the first second or so. Estimate how fast the surface temperature changes with the semi-infinite solution first, then give the coefficient as a function of temperature; the gap to measurements closes much more easily.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Using the semi-infinite solution after the penetration depth exceeds the thickness	The back face is ignored and temperatures are wrong	Check $Fo = \alpha t/L^2$; if large, use the plane-wall solution or a numerical model
Confusing conductivity $k$ with diffusivity $\alpha$	Time scales off by orders of magnitude	Compute $\alpha = k/(\rho c)$ each time and check it is in m²/s
Surface elements too coarse	Early surface heat flux badly underestimated	Put several elements inside the early penetration depth
Assuming perfect contact at an interface	Contact resistance changes the real interface temperature	Give a contact conductance based on roughness and pressure
Fixing properties at room-temperature values	Large errors for big temperature differences	Use the analytical solution for estimates and temperature-dependent FEM for detail

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I'd like to see how the temperature profile spreads as I change the parameters.

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With the finite-difference heat conduction simulator you can step the surface temperature and watch the profile spread as $\sqrt{t}$; the transient heat conduction simulator shows the same in two dimensions. If the body is thin enough to ignore internal gradients, use the Biot number simulator and the lumped capacitance method; once the penetration depth reaches the thickness, switch to the Heisler chart calculator and Heisler charts. Surface temperatures that vary periodically are covered in periodic heating.

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