# In order to solve this problem it is first necessary to use Equation 5.5: x 2 Dt. = 1 erf. = 1.30, and x = 2 mm = m. Thus,

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1 5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Solution (a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism. (b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.

2 5.10 Show that C x = B exp x2 Dt 4 Dt is also a solution to Equation 5.4b. The parameter B is a constant, being independent of both x and t. Solution It can be shown that C x = B Dt exp x2 4 Dt is a solution to C t = D 2 C x 2 simply by taking appropriate derivatives of the C x expression. When this is carried out, C t = D 2 C x 2 = B x 2 2D 1/2 t 3/2 2Dt 1 x2 exp 4Dt

3 5.11 Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt% at a position 2 mm into an iron carbon alloy that initially contains 0.20 wt% C. The surface concentration is to be maintained at 1.30 wt% C, and the treatment is to be conducted at 1000 C. Use the diffusion data for γ-fe in Table 5.2. Solution In order to solve this problem it is first necessary to use Equation 5.5: C x C 0 C s C 0 = 1 erf x 2 Dt wherein, C x = 0.45, C 0 = 0.20, C s = 1.30, and x = 2 mm = m. Thus, C x C 0 C s C 0 = = = 1 erf x 2 Dt or x erf = = Dt By linear interpolation using data from Table 5.1 z erf(z) z z = From which z = = x 2 Dt Now, from Table 5.2, at 1000 C (1273 K)

4 D = ( m 2 148, 000 J/mol /s) exp (8.31 J/mol- K)(1273 K) Thus, = m 2 /s = m (2) ( m 2 /s) (t) Solving for t yields t = s = 19.7 h

5 5.D3 The wear resistance of a steel shaft is to be improved by hardening its surface. This is to be accomplished by increasing the nitrogen content within an outer surface layer as a result of nitrogen diffusion into the steel. The nitrogen is to be supplied from an external nitrogen-rich gas at an elevated and constant temperature. The initial nitrogen content of the steel is wt%, whereas the surface concentration is to be maintained at 0.50 wt%. For this treatment to be effective, a nitrogen content of 0.10 wt% must be established at a position 0.40 mm below the surface. Specify appropriate heat treatments in terms of temperature and time for temperatures between 475 C and 625 C. The preexponential and activation energy for the diffusion of nitrogen in iron are m 2 /s and 76,150 J/mol, respectively, over this temperature range. Solution This is a nonsteady-state diffusion situation; thus, it is necessary to employ Equation 5.5, utilizing the following values for the concentration parameters: C 0 = wt% N C s = 0.50 wt% N C x = 0.10 wt% N Therefore C x C 0 C s C 0 = x = = 1 erf 2 Dt And thus x = = erf 2 Dt Using linear interpolation and the data presented in Table 5.1 z erf (z) y

6 = y From which y = x 2 Dt = The problem stipulates that x = 0.40 mm = m. Therefore m 2 Dt = Which leads to Dt = m 2 Furthermore, the diffusion coefficient depends on temperature according to Equation 5.8; and, as stipulated in the problem statement, D 0 = m 2 /s and Q d = 76,150 J/mol. Hence Dt = D 0 exp Q d (t) = m 2 RT ( m 2 76,150 J/mol /s)exp (t) = m 2 (8.31 J/mol - K)(T) And solving for the time t t (in s) = exp T Thus, the required diffusion time may be computed for some specified temperature (in K). Below are tabulated t values for three different temperatures that lie within the range stipulated in the problem.

7 Temperature Time ( C) s h , ,

8 5.22 The diffusion coefficients for silver in copper are given at two temperatures: T ( C) D (m 2 /s) (a) Determine the values of D 0 and Q d. (b) What is the magnitude of D at 875 C? follows: Solution (a) Using Equation 5.9a, we set up two simultaneous equations with Q d and D 0 as unknowns as ln D 1 = lnd 0 Q d R 1 T 1 ln D 2 = lnd 0 Q d R 1 T 2 Solving for Q d in terms of temperatures T 1 and T 2 (923 K [650 C] and 1173 K [900 C]) and D 1 and D 2 ( and m 2 /s), we get Q d = R ln D 1 ln D 2 1 T 1 1 T 2 = (8.31 J/mol- K) [ ln ( ) ln ( )] K K = 196,700 J/mol Now, solving for D 0 from Equation 5.8 (and using the 650 C value of D)

9 D 0 = D 1 exp Q d RT 1 = ( m 2 196, 700 J/mol /s)exp (8.31 J/mol - K)(923 K) = m 2 /s (b) Using these values of D 0 and Q d, D at 1148 K (875 C) is just D = ( m 2 196, 700 J/mol /s)exp (8.31 J/mol - K)(1148 K) = m 2 /s Note: this problem may also be solved using the Diffusion module in the VMSE software. Open the Diffusion module, click on the D0 and Qd from Experimental Data submodule, and then do the following: 1. In the left-hand window that appears, enter the two temperatures from the table in the book (converted from degrees Celsius to Kelvins) (viz. 923 (650ºC) and 1173 (900ºC), in the first two boxes under the column labeled T (K). Next, enter the corresponding diffusion coefficient values (viz. 5.5e-16 and 1.3e-13 ). 3. Next, at the bottom of this window, click the Plot data button. 4. A log D versus 1/T plot then appears, with a line for the temperature dependence for this diffusion system. At the top of this window are give values for D 0 and Q d ; for this specific problem these values are m 2 /s and 196 kj/mol, respectively

10 5. To solve the (b) part of the problem we utilize the diamond-shaped cursor that is located at the top of the line on this plot. Click-and-drag this cursor down the line to the point at which the entry under the Temperature (T): label reads 1148 (i.e., 875ºC). The value of the diffusion coefficient at this temperature is given under the label Diff Coeff (D):. For our problem, this value is m 2 /s.

11 5.23 Below is shown a plot of the logarithm (to the base 10) of the diffusion coefficient versus reciprocal of the absolute temperature, for the diffusion of iron in chromium. Determine values for the activation energy and preexponential. Solution This problem asks us to determine the values of Q d and D 0 for the diffusion of Fe in Cr from the plot of log D versus 1/T. According to Equation 5.9b the slope of this plot is equal to Q d 2.3R (rather than Q d R since we are using log D rather than ln D) and the intercept at 1/T = 0 gives the value of log D 0. The slope is equal to slope = Δ (log D) Δ 1 T = log D 1 log D 2 1 T 1 1 T 2 Taking 1/T 1 and 1/T 2 as and K -1, respectively, then the corresponding values of D 1 and D 2 are and , as noted in the figure below.

12 The values of log D 1 and log D 2 are and 14.74, and therefore, Q d = 2.3 R (slope) Q d = 2.3 R log D 1 log D 2 1 T 1 1 T ( 14.74) = (2.3)(8.31 J/mol- K) ( ) K 1 = 329,000 J/mol Rather than trying to make a graphical extrapolation to determine D 0, a more accurate value is obtained analytically using Equation 5.9b taking a specific value of both D and T (from 1/T) from the plot given in the problem; for example, D = m 2 /s at T = 1626 K (1/T = K -1 ). Therefore D 0 = D exp Q d RT = ( m 2 329, 000 J/mol /s)exp (8.31 J/mol - K)(1626 K) = m 2 /s

13 7. Core 9.01

14 8. Core 9.02

15 9. (A) Conc. of Al 10E-18/cm 3 0 t

16 E (B) D = D0 exp( ) RT ln D = ln D0 + ( E / RT ) Solve this equation at both 1100 o C and 1000 o C, 5 E = J / mol Here we use E and provided diffusivity at 1300 o C to calculate D at 900 o C 16 D 900 = cm 2 / sec

17 9.1 Consider the sugar water phase diagram of Figure 9.1. (a) How much sugar will dissolve in 1500 g water at 90 C (194 F)? (b) If the saturated liquid solution in part (a) is cooled to 20 C (68 F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20 C? (c) How much of the solid sugar will come out of solution upon cooling to 20 C? Solution (a) We are asked to determine how much sugar will dissolve in 1000 g of water at 90 C. From the solubility limit curve in Figure 9.1, at 90 C the maximum concentration of sugar in the syrup is about 77 wt%. It is now possible to calculate the mass of sugar using Equation 4.3 as C sugar (wt%) = m sugar m sugar + m water wt% = m sugar m sugar g 100 Solving for m sugar yields m sugar = 5022 g (b) Again using this same plot, at 20 C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. (c) The mass of sugar in this saturated solution at 20 C (m' sugar ) may also be calculated using Equation 4.3 as follows: 64 wt% = m' sugar m' sugar g 100 which yields a value for m' sugar of 2667 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that precipitated out of the solution upon cooling m" sugar ; that is m" sugar = m sugar mõ sugar = 5022 g 2667 g = 2355 g

18 9.11 A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300 C (2370 F). (a) At what temperature does the first liquid phase form? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting? Solution Shown below is the Cu-Ni phase diagram (Figure 9.3a) and a vertical line constructed at a composition of 70 wt% Ni-30 wt% Cu. (a) Upon heating from 1300 C, the first liquid phase forms at the temperature at which this vertical line intersects the α-(α + L) phase boundary--i.e., about 1345 C. (b) The composition of this liquid phase corresponds to the intersection with the (α + L)-L phase boundary, of a tie line constructed across the α + L phase region at 1345 C--i.e., 59 wt% Ni; (c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt% Ni with the (α + L)-L phase boundary--i.e., about 1380 C;

19 (d) The composition of the last solid remaining prior to complete melting corresponds to the intersection with α-(α + L) phase boundary, of the tie line constructed across the α + L phase region at 1380 C--i.e., about 79 wt% Ni.

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