reactor physics tutorials 2013 2
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NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
Neutron Diffusion and Moderation
Simon Cöster
April 11, 2013
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
1 Fick’s Law
2 The Equation of Continuity
3 The Diffusion Equation
4 Solutions to The Diffusion Equation
5 The Group Diffusion Method
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
Fick’s Law
Diffusion theory is based on Fick’s Law
Solute will diffuse from high concentration to low
Fick’s Law
J = −D∇φ,
where D is the diffusion coefficient, φ is the neutron flux and Jis the neutron current density vector.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Equation of Continuity
Since neutrons do not disappear (β-decay neglected) thefollowing must be true for an arbitrary volume V .
[Rate of change in number of neutrons inV ] =
[rate of production of neutrons inV ]
− [rate of absorption of neutrons inV ]
− [rate of leakage of neutrons fromV ]
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Equation of Continuity
In mathematical terms the Equation of Continuity can beexpressed as
Neutron change rate =
∫V
∂n
∂tdV
Production rate =
∫VsdV
Absorption rate =
∫V
ΣaφdV
Leakage rate =
∫V∇JdV
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Equation of Continuity
This gives the general Equation of Continuity
The Equation of Continuity
∂n
∂t= s − Σaφ−∇J,
where n is the density of neutrons, s is the rate at whichneutrons are emitted from sources per cm3, Σa is themacroscopic absorption cross-section, J is the neutron currentdensity vector and φ is the neutron flux.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Diffusion Equation
Two unknowns; the neutron density n and the neutroncurrent density vector J.
Substitute Fick’s law into the equation
The Diffusion EquationGeneral:
D∇2φ− Σaφ+ s =∂n
∂t
Time-independent:
D∇2φ− Σaφ+ s = 0
or∇2φ− 1
L2φ+s
D= 0,
where L2 = DΣa
. L is called the diffusion length.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
Solutions to The Diffusion Equation
Infinite Planar Source
φ =SL
2De−|x |/L
Point Source
φ =S
4πDre−r/L
Bare Slab, width 2a (a = a + d is called extrapolatedboundary)
φ =SL
2Dsinh[(a− |x |)/L]
cosh(a/L)
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
Neutrons emitted with a continuous energy spectrum.
Divided into N energy intervals.
Averaged diffusion coefficients and cross-section.
The flux of neutrons in a group g is described by
φg =
∫gφ(E )dE ,
where φ(E ) is the energy-dependent neutron flux.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
The absorption rate in a specific group is given by
Absorption rate =
∫g
Σa(E )φ(E )dE
We can define the macroscopic group absorption cross-section,Σag , as
Σag =1φg
∫g
Σa(E )φ(E )dE
Then the absorption rate can be written as
Absorption rate = Σagφg
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
The rate at which neutrons transfers from group g to h is givenby
Transfer rate = Σg→hφg ,
where Σg→h is called the group transfer cross-section.
Total transfer rate out of g =N∑
h=g+1
Σg→hφg
Analogy, the rate at which neutrons transfers from group h intog is given by
Total transfer rate into g =
g−1∑h=1
Σh→gφh
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
This gives the steady-state diffusion equation for group g
The Diffusion Equation for Groups
Dg∇2φg − Σagφg −N∑
h=g+1
Σg→hφg +
g−1∑h=1
Σh→gφh + sg = 0
where the group-diffusion coefficient Dg is defined by
Dg =1φg
∫gD(E )φ(E )dE
These calculations are done by computers.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
At least two groups must be used to obtain reasonableresult
Thermal neutrons and fast neutrons
For a point source emitting S fast neutrons per second, theDiffusion Equation can be written (Σ1 = Σ1→2)
∇2φ1 −Σ1
D1φ1 = 0
Σa1 ≈ 0 above thermal energies.
Only two groups→ only Σ1→2 is non-zero in the third term
No thermal neutrons are scattered into the fast group.
NeutronDiffusion andModeration
SimonCöster
Outline
Fick’s Law
TheEquation ofContinuity
TheDiffusionEquation
Solutions toTheDiffusionEquation
The GroupDiffusionMethod
The Group Diffusion Method
For neutrons in the thermal group, the diffusion equation canbe written
∇2φT −1L2T
φT =Σ1φ1
D
Necessary to solve for fast neutrons first
φ1 =Se−r/
√τT
4πD1r.
Then
φT =SL2
T
4πD(L2T − τT )
(e−r/LT − e−r/√τT ),
where τT = D1Σ1
and L2T = D
Σa
One-group reactor equation
Time-dependent diffusion equation
D∇2φ− Σaφ+ s = −1v∂φ
∂t(1)
where D and Σa are the one-group diffusion coefficient andmacroscopic absorption cross-section for fuel-coolant mixture.
s = νΣf φ (2)
If the source term is to balance the leak and absorption in (1), weget
D∇2φ− Σaφ+1kνΣf φ = 0 (3)
Ola Håkansson Reactor theory
One-group reactor equation
By letting the Buckling B be defined by
B2 =1D
(νk
Σf − Σa
)(4)
we get the one-group reactor equation
−DB2φ− Σaφ+1kνΣf φ = 0 (5)
or∇2φ+ B2φ = 0 (6)
From (5), we have the multiplication factor k
k =νΣf
DB2 + Σa(7)
Ola Håkansson Reactor theory
One-group reactor equation
Source term for the one-group equation
s = ηΣaFφ (8)
where η is the average number of fission neutrons emitted perabsorbed neutron in the fuel and ΣaF is the cross-section for thefuel.This can be written as
s = ηf Σaφ (9)
wheref =
ΣaF
Σa(10)
Ola Håkansson Reactor theory
One-group reactor equation
For an infinite reactor all neutrons are absorbed, the multiplicationfactor k∞ is
k∞ =ηf Σaφ
Σaφ= ηf (11)
and the source term can now be written as
s = k∞Σaφ (12)
and we now have
−DB2φ− Σaφ+k∞k
Σaφ = −1v∂φ
∂t(?) (13)
For a critical reactor (k = 1), we get
B2 =k∞ − 1
L2 , L2 =DΣa
(14)
Ola Håkansson Reactor theory
The slab reactor
For a critical, infinite bare slab of thickness a the reactor equation is
d2φ
dx2 + B2φ = 0 (15)
Boundary conditions: φ vanishes at x = a/2 and at x = −a/2where a = a + 2d . Note symmetry and dφ
dx = 0|x=0.General solution to (15) is
φ(x) = A cosBx + C sinBx (16)
which reduces toφ(x) = A cosBx (17)
when making use of the condition on the derivative.
Ola Håkansson Reactor theory
The slab reactor
The boundary conditions now gives
φ
(a2
)= φ
(−a2
)= A cos
Ba2
= 0 (18)
For the non-trivial solution, B can take any of the values
Bn =πna
(19)
The flux in the critical reactor then is
φ(x) = A cosπxa
(20)
Ola Håkansson Reactor theory
The slab reactor
A can be found by calculating the power of the reactor P .
P = ERΣf
∫ a/2
−a/2φ(x)dx (21)
where ER is the recoverable energy per fission and Σf φ(x) are thenumber of fissions at the point x . Introducing
φ(x) = A cosπna
(22)
and solve (21) for A, we get
A =πP
2aERΣf sin πa2a
(23)
Ola Håkansson Reactor theory
The spherical reactor
Critical, spherical reactor with radius R - The flux only depends onr . The reactor equation is
1r2
ddr
r2 dφdr
+ B2φ = 0 (24)
with the boundary condition φ(R) = 0 as well as the flux must befinite.Solution to (24) is given by
φ = AsinBr
r+ C
cosBrr
(25)
and reduces toφ = A
sinBrr
(26)
since the flux must be finite when r = 0.
Ola Håkansson Reactor theory
The spherical reactor
As earlier, introducing the boundary conditions and calculating thereactor power yields the flux
φ =P
4ERΣf R2sinπr/R
r(27)
Ola Håkansson Reactor theory
The infinite cylinder reactor
Critical, cylindrical reactor with radius R
1r
ddr
rdφdr
+ B2φ = 0 (28)
ord2φ
dr2 +1r
dφdr
+ B2φ = 0 (29)
This is a special case of Bessel’s equation
d2φ
dr2 +1r
dφdr
+
(B2 − m2
r2
)= 0 (30)
where m = 0.
Ola Håkansson Reactor theory
The infinite cylinder reactor
The solution can thus be written as
φ = AJ0(Br) + CY0(Br) (31)
Since Y0 is not finite at the origin, C = 0 and
φ = AJ0(Br) (32)
The boundary conditions specify that
φ(R) = AJ0(BR) = 0 (33)
so thatB =
xn
R(34)
where xn is the values where J0(x) is zero.
Ola Håkansson Reactor theory
The infinite cylinder reactor
For the critical reactor, the flux can now be written as
φ = AJ0
(2.405r
R
)(35)
A is calculated from the reactor power, resulting in
φ =0.738PERΣf R2 J0
(2.405r
R
)(36)
Ola Håkansson Reactor theory
The finite cylinder reactor
Finite cylindrical reactor with height H and radius R . The flux heredepends on the distance r from the axis and the distance z fromthe midpoint of the cylinder. The reactor equation takes the form
1r∂
∂rr∂φ
∂r+∂2φ
∂z2 + B2φ = 0 (37)
The boundary conditions in this case are
φ(R, z) = φ(r , H/2) = 0 (38)
Ola Håkansson Reactor theory
The finite cylinder reactor
Assuming the solution can be obtained by separation of variables
φ(r , z) = R(r)Z (z) (39)
we now get1R1r∂
∂rr∂R∂r
+1Z∂2Z∂z2 = −B2 (40)
This implies that the first and second term of (40) must beconstants. This gives that
d2Rdr2 +
1r
dRdr
+ B2r R = 0,
d2Zdz2 + B2
z = 0 (41)
where B2r + B2
z = B2. Both of the equations in (41) have beensolved earlier.
Ola Håkansson Reactor theory
Maximum-to-average flux and power
The ratio between φmax and φaverage , Ω, is in some cases ofinterest. φmax , in a uniform bare reactor is always found at thecenter pf the reactor. In the case of a bare spherical reactor, themaximum flux is obtained from the limit
φmax =P
4ERΣf R2 limr→0
sin (πr/R)
r=
πP4ERΣf R3 (42)
The average flux is given by
φaverage =1V
∫φdV (43)
Ola Håkansson Reactor theory
Maximum-to-average flux and power
Making use of
P = ERΣf
∫φdV (44)
we can write φaverage as
φaverage =P
ERΣf V(45)
and Ω now becomes
Ω =φmax
φaverage=π2
3≈ 3.29 (46)
Ola Håkansson Reactor theory
The one-group critical equation
The equationk∞
1 + B21L2 = 1 (47)
determines the conditions under which a bare reactor is critical.
Ola Håkansson Reactor theory
Thermal reactors
an infinite reactor composed of a homogeneous fuel-moderatormixture. Σa is the macroscopic cross-section of the mixture so that
Σa = ΣaF + ΣaM (48)
Letting
f =ΣaF
Σa(49)
it is clear that f ΣaφT neutrons are absorbed per cm3/sec in thefuel. If ηT is the average number of neutrons emitted per thermalneutron absorbed in the fuel, ηT f ΣaφT neutrons are emitted percm3/sec.
Ola Håkansson Reactor theory
Thermal reactors
The multiplication factor of the reactor is given by the four-factorformula
k∞ = ηT fpε (50)
where ε is defined as the ratio of the total number of fissionneutron produces by both fast and thermal fission to the numberproduced by only thermal fission and p is the probability that afission neutron is not absorbed at any other energies than thermal.
Ola Håkansson Reactor theory
Thermal reactors, criticality calculation
Two-group calculation with fast and thermal neutrons.ηT f εΣaφT = (k∞/p)ΣaφT neutrons are emitted to the fast groupand Σ1φ1 are scattered out of the group. The diffusion equation forthe fast group is
D1∇2φ1 − Σ1φ1 +k∞p
ΣaφT = 0 (51)
With pΣ1φ1 neutrons entering the thermal group (i.e the source)the diffusion equation for the thermal group is
D∇2φT − ΣaφT + pΣ1φ1 = 0 (52)
Ola Håkansson Reactor theory
Thermal reactors
Moreover, the two fluxes may be written as
φ1 = A1φ, φT = A2φ (53)
Now we get the two equations
−(D1B2 + Σ1)A1 +k∞p
ΣaA2 = 0 (54)
pΣ1A1 − (DB2 + Σa)A2 = 0 (55)
If these equations are to have non-trivial solutions, the determinantis 0.
Ola Håkansson Reactor theory
Thermal reactors
Calculating the determinant, and setting it to 0, one gets themultiplication factor (in this case k = 1)
k∞(1 + B2L2
T )(1 + B2τT )= 1 (56)
where
L2T =
DΣa, τT =
D1
Σ1(57)
Ola Håkansson Reactor theory
Reflected reactors
For a spherical reactor with a core and infinite reflector, there aretwo reactor equations - One for the core and one for the reflector.In this case,
∇2φc + B2φc = 0 (58)
and∇2φr −
1L2
rφc = 0 (59)
These must be solved and satisfy continuity of the neutron flux atthe boundary between the core and reflector (quite lengthycalculations).
Ola Håkansson Reactor theory
Multigroup calculation
One-group method is a rough estimate. More accurate results areobtained by multigroup calculations.
Σfg is the group-averaged macroscopic fission cross-sectionνg is the average number of fission neutron from fissioninduced by group gXg is the fraction of fission neutrons emitted with energies inthe group g
The multigroup equation for group g is then
Dg∇2φg−Σagφg−N∑
h=g+1
Σg→hφg +
g−1∑h=1
Σh→gφh+Xg
N∑h=1
νhΣfhφh = 0
(60).
Ola Håkansson Reactor theory
Reactor Physics tutorial
Heterogeneous reactors
Quasi-homogeneous vs. heterogeneous reactors
Quasi-homogeneous vs. heterogeneous reactors
I Most reactors are non-homogeneous: fuel (rods), coolant,moderator (if thermal reactor) are separated
I Even such a reactor may be considered to bequasi-homogeneous
I Mean free path λ larger than fuel rod dimensions at all En
I > 1 collision in fuel rod unlikely
I If λ . fuel rod dimensions at some energy: multiple collisionsprobable ⇒ Heterogeneous reactor
Examples
I Highly enriched fuel ⇒ thin fuel rods ⇒ quasi-homogeneous
I Slightly enriched fuel ⇒ thicker fuel rods ⇒ heterogeneous
Reactor Physics tutorial
Heterogeneous reactors
Heterogeneous reactor parameters
Heterogeneous reactor parameters
ηTI Average number of fission neutrons produced per neutron
absorbed by fuel (thermal neutrons)
I Example fuel rod contents: 235U, 238U, 16OI Average number of fission neutrons produced: νΣf
I Σf ,238 = 0 at thermal energies
I Absorption cross section for 16O ≈ 0
ηT =νf ,235Σf ,235
Σa,235 + Σa,238
Reactor Physics tutorial
Heterogeneous reactors
Heterogeneous reactor parameters
Heterogeneous reactor parameters
f - Thermal utilizationI Probability that neutron absorbed in core is absorbed in the
fuelI Number of neutrons absorbed in volume (fuel/moderator) per
second: ∫V
ΣaφTdV = ΣaφTV
f =ΣaFVF
ΣaFVF + ΣaMVMζ
I ζ = φTMφTF
: thermal disadvantage factor. Generally, ζ > 1 in
heterogeneous reactorI f is calculated numerically. Analytical solutions (Wigner-Seitz
method) only rough approximation in most cases
Reactor Physics tutorial
Heterogeneous reactors
Heterogeneous reactor parameters
Heterogeneous reactor parameters
k∞ - Multiplication factor in infinite reactor
I Four-factor formula: k∞ = ηT fpε
I Thermal utilization: fhetero < fhomo
I Resonance escape probability: phetero > phomo . Increases morethan f decreases ⇒ (fp)hetero > (fp)homo
I Fast fission factor: εhetero > εhomo
k∞|hetero > k∞|homo
Homogeneous reactor containing natural uranium and graphite:k∞ ≤ 0.85⇒ non-critical. Rods of same fuel (heterogeneousreactor) ⇒ critical reactor possible.
Reactor Physics tutorial
Classification of time problems
Classification of time problems
Time-dependent neutron populationI Short Time Problems (seconds - tens of minutes)
I Reactor conditions altered ⇒ change in k
I Intermediate Time Problems (hours - 1 or 2 days)I Radioactive decay of fission products ⇒ change in
concentrationI Fission product concentration affects absorption term
I Long Time Problems (days - months)I Variation of neutron flux over long periods.I Assume system in series of stationary states. Solve diffusion
equation for each configuration:
D∇2φ− Σaφ = λνΣf φ
I Change design parameters (buckling, absorption/fissioncross-sections) so that λ = 1
Reactor Physics tutorial
Reactor kinetics
Prompt Neutron Lifetime
Prompt Neutron Lifetime
I Produced directly at fission
I Average time between emission and absorption of promptneutron: lp (prompt neutron lifetime)
I Average time spent as thermal neutron before absorption: td(mean diffusion time)
I For infinite thermal reactor: lp ' td
td =
√π
2vT (ΣaF + ΣaM)
I For thermal reactor: lp ' 1 · 10−4 s
I For fast reactor: lp ' 1 · 10−7 s
Reactor Physics tutorial
Reactor kinetics
Reactor with No Delayed Neutrons
Reactor with No Delayed Neutrons
I 100% of neutrons are prompt neutrons
I Infinite thermal reactor
I Number of fissions at time t, NF (t):
NF (t) = NF (0) exp( t
T
)I Reactor period T :
T =lp
k∞ − 1
I lp = 1 · 10−4 s ⇒ T = 0.1 s ⇒ Power increase by factor22000 after 1 second. Delayed neutrons needed!
Reactor Physics tutorial
Reactor kinetics
Reactor with Delayed Neutrons
Reactor with Delayed Neutrons
I Simplification: single delayed-neutron precursor (in reality: 6)
I Diffusion equation for homogeneous reactor:
sT
Σa
− φT = lpdφTdt
I Pure prompt-neutron source term: sT = k∞ΣaφTI If fraction β are delayed, this becomes
sT |prompt = (1− β)k∞ΣaφTI Delayed-neutron source term depends on resonance escape
probability p, precursor decay constant λ and precursorconcentration C : sT |delayed = pλC
Reactor Physics tutorial
Reactor kinetics
Reactor with Delayed Neutrons
Reactor with Delayed Neutrons
Two coupled differential equations:
I Thermal neutron flux φT
(1− β)k∞φT +pλC
Σa
− φT = lpdφTdt
I Precursor concentration CdC
dt=βk∞ΣaφT
p− λC
I Assume solutions of forms
φ = A exp(ωt) C = C0 exp(ωt)
Reactor Physics tutorial
Reactor kinetics
Reactor with Delayed Neutrons
Reactor with Delayed Neutrons
Solution for the flux:
φT = A1 exp(ω1t) + A2 exp(ω2t)
I Define reactivity ρ = k−1k
I k > 1⇒ ρ > 0
I k < 1⇒ ρ < 0
I k = 1⇒ ρ = 0
I ρ depends on ω ⇒ evolution of flux for specific ρ
I In general, φT → exp(ω1t)⇒ φT → exp(tT
)I Example reactor period with delayed neutrons: 57 s (0.1 s
without)
Reactor Physics tutorial
Reactor kinetics
The Prompt Critical State
The Prompt Critical State
I If (1− β)k = 1, the prompt neutrons are enough to make thereactor critical
I Corresponding reactivity:
ρ =k − 1
k=
11−β − 1
11−β
= β
I Short periods when prompt critical ⇒ restrict additions toreactivity to < β
TABLE OF CONTENTS 1. Thermodynamic Considerations 2. Heat Generation in Reactors 3. Fission Product Decay Heating 4. Heat Flow by Conduction 5. Fuel Elements 6. Heat Transfer to Coolants 7. Boiling Heat Transfer
5/17/2013 Sebastian Thor
2
THERMODYNAMIC CONSIDERATIONS No change in phase of the coolant Temperature increases, pressure invariant The rate of heat absorbed in the coolant is given by:
𝑞 = 𝑤 𝑐𝑝 𝑇 𝑑𝑇𝑇𝑜𝑜𝑜
𝑇𝑖𝑛
The enthalpy:
ℎ = 𝑢 + 𝑃𝑃
ℎ𝑜𝑜𝑜 = ℎ𝑖𝑖 + 𝑐𝑝 𝑇 𝑑𝑇𝑇𝑜𝑜𝑜
𝑇𝑖𝑛
Change in phase of the coolant Up to the saturation temperature it acts the same:
ℎ𝑓 = ℎ𝑖𝑖 + 𝑐𝑝 𝑇 𝑑𝑇𝑇𝑠𝑠𝑜
𝑇𝑖𝑛
Once saturation temperature is achieved, the coolant has to absorb an amount of heat equal to the heat of vaporization ℎ𝑓𝑓per unit mass to change phase.
ℎ𝑜𝑜𝑜 = ℎ𝑖𝑖 + 𝑐𝑝 𝑇 𝑑𝑇𝑇𝑠𝑠𝑜
𝑇𝑖𝑛+ ℎ𝑓𝑓
5/17/2013 Sebastian Thor
3
HEAT GENERATION IN REACTORS Fission fragment, 𝛽-ray and about 1/3 of the 𝛾-ray energy is absorbed in the fuel. This is about 90% of the recoverable fission energy. The rate of heat production per unit volume at the point 𝒓 is given by:
𝑞′′′ 𝒓 = 𝐸𝑑 Σ𝑓𝑓 𝐸 𝜙 𝒓,𝐸 𝑑𝐸∞
0
For the thermal reactor this reduces to: 𝑞′′′ 𝒓 = 𝐸𝑑Σ𝑓𝑓𝜙𝑇(𝒓)
Where 𝐸𝑑 is the energy deposited locally in the fuel per fission, Σ𝑓𝑓 is the thermal cross-section of the fuel and 𝜙𝑇(𝒓) is the thermal flux. Derivations and assumptions then leads to
𝐸𝑞 (8.12) No significant errors when used in heat transfer calculations.
5/17/2013 Sebastian Thor
4
FISSION PRODUCT DECAY HEATING After a few days of reactor operation, the fission products accumulates and together stand for about 7% of the total thermal power output through 𝛽 and 𝛾 decays. This is something that has to be dealed with in the event of a shut down. If not, the temperature may rise to a point where the integrity of the fuel might be compromised. (Fukushima).
5/17/2013 Sebastian Thor
5
HEAT FLOW BY CONDUCTION Fourier’s law
𝒒′′ = −𝑘 𝛻𝛻 Steady-state equation of conductivity
𝛻 ⋅ 𝒒′′ − 𝑞′′′ = 0 Steady-state heat conduction equation:
𝛻2𝑇 +𝑞′′′
𝑘 = 0
Where no heat sources exist (i.e. 𝑞′′′ = 0); Laplace’s equation: 𝛻2𝑇 = 0
These equations are then for example used to calculate how the heat transferes from a fuel rod to a coolant.
5/17/2013 Sebastian Thor
6
FUEL ELEMENTS Plate-type fuel In the fuel:
𝑞 =𝑇𝑚 − 𝑇𝑠𝑎 2𝑘𝑓𝐴⁄
cf. 𝐼 =
𝑉𝑅
With the cladding:
𝑇 = 𝑇𝑠 −𝑥 − 𝑎𝑏
𝑇𝑠 − 𝑇𝑐 Using Fourier’s law:
𝑞 =𝑇𝑚 − 𝑇𝑐𝑎
2𝑘𝑓𝐴+ 𝑏𝑘𝑐𝐴
=𝑇𝑚 − 𝑇𝑐𝑅𝑓 + 𝑅𝑐
This shows that the thermal resistances behaves like two electrical resistors in series. The last part also applies for cylindrical fuel, however 𝑅𝑓 and 𝑅𝑐 are calculated differently.
5/17/2013 Sebastian Thor
7
HEAT TRANSFER TO COOLANTS Continues along the lines of the previous slide.
𝑞 =𝑇𝑐 − 𝑇𝑏1 ℎ𝐴⁄
𝑅ℎ =1ℎ𝐴
𝑇𝑐 is the bulk temperature of the coolant, 𝑅ℎ is the thermal resistance for convective heat transfer, h is the heat transfer coefficient, which depends on many factors such as the coolant temperature and the manner in which it flows by the heated surface. A is the area of contact. Coolant channels
𝑇𝑏 = 𝑇𝑏0 +𝑞𝑚𝑚𝑚′′′ 𝑉𝑓𝜋𝑤𝑐𝑝
1 + sin𝜋𝜋𝐻
𝑇𝑏,𝑚𝑚𝑚 = 𝑇𝑏0 +2𝑞𝑚𝑚𝑚
′′′ 𝑉𝑓𝜋𝑤𝑐𝑝
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8
BOILING HEAT TRANSFER Up to this point it has been assumed that the coolant does not change phase. However there are some advantages to permitting the coolant to boil. The fact that one does not need a heat transfer system between the reactor coolant and the turbines for one, and also lower pressure in the reactor. Boiling regimes No boiling: Temperature rises. Nothing significant happens Local boiling: Bubbles form but quickly transfer their heat to the surrounding liquid coolant Bulk boiling: Bubbles persists. Bubbly flow leads to anular flow. Boiling Crisis Partial film boiling: The sides of the coolant channels gets covered with a thin layer of gas. The gas has higher thermal resistance, heat conduction is reduced. Full film boiling: Even though the heat conduction is reduced, the fuel is still going now becoming hotter and hotter due to decreased cooling…
5/17/2013 Sebastian Thor
9
Table of contents
• History
• Reactor licensing
• Nuclear reactor safety principles
• Radiation release
• Data from NPP
History
• First legislation related to nuclear power 1946 McMahon
Act
• In 1974 –Nuclear regulatory Comission (NCR) was
created to manage licensing and regulation of nuclear
power plants.
• DOE – Department of energy, takes responsibility to
sposor recearch and development of Nuclear Energy.
Licensing
• NRC regulates everything from reactor project approval to
fuel transport licensing and disposal of radioactive waste.
• Although all nuclear power plants have to receive from
other institutions as well. (Like coal or gas plants).
• It is more than 40 licensing actions and may take more
than two years.
Licensing
NRC groups:
– Regulatory staff
»Building, regulation of normal working, fuel regulation
etc.
– ACRS (Advisor committee on reactor safeguards)
»Reviews reactor licensing and predicts potential
hazards
– ASLB (Atomic safety and licensing boards)
»Grants, revokes or suspends license of object. At least
two technical members.
Licensing
• Stages
– Construction permit
» Informal Site review
»Application of license
– Includes financial information, technical information,
preliminary safety analysis, Environmental report.
»Submission of AER
»Review of regulatory staff
Licensing
»Review by ACRS
»Public hearings
– Against Atomic safety and Licensing board which
decides if application should be approved.
»Appeals
Licensing
• Operation license
– Submittal for Operating License
– Review by Regulatory staff
»Determine new information after the CP and its impact
– Review by ACRS
– Hearings
– Appeals
Nuclear power plant safety principles
• Three main contamination paths
– Operation
– Refueling
– Shipping of fuel
Nuclear power plant safety principles
• Multiple barriers
– Fuel
– Cladding
– Closed coolant system
– Pressure vessel
– Containment
Nuclear power plant safety principles
• Three levels of safety
• First:
– Accident prevention by safe design, construction and
surveillance.
»Negative void and temperature coefficients.
»Only known property materials should be used.
»Sufficient instrumentation so that operators should have
information at all times.
»High quality construction.
»Continual monitoring of plant.
Nuclear power plant safety principles
• Second level of safety:
– Objective is to protect operators and public from
radiation damage.
»Emergency core cooling system
»Fast shut down ability without control rod insertion
» Independent sources of power from Nuclear power
plant for instrumentation.
• Third level of safety
– Margin of safety for very unlikely events
Radiation release
• Dose sources
– External radiation from emitted plume
– Internal dose from radionuclide inhalation
– External dose from radionuclide deposited on the
ground
– External dose from radionuclide deposited on clothes
and body
– Direct dose from power plant.
Radiation release
• Gamma from released plume
– It is taken that plume is infinitely large – gives
conservative values and simplifies calculation.
– For more than one gamma ray:
– Dose rate:
Radiation release
• β dose:
– Treatment is similar as gamma ray case.
»Surface dose estimation
» Internal dose estimation
Radiation release
• Internal dose
– Function of breathing activity
– Steady state equilibrium equation for dose rate
Radiation release
• Dose from Ground-deposited nuclides
– 80 % of dose form meltdown would be from Cs137
• Release from nuclear power plant
• Population dose:
– Defined by person-rems
Data from NPP
Product Activity Average
Lithuania
Bq/kg
Vicinity of NPP
50 km diameter
Bq/kg
Milk 90Sr 137Cs
alfa
beta
0,02±0,01
0,03±0,01
0,25±0,05
50±1
0,03±0,01
0,04±0,02
0,14±0,06
49±4
Meat 90Sr 137Cs
alfa
beta
0,03±0,02
0,14±0,18
0,39±0,29
117±6
0,03±0,02
0,09±0,03
0,57±0,29
117±3
Cabbage 90Sr 137Cs
alfa
beta
0,06±0,02
0,04±0,01
0,46±0,32
71±6
0,05±0,03
0,07±0,08
0,33±0,23
62±3
Data from NPP
• Average dose to NPP workers in Sweden in year 2010
1.7 mSv per year.
• Maximal dose in 2010 - 16.9 mSv.
• Doses are ~50 % higher in BWR reactors in Sweden.
Nuclide Coal, Lodz
power station
238U 1.1 GBq/year
210Pb 1.2 GBq/year
Data from NPP
• 131mXe, 133mXe, 135Xe – up to 96 % of released
radioactivity.
• 2790 GBq/a from Xenon
• During Fukushima accident 19.0 ± 3.4 Ebq of
Xenon.
References
• www.RSC.lt
• Lamarsh, Introduction to unclear engineering
• Walinder Robert, Radiation doses to Swedish nuclear
workers and cancer incidence in a NPP
• Martin B. Kalinowski, Matthias P. Tuma, Global
radioxenon emission inventory based on nuclear power
reactor reports, Journal of Environmental Radioactivity,
Volume 100, Issue 1, January 2009,
• Andreas Stohl, Petra Seibert, Gerhard Wotawa, The total
release of xenon-133 from the Fukushima Dai-ichi
nuclear power plant accident, Journal of Environmental
Radioactivity, Volume 112, October 2012
Diffusion of Effluents
• Mainly turbulent diffusion
• Spreads out in gaussian
distribution
• Standard deviation:
𝜎𝑦 =2𝑥𝐾𝑦
𝑣
1/2
𝜎𝑧 =2𝑥𝐾𝑧𝑣
1/2
Concentration of effluents
χ =𝑄
2𝜋𝑣𝜎𝑦𝜎𝑧𝑒𝑥𝑝 −
𝑦2
2𝜎𝑦+
𝑧 + ℎ 2
2𝜎𝑧+ 𝑒𝑥𝑝 −
𝑦2
2𝜎𝑦+
𝑧 − ℎ 2
2𝜎𝑧
χ =𝑄
2𝜋𝑣𝜎𝑦𝜎𝑧𝑒𝑥𝑝 −
ℎ2
2𝜎𝑧
h=0 => released at ground level, use if
do not know emission altitude
z=0 => at ground level
y=0 => at centerline, use if
know emission altitude
Deposition and radioactive decay
Depositionrate: 𝑅𝑑 = χ𝑣𝑑 Ci/m2/s
Radioactive decay: 𝑄 = 𝑄0exp (𝑡𝜆)=𝑄0exp (𝑥
𝑣𝜆)
χ =𝑄
2𝜋𝑣𝜎𝑦𝜎𝑧𝑒𝑥𝑝 −
ℎ2
2𝜎𝑧
χ =𝑄
2𝜋𝑣𝜎𝑦𝜎𝑧𝑒𝑥𝑝 − 𝚲 +
𝝀
𝒗 𝒙 −
ℎ2
2𝜎𝑧 𝚲 =
𝒗𝒅
𝒗𝒛
The location of a nuclear reactor has an obvious bearing on the consequences of a reactor accident to the public
construction permit from the NRC (regulations regarding reactor site criteria) -without undue risk to the health and safety of the public -minimal effect on the environment
The NRC evaluation considerations
Reactor itself, its design characteristics, and its proposed mode of operation. Population Considerations the physical characteristics of the site :seismology, meteorology, geology, and hydrology of the area the use of appropriate engineering safeguards
Population Considerations
the NRC has defined two areas in the vicinity of the reactor
An exclusion area, or exclusion zone: is that area surrounding the reactor in which the reactor licensee has the authority to determine all activities including exclusion or removal of personnel and property from the area
A low-population zone (LPZ) is "the area immediately surrounding the exclusion area which contains residents, the total number and density of which are such that there is a reasonable probability that appropriate protective measures could be taken in their behalf in the event of a serious accident
the NRC also defines the population center distance. "the distance from the reactor to the nearest boundary of a densely populated center containing more than 25,000 residents."
total radiation dose to the whole body in excess of 25 rem the population center distance be no less than 1 .33 times the radius of the LPZ.
The assumptions that the NRC makes in calculating the radii of the exclusion area and the LPZ , are used to compute the external and internal dose from the effluent cloud and the direct dose from nuclides
Population Considerations
The amount of a fission product available for release to the atmosphere can be estimated by
where Fp is the fraction of the radionuclide released from the fuel into the reactor containment and Fb is the fraction of this that remains airborne and capable of escaping from the building.
If the cumulative yield of the fission product is Yi atoms per fission, the rate of production of this nuclide is
To begin the computation
rate of production = P Yi atoms/sec.
Reactor power(MW)
Population Considerations
Physical Characteristics of Site
Nuclear power plants must be designed and constructed in such a manner that all structures and systems important to safety can withstand the effects of earthquakes, tornadoes, hurricanes, floods, and other natural phenomena, without a loss of safety function
Seismology: Geologists now believe that the surface of the earth is composed of large structures called tectonic plates.
Figure 1 1 .19 The earth's tectonic plates and earthquake belts (From C. Kissinger, "Earthquake Prediction," Physics Today, March, 1 974.)
the centers of 42,000 earthquakes
To safety-related structures of reactor plant from: hurricanes and tornadoes
Meteorology
Limitations •Hurricanes: up to 600 miles in diameter, with winds from 75 to 200 mi/hr •Tornadoes, Their diameters range from several feet to a mile
Geology :Studies must be made of the geological structure of a proposed site in order to determine whether the area can family support the reactor building with all its internal components.
Hydrology It is necessary to prevent large quantities of water from entering the site of a nuclear power plant, since water could compromise some of the safety-related systems of the plant.
the hydrological phenomena : depends upon the nature and location of the site
Physical Characteristics of Site
Loss-of-Coolant Accident
coolant flow through a reactor core ---- caused by leak in a small coolant pipe -to serious consequences for the plant as a whole -the pressure in the reactor vessel quickly drops to the saturation
pressure -change in the average water temperature
Three Mile Island Accident: The accident at the Three Mile Island nuclear power station (TMI) near Harrisburg, Pennsylvania, in March 1979 is one of the worst that has occurred in a commercial nuclear power plant.
During maintenance operations, the feedwater flow to the steam generator was lost, an event that can be expected to happen two or three times a year in a plant. Because of the sudden loss of heat removal, pressure began to increase in the primary system
control: emergency core cooling system (ECCS): when the pressure has dropped below about 650 psi
The accident at Three Mile Island did seriously damage the core, but did not result in a large release of radioactivity to the atmosphere
The Chernobyl Accident
Chernobyl Nuclear Power Plant
During the shutdown process, the reactor was in an extremely unstable condition. A peculiarity of the design of the control rods caused a dramatic power surge as they were inserted into the reactor The interaction of very hot fuel with the cooling water led to fuel fragmentation along with rapid steam production and an increase in pressure. Where a low power level with an unfavorable power distribution, a high coolant flow rate in the core, a reduced feedwater flow rate to the reactor with increasing coolant temperature at the core inlet, and an unstable xenon spatial distribution
Ukranian City of Kiev April 26, 1 986 The Chernobyl reactor was a graphite moderated boiling water pressure tube reactor of the RBMK
BWR: Steam Pipe Break: The steam in a BWR plant is somewhat radioactive, since it is produced directly in the reactor
In analyzing this accident ( 1 ) the isolation valves close in the maximum time characteristic of the valves (2) all of the coolant in the broken steam line and its connecting lines at the time of the break, plus the
steam passing through the valves prior to closure, is released; (3) the activity (including all the iodine and noble gases that may be present in the steam from leaking fuel
rods) is released to the atmosphere within 2 hrs, at a height of 30 feet, under fumigation conditions.
BWR: Rod Drop : The control rods in a BWR enter from the bottom of the core and are inserted upwards. A number of failures in the control rod drive system: to the release of some activity into the containment.
PWR: Rod Ejection failure of the control rod housing could occur in such a way that high-pressure reactor coolant water might forcibly eject a cluster control rod assembly. -power transient similar to that in a BWR rod drop accident
The Meaning of Risk as the consequence of the event per unit time
the average individual risk is defined as
The risk of an event can be computed in an obvious way from the frequency of the event and the magnitude of the consequences of the event:
However, the public acceptability of a given risk depends not only on the size of the risk, but also on the magnitude of the consequences of the event.
The calculation of the risk associated with accidents in a nuclear power plant is a three-step process: 1- determine the probabilities of the various releases of radioactivity resulting from accidents 2- the consequences to the public of these releases must be evaluated 3- the release probabilities and their consequences are combined to obtain the overall risk.
Risk Determination
event trees :the identification of the accident sequences leading to various releases
The effluent released to the environments: gaseous or liquid form the origin, amount, and composition of this effluent varies from plant to plant,
The NRC has translated its "as low as reasonably achievable
Regulation of Effluents
Doses from Effluents The gaseous effluents emitted to the atmosphere and liquid wastes discharged to bodies of water, and these two cases will be considered separately.
:noble gases and the isotopes of iodine 131I Gaseous Effluents
radiation dose from ingested food
Where Vd is a proportionality constant, has units of 0,01m/sec and is called the deposition velocity, Rd has units of Ci/m2-sec and X is in Ci/m3 Because the emission of radioactive effluent is often stated in Ci/yr
Where Q’ is in Ci/yr, (X / Q') is the dilution factor in sec/m3
Once the Iodine has fallen on the foliage ?
*When the rates of production and decay are equal
*the annual dose rate is
*the Iodine concentration in sample
Liquid Effluents There are several pathways by which man may become exposed to the radioactive waste discharged into bodies of water
the proportionality constant CF is usually called the concentration factor and sometimes the bioaccumulation factor.
The calculation of the radiation dose from contaminated seafood?
1)-the concentration of the radionuclides discharged from the plant is estimated from the discharge rate and dispersion characteristics of the receiving body of water.
2)-the concentration of the radionuclides in seafood is computed
3)-the consumption rate of seafood from waters near the power plant must be estimated
4)-the dose rate can be found by comparing the activity of the seafood Cs in µCi/cm3 and its consumption rate Rs in cm3/day with the dose rate
the dose rate received from the seafood: