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Materials IA > Course G - Materials under extreme conditions > Flashcards

Course G - Materials under extreme conditions Flashcards

1
Q

define pressure and give the expression for pressure at T = 0K

A

Pressure, P = Force, F / Area, A

P = -dU/dV
at 0K
units = Nm^-2 = Pa

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2
Q

what is 1atm and 1bar

A

1atm = 101325 Pa
1bar = 100000 Pa

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3
Q

define Gibbs free energy and give the differential form

A

G = H-TS
dG = dH - TdS - SdT
dH = δq + VdP
δq = TdS at equil.
so
dG = Vdp - SdT

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4
Q

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

A

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)

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5
Q

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

A

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)

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6
Q

what do pressure-temperature phase diagrams show

A
  • 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
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7
Q

derive the Clausius-Clapeyron equation

A

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

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8
Q

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

A
  • 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
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9
Q

what are the 2 main methods for determining a P-T phase diagram experimentally

A

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

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10
Q

how does the piston/cylinder device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.

A
  • original method
  • piston exerts pressure on sample
  • heaters can be added to change temp
  • materials of piston determine max. pres
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11
Q

how does the opposed anvil device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.

A
  • 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
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12
Q

how does the multi-anvil device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.

A
  • 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

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13
Q

how does the diamond anvil cell device work as experimental apparatus for determining P-T diagrams, give advs. disadvs.

A
  • 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

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14
Q

how does pulsed-laser implosions work as experimental apparatus for determining P-T diagrams

A
  • pulsed lasers for short times focused on sample
  • this gives V V high pres. and temps.
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15
Q

how can ΔS be calculated for a transition (2 methods)

A

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

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16
Q

what is the effect of high pressures on molecular materials/solids

A
  • 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

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17
Q

what is a reconstructive phase transition, what does it include/require

A

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
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18
Q

explain what occurs in the reconstructive phase transformation of NaCl and MgO

A
  • 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
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19
Q

what occurs in the reconstructive phase transition of Mg2SiO4

how does this link to earthquakes

A
  • 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
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20
Q

what occurs in the graphite-diamond transformation i.e. give the conditions

A

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
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21
Q

what is the RTP structure of Boron Nitride, what structure does it form under high temps and pressures

A
  • at RTP, BN has hexagonal form like graphite
  • at high pressures and temps it transforms to sphalerite structure like diamond
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22
Q

what is the asymmetry between freezing and melting

A
  • when freezing there is a barrier to nucleation which allows for supercooling
  • when melting there is no such barrier so melting begins immediately
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23
Q

outline the Lindemann and Born theories of melting and explain what actually happens

A

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

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24
Q

why for most materials is surface melting found at temperatures < Tm

A

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
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25
Q

how can we consider the melting of small particles in terms of r*

A

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
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26
Q

give the expression for the supercooling at which melting occurs for a particle of radius r

A

ΔT = -2 γ(ls) / ΔSv r

hence greater r = lower ΔT so higher effective melting point

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27
Q

what are the three things that can be done to suppress surface melting/ observe superheating

A

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

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28
Q

define creep

A

“Creep is defined as time-dependent permanent deformation (plastic) of a material under the action of a stress σ(applied) where
σ(applied) < σy”

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29
Q

give the three factors that creep rate depends on

A
  • material
  • applied stress
  • homologous temperature
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30
Q

define homologous temperature

A

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

31
Q

what are the three regimes of creep

A

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

32
Q

give the equation for strain rate in the secondary stage of creep and from this, define the two types of creep

A

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

33
Q

give the key points about how dislocations avoid obstacles

A
  • 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
34
Q

how does dislocation creep occur, what must diffuse?

A
  • for dislocation creep, atoms must diffuse to or from the dislocation core
  • this occurs in one of two ways depending on temperature
35
Q

what are the two temperature dependent mechanisms for dislocation creep, briefly explain them

A

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

36
Q

how can we change equation for strain rate in the secondary regime of creep to account for diffusion of atoms in dislocation creep

A

dε/dt = Aσ^n = A’ σ^n D = A’’ σ^n exp(-Q/RT)

  • can plot different log-log graphs of this
37
Q

what is diffusion creep, what are the two types

A

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
38
Q

Explain the difference between Nabarro-Herring and Coble creep and when they occur

A
  • 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)

39
Q

give the equation that describes diffusion creep strain rate

A

dε/dt = B D σ/d^2 = B’ σ/d^2 exp(-Q/RT)

d = grain diameter

40
Q

how can creep rates be minimised

A
  • high mpt
  • low dislocation motion
  • stable ppts
  • larger grains
41
Q

explain the 4 stages of the (idealised) Carnot cycle

A

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

42
Q

give the efficiency equation for the carnot cycle, at what temperature is it most efficient

A

Efficiency = 1 - TC/TH

more efficient at higher temps

43
Q

briefly explain the brayton cycle (used in jet engines)

A

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

44
Q

explain why jet engines blades have significant materials challenges

A

in operation:

  • T > 0.75Tm
  • 10000rpm, high centripetal force
  • lifetime = 3 years
  • extracts 500KW
45
Q

what are some things that have been done to blades to try to keep up with rising operation temperatures in jet engines

A
  • superalloys created
  • cold air pushed through blades
  • thermal coatings
46
Q

explain what properties possible jet engine blades must have and must not have

what metal does this leave

A

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

47
Q

what are some of the elements that are added to Nickel to make it a superalloy

A

Cr (corrosion help)
Co, W (SS strengthening)
Al, Ti, Ta (all help form γ’)

48
Q

what microstructure does a nickel superalloy have

A

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

49
Q

give the structure/ key properties of the γ phase

A

γ:
- cubic F
- ccp
- each lattice site has same average composition
- solid solution

50
Q

give the structure/ key properties of the γ’ phase

A

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)

51
Q

what is the interface between γ and γ’

A

coherent
- low interfacial energy
- prevents ppt coarsening

52
Q

what is the advantage of directionally solidified crystals

A
  • 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
53
Q

what are single crystal blades, why are they useful

A
  • 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

54
Q

state the main strengthening mechanism due to the γ - γ’ precipitate structure

A
  • order hardening
55
Q

explain order hardening in the γ, γ’ structure and how this gives strengthening

A

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

56
Q

explain the need for thermal barrier coating on blades in jet engines and state their material requirements

A

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

57
Q

what are the three layers of thermal barrier coatings

A

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

58
Q

what are the mechanical properties of ice

A

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
59
Q

what is the bonding in ice

A

ice formed of H2O molecules:
- these are non-linear 109.5 degree bond angle
- these pack together with hydrogen bonds roughly tetrahedrally

60
Q

what is the most common structure for ice

A

Ih:
- hexagonal, stable under ambient conditions
- tetrahedrally linked framework, hexagonal symmetry
- arrangement of oxygen like Wurtzite

61
Q

State the 2 ice rules and explain proton disorder

A

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
62
Q

can surface melting occur in ice?
what properties doe this give?

A
  • yes it can
  • it gives low friction coeff., high adhesion of 2 surfaces, good compaction
63
Q

explain regelation (regarding ice)

A
  • 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”

64
Q

give typical energies of neutrons in fission/fusion reactors

A

Fission, fast neutron = 1MeV
fusion = 14 MeV

65
Q

explain the mechanism of how radiation damage occurs, i.e. explain the steps of the displacement cascade

A

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

66
Q

what does each knock on event in the displacement cascade produce

A

Each knock-on-event produces a vacancy and an interstitial
- this is a Frenkel defect

67
Q

why doesn’t recombination of the vacancies and interstitials occur after a displacement cascade

A
  • some recombination occurs but not fully due to the fast diffusion of interstitials
  • interstitials can also dissapear at dislocations and grain boundaries
68
Q

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

A

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

69
Q

what are the two main types of dislocation damage, when does each occur

A

1) Dislocation loops, T < 0.2Tm
2) Voids T > 0.2 Tm

70
Q

explain how/why dislocation loops form after radiation damage, what do they affect

A
  • 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
71
Q

what happens to the populations of dislocation loops during continued radiation damage

A
  • 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
72
Q

explain how voids form during radiation damage/ what effects they have

A
  • 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
73
Q

what can we say about size/number of voids at high vs low temps

A

high temps = fewer, larger voids
low temps = more, smaller voids

Materials IA (9 decks)

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