Shielding of 10 kWe 14 Fast Neutron Flux

7/30/2019 Shielding of 10 kWe 14 Fast Neutron Flux

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Shielding of 10kWe 14 fast

neutron flux

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Shielding of 10 kWe 14 fast neutron

flux reactor

PRINCIPLES OF REACTOR SHIELDING

The design of the biological shields placed

around a nuclear reactor and at various points

in a nuclear power plant to protect operating

personnel and the public at large from the

radiations

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sources of radiation Prompt Fission Neutrons:Emitted from the core, in designing a biological

shield be given foremost thought

Delayed Fission Neutrons:Ordinarily not an important considerationsthere are a few of them and their energies are so low .

Prompt Fission yRays:Emitted in the core, these y-rays are largely

attenuated by the materials in the core.

Fission Product Decay yRays:Emitted from the fuel, these y -rays are

a continuing source of radiation after the shutdown of a reactor

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Inelastic yRays:

emitted due to inelastic neutron scattering, occurs only withenergetic neutrons

Capture yRays:

radiative capture reaction ,most probable at resonance andthermal energies

( thermal neutrons are usually present in a reactor shield, the

shield may be a source of capture y-rays)

Activation yRays:

These y -rays are emitted by radioactive nuclides formed as theresult of neutron absorption(Much of the internal structure of a reactor becomes radioactive in this

way, so does the coolant and any extraneous atoms in the coolant)

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Among all of these radiation sources, the promptfission neutrons are the most difficult to shieldagainst.

fast neutrons are first slowed down to thermalenergies and then absorbed

Neutrons lose , on average, 50% of energy inelastic collision with hydrogenousmaterial

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Most of the shielding in stationary nuclear

power plant is ordinary concrete, which

contains approximately 10 w /o water and has

a hydrogen atom density equal to about one

quarter that of water.

Concrete is inexpensive, structurally sound,

and readily formed into any desired shape

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Very fast neutrons are slowed down by inelasticscattering(often shields of moderately heavy or heavy materials)

the average energy E' of a neutron emerging froman inelastic collision with a nucleus of massnumber A is:

E =6.4sqr E/A

(E is the energy of the incident neutron and both E' and E arein MeV)

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In a nuclear reactor Iron and other heavymaterials are used in for attenuating y-rays( as the average energy of fission neutrons is too low to have inelastic scattering

play a significant role in slowing down)

shield divided into alternating regions of, say,iron and a hydrogenous material (concrete orpolyethylene) attenuates y-rays and neutrons atthe same rate

This is done for mobile nuclear power systems

for a stationary power plant, simple concreteshielding is cheaper

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Once the fast neutrons thermalize in a shield,they must then be absorbed.

when thermal neutrons are captured in water,a 2.2-MeV y-ray is emitted as the result of the1H(n,y )2H reaction.

When the thermal neutrons are captured byiron, however, 7.6-MeV and a 9.3-MeV y-rayare emitted.

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To reduce the intensity of such y-rays, boron-

either natural or enriched is sometimes added

to reactor shields

Boron has a high absorption cross-section

(759 b), due mostly to the 10B (n, a) 7Li

reaction

A 0.5 MeV yray is emitted in this reaction,

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REACTOR SHIELD DESIGN:

REMOVAL-ATTEN UATION CALCULATIONS:

The preliminary design of a reactor shield can

be carried out using the point kernel [The flux G (r)at a distance r from a point fission source emitting one fission neutron per sec

isotropically.] andthe removal cross-sections

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Consider a spherical reactor consisting of a core of radius Rsurrounded by a hydrogenous shield of thickness a

The core consists of a mixture of metal and water, with the

volume fraction of metal equal to f let the fission neutrons are being produced uniformly

throughout the core at the rate ofS neutrons/cm3 -sec.

The fast neutron flux at P at the surface of the shield fromdV within the core:

dV = 2 r 2 sin d dr.

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flux reactor in the case of a small core and a thick shield:

sin

and max R/(R+a)

rs a hence rmax 2R +a

yields the final result:

(P) = S * A/4 ( R/R+a)2 e-RS a(1-e-2R)..1

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flux reactor In equation 1 above:

= neutron flux at the surface of the shield(neutrons/cm2 . sec)

S = No. of neutrons produced /cm3 .sec

A= a constant for water and has a value of .12 forwater

= (1-f)RW +fRM

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where:

f=metal volume fraction of core

1-f=water volume fraction of core(refector)

RW=Macroscopic removal x-section of water(cm-

)RM = Macroscopic removal x-section of metal (cm-)

RS= Macroscopic removal x-section of shield (cm-)

R= core radius (in case of cylindrical core)

a = shield radius( radius of reflector+shield)

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Now we are given that:Reactor Power =10 Kw (e)

= 14 fast neutrons

Calculate:

Reactor shield dimensions for this condition;

Solution:

Let 1. Core is made up of:

An assembly of uraniumrods & water with metal volume fraction of 75% i.ef=.75

Volume of the core =32 l => 32 000 cc

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flux reactor Using eq-1 as above we have:

14 = S * 0.12/4 ( R/R+a)2 e-RS a(1-e-2R)

As =0.25 x 0.103 + 0.75 x 0.174Or = 0.156 cm-

And 2R=0.312R

And RS a= 0.103a

So S = 2.42 x fission density

And fission density = No. of fissions / sec /reactor volume

No. of fissions/sec =energy produced by reactor/energyproduced per fission

No. of fissions /sec = 10000 J/sec / 3.2 x 10-11 j/fission

Note; there are 200 MeV/fission = 3.2 x 10 -11 J

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flux reactorSo No. of fissions/sec = 3125 x 1011 fissions/sec

Fission density = No. of fissions/sec / reactor vol=97.65 x 108 fissions/cm3.sec

Hence S=2.42 x fission density neutrons/cm3.sec

Or S= 236.313 x 108 neutrons/cm3.sec

For a 32 liter(l) reactor core:R= 3 20 cm

Then from eq -1 :

14 = 236.313 x .12 x 108/4 x 0.156 (20/20+a)2e-0.103a(1-e-6.24)

From which;

a =272 cmKeeping reflector thickness= 15 cm

Shield thickness = 257 cm

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Shielding of 10 kWe 14 Fast Neutron Flux

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