Course G - Materials under extreme conditions Flashcards
define pressure and give the expression for pressure at T = 0K
Pressure, P = Force, F / Area, A
P = -dU/dV
at 0K
units = Nm^-2 = Pa
what is 1atm and 1bar
1atm = 101325 Pa
1bar = 100000 Pa
define Gibbs free energy and give the differential form
G = H-TS
dG = dH - TdS - SdT
dH = δq + VdP
δq = TdS at equil.
so
dG = Vdp - SdT
from the differential form of G, what can we say about a phase transformation at a constant pressure (varying temp.)
hence what can we say about which phase is favoured at higher temperatures
dG = VdP - SdT
at const. pres.
dP = 0
so
dG = -SdT
(∂G / ∂T)p = -S
hence at low temps, entropy –> 0
slope of G-T graph always < 0 as S always > 0
“The phase of higher entropy is favoured at higher temperatures” (to minimise G)
at a constant temperature, what can we say about the gradient of a G-P graph
hence what can we say about which phase is favoured at higher pressures
at constant temperature dT = 0
dG = Vdp - SdT = VdP
so
(∂G/∂P)T = V
hence the phase of lower volume is favoured by higher pressure (to minimise G)
what do pressure-temperature phase diagrams show
- G,P,T can all be plotted on a 3D plot by a curved plane
- G planes for 2 phases intersect on a line
- the projection of this line onto the P-T plane gives the phase boundary on a P-T phase diagram
- they show which phase is thermodynamically most stable for a given composition
derive the Clausius-Clapeyron equation
at all points on a phase boundary of a P-T diagram, the free energies of the two phases will be equal
the changes in free energy for a given change in temp/pres will also be eqaual
dGα = Vα dP - Sα dT
dGβ = Vβ dP - Sβ dT
Δ(dG) = ΔVdp - ΔSdt = 0
dp/dt = ΔS/ΔV
ΔS = Sα - Sβ
ΔV = Vα - Vβ
explain how the Clausius-Clapeyron eq. can be used to help predict phase boundary lines on a P-T phase diagram, what assumptions are we making
- the Clausius - Clapeyron eq. gives us the gradient of the phase boundary on the P-T diagram
- if we know any point on the phase boundary, we can deduce the equation of the line
- this assumes ΔS and ΔV are constants over different temps and pressures
- this is a reasonable estimate
what are the 2 main methods for determining a P-T phase diagram experimentally
1) Hold sample at given P,T and determine phase ‘in situ’ - difficult
2) Equilibrate sample at given T,S, quench to RTP and then determine/identify phase
- easier
- cannot be used for kinetically fast transitions
how does the piston/cylinder device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.
- original method
- piston exerts pressure on sample
- heaters can be added to change temp
- materials of piston determine max. pres
how does the opposed anvil device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.
- Anvils (Tungsten carbide) concentrate force onto a small area (high pres.)
- a gasket applies an external constraint to prevent extrusion
- they can be used with a neutron/synchotron x-ray sourse
- heating through inserted heating elements
how does the multi-anvil device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.
- similar to opposed anvil devices
- but now 4,6 or 8 anvils applying pressure
- truncated tungsten carbide or sintered diamond anvils
- heater encoportated
Advs:
- large sample sizes can be used
- good control
Disadvs:
- difficult to observe samples in situ
how does the diamond anvil cell device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.
- sample squeezed betwen 2 diamonds, heated with IR laser for high temps or heater for moderate temps
Advs:
- V high temps
- good in situ measurements
Disadvs:
- sample sizes limited by size of diamonds
how does pulsed-laser implosions work as experimental apparatus for determining P-T diagrams
- pulsed lasers for short times focused on sample
- this gives V V high pres. and temps.
how can ΔS be calculated for a transition (2 methods)
1) use ΔS = ΔH / Teq
for transition
2) use heat capacity from 0K —> T using
Cp = T dS/dT
and remembering to account for phase transitions’ latent heat
what is the effect of high pressures on molecular materials/solids
- molecules are bound by VDW forces
- under high pressure it can become solid, this is where it is still molecular but just ordered
- under V high pressure the intermolecular distances can become similar to the intramolecular distances so its molecular identity is lost and it become more like a metal or polymer
examples include iodine at high temps
what is a reconstructive phase transition, what does it include/require
Reconstructive phase transitions are phase transitions under pressure between two structures that are not closely related:
- requires breaking/reforming bonds
- thermally activated, leads to metastable phases
- require nucleation/growth of new phase
- kinetics are important
explain what occurs in the reconstructive phase transformation of NaCl and MgO
- at RTP, both NaCl and MgO are fcc with an NaCl structure
- at higher pressures, both become the cubic p CsCl structure
- this increases number of nearest neighbours from 6 to 8, higher packing efficiency
what occurs in the reconstructive phase transition of Mg2SiO4
how does this link to earthquakes
- at RTP rocks = brittle, this can cause slip and earthquakes
- deep in mantle, there are high temps and pressures, this means plastic flow of rocks and deep-focus earthquakes cause by a pressure-induced phase transition
- this is because there is a volume contraction of 8%
- olivine exists metastably far beyond equil temp/pres.
- in the uppermost layer the stable rock (Mg2SiO4) form is olivine
- at high pressures and temps in mantle, the spinel phase is more stable
- both are made of SiO4 tetrahedra and MgO6 octahedra with Si cations in tetrahedral sites and Mg cations in octahedral sites
- Olivine = hcp oxygen lattice, orthorhombic
- Spinel = ccp oxygen lattice, cubic F
what occurs in the graphite-diamond transformation i.e. give the conditions
Diamond is metastable but the activation energy is very high so at RTP the transformation back to graphite takes a long time
- at very high temps and pressures (5GPa,1500 celcius) diamonds are thermodynamically stable
- this generally occurs at about 300km depths but they can be brought to the surface by volcanoes
what is the RTP structure of Boron Nitride, what structure does it form under high temps and pressures
- at RTP, BN has hexagonal form like graphite
- at high pressures and temps it transforms to sphalerite structure like diamond
what is the asymmetry between freezing and melting
- when freezing there is a barrier to nucleation which allows for supercooling
- when melting there is no such barrier so melting begins immediately
outline the Lindemann and Born theories of melting and explain what actually happens
Lindemann:
- melting occurs when avr. amplitude of atomic vibrations reaches critical fraction of interatomic spacing
Born:
- melting when elastic shear modulus = 0
- this predicts mechanical melting which is wrong
Actual:
- melting begins at solid surface and works inwards
- at Tm, solid/liquid coexist with well-defined interface
why for most materials is surface melting found at temperatures < Tm
because:
γsv > γls + γlv
i.e. it is more energetically favourable to have a solid-liquid interface AND a liquid-vapour interface than just a solid-vapour interface
- this means the crystal is wet by its own melt so there is no barrier to nucleation
how can we consider the melting of small particles in terms of r*
r* = -2 γ(ls) / ΔSv(Tm-T)
this means:
- Well below Tm the crystal/nucleus is stable as the value of r* at which melting occurs (i.e. not stable nucleation) is V small
- as T approaches Tm, the value of r* that is stable increases i.e. the value of r* at which melting occurs increases
- when r* > r, the particle melts
give the expression for the supercooling at which melting occurs for a particle of radius r
ΔT = -2 γ(ls) / ΔSv r
hence greater r = lower ΔT so higher effective melting point
what are the three things that can be done to suppress surface melting/ observe superheating
1) ensure the centre of the crystal is hotter than the surface:
- internal melting
- dendritic melting can occur
2) coat crystal with higher mpt material:
- ensures no surface liquid layer
- this means the limit to the stability of the crystal is when liquid starts to nucleate within the crystal
3) superfast heating (about 10^12 Ks^-1)
- nucleation is kinetic so takes time
- so it can be prevented through superfast heating
define creep
“Creep is defined as time-dependent permanent deformation (plastic) of a material under the action of a stress σ(applied) where
σ(applied) < σy”
give the three factors that creep rate depends on
- material
- applied stress
- homologous temperature
define homologous temperature
homologous temp = T/Tm
in kelvin
(generally) creep is significant at:
T > 0.3 Tm for pure metals
T > 0.4 Tm for alloys/ceramics
what are the three regimes of creep
1) primary = dε/dt high initially, but slows from work hardening
2) secondary = balance of work hardening and annealing, ε remains approx. const.
3) tertiary = Necking and failure
give the equation for strain rate in the secondary stage of creep and from this, define the two types of creep
dε/dt = A σ^n
the two types of creep depend on the value of n:
1) Dislocation creep, 3<n<10 (approx.)
- high stress
- dislocation motion
2) Diffusion creep, n = 1 (approx.)
- low stress
- atomic diffusion
give the key points about how dislocations avoid obstacles
- dislocations glide on slip planes, obstacles can block the glide causing σy to increase
- dislocations can move to different slip planes to avoid obstacles in a process called climb, both +ve and -ve climb can occur
- climb driven by local forces on dislocation acting perpendicular to the slip plane
- only some of a dislocation must undergo climb to avoid an obstacle
how does dislocation creep occur, what must diffuse?
- for dislocation creep, atoms must diffuse to or from the dislocation core
- this occurs in one of two ways depending on temperature
what are the two temperature dependent mechanisms for dislocation creep, briefly explain them
1) Bulk Diffusion
- this occurs at higher temps
- when high enough temps, there is sufficient mobility in the bulk lattice around the dislocations for atoms to join/leave via the lattice
2) Core diffusion
- dominates at lower temperatures (but still occurs at higher temperatures)
- this is when the only significant diffusion occurs along the dislocation cores
how can we change equation for strain rate in the secondary regime of creep to account for diffusion of atoms in dislocation creep
dε/dt = Aσ^n = A’ σ^n D = A’’ σ^n exp(-Q/RT)
- can plot different log-log graphs of this
what is diffusion creep, what are the two types
Diffusion creep is the diffusion of atoms biased in the direction of the applied stress, causing grains to change shape
- consider a hexagonal grain, if the stress is applied along its length then it’ll stretch out
- the two types are Nabbaro-Herring creep and Coble creep
Explain the difference between Nabarro-Herring and Coble creep and when they occur
- both types of diffusion creep
1) Nabarro-Herring creep:
- occurs at high temps
- diffusion of atoms (to change grain shape) in bulk lattice
2) Coble Creep:
- dominates at lower temps (but still occurs at higher temps)
- diffusion on grain boundaries (lower energy route)
give the equation that describes diffusion creep strain rate
dε/dt = B D σ/d^2 = B’ σ/d^2 exp(-Q/RT)
d = grain diameter
how can creep rates be minimised
- high mpt
- low dislocation motion
- stable ppts
- larger grains
explain the 4 stages of the (idealised) Carnot cycle
TH - hot temp, TC = cold temp
1) isothermal expansion of gas at TH, ΔT = 0, ΔS>0
2) adiabatic expansion of gas, cools/expands to TC, ΔT < 0, ΔS = 0
3) isothermal compression of gas at TC, work done ON gas, ΔT = 0, ΔS < 0
4) Adiabatic compression of gas, heats to TH, ΔS = 0, ΔT>0
NOT SURE IF THIS NEEDS MEMORISING
give the efficiency equation for the carnot cycle, at what temperature is it most efficient
Efficiency = 1 - TC/TH
more efficient at higher temps
briefly explain the brayton cycle (used in jet engines)
1) Adiabatic compression of air, ΔS = 0, ΔT > 0
2) Heating in combustion chamber, ΔP = 0, ΔT>0
3) Adiabatic expansion of gas ΔS = 0, ΔT<0
4) dissipation of hot gases
explain why jet engines blades have significant materials challenges
in operation:
- T > 0.75Tm
- 10000rpm, high centripetal force
- lifetime = 3 years
- extracts 500KW
what are some things that have been done to blades to try to keep up with rising operation temperatures in jet engines
- superalloys created
- cold air pushed through blades
- thermal coatings
explain what properties possible jet engine blades must have and must not have
what metal does this leave
MUST:
- be tough
- have high Tm to prevent creep
MUST NOT:
- be polymorphic (change phase on temperature range of use)
- have high intrinsic diffusivity (ccp>hcp»bcc due to interstices)
- be brittle
- have high density
- oxidise easily
- have too high cost
this leaves Nickel as the only really suitable metal
what are some of the elements that are added to Nickel to make it a superalloy
Cr (corrosion help)
Co, W (SS strengthening)
Al, Ti, Ta (all help form γ’)
what microstructure does a nickel superalloy have
a γ-γ’ microstructure
- this forms a dense dispersion of ppts in a matrix
γ = Ni based ccp solid solution
γ’ = Ni3Al (approx.) - ordered atomic layout
give the structure/ key properties of the γ phase
γ:
- cubic F
- ccp
- each lattice site has same average composition
- solid solution
give the structure/ key properties of the γ’ phase
γ’:
- cubic P
- 4 atoms per lattice point (3 nickel, 1 aluminium)
- Fully ordered, no Al-Al nearest neighbours
- more like Ni3(Al,Ti,Ta)
what is the interface between γ and γ’
coherent
- low interfacial energy
- prevents ppt coarsening
what is the advantage of directionally solidified crystals
- the detrimental effect of grain boundaries due to atomic diffusion/ lower creep resistance is minimised when they are // to the main tensile stress
- blades can be formed like this to reduce creep
what are single crystal blades, why are they useful
- grain boundaries contribute to atomic diffusion and reduce other creep resistance
- the growth of the solid along a helical path gives 1 grain growing into main blade cavity
<100> // blade length
state the main strengthening mechanism due to the γ - γ’ precipitate structure
- order hardening
explain order hardening in the γ, γ’ structure and how this gives strengthening
For dislocations, b is always a lattice vector and is generally the shortest possible
- in γ, b = a/2 <1 bar1 0> {111}
- in γ’ this is NOT a lattice vector
If the γ dislocation goes through γ’ it makes Al-Al nearest neighbours, this creates an antiphase boundary (APB) which is higher in energy
To fix this, a second dislocation passes through the precipitate, removing the APB and restoring the initial γ’:
- first dislocation feels large drag on entering ppt, this gives strengthening
- second dislocation restores, APB exists between
- called superdislocations and superpartials
this energy dissipation and drag = hardening and creep resistance
explain the need for thermal barrier coating on blades in jet engines and state their material requirements
as temps increase, creep resistance decreases so more protection is needed so thermal barrier coatings developed
Requirements:
- Low thermal conductivity
- high melting point
- good strength
what are the three layers of thermal barrier coatings
1) columnar YSZ:
- best option for thermal coating
- zirconia = lower coeff of thermal expansion than Ni superalloy
- columnar structure alloys for slight separation to prevent breaking/cracking
2) dense alumina layer:
- YSZ has good O2 diffusion which can lead to nickel oxidation
- alumina layer stops this
3) bond coat:
- Ni, Cr, Al, Y alloy
- the ‘glue’ between the superalloy and coating
what are the mechanical properties of ice
It remains brittle up to its melting temperature due to proton disorder:
- proton disorder impedes dislocation motion as any dislocation that passes through would give either 0 or 2 H’s between O’s
- creep is possible at low strain rates
what is the bonding in ice
ice formed of H2O molecules:
- these are non-linear 109.5 degree bond angle
- these pack together with hydrogen bonds roughly tetrahedrally
what is the most common structure for ice
Ih:
- hexagonal, stable under ambient conditions
- tetrahedrally linked framework, hexagonal symmetry
- arrangement of oxygen like Wurtzite
State the 2 ice rules and explain proton disorder
Ice rules:
1) must be 2 hydrogens adjacent to each oxygen
2) only 1 hydrogen per bond
- even with these rules, the positions of protons are not determined leading to proton disorder
can surface melting occur in ice?
what properties doe this give?
- yes it can
- it gives low friction coeff., high adhesion of 2 surfaces, good compaction
explain regelation (regarding ice)
- ice has unique property of solid being less dense than liquid
ΔV(l–>s) < 0
so grad of P-T phase boundary < 0 (from clausius clapeyron)
this gives phenomena of regelation:
“Regelation is where a material melts under pressure but refreezes when pressure is reduces”
give typical energies of neutrons in fission/fusion reactors
Fission, fast neutron = 1MeV
fusion = 14 MeV
explain the mechanism of how radiation damage occurs, i.e. explain the steps of the displacement cascade
1) Energetic incident particle (e.g. fast neutron) hits an atom in a crystal
2) KE transfer enough to displace the hit atom from lattice position
- this becomes a primary knock-on atom (PKA)
- it leaves a vacant site
3) PKA moves through lattice with significant energy, creates further knock-on-atoms
- mean free path between displacement collisions depends on energy of knock on atoms
4) PKA and any addition knock on atoms end up interstitially
what does each knock on event in the displacement cascade produce
Each knock-on-event produces a vacancy and an interstitial
- this is a Frenkel defect
why doesn’t recombination of the vacancies and interstitials occur after a displacement cascade
- some recombination occurs but not fully due to the fast diffusion of interstitials
- interstitials can also dissapear at dislocations and grain boundaries
what is the maximum transferable energy of an atom to another atom on collision, what is the average transferred energy, what is the average number of atoms displaced per PKA
Ed = lattice displacement energy
En = energy of incident particle (neutron)
max. transferable energy = ζEn
av. transferable energy = ζEn / 2
ζ = 4A / (1+A)^2
A = atomic mass number
Av. no. of atoms displaced per PKA = Ep / 2Ed
Ep = expected PKA energy
what are the two main types of dislocation damage, when does each occur
1) Dislocation loops, T < 0.2Tm
2) Voids T > 0.2 Tm
explain how/why dislocation loops form after radiation damage, what do they affect
- interstitials can condense as monolayer discs between close packed planes, an interstitial loop
- vacancies can do the same, a vacancy loop
- driving force to form them is lower strain energy
- both create stacking faults
- they are also sessile as b = a/3<111> which is perp. to the slip plane
what happens to the populations of dislocation loops during continued radiation damage
- initially the number increases
- then a steady state forms when the rate of removal due to annealing from local heating is equal to the rate of formation
explain how voids form during radiation damage/ what effects they have
- higher temperature irradiation causes cavities and voids to form, once they form they grow easily
- this is because the interstitials formed on radiation damage are absorbed easily at dislocations but vacancies join voids due to different diffusion rates
- vacancies are continually produced
- they cause swelling
what can we say about size/number of voids at high vs low temps
high temps = fewer, larger voids
low temps = more, smaller voids
