MSE 227
Homework Answers
Chapter 6 – Diffusion
6.5 A sheet of BCC iron 2 mm thick was exposed to a carburizing gas atmosphere on one
side and a decarburizing atmosphere on the other side at 675°C. After having reached
steady state, the iron was quickly cooled to room temperature. The carbon
concentrations at the two surfaces of the sheet were determined to be 0.015 and 0.0068
wt%. Compute the diffusion coefficient if the diffusion flux is 7.36 × 10–9 kg/m2-s. Hint:
Use Equation 5.12 to convert the concentrations from weight percent to kilograms of
carbon per cubic meter of iron.
This problem is steady state diffusion, so use Fick’s first law.
First, convert the concentration in wt% to the concentration in kg/m3 using Equation
5.12a:
0.15 wt% = 1.18 kg C/m3
0.0068 wt% = 0.535 kg C/m3
J = -D C/x for D:
D = 2.3  10-11 m2/s
6.6 Determine the carburizing time necessary to achieve a carbon concentration of 0.30
wt% at a position 4 mm into an iron–carbon alloy that initially contains 0.10 wt% C.
The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be
conducted at 1100°C. Use the diffusion data for γ-Fe in Table 6.2.
Then, solve
This is non-steady state diffusion, so use Fick’s second law. You are given Cs, Cx, Co
and T, and the data in Table 6.2. Solve the diffusion equation D = D0exp(-Qd/RT) for D
and equation 6.5 for t:
D = 5.35  10-11 m2/s
t = 1.13  105 s = 31.3 h
6.8 For a steel alloy it has been determined that a carburizing heat treatment of 15 h
duration will raise the carbon concentration to 0.35 wt% at a point 2.0 mm from the
surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm
position for an identical steel and at the same carburizing temperature.
Use the relationship x = (Dt)1/2 given x1, t1, and x2, knowing that D1 = D2
t2 = 135 h
6.12 The diffusion coefficients for carbon in nickel are given at two temperatures:
T (°C)
D (m2/s)
600
700
5.5 × 10–14
3.9 × 10–13
(a) Determine the values of D0 and Qd.
(b) What is the magnitude of D at 850°C?
Solution
(a) Solve the diffusion equation, D = D0exp(-Qd/RT) at temperatures T1 and T2 to
get Qd and D0, then use these values to solve for D850oC
a.
b.
c.
Qd = 138,300 J/mol
D0 = 1.05  10-5 m2/s
D = 3.8  10-12 m2/s
6.D2 The wear resistance of a steel gear is to be improved by hardening its surface, as
described in Design Example 6.1. However, in this case the initial carbon content of the
steel is 0.15 wt%, and a carbon content of 0.75 wt% is to be established at a position 0.65
mm below the surface. Furthermore, the surface concentration is to be maintained
constant, but may be varied between 1.2 and 1.4 wt% C. Specify an appropriate heat
treatment in terms of surface carbon concentration and time, and for a temperature
between 1000°C and 1200°C.
This problem is non-steady state diffusion, so use the equation. You are given Cs, Co and
Cx, the depth x and the temperatures 1000 < T < 1200 oC. Use Fick’s second law and a
spreadsheet to make a table of temperature/concentration/time as shown below:
Table for Cx = 1.2%
Temperature
(C)
Time
s
h
1000
1100
34,100
12,300
9.5
3.4
1200
5,100
1.4
Repeat for two other values of the surface concentration, say 1.3 wt% C and 1.4 wt% C.
Below is a tabulation of the results, again using temperatures of 1000C, 1100C, and
1200C.
Cs
Temperature
Time
(wt% C)
(C)
s
h
1.3
1000
1100
1200
26,700
9,600
4,000
7.4
2.7
1.1
1.4
1000
1100
1200
21,100
7,900
1,500
6.1
2.2
0.9