Module 17 Diffusion in solids III Lecture 17 Diffusion in solids III 1 NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
Keywords: Numerical problems in diffusion, correlation between melting point and self diffusion coefficient, estimation of inter diffusion coefficients from concentration profile, Introduction In the last two modules the mechanism of diffusion in solids has been introduced. It is a solid state mixing process where vacancy plays an important role. It is governed by a set of laws. We have looked at the nature of its solution under different initial and boundary conditions. However in order to appreciate its importance and have a deeper insight it is necessary that we try to solve a few numerical problems. In this module we therefore shall look at a few problems and their solutions. Problem 1: Melting point, crystal structure, and diffusivity in terms of D & Q (molar activation energy) are given in the following table. Estimate D at a given temperature plot D versus melting point. What do you conclude from this? FCC mp C D m 2 /sec Q kj/mole Cu 183 2.E 5 197 Ag 961 4.4E 5 185 Pb 327 1.4E 4 19 Al 66 1.74E 4 142 Ni 1455 1.9E 4 284 BCC mp K D m 2 /sec Q kj/mole W 3683 5.6E 4 585 V 2173 3.6E 5 38 Cr 2163 2.8E 5 36 Nb 2741 1.1E 4 41 Mo 2883 5.E 5 413 Fe 181 2.E 4 251 This problem illustrates the effect of melting point and crystal structure of self diffusion coefficient at any given temperature. From the previous lectures we may recall that diffusivity at a given temperature T is given by. The two tables give the values of D & Q for a few metals having FCC a few others having BCC structure. Remember that R is the universal gas constant. It is equal to 8.31J/mole/ K. Copy the values to an excel sheet and calculate diffusivity at any given temperature (say 1 C). Do not forget to convert temperature to K. The melting points of FCC metals are given in C. 2 D cm 2 /sec 1E-15 1E-3 1E-45 1E-6 1E-75 1E-9 1 2 3 4 mp deg C FCC BCC Fig 1: Shows the effect of crystal structure & melting point of metals on its self diffusion coefficient. The diffusivity at a given temperature for all metals in the two tables has been plotted against their melting points in C. The data for a given crystal structure follows a linear trend. Higher the melting Q kj/mole 8 6 4 2 FCC BCC 1 2 3 4 mp deg C Fig 2: Shows the effect of crystal structure & melting point of metals on its activation energy for self diffusion. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
point lower is its diffusivity. Most of the metals included in table for FCC structure have relatively low melting point. Nevertheless if you extend the plot for FCC metal to higher temperature you find that diffusivity in BCC structure is higher than that in FCC structure. If you look at fig 2 you find that activation energy increases with melting point for both BCC & FCC metals. Here too the trend is linear. Note that the slopes of the two plots are nearly the same. This shows that in general activation energy for self diffusion of FCC structure is higher than that of BCC metals. We know that packing density of BCC metals is a little lower than that of FCC metals. It has more vacant space therefore thermal activation needed for diffusion is lower. Consequently diffusivity of BCC structure is higher than that of FCC structure. This also suggests if you need diffusivity data for a metal and you do not have access to diffusivity data base for all metals you could use plots as above to find diffusivity if its melting point is known. Problem 2: In diffusion experiment the markers between two metals A & B moves through a distance of.144 cm in 2 hrs with respect to the Matano interface. If inter diffusion coefficient for atom fraction A =.4 is 1 7 cm 2 /sec and the slope of the concentration profile at this point is 2. /cm estimate intrinsic diffusivities of A & B..144 cm Fig 3: A sketch to illustrate how the diffusion couple with markers looks like and the concentration of B in atom fraction as a function of distance at a time t. N B.4 Slope = 2. t = 2 hr x Marker velocity. 21/ 2.21 11 cm 2 /sec (1) 11 cm 2 /sec (2) 1 (3) Since.6 &.4 are known; it is possible to solve equation 1 & equation 2. 1.41 & 9.41 In such problems you need to pay proper attention to the direction in which the marker moves. If it shifts along the positive direction velocity is positive. If it moves in the opposite direction it is negative. In this case D B > D A therefore more number of B atoms move towards left than that of A towards right. 3 Problem 3: The concentration profile in a diffusion experiment conducted couple of consisting of metal A & B is given in the following Fig 4 as a function of distance measured from the Matano interface after t = 5hrs. Calculate inter diffusion coefficient as a function of composition. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
The aim of this example is to illustrate how diffusivity can be estimated from composition versus distance plot at a given time. The first step is to find the location of the Matano interface. If there is no void formation it refers to the original interface between the two metals A & B. 1 d e composition 8 6 4 c Fig 4: Concentration of B as a function of distance one end of the diffusion couple Slide 1 illustrates the procedure for the evaluation of diffusivity from the concentration versus distance plot. Often the composition is reported in weight percent. However it would be more appropriate to convert it into atom fraction. Let us assume the atomic weights of A & B are nearly the same. In such a case weight fraction and atom fraction are approximately the same. Problem 3 : Solution method 1 a -.3 2 a b..2.4.6.8 1. c d e b.5 distance from one end Slide 1: This explains the method to be followed to estimate diffusivity. We need to find the area under the plot defined by. This is done either graphically or numerically if the values could be read from the graph. We also need to find the slope of the plot at a given value of c as shown. If these two are known diffusivity can be calculated since time t is known. The distance x is to be measured from the Matano plane. This divides the plot into two parts having the same area. A look at fig 4 suggests that it could be given by the dotted line. 4 The plot in fig 4 can be divided into two parts of equal area. In this case the area enclosed by abc is nearly same as that within cde. Therefore the dotted line may be taken to the Matano plane. With respect to this find the distances at which the compositions are given. Those who are conversant with spreadsheet can do this easily by entering the composition as a function of distance. Table 1 shows how a spreadsheet would look like. Note that c is the same for all the elemental strips whose areas are to be estimated. Here x denotes the length of the strip along x axis. Find the average values of x for each of the strips. These are entered as xav in table 1. The product gives the area of a strip. Table 1 NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
shows areas of each of these strips separately. Note that some of these are positive and some are negative. If the location of Matano plane is correctly entered the sum total area should be zero. In this case the location of this plane is.393. The column after area in table 1 gives the total areas on the two sides of Matano plane. Since these two are equal it may be concluded that it is the correct location of the Matano plane. Figure 5 gives the concentration profile with respect to the distance from the Matano plane. Matano Plane Table 1: Spread sheet table illustrating how to estimate diffusivity at different concentrations.393 dist. x % comp c xav area x cx mod area sum area D.9.51 1 6.25.41 2.56.19 32.22-2.56.71.31 93.75 6.25.25 1.58.12 51.65-1.58 2.56 1.37E-7.59.19 87.5 6.25.15.92.9 69.44 -.92 4.13 1.65E-7.5.1 81.25 6.25.8.48.5 12.19 -.48 5.5 1.17E-7.44.5 75 6.25.3.21.3 189.39 -.21 5.53 8.11E-8.41.2 68.75 6.25..3 5.77.3 25. -.3 5.74 6.38E-8.39 -.1 62.5 6.25 -.2 -.11.2 312.5.11 5.77 5.13E-8.37 -.3 56.25 6.25 -.3 -.21.1 52.83.21 5.65 3.2E-8.35 -.4 5 6.25 -.5 -.29.1 48.77.29 5.44 3.14E-8.34 -.5 43.75 6.25 -.6 -.36.1 625..36 5.15 2.29E-8.33 -.6 37.5 6.25 -.7 -.43.1 625..43 4.79 2.13E-8.32 -.7 31.25 6.25 -.8 -.5.2 416.67.5 4.36 2.91E-8.31 -.9 25 6.25 -.1 -.61.2 312.5.61 3.86 3.43E-8.29 -.11 18.75 6.25 -.12 -.76.3 223.21.76 3.25 4.4E-8.26 -.14 12.5 6.25 -.16-1..5 132.98 1. 2.48 5.19E-8.21 -.18 6.25 6.25 -.24-1.49-5.77.11 56.82 1.49 1.49 7.27E-8.1 -.29 1 8 composition 6 4 2 -.4 -.2..2.4.6.8 Distance from Matano plane 5 Fig 5: Composition of B as a function of distance from the Matano plane. We also need to find the slope as a function of concentration. The column c/x gives the slope at different values of c. This is obtained by dividing the numerical values under the respective columns. The NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
integral is evaluated by adding the areas of elemental strips in column mod area. The expression which is to be used to find diffusivity is as follows: (4) The area is to be evaluated from c= when x is negative. Therefore initially it is positive (note there is a negative sign in the equation 4) and it keeps increasing as it approaches Matano plane. This is where it attains its maximum value. Thereafter it should keep decreasing. The values under column sum area in table 1 show such a trend. Now that we know the time (t = 5hrs), integral (with sign), and the slope as function of composition(c) the diffusivity can be found using the equation 4. The values thus obtained are given in column D. Note that the slope at the both ends of the composition distance plot are nearly zero. Therefore the error in the estimation of diffusivity in these regions is likely to be high. This is therefore not suitable for the estimation of diffusivity at very low and very high concentration. Figure 6 gives diffusivity as a function of concentration in this system. 2.E-7 1.5E-7 D 1.E-7 5.E-8-1.6E-22 2 4 6 8 1 % Composition Fig 6: Shows diffusivity D as a function of composition Radioactive tracers have been used to find self diffusion coefficients. It is customary to denote tracer diffusivity of an element with asterisk sign. In this case tracer diffusivity of A & B can be denoted as &. The relation between self diffusion coefficient and intrinsic diffusivities are given by: 6 1 (5) 1 (6) NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
D A & D B are intrinsic diffusion coefficients, A & B are activity coefficients and N A & N B are atom fractions. The well known Gibbs Duhem equation gives a relation between activity coefficients of the two constituents in a binary solution. This is given by: ln ln (7) In a binary alloy the relation between atom fractions is. Therefore 1 1 1 (8) Therefore the expression for inter diffusion coefficient is given by: 1 1 (9) Or 1 (1) The inter diffusion coefficient can therefore be evaluate from self diffusion coefficient of the two components and their activity coefficient. This has been found to be valid for several systems (example Au Ni alloy where the methods give similar result till N Ni =.7). For ideal solution activity coefficient is equal to one. Therefore in such a system the following relation would be valid: (11) This shows that when N B approaches zero inter diffusion coefficient is equal to D B and at N A = ;. Diffusion is a process of mixing or alloying. If it is an ideal system where activity is equal to atom fraction the inter diffusion coefficient versus atom fraction plot should be linear. This is illustrated in slide 2. 7 Thermodynamic factor: non ideal solution D = N B D A + N A D B = N B D A +(1-N B )D B D B N B 1 d ln A D DANB DBNA1 dln NA D A Slide 2: Illustrates effect of composition on inter diffusion coefficient. In an ideal system the plot is linear. However in a non ideal system the plot is expected to deviate from linearity. Depending on the nature of deviation it may be higher or lower than the inter diffusion coefficient of an ideal system as shown in the sketch. If The problem that we solved exhibits negative deviation. In this case;. whereas if. The term 1 is the thermodynamic factor. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
Problem 4: Consider a 1D (one dimensional) lattice. Simulate diffusion as random walk along positive & negative directions with equal probability (=.5 each). Find the mean distance covered & its standard deviation as a function of time (number of steps). Compare the results with normal distribution. Fig 7: A sketch representing a one dimensional lattice. An atom at position can move forward or backward with equal probability We have seen that the movement of atoms in diffusion can be viewed as a random walk process. The purpose of this problem is to illustrate how to find the location of the atom after certain number of jumps along the line in fig 7 using this concept. In each jump it moves a unit distance in any of the two directions, forward or backward. Since both are equally likely the probability of moving forward or backward may be taken as.5. A jump may be assumed to represent a time step. Therefore it denotes both time and the total distance covered by the atom. However if you find its location after t jumps and measure its distance along the line you would notice that it is much less than t. We would try to estimate this. Such problems can easily be simulated using electronic spread sheet such as excel. It has a random number generator. This can be used to decide whether the atom from its present location would move forward or backward. Repeat the process for a large number of times and find its final location. Each movement of atom is one step. Let the process be repeated for a set of n atoms considering their movements to be independent of each other. Slide 3 illustrates how this could be setup in excel. 8 Slide 3: Describes how to setup an electronic Construction of matrix X(t,n) in Excel X(t,n) = location of atom at time t in nth trial spreadsheet to solve such a problem. It has several cells arranged in rows and columns. A B C.. Enter integers 1, 2, 3 till N in the first row. N columns represent the state of each of the N 1 1 2 3. N atoms in N sites in a 1D lattice. The second 2 row denotes initial positions of N atoms in 3-1 1-1 1 respective lattice. They are all zero. This 4 2-2 1 indicates that these are at their initial location. The third row gives the locations of N atoms after the first time step. The values in t respective cells are decided using random number generator. Each column of the spreadsheet simulates the locations of respective atoms after every time step till t. Except for the first two rows all the cells are filled up by integers representing the locations of the atoms. The second row denotes initial locations of N atoms in respective lattice. The X(t, n) in general represents the location of nth atom after time step t and X(2, 1) denotes initial position of the first atom at t=. In order to find X(3, 1), the location of the first atom in time step 1 we need to find a random NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
number. This is done in excel by RAND() a standard in built function. It is a pseudo random number generator. It generates a number and 1. If this is greater than.5 the atom would move forward (from its initial position to 1). If it is less than.5 the atom moves backward (from to 1). This is done by entering the following statement in cell A2 of the excel sheet. <cell A3> = <cell A2> + IF(RAND() >.5, 1, 1) (12) Before you do this make sure to set the excel sheet calculation mode to manual. Once this is done enter the statement in equation 12 in cell A3 and copy the same to all the cells that form a part of the matrix X(t,n). Such a statement generates an integer denoting the location of the atom at a given time step by adding either +1 or 1 to the contents of the cell in the same column but in the previous row. The decision is based on the outcome of the IF statement involving RAND() in equation 12. Enter excel function AVERAGE() in column marked mean () to calculate the average of all the values from column 1 to N in the same row. Enter excel function STDV() in the column marked to calculate standard deviation of the same data set in respective rows. Now press F9 to perform calculation. This is done row wise. The column now gives the most likely location of the atoms with respect to their initial positions. The column gives the corresponding standard deviation (see Table 2). Table 2: A typical output of an excel spread sheet 1 2 3 4 5 1 1 1 1 1.6.894427 2 2 1.414214 3 1 1 1 1.6 1.67332 4 2 2 2 1.2 2.28351 3 3 3 1 3 1 2.828427 4 4 2 2 1.6 2.67681 3 5 3 1 3 1.8 3.3315 4 6 4 2 2.4 3.286335 Table 2 gives a typical output of the spreadsheet calculation. Note that with increasing time step the mean does not change significantly. However standard deviation keeps increasing. You may try to perform this exercise for a much larger matrix to find the trend. Figure 7 gives a typical plot of mean as function of time. 9 NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
mean 1.5 -.5-1 -1.5-2 2 4 6 8 1 time Fig 7: Shows the mean position of an atom as a function of time. Mean denotes average of 25 trials. 35 3 25 stdv 2 15 1 5 2 4 6 8 1 time Fig 8: Shows standard deviation of the mean positions of atoms as a function of time steps. 1 8 6 4 2 1 2 4 6 8 1 time Fig 9: Shows that the square of standard deviation (variance) of the mean positions of atoms varies linearly with time. NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
The mean denotes average or the most probable location of the atom. This has been obtained on the basis of 25 trials. The plot in fig 7 keeps changing every time you press F9 button. It is not unexpected since the process is one of random walk. Note that the mean location after around 1 time steps varies within approximately ±1.5. This represents that in spite of such a large number of jumps the average displacement is so less. As against this the trend shown by standard deviation is extremely stable. Figure 8 gives a typical plot of standard deviation a function of time. The plot does not change significantly irrespective of the number of times you repeat the calculation by pressing F9 button. The trend is parabolic. This is why the plot 2 versus time is linear (fig 9). Nevertheless on the basis of 25 trials it may be concluded the final location after nearly 1 time steps is given by a distribution having a mean and a standard deviation. The question that comes up is what is the nature of this distribution? Excel has a function called frequency which has been used to establish this. Create a column named bin. It has numbers ranging from 9 to +9. This represents class interval. Use the frequency function of excel to count the number of times the final location after 1 time steps lies between each of these. Table 3 Typical output from excel sheet having all the data on the simulation of the 1D random walk problem. Column RAND gives predicted result. Column Norm gives normal distribution plot. Bin Rand Cum Freq Norm 9.63783.63783 7 2 3.58664 2.98228 5 1 14.92371 11.33765 3 33 44.24469 29.3298 1 53 95.8522 51.675 1 63 157.6893 61.83714 3 45 28.1356 5.44621 5 22 236.1515 28.1592 7 2 246.744 1.58888 9 2 249.4628 2.722468 25 Mean.592 stdv 31.72588 11 NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
Frequency 7 6 5 4 3 2 1 1 5 5 1 Location, x Rand Norm Fig 1: A comparison of the frequency distribution obtained as a result of 1D simulation (RAND) and the corresponding normal distribution plot (Norm) obtained on the basis of the mean position and standard deviation after 1 time step. Frequency is a standard array function in excel. It looks up an array having all the data. In this it refers to the row corresponding to say 1th time step (A11:IP11). Subsequently it divides the same into the specified class interval stored the column named BIN. Here let this be A16:A115. The statement to be entered in the column marked RAND is = {= (Frequency (A11:IP11, A16:A115)}. Note that in order to enter an array function you need to press Ctrl, Shift & Enter together. Use normal distribution function to convert the mean & standard deviation to frequency distribution as shown in Table 3 under column marked NORM. Figure 1 compares the predictions based on random walk process and those calculated using normal distribution function. The standard deviation is proportional to the square root of time. The effective depth of penetration due to diffusion is also proportional to the square root of time. In short and. The constant of proportionality is related to the diffusivity. Summary 12 In the last three modules an attempt has been made to give an elementary concept of the solid state mixing process known as diffusion. This is governed by a set of laws expressed in the form of a differential equation. We have looked a 1D solution only. However the concept can easily be extended to 3D. In this lecture we have solved a few numerical problems in detail to give a deeper insight into this process. Diffusion is associated with the random movement of atoms. It has a direct correlation with several material parameters like crystal structure, melting point & enthalpy. A first estimate of diffusivity can be obtained from such data. We have also seen that diffusivity in an alloy is a function of composition. It is related to the movement of all the species present in an alloy. We however looked at binary systems only. We have solved a problem to find the diffusivities of two species where the NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering
markers move with respect to the initial interface of the diffusion couple called the Matano plane. Finally we have tried to simulate the process of diffusion as a random walk problem. It has been shown that the standard deviation truly represents effective depth of penetration in a diffusion couple. 13 NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering