Subtitle or main author’s name

Materials of EngineeringMaterials of Engineering

Lecture 8Lecture 8- Diffusion Review - Diffusion Review

& Thermal & Thermal PropertiesProperties

Engineering 45Engineering 45

Carlos Casillas, PE 16-Jan-12Licensed Chemical Engineer Spring 2012CCasillas@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

A

dq

dt1

A

d(Ne)

dt

Q 1

A

dH

dt1

A

d(N )dt

e = electric chg.N = net # e- cross A

N = # of phonons with avg. energy

• Heat Flux – (by phonons)

Defining thermal conductivity (a material property)

Q dT

dxTx Ts

T0 Ts

erfx

2 ht

Solution: Fick’s 2nd Law

Defining conductivity (a material property) j

dV

dxVx Vs

V0 Vs

erfx

2 t

Solution: Fick’s 2nd Law

jq

I

A1

A

V

R

V

l

Ohm’s Law

Or w/ Thermal Diffusivity: h

cVQ cVT

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

cin

iNe

EkB

T

• kBT gives eV

Number (or concentration*) of Vacancies at T

E is an activation energy for a particular process (in J/mol, cal/mol, eV/atom).

• applies to substitutional impurities• atoms exchange with vacancies• rate depends on (1) number of vacancies;

(2) activation energy to exchange.

ExampleThe 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.

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.

Where can we use Fick’s Law?

CO2 in air has D~16 mm2/s, and, in water, D~ 0.0016 mm2/s

Gas Diffusion in Polymers

Polymer Membrane Structure

• Flux:

J

1A

dMdt

kg

m2s

or

atoms

m2s

• Directional Quantity

• Flux can be measured for: - vacancies - host (A) atoms - impurity (B) atoms

• Empirically determined: – Make thin membrane of known surface area – Impose concentration gradient – Measure how fast atoms or molecules diffuse

through the membrane

J x

J y

J z x

y

z

x-direction

Unit area A through which atoms move.

Modeling rate of diffusion: flux

A = Area of flow

J slopediffused

massM

time

• Concentration Profile, C(x): [kg/m3]

• Fick's First Law: D is a constant

Concentration of Cu [kg/m3]

Concentration of Ni [kg/m3]

Position, x

Cu flux Ni flux

• The steeper the concentration profile, the greater the flux

Adapted from Fig. 6.2(c)

J x D

dCdx

Diffusion coefficient [m2/s]

concentration gradient [kg/m4]

flux in x-dir. [kg/m2-s]

Steady-state Diffusion: J ~ gradient of C

• Steady State: concentration profile not changing with time.

• Apply Fick's First Law:

• Result: the slope, dC/dx, must be constant (i.e., slope doesn't vary with position)

J x(left) = J x(right)

Steady State:

Concentration, C, in the box doesn’t change w/time.

J x(right)J x(left)

x

J x D

dCdx

dCdx

left

dCdx

right

• If Jx)left = Jx)right , then

Steady-State Diffusion

Steady-State Diffusion

dx

dCDJ

Fick’s first law of diffusionC1

C2

x

C1

C2

x1 x2 D diffusion coefficient

Rate of diffusion independent of time J ~dx

dC

12

12 linear ifxx

CC

x

C

dx

dC

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– surface concentrations:– Diffusion distance:

C2 = 0.02 g/cm3C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Dtb 6

2

gloveC1

C2

skinpaintremover

x1 x2

= 1.16 x 10-5

g

cm2s J

x= -D

C2

- C1

x2

- x1

• Concentration profile, C(x), changes w/ time.

• To conserve matter: • Fick's First Law:

• Governing Eqn. Fick’s 2nd Law:

Concentration, C, in the box

J (right)J (left)

dx

dCdt

=Dd2C

dx2

dx

dC

dtJ D

dC

dxor

J (left)J (right)

dJ

dx

dC

dt

dJ

dx D

d2C

dx2

(if D does not vary with x)

equate

Non-Steady-State Diffusion

• 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.

to

t1

t2

t3

xo x1 x2 x3time (s)

x (mm)

Simple Diffusion expt

• Concentration profile, C(x), changes w/ time.

Concentration, C, in the box

J (right)J (left)

dx

Non-Steady-State Diffusion: another look

C

tdx Jx Jxdx

• Rate of accumulation C(x)

C

tdx Jx (Jx

Jx

xdx)

Jx

xdx

• Using Fick’s Law: C

t

Jx

x

x

DC

x

Fick's Second"Law" c

t

xDc

x

Dc 2

x2

• If D is constant:

Fick’s 2nd Law

Non-Steady-State Diffusion: C = c(x,t)

Fick's Second"Law" c

tDc 2

x2

concentration of diffusing species is a function of both t and position x

pre-existing conc., Co of copper atoms

Surface conc., Cs of Cu atoms bar

• Copper diffuses into a bar of aluminum.

Cs

Adapted from Fig. 6.5, Callister & Rethwisch 3e.

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 =

• Cu diffuses into a bar of Al.

• Solution:

"error function” Values found in Table 5.1

C(x,t) Co

Cs Co

1 erfx

2 Dt

Non-Steady-State Diffusion

erf (z)

2

0

z

e y 2

dy

CS

Co

C(x,t)

Fick's Second"Law": c

tDc 2

x2

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?

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

CC

CtxC

os

o

erf(z) = 0.8125

z erf(z)

0.90 0.7970z 0.81250.95 0.8209

Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation.

7970.08209.0

7970.08125.0

90.095.0

90.0

z z 0.93So,

Solution

Now solve for DDt

xz

2

tz

xD

2

2

4

D x2

4z2t

(4 x 10 3m)2

(4)(0.93)2(49.5 h)

1 h

3600 s2.6 x 10 11 m2 /s

• 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))lnln( DDR

QT

o

d

T

148,000 J/mol

(8.314 J/mol-K)(ln 2.3x10 5 m2 /s ln 2.6x10 11 m2 /s)

Solution (cont.):

T = 1300 K = 1027°C

• From Table 6.2, for diffusion of C in FCC Fe

Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol D 2.6 x 10 11 m2 /s

• 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?

(Dt)500ºC =(Dt)600ºC

• Result: Dt should be held constant.

• Answer:

Note D(T) are T dependent

Values of D are provided.

Key point 1: C(x,t500C) = C(x,t600C).

Key point 2: Both cases have the same Co and Cs.

t500(Dt)600

D500

110hr

4.8x10-14m2/s

5.3x10-13m2/s 10hrs

Example: Processing

C(x,t) Co

Cs C

o

1 erfx

2 Dt

22

Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are

D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

101

202

1lnln and

1lnln

TR

QDD

TR

QDD dd

121

212

11lnlnln

TTR

Q

D

DDD d

transform data

D

Temp = T

ln D

1/T

23

Example (cont.)

K 573

1

K 623

1

K-J/mol 314.8

J/mol 500,41exp /s)m 10 x 8.7( 211

2D

1212

11exp

TTR

QDD d

T1 = 273 + 300 = 573 K

T2 = 273 + 350 = 623 K

D2 = 15.7 x 10-11 m2/s

VMSE: Student Companion SiteDiffusion Computations & Data Plots

24

25

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

Dt

x

CC

CtxC

os

o

2erf1

),(

26

Solution (cont.):

– t = 49.5 h x = 4 x 10-3 m

– Cx = 0.35 wt% Cs = 1.0 wt%

– Co = 0.20 wt%

Dt

x

CC

C)t,x(C

os

o

2erf1

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

CC

CtxC

os

o

erf(z) = 0.8125

27

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 0.7970z 0.81250.95 0.8209

7970.08209.0

7970.08125.0

90.095.0

90.0

z

z 0.93

Now solve for D

Dt

xz

2

tz

xD

2

2

4

/sm 10 x 6.2s 3600

h 1

h) 5.49()93.0()4(

m)10 x 4(

4

2112

23

2

2

tz

xD

28

• To solve for the temperature at which D has the above value, we use a rearranged form of Equation (5.9a);

)lnln( DDR

QT

o

d

from Table 5.2, for diffusion of C in FCC Fe

Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(

J/mol 000,14821125

T

Solution (cont.):

T = 1300 K = 1027ºC

29

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

30

CPC Example (cont.)

Time required for breakthrough ca. 4 min

gloveC1

C2

skinpaintremover

x1 x2

• Solution – assuming linear conc. gradient

Dtb 6

2

Equation from online CPC Case Study 5 at the Student Companion Site for Callister & Rethwisch 8e (www.wiley.com/college/callister)

cm 0.04 12 xx

D = 110 x 10-8 cm2/s

min 4 s 240/s)cm 10 x 110)(6(

cm) 04.0(28-

2bt

Breakthrough time = tb

31

Diffusion FASTER for...

• open crystal structures

• materials w/secondary bonding

• smaller diffusing atoms

• lower density materials

Diffusion SLOWER for...

• close-packed structures

• materials w/covalent bonding

• larger diffusing atoms

• higher density materials

Summary

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)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)

• The specific heat (lower case c) describes the heat capacity per unit mass (Joules/kg-K)

C

dQdT

heat capacity(J/mol-K)

energy input (J/mol or J/kg)

temperature change (K)

Materials Comparison of cp and C @ 298K

Battery

Insulation

(Lab3 set up)

CCvv as Function of Temperature 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

CCvv as Function of Temp cont as Function of Temp cont

• For Many Crystalline Solids

D

pTHivv

Dv

TTconst

CC C

TTAT C

:

:

,

3

• Where– A Material

Dependent CONSTANT

– TD Debye Temperature, K

Thermal Physics

• Energy is Stored in Lattice Vibration Waves Called Phonons– Analogous to Optical

PHOTONS

37

Atomic Vibrations

Atomic vibrations are in the form of lattice waves or phonons

Adapted from Fig. 19.1, Callister & Rethwisch 8e.

• 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

38

incr

eas

ing

cp • Why is cp significantly

larger for polymers?

Selected values from Table 19.1, Callister & Rethwisch 8e.

• PolymersPolypropylene Polyethylene Polystyrene Teflon

cp (J/kg-K)

at room T

• CeramicsMagnesia (MgO)Alumina (Al2O3)Glass

• MetalsAluminum Steel Tungsten Gold

1925 1850 1170 1050

900 486 138 128

cp (specific heat): (J/kg-K)

Material

940 775 840

Specific Heat: Comparison

Cp (heat capacity): (J/mol-K)

Thermal ExpansionThermal Expansion• Concept Materials Change Size When

HeatedTinit

TfinalLfinal

Linit initfinalinit

initfinal TT L

LL

Coefficient of Thermal Expansion

due to Asymmetry of PE InterAtomic Distance Trough• T↑ E↑

• ri is at the Statistical Avg of the Trough Width

Bond energy

Bond length (r)

incr

easi

ng

T

T1

r(T5)

r(T1)

T5Bond-energy vs bond-lengthcurve is “asymmetric”

40

Coefficient of Thermal Expansion: Comparison

• Q: Why does

generally decrease with increasing bond energy?

Polypropylene 145-180 Polyethylene 106-198 Polystyrene 90-150 Teflon 126-216

• Polymers

• CeramicsMagnesia (MgO) 13.5Alumina (Al2O3) 7.6Soda-lime glass 9Silica (cryst. SiO2) 0.4

• MetalsAluminum 23.6Steel 12 Tungsten 4.5 Gold 14.2

(10-6/C)at room T

Material

Selected values from Table 19.1, Callister & Rethwisch 8e.

Why do Polymers have

larger values?

incr

eas

ing

Thermal ConductivityThermal 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

Heat Flux given by

T2 > T1 T1

x1 x2heat flux

q k

dTdx

Thermal conductivity (W/m-K) or (J/m-K-s)

Heat flux(W/m2) or (J/m2-s)

• Q: Why theNEGATIVE Sign before k?

TemperatureGradient (K/m)Fourier’s Law

Thermal Conductivity: ComparisonThermal Conductivity: Comparisonin

crea

sin

g k

• PolymersPolypropylene 0.12Polyethylene 0.46-0.50 Polystyrene 0.13 Teflon 0.25

By vibration/ rotation of chain molecules

• CeramicsMagnesia (MgO) 38Alumina (Al2O3) 39 Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4

By vibration of atoms

• MetalsAluminum 247Steel 52 Tungsten 178 Gold 315

By vibration of atoms and motion of electrons

k (W/m-K) Energy TransferMaterial

Thermal StressesThermal 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 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)

ollE /

Thermal Stresses Thermal Stresses con’tcon’t

Thermal Stresses cont.Thermal Stresses cont.

• From Before Sub l/l into Young’s Modulus Eqn To

Determine the Thermal Stress Relation

Tll o /

TE 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 ExampleThermal Stress Example

• Find Stress

MPapsiA

F

in

lb

d

lb

A

F

7244010

41

8200

4

820022

Recall the Thermal Stress Eqn

E = 10.4 Mpsi = 71.7 GPa

α = 13.5 µin/in-°F = 13.5 µm/m-°F

8200 lbs

0 lbs

TE Need E & α

• Consult Matls Ref

Thermal Stress Example contThermal Stress Example cont

• Solve Thermal Stress Reln for ΔT

Since The Bar was Originally at Room Temp

8200 lbs

0 lbs

KFTFPa

PaT

ET

3.414.74/105.13107.71

107269

6

FCT

KT

TTT

f

f

if

15065

33841297• Heating to Hot-Coffee

Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)

Thermal Shock ResistanceThermal Shock Resistance

• Occurs due to: uneven heating/cooling.

• Ex: Assume top thin layer is rapidly cooled from T1 to T2:

rapid quench

doesn’t want to contract

tries to contract during coolingT2T1

Tension develops at surface

)( 21 TTE

E

TT ffracture )( 21

Temperature difference thatcan be produced by cooling:

k

ratequenchTT )( 21

set equal

Critical temperature differencefor fracture (set = f)

10

• Result: TSRE

kratequench f

fracturefor

)(

Thermal Shock Resistance Thermal Shock Resistance con’tcon’t

TSR – Physical MeaningTSR – Physical Meaning

• The Reln

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

E

kTSR f

51

• Application: Space Shuttle Orbiter

• Silica tiles (400-1260ºC):-- large scale application -- microstructure:

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.)

Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.)

Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace CeramicsSystems, Sunnyvale, CA.)

Thermal Protection System

reinf C-C (1650ºC)

Re-entry T Distribution

silica tiles(400-1260ºC)

nylon felt, silicon rubbercoating (400ºC)

~90% porosity!Silica fibersbonded to oneanother duringheat treatment.

100 m

Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.)

WhiteBoard WorkWhiteBoard Work

• Problem 19.5 – Debye Temperature

Charles Kittel, “Introduction to Solid State Physics”, 6e, John Wiley & Sons, 1986. pg-110