Engineering Materials vol 2 Part 4 doc

66 Engineering Materials 2 from the solid–liquid interface (for the same sort of reason that your hands lose heat much more rapidly if you wear gloves than if you wear mittens). And the faster solidification that we get as a consequence “pays for” the high boundary energy. Large driving forces produce fine dendrites – which explains why one can hardly see the dendrites in an iced lollipop grown in a freezer (−10°C); but they are obvious on a freezing pond (−1°C). To summarise, the shapes of the grains and phases produced during transforma- tions reflect a balance between the need to minimise the total boundary energy and the need to maximise the speed of transformation. Close to equilibrium, when the driving force for the transformation is small, the grains and phases are primarily shaped by the boundary energies. Far from equilibrium, when the driving force for the transforma- tion is large, the structure depends strongly on the mechanism of the transformation. Further, even the scale of the structure depends on the driving force – the larger the driving force the finer the structure. Further reading D. A. Porter and K. E. Easterling, Phase Transformations in Metals and Alloys, 2nd edition, Chapman and Hall, 1992. M. F. Ashby and D. R. H. Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann, 1996. G. A. Chadwick, Metallography of Phase Transformations, Butterworth, 1972. P. G. Shewmon, Diffusion in Solids, 2nd edition, TMS Publishers, 1989. Problems 6.1 The solidification speed of salol is about 2.3 mm min –1 at 10°C. Using eqn. (6.15) estimate the energy barrier q that must be crossed by molecules moving from liquid sites to solid sites. The melting point of salol is 43°C and its latent heat of fusion is 3.2 × 10 –20 J molecule –1 . Assume that the molecular diameter is about 1 nm. Answer: 6.61 × 10 –20 J, equivalent to 39.8 kJ mol –1 . 6.2 Glass ceramics are a new class of high-technology crystalline ceramic. They are made by taking complex amorphous glasses (like SiO 2 –Al 2 O 3 –Li 2 O) and making them devitrify (crystallise). For a particular glass it is found that: (a) no devitrification occurs above a temperature of 1000°C; (b) the rate of devitrification is a maximum at 950°C; (c) the rate of devitrification is negligible below 700°C. Give reasons for this behaviour. 6.3 Samples cut from a length of work-hardened mild steel bar were annealed for various times at three different temperatures. The samples were then cooled to room temperature and tested for hardness. The results are given below. Kinetics of structural change: I – diffusive transformations 67 Annealing temperature Vickers hardness Time at annealing temperature (°C) (minutes) 600 180 0 160 10 135 20 115 30 115 60 620 180 0 160 4 135 9 115 13 115 26 645 180 0 160 1.5 135 3.5 115 5 115 10 Estimate the time that it takes for recrystallisation to be completed at each of the three temperatures. Estimate the time that it would take for recrystallisation to be completed at an annealing temperature of 700°C. Because the new strain-free grains grow by dif- fusion, you may assume that the rate of recrystallisation follows Arrhenius’ law, i.e. the time for recrystal-lisation, t r , is given by t r = Ae Q/RT , where A is a constant, Q is the activation energy for self-diffusion in the ferrite lattice, R is the gas constant and T is the temperature in kelvin. Answers: 30, 13, 5 minutes respectively; 0.8 minutes. 68 Engineering Materials 2 Chapter 7 Kinetics of structural change: II – nucleation Introduction We saw in Chapter 6 that diffusive transformations (like the growth of metal crystals from the liquid during solidification, or the growth of one solid phase at the expense of another during a polymorphic change) involve a mechanism in which atoms are attached to the surfaces of the growing crystals. This means that diffusive transforma- tions can only take place if crystals of the new phase are already present. But how do these crystals – or nuclei – form in the first place? Nucleation in liquids We begin by looking at how crystals nucleate in liquids. Because of thermal agitation the atoms in the liquid are in a state of continual movement. From time to time a small group of atoms will, purely by chance, come together to form a tiny crystal. If the liquid is above T m the crystal will, after a very short time, shake itself apart again. But if the liquid is below T m there is a chance that the crystal will be thermodynamically stable and will continue to grow. How do we calculate the probability of finding stable nuclei below T m ? There are two work terms to consider when a nucleus forms from the liquid. Equa- tions (6.1) and (6.2) show that work of the type ∆H (T m − T)/T m is available to help the nucleus form. If ∆H is expressed as the latent heat given out when unit volume of the solid forms, then the total available energy is (4/3) π r 3 ∆H (T m − T)/T m . But this is offset by the work 4 π r 2 γ SL needed to create the solid–liquid interface around the crys- tal. The net work needed to form the crystal is then W f = 4 π r 2 γ SL − 4 3 3 πrH TT T m m ∆ ( ) . − (7.1) This result has been plotted out in Fig. 7.1. It shows that there is a maximum value for W f corresponding to a critical radius r*. For r < r* (dW f /dr) is positive, whereas for r > r* it is negative. This means that if a random fluctuation produces a nucleus of size r < r* it will be unstable: the system can do free work if the nucleus loses atoms and r decreases. The opposite is true when a fluctuation gives a nucleus with r > r*. Then, free work is done when the nucleus gains atoms, and it will tend to grow. To summar- ise, if random fluctuations in the liquid give crystals with r > r* then stable nuclei will form, and solidification can begin. To calculate r* we differentiate eqn. (7.1) to give Kinetics of structural change: II – nucleation 69 Fig. 7.1. The work needed to make a spherical nucleus. d d SL W r rrH TT T f m m ( ) .=− − 84 2 ππγ ∆ (7.2) We can then use the condition that dW f /dr = 0 at r = r* to give r T HT T m m * ( ) = − 2 γ SL ∆ (7.3) for the critical radius. We are now in a position to go back and look at what is happening in the liquid in more detail. As we said earlier, small groups of liquid atoms are continually shaking themselves together to make tiny crystals which, after a short life, shake themselves apart again. There is a high probability of finding small crystals, but a small probabil- ity of finding large crystals. And the probability of finding crystals containing more than 10 2 atoms (r Պ 1 nm) is negligible. As Fig. 7.2 shows, we can estimate the tem- perature T hom at which nucleation will first occur by setting r* = 1 nm in eqn. (7.3). For typical values of γ SL , T m and ∆H we then find that T m − T hom ≈ 100 K, so an enormous undercooling is needed to make nucleation happen. This sort of nucleation – where the only atoms involved are those of the material itself – is called homogeneous nucleation. It cannot be the way materials usually solidify because (usually) an undercooling of 1°C or less is all that is needed. Homogeneous nucleation has been observed in ultraclean laboratory samples. But it is the exception, not the rule. Heterogeneous nucleation Normally, when a pond of water freezes over, or when a metal casting starts to solidify, nucleation occurs at a temperature only a few degrees below T m . How do we explain 70 Engineering Materials 2 Fig. 7.2. Homogeneous nucleation will take place when the random crystals can grow, i.e. when r > r *. this easy nucleation? Well, liquids like pond water or foundry melts inevitably contain solid particles of dirt. These act as catalysts for nucleation: they give a random crystal a “foothold”, so to speak, and allow it to grow more easily. It is this heterogeneous nucleation which is responsible for solidification in all practical materials situations. Heterogeneous nucleation is most likely to occur when there is a strong tendency for the crystal to stick to the surface of the catalyst. This sticking tendency can be described by the angle of contact, θ , shown in Fig. 7.3: the smaller θ , the better the adhesion. Anyone who has tried to get electronic solder to stick to a strip of copper will understand this well. If the copper is tarnished the solder will just roll around as a molten blob with θ = 180°, and will not stick to the surface at all. If the tarnished Fig. 7.3. Heterogeneous nucleation takes place on the surface of a solid catalyst. For the catalyst to be effective there must be a strong tendency for it to be “wetted” by the crystal, i.e. q must be small. Kinetics of structural change: II – nucleation 71 copper is fluxed then θ may decrease to 90°: the molten solder will stay put on the copper but it will not spread. Only when the copper is both clean and fluxed will θ be zero, allowing the solder to “wet” the copper. If we know the contact angle we can work out r* quite easily. We assume that the nucleus is a spherical cap of radius r and use standard mathematical formulae for the area of the solid–liquid interface, the area of the catalyst–solid interface and the vol- ume of the nucleus. For 0 ഛ θ ഛ 90° these are: solid–liquid area = 2 π r 2 (1 − cos θ ); (7.4) catalyst–solid area = π r 2 (1 − cos 2 θ ); (7.5) nucleus volume = 2 3 1 3 2 1 2 3 3 πr cos cos −+       θθ . (7.6) Then, by analogy with eqn. (7.1) we can write W f = 2 π r 2 (1 − cos θ ) γ SL + π r 2 (1 − cos 2 θ ) γ CS − π r 2 (1 − cos 2 θ ) γ CL −−+       − cos cos ( ) . 2 3 1 3 2 1 2 3 3 πr H TT T m m θθ ∆ (7.7) Note that this equation has two energy terms that did not appear in eqn. (7.1). The first, π r 2 (1 − cos 2 θ ) γ CS , is the energy needed to create the new interface between the catalyst and the solid. The second, – π r 2 (1 − cos 2 θ ) γ CL , is the energy released because the area of the catalyst–liquid interface is smaller after nucleation than it was before. As it stands, eqn. (7.7) contains too many unknowns. But there is one additional piece of information that we can use. The interfacial energies, γ SL , γ CS and γ CL act as surface tensions in just the way that a soap film has both a surface energy and a surface tension. This means that the mechanical equilibrium around the edge of the nucleus can be described by the triangle of forces γ CL = γ CS + γ SL cos θ . (7.8) If we substitute this result into eqn. (7.7) we get the interfacial energy terms to reduce to 2 π r 2 (1 − cos θ ) γ SL + π r 2 (1 − cos 2 θ )(– γ SL cos θ ), or 2 π r 2 {1 − 3 2 cos θ + 1 2 cos 3 θ } γ SL . (7.9) The complete result for W f then becomes W f = 1 3 2 1 2 2 2 3 32 3 cos cos ( ) −+       − −       θθπ π r r H TT T m m γ SL ∆ . (7.10) Finally, if we use the condition that dW f /dr = 0 at r = r* we get r* = 2 γ SL T HT T m m ∆ ( )− (7.11) for the critical radius in heterogeneous nucleation. 72 Engineering Materials 2 If we compare eqns (7.11) and (7.3) we see that the expressions for the critical radius are identical for both homogeneous and heterogeneous nucleation. But the expressions for the volume of the critical nucleus are not. For homogeneous nucleation the critical volume is Vr * ( * ) hom hom = 4 3 3 π (7.12) whereas for heterogeneous nucleation it is Vr het het * ( * ) { cos cos }=−+ 2 3 3 3 2 1 2 3 1 πθθ . (7.13) The maximum statistical fluctuation of 10 2 atoms is the same in both homogeneous and heterogeneous nucleation. If Ω is the volume occupied by one atom in the nucleus then we can easily see that VV * * . hom ==10 2 Ω het (7.14) Equating the right-hand terms of eqns (7.12) and (7.13) then tells us that r* het = r hom / * ( { cos cos }) . 1 2 3 2 1 2 313 1 −+ θθ (7.15) If the nucleus wets the catalyst well, with θ = 10°, say, then eqn. (7.15) tells us that r* het = 18.1r* hom . In other words, if we arrange our 10 2 atoms as a spherical cap on a good catalyst we get a much bigger crystal radius than if we arrange them as a sphere. And, as Fig. 7.4 explains, this means that heterogeneous nucleation always “wins” over homogeneous nucleation. It is easy to estimate the undercooling that we would need to get heterogeneous nucleation with a 10° contact angle. From eqns (7.11) and (7.3) we have 2 18 1 2 γγ SL het SL hom T HT T T HT T m m m m ∆∆( ) . ( ) , − =× − (7.16) which gives T m − T het = TT m . hom − 18 1 ≈ 10 18 1 2 K . ≈ 5K. (7.17) And it is nice to see that this result is entirely consistent with the small undercoolings that we usually see in practice. You can observe heterogeneous nucleation easily in carbonated drinks like “fizzy” lemonade. These contain carbon dioxide which is dissolved in the drink under pres- sure. When a new bottle is opened the pressure on the liquid immediately drops to that of the atmosphere. The liquid becomes supersaturated with gas, and a driving force exists for the gas to come out of solution in the form of bubbles. The materials used for lemonade bottles – glass or plastic – are poor catalysts for the heterogeneous nucleation of gas bubbles and are usually very clean, so you can swallow the drink before it loses its “fizz”. But ordinary blackboard chalk (for example), is an excellent former of bubbles. If you drop such a nucleant into a newly opened bottle of carbon- ated beverage, spectacular heterogeneous nucleation ensues. Perhaps it is better put another way. Chalk makes lemonade fizz up. Kinetics of structural change: II – nucleation 73 Fig. 7.4. Heterogeneous nucleation takes place at higher temperatures because the maximum random fluctuation of 10 2 atoms gives a bigger crystal radius if the atoms are arranged as a spherical cap. Nucleation in solids Nucleation in solids is very similar to nucleation in liquids. Because solids usually contain high-energy defects (like dislocations, grain boundaries and surfaces) new phases usually nucleate heterogeneously; homogeneous nucleation, which occurs in defect-free regions, is rare. Figure 7.5 summarises the various ways in which nucleation can take place in a typical polycrystalline solid; and Problems 7.2 and 7.3 illustrate how nucleation theory can be applied to a solid-state situation. Summary In this chapter we have shown that diffusive transformations can only take place if nuclei of the new phase can form to begin with. Nuclei form because random atomic vibrations are continually making tiny crystals of the new phase; and if the temper- ature is low enough these tiny crystals are thermodynamically stable and will grow. In homogeneous nucleation the nuclei form as spheres within the bulk of the material. In 74 Engineering Materials 2 Fig. 7.5. Nucleation in solids. Heterogeneous nucleation can take place at defects like dislocations, grain boundaries, interphase interfaces and free surfaces. Homogeneous nucleation, in defect-free regions, is rare. heterogeneous nucleation the nuclei form as spherical caps on defects like solid surfaces, grain boundaries or dislocations. Heterogeneous nucleation occurs much more easily than homogeneous nucleation because the defects give the new crystal a good “foothold”. Homogeneous nucleation is rare because materials almost always contain defects. Postscript Nucleation – of one sort or another – crops up almost everywhere. During winter plants die and people get frostbitten because ice nucleates heterogeneously inside cells. But many plants have adapted themselves to prevent heterogeneous nucleation; they can survive down to the homogeneous nucleation temperature of −40°C. The “vapour” trails left by jet aircraft consist of tiny droplets of water that have nucleated and grown from the water vapour produced by combustion. Sub-atomic particles can be tracked during high-energy physics experiments by firing them through super- heated liquid in a “bubble chamber”: the particles trigger the nucleation of gas bubbles which show where the particles have been. And the food industry is plagued by nucleation problems. Sucrose (sugar) has a big molecule and it is difficult to get it to crystallise from aqueous solutions. That is fine if you want to make caramel – this clear, brown, tooth-breaking substance is just amorphous sucrose. But the sugar refiners have big problems making granulated sugar, and will go to great lengths to get adequate nucleation in their sugar solutions. We give examples of how nucleation applies specifically to materials in a set of case studies on phase transformations in Chapter 9. Further reading D. A. Porter and K. E. Easterling, Phase Transformations in Metals and Alloys, 2nd edition, Chapman and Hall, 1992. G. J. Davies, Solidification and Casting, Applied Science Publishers, 1973. G. A. Chadwick, Metallography of Phase Transformations, Butterworth, 1972. Kinetics of structural change: II – nucleation 75 Problems 7.1 The temperature at which ice nuclei form homogeneously from under-cooled water is –40°C. Find r* given that γ = 25 mJ m –2 , ∆H = 335 kJ kg –1 , and T m = 273 K. Estimate the number of H 2 O molecules needed to make a critical-sized nucleus. Why do ponds freeze over when the temperature falls below 0°C by only a few degrees? [The density of ice is 0.92 Mg m –3 . The atomic weights of hydrogen and oxygen are 1.01 and 16.00 respectively.] Answers: r*, 1.11 nm; 176 molecules. 7.2 An alloy is cooled from a temperature at which it has a single-phase structure ( α ) to a temperature at which the equilibrium structure is two-phase ( α + β ). During cooling, small precipitates of the β phase nucleate heterogeneously at α grain boundaries. The nuclei are lens-shaped as shown below. Show that the free work needed to produce a nucleus is given by WrrG f cos cos =− +       −       1 3 2 1 2 4 4 3 32 3 θθπγπ αβ ∆ where ∆G is the free work produced when unit volume of β forms from α. You may assume that mechanical equilibrium at the edge of the lens requires that γ GB = 2 γ αβ cos θ . Hence, show that the critical radius is given by rG* / .,= 2 γ αβ ∆ 7.3 Pure titanium is cooled from a temperature at which the b.c.c. phase is stable to a temperature at which the c.p.h. phase is stable. As a result, lens-shaped nuclei of the c.p.h. phase form at the grain boundaries. Estimate the number of atoms needed to make a critical-sized nucleus given the following data: ∆H = 3.48 kJ mol –1 ; atomic weight = 47.90; T e – T = 30 K; T e = 882°C; γ = 0.1 Jm –2 ; density of the c.p.h. phase = 4.5 Mg m –3 ; θ = 5°. Answer: 67 atoms. αβ αβ γ γ Spherical cap of radius r θ γ θ GB α α β [...]... = 910 2 αβ (9 14 + 27 3) 2 αβ = × 29 7 ∆ H (9 14 − 910) ∆H (8 .4) But at 900°C the critical radius is r* = 900 2 αβ (9 14 + 27 3) 2 αβ = × 85 ∆ H (9 14 − 900) ∆H (8.5) Thus (r*10 /r*00) = (29 7/85) = 3.5 9 9 (8.6) As Fig 8.3 shows, grain boundary nuclei will be geometrically similar at all temperatures The volume V* of the lens-shaped nucleus will therefore scale as (r*)3, i.e (V* /V* ) = 3.53 = 43 910... We have already looked at grain boundary nucleation in Problems 7 .2 and 7.3 Problem 7 .2 showed that the critical radius for grain boundary nucleation is given by r* = 2 αβ / ∆G (8 .2) Since ∆G = ∆ H (Te − T )/Te (see eqn 6.16), then r* = 2 αβ Te ∆ H (Te − T ) (8.3) 78 Engineering Materials 2 Fig 8 .2 In a diffusive transformation the volume transforming per second increases linearly with the number... Transformations in Metals and Alloys, 2nd edition, Chapman and Hall, 19 92 R W K Honeycombe and H K D H Bhadeshia, Steels: Microstructure and Properties, 2nd edition, Arnold, 1995 K J Pascoe, An Introduction to the Properties of Engineering Materials, Van Nostrand Reinhold, 1978 R E Reed-Hill, Physical Metallurgy Principles, Van Nostrand Reinhold, 19 64 88 Engineering Materials 2 Problems 8.1 Compare and contrast... transformation Semi-schematic only Fig 8.6 If we quench f.c.c iron from 9 14 C to room temperature at a rate of about 105°C s−1 we expect to prevent the diffusive f.c.c → b.c.c transformation from taking place In reality, below 550°C the iron will transform to b.c.c by a displacive transformation instead  82 Engineering Materials 2 Fig 8.7 The displacive f.c.c → b.c.c transformation in iron B.c.c lenses... of the lattice The crystallographic relationships shown here are for pure iron  84 Engineering Materials 2 Fig 8.9 (a) The unit cells of f.c.c and b.c.c iron (b) Two adjacent f.c.c cells make a distorted b.c.c cell If this is subjected to the “Bain strain” it becomes an undistorted b.c.c cell This atomic “switching” involves the least shuffling of atoms As it stands the new lattice is not coherent with... hard you can hear a faint “pinging” sound as * This may seem a strange result – after all, only 68% of the volume of the b.c.c unit cell is taken up by atoms, whereas the figure is 74% for f.c.c Even so, the largest holes in b.c.c (diameter 0.0 722 nm) are smaller than those in f.c.c (diameter 0.1 04 nm) Kinetics of structural change: III – displacive transformations 87 Fig 8.13 Displacive transformations... nucleus containing only two or three atoms! To define a b.c.c crystal we would have to assemble at least 20 or 30 atoms But it will still be far easier to fluctuate 30 atoms into position than to fluctuate 100 Our argument is thus valid qualitatively, if not quantitatively 80 Engineering Materials 2 Fig 8 .4 The diffusive f.c.c → b.c.c transformation in iron: overall rate of transformation as a function of... gives a different result for the driving force 90 Engineering Materials 2 Fig 9.1 Rain falls when the water droplets in clouds turn to ice This can only happen if the clouds are below 0°C to begin with If the droplets are clean, ice can form only in the unlikely event that the clouds cool down to the homogeneous nucleation temperature of 40 °C When dust particles are present they can catalyse nucleation... understand how pure iron transforms we will have no problem in generalising to iron–carbon alloys) Now, as we saw in Chapter 2, iron has different crystal structures at different temperatures Below 9 14 C the stable structure is b.c.c., but above 9 14 C it is f.c.c If f.c.c iron is cooled below 9 14 C the structure becomes thermodynamically unstable, and it tries to change back to b.c.c This f.c.c → b.c.c transformation... α + Fe3C) The eutectoid reaction can only start when the steel has been cooled below 723 °C The nose of the C-curve occurs at ≈ 525 °C (Fig 8.11), about 175°C lower than the nose temperature of perhaps 700°C for pure iron (Fig 8.5) Diffusion is much slower at 525 °C than it is at 700°C As a result, a cooling rate of ≈ 20 0°C s−1 misses the nose of the 1% curve and produces martensite Pure iron martensite . is r* 910 = 2 9 14 27 3 9 14 910 γ αβ ∆H ( ) ( ) + − = 2 γ αβ ∆H × 29 7. (8 .4) But at 900°C the critical radius is r* 900 = 2 9 14 27 3 9 14 900 γ αβ ∆H ( ) ( ) + − = 2 γ αβ ∆H ×. the vol- ume of the nucleus. For 0 ഛ θ ഛ 90° these are: solid–liquid area = 2 π r 2 (1 − cos θ ); (7 .4) catalyst–solid area = π r 2 (1 − cos 2 θ ); (7.5) nucleus volume = 2 3 1 3 2 1 2 3 3 πr . (7.1) we can write W f = 2 π r 2 (1 − cos θ ) γ SL + π r 2 (1 − cos 2 θ ) γ CS − π r 2 (1 − cos 2 θ ) γ CL −−+       − cos cos ( ) . 2 3 1 3 2 1 2 3 3 πr H TT T m m θθ ∆ (7.7) Note