Course G - Materials under extreme conditions

Question 1

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

Answer 1

Pressure, P = Force, F / Area, A

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

Question 2

what is 1atm and 1bar

Answer 2

1atm = 101325 Pa
1bar = 100000 Pa

Question 3

define Gibbs free energy and give the differential form

Answer 3

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

Question 4

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

Answer 4

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)

Question 5

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

Answer 5

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)

Question 6

what do pressure-temperature phase diagrams show

Answer 6

• 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

Question 7

derive the Clausius-Clapeyron equation

Answer 7

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

Question 8

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

Answer 8

• 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

Question 9

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

Answer 9

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

Question 10

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

Answer 10

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

Question 11

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

Answer 11

• 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

Question 12

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

Answer 12

• 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

Question 13

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

Answer 13

• 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

Question 14

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

Answer 14

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

Question 15

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

Answer 15

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

Question 16

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

Answer 16

• 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

Question 17

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

Answer 17

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

Question 18

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

Answer 18

• 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

Question 19

what occurs in the reconstructive phase transition of Mg2SiO4

how does this link to earthquakes

Answer 19

• 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

Question 20

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

Answer 20

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

Question 21

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

Answer 21

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

Question 22

what is the asymmetry between freezing and melting

Answer 22

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

Question 23

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

Answer 23

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

Question 24

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

Answer 24

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

Question 25

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

Answer 25

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

Question 26

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

Answer 26

ΔT = -2 γ(ls) / ΔSv r hence greater r = lower ΔT so higher effective melting point

Question 27

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

Answer 27

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

Question 28

define creep

Answer 28

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

Question 29

give the three factors that creep rate depends on

Answer 29

- material - applied stress - homologous temperature

Question 30

define homologous temperature

Answer 30

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

Question 31

what are the three regimes of creep

Answer 31

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

Question 32

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

Answer 32

dε/dt = A σ^n the two types of creep depend on the value of n: 1) Dislocation creep, 3

Question 33

give the key points about how dislocations avoid obstacles

Answer 33

- 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

Question 34

how does dislocation creep occur, what must diffuse?

Answer 34

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

Question 35

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

Answer 35

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

Question 36

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

Answer 36

dε/dt = Aσ^n = A' σ^n D = A'' σ^n exp(-Q/RT) - can plot different log-log graphs of this

Question 37

what is diffusion creep, what are the two types

Answer 37

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

Question 38

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

Answer 38

- 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)

Question 39

give the equation that describes diffusion creep strain rate

Answer 39

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

Question 40

how can creep rates be minimised

Answer 40

- high mpt - low dislocation motion - stable ppts - larger grains

Question 41

explain the 4 stages of the (idealised) Carnot cycle

Answer 41

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

Question 42

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

Answer 42

Efficiency = 1 - TC/TH more efficient at higher temps

Question 43

briefly explain the brayton cycle (used in jet engines)

Answer 43

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

Question 44

explain why jet engines blades have significant materials challenges

Answer 44

in operation: - T > 0.75Tm - 10000rpm, high centripetal force - lifetime = 3 years - extracts 500KW

Question 45

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

Answer 45

- superalloys created - cold air pushed through blades - thermal coatings

Question 46

explain what properties possible jet engine blades must have and must not have what metal does this leave

Answer 46

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

Question 47

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

Answer 47

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

Question 48

what microstructure does a nickel superalloy have

Answer 48

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

Question 49

give the structure/ key properties of the γ phase

Answer 49

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

Question 50

give the structure/ key properties of the γ' phase

Answer 50

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)

Question 51

what is the interface between γ and γ'

Answer 51

coherent - low interfacial energy - prevents ppt coarsening

Question 52

what is the advantage of directionally solidified crystals

Answer 52

- 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

Question 53

what are single crystal blades, why are they useful

Answer 53

- 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

Question 54

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

Answer 54

- order hardening

Question 55

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

Answer 55

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

Question 56

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

Answer 56

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

Question 57

what are the three layers of thermal barrier coatings

Answer 57

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

Question 58

what are the mechanical properties of ice

Answer 58

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

Question 59

what is the bonding in ice

Answer 59

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

Question 60

what is the most common structure for ice

Answer 60

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

Question 61

State the 2 ice rules and explain proton disorder

Answer 61

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

Question 62

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

Answer 62

- yes it can - it gives low friction coeff., high adhesion of 2 surfaces, good compaction

Question 63

explain regelation (regarding ice)

Answer 63

- 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"

Question 64

give typical energies of neutrons in fission/fusion reactors

Answer 64

Fission, fast neutron = 1MeV fusion = 14 MeV

Question 65

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

Answer 65

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

Question 66

what does each knock on event in the displacement cascade produce

Answer 66

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

Question 67

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

Answer 67

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

Question 68

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

Answer 68

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

Question 69

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

Answer 69

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

Question 70

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

Answer 70

- 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

Question 71

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

Answer 71

- 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

Question 72

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

Answer 72

- 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

Question 73

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

Answer 73

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