ENGR 321: Homework Set 5
WINTER 1997

Prof. W. H. Warnes
Office: Rogers Hall 308
Phone and Voice Mail: 737-7016


This page last updated March 3, 1997

email: warnesw@engr.orst.edu


HOMEWORK ASSIGNMENT 5
DUE IN CLASS MARCH 3

REACTION RATES

1) BNT 5-2
2) BNT 5-5

DIFFUSION

3) BNT 5-6
4) Determine the carburizing time needed to achieve a carbon concentration of 0.30 wt% C at a position 4 mm into an iron-carbon alloy that initially contains 0.10 wt% C. The surface concentration will be maintained at 0.90 wt% C, and the treatment will be performed at 1100 C. The diffusion data for austenite (GAMMA-Fe) is: Do = 2.3e-5 m^2/s, and Qd = 148 kJ/mole.
5) For a steel alloy, it has been determined that a carburizing heat treatment of 15 hrs. will raise the carbon content to 0.35 wt% C at a point 2.0 mm from the surface. Estimate the time needed to produce the same concentration at 6.0 mm from the surface in an identical steel at the same carburizing temperature.

TABULATION OF ERROR FUNCTION VALUES

z

erf(z)

z

erf(z)

z

erf(z)

0

0

0.55

0.5633

1.3

0.9340

0.025

0.0282

0.60

0.6039

1.4

0.9523

0.05

0.0564

0.65

0.6420

1.5

0.9661

0.10

0.1125

0.70

0.6778

1.6

0.9763

0.15

0.1680

0.75

0.7112

1.7

0.9838

0.20

0.2227

0.80

0.7421

1.8

0.9891

0.25

0.2763

0.85

0.7707

1.9

0.9928

0.30

0.3286

0.90

0.7970

2.0

0.9953

0.35

0.3794

0.95

0.8209

2.2

0.9981

0.40

0.4284

1.0

0.8427

2.4

0.9993

0.45

0.4755

1.1

0.8802

2.6

0.9998

0.50

0.5205

1.2

0.9103

2.8

0.9999



SOLUTIONS



1)


2)


3) From the problem statement, we can see that we can use Fick's First Law (equ. 5-8) because a) the concentrations of hydrogen in the iron at the two surfaces do not change with time (steady state diffusion), and b) because we want to find the FLUX. The first step is to find the concentrations of H in Fe at the two surfaces using the equation given in the problem. Given T = 200C = 473K, and R = 1.98 cal/moleK, we find the two concentrations are:

C1 = 0.582 ppm, and C2 = 0.0584 ppm (ppm = parts per million), or C1 = (100 X 0.582e-6=) 5.82e-5 wt% H, and C2 = 5.84e-6 wt% H. First, convert the wt% compositions into at% compositions using the usual technique:

C1 = Fe 0.323 at% H, and C2 = Fe 0.0323 at% H.

We want to find these concentrations in terms of the number of moles of H per cm^3 of Fe (since the problem asks for the flux in terms of the number of moles of H/cm^2 s):

So, C1 = 0.0455 moles/cm^3, and C2 = 0.00455 moles/cm^3.

Next, we can plug into the problem equation to find the diffusivity at 200C:

D = 4.59e-5 cm^2/s.

FFL gives us J(flux) = -D dC/dx = -(4.59e-5 cm^2/s)(0.0455 moles/cm^3 - 0.00455 moles/cm^3)/0.01 cm), or

FLUX = 1.88e-4 moles/cm^2 sec.


4) Since the composition is changing as a function of time, we will need to use the Fick's Second Law equation for non-steady state diffusion. In this problem, we can apply the semi-infinite solid boundary conditions to use the solution to FSL given in class and equ. 5-14. The problem statement gives us:

Cx = 0.3 wt% C, Co = 0.10 wt% C, Cs = 0.90 wt% C;

(Cx - Co)/(Cs - Co) = (0.3 - 0.1)/(0.9 - 0.1) = 0.25 = 1-erf(Z). This gives us that the erf = 0.75. From the error function table we have that when erf(Z) = 0.75, then Z = 0.814 (interpolating from the table).


5) Since the concentrations for the two heat treatments are the same, we know that the erf(Z) is the same, and therefore Z is the same:


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