Workshop 10: Transient mass diffusion

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Attempt these problems before opening the worked solutions.

1 Carbon diffusion into steel

Time for case hardening

The surface of a mild steel component is commonly hardened by packing the component in a carbonaceous material in a furnace at a high temperature for a predetermined time. Consider such a component with a uniform initial carbon concentration of 0.15% by mass. The component is packed in a carbonaceous material and is placed in a high‐temperature furnace. The diffusion coefficient of carbon in steel at the furnace temperature is given to be 4.8 × 10−10$10^{-10}$ m2/s$\mathrm{m^2/s}$, and the equilibrium concentration of carbon in the iron at the interface is determined from equilibrium data to be 1.2% by mass. Determine how long the component should be kept in the furnace for the mass concentration of carbon 0.5 mm below the surface to reach 1%.

Worked solution

Figure 1: Carbon enters the steel through a maintained surface concentration. The target is at a depth measured inward from that surface; the far boundary remains unaffected.

Data

Quantity	Symbol	Value
Initial, surface and target mass fractions	𝑤𝑖,𝑤𝑠,𝑤$w_i,w_s,w$	0.0015,0.012,0.010$0.0015,0.012,0.010$
Target depth	𝑥$x$	0.00050 m$0.00050\ \mathrm m$
Diffusivity at furnace temperature	𝐷$D$	4.8 ×10−10 m2 s−1$4.8\times10^{-10}\ \mathrm{m^{2}\,s^{-1}}$

Assumptions

Assume constant diffusivity, no bulk motion or reaction inside each solid, uniform initial concentration and a maintained surface concentration. The surface condition changes at 𝑡 =0$t=0$. For constant density, carbon mass fraction obeys the same diffusion equation as concentration.

Calculations

Define the normalised concentration

Θ=𝐶−𝐶𝑠𝐶𝑖−𝐶𝑠.$$\Theta=\frac{C-C_s}{C_i-C_s}.$$(1)

Use the same ratio with mass fraction 𝑤$w$ replacing 𝐶$C$ in problem 1. The error function is

erf⁡(𝑧)=2√𝜋∫𝑧0𝑒−𝑢2 𝑑𝑢.$$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-u^2}\,du.$$

For interpolation, erf⁡(0.170) =0.189992$\operatorname{erf}(0.170)=0.189992$ and erf⁡(0.171) =0.191088$\operatorname{erf}(0.171)=0.191088$. A calculator with an inverse error function can also be used.

Treat the solid as a semi-infinite medium with a planar surface at 𝑥 =0$x=0$, where 𝑥$x$ increases into the steel. Other surfaces are far beyond the diffusion penetration depth during the treatment. Assume no external transport resistance and constant steel density. Use fractions 0.0015, 0.012 and 0.010, or use percentages consistently. The semi-infinite solution is

𝑤−𝑤𝑖𝑤𝑠−𝑤𝑖=1−erf⁡(𝑥2√𝐷𝑡).$$\frac{w-w_i}{w_s-w_i}=1-\operatorname{erf}\left(\frac{x}{2\sqrt{Dt}}\right).$$(2)

The steel treatment time follows from the inverse error function at the specified depth.

The required fraction of the surface concentration change is

0.010−0.00150.012−0.0015=0.809524.$$\frac{0.010-0.0015}{0.012-0.0015}=0.809524.$$

Hence Eq. 2 gives erf⁡(𝑧) =0.190476$\operatorname{erf}(z)=0.190476$, where

𝑧=𝑥2√𝐷𝑡.$$z=\frac{x}{2\sqrt{Dt}}.$$

Inversion gives 𝑧 =0.170441$z=0.170441$.

Using 𝑥 =5.0 ×10−4 m$x=5.0\times10^{-4}\,\mathrm m$,

𝑡=𝑥24𝐷𝑧2=4482.18 s=74.703 min.$$t=\frac{x^2}{4Dz^2} =4482.18\,\mathrm s=74.703\,\mathrm{min}.$$

Interpolation of the supplied table gives about 4482 s$4482\,\mathrm s$.

The source’s rounded 𝑧 =0.17$z=0.17$ gives about 4505 s$4505\,\mathrm s$.

The characteristic penetration distance 2√𝐷𝑡 =2.93 mm$2\sqrt{Dt}=2.93\,\mathrm{mm}$ must be small relative to the component dimensions for this planar semi-infinite model to apply.

The answer excludes furnace heating time.

Answer

The required diffusion time at furnace temperature is 4.48 ×103 s$4.48\times10^3\ \mathrm s$, approximately 74.7 min$74.7\ \mathrm{min}$. Furnace warm-up time is additional.

2 Hydrogen leaving a sheet

Centre concentration in a finite slab

A large sheet of material 40 mm thick contains dissolved hydrogen at a uniform concentration of 3 kmol/m3$\mathrm{kmol/m^3}$. The sheet is exposed to a fluid stream, causing the concentration of the dissolved hydrogen to be reduced suddenly to zero at both surfaces. This surface condition is maintained constant thereafter. If the mass diffusivity of hydrogen is 9 × 10−7$10^{-7}$ m2/s$\mathrm{m^2/s}$, how much time is required to bring the density of dissolved hydrogen to 1.2 kg/m3$\mathrm{kg/m^3}$ at the centre of the sheet?

Worked solution

Figure 2: Hydrogen leaves both faces of the sheet under equal surface conditions. Use half-thickness L between the centre plane and either face.

Data

Quantity	Symbol	Value
Sheet thickness and half-thickness	2𝐿,𝐿$2L,L$	0.040,0.020 m$0.040,0.020\ \mathrm m$
Initial molar concentration	𝐶𝑖$C_i$	3.0 kmol m−3$3.0\ \mathrm{kmol\,m^{-3}}$
Target centre mass concentration	𝜌𝐻2,0$\rho_{H_2,0}$	1.2 kg m−3$1.2\ \mathrm{kg\,m^{-3}}$
Surface concentration	𝐶𝑠$C_s$	0$0$
Diffusivity	𝐷$D$	9.0 ×10−7 m2 s−1$9.0\times10^{-7}\ \mathrm{m^{2}\,s^{-1}}$
Molecular hydrogen molar mass	𝑀𝐻2$M_{H_2}$	2.0 kg kmol−1$2.0\ \mathrm{kg\,kmol^{-1}}$

The worked solution uses molecular hydrogen molar mass 𝑀𝐻2 =2.0 kg kmol−1$M_{H_2}=2.0\,\mathrm{kg\,kmol^{-1}}$.

Assumptions

Assume one-dimensional diffusion normal to the faces, negligible edge effects and no external transport resistance. Set 𝑥 =0$x=0$ at the centre, with faces at 𝑥 = ±𝐿$x=\pm L$ and half-thickness 𝐿 =20 mm$L=20\,\mathrm{mm}$. Define 𝐹𝑜 =𝐷𝑡/𝐿2$Fo=Dt/L^2$.

Calculations

The exact centre concentration ratio is

Θ0=4𝜋∞∑𝑛=0(−1)𝑛2𝑛+1exp⁡[−(2𝑛+1)2𝜋2𝐹𝑜4].$$\Theta_0=\frac4\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \exp\left[-\frac{(2n+1)^2\pi^2Fo}{4}\right].$$

For 𝐹𝑜 ≳0.2$Fo\gtrsim0.2$, use the first-term approximation

Θ0≃4𝜋exp⁡(−𝜋2𝐹𝑜4).$$\Theta_0\simeq\frac4\pi\exp\left(-\frac{\pi^2Fo}{4}\right).$$(3)

Use Eq. 1 with 𝐶𝑠 =0$C_s=0$. Convert the target mass concentration to molar concentration before forming the ratio. Check that the calculated 𝐹𝑜$Fo$ is consistent with the approximation.

The sheet target is one fifth of its initial hydrogen concentration after unit conversion.

Convert the centre target to

𝐶0=1.22.0=0.60 kmol m−3.$$C_0=\frac{1.2}{2.0}=0.60\,\mathrm{kmol\,m^{-3}}.$$

Then

Θ0=0.603.0=0.20.$$\Theta_0=\frac{0.60}{3.0}=0.20.$$

Solving Eq. 3 gives

𝐹𝑜=4𝜋2ln⁡(4𝜋0.20)=0.750183.$$Fo=\frac4{\pi^2}\ln\left(\frac{\frac{4}{\pi}}{0.20}\right)=0.750183.$$

The time uses the half-thickness, not the total thickness:

𝑡=𝐹𝑜𝐿2𝐷=333.415 s=5.55691 min.$$t=Fo\frac{L^2}{D} =333.415\,\mathrm s=5.55691\,\mathrm{min}.$$

Since 𝐹𝑜 >0.2$Fo>0.2$, the first-term approximation is appropriate.

The next term at this time is about −2.47 ×10−8$-2.47\times10^{-8}$ in Θ0$\Theta_0$, so retaining it does not change the reported time.

The source chart gives 𝐹𝑜 ≈0.75$Fo\approx0.75$ and 𝑡 ≈333 s$t\approx333\,\mathrm s$.

Using the full 40 mm thickness in place of 𝐿$L$ would overestimate time by a factor of four.

Answer

The centre reaches 1.2 kg m−3$1.2\ \mathrm{kg\,m^{-3}}$ of molecular hydrogen after 333 s$333\ \mathrm s$, or 5.56 min$5.56\ \mathrm{min}$. The resulting Fourier number supports the first-term approximation.

3 Drug release from a bead

Drug release from a spherical bead

Consider a small spherical bead (D = 0.1 cm) that initially contains 0.2 mmol/L of the drug griseofulvin (species A). Determine the time it will take for the concentration of griseofulvin to reach 10% of its initial value, if its diffusivity within the bead material is 1.5 × 10−7$10^{-7}$ cm2/s$\mathrm{cm^2/s}$. Upon release from the bead, the drug is immediately consumed so that the surface concentration can be assumed to be zero.

Worked solution

Figure 3: Drug diffuses radially from an initially uniform spherical bead to a sink. The centre concentration and total amount remaining are different measures.

Data

Quantity	Symbol	Value
Diameter and radius	𝑑,𝑎$d,a$	0.10,0.050 cm$0.10,0.050\ \mathrm{cm}$
Initial concentration	𝐶𝑖$C_i$	0.20 mmol L−1$0.20\ \mathrm{mmol\,L^{-1}}$
Target centre concentration	𝐶0$C_0$	0.020 mmol L−1$0.020\ \mathrm{mmol\,L^{-1}}$
Surface concentration	𝐶𝑠$C_s$	0$0$
Diffusivity	𝐷$D$	1.5 ×10−7 cm2 s−1$1.5\times10^{-7}\ \mathrm{cm^{2}\,s^{-1}}$

The source question does not specify the location of the target concentration. This worked solution uses the centre, following the supplied solution.

Assumptions

Assume spherical symmetry, a fixed bead radius 𝑎 =𝑑/2$a=d/2$, constant temperature and no reaction inside the bead. The centre has zero radial concentration gradient; the surface at 𝑟 =𝑎$r=a$ has 𝐶𝑠 =0$C_s=0$. Define 𝐹𝑜 =𝐷𝑡/𝑎2$Fo=Dt/a^2$.

Calculations

The exact centre concentration ratio is

Θ0=2∞∑𝑛=1(−1)𝑛+1exp⁡(−𝑛2𝜋2𝐹𝑜).$$\Theta_0=2\sum_{n=1}^{\infty}(-1)^{n+1}\exp(-n^2\pi^2Fo).$$

For 𝐹𝑜 ≳0.2$Fo\gtrsim0.2$, use

Θ0≃2exp⁡(−𝜋2𝐹𝑜).$$\Theta_0\simeq2\exp(-\pi^2Fo).$$(4)

Check the approximation after computing 𝐹𝑜$Fo$. Centre concentration differs from the volume-averaged concentration. This question uses the centre, matching the coordinator’s worked solution.

The bead centre reaches the specified fraction when its Fourier number is about 0.304.

The target is 𝐶0 =0.020 mmol L−1$C_0=0.020\,\mathrm{mmol\,L^{-1}}$, giving Θ0 =0.10$\Theta_0=0.10$.

Using Eq. 4 yields

𝐹𝑜=ln⁡(20.10)𝜋2=0.303531.$$Fo=\frac{\ln\left(\frac{2}{0.10}\right)}{\pi^2}=0.303531.$$

Use

𝑎=0.050 cm=5.0×10−4 m$$a=0.050\,\mathrm{cm}=5.0\times10^{-4}\,\mathrm m$$

and 𝐷 =1.5 ×10−11 m2 s−1$D=1.5\times10^{-11}\,\mathrm{m^2\,s^{-1}}$.

Then

𝑡=𝐹𝑜𝑎2𝐷=5058.85 s=84.3142 min=1.40524 h.$$t=Fo\frac{a^2}{D} =5058.85\,\mathrm s =84.3142\,\mathrm{min}=1.40524\,\mathrm h.$$

The Fourier number exceeds 0.2.

At this time the second term of the exact centre series is −1.25 ×10−5$-1.25\times10^{-5}$, so the first-term ratio is accurate to about 0.013%.

The exact series gives about 5058.64 s$5058.64\,\mathrm s$.

The source rounds 𝐹𝑜$Fo$ to 0.3 and obtains 5000 s$5000\,\mathrm s$ or 1.39 hours.

The diameter must first be halved.

Initial concentration cancels from the ratio because this constant-property diffusion model is linear.

Answer

The bead centre reaches ten percent of its initial concentration after approximately 5.06 ×103 s$5.06\times10^3\ \mathrm s$, or 84.3 min$84.3\ \mathrm{min}$. This is a centre-concentration result, not a volume-average result.