Reactor Physics Tutorials 2013 2

NeutronDiffusion andModeration

SimonCöster

Outline

Fick’s Law

TheEquation ofContinuity

TheDiffusionEquation

Solutions toTheDiffusionEquation

The GroupDiffusionMethod

Neutron Diffusion and Moderation

Simon Cöster

April 11, 2013


SimonCöster

Outline

Fick’s Law





1 Fick’s Law

2 The Equation of Continuity

3 The Diffusion Equation

4 Solutions to The Diffusion Equation

5 The Group Diffusion Method


SimonCöster

Outline

Fick’s Law





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.


SimonCöster

Outline

Fick’s Law





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 ]


SimonCöster

Outline

Fick’s Law






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


SimonCöster

Outline

Fick’s Law






This gives the general 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.


SimonCöster

Outline

Fick’s Law





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.


SimonCöster

Outline

Fick’s Law





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)


SimonCöster

Outline

Fick’s Law





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.


SimonCöster

Outline

Fick’s Law






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


SimonCöster

Outline

Fick’s Law






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


SimonCöster

Outline

Fick’s Law






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.


SimonCöster

Outline

Fick’s Law






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.


SimonCöster

Outline

Fick’s Law






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

Reactor theory

Ola Håkansson

April 15, 2013

Ola Håkansson Reactor theory

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)



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)



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)



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)


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.


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)


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)


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.


The spherical reactor

As earlier, introducing the boundary conditions and calculating thereactor power yields the flux

φ =P

4ERΣf R2sinπr/R

r(27)


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.



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.



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)


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)


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.


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)


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)


The one-group critical equation

The equationk∞

1 + B21L2 = 1 (47)

determines the conditions under which a bare reactor is critical.


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.


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.


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)


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.


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)


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).


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).


Reactor Physics tutorial


Markus Preston

April 22, 2013


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



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





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





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.


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 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 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 kinetics

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 kinetics



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 kinetics



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 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

HEAT REMOVAL FROM NUCLEAR REACTORS Sebastian Thor

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


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.


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

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.


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.


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


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…


9

Nuclear reactor licensing and

regulation BENJAMINAS MARCINKEVICIUS

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

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


• Multiple barriers

– Fuel

– Cladding

– Closed coolant system

– Pressure vessel

– Containment


Containment. Left –PWR, Right BWR. [Lamarsh]


• 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.


• 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

Milk

Meat

Fish

Veggies

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

http://www.rsc.lt/

Dispersion of Effluents Reactor physics 2013

SANDRA ANDERSSON

Atmospheric

structure

Themperature profile of the lowermost

troposphere

Atmospheric stabillity

Dispersion of a plume

Modelling the dispersion of pollutants

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


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

[X]/Q= dilution factor

Deposition and radioactive decay

Depositionrate: 𝑅𝑑 = χ𝑣𝑑 Ci/m2/s

Radioactive decay: 𝑄 = 𝑄0exp (𝑡𝜆)=𝑄0exp (𝑥

𝑣𝜆)

χ =𝑄


ℎ2

2𝜎𝑧

χ =𝑄

2𝜋𝑣𝜎𝑦𝜎𝑧𝑒𝑥𝑝 − 𝚲 +

𝝀

𝒗 𝒙 −

ℎ2

2𝜎𝑧 𝚲 =

𝒗𝒅

𝒗𝒛

Releases from Buildings

Releases from Buildings

𝐷𝐵 = 𝑐𝐴𝑣 Building dilution factor

The wedge model

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


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)


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

the NRC has divided the spectrum of possible accidents into nine classes,

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:

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Reactor Physics Tutorials 2013 2

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