CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS

Transcription

1 CHAPTER 5 DIFFUSION PROBLEM SOLUTIONS 5.5 (a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion? (a) The driving force is that which compels a reaction to occur. (b) The driving force for steady-state diffusion is the concentration gradient.

2 5.6 The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m 2 at 500 C. Assume a diffusion coefficient of m 2 /s, that the concentrations at the highand low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steadystate conditions have been attained. This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass yields M = JAt = DAt C x = ( m 2 /s)(0.20 m kg /m3 ) (3600 s/h) m = kg/h

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 -Fe in Table 5.2. 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 Uerf(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 5.15 For a steel alloy it has been determined that a carburizing heat treatment of 10-h duration will raise the carbon concentration to 0.45 wt% at a point 2.5 mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature. This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0.45 wt% at a point 5.0 mm from the surface. From Equation 5.6b, x 2 Dt = constant But since the temperature is constant, so also is D constant, and or x 2 t = constant x2 1 = x 2 2 t 1 t 2 Thus, (2.5 mm) 2 10 h = (5.0 mm)2 t 2 from which t 2 = 40 h

6 Factors That Influence Diffusion 5.16 Cite the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900 C. Which is larger? Explain why this is the case. Table 5.2, We are asked to compute the diffusion coefficients of C in both and iron at 900 C. Using the data in D = ( m 2 80, 000 J/mol /s) exp (8.31 J/mol- K)(1173 K) = m 2 /s D = ( m 2 148, 000 J/mol /s) exp (8.31 J/mol- K)(1173 K) = m 2 /s The D for diffusion of C in BCC iron is larger, the reason being that the atomic packing factor is smaller than for FCC iron (0.68 versus 0.74 Section 3.4); this means that there is slightly more interstitial void space in the BCC Fe, and, therefore, the motion of the interstitial carbon atoms occurs more easily.

7 5.20 The activation energy for the diffusion of carbon in chromium is 111,000 J/mol. Calculate the diffusion coefficient at 1100 K (827 C), given that D at 1400 K (1127 C) is m 2 /s. To solve this problem it first becomes necessary to solve for D 0 from Equation 5.8 as D 0 = D exp Q d RT = ( m 2 111,000 J /mol /s)exp (8.31 J/mol- K)(1400 K) = m 2 /s Now, solving for D at 1100 K (again using Equation 5.8) gives D = ( m 2 111, 000 J/mol /s)exp (8.31 J/mol- K)(1100 K) = m 2 /s

8 5.21 The diffusion coefficients for iron in nickel are given at two temperatures: T (K) D (m 2 /s) (a) Determine the values of D 0 and the activation energy Q d. (b) What is the magnitude of D at 1100ºC (1373 K)? (a) Using Equation 5.9a, we set up two simultaneous equations with Q d and D 0 as unknowns as follows: ln D 1 lnd 0 Q d R 1 T 1 ln D 2 lnd 0 Q d R 1 T 2 Now, solving for Q d in terms of temperatures T 1 and T 2 (1273 K and 1473 K) 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 = 252,400 J/mol Now, solving for D 0 from Equation 5.8 (and using the 1273 K value of D) D 0 = D 1 exp Q d RT 1

9 = ( m 2 252, 400 J/mol /s)exp (8.31 J/mol- K)(1273 K) = m 2 /s (b) Using these values of D 0 and Q d, D at 1373 K is just D = ( m 2 252, 400 J/mol /s)exp (8.31 J/mol - K)(1373 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 (viz and 1473, in the first two boxes under the column labeled T (K). Next, enter the corresponding diffusion coefficient values (viz. 9.4e-16 and 2.4e-14 ). 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 252 kj/mol, respectively 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 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.