Phase Transformations in Materials: Nucleation, Growth, and Heat Treatment - PDF

Okay, here is the transcription of the images provided in a structured markdown format. ### Phase Transformation The first image is a block diagram depicting phase transformations: * Liquid (L) to Gas (G) via Evaporation * Gas (G) to Liquid (L) via Condensation * Liquid (L) to Solid (S) via...

Okay, here is the transcription of the images provided in a structured markdown format. ### Phase Transformation The first image is a block diagram depicting phase transformations: * Liquid (L) to Gas (G) via Evaporation * Gas (G) to Liquid (L) via Condensation * Liquid (L) to Solid (S) via Freezing/Casting/Single Crystal Growth * Solid (S) to Liquid (L) via Melting * Solid (S) to Gas (G) via Sublimation * Gas (G) to Solid (S) via Deposition * Solid (S) to Solid (S') via Solid State Transformation *Heat Treatment: Solid $\leftrightarrow$ Solid"* ### Thermodynamic Driving Force What is the thermodynamic driving force for a phase transformation? $\alpha \rightarrow \beta$ $G_{\alpha} \quad G_{\beta}$ If $G_{\beta} < G_{\alpha} \rightarrow$ Transformation feasible (thermodynamically) Driving force $\Delta G = G_{\beta} - G_{\alpha}$ ### Free Energy Diagram Graph showing Free energy (G) vs Temperature (T): * Liquid (L) phase: $G_L(T)$ * Solid (S) phase: $G_S(T)$ * $\Delta G = G_{S}(T) - G_{L}(L)$ * $G_L(T_m) = G_S(T_m)$ * $\Delta G = 0$ * Where $T_m$ is the melting temperature ### Estimation of $\Delta G$ for Freezing $\Delta G = G_S - G_L$ $\Delta G(T) = G_S(T) - G_L(T)$ $\Delta G(T_m) = G_S(T_m) - G_L(T_m) = 0$ $G_S = H_S - TS_S$ $G_L = H_L - TS_L$ $\Delta G(T) = (H_S - H_L) - T(S_S - S_L)$ $\Delta G(T) = \Delta H(T) – T\Delta S(T_m)$ $\Delta G(T_m) = \Delta H(T_m) -T_m \Delta S(T_m)$ $0 = \Delta H(T_m) - T_m \Delta S(T_m)$ $\Delta S(T_m) = \frac{\Delta H(T_m)}{T_m}$ Latent heat of melting $\Delta S_m = \frac{\Delta H_m}{T_m}$ $\Delta S_m \equiv \Delta S(T_m)$ $\Delta H_m \equiv \Delta H(T_m)$ ### Assuming $\Delta H(T)$ and $\Delta S(T)$ to be independent of T: $\Delta H(T) = \Delta H(T_m) = \Delta H_m$ $\Delta S(T) = \Delta S(T_m) = \Delta S_m$ $\Delta G(T) = \Delta H(T) - T\Delta S(T)$ $= \Delta H_m - T\Delta S_m$ $= \Delta H_m - T (\frac{\Delta H_m}{T_m}) $ $\Delta G(T) = \Delta H_m - \frac{T}{T_m} \Delta H_m$ $\Delta G(T) = (1 - \frac{T}{T_m}) \Delta H_m$ $\Delta G(T) = \frac{T_m - T}{T_m} \Delta H_m$ $\Delta G(T) = \frac{\Delta T}{T_m} \Delta H_m$ $\Delta T = T_m - T = \text{undercooling}$ $\Delta G(T) = \Delta H_m - \frac{T}{T_m} \Delta H_m$ $\Delta G(T) = (1 - \frac{T}{T_m}) \Delta H_m$ $\Delta G(T) = \frac{T_m - T}{T_m} \Delta H_m$ $\Delta G(T) = \frac{\Delta T}{T_m} \Delta H_m$ Driving force for freezing. ### Phase Transformation: Nucleation + Growth Phase transformation = nucleation + growth * nucleation - first formation of a new phase (nucleus) * growth - further increase in the size of the nucleus What is the size of the nucleus? Diagram of Liquid $\rightarrow$ Solid Solidification ### Nucleation in a Liquid A solid ball of radius r nucleates in a liquid. $G_S$ = Gibbs free energy change per unit volume of solid. $G_L$ = Gibbs free energy per unit volume of liquid Assume Density of Solid = Density of liquid $\Rightarrow$ No volume change on transformation. $\Delta G_f$ = free energy change due to formation of a solid ball of radius r in a liquid. Diagram showing a sphere of radius $r$ nucleating within a liquid. ### Equation for $\Delta G$ $T < T_m$ $\Delta G_f = \frac{4}{3} \pi r^3 (G_S - G_L) + 4 \pi r^2 \gamma$ $-ve \qquad +ve$ $\gamma$ = Energy per unit area of the solid/liquid interface Diagram showing $\Delta G_f$ vs $r$, and the contributions of the volume and surface terms. The volume free energy is negative while the surface free energy is positive. Diagram also shows max($\Delta G_f$). $4\pi r^2\gamma$ $\Delta G_f = \frac{4}{3} \pi r^3 (G_S - G_L) + 4\pi r^2\gamma$ $\frac{4}{3} \pi r^3 (G_S - G_L)$ ### Critical Radius for Nucleation For $r < r^* \quad \Delta G_f \uparrow$ as $r \uparrow \Rightarrow$ Growth not possible. For $r > r^* \quad \Delta G_f \downarrow$ as $r \uparrow \Rightarrow$ Growth is possible. $r^*$= CRITICAL RADIUS FOR NUCLEATION At $r^*$, $\Delta G_f$ is max. $\frac{\partial \Delta G_f}{\partial r}|_{r=r^*} = 0$ $\Delta G_f = \frac{4}{3} \pi r^3 (G_S - G_L) + 4 \pi r^2 \gamma$ $\frac{\partial \Delta G_f}{\partial r} = \frac{4}{3} \pi 3r^2 (G_S - G_L) + 4 \pi 2 r \gamma = 4 \pi r^2 (G_S - G_L) + 8 \pi r \gamma$ Condition for maximum $\frac{\partial \Delta G_f}{\partial r}|_{r=r^*} = 0$ $4 \pi {r^*}^2 (G_S - G_L) + 8 \pi {r^*}\gamma = 0$ $r^* = \frac{-2\gamma}{G_S - G_L}$ driving force ### Critical Radius for Nucleation Equation $\Delta G_f = \frac{4}{3} \pi r^3 (G_S - G_L) + 4 \pi r^2 \gamma$ $\frac{\partial \Delta G_f}{\partial r} = \frac{4}{3} \pi 3r^2 (G_S - G_L) + 4 \pi 2 r \gamma$ $= 4 \pi r^2 (G_S - G_L) + 8 \pi r \gamma $ Condition for maximum $\frac{\partial \Delta G_f}{\partial r}|_{r=r^*} = 0$ $ 4 \pi {r^*}^2 (G_S - G_L) + 8 \pi {r^*}\gamma = 0$ $ r^* = \frac{-2\gamma}{G_S - G_L}$ Obstacle to nucleation(barrier) driving force $G_S - G_L = \Delta G_V = \frac{\Delta H \Delta T}{T_m}$ Latent heat of fusion $\Delta T = T-T_m$ undercooling. melting point. $r^* = \frac{-2\gamma}{G_S - G_L} = \frac{-2\gamma}{\Delta Gv} $ $r^* = \frac{-2\gamma}{ \Delta H_m \frac{\Delta T}{T_m} }$ Higher undercooling (lower T of solidification) $\Rightarrow$ smaller $r^*$ ### Nucleation and Capillary Rise | Property | Nucleation | Capillary Rise | | :---------------- | :------------------ | :----------------- | | Driving Force | Volume free energy | | | Opposing Force | Surface energy | | | Equilibrium | Unstable | | | Diagram | Sphere in a liquid | Liquid in a Column | | Formula for Ratio | | | $ \gamma_{GL}$ = Energy per unil area of ize $TGA = Energy per unit area of the glanslait interfure. $AEH = _1 TT 2/gH2 – 2ARH(IGA GL)H 2 ан H21=0 $H$ = 784 – -742)pg R. R. Prasad “On Capillary Rise and Nucleation” Journal of Chemical Education Vol. 85 , No. 10, Oct 2008, 1389. ### Nucleation Rate and Growth Rate $x$: 1 less no. of nuclei ` slow down due to 2 impingements, reduced volume { for transformi aTima Nucleation rate, I (m35'): No. of nucleation events per unit volume per second. Growth rate, G=dR (ms'): Rate of increase dt of the size of growing particle (R: particle size) Overall transformation rate dx (5"): Fraction transformed per second. >X= fraction transformed. ## Temerature Dependence of $I$ and $G$ Images are Graphs showing the effect of temperature on Nucleation rate ($I$) and Growth Rate ($G$) d = f (I,G) TG T! ## Time-Temperature-Transformation (TTT) Diagram L>S transformation 1. Diagram or graph showing transformation over time. 2. x and T on axis indicating what transformation has happened at given time and temp. TTT Diagram for L>S transformation 1. Graph diagram 2. High nucleation rate is dominant at low temperature 3. High growth rate is dominant at high temperature Liquid Solid Supressed diagram 1. transformation to solid is supessedlUndercooled liquid = Very small/no crystal graias 2. At very low T visoosity has valuable properties like soild Homogeneous and heterogenerous Nucleation . Homogeneous nucleation: 1. If solid is added to Liquid both can produce nuclei on eachother. 2. If solid is solid can make nucleus on solid. Heterogeneous Nucleation al ab contact angle ab spherical cap of in 4 on 9 = Vi aR*. -52 - S6) = 2-3 cose + cos" T+B a . Heterogeneous nucleation is favoured over homogeneous nueleation for smaller contacto angles 0. Heat Treatment of Steels Hardness w4.C Microstrictore Heat Stes +C Microstructure Hardess ABC . Treatment 1 . Coarse Peartife Tine Pearlite Tempered Martinsite Anseating Normalizing C. * . Bainite Alustexpering Tempering Quenching . * . Marten site Non equilibrium plane forms on quenching of austerite Interstiial solid solution of C in Fe eith Body-centred tetragomal structure. d 8=1 Tvery hard and brittle. 6 Martensitic Transformation - Martensite - transformation is expert extremely vapid. Amount transformed depends on temperature. > Athermo: Rate of transformatioe that hasnt HUSTENPERING AND BAINITE. Phase us Micoeconstituant. Peaalite and Bailnile ave mixtures ot the same two phases lat Tes() but are different microConstituent, due to different shapes and distribution ot these phases. Tempering ie heating a quenched specimen = steel & below che entectord temperature = Quenching e Martensite Volume change during Austenite «- Martersits transformation COP ser Density P, > Pan . M Velene increases. tion. Tempering produces tempered martensite - uith lower handness than maxtensite but = better coughness and duchlity = Residual Stressee = and Quenrh Cracke - by Stessee. . TTTT Diagram of the Following Alleying additions: Mas, Cr, Mo, Si, - separale rete, for Pand Bo - Shifted to the - right Substiutional Alleyngby shifting the nose of a curve to the - right, enharves the hardenabily Extenption to the . e