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The Grind > Nuclear design > Flashcards

Nuclear design Flashcards

1
Q

average volumetric power of typical reactors

A

AGR = 3 kW/l
PHWR = 10 kW/l
RBMK = 20 kW/l
BWR = 50 kW/l
PWR = 100 kW/l
LMFR = 300 kW/l

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2
Q

what is the equivalent radiative heat transfer coefficent for the gap

A

h_irr = 4 * sigma * T_fo^3 / (1/eps_f + 1/eps_cl -1)

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3
Q

Thermal resistance of vod fuel

A

R_fuel = 1 / (4 * pi * avgK) * (void factor)
void factor = 1 - ln( (R_fo/R_fi)^2 ) / ((R_fo/R_fi)^2 -1)

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4
Q

hot channel coefficents

A

FzN = average hf hot ch / average hf average ch = 1.5 (Bessel)
FmaxN = max hf hot ch / average hf hot ch = 2.3 (cosine)

hentalpy hot ch factor
Fh = hentalpy rise in hot ch / hentalpy rise in avergae ch = 1.1

engineering safety factor
Feng = 1.05 (5%)

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5
Q

Robertson factor

A

RF = 4 / (chi^2R_fo^2) * I(chiR_fo) -1)

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6
Q

How can we categorize different tyoe of reactor (U, BR, cycle)

A

U = fuel fissioned/resource input = 0.5%(LWR) - 5% (LMFR)
BR = fission produced/atoms consumed
open or closed cycle

Burner: BR<1, open
Breeder: BR>1, open
Converter: BR<1, closed
Incinerators: BR=1

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7
Q

Melting temperature for fuels

A

Tmelt:
UO2 = 2800 °C
UN = 2300°C
Umetal = 110°C

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8
Q

What phenomena must be descirbed by fuel performance codes?

A

-Thermo-mechanical behaviour
-Neutron flux
-Fission gas and Helium
-Microstructural change (high burnup and restructuring)
-Radial temperature gradient (hourglassing, …)
-Chemical phenomena

Very different scales in space and time

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9
Q

Which phenomena are involved in fuel restructuring?

A

-Densification
-Grain growth
-Formation of columnar and equiaxed
-Formation of central void

Other related phenomena are:
-crack healing
-plutonium, hence power redistribution

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10
Q

Which phenomena contribute, negatively or positively, to restructuring?

A
  • temperature caused evaporation and condensation in lenticular pores (high temperature vapours)
  • diffusion due to temperature (Soret) and concentration gradients
  • irradiation at EoL
  • cracks increase the migration
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11
Q

Which are the thermal effects of restructuring?

A
  • Better geometry
  • Less porosity
  • Pu redistribution increase power production towards the center
  • O/M = 2 at perphery, lower at center, due to high volatility
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12
Q

Why a O/M ratio higher than 2 is dangerous?

A

High oxygen concentration can react and corrode the cladding if there is contact. For this reason pelets are always hypostoichiometric

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13
Q

What are the effects of plutonium redistribution?

A

-Reduction of melting temperature
-Increase of power production
-Reduction of thermal conductivity

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14
Q

What are the O/M stoichiometry effects?

A

-Fabrication with hypostochiometry to reduce risk of cladding corrosion so that conductivity increases during operation (pyramidal shape)
- T melting is maximum at O/M = 2

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15
Q

What is the effect of porosity?

A

It decreases fuel conductivity

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16
Q

fuel conductivity formula with all dependences?

A

K = (1/(A+BT) + CT^3 + D/T^2 * exp(-E/T) ) * (1-p)^alpha
Also it decrases asintotically with burnup

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17
Q

When densification and swelling are important in LWR?

A

Densification is dominant for Bu< 10 GWd/ton. It is due to frenkel pair created from fission. 5 MeV create 25000 Frenkel pair per each fission but only 5000 do not recombine and create densification. Interstitals are absobed and vacancies diffuse to grain boundary.
To limit densification, a limit on fuel porosity at fabrication must be respected (in LWR)

Above Bu>10 GWd/ton swelling dominate, due to fission gas

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18
Q

What happens in a LWR if p_gap > p_coolant ?

A

We have outward creep, cause Zry it’s not very resistant to creep -> the gap increase -> fuel temperature increases -> more gas are released -> p_gap increases
Also k_fuel decreases and D_hyd decreases -> T_coolant increases

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19
Q

What about densification in LMFR?

A

Densification due to restructuring is dominant respect to the one due to fission. Because of this porosity is actually wanted in FR fuel becuase it enhances restructuring

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20
Q

Void swelling

A

Happens at EoL for FR and causes both fuel and cladding to expand. The rate is almost the same.

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21
Q

Which are the main Pellet-Cladding interaction

A

Hourglassing, negligible for hollow pellets. This leades to
-high stresses in cladding
-chemical reaction between iodine and Zry (not with SS)

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22
Q

What is the burn up in MWd/ton?

A

Bu is the energy extracted per unit mass.
Bu = releases energy (MWd) / mass of uranium fuel (kg_U)
Tipically 1MWd/1.05 g_U235

We can also measure it as (released energy / mass of oxide fuel) which will be smaller of a factor 0.88.

The teoretical limit is 950’000 MWd/ton_U but in reality we reach around 50’000 - 70’000 MWd/ton_U

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23
Q

What is the burnup in FIMA?

A

FIMA = Fission per metal atom = fissions / initail metal atoms

So 1% FIMA = 9.5 MWd/kg_U

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24
Q

What are the effects of burnup ?

A
  • may lead to a high burnup structure
  • decrease young modulus
  • decrease thermal conductivity
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25
Q

Bateman equations for burnup?

ciao

A

They are used for estimating the burnup. The complete form is an integral one but with the hp of one group cross section and no energy dependence:
dNm(r,t)/dt = -sigma_a,m N_m phi + sum_j(sigma_c,j N_j phi) + sum_k(lambda_k N_k) - lambda_m

We can also neglect isotopes with low concentration or reaction with low cross section.
We may understimate He production

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26
Q

Generalized plain strain

A

eps_zz = deltaZ/Z = constant

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27
Q

pure plain strain

A

eps_z = deltaZ/Z = 0

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28
Q

pure plain stress

A

sigma_z = 0

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29
Q

What are the consequences of helum embrittlement?

A

On metals is a loss of ductility not recoverable with annealing

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30
Q

How is He embrittlement generated?

A
  • Due to n,alfa reactions on metals
  • He migrate only for T/Tmelt > 0.5
  • He migrate to grain boundaries
  • At grain boundary it nucleate due to presence of nucleation sites (M_23 C_6) for T/Tmelt < 0.8
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31
Q

Why He can not be annealed?

A

If T increases M23C6 dissolve but the bubbles stay there and they will even expand. He embrittlement cannot be recovered

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32
Q

Bateman equation for helium production?

A

Scrivile scemo (pag.118)

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33
Q

Why He production is worse in fusion reactor?

A

Because in fusion reactor the neutrons have higher energy (14MeV) and the crossection of the reaction (n,alfa) increases with energy up to (10MeV)

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34
Q

What are the Helium embrittlement units of measure?

A

We use the ppm_He, the DPA (Displacement Per Atom) and the ration ppm/DPA.
fusion: ppm/DPA = 100
fast: ppm/DPA = 10
thermal: ppm/DPA = 1

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35
Q

What is the fission yield for gas?

A

Total is around 0.3.
Xe = 0.27
Kr = 0.03
Ar and He yield is very low

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36
Q

What is the trade off in fission gas release and retention?

A

If (deltaV/V)_gas is higher I close gap faster, Tfuel decreases but higher contact pressure
If FGR is higher the conductivity in gap decreases so Tfuel increases, p_gap increases but less contact pressure

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37
Q

FG in crystals in oxide vs metallic fuel?

A

Xenon is a very big atom.
Oxides fuel have fluorite structure that can accomodate Xe with low deformation (1%), while metallic fuel expand a lot (10%)

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38
Q

What is a-thermal release of fission gas?

A

If a fission happens in a range mu=6-7 micrometri to fuel boundary the particle is ejected directly outside the fuel.
fraction of a-thermal release = (piR^2-pi(R-mu)^2) / (piR^2) = 2pi muR / (piR^2) = 2mu/R
but actually mu/R due to simmetry of fission product emission
= 0.1%

This percentage is increased from
-porosity connected to the outside
-fuel cracks
-not uniform fission rate in fuel ( in LWR the fraction is higher due to selfshielding)

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39
Q

FGR magnitude in LWR and FR? How can FGR be measured?

A

LWR : 1%
FR : 70-90%

It can be measured from pressure in experimental setup, post irradiation examination with fluorescence or puncturing

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40
Q

What are the hp for FGR model?

A

Xe with null solubility
Xe precipitate at grain boundary or in bubbles
T and fission rate F are uniform in grain because grains are very small

Then we consider a spherical grain, trapping and re-solution faster than diffusion

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41
Q

What is the FGR model?

A

dP/dt = y * F
dC/dt = D* div^2 C - gC + bM + yF
dM/dt = gC - bM

P = fission gas produced
y = yield
F = fission rate
C = gas in fuel matrix, moving
M = gas in bubbles

with G = C+M
g,b&raquo_space; D/a^2
b/(b+g)D = D_eff = 5e-8 * exp( -40262/T ) is very low

So we get: dG/dt = D_eff div^2 G +yF = - D_eff * pi^2/a^2 * G + yF

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42
Q

What happens when FG accumulate at boundary?

A

Gas accumulate, nucleate in bubbles, pressure increases, bubbles grow and connect. A path towards the gap is created by merging of the bubbles.
Swelling rate decreases after venting, togheter with FG released

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43
Q

What is the Vitanza threshold?

A

In a graph Bu - Tfuel the Vitanza threshold is the line below where the FGR is less than 1%.
It stops at 50GWd/ton of Burnup since this is when high burnup structure is formed

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44
Q

How to design the plenum for FG pressure?

A

Determine the FG produced, P, evaluate the released part, R, and use pV = nRT for calculating the new plenum pressure.
Plenum pressure will increase due to thermal expansion at startup then, less, due to FGR

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45
Q

What is the displacement vector?

A

u = ux + uy + uz
It’s the vector for each point that goes from the undeformed to the deformed configuration

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46
Q

fundamental Hps for mechanical analysis

Infinitesimal strain theory
Continuum deformable body

A

Infinitesimal strain theory. Means that displacements are smaller than the body so that geometry doesn’t change
Continuum deformable body, displacements are C^2 functions, acceptable only in infinitesimal strain theory.

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47
Q

What is the displacement gradient tensor?

A

It expresses the rigid motion + the local deformation. It’s U_ij = du_i/dx_j
It can be factored into a skew symmetric component (rigid rotations) omega_ij and a symmetric component (local strains) eps_ij

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48
Q

What is the local strain tensor?

A

It’s a connection between the displacement field and the stress field.
It’s normalized.
It can always be diagonalized.
The sum of the principal strains is invariant (deltaV/V).

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49
Q

Compatibility equation?

A

eps_ij = 1/2(u_ij + u_ji)
where u_ij = du_i / dx_j

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50
Q

What is the stress tensor?

A

It’s a tensor that take into account for all the forces acting in a single point. It has 6 unknowns since we’ll never consider polar materials.
So it’s symmetric.
If expressing tension, is positive.
It can be diagonalized and the sum of the trace is invariant.
1/3*tr(sigma) = sigma_hyd: hydrostatic pressure.

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51
Q

Which are the unknown of the mechanical problem?

A

3 u (u_x, u_y, u_z), displacements
6 eps (eps_ij), deformations
9 sigma (sigma_ij), stresses

+ the temperature

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52
Q

static equilibrium equation?

A

sigma_ij = sigma_ji (rotation equilibrium) per ogni i, j
sum_i (dsigma_ij/dx_i) + F_j = 0 (transaltion equilibrium) per ogni j

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53
Q

What equations describe a rigid body in equilibrium?

A

sum of moments and sum of forces = 0 if it has no costrains

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54
Q

what is a isostatic, hyperstatic or hypostatic problem?

A

isostatic: # costraint = # of degree of freedom

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55
Q

postulate of additional costraint, of stiffening and postulate of internal pressure?

A
  • add. costraint. If I add a costrain to a body in equlibrium it will remain like that
  • stiffening: If a deformable body is in equilibrium and it stiffens it will remain like that
  • internal pressure: the equilibrium of each part of a body is sufficent for the equilibrium of the all body
56
Q

What’s a uniaxial tensile test?

A

Pull a sample from extremities with force F, consider L0, length with no change in section so that tension is uniform and same way A0, D0.
eps_engL = deltaL/L0
deps_trueL = dL/L => eps_trueL = ln(L/L0) = ln(1+eps_engL)
There is also a transversal deformation: eps_T = deltaD/D0

for z = axial direction (same as L)
sigma_eng,z = P/A0
sigma_true,z = P/A

PLOTS diverge because of necking and sigma_true increases due to A decrease

57
Q

stress - strain plot for mild vs austenitic steels

A

MILD: has a peak at yield strength
austenitic, then a decrease in stress due to precipitation of carbon in lattice to grain boundary, then hardening then rupture
AUSTENITIC: yield strength, hardening and rupture

58
Q

What is the young modulus?

A

E = sigma_z/eps_L
engineering stress and starin considered
It’s the inclination of the line in the elastic domain also callled Stifness

59
Q

What’s the Poisson ratio?

A

nu = -eps_T/eps_L
In the elastic domain
it ranges (-1, 0.5)
metals and cermaic have nu=0.25 - 0.35

60
Q

What is the UTS?

A

The Ultimate Tensile Stress is the maximum value of the engineering stress. After that material fail

61
Q

How do brittle and ductile materials get to rupture in axial test?

A

Ductile materials form a cone at a 45° degree inclination. This is because it’s a caused by shear
Brittle material break flat due to normal stress. Small deformation is seen so the curve is very short

62
Q

Difference between elastic and plastic deformation

A

In the elastic domain all trasnsformation are reversible while plastic domain transformation are permanent

63
Q

How is yielding point find?

A

with the 0.2% offset in the stress-strain diagram

64
Q

bulk modulus

A

It’s another constant to express the stifnesss tensor as E, nu.
B = K = E/(3(1-2nu)) = -p/(deltaV/V)
It express the pressure change in the volume

65
Q

Shear modulus

A

It’s another constant to express the stifnesss tensor as E, nu.
G = E/(2(1+nu))

66
Q

Charpy test?

A

A specimen is hit with an hammer and we measure the final height of the hammer. The energy absorbed by the sample is called notch impact energy or toughness and it’s the total integral of the stress-strain diagram. The integral of the elastic domain is called resilience

67
Q

Ductile vs brittle in charpy test?

A

DUCTILE: big deformation
BRITTLE: small to non deformation

68
Q

What is the behaviour of steels ductility with temperature?

A

fcc aka austenitic are always ductile, even at low temperatures
bcc aka ferritic are brittle at low temperature, have a steep increase of nocth impact energy and are ductile at high T. We call the transition point Nihil Ductility Transition Temeperature (-20/-10 °C) and it increases with irradiation

69
Q

How to avoid creep?

A

T/Tmelt<0.3

70
Q

What is the behaviour of steel constitutive diagram with irradiation?

A

FCC: at high T irradiation decrease UTS, at low T irradiation increase stifness and strength(UTS).
BCC: more stifness but lower ductility and more strength.
more stifness means higher E

71
Q

Which hp are made in the elastic domain?

A
  • small strains (undeformed geometry)
  • linear relation between sigma and eps
  • from previous two we get the superposition principle
72
Q

Which hp are made to reduce stifness tensor to 2 variables?

A

-macroscopically homogeneous material
-material perfectly isotropic
- loading history is reversible
- symmetric stress and strain tensors

73
Q

Constitutive eq. with E and nu?

A

eps_ii = 1/E * (sigma_ii - nu*sum_jj sigma_jj) + alfa(T-T_ref)
eps_ij = 1/(2G) * sigma_ij = 1+nu/E * sigma_ij
with i =! j

74
Q

Constitutive equation with lame parameters?

A

Lame parameters are mu and lambda (harder to find experimentally):
sigma_ii = 2 * mu * eps_ii + lambda * sum_j eps_jj
sigma_ij = 2 * mu * eps_ij

75
Q

Why are thermal and mechanical problems decoupable?

A

Because deltaT causes a big deltaV/V, but deltaV/V causes a small deltaT.
This can be checked with Gruneisen parameter:
deltaT = - gammaTdeltaV/V = 0.1 K

76
Q

What change in temperature is caused by changing the volume?

A

In elastic region an expansion lead to a lower T, while in plastic region T will increase because moving dislocations dissipate energy

77
Q

What are the stresses on an object with uniform temperature no mech loads and no costraints?

A

stresses are zero, eps_ii>0 due to expansion

78
Q

What are the stresses on an object with uniform temperature no mech loads and full costrained?

A

eps_i = 0 = 1/E(sigma_i-nu(sigma_j+sigma_k) +alfa(T-Tref)
=> sigma_i = -alfaE/(1-2nu)deltaT = Mpa

79
Q

What are the hp for cylindrical geometry for the mechanical analysis?

A
  • Axial symmetry of geometry and load -> tau_theta,z = tau_theta,r 0 -> eps_theta,z = eps_theta,r = 0
  • Orthocilindricity (geometry is preserved) -> eps_r,z = 0 -> tau_r,z = 0
  • No volume forces
80
Q

Equilibrium eq. for a cylinder section?

A

sigma_theta = sigma_r + r*dsigma_r/dr

81
Q

compatibility equation for a cylinder?

A

eps_r = du_r/dr
eps_theta = u_r/r = deltaD/D
eps_z = du_z/dz

82
Q

pipe equation?

A

It’s derived from compatibility, equilibrium and constitutive equations:
d(r^3 * dsigma_r/dr)/dr + r^2 *alfa * E / (1-nu) *dT/dr = 0
Contains mech and thermal part. We write it with sigma_r because boundary conditions are easier to implement

83
Q

alfa*E/(1-nu) is hgher in bcc or fcc?

A

fcc

84
Q

pure thermal problem for pipe?

A

T(r) from thermal analysis.
from pipe eq I find sigma_r then sigma_theta and then sigma_z:
sigma_r (r) = alfaE/(1-nu) * 1/r^2 * ( (r^2-a^2)/(b^2-a^2) * int_ab(Trdr)- int_ar(Trdr) )
sigma_theta (r) = alfaE/(1-nu) * (T_mean - T(r))

85
Q

thermo-mechanical problem for a pipe?

A

Using superposition principle we first solve the thermal problem than the mechanical one.

86
Q

What are the boundary condition for pure thermal problem?

A

for pipe:
- sigma_r(a) = sigma_r(b) = 0
- sigma_theta = sigma_r + r*dsigma_r/dr

for cylinder:
- dsigma_r(a)/dr = 0
- sigma_r(b) = 0
- sigma_theta = sigma_r + r*dsigma_r/dr

87
Q

pure mech problem for pipe?

A

solving d(r^3*dsigma_r/dr)/dr = 0
we get sigma_r (r) = A/r^2 + B
If the BC are in term of displacements we can write
u_r (r) = A/r + Br
from equilibrium equation we get hoop stress.
For sigma_z we need an Hp

88
Q

what is the hp of plane stress?

A

In mech problem sigma_z = 0
from constitutive eq. we get eps_z

89
Q

what is the hp of plane strain?

A

In mech problem eps_z = 0
from constitutive eq. we get sigma_z

90
Q

what is the hp of generalized plane strain?

A

In mech problem eps_z = const
from constitutive eq. we get sigma_z, qhich is lower than the pure plain strain condition

91
Q

Thermal shock?

A

very hot thing in a heat sink. Tmean is high while T(surface) is low. for thermal problem
sigma_theta(surface) = alfaE/(1-nu)(Tmean-T(surface)) is high

92
Q

Mech problem of pipe for internal pressure (lame solution)?

A

sigma_r(Ri) = -p
sigma_r(Re) = 0
sigma_r(r) A/r^2 + B

->A = -p * Re^2*Ri^2 / (Re^2-Ri^2)
B = p * Ri^2 / (Re^2-Ri^2)
sigma_r(r) = - Ri^2/(Re^2-Ri^2) * (Re^2/r^2 - 1) * p
from equilibrium:
sigma_theta = R^2/(Re^2-Ri^2) * (Re^2 /r^2 + 1) * p
These are called the Lamè solutions in cylindrical coordinates
It’s similar for external pressure

93
Q

What is ovality?

A

W = (D_max-D_min)/D_av
It’s the ovality of the “base” of the cyilinder.
To avoid buckling it’s necessary that this and eccentricity are below a limit

94
Q

What is eccentricity?

A

ecc = (t_max-t_min)/t_av
It’s the variation in thickness
To avoid buckling it’s necessary that this and ovality are below a limit

95
Q

pro and cons of outer pressure for cladding?

A

pros: reduce crack propagation
cons: can lead to buckling

96
Q

What is buckling?

A

It’s a collapse in the elastic domain

97
Q

mariotte solution for inner load?

A

Mariotte suppose that if R/t > 5 - 7 then Re = Ri = r = R. It’s a sort of average
Re^2-Ri^2 = (Re - Ri) * (Re +Ri) = t *2R

so sigma_theta = R/tp
sigma_r = -p/2
sigma_z = r/2tp (generalized plain strain)

98
Q

How was mariotte stresses derived?

A

Calculating an eq for half pipe.
With this approach we can derive also sigma_z (same as generalized plain strain)

99
Q

Mariotte stress for a sphere?

A

sigma_r = -p/2
sigma_theta = sigma_phi = R/(2t)*p

100
Q

Why using a dome over a cyilinder is a problem?

A

Because of the junction where thickness changes. The displacements, u, will be different and this will induce stresses

101
Q

Which stresses are more conservative towards what?

A

Lame is more conservative towards yielding
Mariotte is more conservative towards failure

102
Q

Galileo - Rankine criterion

A

Used for brittle materials. The failure is due to normal stresses so we impose:
for tensile stress
sigma_c = max (S1, S2, S3) < sigma_a,t
for compressive stress
sigma_c = min (S1, S2, S3) > sigma_a,c
tipically sigma_a,t < sigma_a,c ( in modulo)
Tipically sigma allowable consider a safety margin design.
It’s a square

103
Q

Mohr-cCoulomb generalization in GR criterion?

A

The shear in overestimated so we use some lines

104
Q

Tresca criterion?

A

For ductile materials. The failure is due to shear stress due to 45° rupture.
We impose max (abs(S1-S2), abs(S2-S3), abs(S3-S1)) < sigma_a (S_y)

S_yield = 2*tau_yield due to geometry so

We impose 0.5 * max (abs(S1-S2), abs(S2-S3), abs(S3-S1)) < sigma_a (tau_y)
It’s a rombo

105
Q

Von Mises criterion?

A

Since from observation Tresca is overconservative we use von mises that compute the deviatoric energy to calculate the limit stresses:
E_dev = 1/2simga_deveps_dev

from constitutve eq.
sigma_dev = 2mueps_dev
so the equivalent stress results:
sigma_c = sqrt( 0.5* ((S1-S2)^2 + (S2-S3)^2 + (S3-S1)^2 ) < sigma_a

It’s an oval

106
Q

When are Tresca or Von Mises better?

A

Tresca is better for being conservative
Von Mises is better when shape changes like creep

107
Q

What is the Ros Eichinger criterion?

A

Same results as Von mises but valid also in plastic domain

108
Q

When is the mechanical bearing capacity independent from thermal load?

A

If
- no thermal shock
- no impact
- no buckling
- no creep
- no fatigue
- no cracks
- symmetry of consistutive diagram (ductile)

109
Q

Which vategory of stresses ASME ask to check?

A

-Membrane, or primary, stresses, P_m
-Local stresses P_l including bending stresses P_l+P-b
-Secondary stresses P_l + P_b + Q, where Q are the thermal stresses
The limit tipically is S_m = 2/3 S_yield

110
Q

What are membrane stresses (ASME)?

A

They are calculated with Mariotte and Tresca criterion
P_m < S_m

111
Q

What are the local stresses (ASME)?

A

They are calculated with Lamé + the bending stresses which are associated with irregularities like junctions.

112
Q

What are the secondary stresses?

A

The sum of local and thermal stresses.

113
Q

What is the behaviour of constitutive diagram with temperature?

A

It decreases both yield and UTS

114
Q

Which apporaches can we use towards crack propagation?

A
  • T > FTP = NDT + 33°C never propagate cracks
  • FTP > T > FTE = NDT + 20°C doesn’t propagate cracks in elsatic domain
  • FTE > T > NDT + 17°C doesnt’ propagate for stresses lower than S_y/2
    note that bcc increase their NDT with irradiation
115
Q

What is ratcheting?

A

It’s the accumulation of plastic deformation if the stresses (thermal for simplicity) gets loaded and unloaded many times with sigma > 2S_yield.
The general limit is Q < 3S_m

116
Q

What influences creep?

A

Temperature, irradiation, stress, time

117
Q

What temperature range thermal creep can happen?

A

T/T_m > 0.4 for steels (0.3 for Zry)

118
Q

What are primary, secondary and tertiary creep?

A

They are obtained in the uniaxial imposed stress test.
Primary: creep rate decrease with time since work hardening decrease with time
secondary: creep rate is constant, work hardening and recovery balances out
Tertiary: corresponds to necking and it happens only for T/Tmelt>0.5. Needs to be avoided

119
Q

Temperature effect on creep?

A

It gets faster, istantaneous strain increase due to E decrease and time to failure reduces

120
Q

Stress effect on creep?

A

same as temperature?

121
Q

What are the main charateristic of creep?

A

It’s
-time dependet
-permament
-isotropic
-isochoric

122
Q

What’s an Asbhy map?

A

A material dependet map that show most critical failure phenomena for each working condition, plotting normalized stress on normalized temperature

123
Q

Describe an Ashby map for creep!

A
  • At high T, low sigma I have lattice diffusion (Nabarro Herring zone). eps’_NH = f(D_vacancies)*sigma
  • lower T, low sigma I have grain boundary diffusion Coble creep zone). eps’_C = f(D_v, grain bound)*sigma
  • high sigma I have dislocation dynamics (superlinear or creep power law). eps’ = sigma^4 (or 5)
124
Q

Larson Miller Parameter?

A

LMP = T* (C + log(t_rupt) ) , C = 20 for almost all materials
It condenses time and temperature in a single parameter to describe rupture stresses
We can also use rupture strain instead of rupture time

125
Q

imposed strain creep test?

A

The imposed strain can be decomposed in elastic and creep component:
eps = sigma/E + eps_creep
eps’_creep = A * sigma
-> sigma = sigma0 * exp(-A * E * t)
Stress relaxes while a strain is generated. Stress relaxation isn’t always good

126
Q

is uniaxial creep conservative?

A

No because multidimensional creep is faster

127
Q

how is creep influenced by irradiation?

A

Irradiation can enhance:
* - The enhancement is due to the fact that creep require defects movements and irradiation creates defects
or can induce creep (T/Tmelt<0.3):
* - eps’_irr = const * phi_f * sigma
where phi_f is fast neutron flux

128
Q

How is creep verification made?

A

We need to check T, sigma and rupture time.
T and sigma depends on the coordinates (r,z) so we need to choose a worst case scenario and consider the average on r to avoid failure in the all section.
We use sigma_des, T_des , t_des design values representative and conservative but not too much. t_des is the design rupture time.
Then we use LMP entering 2 parmeters at a time and calculating the third one.
we calculate zi_i = i_rupt/i_des and each zi must be greater than 1

129
Q

Which is the most senstitve parmeter to creep failure to verify?

A

Temperature

130
Q

What is the Cumulative Damage Function?

A

The CDF is a cumulative function that goes from 0 to 1 and compute for each time step the accumulated damage from creep

131
Q

What kind of damage does radiation do to metals?

A
  • vacancies
  • interstitals
  • frenkel pair
  • dislocation
  • voids
  • heat deposition
  • impurities
132
Q

What happens when a fast neutron impinges on a metal?

A

It first: Primary Knock-on Atom (PKA). Then it generates a collisional casade.
In the path there is vacancy creation due to energy given to atoms
At the end there is thermal spike due to energy given to electrons

133
Q

What happens when a thermal neutron impinges on a metal?

A

It’s absorbed and generate gamma + impurity. The recoil from gamma of the impurity is:
momentum balance: AV = E_gamma/c
energy balance: E_rec = 1/2Av^2
-> E_rec = E_gamma^2/(2Ac^2) = 10 keV

134
Q

What happens after a collisional cascade?

A
  • Recombination of interstitials and vacancies
  • clustering of vacancies (void formation)
  • clustering of interstitials (dislocation formation)
135
Q

How to account for damage in steels in time during irradiation? Kinchin Pease model?

A

We use the fluence (int_time flux dt).
dpa = R/Nt where R is rate of displacements per volume and N atomic density
R = N int_energy(sigma_dflux dE)

According to kinchin pease model we have the hp:
- monoenergetic neutrons
- elastic scattering (sigma_d)
- monoatomic metal (A)
- nu(T) = T/(2E_d) (T is energy transferred)
- sigma_s (E,T) = A/4E * sigma_s(E)

dpa = sigma_s * E / (A * E_d) * fluence

136
Q

void swelling

A

in T = [0.3, 0.6]
fluence > soglia per non ricombinazione perchè troppi neutroni

The Grind (7 decks)

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