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DIFFUSION
Used To enhance Physical/Mechanical properties
DIFFUSION is involved in:
•GRAIN GROWTH (softening, magnetic, …)
•Equil # of Vacancies at a given T.
•Precipitation (to form or dissipate )
•Surface hardening
DIFFUSION (topics)
•I –MECHANISM
•II- PATH
•III- FORMULA
DIFFUSION, MECHANISM
Inter diffusion: Vacancies and Interstitial
Self Diffusion: for pure metals
https://www.youtube.com/watch?v=cXopIVWtNL8
See simulation in publisher’s website.
DIFFUSION, MECHANISMDiffusion by vacancy
Grain Growth
https://www.youtube.com/watch?v=HsVnswfHGXM
https://www.youtube.com/watch?v=8WOPLUBq4l0
https://www.youtube.com/watch?v=1Ou1bIDn1xE
Powder Metallurgy
https://www.youtube.com/watch?v=n_FW7Q2xO5o
https://www.youtube.com/watch?v=O7U4HWjYcqo
https://www.youtube.com/watch?v=5VmeIunoyKw&list=PLP-8P6M3G8FZ8Z-BqIK_yfm4UIaribT_I
Grain Growth
Time or Temperature?
Also, Metal Injection Molding, MIM
https://www.youtube.com/watch?v=SufKCjYRqh4
https://www.youtube.com/watch?v=zCfUd7ZEOjg
DIFFUSION, PATH
• Steel plate
at
700C with
geometry
shown:
• Q: How much
carbon transfers
from the rich to
the deficient side?
EX: STEADY STATE DIFFUSION
EX: STEADY STATE DIFFUSION
For diffusion of Al into Cu, D500=2.6x10 b-17 m2/S and D1000 = 1.6x10 -12 m2/s. Calculate QD and D0
For the previous example, calculate D for RT.
Steady state, example
In a H2 purification apparatus, the thickness of the container is 5 mm with a total surface area of 0.2 m2. The Hydrogen concentration outside the container is 0.6 Kg/m3 and the pressurized container has a H2 concentration of 2.4 Kg/m3. How much H2 is purified in I hr if the operation is conducted at 1000 C and D1000=1x10 -8 m2/s.
CASE 2: NON-STEADY STATE (THICK SECTIONS)
Fick’s 2nd law
Cs C0X
Cx
{Cs-Cx}/{Cs-C0} = erf [ X/(2 Dt )]
DIFFUSION , FORMULA
• Copper diffuses into a bar of aluminum.
• General solution:
Co
Cs
position, x
C(x,t)
tot1
t2t3
Adapted from
Fig. 5.5,
Callister 6e.
NON STEADY STATE DIFFUSION
Preferred equation:
Error function table
NON-STEADY, EXAMPLE 1
Ag Cu
Given D900=1.3E-13 m2/s, how long
will it take to have a concentration of
%5Ag at 1 cm from interface at 900C?
{Cs-Cx}/{Cs-C0} = erf [ X/(2 Dt )]
(100-5)/(100-0) = erf [ X/(2 Dt )]
0.95 = erf [ X/(2 Dt )]
0.9340 1.3
0.95 ?
0.9523 1.4
?=1.387
erf (1.387) = erf [ X/(2 Dt )]
1.387 = (.01m)/(2*{1.3E-13}^0.5*t^.5)
t=1E8 Sec. = 3.2 years!
NON-STEADY, Example 2
C0=0.02% CCs=1.2%C
Cx=?
In a Carburizing process, what would
be the C concentration at 0.15 cm
beneath the surface if sample was
heated at 912C for 24 hrs?
D912=1.54E-6 cm2/S
{Cs-Cx}/{Cs-C0} = erf [ X/(2 Dt )]
(1.2-Cx)/(1.2-0.05)=erf [0.15/(2*{1.5E-6*24*3600}^0.5)
(1.2-Cx)/(1.2-0.05)= erf 0.208
X=0.15 cm
.2227 .2
.? .208
.2763 .25
?=.2313=.23
(1.2-Cx)/(1.2-0.05)=0.23
Cx=0.94 %C
Example 3
How long would it take to have o.5% carbon concentration at point X at temperature T? Assume DT=1.6x10 -11 m2/s
Continued…
What is %C at X=1 mm after 20 hours at the same temperature?
MISC
Bubble raft sintering model: https://www.youtube.com/watch?v=ah1Q6yqTdpA
Isostatic Pressing : https://www.youtube.com/watch?v=p3qnsP17lYs
Hot Isostatic Pressing (HIP) : https://www.youtube.com/watch?v=BsnzgsEXT_A
Mathematical proof of Fick’s 2nd law: https://www.youtube.com/watch?v=TzsYtVpcmc8
