4.4 SOLID-STATE DIFFUSION
4.4.7 Fick’s Second Law
Thus far we have considered the fundamentals of diffusion. In particular, the effects of point defects on atom movements in solids have been studied, and Fick’s first law has been developed for the rate of mass transport in a constant composition gradient. In the following sections some important engineering examples of diffusion are described in quantitative detail.
4.4.7 Fick’s Second Law
Recall that Fick’s first law assumes that the concentration gradient is independent of time.
During many commercially important diffusion processes, however, the concentration of the diffusing species within the host solid is changing with time at any given location.
Thus, we need to extend our mathematical model to account for the changes in concentra-tion with time (i.e., we must include the term ⭸C兾⭸t).
The rate at which the concentration of an atomic species varies with time and position can be developed by use of Fick’s first law. Consider the geometry shown in Figure 4.4–9.
How does the number of “A” atoms within a thin slice of material of thickness dx and area dA change with time? The rate of change of the number of atoms in the slice is equal to the rate that atoms are entering the slice minus the rate that atoms are leaving the slice.
Since concentration is just number divided by volume, and flux is rate divided by area,
FIGURE 4.4–9 Diffusion through a solid bar. The derivation of Fick’s second law involves the relationship between the time rate of change in the concentration of the diffusing species in the volume dV and the fluxes into and out of this differential volume.
we can write the equation4:
冉⭸C⭸t冊dV ⫽ 共Jin⫺ Jout兲 dA (4.4–4)
Recognizing that dV⫽ dA · dx and rearranging yields:
⭸C
⭸t ⫽Jin⫺ Jout
⭸x ⫽ ⫺⭸J
⭸x (4.4–5)
Substituting the expression for flux given in Fick’s first law,共J ⫽ ⫺D共dC兾dx兲兲, yields:
⭸C
⭸t ⫽ D冉⭸⭸x2C2冊 (4.4–6a)
where we have made the reasonable assumption that the diffusion coefficient is indepen-dent of position, which is equivalent to saying that the diffusion coefficient does not depend on composition. Equation 4.4–6a is known as Fick’s second lawof diffusion.
This equation is quite general for diffusion taking place in one direction and can be generalized for diffusion in three dimensions.
Just as there is a heat transfer analogy for Fick’s first law of diffusion, there is an exact heat transfer analogy for Fick’s second law of diffusion. The change in temperature for non-steady-state conditions is governed by the equation:
⭸T
⭸t ⫽ Dth冉⭸⭸x2T2冊 (4.4–6b)
where Dthis the thermal diffusivity. This equation is known as the heat equation.
Solving Fick’s second law yields solutions in which the concentration is a function of position and time, that is to say, C⫽ C共x, t兲. The form of the solution to Equation 4.4–6a depends on the geometry of the problem (more precisely, on the initial and boundary conditions for the differential equation). For diffusion into a thick plate (see Fig-ure 4.4–10), the boundary conditions are that the concentration at the surface, Cs, and the initial bulk concentration, C0, are constants. A constant concentration of C at the surface can be maintained through a balance between the rate at which C is deposited at the surface and the rate at which C diffuses into the base metal. Under the condition of a continuously replenished source, solution of Fick’s second law yields:
C共x, t兲 ⫺ C0
Cs⫺ C0
⫽1 ⫺erf冉2兹Dtx 冊 (4.4–7a)
where C共x, t兲 is the concentration at position x at time t. The term erf refers to a mathematical function called the error function. Values of the error function are given in Table 4.4–2 and in Figure 4.4–11. On the other hand, if the supply of the diffusing species ( grams per cm2) is consumed (i.e., not continuously replenished), as is the case for doping in the semiconductor industry, then the solution has the form:
C共x, t兲 ⫽冉2兹Dt冊exp冉⫺4Dtx2冊 (4.4–7b)
4Please note that in this discussion partial derivatives must be used since the concentration is a func-tion of the variables, x and t.
Cs Carbon concentration at the surface
Co Initial bulk concentration in Fe sample
C(x, t) Carbon concentration at a distance x below the surface at time t
xeff Co = 0.2
Position x
Carbon concentration
Cs = 0.4 Carbon atmosphere (CO/CO2 gas mixture)
x
Fe slab
From these examples we see that the concentration depends on both location and time, and that the exact form depends on the initial conditions and geometry.
As mentioned earlier in this chapter, carburizing of steel is frequently done to produce surfaces that are wear-resistant. This process is used extensively and is perhaps one of the most important examples of diffusion. The goal of gas carburizing is to produce a concentration profile of Cthat will yield the desired surface properties. Modeling this process requires the use of Fick’s second law, as illustrated in the following example.
FIGURE 4.4–10 Diffusion of C into a large (i.e., semi-infinite) slab of Fe. Schematic concentration profile of C as a function of position at a particular instant in the carburizing cycle is also indicated.
TABLE 4.4–2 Selected values of the error function.
z erf(z) z erf(z)
0 0 0.55 0.563
0.05 0.056 0.60 0.604
0.10 0.113 0.65 0.642
0.15 0.168 0.70 0.678
0.20 0.223 0.75 0.711
0.25 0.276 0.80 0.742
0.30 0.329 0.85 0.771
0.35 0.379 0.90 0.797
0.40 0.428 0.95 0.821
0.45 0.476 1.00 0.843
0.50 0.521 1.05 0.862
0.0 0.2 0.4 0.6 0.8 1 z
0.0 0.2 0.4 0.3
0.1 0.5 0.6 0.7 0.8 0.9
erf (z)
...
EXAMPLE 4.4–7
Determine the time it takes to obtain a carbon concentration of 0.24% at a depth 0.01 cm beneath the surface of an iron bar when carburizing at 1273 K. The initialconcentration of carbon in the iron bar is 0.20% and the surface concentration in a CO/CO2gas environment is maintained at 0.40%. In this example, the Fe has the FCC structure and from Table 4.4–1 the diffusion coefficient is D⫽ 共2 ⫻ 10⫺5m2/s兲兵exp关⫺(142,000 J/mol)兾RT兴其.
Solution
The first step is to obtain the value of the diffusion coefficient at the temperature of interest. Direct substitution into the given equation yields:
D共1273 K兲 ⫽ 2.98 ⫻ 10⫺11m2/s⫽ 2.98 ⫻ 10⫺7cm2/s
Equation 4.4–7a may be used to solve this problem, since the boundary conditions are satisfied. The left side of Equation 4.4–7a is equal to 0.2. From this, we find using simple algebra that the value of the error function is 0.8. Using Table 4.4–2 (or Figure 4.4–11), we find that when the error function is 0.8, then the argument is approximately 0.90, that is,
x
2兹Dt⫽ 0.9
In the above equation all terms are known except for t, the time. Solving for time gives:
t⫽兵x兾关2共0.9兲兴其2
D ⫽ 共0.01兾1.8兲2 2.98⫻ 10⫺7
⫽ 104 s ⫽ 1.73 min
Thus it will take just 1.73 minutes for the carbon concentration to reach 0.24% at a depth of 0.01 cm beneath the surface when carburizing at 1273 K.
...
EXAMPLE 4.4–8
The strength of many oxide glasses can be improved tremendously by exchanging larger ions for smaller ions in the surface of the glass, a process called ion stuffing. For example, immersion of a lithia-alumina-silicate glass body in molten potassium nitrate (KNO3) at about 500⬚C for four hours FIGURE 4.4–11 Graph of the error function.
allows the potassium ions to replace the smaller lithium ions, putting the surface in compression.
Composition profiles show that the concentration of potassium at a depth of 20m increases from 0 to about 10 weight percent. During immersion in the salt the surface concentration of potassium is about 16 weight percent. Estimate the effective diffusion coefficient.
Solution
This is a Fick’s second law problem with solution C共x, t兲 ⫺ C0
Cs⫺ C0
⫽ 1 ⫺ erf冉2兹Dtx 冊
The initial concentration of potassium in the glass is 0 weight percent (wt. %) so that C0⫽ 0. The surface concentration is maintained at Cs⫽ 16 wt. %. After four hours (t ⫽ 14,400 s) the concen-tration at a depth of 20 m 共x ⫽ 20 ⫻ 10⫺6 m兲 is 10 wt. %, so that C共x, t兲 ⫽ 10 wt. %.
Rearranging the above equation to solve for the erf yields:
erf冉2兹Dtx 冊⫽ 1 ⫺C共x, t兲 ⫺ C0
Cs⫺ C0
⫽ 1 ⫺10⫺ 0
16⫺ 0⫽ 0.625
Linear interpolation between the values for z⫽ 0.60 and z ⫽ 0.65 in Table 4.4–2 shows that for erf共z兲 ⫽ 0.625, z ⫽ 0.6277. Thus,
x
2兹Dt⫽ 0.6277 Solving this equation for D yields:
D⫽关共x兾2兲兾0.6277兴2 t
⫽兵关共20 ⫻ 10⫺6m兲兾2兴兾0.6277其2 14,400 s
⫽ 1.76 ⫻ 10⫺14m2/s⫽ 1.76 ⫻ 10⫺10cm2/s
...
In the preceding examples we have carried out rather precise calculations. However, engineers frequently find it necessary to make good estimates using less complicated approaches. One of the most useful approximations when dealing with diffusion involves the concept of an effective penetration distance. Let us define theeffective penetration distanceas the point at which the concentration of the diffusing species has a value equal to the average of the initial concentration and the surface concentration:
C共xeff, t兲 ⫽ C0⫹ Cs
2 (4.4–8)
Substituting this value into Equation 4.4–7 yields the result:
0.5⫽ erf冉2兹Dtxeff 冊 (4.4–9)
From Figure 4.4–11 we find that erf (0.5)⬇ 0.5. Therefore, we have
xeff⬇ 兹Dt (4.4–10)
The utility of this expression is that for a given effective penetration distance, it provides a simple method for determining diffusion times and temperatures (temperatures are
determined through the diffusion coefficient). Equation 4.4–10 can be generalized to most diffusion problems with the introduction of a geometry-dependent constant:
xeff⫽␥兹Dt (4.4–11)
The constant␥is 1 for a plate geometry and 2 for cylinders.
...
EXAMPLE 4.4–9
Suppose that an effective diffusion distance of 0.05 cm is required for carburizing a steel whose initial bulk concentration is 0.04% carbon. Suppose also that because of economic and time constraints, the time in the furnace cannot exceed one hour. Estimate the temperature at which the process must be carried out, assuming that the carburizing atmosphere is such that a surface concentration of 0.4% is maintained. Further assume that the parts being carburized are cylinders 5 cm in diameter.
Solution
We may use Equation 4.4–11 with a␥ value of 2 (since this is the value corresponding to a cylindrical geometry) and xeff⫽ 0.05 cm. Solving the equation
xeff⫽␥ 兹Dt
for the diffusion coefficient yields:
D⫽冉x␥eff冊2冉1t冊
⫽冉0.05 cm2 冊2冋共1 h兲共3600 s兾1 h兲1 册
⫽ 1.736 ⫻ 10⫺7cm2/s⫽ 1.736 ⫻ 10⫺11m2/s From Table 4.4–1,
D⫽ 共2 ⫻ 10⫺5cm2/s兲 exp冉⫺142,000 J/mol
RT 冊
Solving this expression for T and substituting the appropriate value for D yields:
T ⫽ ⫺142,000 J/mol
8.314 J/mol-K冒ln冉1.7362⫻ 10⫻ 10⫺5⫺11m2m/s2/s冊
⫽ 1224 K ⫽ 951⬚C
...
It is worth noting that diffusion problems can be solved by analogy to heat transfer.
Solutions to the heat equation (Equation 4.4–6b) can be and are applied to diffusion situations with appropriate modifications. Thus whole classes of solutions and techniques are readily available for solving diffusion problems.
SUMMARY
...
In this chapter we have seen that point defects occur in all crystalline engineering materials. Although there are general rules that guide the formation of defects in all crystals, it is important to remember the additional requirement of electroneutrality in ionic crystals.
Point defects such as vacancies and interstitials are thermodynamically stable and usually have a significant influence on the mechanical, chemical, and electrical properties.
Their equilibrium concentrations can be determined using an analysis of the competing effects of changes in internal energy and configurational entropy with variations in the point-defect concentration.
We have seen that vacancies play an important role in diffusion of substitutional solid solutions, since atom movement occurs by atoms and vacancies exchanging positions. By considering random atomic jumps, it was demonstrated that diffusion will occur down a concentration gradient. Under steady-state conditions, the net flux of atoms is given by Fick’s first law. In non-steady-state conditions, diffusion is governed by Fick’s second law. The diffusion coefficient, knowledge of which is essential for solving problems involving mass transport, can be expressed in terms of a structure-related constant and an activation energy. Since diffusion is thermally activated, the diffusion coefficient in-creases exponentially with temperature. Diffusion in polymer melts is somewhat differ-ent. Polymer molecules diffuse or move most easily along their contour length. Movement in other directions is strongly hindered by a multiplicity of entanglements.
Impurities exist in all solids. Their locations within the crystal depend on several factors, including the size, crystal structure, and bond characteristics of both the solute and solvent atoms. The impurities can occupy interstitial positions if they are much smaller than the solvent atoms, or they can substitute for solvent atoms if they satisfy the Hume-Rothery rules. Impurities can interact with other defects so as to profoundly influence the engineering properties. Thermodynamic arguments show that it is impossi-ble to obtain engineering quantities of perfectly pure materials.
The fundamental point of this chapter is that defects and impurities are always present in engineering materials and the properties of such materials can be understood largely in terms of the defect state. Through intelligent use of defects and impurities, materials can be engineered to achieve combinations of desirable properties.
KEY TERMS
...
diffusion Fick’s second law interstitial solute
diffusion coefficient Frenkel defect interstitial solid solution solvent
effective penetration Hume-Rothery rules point defect substitutional solid solution
distance impurities Schottky defect vacancy
Fick’s first law impurity diffusion self-diffusion
HOMEWORK PROBLEMS
...
1. Provide an example of a material property that is dramatically affected by the presence of a small concentration of defects.
2. In a certain crystalline material, the vacancy concentration at 35⬚C is twice that at 25⬚C. At what temperature would the vacancy concentration be one-half that at 25⬚C?
关Note: An alternative statement of the problem is, Given Cv共35⬚C兲 ⫽ 2Cv共25⬚C兲, find T such that Cv共T兲 ⫽ 0.5Cv共25⬚C兲.兴
3. In a certain crystalline material the vacancy concentration at 25⬚C is one-fourth that at 80⬚C. At what temperature would the vacancy concentration be 3 times that at 80⬚C?
4. When Ge is the solvent and N is the solute, an interstitial solid solution results. When Ge and Si are mixed in any proportion, a substitutional solid solution results. Using only the information given in this problem statement, determine what type of solid solu-tion will result when N is the solvent and Si is the solute. Explain your answer.
SECTION 4.2 Point Defects
SECTION 4.3 Impurities
5. Al-Zn alloys are used as corrosion-resistant coatings on steel sheets. In the solid state, Al and Zn form a metallic solid solution. Are the two elements completely soluble in this solid solution?
6. Using the Hume-Rothery rules for the formation of substitutional solid solutions, indi-cate whether the following systems would be expected to exhibit extensive solid solubility:
a. Al in Ni b. Ti in Ni c.Zn in Fe d. Si in Al e. Li in Al f. Cu in Au g. Mn in Fe h. Cr in Fe i. Ni in Fe
7. Suppose one attempts to add a small amount of Ni to Cu in an effort to create a solid solution.
a. Which element is the solvent and which is the solute?
b. Do you think the resulting solid solution (S.S.) will be a substitutional S.S. or an interstitial S.S. Why?
c. Do you think Ni and Cu will be completely soluble in each other? Why or why not?
8. Cu can be used to increase the strength of Al. How does Cu dissolv e in Al?
9. What type of solid solution is likely to be formed when C is added to Fe?
10. In the text, we stated that the equilibrium concentration of C in Fe is 0.022% at 727⬚C.
Calculate Qfv.
11. Use a sketch to show both a Schottky defect and a Frenkel defect in the MgF2structure.
12. Explain why the following statement is incorrect: In ionic solids, the number of cation vacancies is equal to the number of anion vacancies.
13. What type of defects and how many of them are likely to be created when 2 moles of NiO are added to 98 moles of SiO2? Assume the concentration of interstitials in the system is low enough to be neglected.
14. What type of defects and how many of them are likely to be created when 1 mole of MgO is added to 99 moles of Al2O3? Assume the concentration of interstitials in the system is low enough to be neglected.
15. An ionic solid contains cation vacancies, anion vacancies, and cation interstitials. The energy of formation of the cation vacancies is 20 kJ/mol. For anion vacancies it is 40 kJ/mol, and for the cation interstitials it is 30 kJ/mol. Compute the relative con-centrations of all of the defect species. Assume that the cations and anions are monovalent.
16. Consider the ionic compound UO2.
a. Describe what a Schottky defect would look like in this compound.
b. Would you expect to find more cation or anion Frenkel defects in this compound?
Why?
17. Consider the ionic compound Li2O.
a. Describe what a Schottky defect would look like in this compound.
b. Describe what a Frenkel defect would look like in this compound.
18. Consider the ionic compound Na2O关with r共Na⫹兲 ⫽ 0.098 nm and r共O2⫺兲 ⫽ 0.132 nm兴.
Suppose a small sample of this material contains one Schottky defect and one Frenkel defect. Determine the most likely number of each of the following types of point defects:
a. Interstitial Na⫹ions b. Interstitial O2⫺ions c. Vacant Na⫹sites d. Vacant O2⫺sites
19. Consider the possibility of solid solutions with Au acting as the solvent.
a. Which element (N, Ag, or Cs) is most likely to form an interstitial solid solution with Au?
b. Which element (N, Ag, or Cs) is most likely to form a substitutional solid solution with Au?
20. Under what condition can Fick’s first law be used to solve diffusion problems?
21. Suppose that 1 weight percent of B is added to Fe.
a. Would the B be present as an interstitial or a substitutional impurity?
b. Compute the fraction of sites (either interstitial or substitutional) occupied bythe B atoms.
c. If the Fe containing the B were to be gas-carburized, would the process be faster or slower than for Fe that has no B? Explain.
22. Which type of diffusion do you think will be easier (have a lower activation energy)?
a. C in HCP Ti b. N in BCC Ti c. Ti in BCC Ti Explain your choice.
23. At one instant in time there is 0.19 atomic % Cu at the surface of Al and 0.18 atomic
% Cu at a depth of 1.2 mm below the surface. The diffusion coefficient of Cu in Al is 4⫻ 10⫺14m2/s at the temperature of interest. The lattice parameter of FCC Al is 4.049`.What is the flux of Cu atoms from the surface to the interior?
24. SiO2can be either a glass or a crystalline solid at room temperature. Is diffusion faster in vitreous (glassy) silica or crystalline silica? Why?
25. Most textile fibers are semicrystalline. That is, they contain both crystalline and non-crystalline regions. Do you expect dyes to penetrate faster in the non-crystalline or in the noncrystalline regions? (In fact, analysis of dyed fibers shows the dye has penetrated onlyone type of region.)
26. Bywhat factor does the diffusion coefficient of Al in Al2O3change when the tempera-ture is increased from 1800 to 2000⬚C?
27. Consider the diffusion of C into Fe. At approximatelywhat temperature would a speci-men of Fe have to be carburized for 2 hours to produce the same diffusion result as at 900⬚C for 15 hours?
28. Helium gas is stored at 20⬚C under a pressure of 5 ⫻ 105Pa in a glass cylinder 50 cm in diameter and 10 cm long. The glass walls are 3 mm thick. What is the initial rate of mass loss through the cylinder walls? (The diffusion coefficient of He through oxide glass at 20⬚C is about 4 ⫻ 10⫺4m2/s and the concentration of He on the inner surface is about 2.2⫻ 10⫺3kmol/m3.)
29. When food is wrapped in polyethylene, a nonpolar polymer, and placed in the freezer, the amount of ice within the package increases with time in the freezer. This observa-tion is not true for Saran Wrap, which is made using a polymer that contains highly
29. When food is wrapped in polyethylene, a nonpolar polymer, and placed in the freezer, the amount of ice within the package increases with time in the freezer. This observa-tion is not true for Saran Wrap, which is made using a polymer that contains highly