ME 210 Exam 2 Review Session Dr. Aaron L. Adams, Assistant Professor

Chap 6 Fick’s Laws  What are Fick’s Laws of Diffusion? Can you define the terms and units? (Know how to apply to solve mathematical problems)  How can the rate of diffusion be predicted for some simple cases?  How does diffusion depend on structure and temperature? Chap 6 Diffusion in Solids  How does diffusion occur?  How is activation energy calculated? (Know how to apply to solve mathematical problems)  How is diffusion used in processing of materials? Chap 10 Phase Diagrams  What is a phase?  What is thermodynamic equilibrium?  What three components need to be used (established) to define the equilibrium?  How to determine the composition and fraction of a phase in a two-phase regime (Lever Rule). Chap 10 Microstructural development  How can we use a phase diagram to predict microstructure?  What is the consequence of solidification in an alloy (coring)? Re-cap of topics for Test #2

Fick’s 1st Law for steady-state diffusion ►It tells you the flow rate (i.e., “flux”) of a diffusing species due to a concentration gradient. ►J is the flux ►D is the diffusivity (e.g., m 2 /s, cm 2 /s, etc…) ►(dc/dx) is the concentration gradient – i.e., change in amount/distance (e.g., g/m, %/mm, etc…)

Flux ►Flux is essentially the amount or rate of diffusion

Refers to the amount of material flowing through an area over a period of time. This is an Arrhenius equation (very common to materials science; used to describe the statically probability of an event) Diffusion Coefficient = pre-exponential [m 2 /s] = diffusion coefficient [m 2 /s] = activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K] = absolute temperature [K] D DoDo QdQd R T

A differential nitrogen pressure exists across a 2-mm-thick steel bulkhead. After some time, steady-state diffusion of the nitrogen is established across the wall. If the nitrogen concentration on the high-pressure surface of the wall is 2 kg/m 3 and on the low-pressure surface is 0.2 kg/m 3, what is the flow of nitrogen through the wall (in kg/m 2  h)? The diffusion coefficient for nitrogen in this steel is 1.0  m 2 /s at its operating temperature. Make sure units are consistent Substitute D into flux equation Low 0.2 kg/m 3 High 2 kg/m 3 2 mm This is a flux problem… Draw the system… Then solve… A relevant problem:

Diffusion and Temperature Adapted from Fig. 5.7, Callister & Rethwisch 8e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.) D has exponential dependence on T D interstitial >> D substitutional C in  -Fe C in  -Fe Al in Al Fe in  -Fe Fe in  -Fe 1000 K/T D (m 2 /s) C in  -Fe C in  -Fe Al in Al Fe in  -Fe Fe in  -Fe T(  C) Interstitial atoms are smaller and more mobile. Also there are more empty interstitial positions than vacancies. This increases the probability of interstitial diffusion. Interstitial diffusion is much faster. WHY?

Diffusion and Crystal Structure Adapted from Fig. 5.7, Callister & Rethwisch 8e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.) D has exponential dependence on T D BCC >> D FCC C in  -FeC in  -Fe 1000 K/T D (m 2 /s) C in  -Fe C in  -Fe Al in Al Fe in  -Fe Fe in  -Fe T(  C) BCC structures are less dense (i.e., less close packed). There is more room for interstitials to move. Interstitials diffuse is much faster in BCC than FCC. WHY?

Diffusion Data Constant (aka ‘fixed’) Energy to cause diffusion Diffusivity Depends on T (not fixed)

10 Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are: D(300ºC) = 7.8 x m 2 /s Q d = 41.5 kJ/mol What is the diffusion coefficient for Cu in to Si at 350ºC? transform data D Temp = T ln D 1/T Transform data for each temperature: 5 Plot and determine slope of line: collect data

11 Plot and determine slope of line: Y = ln D X = 1/T Y = MX + B M = slope =  Y/  X 2 1 YY XX BASIC MATH

12 Example (cont.) T 1 = = 573 K T 2 = = 623 K D 2 = 15.7 x m 2 /s 3 Solve for unknown value: Remember to convert from °C to K

13 wt% Ni T( o C) L (liquid)  (solid) L +  liquidus solidus L +  Cu-Ni system Phase Diagrams : Determination of phase compositions If we know T and C 0, then we can determine: -- the compositions of each phase. (Just read them) Examples: TATA A 35 C0C0 32 CLCL At T A = 1320°C: Only Liquid (L) present C L = C 0 ( = 35 wt% Ni) At T B = 1250°C: Both  and L present CLCL = C liquidus ( = 32 wt% Ni) CC = C solidus ( = 43 wt% Ni) At T D = 1190°C: Only Solid (  ) present C  = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni D TDTD tie line 4 CC 3 Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). B TBTB CC = C alloy ( = 35 wt% Ni)

14 If we know T and C 0, then can determine: -- the amount (i.e., weight fraction) of each phase. (Calculate them) Examples: At T A : Only Liquid (L) present W L = 1.00, W  = 0 At T D : Only Solid (  ) present WL WL = 0, W  = 1.00 Phase Diagrams : Determination of phase weight fractions (We use the Lever Rule) wt% Ni T( o C) L (liquid)  (solid) L +  liquidus solidus L +  Cu-Ni system TATA A 35 C0C0 32 CLCL B TBTB D TDTD tie line 4 CC 3 R S At T B : Both  and L present = 0.27 WLWL  S R+S WW  R R+S Consider C 0 = 35 wt% Ni Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).

15 ►Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm The Lever Rule (gives us the ‘amounts’ of phases that are present) What fraction of each phase? Think of the tie line as a lever (i.e., teeter-totter) MLML MM RS wt% Ni T(oC)T(oC) L (liquid)  (solid) L +  liquidus solidus L +  B T B tie line C0C0 CLCL CC S R Adapted from Fig. 10.3(b), Callister & Rethwisch 4e.

Binary Phase Solidification Alloys above solubility limit but below max solubility (far from the eutectic). The solidification path is: L  L +     Adapted from Fig , Callister & Rethwisch 4e. Pb-Sn system L +  200 T( o C) C,wt% Sn C0C L  30  +  400 (Max sol. limit at T E ) TETE 2 (sol. limit at T room ) L  L: C 0 wt% Sn    : C 0 wt% Sn 16 liquidus solidus solvus Solubility limits can change as function of temperature. This affects microstructure.

Adapted from Fig , Callister & Rethwisch 4e. Pb-Sn system LL  200 T( o C) C, wt% Sn L   L+  183°C 40 TETE 18.3  : 18.3 wt%Sn 97.8  : 97.8 wt% Sn CECE 61.9 L: C 0 wt% Sn L   +    saturated with Sn and  saturated with Pb must form at the same time. Requires diffusion. Mix of A & B has lower melting point than pure A or pure B. Eutectic Solidification Pure Elements Eutectic Pure Elements 17 PbSn

18 For alloys for which 18.3 wt% Sn < C 0 < 61.9 wt% Sn Result:  phase particles and a eutectic microconstituent Microstructural Developments SR 97.8 S R primary  eutectic   WLWL = (1-W  ) = 0.50 CC = 18.3 wt% Sn CLCL = 61.9 wt% Sn S R +S WW = = 0.50 Just above T E : Just below T E : C  = 18.3 wt% Sn C  = 97.8 wt% Sn S R +S W  = = 0.73 W  = 0.27 Adapted from Fig , Callister & Rethwisch 4e. Pb-Sn system L+ β 200 T( o C) C, wt% Sn L   L+  40  +  TETE L: C 0 wt% Sn L  L α Pb

19 L+  L+   +  200 C, wt% Sn L   TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic Adapted from Fig. 10.8, Callister & Rethwisch 4e. (Fig adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) 160  m eutectic micro-constituent Adapted from Fig , Callister & Rethwisch 4e. hypereutectic: (illustration only)       Adapted from Fig , Callister & Rethwisch 4e. (Illustration only) (Figs and from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 175  m       hypoeutectic: C 0 = 50 wt% Sn Adapted from Fig , Callister & Rethwisch 4e. T( o C) 61.9 eutectic eutectic: C 0 = 61.9 wt% Sn

Hypo-eutectoid alloys form primary  on prior grain boundaries. These alloys have low to medium strength, and good ductility. Hyper-eutectoid alloys form primary Fe 3 C on prior grain boundaries. These alloys are brittle, but strong. They also have excellent wear resistance. Understand how Microstructures Evolve

Summary ►Study your notes, assessments, and in book example problems. ►Exam #2 will have 7 questions. ►There will be NO true/false or multiple choice questions. ►All calculations AND short answer. ►Know how to do the types of problems that we’ve covered and you can do quite well. ►Good luck!