Drop a block of clear gel into a dye bath and a colored front creeps inward. For a small dye with $D = 1.0 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$, the front reaches a depth of about 6.6 mm in the first hour, but after a full day it is only about 32 mm deep: 24 times the time makes it only about five times deeper, because the depth grows with $\sqrt{t}$ and $\sqrt{\frac{24\,\text{h}}{1\,\text{h}}} = 4.90$. The same slow, square-root-of-time spreading controls how a drug leaches out of a tablet coating, how carbon hardens the surface of a steel gear and how quickly a solvent swells a polymer film. In every case the solid is thick compared with how far the solute has traveled so far, so it behaves as a semi-infinite solid.

In this lesson you will write the diffusion equation for that case, find the concentration at any depth and time from the error function, work backward to the time needed to reach a target, add a liquid film and a partition coefficient at the surface, and check when the semi-infinite picture stops being valid (Figure 1).

1. InputsD, x, t; for a film also k_c and K
2. $\eta = \dfrac{x}{2\sqrt{Dt}}$similarity variable
3. Profile$\dfrac{c}{Kc_1} = 1-\operatorname{erf}\eta = \operatorname{erfc}\eta$erf solution, or the film solution with $k^{\prime} = \dfrac{k_c}{K}$
4. $\operatorname{erf}\eta \rightarrow \eta \rightarrow t$backward: time to reach a target
5. Check$\delta \approx \sqrt{12Dt}$still semi-infinite?

Learning Objectives

After this lesson, you should be able to:

1. Write Fick’s second law for a semi-infinite solid and use the similarity variable to relate depth and time.
2. Find the concentration at a given depth and time with the error function, and the time needed to reach a target concentration.
3. Estimate the penetration depth and decide whether a solid of finite size can be treated as semi-infinite.
4. Include a surface film and a partition coefficient in the concentration profile.

Key Ideas

1. Fick’s Second Law and the Similarity Variable

When the concentration inside a solid changes with time, a mass balance on a thin slice combined with Fick’s law gives Fick’s second law, the mass-transfer form of Perry’s Eq. 5-24 with the diffusivity $D$ in place of the thermal diffusivity:

Take a thick solid with a uniform initial concentration $c_0$. At time zero its surface ($x = 0$) is brought to a new concentration and held there, and far inside ($x \to \infty$) nothing has changed yet. For these conditions the two variables $x$ and $t$ always appear together in one group, the similarity variable:

Any two points with the same $\eta$ have the same concentration. Doubling the depth gives the same $\eta$ only after four times the time, because $x$ is divided by $\sqrt{t}$. Profiles taken at different times all fall onto one curve when they are plotted against $\eta$ (Figure 2).

$\eta = \dfrac{x}{2\sqrt{Dt}}$

$x = 1.0\,\text{mm}, \quad t = 1.0\,\text{h}$ gives $\eta = 0.2635$, so $\dfrac{c}{Kc_1} = 0.709$

$x = 2.0\,\text{mm}, \quad t = 4.0\,\text{h}$ gives the same $\eta = 0.2635$ and the same ratio

Twice the depth, four times the time: profiles at every time collapse onto one curve against the similarity variable.

Key idea: in a semi-infinite solid the concentration depends on $x$ and $t$ only through $\eta = \frac{x}{2\sqrt{Dt}}$. Twice the depth needs four times the time.

2. Surface at Equilibrium: the Error Function

Suppose the solid is placed in a well-stirred solution of solute A at concentration $c_1$, and the surface reaches equilibrium with the liquid at once. With a partition coefficient $K$ (solid concentration over liquid concentration at equilibrium), the surface concentration is $Kc_1$. The solution, the analogue of Perry’s Eq. 5-30, is

Here $Y$ is the fraction of the change that has not happened yet: it is 0 at the surface and 1 deep inside. For a solid that starts free of solute ($c_0 = 0$), the concentration ratio is therefore

The error function rises from 0 at $\eta = 0$ toward 1, and it is already 0.9953 at $\eta = 2$ (Figure 3). Most calculators and spreadsheets evaluate it directly; to go backward from a target, find the $\eta$ that gives the required value of $\operatorname{erf}\eta$ and solve $\eta = \frac{x}{2\sqrt{Dt}}$ for $t$:

$\dfrac{c}{Kc_1} = 1-\operatorname{erf}\eta = \operatorname{erfc}\eta$

$\operatorname{erf}(0.1) = 0.1125$; $\operatorname{erf}(0.2) = 0.2227$; $\operatorname{erf}(0.3) = 0.3286$; $\operatorname{erf}(0.4) = 0.4284$; $\operatorname{erf}(0.5) = 0.5205$

$\operatorname{erf}(0.6) = 0.6039$; $\operatorname{erf}(0.8) = 0.7421$; $\operatorname{erf}(1.0) = 0.8427$; $\operatorname{erf}(1.5) = 0.9661$; $\operatorname{erf}(2.0) = 0.9953$

Backward: $\operatorname{erf}\eta \rightarrow \eta \rightarrow t$, with $t = \dfrac{x^{2}}{4D\eta^{2}}$

Key idea: with the surface at equilibrium, $\frac{c}{Kc_1} = 1-\operatorname{erf}\eta = \operatorname{erfc}\eta$. Use $x$ in m, $t$ in s and $D$ in $\text{m}^{2}\,\text{s}^{-1}$ so that $\eta$ has no units.

Try it: watch the front penetrate

Choose the semi-infinite geometry and run the simulator. Watch two things: the profile keeps the same shape as it moves inward, and the time needed for the front to reach a given depth grows four times when that depth doubles. Then lower the film coefficient and see the surface concentration lag behind its equilibrium value.

Open the unsteady diffusion simulator full screen

3. Penetration Depth: When Is a Solid Semi-Infinite?

The solute front has no sharp edge, but the profile becomes very small beyond a certain depth. Perry’s p. 5-7 gives the depth to which the change penetrates in time $t$ as approximately

At $x = \delta$, $\eta = \sqrt{3} = 1.732$ and $\frac{c}{Kc_1} = \operatorname{erfc}(1.732) \approx 0.014$, so only about 1.4% of the surface value has arrived there (Figure 4). A real body of finite size behaves as a semi-infinite solid as long as $\delta$ is smaller than the distance to the next boundary. For a sheet of thickness $L$ wetted on both faces, the two fronts meet at the mid-plane, so each face behaves as a semi-infinite solid until

Integrating the error-function profile over the depth gives the amount of solute taken up per unit area of surface while the solid is still semi-infinite:

The uptake also grows as $\sqrt{t}$: the first minutes take up far more solute than any later minute.

1. Surface$\dfrac{c}{Kc_1} = 1$x = 0
2. $\eta = 0.2635$Inside: at 1.0 mm
3. Inside$\dfrac{c}{Kc_1} = 0.709$Worked Example 1(a)
4. $\delta \approx \sqrt{12Dt}$Edge: at 6.57 mm after 1.0 h
5. Penetration depth$x = \delta: \quad \eta = \sqrt{3} = 1.732$, $\dfrac{c}{Kc_1} \approx 0.014$about 1.4% of the surface value

Key idea: use $\delta \approx \sqrt{12Dt}$ to check that the solid is still semi-infinite. For a sheet wetted on both faces, the limit is reached when $\delta$ equals half the thickness.

Worked Example 1: Solute Soaking into a Thick Gel

Problem: A thick gel slab contains no solute A at first. At time zero its surface is brought to the equilibrium concentration $Kc_1$ and held there. The diffusivity of A in the gel is $D = 1.0 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$, about the size of a small solute diffusing in water. (a) Find $\frac{c}{Kc_1}$ at a depth of 1.0 mm after 1.0 h. (b) Estimate the penetration depth after 1.0 h. (c) How long does it take for the concentration at a depth of 2.0 mm to reach 50% of $Kc_1$? (d) The same gel is cast as a sheet 10 mm thick and both faces are wetted. For how long does each face behave as the surface of a semi-infinite solid?

Step 1. (a) Similarity variable with $x$ in m and $t = 3600$ s.

Step 2. (a) Error function, then subtract from 1.

Step 3. (b) Penetration depth, converted from m to mm.

Step 4. (c) Invert for the time. A target of 50% means $1-\operatorname{erf}\eta = 0.50$, so $\operatorname{erf}\eta = 0.50$ and $\eta = 0.4769$. Then solve for $t$ and convert s to h.

Step 5. (d) Validity of the semi-infinite model. Each front only has to reach the mid-plane, so $\delta = \frac{L}{2} = 5.0$ mm. Solve the penetration depth for $t$ and convert s to min.

Final Answer: (a) $\frac{c}{Kc_1} \approx 0.709$; (b) $\delta \approx 6.57\,\text{mm}$; (c) $t \approx 1.22\,\text{h}$; (d) about $34.7\,\text{min}$

Check: at 1 h the 2.0 mm point has $\eta = 0.527 \gt 0.477$, so $t \gt 1$ h; and since 50% is below the 0.709 reached there at 4 h (the same $\eta$ as part (a)), $t \lt 4$ h. The answer, 1.22 h, lies between. Part (d) shows why a 10 mm sheet soaked for 1 h is not semi-infinite: after about 34.7 min the two fronts meet and the erf solution no longer applies.

Your turn: concentration at a depth

Same steps as Worked Example 1(a): find $\eta$ with $x$ in m and $t$ in s, evaluate $\operatorname{erf}\eta$ to 4 significant figures, then subtract from 1.

Your turn: time to reach a target

Same steps as Worked Example 1(c): turn the target into a value of $\operatorname{erf}\eta$, read $\eta$ from the inverse value given in the problem, then solve $t = \frac{x^{2}}{4D\eta^{2}}$ and convert s to min.

Your turn: how long is a sheet semi-infinite?

Same steps as Worked Example 1(d): set $\delta$ to half the thickness, solve $t = \frac{\delta^{2}}{12D}$ with $\delta$ in m, and convert s to min.

4. Surface Film and Partition Coefficient

If the liquid is not stirred hard enough, a film resistance at the surface slows the transfer, and the surface concentration only approaches $Kc_1$ gradually. With a liquid film coefficient $k_c$ and a partition coefficient $K$, the flux into the solid at its surface is $k_c\left(c_1-\frac{c_s}{K}\right)$. Written in terms of the solid concentration, the film coefficient becomes $k^{\prime} = \frac{k_c}{K}$, and the solution, the analogue of Perry’s Eq. 5-31, is

The group $h$ measures how strong the film is compared with diffusion inside the solid. A very large $k_c$ makes $h$ large, the second term vanishes and the error-function solution of Key Idea 2 returns. At the surface ($x = 0$, so $\eta = 0$ and $a = 0$) the solution reduces to

The concentration inside is always lower than the equilibrium-surface answer at the same depth and time (Figure 5). Keep all the groups in SI units: $k_c$ in $\text{m}\,\text{s}^{-1}$, $D$ in $\text{m}^{2}\,\text{s}^{-1}$, $x$ in m and $t$ in s.

$\dfrac{c}{Kc_1} = \operatorname{erfc}\eta-e^{\,a + h^{2}}\operatorname{erfc}(\eta + h)$

$k^{\prime} = \dfrac{k_c}{K}, \qquad h = k^{\prime}\sqrt{\dfrac{t}{D}}, \qquad a = \dfrac{k^{\prime}x}{D}$

Surface: $\dfrac{c_s}{Kc_1} = 1-e^{\,h^{2}}\operatorname{erfc}h$

Worked Example 2, 0.80 mm after 30 min: no film $\dfrac{c}{Kc_1} = 0.731$; with the film $\dfrac{c}{Kc_1} = 0.478$

At the surface: no film $\dfrac{c_s}{Kc_1} = 1$; with the film $\dfrac{c_s}{Kc_1} = 0.700$

Key idea: with a film resistance, divide $k_c$ by $K$ first, evaluate $h$ and $a$, then subtract $e^{\,a + h^{2}}\operatorname{erfc}(\eta + h)$ from $\operatorname{erfc}\eta$.

Worked Example 2: Soaking Through a Liquid Film

Problem: A thick gel slab contains no solute A at first. At time zero it is placed in a solution of A at concentration $c_1$, with a liquid film coefficient $k_c = 3.0 \times 10^{-6}\,\text{m}\,\text{s}^{-1}$ and a partition coefficient $K = 2.0$ (gel concentration over liquid concentration). The diffusivity of A in the gel is $D = 1.5 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$. After 30 min, find (a) $\frac{c}{Kc_1}$ at a depth of 0.80 mm; (b) the same ratio if the film resistance were negligible; (c) the surface ratio $\frac{c_s}{Kc_1}$.

Step 1. Groups, with $t = 30 \times 60$ s $= 1800$ s.

Step 2. (a) The two terms.

Step 3. (a) Combine.

Step 4. (b) No film resistance. Only the first term remains: $\frac{c}{Kc_1} = \operatorname{erfc}(0.2434) = 0.7307 \approx 0.731$.

Step 5. (c) Surface, with $\eta = 0$ and $a = 0$.

Final Answer: (a) $\frac{c}{Kc_1} \approx 0.478$; (b) 0.731 without the film; (c) $\frac{c_s}{Kc_1} \approx 0.700$

Check: the film lowers the concentration at 0.80 mm by about 35%, and after 30 min the surface is still only at 0.700 of its equilibrium value. The penetration depth is $\sqrt{12Dt} = 5.69\,\text{mm}$, so the slab must be much thicker than that for the semi-infinite model to hold.

Your turn: soaking through a film

Same steps as Worked Example 2(a): find $k^{\prime} = \frac{k_c}{K}$, then $\eta$, $h$ and $a$ to 4 significant figures, and combine the two complementary error functions.

Common Mistakes

Common mistake: Swapping erf and erfc. $Y = \operatorname{erf}\eta$ is the fraction of the change still to come, not the concentration ratio. In Worked Example 1(a), $\frac{c}{Kc_1}$ is $1-0.2906 = 0.709$; reporting 0.2906 says the gel at 1.0 mm is far emptier than it is.

Common mistake: Leaving out the 2 in the similarity variable. The denominator of $\eta$ is $2\sqrt{Dt}$. Dropping the 2 doubles $\eta$ and makes the front look much shallower than it is.

Common mistake: Mixing minutes, seconds and millimeters. Put $t$ in s, $x$ in m and $D$ in $\text{m}^{2}\,\text{s}^{-1}$ before you evaluate $\eta$, $\delta$, $h$ or $a$. A formula solved for $t$ gives seconds: divide by 60 for min or by 3600 for h.

Common mistake: Misusing the penetration depth. The penetration depth is $\sqrt{12Dt}$, not $2\sqrt{Dt}$ or $\sqrt{Dt}$. For a sheet wetted on both faces, compare $\delta$ with half the thickness: using the full thickness in Worked Example 1(d) gives four times the correct time.

Common mistake: Using c₁ instead of Kc₁, or k_c instead of k′. The surface of the solid is in equilibrium with $Kc_1$, not $c_1$, and the film coefficient on the solid side is $k^{\prime} = \frac{k_c}{K}$. Multiplying by $K$ instead of dividing, or leaving $K$ out, changes $h$ and $a$ and gives a wrong profile.

Common mistake: Dropping the film term. With a film resistance, $\operatorname{erfc}\eta$ alone is wrong: it is the answer for a surface at equilibrium (0.731 instead of 0.478 in Worked Example 2). At the surface, $e^{\,h^{2}}\operatorname{erfc}h$ is the part still missing; the surface ratio is 1 minus it.

Key Takeaway

A thick solid whose surface concentration changes suddenly is a semi-infinite solid: its profile depends only on $\eta = \frac{x}{2\sqrt{Dt}}$, so twice the depth takes four times the time. With the surface at equilibrium, $\frac{c}{Kc_1} = \operatorname{erfc}\eta$; with a liquid film, use $k^{\prime} = \frac{k_c}{K}$ and subtract $e^{\,a + h^{2}}\operatorname{erfc}(\eta + h)$. The model holds while the penetration depth $\sqrt{12Dt}$ is smaller than the distance to the next boundary.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson, including penetration depths and surface concentrations with a film: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Methods: Perry’s Chemical Engineers’ Handbook, 9th edition, Section 5: the unsteady conduction equation and its semi-infinite solid solutions with and without a surface film, and the penetration depth, applied to diffusion by the heat and mass transfer analogy. The practice problems use illustrative conditions and are audited; see the lesson audit log.