Course D - Microstructure Flashcards
how should you prepare a sample for reflected light microscopy
1) mount, grind, polish the sample to achieve a flat, level surface
2) etch the surface to highlight important features - usually done using chemicals
what does etching do
- etching is treating a sample with a chemical to highlight features, the etching process occurs quicker at grain boundaries and other impurities so they will reflect light differently
what sort of samples are required for transmitted light microscopy
- Transmitted light microscopy relies on detecting the light that has passed through a sample
- very thin samples required
- good for showing grain sizes
what causes contrast in transmitted light microscopy
- differences in absorption
- differences in birefringence
what properties must a material sample have for SEM
- should be prepared through polishing and etching
- cannot be conductive
what is atomic force microscopy (AFM)
- a very sharp tip is mounted on a cantilever and moved across the surface of a material
- this measures the height of the sample
- atomic level resolution can be achieved
what do pressure-temperature phase diagrams show
- for a given composition they show which phase is most THERMODYNAMICALLY stable at some temperature and pressure
why is the most thermodynamically stable phase not necessarily the phase present at a given temp/pres.
- kinetics can prevent a phase from forming
what do temperature-composition phase diagrams show
- they show which phase is most stable at some temperature and composition for a constant, given pressure
what do we mean when we talk about a system
“The subject of a thermodynamic analysis”
what are the 4 types of equilibrium (or not equilibrium) that a phase can be in at any given point
1) stable equilibrium - in a potential energy minimum, the overall minimum for the system
2) unstable equilibrium - at a potential energy maximum, unstable to any perturbations
3) not in equil - not at a max/min, there’s a driving force
4) metastable equil. - stable to small perturbations but not lowest energy state, local potential energy min. but not overall min.
what are the first and second laws of thermodynamics
1st law of thermodynamics: Total energy of universe is conserved
2nd law of thermodynamics: Entropy of the universe cannot decrease
what is the internal energy, U of a system, give the differential form of the equation for U
internal energy of a system, U = potential energy + kinetic energy
dU = δq + δw
q = heat
w = work done
define heat, give a differential equation for it
“The energy that ‘flows’ across a system boundary in response to a temperature gradient”
δq = CdT
C = heat capacity
define work done, give a differential equation for it
“the energy that flows across a system boundary in response to a force moving through a distance”
δw = -p dV
p = pres.
V = vol.
using the differential definitions for heat and work done, redefine the differential equation for internal energy
we know
dU = δq + δw
so
dU = CdT - pdV
give the overall equation for enthalpy, H
H = U +PV
give a differential form of the equation for enthalpy
dH = dU + PdV + VdP
and dU = CdT - pdV
so
dH = CdT + VdP
i.e. enthalpy is the heat transferred at a constant pressure
define entropy, how does it link to the 2nd law of thermodynamics
entropy, S, is a measure of disorder
from 2nd law of TD we get
dS(univ) > 0
what is entropy at equilibrium
dS = δqrev / T
define Gibbs free energy and give both differential and non-differential eq.’s
Gibbs free energy, G, is the energy available to do useful work
- G allows us to find the equil. state of a system, using only properties of the system, not the surroundings
G = H -TS
dG = dH - TdS - SdT
we know dH = δq + VdP
so
dG = δq + VdP - TdS - SdT
at const. temp. and pres.
dG = δq -TdS
what can we say about G at equil for a single phase
dG = 0
Gibbs free energy, G, tends to a minimum at equil.
for spontaneous processes dG<0
for a single phase of constant composition, what can we say about the variation of G with temp
- it will ALWAYS decrease
G = H-TS
its a straight line
y-int = H
grad at any point (draw tangent) = -S
what can we say about which phase is most stable at a temperature for a system of two phases of const. identical comp (e.g. liquid, solid of same phase)
link to G-T graph
- we have two curves on a G-T graph
- whichever is the lowest G phase at any point is the more stable phase
- the point at which the two lines cross is the equil. temp
phase 1: G1 = H1 - TS1
phase 2: G2 = H2 - TS2
difference in G:
ΔG = ΔH - T ΔS
we are interested in the sign of ΔG
how does G change with composition for a mechanical mixture of two phases A and B
- it simply changes as a weighted average of the two phases
- i.e. at 100% A, mark Ga on the y-axis, at 100% B, mark Gb on the y-axis, draw a straight line between them
what are the three things we must consider when determining the free energy curve for a single phase of A and B (or the change in free energy of mixing)
1) an enthalpy change associated with the fact that A-B interactions are different to A-A or B-B interactions, ΔHmix
2) an enthalpy change due to the random mixing of A and B atoms, ΔSmix
3) combining these to form a change in free energy when mixing
derive the expression for ΔHmix when forming a solution of A and B from a solid solution for 1 mole of atoms
ΔHmix = Hs - Hmm
let:
XA = fraction of A atom
XB = 1-XA = fraction of B atoms
EAA = interaction energy between A-A as nearest neighbours
EBB = “”
EAB = “”
coordination number of A,B = Z
Hmm = (NAZ / 2) (XA EAA + XB EBB)
Hs = (NAZ / 2) ( XA^2 EAA + XB^2 EBB + 2 XA XB EAB)
combining and rearranging using XA + XB = 1 gives
ΔHmix = Hs - Hmm = (NA*Z /2) XA XB (2EAB - EAA - EBB) = XA XB ψ
ψ = (NA*Z /2) (2EAB - EAA - EBB)
what 3 cases does the expression for ΔHmix imply (regarding the sign of ψ)
1) ψ = 0 , EAA = EBB = EAB, totally randomly mixed solid solution
2) ψ < 0, EAB < EAA, EBB, totally ordered alternating A-B structure, ΔHmix < 0
3) ψ > 0, EAA,EBB < EAB, totally separated, unmixed, ΔHmix > 0
derive the expression for ΔSmix when forming a solution of A and B from a solid solution for one mole
we start with
S = kln(ω)
where ω is the number of distinguishable ways of arranging atoms on the available sites
Ωmm = 1
(A atoms on A sites, B atoms on B)
Ωs = NA! / ((XANA)! ((1-XA)NA)!)
ΔSmix = kln(Ωs) - kln(Ωmm) = kln(Ωs)
once the Stirling approximation has been used this gives
ΔSmix = -R(XA ln(XA) + XB ln(XB))
derive the expression for ΔGmix when forming a solution of A and B from a mechanical mixture
ΔGmix = ΔHmix - TΔSmix
= XA XB ψ + RT(XA ln(XA) + XB ln(XB))
consider the same 3 cases regarding the sign of ψ, explain what this means for when forming a solution is favourable
1,2) ψ = 0 or ψ < 0, ΔHmix <= 0, ΔSmix, >0, ΔGmix<0 at ALL TEMPS – always favourable to form solid solution
3) ψ > 0, ΔHmix, ΔSmix >0, ΔGmix sometimes<0 depending on temp, shape of curve depends on temp
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in case 3) (ψ > 0, ΔHmix, ΔSmix >0, ΔGmix sometimes<0 depending on temp, shape of curve depends on temp), how do we know when it’s better in 2 phases or a solid solution
- at high T, the curve will have 1 minimum, it is always favourable to form a solid solution in this case
- at lower T, the curve will have two minimums, between the two minimums it is NOT favourable to form a solid solution, elsewhere it is
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what process can we apply to a two phase mixture with lines plotted on a G against comp diagram, to work out whether it is more favourable to have one of the two phases or a mixture
- draw a common tangent between the two curves of interest
- mark the compositions of the two phases where the common tangent meets their curve
- if the composition is between these two then it will split into two different phases of the compositions marked, the proportions are just a weighted average
- if it is outside of this region, then it will remain as the relevant phase
NOTE: the compositions that the common tangent touches are NOT necessarily the minimum of the curve
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Define a phase
A phase can be defined as a portion of a system whose structure, properties and
composition are homogeneous and which is physically distinct from other parts of the
system.
how do G curves relate to a phase diagram
- G-composition plots can be obtained for a range of temperatures
- for each plot the points where there is the boundary between two phases and one phase can be determined then plotted on a temperature- composition plot as points
- this can be repeated and phase boundaries determined
- or the same can be done in reverse to determine the G-Composition plot
using a Temp-comp. phase diagram, if you’re in a two phase region, how can the proportions of each phase be determine
- use the lever rule
- basically take weighted averages of each phase to give the correct overall composition
what can we say about the free energy curves of the phases in a ‘complete solubility in liquid and solid phases’ phase diagram
give some features of the phase diagram
- this is the case where 3) ψ <= 0 for both liquid and solid phases
- hence the curves for both phases only have 1 minimum
- the phase diagram has a solid phase section at bottom, a liquid phase section at top and a tilted disc shape section of (L+S) spanning the whole composition range midway
what is the key feature of a Eutectic system
what does the phase diagram look like
- the defining feature of a Eutectic system is that at some (Eutectic) composition, solidification occurs at a single temp. and we have L —> α + β
- the phase diagram has thin(ish) p-shaped sections at either extreme of composition, a large liquid section at top, a large α + β section at bottom and then two sections which meet at the Eutectic point of L+α and L+β
describe the process of Eutectic solidification
- as the alpha phase forms in the liquid, B-atoms are rejected out of the α
- a B-rich section of liquid then surrounds the α phase
- this causes β to form, generally adjacent to the α
- α + β forms
describe how the Eutectic microstructure forms
- α, β form simultaneously, forming a microstructure of alternating lamallae (plates) of α, β
As α, β form:
- B-atoms are pushed out the front of the α phase, and vice versa as they advance
- the atoms must redistribute in the liquid to allow solid phases to grow correctly
- the extent to which they can redistribute determines the length-scale (thickness) of the plates
give an example of a Eutectic system and a common use of eutectic alloys
Pb-Sn is a eutectic system
Eutectic systems (but not necessarily Pb-Sn) are often used as solders because if not at eutectic comp, they will go mushy before solidifying fully so can still be manipulated on solidifying
how do eutectic systems (once solidified) tend to appear under a microscope
stripy due to the lamallae
describe what microstructure forms and how it does when cooling a eutectic system not at the eutectic composition
- initially liquid single phase
- moves into two phase region e.g α+L
- on cooling through α+L, some α ‘nuclei’ form, drawing a tie line generally shows more liquid than α is present
- as the α forms, it pushes more β into solution, the composition of the liquid changes/ becomes more B-rich
- the α that forms in the α+L section is called primary alpha
- as the solution approaches the eutectic temp, the liquid approaches the eutectic comp.
- on passing through the eutectic comp, the remaining liquid undergoes a eutectic transformation and the standard lamallae structure forms between the primary alpha sections
how can we determine a phase diagram from a cooling curve
Obtain cooling curves for samples at different compositions:
- for a pure sample (or eutectic), phase transformation occurs at a specific temperature
- for a mixture, phase transformation occurs over a range of temperatures
- any change in gradient represents passing over a phase boundary
define equilibrium cooling, what compositions will be present
equilibrium cooling is when the composition of the solid remains uniform and changes as solidification progresses:
- the composition of both the solid and liquid at a temperature is directly given by the tie line that can be drawn
what must occur to have equilibrium cooling
- a very slow cooling rate (to allow for atomic diffusion)
what are the 4 things that can occur when non-equilibrium cooling occurs (2 are only for Eutetic)
1) Coring
2) dendrite formation
3) finer Eutectic pattern
4) more Eutectic phase than the lever rule predicts
explain how coring occurs/ the effects of it
- if equil. is not maintained, the first solid that forms will retain a higher conc. of B atoms (assuming the β phase is forming)
- this means the fractions of liquid/solid are no longer given by simple tie lines on the phase diagram
- as more B is trapped inside the solid, the liquid is more A-rich than is expected at equil.
- true compositions depend on diffusion rates within the solid
- if slow diffusion, we have significant conc. grad in solid
explain the formation of dendrites in non-equilibrium cooling
- as a solid grows on cooling through the α+l region, it rejects β atoms
- this gives a B-rich region surrounding the α, there’s a conc. grad.
- If there are any small, random pertubances on the advancing solid, they find themselves in a less B-rich region
- this encourages growth and forms a +ve feedback system
in what direction do dendrites tend to grow in
- some preferred crystallographic directions
state the two main diffusion mechanisms
- interstitial
- substitutional
explain how interstitial diffusion occurs, what can we say about its rate
- smaller atoms can sit in interstitial sites and ‘jump’ from one interstice to another to diffuse
- generally fast as the atom can move in any direction at any time
explain how substitutional diffusion occurs, what can we say about its rate
- atoms move from one site to an adjacent vacant one
- requires vacancies so generally slower
what does Fick’s first law show
- in a concentration gradient, there’s a net flux of solute through the solid
- Fick’s first law describes this as
J = -D (∂^2C / ∂x^2)
D(interstitial) > D(substitutional)
what does Fick’s second law show
- it shows that when the system is not in a steady state, conc. grad. and flux vary with time, we use Fick’s second law for this
∂C/∂t = D (∂^2C/ ∂x^2)
what is the solution to find distance diffused from diffusivity and time
x = sqrt(Dt)
note this is only approximate
how does the diffusion constant vary with temperature
- it follows an Arrhenius relationship
D = Do exp(-Q/KT)
OR
D = Do exp(-Q/RT)
depending on units
Q = activation energy
T = temp
R = ideal gas const.
K = Boltzmann constant
what is the driving force for the nucleation of a new phase, give the equation that describes it
“The driving force for the nucleation of a new phase is the change in free energy ΔG”
at equil, Te
ΔG = ΔH-TeΔS = 0
ΔH = TeΔS
so at a given temp
ΔG = (Te-T)ΔS
i.e. driving force is proportional to deviation from equil. temp.
what are the two main factors that we should consider when thinking about nucleation (either type)
1) the change in energy per unit vol.on transformation to a new phase, ΔGv
2) γ, the surface energy per unit area between phases
give the expression for the energy of homogenous nucleation (sphere)
Wn = 4/3 π r^3 ΔGv + 4π r^2 γ
what is the critical radius for homogenous nucleation, how can we find the expression for it, give the equation
we know the graph of Wn against r has a maximum, this is at r*
- for r > r, the nucleus naturally grows
- for r < r the nucleus naturally shrinks
doing dWn / dr = 0 yields
r* = -2γ /ΔGv
and
Wn* = (16π/ 3)(γ^3 /ΔGv^2)
what is the issue with the model for homogenous nucleation, how can we edit it
- it assumes no strain term when forming a new phase, this is unlikely, we can add a strain term U
Wn = (4πr^3 / 3)(ΔGv + U) + 4πr^2 γ
- both r* and Wn* increase
NOTE: in the case of ferromagnetism or ferroelectric materials, this term could represent stray field energy
what are the two factors we must consider when thinking about the rate of homogenous nucleation
1) population of critical nuclei (r<r), proportional to exp(-Wn / KT)
2) rate at which particles can add to make nuclei post-critical, proportional to exp(-Q/KT)
give the overall equation for nucleation frequency (the homogenous case), what can we say about the nucleation frequency close to or far away from Te
nucleation frequency = I
I = Cn exp( - (Wn*+Q) / KT)
- no nucleation at Te
- increases (due to increased driving force) moving away from Te
- if too far below Te atomic motion is too low so nucleation rate drops
what is heterogenous nucleation
- this is where nucleating is catalysed by the presence of defects in the original phase, spherical ‘caps’ form on defects such as the wall
what are the relationships between critical radius and critical work done in homogenous and heterogenous nucleation
r(hetero) = r(homo)
Wn(hetero) = Wn(homo) ((2+cosθ)(1-cosθ)^2 /4)
θ = angle between surface and cap at base
what can we say about the ease with which heterogenous and homogenous nucleation can occur, which occurs faster
- heterogenous nucleation has a lower critical work done so will generally occur quicker
- often occurs at grain boundaries, triple points, container surfaces and at defects
what do we need to consider when deriving the growth rates of a new phase
- the energy barrier to forming the new vs. old phase / the asymmetry between them
Forward jump rate (old –> new) proportional to exp(-Q/KT)
Backwards jump rate (new –> old) proportional to exp( -Q+VAΔGv) / KT)
hence growth rate, ν, is proportional to the difference between the two
ν = Cg exp(-Q/KT) [1-exp(VAΔGv/KT)]
VA = vol. per atom
what can we say about the growth rate at different deviations from Te
- no growth at Te
- max growth near Te
- low atomic mobility so little growth a distance from Te
what are the 4 types of solid-solid interface
1) coherent
2) strained coherent
3) semi-coherent
4) incoherent
give the features of a coherent solid-solid interface
- perfect alignment of lattices
- ideal, lowest energy
give the features of a strained coherent solid-solid interface
- it is likely that any coherent interface will have some elastic strain
- higher energy, strain energy increases with size of particle
- this causes a transition to semi-coherent
give the features of a semi-coherent solid-solid interface
- introduction of dislocations
- reduces overall elastic strain
- contributes to energy in other ways
- occurs where there’s more mismatch or a larger precipitate
give the features of an incoherent inteface
- Large mismatch
- V high energy
we tend to model precipitates forming as being spherical, why is this not usually the case
- spherical precipitates minimise surface area but we must consider interfacial energy too
- atoms are more likely to add to a higher energy incoherent interface so they tend to grow faster
- this often leads to disk like shapes
what are Widmanstätten microstructures
- plate-like precipitates often lie in particular crystallographically related orientations
- e.g. precipitation of bcc (110) plane on an fcc close packed (111) plane
- this allows for the matching of close packed lattice rows/planes and is thus lower energy
give the main features of the Fe-C system, what phases of steel are (that we are interested in)
- Eutectoid transformation (723°C, 0.8wt% C)
- Eutectic transformation (1130°C, 4.3%C)
4 main phases:
- α-Fe, Ferrite, bcc, V low C%, up to 910°C
- γ-Fe, Austenite, fcc, up to 1.7wt%C, 723-1492°C
- δ-Ferrite, bcc, higher temp, 1400-1535°C, low wt%C
- Cementite, Fe3C (higher carbon content)
describe the Eutectoid transformation of steel, what phases form, what is the resulting microstructure
Eutectoid transformation = Fe-0.8wt%C, 723°C
Austenite –> Ferrite + Cementite
γ —> α + Fe3C
- on the eutectoid transformation, Ferrite rejects C atoms, Fe3C accepts them
- lamallae structure forms
- this is called Pearlite
explain the hypereuctectoid transformation in steels, give the features of the resulting microstructure
Hypereuctectoid transformation from about 1000°C, Fe-1wt%C
1) initially we have large grains of γ
2) on the γ–> γ + Fe3C transformation (about 740°C), Fe3C precipitates at grain boundaries
3) At the eutectoid temp, the remaining γ undergoes γ–> α + Fe3C forming pearlite
explain the hypoeuctectoid transformation in steels, give the features of the resulting microstructure
Fe - 0.1wt%C
1) Ferrite produced below about 850°C
2) about 10% γ still remains at Eutectoid temp
3) at eutectoid temp, remaining γ undergoes γ–> α + Fe3C forming pearlite
what is a time-temperature-transformation (TTT) diagram
- for first-order phase transformation you need nucleation and growth
- we can describe the overall rate of transformation using t(y), the time taken to transform a fraction y of the material from one phase to another at a given temp.
- an isothermal transformation diagram shows t(y) as a func. of temperature
- contours are plotted for given y values, C-shaped curves
- generally plotted on log-log axis
what occurs if the cooling rate of a substance is sufficiently fast to avoid the ‘nose’ of the TTT diagram
- Quenching occurs
- none of the new phase forms, often a new metastable phase will form
- technically TTT diagrams are isothermal so should not be used for this purpose but a rough cooling rate can be estimated using
dT/dt = ΔTnose/tnose
what is martensite
- martensite is a metastable phase of steel which forms on rapid cooling, sufficiently quick that no pearlite forms from austenite
- it forms through a diffusionless mechanism
what occurs in a martenistic transformation
- it is a displacive phase transformation, it does not require diffusion
- in fcc austenite a carbon atom sits in an octahedral interstice
- on slow cooling usually pearlite is formed so the carbon forms Fe3C precipitates but on rapid cooling a displacive phase transition occurs and a bct structure forms
- the carbon atom now sits in the centre of the bct martensite
- there is much more strain
explain why martensite forms and is brittle
- austenite and ferrite have massively different solubilities for carbon so when austenite is cooled rapidly, it forms a very super-saturated solution
- it effectively forms a ferrite structure but with too much carbon contained
- the carbon sits in the centre of the unit cell and changes it from bcc to bct
- this causes strain fields and locks dislocations
- making it very hard and brittle
what can occur when martensite is tempered/ annealed
- the carbon diffuses and forms precipitates
i.e. ferrite and cementite form
NOTE: the cementite does not have the pearlite structure in this case
- the resultant structure after tempering is still hard but much less brittle
what are the 3 categorisations of cooling rate, give examples of how each is achieved
1) slow cooling e.g. leave to cool in air
2) moderate cooling e.g. cool in oil
3) rapid/ fast cooling e.g. cool in water
what are the 2 general consequences of faster cooling rates
- stronger driving force for nucleation so many smaller nuclei form
- diffusion is very slow at lower temperatures so smaller grains form
explain how casting works, describe the 3 ‘sections’/ grain forms of a cast structure
- it is a common metal processing route, liquid metal is poured into a mould
- solid nucleates and grows, often heterogenously from the walls of the container
1) small equiaxed grains
- occur at edge
- fast cooling rates near edges and sites for heterogenous nucleation means nucleation is favoured over growth so lots of small grains form
2) columnar grains
- results from grain growth along preferred directions
- along temperature gradient
3) large equiaxed grains
- slowest cooling rate in centre means growth is favoured over nucleation
what occurs in glass formation
- if a fast enough cooling rate can be achieved then crystallisation can be avoided altogether
- for metals it is difficult to form but small quantities can be made by dropping metal onto a spinning, cooled wheel
what can we say about the effect of cooling rate on eutectic and eutectoid microstructures
- faster cooling rate leads to finer lamallae structure, finer spacing λ
- the rate at which a eutectic/oid phase grows is dependent on how fast diffusion in front of growing lamallae can occur
- at low temps, this rate is low so λ is smaller
how can we estimate the minimum lamallae spacing λ at a given temperature
- consider the solidification on 1m^3 of liquid
- forms eutectic lamallae
- driving force ΔGv at T
- 2/λ m^2 of α-β interface for each m^3 of eutectic, let γ(α-β) be the surface energy for this interface
hence
ΔGtot = ΔGv + 2γ(α-β)/λ
min is at ΔGtot=0
so
λmin = -2γ(α-β)/ΔGv
what is hot rolling
- material is pushed through rollers whilst still hot
- deformation process
what is forging
- deformation process
- the material is squashed in a press between die
what is extrusion
- deformation process
- the material is pumped through a shape
what is deep drawing
- deformation process
- a sample is held between die and ‘punched’
in the Al-Cu system (important in aircraft) what is the general method to make commercial alloys (4wt% Cu)
1) anneal at 540°C to have all Cu in solution
- homogenises solution, removes coring
2) quench to room temperature to give supersaturated solid solution (SSSS)
- forms SSSS, prevents θ phase from forming
3) anneal at < 200°C to form ppts
- low temps, more nucleation, less growth
focusing on the annealing part of forming Al-Cu alloys, what is the first phase to form
- a metastable phase
- Guinier-Preston Zones (GP zones) form first
- these are generally single atomic layers of Cu-rich material
- form by homogeneous nucleation
- coherent interface
why doesn’t the θ phase form immediately when annealing a SSSS of an Al-Cu alloy
θ = tetragonal
α = fcc
- incoherent faces, large barrier to nucleation
focusing on the annealing part of forming Al-Cu alloys, what is the second phase(s) to form
from GP zones and α1, θ” and α2 forms:
- θ” = tetragonal crystal structure
- layers of Cu atoms separated by 3 atomic layers of Al
- Remains coherent but (001) face is least strained
- (100) and (010) are more strained, grow faster, leads to disc shapes
focusing on the annealing part of forming Al-Cu alloys, what is the third phase(s) to form
- if annealing continues then the α2 and θ” from α3 and θ’
- the (001) is still semicoherent
- the (010) and (100) now incoherent
- θ’ nucleates heterogeneously on crystal defects
focusing on the annealing part of forming Al-Cu alloys, what is the fourth/final phase to form
α3 and θ’ form α4 and θ
- it’s bct so all interfaces are incoherent
- particles nucleate heterogenously
what is the general process under annealing to form θ
GP zones —> θ” —> θ’ —> θ
why does the formation of θ occur in 4 different stages and not in one
- the individual transformations have low Ea compared to a straight change from α to θ
- so the rate via the intermediates is greater
when calculating the compositions and proportions of a hypoeutectic microstructure what temperature should we do it for
- the eutectic temperature
what does the cooling curve for a hypoeutectic (or hyper) alloy look like
- initially linear negative
- change of gradient, still linear negative as entering L+α region
- then horizontal section where liquid undergoes eutectic transformation
- then decreasing again
