nuclear reactors, bau, 1st semester, 2008-2009 (saed dababneh). 1 review test we have done today the...

Nuclear Reactors, BAU, 1st Semester, 2008-2009 (Saed Dababneh).

1

Review Test

• We have done today the review test.

• We will discuss it next time.• Please study the review test of last year (2007-2008).

http://nuclear.bau.edu.jo/Reactors/Notes/2007/ReviewTest.ppt




2

Controlled Fission

• 235U + n X + Y + (~2.4)n• Moderation of second generation neutrons Chain reaction.• Water, D2O or graphite moderator.• Ratio of number of “neutrons” (fissions) in one generation to the preceding k (neutron reproduction or multiplication factor).

• k 1 Chain reaction.• k < 1 subcritical.• k = 1 critical system.• k > 1 supercritical.For steady release of energy (steady-state operation) we need k =1.

Fast second generation neutrons

Infinite medium (ignoring leakage at the surface).

Chain reacting pileChain reacting pile


Controlled Fission• Average number of allall neutrons released per fission

(for thermal neutrons, 0.0253 eV).• 233U : 2.492• 235U : 2.418• 239Pu : 2.871• 241Pu : 2.927

• Reactor is critical (k = 1): rate of neutrons produced by fission = rate of neutrons absorbed absorbed + leaked. Size and composition of the reactor. 3Nuclear Reactors, BAU, 1st Semester, 2008-2009

(Saed Dababneh).


4

Probability for a thermal neutron to cause fission on 235U is

Controlled Fission

235U thermal cross sectionsfission 584 b.scattering 9 b.radiative capture 97 b.

1

1

f

f

If each fission produces an average of neutrons, then the mean number of fastfast fission neutrons produced per thermal neutron =

1f

f

a

f <

Check Check numbers!numbers!

5

Controlled Fission235U

238U

• Assume Assume natural natural uranium:uranium:99.2745% 238U, 0.7200% 235U.

• f / N = (0.992745)(0) + (0.0072)(584)

= 4.20 b.• / N = (0.992745)(2.75) +

(0.0072)(97) = 3.43 b.

Thermal f = 0 b 584 bThermal = 2.75 b 97 b


N

NN

yyxx

yyxxyx

)(

24 R

24 R

Using the experimental elastic scattering data the radius of the nucleus can be estimated.

Doppler effect?Doppler effect?

Why?

Moderation (to compare x-section)

1H

(n,) (n,)

(n,n)(n,n)2H

• Resonances?

6Nuclear Reactors, BAU, 1st Semester, 2008-2009 (Saed Dababneh).

7

• Probability for a thermal neutron to cause fission in natural natural uraniumuranium

• If each fission produces an average of = 2.4 neutrons, then the mean number of fast fission neutrons produced per thermal neutron = = 2.4 x 0.55 1.3• This is close to 1. If neutrons are still to be lost, there is a danger of losing criticality. (Heavy water?).• For enriched uraniumenriched uranium (235U = 3%) = ????? (> 1.3). (Light water?).• In this case is further from 1 and allowing for more neutrons to be lost while maintaining criticality.

55.043.320.4

20.4

Controlled Fission


Compare to pure 235U.


8

Controlled Fission

i

fa

ii )()(1 • Verify

• Comment on the calculation for thermal neutrons and a mixture of fissile and non-fissile materials, giving an example.• Comment for fast neutrons and a mixture of fissionable materials, giving an example.

HW 11HW 11

9

Converters: Convert non-thermally-fissionable material to a thermally-fissionable material.

_2393.2

_239min23239238

Pu

NpUnU

d

f,th = 742 b


_23327

_233min22233232

U

PaThnTh

d

f,th = 530 b

Conversion and Breeding


India

10

• If = 2 Conversion and fission.• If > 2 Breeder reactor.• 239Pu: Thermal neutrons ( ~ 2.1) hard for breeding.

Fast neutrons ( ~ 3) possible breeding fast breeder reactors.

• After sufficient time of breeding, fissile material can be easily (chemically) separated from fertile material.Compare to separating 235U from 238U.• Reprocessing.



Delicate neutron economy…!



11


12

Controlled Fission• N thermalthermal neutrons in one generation have produced have produced so far so far NN fastfast neutrons. neutrons.• Some of these fastfast neutrons can cause 238U fission more fast neutrons fast fission factor fast fission factor = (= 1.03 for natural uranium).• Now we have Now we have NN fastfast neutrons. neutrons.• We need to moderate these fast neutrons use graphite as an example for 2 MeV neutrons we need ??? collisions. How many for 1 MeV neutrons?• The neutron will pass through the 10 - 100 eV region during the moderation process. This energy region has many strongstrong 238U capture resonances (up to ????? b) Can not mix uranium and moderator.• In graphite, an average distance of 19 cm is needed for thermalization the resonance escape probability resonance escape probability p ( 0.9).Nuclear Reactors, BAU, 1st Semester, 2008-2009

(Saed Dababneh).Reactor design.

13

Controlled Fission• Now we have Now we have ppN N thermalthermal neutrons. neutrons.• Moderator must not be too large to capture thermal neutrons; when thermalized, neutrons should have reached the fuel.• Graphite thermal cross section = 0.0034 b, but there is a lot of it present.• Capture can also occur in the material encapsulating the fuel elements.• The thermal utilization factor thermal utilization factor f ( 0.9) gives the fraction of thermal neutrons that are actually available for the fuel.• Now we have Now we have fpfpNN thermalthermal neutrons neutrons, could be > or < N thus determining the criticality of the reactor.

k = fp The four-factor formula.The four-factor formula.

k = keff = fp(1-lfast)(1-lthermal)Fractions lost at surfaceNuclear Reactors, BAU, 1st Semester, 2008-2009

(Saed Dababneh).


14

k = fp, leaknoneff Pfk

• Fast from thermal, as defined in HW 11.

• Fast from fast, .• Thermal from fast, p.• Thermal available for fission

Thinking QUIZThinking QUIZ• For each thermal neutron absorbed, how many fast neutrons are produced? Will need this when discuss two-group diffusion.

i

fa

ii )()(1

poisona

eratora

clada

fuela

fuelaf

mod

Controlled Fission

15

x 1.03x 1.03Fast fission Fast fission factor “factor “””

x 0.9x 0.9Resonance Resonance

escape escape probability ”p”probability ”p”

x 0.9x 0.9Thermal Thermal

utilization utilization factor “f”factor “f”

x

What is:• Migration length?• Critical size?How does the geometry affect the reproduction factor?

Neutron reproduction

factork = 1.000



Neutron Life Cycle

Why should we worry about these?

How?


17

k = fp(1-lfast)(1-lthermal)

Controlled Fission

• Thermal utilization factor f can be changed, as an example, by adding absorber to coolant (PWR) (chemical shim, boric acid), orby inserting movable control rods in & out.• Reactors can also be controlled by altering neutron leakages using movable neutron reflectors.• f and p factors change as fuel is burned.• f, p, η change as fertile material is converted to fissilematerial.

Not fixed…!


18

Controlled Fission

• Attention should be paid also to the fact that reactor power changes occur due to changes in resonance escape probability p. If Fuel T↑, p↓ due to Doppler broadening ofresonance peaks.

Under-moderation and

over-moderation.


19

Controlled Fission• Note that is greater than 2 at thermal energies andalmost 3 at high energies.• These “extra” neutrons are Used to convert fertile into fissile fuel.• Plutonium economy.• India and thorium.• Efficiency of this process is determined by neutronenergy spectrum.

Variations in Variations in

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