Solidification Most metals are melted and then cast into semifinished or finished shape. Solidification of a metal can be divided into the following steps: •Formation of a stable nucleus •Growth of a stable nucleus.
Formation of stable Growth of crystals Grain structure nuclei
Polycrystalline Metals In most cases, solidification begins from multiple sites, each of which can produce a different orientation The result is a “polycrystalline” material consisting of many small crystals of “grains” Each grain has the same crystal lattice, but the lattices are misoriented from grain to grain Driving force: solidification ⇒ For the reaction to proceed to the right ∆G AL AS V must be negative. • Writing the free energies of the solid and liquid as: S S S GV = H - TS L L L GV = H - TS ∴∴∴ ∆GV = ∆H - T∆S • At equilibrium, i.e. Tmelt , then the ∆GV = 0 , so we can estimate the melting entropy as:
∆S = ∆H/Tmelt where -∆H is the latent heat (enthalpy) of melting. • Ignore the difference in specific heat between solid and liquid, and we estimate the free energy difference as: T ∆H × ∆T ∆GV ≅ ∆H − ∆H = TMelt TMelt NUCLEATION The two main mechanisms by which nucleation of a solid particles in liquid metal occurs are homogeneous and heterogeneous nucleation. Homogeneous Nucleation Homogeneous nucleation occurs when there are no special objects inside a phase which can cause nucleation. For instance when a pure liquid metal is slowly cooled below its equilibrium freezing temperature to a sufficient degree numerous homogeneous nuclei are created by slow-moving atoms bonding together in a crystalline form. Consider the free energy changes when some atoms in the liquid collapse and agglomerate to form a solid of radius r. The energy changes involve two terms: (a)The chemical free energy change associated with the transfer of
atoms from liquid to solid state (∆∆∆Gv); (b)the interfacial energy ( γγγ)due to the creation of new interface (liquid-solid interface)!
Assume that ∆∆∆Gv is the change in free energy per unit volume and ∆∆∆GT is the total Free energy change, r is the radius of the nucleus 4πr 3 ⇒ ∆G = ∆G + 4πr 2γ T 3 V d∆G 2γ T = 4πr 2 ∆G + 8πr γ = 0 ⇒ r = − dr 0 V 0 0 ∆G r = r0 V 4πγr 2 16πγ 3 ⇒ 0 ∆GT (r0 ) = = 2 3 3∆GV r0 : critical radius; * for r < r0 : the growth of the droplet ⇒ ∆∆∆GT ↑↑↑ ⇒ the embryos should shrink and disappear!
* for r > r0 : the growth of the droplet ⇒ ∆∆∆GT ↓↓↓ ⇒ the nuclei could steadily grow!
γ ∆H f ∆T 2γT ∆G = ⇒ r = m V T 0 m ∆H f ∆T Where:
-∆∆∆Hf = Latent Heat of fusion ∆∆∆T = amount of undercooling at which the nucleus form Heterogeneous Nucleation Heterogeneous nucleation is the nucleation that occurs in a liquid on the surfaces of its container, insoluble impurities or other structural material (catalyst) which lower the critical free energy required to form a stable nucleus.
γγγ Liquid L-S γγγL-C = γγγS-C + γγγL-S Cos θθθ γγγ L-C θθθ Solid
γγγS-C
ro
Catalyst Supersaturated Solutions • If the liquid is just at the freezing point, only a few molecules stick, because they have comparatively high energy • As the liquid is cooled, more molecules can form into nuclei. • When the nucleus is big enough (because of undercooling) the supercooled liquid suddenly changes to a solid. • Metals often experience undercooling of 50 to 500 oC • Homogeneous nucleation usually only occurs in the lab. • Impurities provide a “seed” for nucleation • Solidification can start on a wall. • It’s like cloud seeding, or water condensing on the side of a glass. • Adding impurities on purpose is called inoculation • Nucleation begins • Chill Zone • Columnar Zone (a) There may be dendrites in the columnar zone (b) Grains grow in preferred directions • Equiaxed Zone
Defects in Solids Imperfections in Solids Materials are often stronger when they have defects. The mechanical and electrical properties of a material are affected by the presence of defects. The study of defects is divided according to their dimension: Examples of the large impact of defects : Gemstones – Hope Diamond blue color due to boron impurities (ppm)
• Metals - ductility, stiffness, brittleness, etc. drastically affected Bonding The defects have a profound + effect on the macroscopic Structure Properties properties of materials + Defects
The processing determines the defects
Type of Bonding Microstructure
Chemical Thermomechanical Composition Processing
Crystal Structure 0D – point defects: vacancies and interstitials impurities. 1D – linear defects: dislocations (edge, screw, mixed) 2D – planar defects: grain boundaries, surfaces. 3D – extended defects: pores, cracks. Point Defects (0D) Vacancies and Self-Interstitials A vacancy is a lattice position that is vacant because the atom is missing. It is created when the solid is formed. They occur naturally as a result of thermal vibrations. An interstitial is an atom that occupies a place outside the normal lattice position. It may be the same type of atom as the others (self interstitial) or an impurity atom. In the case of vacancies and interstitials, there is a change in the coordination of atoms around the defect. This means that the forces are not balanced in the same way as for other atoms in the solid (lattice distortion). Vacancy - a lattice position that is vacant because the atom is missing. Interstitial - an atom that occupies a place outside the normal lattice position. It may be the same type of atom as the others (self interstitial) or an impurity interstitial atom. The number of vacancies formed by thermal agitation follows an Arrhenius type of equation: QV QV NV = NA exp− NV = NA exp− kBT RT where NA is the total number of atoms in the solid, QV is the energy required to form a vacancy (per atom or per mole), kB is Boltzmann constant, R is the gas constant and T the temperature in Kelvin. Note that kT(300 K) = 0.025 eV (room temperature) is much smaller than typical vacancy formation energies.
For instance, QV(Cu) = 0.9 eV/atom . -16 This means that NV/N A at room temperature is exp(-36) = 2.3 × 10 , an insignificant number. Thus, a high temperature is needed to have a high thermal concentration of vacancies . Even so, NV/N A is typically only about 0.0001 at the melting point. Calculate equilibrium number of vacancies per cubic meter for copper at 1000 oC Given: Activation Energy per vacancy = 0.9 eV/atom atomic weight of copper = 63.5 g/mol and density at 1000 oC = 8.40 g/cm 3 -23 -5 Boltzmann’s constant kB=1.38 × 10 J/atom-K=8.62 × 10 eV/atom-K Solution: Determine N, number of atomic sites per cubic meter for Cu
N ρ (6.023 X 10 23 atoms / mol )(8.40 g / cm 3 )(10 6 cm 3 / m 3 ) N = a = ACu 63.5g / mol N =8.0 X 10 28 atoms / m 3 Thus ,the number of vacancies at1000 0 C (1273 K )is equal to Q N =N exp − v v kT
28 3 (0.9eV ) NV =(8.0 X 10 atoms / m )exp (8.62 X 10 −5 eV / K )(1273 K )
25 3 NV = 2.2 X 10 vacancies / m Imperfections in Ceramics Point defects in ionic crystals are charged . Coulombic forces are large and any charge imbalance wants to be balanced. Charge neutrality --> several point defects created: Frenkel defect : a cation vacancy and a cation interstitial or an anion (negative ion) vacancy and anion interstitial. (Anions are larger so it is not easy for an anion interstitial to form).+ Schottky defect: pair of anion and cation vacancies-
Schottky defect Frenkel defect Imperfections in Ceramics • Frenkel or Schottky defects: no change in cation to anion ratio →→→ compound is stoichiometric • Non-stoichiometry (composition deviates from the one predicted by chemical formula) may occur when one ion type can exist in two valence states, (e.g. Fe 2+ , Fe 3+ ). In FeO, usual Fe valence state is 2+. If two Fe ions are in 3+ state, then a Fe vacancy is required to maintain charge neutrality →→→ fewer Fe ions →→→ non- stoichiometry Impurities in Ceramics Impurity atoms can exist as either substitutional or interstitial solid solutions Substitutional ions substitute for ions of like type Interstitial ions are small compared to host structure (anion interstitials are unlikely). Solubility high if ion radii and charges match Incorporation of ion with different Interstitial impurity atom charge state requires compensation by point defects. Substitutional impurity ions Impurities in Solids All real solids are impure. A very high purity material, say 99.9999% pure (called 6N – six nines) contains ~ 6 × 10 16 impurities per cm 3. Impurities are often added to materials to improve the properties. For instance, carbon added in small amounts to iron makes steel, which is stronger than iron. Boron impurities added to silicon drastically change its electrical properties. Solid solutions are made of a host, the solvent or matrix) which dissolves the solute (minor component). The ability to dissolve is called solubility . Solid solutions are: •homogeneous •maintain crystal structure •contain randomly dispersed impurities (substitutional or interstitial) SolidSolid SolutionSolution Solids dissolve other solids
Solid solutions are made of a host (the solvent or matrix) which dissolves the minor component (solute). The ability to dissolve is called solubility. So, most of engineering materials are solid solutions, i.e., alloys: solvent and solutes Substitutional - solvent : usually the element present in greatest amount (sometimes referred to as “host Ni/Cu atoms”) - solute : usually the element present in minor concentration.
Two ways : depending on the size and host structure Interstitial
C/Fe
For fcc, bcc, hcp structures the voids (or interstices) between the host atoms are relatively small ⇒ atomic radius of solute should be significantly less than solvent. Normally, max. solute concentration ≤ 10%, (2% for C-Fe) Second Phase: as solute atoms are added, new compounds/structures are formed, or solute forms local precipitates . Whether the addition of impurities results in formation of solid solution or second phase depends the nature of the impurities, their concentration and temperature, pressure… Factors for high solubility in Substitutional alloys (Hume-Rothery Solubility Rules) •Similar atomic size (to within 15%) •Similar crystal structure •Similar electronegativity (otherwise a compound is formed) •Similar valence Composition can be expressed in weight percent, useful when making the solution, and in atomic percent, useful when trying to understand the material at the atomic level.
Example Ni is completely miscible in Cu (all rules apply) Zn is partially miscible in Cu (different valence, different crystal structure) Ni - Cu binary isomorphous Limited solubility alloy (eutectic) alloys Ni Cu Pb Cu crystal structure FCC FCC FCC FCC atomic radius 0.125 0.128 0.175 0.128 2.4% 36.7% electronegativities 1.8 1.8 1.6 1.8 valence 2 + 2 + 2+, 4+ 2 +
Solubility Cu in Ni 100% Solubility Cu in Pb 0.1% Solid Solution : homogeneous maintain crystal structure contain randomly dispersed impurities (substitutional or interstitial) Second Phase: as solute atoms are added, new compounds / structures are formed, or solute forms local precipitates Examples (A)Calculate the critical radius (in nanometers) of a homogeneous nucleus that forms when pure liquid copper
solidifies. Assume ∆∆∆T(undercooling) = 0.2T melt . o 3 For Cu Tm=1083 C; Heat of fusion ( ∆∆∆Hf) = 1826J/cm ; Surface Energy ( γγγ) = 177x10 -7 J/cm 2; Lattice parameter of FCC copper = a=0.361nm Calculate the number of atoms in the critical-sized nucleus at this undercooling. (A) We make use of the equation for a spherical nucleus to calculate the size of the critical nucleus −7 −2 2γTm 2(177 x10 J .cm )Tm −8 r0 = = −3 = .9 69 x10 cm = .0 969 nm ∆H f ∆T ()1826 J .cm ()2.0 Tm Then, the volume of the critical nucleus is 4 4 Vol − crit − nucleus = πr 3 = π ().0 97 3 = .3 82nm3 3 3 The volume of an FCC cell is Vol − cell = a 3 = ( .0 361nm)3 = .0 047nm3 The number of cells in the critical nucleus is Vol − nucleus .3 82nm3 = = 81.34cells Vol − cell 0.047nm3 As the number of atoms in an FCC cell is 4 then the total number of atoms is TotalNumbe rofAtoms = 4x81.34 = 325 atoms (B) Calculate (a) the equilibrium number of vacancies per cubic meter in pure magnesium at 450 oC. Assume that the energy of formation of a vacancy in pure magnesium is 0.89eV. (b) What is the vacancy fraction at 600 oC? -5 (Boltzmann’s constant kB=8.62x10 eV/K) Atomic weight of Mg = 24.31g.mol -1 Density of Magnesium = 1.74g.cm -3 (B) First, we calculate the number of magnesium atoms in a cubic meter N ρ (6.023 X 10 23 atoms / mol )(1.74 g / cm 3 )(10 6 cm 3 .m −3 ) N = a = AMg 24.31g / mol N = 4.31x10 28 atoms / m 3 Thus ,the number of vacancies at 450 0 C (723K )is equal to Q v Nv =N exp − k BT
28 −3 (0.89eV ) NV =(4.31x10 atoms .m )exp − (8.62x10 −5 eV .K −1 )(723K ) 22 −3 NV = 2.71x10 vacancies .m Vacancy fraction at 600 oC N .0( 89eV ) V = exp − = .7 36x10 −6 N .8( 62x10 −5 eV .K −1 )(873 K )