diffusion chapter 8
DESCRIPTION
Diffusion Chapter 8. Selection of steel for gears. Wear resistance. 1410. 1150. 910. 725. 0.8. 0.02. Fraction of cementitite by LEVER rule:. More carbon. More cementite. More wear resistance. Cementite is hard as well as brittle. Hardness or strength is desirable. - PowerPoint PPT PresentationTRANSCRIPT
Diffusion
Chapter 8
Selection of steel for gears
Wear resistance
0.8
1410
910
1150
725
0.02
67.6067.6
0 003
CCf CF
Fraction of cementitite by LEVER rule:
More carbon
More cementite
More wear resistance
Cementite is hard as well as brittle
Hardness or strength is desirable.
But brittleness is not.
Silica Glass is also hard and brittle
Selection of steel for gears
High Carbon Steel: Good wear resistance
but
Btittle
Low Carbon steel: Good ductility
but
poor wear resistance
Wear resistance is required only at the surface
High C steel on the surface
Mild steel inside
Q: How do you achieve this?
Ans: By case carburization
Case carburization
Pack a mild steel gear in carbon and heat at a high temperature in the austenite phase field for some time.
Carbon will enter into the mild steel to give a high-carbon wear resistant surface layer called case.
How do carbon enter into solid steel?
At what temperature and how long should we do the carburization?
The process why which Carbon enters into solid steel during Case carburization is an example of DIFFUSION
Diffusion is relative movement of atoms inside a solid
We can find appropriate time and tempearture for case carburization by solution of Fick’s second law
How do we create an n-p junction in silicon chip?
Ans: by DIFFUSION
Deposit n type element
Deposit p type element
Heat
Si substrate Si substrateSi substrate
Diffusion: flow of matter
Heat: flow of thermal energy
Electric current: flow of electric charge
Different kinds of flows in material
Heat: flow of thermal energy
Fourier’s law of heat conduction (1811)
x
Tq
q: heat flux (J m-2 s-1)
x
T
Gradient?
Temperature gradient
Thermal conductivity
Joseph Fourier (1768-1830)
Electric current: flow of charge
Ohm’s law of electrical conduction
(1827)
x
VEj
j : charge flux (C m-2 s-1), current density
x
V
Gradient?
Electric potential gradient, electric field E
electrical conductivity
Georg Simon Ohm
(1787-1854)
Diffusion: flow of mass
Fick’s first law of diffusion1855
x
cDj
j : mass flux (kg m-2 s-1, moles m-2 s-1)
x
c
Gradient?
concentration gradient, kg m-4
D: Diffusivity, m2 s-1
1829-1901
Temperature dependence of Diffusivity
RT
QDD exp0
D0 = preexponential factor
Q = activation energy
Empirical constants
Self-Diffusion in Amorphous Se (Problem 8.3)
0.00305 0.00315 0.00325
-34
-32
-30
-28
ln D
1/T
T (ºC) D (m2s-1)
35 7.7 x 10-16
40 2.4 x 10-15
46 3.2 x 10-14
56 3.2 x 10-13
D0 = 2 x 1027 m2 s-
1
Q = 250 kJ mol-1
RT
QDD exp0
TR
QDD
1lnln 0
x x+x
c c+c
j j + j
Mass in at x: min = A t j
Mass out at x+ x: mout = A t (j + j)
Mass accumulation between x and x+ x
m = min-mout
= A t ( j – j - j ) = -A t j
Change in concentration in a volume V = A x and time interval t :
m = -A t j
xA
mc
x x+x
c c+c
j j + j
Average rate of change of concentration between x and x + x in time interval t:
x
j
t
c
xA
jtA
x
jt
x
j
t
c
Instantaneous change in concentration at a time t, at a point x:
x
j
t
c
t
x
t
x
0
0
0
0 limlim
x
j
t
c
Fick’s 2nd Law
x
j
t
c
x
cD
xt
c
2
2
x
cD
t
c
If D is independent of x
Fick’s 2nd law
Using Fick’s First Law
Solution to Fick’s 2nd law:
2
2
x
cD
t
c
Solution depends on the boundary condition.
tD
xBAtxc
2erf),(
A and B : constants depending on the boundary conditions
erf (z) : Gaussian error function
The Gaussian Error Function
z
dzerf0
2 )(exp2
)(
0.2
0.4
0.6
0.8
1
exp (-2)
-3 -2 -1 1 2 3
0 z
Hatched area (2/) = erf (z)
erf (0) = 0, erf (-z) = - erf (z), erf (+ ) = +1, erf (- ) = -1
-3 -2 -1 1 2 3
-1
-0.5
0.5
1
z
dzerf0
2 )(exp2
)(
z
erf (z)0.85 0.770668
0.9 0.796908
0.95 0.820891
1. 0.842701
1.1 0.880205
1.2 0.910314
1.3 0.934008
1.4 0.952285
1.5 0.966105
1.6 0.976348
1.7 0.98379
1.8 0.989091
1.9 0.99279
2. 0.995322
2.2 0.998137
2.4 0.999311
2.6 0.999764
2.8 0.999925
0. 0.
0.025 0.0282036
0.05 0.056372
0.1 0.112463
0.15 0.167996
0.2 0.222703
0.25 0.276326
0.3 0.328627
0.35 0.379382
0.4 0.428392
0.45 0.475482
0.5 0.5205
0.55 0.563323
0.6 0.603856
0.65 0.642029
0.7 0.677801
0.75 0.711156
0.8 0.742101TABLE 8.1
Mistakes in the textbook in the boxed values
Carburisation of steel
cs
Distance in steel from surface
c (wt% C)
xc0
Surface concentration
initial concentration
Concentration profile after carburization for time t at a temperature T
Boundary conditions:
1. c=c0 at x>0 , t=0
2. c=cs at x=0 , t>0
Carburising atmosphere Steel
Carburisation of steel (contd.)
tD
xBAtxc
2erf),(
Boundary conditions:
2. c(x,t) = c0 at x > 0, t = 0
1. c(x,t) = cs at x = 0, t > 0
B.C. 1 cs = A – B erf(0) = A
B.C. 2 c0 = A – B erf(+) = A-B
A = cs
B = cs – c0
tD
xccctxc ss
2erf)(),( 0
Case carburization of steel Problem 8.4
Initial concentration c0 = 0.2 wt% C
Surface concentration cs = 1.4 wt % C
Temperature = 900 ºC = 1173 K
Desired concentration c = 1.0 wt% C at x = 0.2 mmAt 900 ºC the equilibrium phase of steel is austenite ()
Diffusivity data for C in austenite:
D0 = 0.7 x 10-4 m2s-1
Q = 157 kJ mol-1
RT
QDD exp0 = 7.13688 x 10-12
m2s-1
Carburization of steels
0.1 0.2 0.3 0.4 0.5
0.2
0.4
0.6
0.8
1.0
1.2
1.4
100 s
1000 s
10000 s
Distance in steel from surface, mm
wt% C
tD
xccctxc ss
2erf)(),( 0
tD
x
2erf)2.04.1(4.10.1
3333.02.1
4.0
2.04.1
0.14.1
2erf
tD
x
z erf(z)
0.30 0.3286
0.35 0.3794
0.305 0.3333
)305.0(erf3333.02
erf
tD
x
305.02
tD
x
305.0 10 7.136882
102.012-
3
t
t = 15062 s = 4 h 11 min
This is reasonable. If not, change D by changing T
Ans
Atomic Mechanism of Diffusion
How does C enter into solid steel?
is INTERSTITIAL solid solution of C in FCC Fe
C occupies octahedral voids in FCC Fe
Maximum solubility of C in austenite () is 2.14 wt% at 1150 ºC.
C at% 9
85.5598
122
122
Cwt%2
Thus 9 out of 91 OH voids are occupied.
90 % of OH voids are empty
C atoms can jump from one interstitial site to another vacant interstitial site.
This is interstitial diffusion.
For OH voids the void size is 0.414 R but the window through which C atoms can jump outside is only 0.155 R.
Thus to jump out of an interstitial OH site the C atoms will have to displace neighbouring Fe atoms. This will increase the energy of the system
OH void
OH void
Potential energy
Hm
A carbon atom can jump to a neighbouring site if it has sufficient energy Hm. It can gain this energy only through random thermal vibration.
If thermal vibration frequency is then it makes attempts per second.
Only a fraction
RT
Hmexp of these attempts
will have an energy Hm and will be successful.
1 2
C1 C2
No. of successful jumps per second from plane 1 to plane 2,
n1->2 = A c1 exp(- Hm/RT) pNo. of successful jumps per second from plane 2 to plane 1,
n2->1 = A c2 exp(- Hm/RT) pNet jumps per second from
plane 1 to plane 2
n = n1->2 - n2->1 = A (c1-c2) exp(- Hm/RT) p
122 exp
cc
RT
HpA m
Flux:x
c
RT
Hp
A
nj m
exp2
x
c
RT
Hpj m
exp2 Fick’s 1st Law
RT
HpD mexp2
20 pD mHQ
An atom making a successful jump may remain in plane 1, go to the back plane or jump to forward plane 2. Thus only a fraction p of successful jumps are from plane 1 to plane 2. This factor has been omitted in the textbook.
Initially After some time
Adapted from Figs. 5.1 and 5.2, Callister 6e.
100%
0
Ni
x
Wt % Ni
100%
0
increasing elapsed time
How is diffusion taking place in a substitutional solid solution ?
Mechanism of substitutional diffusion
Vacancy mechanism of substitutional diffusion
However, only a very small fraction of the order of 10-4 to 10-30 are vacant.
A jump can only be successful if the neighbouring site is vacant.
Probability of finding a vacant site
= fraction of vacant site
RT
H
N
n fexp
x
c
RT
H
RT
Hpj fm
expexp2
fmsubs HHQ 2,0 pD subs
Sbstitutional diffusion is usually slower than interstitial diffusion due to difficulty of finding a vacant site.
Lattice diffusion
Grain boundary diffusion
Surface diffusion
Experimentally
Qsurface < Qgrain boundary < Qlattice
Other Diffusion Paths