undamentals of Crystal Structure: BCC, FCC and HCP Structures, coordination number and atomic packing factors, crystal imperfections -point line and surface imperfections. Atomic Diffusion: Phenomenon, Fick’s laws of diffusion, factors affecting diffusion.

1. Introduction to Materials Science
and Engineering
Department of IE & M
JSS Academy of Technical Education, Bangalore-560060

2. Diffusion
• Materials of all types are often heat-treated to improve their properties.
• The phenomena that occur during a heat treatment almost always
involve atomic diffusion.
• The migration of atoms from their original lattice site in a crystal structure
to other sites is referred as diffusion.

3.  DIFFUSION MECHANISMS
Diffusion is just the stepwise migration of atoms from lattice site to lattice site.
For an atom to make such a move, two conditions must be met:
(1) there must be an empty adjacent site / vacancy, and
(2) the atom must have sufficient energy to break bonds with its neighbor
atoms and then cause some lattice distortion during the displacement.

4.  DIFFUSION MECHANISMS
• Vacancy Diffusion
• Interstitial Diffusion

5. • The interchange / migration of an atom from a normal lattice position to an
adjacent vacant lattice site or vacancy.
 Vacancy Diffusion

6.  Interstitial Diffusion
• Diffusion involves atoms that migrate from an interstitial position to a
neighboring position that is empty.

7. The basic mathematical expression governing atomic diffusion are known as
Fick’s laws of diffusion
Fick’s laws of diffusion
• Diffusion Flux (J), defined as the mass (or, the number of atoms) ‘M’
'diffusing through and perpendicular to a unit cross-sectional area of solid
per unit time.
In mathematical form, this may be represented as
• A - denotes the area across which diffusion is occurring
• t - is the elapsed diffusion time
• units for J = kg/m2.s or atoms/m2.s

8.  FICK’S FIRST LAW
It states that, the flux of atoms ‘J’ moving across a unit surface area in unit time is
Proportional to the concentration gradient.
Under steady state flow,
J = No. of atoms/unit area of diffusion per unit time. (kg/m2.s or atoms/m2.s)
C= Volume concentrations of atoms (atoms/m3 )
X=distance between the planes in the direction of flow of atoms (m)
D = diffusion co-efficient or diffusivity (m2/s)
-ve sign = flux moves from higher to lower concentration

9. The geometry of Fick’s I law is illustrated

10.  FICK’S SECOND LAW—NONSTEADY-STATE DIFFUSION
• This law relates to the rate of change in concentration with time.
• This non-steady state is represented by a second order differential
equation;
Rate of compositional change (rate of accumulation of diffusing atoms at a point
where the concentration gradient is dc/dx.

11. Arrhenious equation
D = D0 e(-Q/RT)
Arrhenious equation for atomic diffusion is given by;
D = Diffusion coefficient (m2/s)
D0 = constant (m2/s)
Q = Activation energy for diffusion (J/mol)
R = universal gas constant (8.314 J/mol-k)
T = absolute temperature (K)

12.  Factors affecting diffusion
1. Temperature
2. Crystal structure
3. APF
4. Grain boundaries
5. Grain size
6. Atomic Size
7. Concentration Gradient

13. List of Formulae

14. List of Formulae

15. Examples
1. Calculate the value of the diffusivity D in m2/s for the diffusion of carbon in
ɣ- iron (FCC) at 972°C. Use values D0 = 2.0x10-5 m2/s, Q = 142 KJ/mol and
R =8.314 J/mol.k
2. The diffusivity of copper atoms in FCC copper lattice is 8.0x10-21 m2/s at 400°C and
6.0x 10-15 m2/s at 800°C. calculate the activation energy in J/mol for the diffusion of
copper atoms in FCC copper lattice in this temperature range, given R=8.314 J/mol.k
3. Calculate the diffusion rate of carbon in iron at 700°C assuming the constants
A=4.9 x 10-5m2/s and E =153.2 KJ/mol.

16. 4. A steel tank in a process industry contains nitrogen at 400K and a constant pressure
of 15 atm. Vacuum exists outside the tank. Nitrogen concentration at the inner
surface of the tank is equal to 12kg/m3. The constant D0 and activation energy for the
diffusion process between nitrogen and steel are 5 x 10-7 m2/s and 75 KJ/mol.
Calculate the rate at which nitrogen escapes through the tank wall. Tank wall is 6mm
thick.
Examples

17. 5. A surface of steel gear is made of 1020 steel (0.2% C) is to be carburized at 927°C.
calculate the time required to increase the carbon content to 0.4% at 1mm
below the surface if the carbon potential at surface is 1.2 wt%.
Take erf(0.9) = 0.8
6. 0.2% C steel component is to be carburized at 920°C. calculate the time required to
Increase the carbon content to 0.4% at 0.5mm below the surface. Assume that the carbon
Content at the surface is 0.9%. Given D920°C = 1.28x10-11 m2/s.
Error function values:
Examples
Z erf Z
0.75 0.7112
0.80 0.7421

18. End of Module 5