Engineering 45
Materials of Engineering
Lecture 8
- Diffusion Review
& Thermal
Properties
Carlos Casillas, PE
16-Jan-12
Licensed
Chemical
Engineer
Spring 2012
Subtitle
or main
author’s name
CCasillas@ChabotCollege.edu
Hayward, CA
Recruitment
• Click to edit Master text styles
– Second level
• Third level
– Fourth level
» Fifth level
Other Types of Diffusion (Beside Mass or Atomic)
• Flux is general concept: e.g. charges, phonons,..
• Charge Flux –
jq 
1 dq 1 d(Ne)

A dt A dt
Defining conductivity 
(a material property)
Ohm’s Law
dV
j  
dx
I 1V
V
jq  

A AR
l
• Heat Flux –
(by phonons)
Q
1 dH 1 d(N  )

A dt
A dt
dT
Q  
dx
Defining thermal conductivity 
(a material property)
Or w/ Thermal Diffusivity:
h

cV 
Q  cV T
e = electric chg.
N = net # e- cross A
Solution: Fick’s 2nd Law
 x 
Vx  Vs
 erf 

V0  Vs
2 t 
N = # of phonons with
avg. energy 
Solution: Fick’s 2nd Law
 x 
Tx  Ts
 erf 

T0  Ts
 2 ht 
Diffusion- Steady and Non-Steady State
Diffusion - Mass transport by atomic motion
Mechanisms
• Gases & Liquids – random (Brownian) motion
• Solids – vacancy diffusion or interstitial diffusion
Substitution-diffusion:vacancies and interstitials
• applies to substitutional impurities
• atoms exchange with vacancies
• rate depends on (1) number of vacancies;
(2) activation energy to exchange.
Number (or concentration*)
of Vacancies at T
 E
ni
ci  e
N
k T
B
• kBT gives eV
E is an activation energy
for a particular process (in
J/mol, cal/mol, eV/atom).
Where can we use Fick’s Law?
Fick's law is commonly used to model transport processes in
• food processing,
• chemical protective clothing,
• biopolymers, synthetic polymer membranes (fluid sepaations),
• pharmaceuticals (controlled-release),
• porous soils & solids, catalysts, nuclear isotope enrichment,
• vapor & liquid thin-film coating & doping process, etc.
Example
The total membrane surface area in the lungs (alveoli)
may be on the order of 100 square meters and have
a thickness of less than a millionth of a meter, so it is
a very effective gas-exchange interface.
CO2 in air has D~16 mm2/s, and, in water, D~ 0.0016 mm2/s
Gas Diffusion in Polymers
Polymer Membrane Structure
Modeling rate of diffusion: flux
• Flux:
• Directional Quantity
• Flux can be measured for:
- vacancies
- host (A) atoms
- impurity (B) atoms
A = Area of flow
• Empirically determined:
– Make thin membrane of known surface area
– Impose concentration gradient
– Measure how fast atoms or molecules diffuse
through the membrane
diffused
mass
M
J  slope
time
Steady-state Diffusion: J ~ gradient of C
• Concentration Profile, C(x): [kg/m3]
Cu flux Ni flux
Concentration
of Cu [kg/m 3]
Adapted from Fig. 6.2(c)
Concentration
of Ni [kg/m 3]
Position, x
• Fick's First Law: D is a constant
• The steeper the concentration profile, the greater the flux
Steady-State Diffusion
• Steady State: concentration profile not changing with time.
• Apply Fick's First Law:
• If Jx)left = Jx)right , then
dC
J x  D
dx
dC 
dC 
 
  
dx left dx right
• Result: the slope, dC/dx, must be constant
(i.e., slope doesn't vary with position)
Steady-State Diffusion
Rate of diffusion independent of time
dC
J~
dx
Fick’s first law of diffusion
C1 C1
C2
x1
x
C2
x2
dC C C2  C1
if linear


dx
x
x2  x1
dC
J  D
dx
D  diffusion coefficient
Example: Chemical Protective Clothing
• Methylene chloride is a common ingredient of paint removers.
Besides being an irritant, it also may be absorbed through skin.
When using, 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:
– D in butyl rubber:
D = 110 x10-8 cm2/s
C1 = 0.44 g/cm3 C = 0.02 g/cm3
– surface concentrations:
2
– Diffusion distance:
x2 – x1 = 0.04 cm
glove
C1
paint
remover
tb 
2
skin
6D
C2
x1 x2
Jx = -D
C2 - C1
x 2 - x1
= 1.16 x 10-5
g
cm2 s
Non-Steady-State Diffusion
• Concentration profile,
C(x), changes w/ time.
• To conserve matter:
• Governing Eqn.
Fick’s 2nd Law:
• Fick's First Law:
Simple Diffusion expt
• Glass tube filled with water.
• At time t = 0, add some drops of ink to one end
of the tube.
• Measure the diffusion distance, x, over some time.
• Compare the results with theory.
Non-Steady-State Diffusion: another look
• Concentration profile,
C(x), changes w/ time.
• Rate of accumulation C(x)
C
dx  J x  J xdx
t
• Using Fick’s Law:
• If D is constant:
C
J x
J x

dx  J x  (J x 
dx)  
dx
t
x
x
C
J x

C 

   D

t
x
x 
x 
Fick's Second "Law"
c
t




x 
Fick’s
2nd Law
D
c 
x









D
c
x
2


2 


Non-Steady-State Diffusion: C = c(x,t)
concentration of diffusing species is a function of both t and position x






c
Fick's Second "Law" c  D
t
x
2


2 


• Copper diffuses into a bar of aluminum.
Cs
B.C. at t = 0, C = Co for 0  x  
at t > 0, C = CS for x = 0 (fixed surface conc.)
C = Co for x = 
Adapted from Fig. 6.5,
Callister & Rethwisch 3e.
Non-Steady-State Diffusion
CS
• Cu diffuses into a bar of Al.
C(x,t)
Fick's Second "Law":
c
t






D
c
x
2


2 


Co
• Solution:
"error function” Values found in Table 5.1
erf (z) 
2


z
e
0
y 2
dy
Example: Non-Steady-State Diffusion
FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated
temperature and in an atmosphere that gives a surface C content 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, what temperature was treatment done?
Solution
C( x, t )  Co 0.35  0.20
 x 

 1  erf 
  1  erf ( z )
Cs  Co
1.0  0.20
 2 Dt 
Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation.
z  0.90
0.8125  0.7970

0.95  0.90 0.8209  0.7970
Now solve for D
x
z
2 Dt
So, z  0.93
D
 erf(z) = 0.8125
z
erf(z)
0.90
z
0.95
0.7970
0.8125
0.8209
x2
4 z 2t
 x2 
(4 x 103 m)2
1h
11
2
D   2  

2.6
x
10
m
/s
2
 4z t  (4)(0.93) (49.5 h) 3600 s
Solution (cont.):
•
To solve for the temperature at which D
has the above value, we use a rearranged
form of Equation (6.9a);
D=D0 exp(-Qd/RT)
• From Table 6.2, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

Qd
T
R(lnDo  lnD)
D  2.6 x 1011 m2 /s
148,000 J/mol
T
(8.314 J/mol-K)(ln 2.3x105 m2 /s  ln 2.6x1011 m2 /s)
T = 1300 K = 1027°C
Example: Processing
• Copper diffuses into a bar of aluminum.
• 10 hours processed at 600 C gives desired C(x).
• How many hours needed to get the same C(x) at 500 C?
Key point 1: C(x,t500C) = C(x,t600C).
Key point 2: Both cases have the same Co and Cs.
• Result: Dt should be held constant.
C(x,t)  Co
Cs  Co
• Answer:
 x 
 1 erf 

 2 Dt 
Note
D(T) are T dependent
Values of D are provided.
Example: At 300ºC the diffusion coefficient and
activation energy for Cu in Si are
D(300ºC) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient at 350ºC?
transform
data
D
Temp = T
1
  and
 T2 
Q
D
 lnD2  lnD1  ln 2   d
D1
R
Qd
lnD2  lnD0 
R
ln D
1/T
Qd
lnD1  lnD0 
R
 1 1
  
 T2 T1 
 1
 
 T1 
22
Example (cont.)
 Qd
D2  D1 exp 
 R
 1 1 
  
 T2 T1 
T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K
D2  (7.8 x 10
11
  41,500 J/mol  1
1 
m /s) exp 



 8.314 J/mol - K  623 K 573 K 
2
D2 = 15.7 x 10-11 m2/s
23
VMSE: Student Companion Site
Diffusion Computations & Data Plots
24
Non-steady State Diffusion
• Sample Problem: An FCC iron-carbon 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 )  Co
 x 
 1  erf 

Cs  Co
 2 Dt 
25
Solution (cont.):
C( x , t )  Co
 x 
 1  erf 

Cs  Co
 2 Dt 
– t = 49.5 h
– Cx = 0.35 wt%
– Co = 0.20 wt%
x = 4 x 10-3 m
Cs = 1.0 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
26
Solution (cont.):
We must now determine from Table 5.1 the value of z for which the
error function is 0.8125. An interpolation is necessary as follows
z
erf(z)
0.90
z
0.95
0.7970
0.8125
0.8209
Now solve for D
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
3
2
 x2 
(
4
x
10
m)
1h


D 

 2.6 x 10 11 m2 /s
 4z 2t  ( 4)(0.93)2 ( 49.5 h) 3600 s


27
Solution (cont.):
• To solve for the temperature 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, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

T
148,000 J/mol
(8.314 J/mol - K)(ln 2.3x105 m2 /s  ln 2.6x10 11 m2 /s)
T = 1300 K = 1027ºC
28
Example: Chemical Protective
Clothing (CPC)
• Methylene chloride is a common ingredient of paint
removers. Besides being an irritant, it also may be
absorbed through skin. When using this paint remover,
protective gloves should be worn.
• If butyl rubber gloves (0.04 cm thick) are used, what is the
breakthrough time (tb), i.e., how long could the gloves be
used before methylene chloride reaches the hand?
• Data
– diffusion coefficient in butyl rubber:
D = 110 x10-8 cm2/s
29
CPC Example (cont.)
• Solution – assuming linear conc. gradient
Breakthrough time = tb
glove
C1
paint
remover
2
tb 
6D
skin
C2
Equation from online CPC
Case Study 5 at the Student
Companion Site for Callister &
Rethwisch 8e (www.wiley.com/
college/callister)
  x2  x1  0.04 cm
x1 x2
D = 110 x 10-8 cm2/s
tb 
(0.04 cm) 2
(6)(110 x 10
-8
2
 240 s  4 min
cm /s)
Time required for breakthrough ca. 4 min
30
Summary
Diffusion FASTER for...
Diffusion SLOWER for...
• open crystal structures
• close-packed structures
• materials w/secondary
bonding
• materials w/covalent
bonding
• smaller diffusing atoms
• larger diffusing atoms
• lower density materials
• higher density materials
31
Chapter 19 Thermal Properties
• Response of materials to the application of heat
• As a material absorbs energy in the form of heat, its
temperature rises and its dimensions increase.
• This dimensional change is different for different
materials.
• Also, the ability to transfer heat by conduction varies
greatly between materials.
• How metals, ceramics, and polymers rank in hi-temp
applications
• Important concepts:
–
–
–
–
heat capacity (C)
thermal expansion (a)
thermal conductivity (k)
thermal shock resistance (TSR)
Heat Capacity (Specific Heat)
• The ability of a material to absorb heat is specified by its
heat capacity, C
• Cp = heat capacity measured @ constant pressure
• Cv = heat capacity measured @ constant volume
• Heat capacity is proportional to the amount of energy
required to produce a unit of temperature change in a
specified amount of material (Will measure in Lab3)
heat capacity
(J/mol-K)
dQ
C
dT
energy input (J/mol or J/kg)
temperature change (K)
• The specific heat (lower case c) describes the heat
capacity per unit mass (Joules/kg-K)
Materials Comparison of cp and C
@ 298K
(Lab3 set up)
Battery
Insulation
Cv as Function of Temperature
• Cv  The vibrational contribution to heat capacity
varies as a function of temperature for a crystalline
solid
– Increases with Increasing T
– Tends to a limiting value of 3R = 24.93 J/mol-k
3R=24.93
Cv as Function of Temp cont
• For Many Crystalline
Solids
Cv  AT 3 : T  TD
Cv  Cv , HiT  C p
 const : T  TD
• Where
– A  Material
Dependent
CONSTANT
– TD  Debye
Temperature, K
 Thermal Physics
• Energy is Stored in
Lattice Vibration Waves
Called Phonons
– Analogous to Optical
PHOTONS
Atomic Vibrations
Atomic vibrations are in the form of lattice waves or
phonons
• Vibrations become coordinated
& produce elastic waves that
propagate through the solid
• Waves have specific
wavelengths and energies
characteristic of the material –
i.e., they are quantized
• A single quantum of vibrational
energy is a phonon
• Phonons are responsible for the
transport of energy by thermal
conduction
Adapted from Fig. 19.1,
Callister & Rethwisch 8e.
37
Specific Heat: Comparison
increasing cp
Material
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
cp (J/kg-K)
at room T
1925
1850
1170
1050
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Glass
940
775
840
• Metals
Aluminum
Steel
Tungsten
Gold
900
486
138
128
cp (specific heat): (J/kg-K)
Cp (heat capacity): (J/mol-K)
• Why is cp significantly
larger for polymers?
Selected values from Table 19.1,
Callister & Rethwisch 8e.
38
Thermal Expansion
• Concept  Materials Change Size When
Heated
L final  Linit
Linit
  T final  Tinit
Linit


Lfinal
Tinit
Tfinal
Coefficient of Thermal Expansion
• T↑  E↑
• ri is at the
Statistical Avg of
the Trough Width
increasing T
Bond energy
r(T1)
r(T5)
  due to Asymmetry of
PE InterAtomic
Distance Trough
T5
T1
Bond length (r)
Bond-energy vs
bond-length
curve is “asymmetric
Coefficient of Thermal Expansion:
Comparison
Material
increasing 
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
• Metals
Aluminum
Steel
Tungsten
Gold
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Soda-lime glass
Silica (cryst. SiO2)
 (10-6/C)
at room T
145-180
106-198
90-150
126-216
23.6
12
4.5
14.2
13.5
7.6
9
0.4
Why do Polymers have
larger  values?
• Q: Why does 
generally decrease
with increasing
bond energy?
Selected values from Table 19.1,
Callister & Rethwisch 8e.
40
Thermal Conductivity
• Concept  Ability of a solid to transport
heat from high to low temperature regions
 atomic vibrations and free electrons
• Consider a Cold←Hot Bar
T2 > T1
T1
x1
heat flux
 Heat Flux given by
Fourier’s Law
Heat flux
(W/m2) or
(J/m2-s)
dT
q  k
dx
Thermal conductivity
(W/m-K) or (J/m-K-s)
x2
Temperature
Gradient (K/m)
• Q: Why the
NEGATIVE
Sign
before k?
Thermal Conductivity: Comparison
Material
k (W/m-K)
Energy Transfer
• Metals
increasing k
Aluminum
Steel
Tungsten
Gold
247
52
178
315
By vibration of
atoms and
motion of
electrons
38
39
1.7
1.4
By vibration of
atoms
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Soda-lime glass
Silica (cryst. SiO2)
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
By vibration/
0.12
0.46-0.50 rotation of chain
molecules
0.13
0.25
Thermal Stresses
• As Noted Previously a Material’s Tendency to
Expand/Contract is Characterized by α
• If a Heated/Cooled Material is Restrained to its
Original Shape, then Thermal Stresses will
Develop within the material
• For a Solid Material
• Where
–
–
–
–
  E l / lo 
  Stress (Pa or typically MPa)
E  Modulus of Elasticity; a.k.a., Young’s Modulus (GPa)
l  Change in Length due to the Application of a force (m)
lo  Original, Unloaded Length (m)
Thermal Stresses con’t
Thermal Stresses cont.
l / lo   T 
• From Before
 Sub l/l into Young’s Modulus Eqn To
Determine the Thermal Stress Relation
  E T 
 Eample: a 1” Round 7075-T6 Al
(5.6Zn, 2.5Mg, 1.6Cu, 0.23Cr wt%’s)
Bar Must be Compressed by a 8200 lb
force when restrained and Heated
from Room Temp (295K)
• Find The Avg Temperature for the Bar
Thermal Stress Example
• Find Stress
F 8200lb
8200lb
 

2
A d 4  1in 2 4
F
   10 440 psi  72MPa
A
 Recall the Thermal
Stress Eqn
  E T 
 Need E & α
• Consult Matls Ref
0 lbs
8200 lbs
 E = 10.4 Mpsi
= 71.7 GPa
 α = 13.5 µin/in-°F
= 13.5 µm/m-°F
Thermal Stress Example cont
• Solve Thermal
Stress Reln for ΔT
T   E
0 lbs
8200 lbs
72 106 Pa
T 
71.7 109 Pa 13.5 10 6 /  F
T  74.4 F  41.3K
 Since The Bar was Originally at Room Temp
T f  Ti  T
T f  297  41  338 K
T f  65C  150 F
• Heating to Hot-Coffee
Temps Produces Stresses
That are about 2/3 of the
Yield Strength (15 Ksi)
Thermal Shock Resistance
• Occurs due to: uneven heating/cooling.
• Ex: Assume top thin layer is rapidly cooled from T1 to T2:
rapid quench
tries to contract during cooling T2
doesn’t want to contract

T1
Temperature difference that
can be produced by cooling:
quenchrate
(T1  T2 ) 
k
Tension develops at surface
  E (T1  T2 )
Critical temperature difference
for fracture (set  = f)
f
(T1  T2 ) fracture 
E
set equal
fk
 TSR
• Result: (quenchrate) for fracture 
E
Thermal Shock Resistance con’t
TSR – Physical Meaning
• The Reln
fk
TSR 
E
 For Improved (GREATER) TSR want
• σf↑  Material can withstand higher
thermally-generated stress before fracture
• k↑  Hi-Conductivity results in SMALLER
Temperature Gradients; i.e., lower ΔT
• E↓ More FLEXIBLE Material so the thermal stress
from a given thermal strain will be reduced (σ = Eε)
• α↓  Better Dimensional Stability; i.e., fewer
restraining forces developed
Thermal Protection System
Re-entry T
Distribution
• Application: Space Shuttle Orbiter
reinf C-C
silica tiles
(1650ºC) (400-1260ºC)
Chapter-opening photograph, Chapter 23, Callister 5e
(courtesy of the National Aeronautics and Space
Administration.)
• Silica tiles (400-1260ºC):
-- large scale application
Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the
National Aeronautics and Space Administration.)
nylon felt, silicon rubber
coating (400ºC)
Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J.
Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The
Shuttle Orbiter Thermal Protection System", Ceramic
Bulletin, No. 11, Nov. 1981, p. 1189.)
-- microstructure:
~90% porosity!
Silica fibers
bonded to one
another during
heat treatment.
100 mm
Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy
Lockheed Aerospace Ceramics
Systems, Sunnyvale, CA.)
51
WhiteBoard Work
• Problem 19.5 – Debye Temperature
Charles Kittel, “Introduction to Solid State
Physics”, 6e, John Wiley & Sons, 1986. pg-110