Physical Metallurgy: Principles and Design (Solutions, Instructor Solution Manual) [1 ed.] 1138627682, 9781138627680

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PHYSICAL METALLURGY Principles and Design

G.N. Haidemenopoulos

SOLUTION MANUAL

CRC Press

Chapter 2 Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is

4R , then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then 2 the density of the (100) plane is



r(100) =

2 = 8.2x1012 atoms/mm 2 4.95x10 -7

In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2 3R After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density than the (100) plane, it is a close-packed plane.

Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at ½. This has a

a 2R = 2 0 + 0 + 22 2 Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms. d(002) =

=

In the same way

a a 4R = = 1+1+1 3 6 and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing. d(111) =

Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is [111] , which corresponds to the diagonal of the cubic unit cell where there is a consecutive contact of spheres (in the model of hard spheres). Furthermore, the number of atoms per unit cell for the BCC structure is 2. The first step is to find the lattice parameter α. The density is

2



 

3

Where   is the Avogadro’s number. Therefore the lattice parameter is

3 

2  50.94  a  3.08 108 cm  3.08 1010 m 23 5.8  6.023 10

The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2 atoms. Hence the atomic density of the close-packed direction of vanadium (V) is

[111] 

2

 3



2  3.75 109 atoms / m 10 3.08 10 3

The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.

4R . The (10 0) plane is the 2 face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of the face. Hence the face consists of 4  (1/ 4)  1  2 atoms. The atomic density of the (10 0) plane is Problem 2.4. The lattice parameter for the FCC structure is  

 (100) 

2 2 1   2 2 a 4R2  4R     2

The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this triangle is 4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which is 2 3R . Thus the area of the triangle is (base  height / 2)  4 3R 2 . The equilateral triangle comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the equilateral triangle consists of 3  (1/ 6)  3  (1/ 2)  2 atoms. The atomic density of the (111) plane is

(111) 

2 1  2 4 3R 2 3R 2

The ratio of the atomic densities is

(111) 2   1.154  1 (100) 3 Therefore  (111)   (100) and specifically the (111) plane has 15% higher atomic density than the (10 0) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-Fe, etc.) is accomplished with dislocation glide on the close-packed planes.

Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have an arrangement as dense as the atoms of the FCC structure. The distance between the (0 0 01) bases of the HCP structure is c. Using the fact that the (0 0 01) planes of HCP structure correspond to the (111) planes of the FCC structure, we get

c  2 d(111)  FCC Where d (111) is the distance between the (111) close-packed planes. We find that

d (111) 

a h k l 2

2

2



a

a 1 1 1 2

2

2



4R 2

a  3 4R     d (111)  FCC  6   

Thus,

8R  c 4   1.63 6   a 6  HCP : a  2 R  c

Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for zinc (Zn) is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the distance between the (0 0 01) planes is longer in Zn than in Ti. This fact affects the plastic deformation in these metals, since the slip on (0 0 01) planes is easier in Zn than in Ti. Indeed the critical shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic deformation in Ti is performed on (10 1 0) plane, where the critical shear stress is approximately 49 MPa. Thus in Ti the slip is not performed on the close-packed planes of the crystal structure. For more details look at the 7.3 paragraph of the book (plastic deformation of single crystals with slip).

Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon) 8R multiplied by the height c. The base of hexagon is A  6 R 2 3 and the height is c  . As a 6 result, the cell volume is V  24 2 R 3

The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is 4  R3   3   0.74 3 24 2 R 3 2 6

APFHCP

Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell volume is a 3 , 4R where a  . Therefore the atomic packing factor of BCC structure is 3 4  R3  3 3    0.68 3 8  4R     3 2

APFBCC

Since the atomic packing density of BCC is less than that of the HCP, the diffusion in BCC (movement of atoms inside the lattice) is faster.

Problem 2.7 The density is



mass of cell atoms cell volume

The structure of copper (Cu) is FCC and the number of atoms per unit cell is 4. The atomic mass  is , where  is the atomic weight and N A is the Avogadro’s number. The cell volume is NA 3

 4R  a   , thus the density is  2 3

 63.57  4 23   6.023 10   9 gr / cm3  3  4 1.276 108    2  

4  R 3 ! It is the corresponding volume of 3 every atom of the structure plus the empty surrounding space inside the cell. Due to the fact that the structure of gold (Au) is FCC, the number of atoms per unit cell is 4. Therefore the atomic volume is

Problem 2.8 Notice that the atomic volume is not

 For the FCC structure we get that  

3 4

4R , hence 2

 4R    2  4

3

After replacing the value of the atomic radius of gold R , we find that   1.7 1029 m3 . Since o

1   1010 m , the atomic volume of gold is o

  17 ( )3

The molar volume Vm is the volume corresponding to one mole of gold and is obtained by multiplying the atomic volume by the Avogadro’s number. Thus the molar volume is Vm   N A  1.02 105 m3 / mol

Problem 2.9 For the atomic radius of the iron atom, RFCC=1.270 and RBCC=1.241 Angtroms (A) FCC has 4 atmos/cell while BCC has 2 atoms/cell. In FCC a = 4R 2 = 3.591A,VFCC = a 3 = 46.34A3 In BCC a = 4R 3 = 2.865A,VBCC = 23.51A3 Taking 4 atoms as a reference, this corresponds to 1 FCC cell and 2 BCC cells, then DV 2VBCC - VFCC = = 0.0144 V 2VBCC or 1.44% volume increase.

Assume that the initial volume is V and the final volume is Vt . Hence the volume change is V V Vt  V Vt V   1  t  1 V V V V V

Vt ( L  L)3 L   (1  )3 3 V L L Using the two previous relations, we get that the respective length change is (1+ DL / L)3 = 1+ DV / V Þ DL / L = 3 1+ DV / V -1 = 0.00477

Therefore there is a 0.477% increase in length.

Chapter 3 Problem 3.1 We consider n= number of divacancies,

N= number of atomic positions

AB and AC divacancies are not distinct, while AB and AD divacancies are distinct. If the coordination number is  , then the number of the distinct divacancies is  N . The change in  Gibbs free energy after the formation of divacancies is:

 

G  n E2v  nT S2 v  T S2 v  conf  S2v  conf   k ln p (configurational entropy  Boltzmann)

P  number of combinations for n divacancies distribution at x 

P

 N positions. 

x! n ! x  n !

ln P  ln x !  ln n !  ln  x  n !   x ln x  x   n ln n  n    x  n  ln  x  n    x  n    ln P n n   ln   ln  n xn N n  Since



n n X   ≫ n , then   Z N n N  

n  S  X 2ev  exp  2 v N   k

We get X 2ev 

E2bv  2 E1v  E2v   E2 v  exp  . Using the following relations:      kT  S2bv  2 S1v  S2v

 S 2bv   E 2bv  e 2 X exp  exp  1v   k   kT    

   6 for FCC structure.  , where  

Problem 3.2 Because the energies are given in kJ/mol we use RT instead of kT in the fraction calculations.

 S   E  X v  exp  vib  exp   v   k   RT   S b   Eb  2 X 2 v   X v  exp  2 v  exp  2 v  from previous problem.  k   RT   S   E  X i  exp  vib  exp   i   k   RT 

After the calculations we get the following results:

27 oC (300  )

1000 oC (1273  )

Xv

7 1015

1.4 10 3

X 2v Xi

3.34 1023

1.17 105

4.111067

7.13 1016

Problem 3.3  E   Gv     exp     exp     kT   kT   Gv  1   Gv   X v         exp     P T  k T  k T   P T S X ve  exp  nb  k

 Gv   G  Generally    V , so    Vv .  P T  P T Where Vv is the volume difference per vacancy ( Vv  0 ).

X  X  Thus,  v    v Vv  0 . kT  P T Therefore the application of hydrostatic pressure reduces the equilibrium vacancy concentration.

Problem 3.4 If we heat a specimen of initial length Lo , then the dilatation comprises two parts:  L    L        o  Lo v  Lo tot

Where



o

is the change of lattice parameter at temperature T with respect to the reference

 L  lattice parameter  o at temperature To , and   is the length change due to the addition of  Lo v L vacancies. We can express the length in cell number N C , so that N C  o . ao

After heating we get N C  N C 

Lo  L , so ao  a

N C  N C  

 Lo

 L  ao  a 

a

2 o

 a

2





 Lo

 L  ao  a  ao2

 Lo  L  a  L a   1  1    NC 1   ao  Lo  ao  Lo ao  

Where a 2  0 and Since X ve  3

L a . Lo ao

0.

 L a  N C , we get X ve  3    . NC ao   Lo

Problem 3.5 The vibration entropy Svib and the vacancy formation energy Ev are both temperature dependent. Ev  Ev T 

H v T 

Svib  S T 

 S T    H v T   So, X ve  exp   exp    kT   k   slope     ln X e T   H T  1 v Thus,     v  1 k kT   T   P

 

    H v T    1  S T   1  k  1 T P T  

 

 

  (0.1)  P

We can observe from (1.1) relation that the slope depends on the temperature, therefore the Arrhenius-type diagram will exhibit curvature.

     H   S T    T  (0.2) Generally, dH  T dS  V dP    1    1     T     T   P P Due to the (1.2) relation, the (1.1) relation becomes:

  e   ln X v T     H v T    Ev T    1 k k T  P

 

Problem 3.6 (a) Consider the reaction: 1 1 1 [110]  [211]  [12 1] 2 6 6

The components of the Burgers vector at the right-hand side are: 1 1 1 [2  1,1  2,1  1]  [3,3, 0]  [110] 6 6 2

Which is equal to the left-hand side. The same holds for the other two reactions. (b) For the same reaction as above: The magnitude of the Burgers vector of the perfect dislocation is Burgers vector of the partial dislocations is

a and the magnitude of the 2

a . 6

For the perfect dislocation:

a2 2 2 2 a2   b  1  1  0   4 2 2

For the partial dislocations:

a2 2 2 2  2  1  1   36  a2 2 2 b2  1  22  12   36

b12 

Then

a2 6 a2 6

a2 a2 a2 a2    and Frank’s rule is obeyed. The same holds for the other two reactions. 2 6 6 3

Problem 3.7 Let A and B be the two partial dislocations. The force F acting on partial B due to the stress field of partial A is: F   r b Where  r 

b

cos  (equation 3.13). 2 (1  ) r

We have r  x 2  y 2  x for y  0 . The angle  between the two partials is   60o

Getting r  d (separation distance), we have F 

b2 . 4 (1  )d

This is a repulsive force between the two partials. The stacking fault has an energy equal to SFE . This can be considered as an attractive force, which tends to reduce the stacking fault between b2 the partials. At equilibrium F  SFE and solving for d we get d  . 4 (1  )( SFE )

a2 2 2 2 a2   0  1  1  Problem 3.8 Right-hand side:  18 36  Left-hand side:

a2 2 2 2 a2 a2 1  2  1   12  12  22   36 36 3

So Frank’s rule is obeyed. 1  011 dislocation forms at the intersection of the two 111 planes of the partials. The 6 Burgers vector and the dislocation line lie on a 10 0 plane, which is not a close-packed plane.

The

1  011 dislocation cannot glide. It is a sessile dislocation. Other dislocations 6 arriving at the lock cannot glide and are immobilized. Thus Lomer-Cottrell locks contribute to strain hardening.

Therefore the

Chapter 4 Problem 4.1 The free energies of the liquid and the solid at temperature T are G L  H L  TS L G S  H S  TS S

Hence the driving force for solidification is G  H  T S

Where H  H L  H S

S  S L  S S

At the melting point Tm the free energy of the liquid is equal to the free energy of the solid G L  G S  G  0

Therefore at the melting point Tm

G  H  Tm S  0   S 

H L  Tm Tm

Where L is the latent heat of fusion and S the melting entropy of the pure metal. For most metals it is experimentally observed that the melting entropy is a constant approximately equal to R , known as universal gas constant. For a small undercooling T the difference of the specific heat capacities of liquid and solid  C pL  C pS  is negligible, so that H and S are independent of temperature. From the above relations, we get G  L  T

 L T  T  L 1    L Tm Tm  Tm 

 4.10 

The relations 4.10 gives the driving force for solidification of a pure metal.

Problem 4.2 (a) The Ag-Cu phase diagram is depicted below.

(b) The Liquidus temperature of the Ag-40Cu (wt%) alloy is 840οC. The other alloy with the same Liquidus temperature is the Ag-20Cu (wt%) alloy. (c) The 26 gr of Ag-7.5Cu (wt%) alloy consists of 26  0.925  24.05 gr Ag and 26  0.075  1.95 gr Cu. The total amount of Cu in the new formed alloy is 1.95  376  377.95 gr Cu and the copper composition is 377.95 gr Cu %Cu   94 24.05 gr Ag  377.95 gr Cu The Liquidus temperature of Ag-94Cu (wt%) alloy is 1060 οC, while the Solidus temperature is 870oC. (d) At the temperature of 600οC the eutectic alloy (28 wt% Cu) consists of phase α and phase β with the following compositions: Composition of phase α: 5 wt% Cu (phase rich in Ag) Composition of phase β: 96 wt% Cu (phase rich in Cu) The phase fraction of α is determined by the lever rule at the 600οC. 0.96  0.28 fa   0.75 0.96  0.05 Therefore 100 gr of eutectic alloy contain 75 gr of phase α. Since phase α consists of 5% Cu, then the 75 gr of phase α contain 75  0.05  3.75 gr Cu. (e) During the eutectic reaction, phases α and β are formed with copper composition of 9 wt% and 92 wt% respectively. The phase fractions are determined by the lever rule at 780 ο C. 0.92  0.28 fa   0.77 0.92  0.09 f   1  0.77  0.23 80 gr of eutectic alloy contain 80  0.77  61.7 gr of phase α and 80  0.23  18.3 gr of phase β.

(f) The Ag-15Cu (wt%) alloy contains 120  0.15  18 gr Cu. Assume that we add x gr Cu. Then the amount of copper will be 18  x gr and the new alloy will weight 120  x gr , while its copper composition will be 30 wt%. x  18  0.3  x  25.7 x  120 Therefore we should add 25.7 gr Cu in order to obtain a Ag-30Cu (wt%) alloy. Problem 4.3 (a) Alloy Ag-7Cu At 1000οC the alloy is liquid. The Liquidus temperature is 905οC. The Solidus temperature is 840οC. Between 840 and 738 οC the alloy exists as a single phase α. Below the Solvus temperature of 738 οC precipitates of β phase are formed. Alloy Ag-15Cu The Liquidus temperature is 850οC. Between 850 and 780 οC a proeutectic α phase is formed. At 780 οC+θ we have 28  15  0.68 28  9 f L  1  0.68  0.32

fa 

At 780 οC the eutectic reaction L(28Cu )   (9Cu )   (92Cu ) is obtained and all the liquid becomes a eutectic α+β mixture with phase fractions 92  98 fa   0.77 92  9 f   1  0.77  0.23 At 780 οC-θ the alloy forms a 68% proeutectic α phase and a 32% eutectic α+β mixture. The eutectic mixture α+β consists of 77% α phase and 23% β phase. Alloy Ag-28Cu This is a eutectic alloy with eutectic composition. The solidification of the alloy begins at 780 οC according to the eutectic reaction L(28Cu )   (9Cu )   (92Cu ) and all the liquid becomes a eutectic α+β mixture with phase fractions 92  98  0.77 92  9 f   1  0.77  0.23

fa 

Alloy Ag-70Cu The Liquidus temperature is 960οC. Between 960 and 780 οC a proeutectic β phase is formed. At 780 οC the remaining liquid is solidified according to the eutectic reaction and a eutectic

α+β mixture is formed. Therefore at room temperature the alloy contains a proeutectic β phase and a eutectic α+β mixture. Phase fractions at 820οC The Ag-7Cu (wt%) alloy has been solidified as a 100% α phase. The Ag-15Cu (wt%) alloy is all liquid. The Ag-28Cu (wt%) alloy is all liquid. The Ag-70Cu (wt%) alloy contains L+β phases. The composition of liquid is 44 wt% Cu and the composition of β phase is 90 wt% Cu. Applying the lever rule we get 90  70  0.434 90  44 f   1  0.434  0.566

fL 

(b) Free energy-composition curves Since phases α and β have the same crystal structure FCC (since Ag and Cu are FCC), we do not draw two different curves G but only one. At the region rich in Ag, the part of curve G is named α and at the region rich in Cu, the part of curve G is named β. It is clear that the alloy exhibits a miscibility gap and so the alloy is separated in two phases with the same crystal structure but different compositions. The temperatures 820, 780 and 760οC correspond to 1093, 1053 and 1033Κ. It is observed that with the decrease of the temperature the curve of liquid increases in contrast to the curve of FCC and that the concave of the FCC curve intensifies. At 780οC the curves have a common tangent. We are at the eutectic temperature and three phases L, α and β coexist.

3000

1:W(CU)*100.0,GMR(LIQUID) 2:W(CU)*100.0,GMR(FCC_A1#1) 3:W(CU)*100.0,GMR(FCC_A1#2)

2500

Gibbs energy J/mol

2000 1500 1000 500 0

2 3 1

-500 -1000

α

β

L

32 1

-1500 0

20

40

60

MASS_PERCENT CU

80

100

2010-12-15 12:00:23.54 output by user Elen from PC205-111

THERMO-CALC (2010.12.15:12.00) :AG CU DATABASE:TCBIN P=1E5, N=1, T=1093.15;

Gibbs energy J/mol

2500

1:W(CU)*100.0,GMR(LIQUID) 2:W(CU)*100.0,GMR(FCC_A1#1) 3:W(CU)*100.0,GMR(FCC_A1#2)

2010-12-15 11:57:50.26 output by user Elen from PC205-111

3000

1:W(CU)*100.0,GMR(LIQUID) 2:W(CU)*100.0,GMR(FCC_A1#1) 3:W(CU)*100.0,GMR(FCC_A1#2)

2010-12-15 12:03:30.98 output by user Elen from PC205-111

THERMO-CALC (2010.12.15:11.57) :AG CU DATABASE:TCBIN P=1E5, N=1, T=1053.15;

2000 1500 1000 500 12

0 -500

3

1 2 3

-1000 0

20

40

60

80

100

MASS_PERCENT CU

THERMO-CALC (2010.12.15:12.03) :AG CU DATABASE:TCBIN P=1E5, N=1, T=1033.15; 3200 2800

Gibbs energy J/mol

2400 2000 1600 1200 800

L 1

400

2

3

0 1

-400

β

32

-800

α

-1200 0

20

40

60

80

100

MASS_PERCENT CU

(c) Due to the separation of the two phases, the enthalpy of mixing should be positive H m  0 . The entropy of mixing is always positive Sm  0 . At 800οC the alloy contains a single α phase, hence Gm  0 . However at 300 οC the free energy of mixing is positive Gm  0 , so the alloy cannot exist as a single α phase and a second β phase is formed as well. (d) At 800 οC the alloy with composition x consists of L(92Cu ) and  (35Cu ) . Applying the lever rule, we find

x  35  0.30  x  52 92  35 Hence the alloy, which contains a 30% proeutectic β phase, has 52 wt% copper composition.

Problem 4.4 (a) At 200 οC the solid solubility limit of Zn in Cu is 34 wt%. The solid solubility of Cu in Zn is almost zero. (b) At the eutectoid temperature of 558οC, the eutectoid reaction is      . (c) The three free energy-composition curves of γ, δ and ε phases have a common tangent. (d) At 200 οC the Cu-40Zn (wt%) alloy contains two phases α+β. The tie line gives that the composition of α phase is 34Zn and of β phase is 46Zn. Applying the lever rule at 200 οC, the phase fractions are 46  40 fa   0.5 46  34 f   1  0.5  0.5 Problem 4.5 (a) The Al-4Cu (wt%) alloy should be heated above the Solvus temperature of 500οC. (b) and (c) The supersaturation ratio of the Al-4Cu alloy and the phase fractions of θ at 450, 400, 350 and 300oC are the following: 4  2.1 4  2.1  0.475 and f   0.037 4 53  2.1 4  1.8 4  1.8 400οC sr   0.55 and f   0.042 4 53  1.8 4 1 4 1 350οC sr   0.75 and f   0.057 4 53  1 4  0.5 4  0.5 300οC sr   0.875 and f   0.066 4 53  0.5 It is observed that with the decrease of the temperature the supersaturation ratio and the fraction of θ phase increase. (d) The Al-4.5Cu (wt%) alloy in α phase exhibits the maximum solid solubility. The applied heat treatment is the following: 1. Solution treatment at 550 οC. 2. Quenching (rapid cooling) in order to form a supersaturated solid solution. 3. Aging at 180οC in order to achieve the precipitation of θ phase.

450οC sr 

Chapter 5 Problem 5.1 Applying the Arrhenius Equations 5

10  Do e 6

10  Do e



H * RT1



H RT2

1

*

 2

We find that the activation energy for diffusion is 10  e



H *  1 1     R  T1 T2 

 2.3  

H *  1 1  *     H  62 KJ / mol R  T1 T2 

The pre-exponential term Do is calculated by replacing H * in Equation (1), so 5

10  Do e



H * RT1

 Do  2.3 103 cm 2 / sec

Problem 5.2 Using Arrhenius Equation, we calculate the temperature in which the diffusion coefficient is Do  5 1011 cm 2 / sec

D  Do e



H * RT

 5 10

11

 1.7e



31500 8.31T

 T  1563 K  1290o C

Problem 5.3 The concentration profile derives from Relation 5.34 in the book. The plot of c ( x, t ) after some annealing time t , is shown in Fig.5.20. Using the experimental validation of the concentration profile and the Relation 5.34, we can determine the self-diffusion coefficient D . Converting into a logarithm the Relation 5.34 n

c  x, t  x2  c  0, t  4 Dt

 5.160 

from which the self-diffusion coefficient of A can be determined. Problem 5.4 Consider two semi-infinite solids (1) and (2) attached at position x  0 . The (1) solid is pure iron, while the (2) solid is steel (Fe-C) with carbon content c . The couple is annealed for time t. We have to determine the carbon concentration profile c ( x, t ) in the couple. The initial conditions are

c( x, 0)  0, x  0 c( x, 0)  c, x  0 The boundary conditions are

c  , t   c  c  , t   0

We consider a thin slice  in side (2) at position  i , and we apply the solution of instantaneous source for S  c   x    2   c i ci  x, t   exp     4 Dt   4 Dt  

Using the superposition method, we sum the solutions of all the thin slices  i .

é ( x - x )2 ù i ú c ( x,t ) = Dxi exp ê å 4Dt ú ê 4p Dt i=1 ë û c¢

n

For   0 and n   , the aboveRelation becomes

é ( x - x )2 ù ú dx c ( x,t ) = exp ê ò 4Dt ê ú 4p Dt 0 ë û ¥

c¢

Substituting    x    / 2 Dt , we get d  2 Dtd . When   0 then   x / 2 Dt , and when    then    . So the above Relation is rewritten as

c  x, t   



c





exp   2  d

x 2 Dt

Reversing the integration limits and separating the integration, we get x

c ( x,t ) =

0

ò

-¥

c¢

p

( )

exp -h 2 dh +

2 Dt

ò 0

By definition the error function is erf   

c¢

p

( )

exp -h 2 dh



exp    d .  

2

2

0

becomes c  x, t  

Therefore c  0, t  

c   x 1  erf  2  2 Dt

c , c  , t   c and c  , t   0 . 2

  

Thus the above Relation

Problem 5.5 Assume co is the concentration at the surface and c is the concentration at 3 μm depth from the surface. Hence,

c  x  0.5  1  erf  co  2 Dt

  x   erf    2 Dt

   0.5 

From the error function table we get

x x2 9 108  0.48  t    12228sec  3.4 hr (0.96)2 D 0.92  8 1012 2 Dt Problem 5.6 The carbon concentration is given by the solution of diffusion equation

c  x   1  erf   co  2 Dt 

c c x x2  const . That means For both depths the ratio has the same value, so  K   K΄ co co t t and K

΄

 0.04  

2

 1.6 104 cm2 / hr , where K΄ is a constant value. The carburization time, in

10 order to reach 800 μm carburization depth, is

 0.08 t K΄

2



 0.08

2

1.6 104

 40 hr

Problem 5.7 (a) The solution of diffusion equation with constant surface concentration on a semi-infinite medium is

 x  c  x, t     erf    2 Dt  The boundary conditions are

c  , t   0.005     c  0, t   0.8   Hence,   x c( x, t )  0.8  0.795erf   12  2 1.14 10 t 

Using Mathematica, we compute the concentration profiles c  x  for diffusion times of 5, 50 and 500 sec.

0.8

0.6

500 sec

0.4

0.2

50 sec 5 sec

0.0 0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

(b) We can observe that the target carbon concentration of 0.4 wt% at 20 μm depth from the surface is achieved for a diffusion time of 500 sec. (c) The concentration profiles of carbon for t  5sec and carburization temperatures 900oC and 1000oC are shown below. 0.8

0.6 D=1.14x10E-12 at 900C

0.4 D=1.14E-11 at 1000C 0.2

0.0 0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

(d) For t  50sec and carburization temperature 1000 oC, the concentration profile is

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

The target carbon concentration of 0.4 wt% at 20 μm depth from the surface is now achieved for diffusion time of 50 sec instead of 500 sec at 900oC.   0.00002 (e) Solving this equation with respect to variable t , 0.4  0.8  0.795erf  , 12  2 1.14 10 t  we find that the diffusion time at 900oC is t  380sec . Accordingly if we solve the   0.00002 0.4  0.8  0.795erf   with respect to variable t , we find that the 11  2 1.14 10 t  diffusion time at 1000oC is t  38sec .

  x  with respect to variable x (f) Solving the equation 0.5  0.8  0.795erf   2 1.14 1012 500    o , we find that the depth at 900 C is x  17  m . Accordingly if we solve the equation

  x  with respect to variable x , we find that the 0.5  0.8  0.795erf   2 1.14 1011 500    depth for a carburization temperature of 1000oC is x  52  m . Problem 5.8 (a) The initial condition is

c  r,0  co  25 ppm  1.95 104 gr / cm2 The first boundary condition is

c  0, t   0 The required flux J can be calculated by Fick’s first law, once we find the concentration profile c  r , t  .For a spherical geometry the concentration profile is

  n  2  c(r , t )  co 2a   n r  n  1 (  1) sin exp       Dt  c f  co  r n 1  a    a   Where co and c f are the initial and final hydrogen concentrations on the steel ball, respectively. In order to calculate the flux J at t  1sec , we have to find the derivative of the above array. However the problem requires to estimate the flux, hence we will do something simpler but less precise. Considering that the net diffusion distance is small, we will assume a planar geometry instead of spherical. Therefore the second boundary condition  r  is c  , t   co . Using the solution of the error function c  r , t     erf   and the  2 Dt  boundary conditions, we get

 r  c(r , t )  co erf    2 Dt  In order to calculate the flux J   D

dc , we have to find the derivative of the error function. dr

derf [ f (r )] derf [ f ] df  dr df dr So the flux is d  J   D   co erf  dr 

 r   2 Dt

 r2  1    2 Dco    exp               4 Dt  2 Dt

 r2  D  co   exp     t   4 Dt  1/2

For D  1104 cm2 / sec and r  0 , we get J  1.1106 gr / cm2 sec . Thus the hydrogen release rate at the surface of the ball is J   3.46 106 gr / sec . (b) In order to estimate the time required to reduce the hydrogen concentration, we need the average concentration as a function of time. For a spherical geometry c 6  2 co 

Since



n 1

 n2 2 Dt  exp   2 r2  

1

n

c   2 Dt  c 6  0.2  0.8 , we use only the first term  2 exp   2  , from which we get co co   r 

2 c  r2 (0.5cm)2 0.2 2 t ln  ln( )  281.6sec .    4 2 D 2  6 co  10 cm / sec 6

We need to examine the contribution of the second term

 4 2 Dt  1 exp   r2  1  3 2 Dt  1  3 2 (104 cm2 / sec)(281.6sec)  2nd 4  3   exp   exp      8.9 10 2 2 1st 4 r 4 (0.5 cm )   2 Dt      exp   2   r 

Thus, we conclude that our estimation is close to the real solution. Problem 5.9 (a) The initial condition is

c  x,0   0.3 wt %

For the short time solution, the boundary conditions are 1 c  0, t   1.2  0.3  0.75 wt % 2 c  4, t   0.3 wt %

 x  Using the error function c  x, t     erf   and the boundary conditions, we get  2 Dt   x  c( x, t )  0.75  0.45erf    2 Dt  The time when the concentration in the right side will reach 0.5 wt% at a distance 0.15 cm from the surface, is given by c  0.15, t   0.5 wt % . So,









0.5  0.75  0.45erf x / 2 Dt  erf x / 2 Dt  0.556  x / 2 Dt  erf 1 (0.556)  0.556 Since we are on the linear area of error function, the diffusion time is 1 x2 t  60660sec  17 hr (0.556)2 4 D (b) Applying the “leak test” with 1% margin 

2 0.01 





exp( 2 )d

L /2 Dt 

2



 exp(

2

)d

 L   erfc    2 Dt 

0

Where L is the limit of the problem, which is the thickness of the plate (2 cm). Thus, the time above which the error function solution is no longer valid is L 1 L  erfc 1 (0.01)  1.82  t  ( ) 2  1.1106 sec  280 hr D 2  1.82 2 Dt

The solution (a) is valid for small diffusion times.

(c) The velocity of the plane can be derived from question (a), where the position of the plane during small times and for a concentration c  0.15, t   0.5 wt % is given that

x dx 2(0.556) 1/2 1/2 dx  0.556  x  2(0.556) Dt   D t   3.05 x104 t 1/2cm / sec dt 2 dt 2 Dt (d) For the case of long times, the planes cannot be considered semi-infinite, so we will use the solution of trigonometric series. The initial condition is c  x,0   0.3 wt %

While the boundary conditions are

c  0, t   0.75 wt % J  L, t   0

Hence,

c( x, t )  0.75 

4(0.75  0.30)



  (2 j  1)  2  1 (2 j  1) x sin exp     Dt  L L j 0 2 j  1    

Considering that the times are long enough so that we can use only the first term, the velocity of the plane expressed on cm/sec is

x

  2 Dt  c( x, t )  0.5  0.75  sin exp   2    L L     2 Dt  dx   2 Dt  x 0.4363 D  sin  0.4363exp   2    exp  2   L L  dt L cos( x / L)   L  4(0.45)

dx 2.06 x107  exp(7.4 x107 t ) dt cos( x / 2) (e) For short times, the position of the plane with 0.5 wt% carbon concentration is 

x  2(0.556) Dt  6.09 x104 t1/2

Accordingly, for long times, the position is

   2 Dt    2 Dt   x L 0.4363exp   2   sin  x  sin 1 0.4363exp   2   L  L  L     The following Table contains the values of the position of the plane with 0.5 wt% carbon concentration at different times.

Time t (hours) 1 10 50 100 150 200 250 316

Position x (cm) Position x (cm) (short times) (long times) 0.036 0.115 0.60 0.255 0.66 0.360 0.78 0.435 0.90 0.503 1.05 0.615 1.60 2.00

Problem 5.10 We will use the Laplace transform to solve the following problem. We consider diffusion in a semi-infinite medium with initial concentration co and a constant flux J   at the surface

 x  0 . We have to solve the following equation

U  2U D for x  0 and t  0 t  x2 The initial condition is

U  x,0  U0 The boundary conditions are

J  0, t      DU x  0, t   5.165  U  , t   U 0  5.166  Where in Equation 5.165 U x 

U , while in the Equation 5.166 U  x, t  means that is x

bounded. Like the previous example, the subsidiary Equation is

U d 2u s  u  0  0  5.167  2 dx D D The general solution of Equation 5.167 is

u  x, s   c1e x

s/ D

 c2e  x

s/ D

 c3  5.168 

Due to Equation 5.166, we get c1  0 . Therefore

u  x, s   c2e  x The Laplace transform of Equation 5.165 is

s/D

 c3  5.169 

D

du    or u x  0, s     5.170  dx s sD

From Equation 5.169 we have ux  x, s   

s x c2e D

s/D

, from which due to Equation 5.170

we get c2 

 D s

1/ 2 3/ 2

In order to determine c3 we use the initial condition, so we find that c3 

U0 s

Thus the general solution 5.168 becomes U  x, s  

 D s

1/2 3/2

e x

s/D



U0 s

 5.171

The inverse Laplace transform of Equation 5.171 gives  2  t     1  L1  3/ 2 e / s   2 exp      erfc    s  2 t   4t 

Substituting  

x , we get D

U  x, t  

 x2    Dt  x 2 exp     xerfc  D    2 Dt  4 Dt 

     U 0  5.172   

In order to validate the Solution 5.172, we need to check the initial and boundary conditions. The initial condition U  x,0  U0 is satisfied. The boundary condition U  , t   U0 is satisfied since exp     0 . For the boundary condition 5.165, U x  x, t  is demanded. The derivative of Equation 5.172 is

U x  x, t  

 x2    Dt  2 x   x 2 exp     erfc    D    4 Dt   2 Dt  4 Dt 

Therefore we get U x  0, t   

d   x   x dx erfc    2 Dt

    

 which is the boundary condition 5.165. For x  0 , the D

Solution 5.172 gives

U  0, t  

2 Dt  U0 D 

Which means that the surface concentration changes with

t.

Problem 5.11 Considering the above assumptions, the Fick’s second law for the single-phase area 0  x  s is

c  2c D 2 t x The boundary conditions are

c(0, t )  cs c ( s , t )  c* The mass balance at the interface is

æ ¶c ö ds -D ç ÷ = c* - co dt è ¶x ø x=s

(

)

Using the error function solution

æ x ö c(x,t) = A + Berf ç è 2 Dt ÷ø A and B are derived from the boundary conditions

c(0, t )  cs  A  Berf (0)  A  s  c( s, t )  c*  cs  Berf    2 Dt  From the second boundary condition we obtain

æ s ö Berf ç = c* - cs ÷ è 2 Dt ø Substituting

æ

b =ç

s

è 2 Dt

ö ÷ø

Since c * and cs are independent of temperature, then  is constant and

ds =b D/t dt The solution is

æ x ö c(x,t) = cs + Berf ç è 2 Dt ÷ø From which occurs that

æ x2 ö æ x ö ö 2B æ 1 ö dc d æ = B ç erf ç = exp çè - 4Dt ÷ø ç ÷ dx dx è è 2 Dt ÷ø ÷ø p è 2 Dt ø Replacing into the main Equation , we get  s2  B D 2 BD  1   dc  D    exp exp    2       dx 4 Dt  2 Dt  t   xs     ds   c*  co    c*  co   D / t dt  B     co  c*  exp   2 

Therefore the Equation becomes

 x  c( x, t )  cs     co  c*  exp   2  erf    2 Dt  For x  s , we get

c* - cs = b p exp b 2 erf ( b ) * co - c

( )

From the above Equation, we can calculate  and the position of the interface s , where s is the decarburization depth s = 2b Dt

For example, we will calculate the decarburization depth of a hypo-eutectoid steel with 0.2 wt% carbon concentration, which is subjected to a temperature of 560oC for a year under extreme decarburization conditions. Assuming that the diffusion coefficient is D  109 cm2 / sec , we have co  0.2 wt % , c*  0.02 wt % (taken from the Fe-C phase diagram at local equilibrium temperature of 560oC), cs  0 (extreme decarburization conditions). Replacing in the above Equation, we get

 exp   2  erf     0.063 Which we solve numerically with respect to variable  , so that   0.016 . Thus from Equation 5.183 the decarburization depth is s  0.0568 cm  568  m . Problem 5.12 Generally D depends on the concentration and since there is a concentration gradient, the D changes with the distance. D ( c ) can be calculated by an experimentally

observed concentration profile. This method is called Matano analysis. Assuming a diffusion couple A-B, we write Fick’s second law

¶c ¶ æ ¶cö = D ¶ t ¶ x çè ¶ x ÷ø The boundary conditions are

c  , t   c1 c  , t   c2 Combining the x and t variables, we define a new variable  where   x / t . In this way the concentration c is only with respect to variable  . Transforming the Fick’s second law into an ordinary differential equation, the partial derivatives in the diffusion equation become

 c   dc  1 x  dc    dc       3/ 2      t  t  d  2 t  d  2t  d   And,

 c   dc  1  dc     x  x  d   t1/ 2  d   Replacing the above Equation in Fick’s second law, we find

-

l æ dc ö ¶ é D æ dc ö ù 1 d é dc ù = D ê ú= 2t çè d l ÷ø ¶ x ë t1/2 çè d l ÷ø û t d l êë d l úû

Thus,



 dc d  dc   D  5.184  2 d  d   d  

Integrating the Equation 5.184 from c  c1 to c  c , we get c



1 dc  dc  D  2 c1 d

 5.185

Since the concentration gradient tends towards zero when c  c1 , the right term of Equation 5.185 is just D ( dc / d l ) . Therefore, c

D

1  dc  5.186   dc  c1 2   d 

From Equation 5.186 we can define Matano interface. Since the concentration gradient tends towards zero as well as when c  c2 , from Equation 5.185 we have c2

  dc  0  5.187 

c1

Since the experimental data (concentration profile) had been taken for a fixed time period of annealing t , the Equation 5.186 and 5.187 can be written as function of x and t . The interdiffusion coefficient is

Here, the Matano interface is determined at the x  0 layer, so that

c2

 xdc  0 .

c1

The method used to estimate D for a fixed concentration.

Chapter 6 Problem 6.1 (a) Calculate cosθ from the relation

g aa = 2g ab cosq and then f(θ) from

f (q ) =

(

1 2 - 3cosq + cos3 q 2

)

We get f(θ)=0.148 and the energy barrier is reduced by about 15%. (b) The energy barrier vanishes when f(θ)=0. This takes place when cosθ=1, then γαα=2γαβ and γαα=1200mJ/m2

Problem 6.2 (a) The liquid is noted by L, the solid by S and the mold by M. The free energy change from the formation of a solid nucleus is

G  VS GSL  ASL SL  AMS  MS  AMS  ML where Vs is the volume of the solid. A are the contact surfaces and γ the respective interfacial energies. Due to θ=π/2, from the equilibrium of surface tensions we get

 MS   ML and the above relation becomes

G  VS GSL  ASL SL Since VS   r 2 / 2 and ASL   r (unit length) we get

G 

r2 2

GSL   r SL

We take G / r  0 and solve for r

r*  

 SL GSL

We see that the critical nucleation size is independent of the contact angle. (b) Now consider the effect of the crack angle α on the energy barrier for nucleation. Substitute the critical size in the free energy and get

G   *

2  SL

2GSL

We see that for a constant temperature T, the energy barrier increases linearly with the crack angle α. (c) Now compare the activation energy for α