Unformatted text preview: ANNOUNCEMENTS #1
Reading:
Chapter 5, all sections
Recitation Problems:
#s 5.3, 5.7, 5.13, 5.16, 5.22, 5.25
This topic will be included in Midterm I Midterm I:
8:008:50 am, Tuesday May 4, Main Auditorium
Accommodations: LeBow 348; 8:009:15 (1.5X) or
8:009:40 am (2X) Be there or…;)
Help Session: Thursday, April 29, 46 pm, Hill
Seminar Room, Room 240, LeBow
Chapter 5  12 ANNOUNCEMENTS #2
Midterm Coverage:
Material from Chapters 1, 2, 3, 4 & 5 Exam Format:
3 Questions (100 Pts.):
o 1 short answer, multipart…choose/circle the correct answer(s);
write in a few words or numbers; show brief calculation…
o 2 longer answer, 23 part…calculation + explanation…similar to
the recitation problems Formula Sheet & Periodic Table provided
Work smartly & efficiently…don’t write essays, show full
working… Chapter 5  13 Diffusion  2 Cases
• SteadyState Diffusion:
– Concentration profile IS NOT a function of
time • Nonsteady State (or Transient)
Diffusion:
– Concentration profile IS a function of time Chapter 5  14 Diffusion  Consider the Following
SS diffusion across thin plate Now let’s plot C vs. distance x Slope = concentration gradient:
Conc. grad. = dC/dx
f04_05_pg113 Chapter 5  15 SteadyState Diffusion
Rate of diffusion is independent of time dC
Flux J proportional to concentration gradient =
dx
Fick’s First Law of
diffusion C1 C1 C2
x1 if linear dC
dx x C2 x2 dC
J= D
dx
D diffusion coefficient
(proportionality constant) C C2 C1
=
x
x2 x1
Chapter 5  16 A Few Words About Units
SINCE: dC
J= D
dx
If the units of Flux J are (something  e.g. mol or kg)/m2s…
Then the units of concentration C HAVE to be (something)/m3
and vice versa! We do not care what that “something” is:
Cats, albatrosses, people, atoms, moles, kg…all fine!
The units of D are ALWAYS, and always will be, m2/s
Chapter 5  17 Example: Chemical Protective
Clothing (CPC)
• Methylene chloride is a common ingredient in paint
removers. Besides being an irritant, it also may be
absorbed through the skin. When using this paint
remover, protective gloves should be worn.
• If butyl rubber gloves (0.04 cm thick) are used, what
is the diffusive flux of methylene chloride through
the glove?
• Data:
– diffusion coefficient in butyl rubber:
D = 110 x108 cm2/s
– surface concentrations: C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
Chapter 5  18 Example (cont).
• Solution – assuming a linear conc. gradient
glove
C1 dC
J =D
dx 2 tb = paint
remover 6D skin Data: D = 110 x 108 cm2/s
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
x2 – x1 = 0.04 cm C2
x1 x2 J = (110 x 10 8 C2 C1
D
x2 x1 (0.02 g/cm3 0.44 g/cm3 )
g
cm /s)
= 1.16 x 10 5
(0.04 cm)
cm2s
2 Chapter 5  19 Diffusion and Temperature
Implicit in the equation for D:
Diffusion coefficient increases with increasing T D = Do exp  Qd
RT D = diffusion coefficient (m2/s)
D0 = preexponential term (m2/s)
Qd = activation energy (J/mol or eV/atom)
R = the gas constant (8.314 J/molK)
T = absolute temperature (K) Chapter 5  20 Activation Energy for Diffusion Qd is the height of
the energy barrier
atoms need to scale
to make a jump…
and thus diffuse Qd
X Chapter 5  21 Diffusion and Temperature
D has an exponential dependence on T:
 log D = log D0 Qd 1
2.3R T  plot log D vs. 1/T 108 300 600 1000 1500  Straight line, slope = Qd/2.3R; intercept = log D0
T(°C)
Dinterstitial >> Dsubstitutional
C in Fe (BCC) Al in Al
Fe in Fe
C in Fe
Fe in Fe D (m2/s)
1014 1020
0.5 1.0 1.5 1000 K/T Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A.
Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th
ed., ButterworthHeinemann, Oxford, 1992.) Chapter 5  22 Example: At 300 ºC the diffusion coefficient D and
activation energy Qd for Cu in Si are:
D(300ºC) = 7.8 x 1011 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient at 350 ºC?
transform
data D ln D Temp = T lnD2 = lnD0 1/T Qd 1
R T2 lnD2 lnD1 = ln and lnD1 = lnD0 Qd 1
D2
=
D1
R T2 1
T1 Qd 1
R T1 Chapter 5  23 Example (cont.)
D2 = D1 exp Qd 1
R T2 1
T1 T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K D2 = (7.8 x 10 11 41,500 J/mol
1
m /s) exp
8.314 J/mol  K 623 K
2 1
573 K D2 = 15.7 x 1011 m2/s
Chapter 5  24 NonsteadyState Diffusion
Most practical cases…nonsteadystate diffusion:
Flux and conc. gradient changing with time
Implies net accumulation or depletion of diffusing species Chapter 5  25 NonSteady State Diffusion
• Concentration C of diffusing species is a function of
both time and position, i.e. C = C(x,t)
• Fick’s first law no longer applicable..
• Now use Fick’s Second Law:
Fick’s Second Law (assuming D independent of comp’n.) 2 C
C
=D 2
t
x
Chapter 5  26 A Few Words About…Differential
Equations
• The Unreasonable Effectiveness of Mathematics
in the Natural Sciences (Eugene Wigner)
“It is difficult to avoid the impression that a miracle
confronts us here, quite comparable in its striking
nature to the miracle that the human mind can string a
thousand arguments together without getting itself into
contradictions, or to the two miracles of the existence of
laws of nature and of the human mind’s capacity to
divine them.”
http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences Chapter 5  27 Differential Equations • You CANNOT solve a DE, e.g. conc. as
fn. (position and time) without…
• BOUNDARY and INITIAL conditions… Chapter 5  28 NonsteadyState Diffusion
• Copper (Cu) diffuses into a bar of aluminum (Al)
Surface conc.,
CS of Cu atoms Al bar
preexisting conc., Co of copper atoms Cs Adapted from
Fig. 5.5,
Callister 7e. Boundary Cond.: at t = 0, C = Co for 0 x at t > 0, C = CS for x = 0 (const. surf. conc.)
C = Co for x =
Chapter 5  29 Solution:
C (x , t ) Co
x
= 1 erf
Cs Co
2 Dt C(x,t) = Conc. at point x at time t
erf(z) = Gaussian error function: CS = 2 z
0 e y2 dy
C(x,t) erf(z) values are given in Table C
o
5.1 for various x/2 Dt values Chapter 5  30 Solution to Differential Equation
given Initial and Boundary
conditions shown in previous
slides is: C (x , t ) Co
x
= 1 erf
Cs Co
2 Dt
This is a very common solution…many diffusion problems
solutions look like this… which leads to:
Chapter 5  31 Some Implications of the Solution
• Suppose it is desired to have a specific conc. of
solute C1 in an alloy, then LH side becomes:
C1 C0
= constant
Cs C0 • If so, then the RH side of the eqn. must also
be a constant, thus…
x
2 Dt or = constant x2
= constant
Dt and…
Chapter 5  32 VERY IMPORTANT SLIDE x Dt Excellent “back of the envelope” calc’n. relating
characteristic diffusion distance x, to time t, and
diffusion coefficient D
Here is what the critics said about it:
Amazing equation!; Very powerful!; Never seen anything like it!
And to think such a complex phenomenon like nonsteadystate
diffusion could be captured so easily. A must use…
Chapter 5  33 EMVISI
1) Temperature: Q. Does higher T lead to higher or lower D?
A. Higher D…
2) Diffusion Mechanism:
Interstitial diffusion is much faster than vacancy diffusion
3) Activation Energy:
High Qd = ????
Ans. = low D
Qd determined by diffusing species and host:
Small diffusing species, open structure leads to
low Qd or high Qd ?
which leads to low D or high D?
Large diffusing species, closepacked structure leads to:
low Qd or high Qd ?
which leads to a low D or high D? *Even More Very Important Slide Indeed! Chapter 5  34 NonsteadyState Diffusion
• Example Problem: An FCC ironcarbon alloy initially
containing 0.20 wt.% C is carburized at an elevated
temperature and in an atmosphere that gives a surface
carbon concentration constant at 1.0 wt.%. If after 49.5
h the concentration of carbon is 0.35 wt.% at a position
4.0 mm below the surface, determine the temperature
at which the treatment was carried out.
• Solution: use Eqn. 5.5 C ( x, t ) C o
x
= 1 erf
Cs Co
2 Dt Chapter 5  35 Solution:
C (x , t ) Co
x
= 1 erf
Cs Co
2 Dt C(x,t) = Conc. at point x at time t
erf(z) = Gaussian error function: CS = 2 z
0 e y2 dy
C(x,t) erf(z) values are given in Table C
o
5.1 for various x/2 Dt values Chapter 5  36 Solution (Cont.): C ( x , t ) Co
x
= 1 erf
Cs Co
2 Dt Boundary Conditions:
– t = 49.5 h
x = 4 x 103 m
Cs = 1.0 wt.%
– Cx = 0.35 wt.%
– Co = 0.20 wt.%
C( x, t ) Co 0.35 0.20
x
= 1 erf
=
= 1 erf ( z )
Cs Co
1.0 0.20
2 Dt erf(z) = 0.8125 Chapter 5  37 Solution (cont.):
We now determine, from Table 5.1 (p.116), the value of z for which
the error function is = 0.8125. A linear interpolation is necessary, as
follows:
z erf(z) 0.90
z
0.95 0.7970
0.8125
0.8209 Now solving for D x2 z 0.90
0.8125 0.7970
=
0.95 0.90 0.8209 0.7970 z = 0.93 x
z=
2 Dt D= x2
4 z 2t ( 4 x 10 3 m)2 1h
D=
=
= 2.6 x 10 11 m2 /s
4 z 2t
( 4)(0.93)2 ( 49.5 h) 3600 s
Chapter 5  38 Solution (cont.):
• Now, to solve for the temperature T at
which D has the above value, we use
a rearranged form of Equation (5.9a): Qd
T=
R(lnDo lnD ) from Table 5.2 (p.119), for the diffusion of C in FCC Fe
Do = 2.3 x 105 m2/s Qd = 148,000 J/mol
T= 148,000 J/mol
(8.314 J/mol  K)(ln 2.3 x10 5 m2 /s ln 2.6 x10 11 m2 /s) T = 1300 K = 1027 °C
Chapter 5  39 ...
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