The Introduction to the Solution Method of Thermal Neutron Effective Proliferation Coefficient of An

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International Journal of Nuclear Energy Science and Engineering Volume 4 Issue 3, September 2014 doi: 10.14355/ijnese.2014.0403.02

The Introduction to the Solution Method of Thermal Neutron Effective Proliferation Coefficient of Any Point in the Non-uniformity Dispersion Reactor with Non-uniformity Dispersion Source Neutrons Deqiang Pei Nuclear Island Branch of Engineering Department of Tai Shan Nuclear Power Joint-Limited Company (A-212 01 Building on Site), Tai Shan city Guang Dong Province People’s Republic of China peideqiang@cgnpc.com.cn Abstract: The quation k eff  keff (1  0 /  ) =1 is valid in any point of '

balance of generation and disappearence of thermal fission thermal neutrons in the reactor.

non-uniformity stable reactor (See referenceⅡ), the distribution of 0 in the above equation could be

However, after source disappearence( 0  2c / Pd ),

calculated by special program. Therefore, in order to get the distribution of flux  of thermal fission thermal

thermal fission thermal neutrons as the parameter: 0  2c f  PPs . These neutrons participate

neutrons in the reactor, that is to say, the power distribution in the reactor, we need to get the value of keff in any point of non-uniformity stable reactor. The

in the process of the generation and disappearence balance of thermal fission thermal neutrons in every point in the reactor. The generating rates of these neutrons with the rates of generation and disappearence of thermal neutron generated by thermal fission thermal neutron in the reactor both influence the changes of thermal fission thermal neutron flux in the reactor and the change of the ' value of keff .

paper introduces the theoretical method to get the value of keff in any point of the reactor, and prove the correctness and effectiveness of the method.

Key Word:

keff ; Nuclear Reactor; The Distribution Of Flux Of Thermal Fission Thermal Neutrons

Introduction We know that neutrons in the reactor can be divided into source neutrons and thermal fission neutrons.

Thermal neutrons in the reactor could be divided into source thermal neutrons and thermal fission thermal neutrons too. We suppose we induce source neutrons that release the number S of fast neutrons at same point every second; the flux of source thermal neutrons is 0 .Then, at that point, source thermal neutrons generating rate is S PPs , disappearing rate is 0  2c / Pd . They generate when they are produced, and disappear when they don’t exist. They all disappear after being generated. The generation and disappearence don’t have any influence on the

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thermal neutrons they will generate

As for non-uniformity or uniformity reactor, only when the generation and disappearence balance of every point in the reactor is stable, then the thermal fission is in balance and stable for the whole reactor.

But in the uniformity reactor with the equation keff  1 , the flux is stable if we use some method to make everywhere take fission reaction.

But after being operation for a while, and because the neutrons’ flux in every point of the reactor has a non-horizontal distribution, the burning depletion of every point is different. The reactor becomes non-uniformity reactor. At the moment by regulating the density of Boron, we can’t make the equation keff  1 established everywhere in the reactor. If we make

keff  1 at somewhere in the reactor by

International Journal of Nuclear Energy Science and Engineering Volume 4 Issue 3, September 2014

regulating the density of Boron, the neutrons’ flux in the reactor here will continue to rise, and can’t be stable. Therefore, in the stable reactor, it is impossible that the equation keff  1 comes into existence. If keff  1 at some point in the reactor, there are keff  1 at other points. As for keff  1 , it is essential to

introduce source neutrons in the reactor, in order to make the reactor operate stably. But after introducing source neutrons in the reactor, neutrons’ flux will increase linearly at the point with the equation keff  1 . So that it can’t be stable. So in the non-uniformity reactor, in order to maintain operate stably, all the values of keff of every point in the reactor must be below than one. Although the reactor operate with the one hundred percent power, the reactor only operate stably after introducing source neutrons in sub-critical condition when keff  1 at every point in the reactor.

When the reactor operate stably, the equation k 'eff  keff (1  0 /  )  1 come into existence at every point in the reactor (See reference Ⅱ). The values of 0 , which is the source neutrons’ flux can be calculated by special program. So if we want to calculate the thermal fission thermal neutrons’ flux  and the powers at every point in the reactor, we need to get the value keff at every point in the reactor.

The Theoretical Solution to Solving the Value Of keff As for non-uniformity reactor, if the reactor is in stable and balanced condition, every point in the reactor is also in stable and balanced condition. At this moment, every point in the reactor takes nuclear reaction only according to the parameter of the point, and takes part in the neutrons’ balance of the point and satisfied neutrons’ balance equation of the point as equation ⑥, ⑦. However, the parameter of the point has no relationship with other points’ parameters of surrounding. As the parameter of the point has no relationship with other points’ parameters, if the surrounding points’ parameters of the point are the same with the point’ parameter, so every point’ parameter in the reactor is the same with the point’, that is to say the reactor is the uniformity-reactor, the point still takes nuclear reaction according to the parameter of the point. So the value of keff of the non uniformity reactor of the

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point could be calculated as the keff

of

the

uniformity reactor whose size is the same with the non-uniformity reactor and whose parameter equates parameter of the point of the non-uniformity reactor.

That is to say in order to get the value of keff of the non-uniformity reactor of the point, we should get the value of keff of the uniformity reactor, whose size is the same with the non-uniformity reactor and whose parameter equates parameter of the point of the non-uniformity reactor.

The Proof of the Theoretical Method to Solve the Value Of keff The Condition of Uniformity Reactor As for rectangular coordinate system (See equation 5.30.1 in referenceⅠ), the cubic volume of small unit at some point in the reactor is dv  dx  dy  dz . See figure 1. The leakage has the following relationship in Z direction. The neutrons’ flow is J Z  dx  dy in the direction into section of dx

and dy . The net

  transudation volume is  D  2  dv in Z direction  Z  in the unit volume. The neutrons outflow volume from the unit volume in the Z direction is J Z  dZ  dx  dy .

2

They have the following relationship.   2  J Z  dx  dy  D  2  dv  J Z  dZ  dx  dy  Z 

(1)

dx dz

dy FIGURE 1

Namely, the neutrons inflowing into the unit volume from Z direction with the net neutrons transudation from the unit volume is fully outflowing the unit volume from the Z direction. All volume of inflow is fully outflowing the unit, the volume of inflow don’t take part in the nuclear reaction in that unit, and don’t take part in the found of neutron balance equation for the unit volume (see equation⑥, ⑦.)

From the description and analysis of all the

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International Journal of Nuclear Energy Science and Engineering Volume 4 Issue 3, September 2014

directions in the cubic volume dv , we can also get the above result.

For the pillar coordinate, see figure 2.

is from ri to ri 1 , and thickness is dri  ri 1  ri , has a net thermal neutrons’ volume of inflow '  D2C J or  2 ri  a in internal interface. That volume of i

inflow and the net thermal neutrons transudation ri

i

inflow is fully outflowing the out surface of the barrel, the volume of inflow don’t take part in the nuclear reaction in the annulus of pillar barrel, whose radius is from ri to ri 1 , and don’t take part in the

FIGURE 2

We suppose that the value of thermal neutrons’ flux at point in diameter direction is ri  2.405  J or  J o  ri  . Its density of thermal neutrons  R   2.405  stream is  D2C J or' i   D2C J o'  ri  . The thermal  R  neutrons’ flux at ri 1 point in diameter direction i

 2.405  is J or  J o  ri 1  . Its density of thermal R    2.405  neutrons stream is  D2C J or'   D2C J o'  ri 1  . If R   we suppose dri is small enough in dri unit, the i 1

i 1

 2.405  thermal neutrons’ flux in it is J or  J o  ri  .  R  From reference Ⅰ equation 5.30.1, 5.32.3 and the above equation ①, we can get the following result(see equation (7.58.2) reference I). i





      D2C r  J ori1 cos  Z   dZ  2 rr 1  H  

②

Supposing H  H0  2  0.7104i , integrate the above  H H  equation in region   0 , 0  . And supposing  2 2 

  cos  Z  dZ  a ③, so equation ② can be H 

changed to:





' D2C J or  2 ri  a  D2C r2 J or  ri21  ri2  a i

'  D2C J or  2 ri 1  a ④ i 1

That means the annulus of pillar barrel, whose radius

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found of neutron balance equation in the annulus of pillar barrel, whose radius is from ri to ri 1 (See equation⑥, ⑦.)

So we can get the result that the net thermal neutrons volume of outflow of pillar section at radius ri equate the net thermal neutrons volume of inflow of pillar section at radius ri 1 plus the net thermal neutrons transudation volume from radius ri 1 to radius ri in the annulus of pillar barrel. Further more, we can get the result that the net thermal neutrons volume of outflow on the out face of pillar barrel, with radius dr , equate the transudation volume of thermal neutrons in pillar barrel with radius dr . Therefore, the net thermal neutrons outflow volume of pillar section with radius ri 1 equate the sum of all the points’ thermal neutrons transudation volume of pillar barrel with radius ri 1 .

So the net thermal neutrons’ outflow volume of pillar section with actual radius R, that is the outflow volume of thermal neutrons equate the sum of transudation volume of all point in R directions in the whole reactor.

     D2C r  J ori cos  Z   dZ  2 ri  D2C r2  H       2 2  J ori cos  Z   dZ ri 1  ri  H   

i



flow out from the out surface of the barrel. The ' outflow volume is  D2C J or  2 ri 1  a , All volume of i 1

z

H0  2H 0  2



volume in dri barrel  D2C r2 J or  ri21  ri2  a , both

ri+1

We can also say the rate of neutrons’ transudation volume at every point in uniformity reactor all outflow out of the reactor. In the process of outflow, except neutrons participate neutrons’ balance of the unit volume (See  D2C 22C in equation ⑥), they don’t participate the found of balance equation of generation and disappearence at every other point for thermal neutrons (that is equations ⑤ and ⑥).

Now we see an actual example. For convenience discussing, we suppose there are not source neutrons in the reactor at this phase. Supposing at some point ri hi the physical parameters are as follow:

1C : fast neutrons slowdown section;

International Journal of Nuclear Energy Science and Engineering Volume 4 Issue 3, September 2014

 2C

thermal neutrons macroscopical absorbing

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diffusing process of neutrons. When keff  1 ,⑦

section;

equation same as 8.40.1 equation reference I.

D2C : thermal neutrons diffuse coefficient;

In fact,  D2C 22C only participates the thermal

f: the absorbing effective rate of thermal neutrons in interior reactor, that is to say, the absorption of fuel to thermal neutrons make the proportion of the absorptions of all the materials to thermal neutrons.

neutrons’ balance at the point. It doesn’t influence the thermal neutrons’ balance for the next unit volume. That is to say, after it flows out of the unit volume, it will flow into the next unit volume with the transudation volume of next unit volume, finally it will flow out of the reactor. But it doesn’t participate and influence thermal neutrons’ balance and quantity relationship of balance equation in other unit volumes.

 : The average producing amount of neutrons after the absorption and fission of thermal neutrons.

 : considering the modified factor after being captured from fission of U-238 fast neutrons.

P: considering the modified factor after the resonance absorption of the slowed process of fast neutrons.

No matter the thermal neutrons’ flux at every point in the reactor how to distribute, the net inflow volume in r direction at the ri surface of the barrel

Ps : The non let out probability in the slowed

' is  D2C J or  2 ri  a , and no matter the net inflow i

process of fast neutrons.

Pd : The non let out probability in the diffusion process of thermal neutrons.

For fast and thermal neutrons in uniformity reactor.

Supposing the flux for fast neutrons and thermal neutrons at the point at some moment are 1C and 2C . Then the balance equation of thermal neutrons at the point is as the following.

The generating rate of the thermal neutrons which are produced from fast neutrons during the process of fast neutrons decelerated to thermal neutrons district( 1C 1C )  the net transudation rate of thermal neutrons(  D2C 22C )  the disappearing rate of thermal neutrons(  2C 2C )  The production rate of thermal neutrons – the disappearing rate of thermal neutrons = the changing rate of the density of thermal neutrons. ⑤ Namely, D2C  2C   2C 2C  1C 1C  2

2C  2C f  PPS 2C  2C / Pd  Also namely, D2C 22C   2C 2C  1C 1C 

1 d2C  keff  1  V2 dt

1 d2C V2 dt

⑥

2C  2C Pd

⑦

The above equation is the thermal neutrons balance equation at some point ri hi in the decelerated and

volume is how large in r direction, the thermal neutrons and fast neutrons still take nuclear reaction according to the physical parameters of the point and the law of equation ⑦.

The Condition for Non-Uniformity Reactor Same as uniformity reactor, the balance equation of thermal neutrons still satisfy the equation ⑤,⑥ and ⑦,in any point of non-uniformity reactor. The net transudation volume is still  D2C 22C .The inflow volume of upstream will fully outflow out the unit with the net transudation, and don’ take part in the nuclear reaction and the found of balance equation of thermal neutrons in that unit, that is, the fission reaction is doing as per the parameter of this point, and is independent of the surrounding physic parameter in the non uniformity reactor.

Conclusion The above content already proves any point in the non-uniformity reactor, no matter the parameters of surrounding round region are what values, it takes reaction according to physical parameters at the point and the law of equations ⑥ and ⑦. So if the surrounding region has the same material with the point, that is to say uniformity reactor, it will take reaction in this way too. Therefore, when we are considering the thermal neutrons’ balance equation of any point in the non-uniformity reactor, we can take it as uniformity reactor that the parameters is the parameters of that point of non-uniformity reactor, and their size equates the size of non-uniformity reactor. We can study them in using

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International Journal of Nuclear Energy Science and Engineering Volume 4 Issue 3, September 2014

the features of uniformity reactor.

Uniformity Dispersion Source Neutrons”. Vol.4.ISS.1 of International Journal of Nuclear Energy Science and

REFERENCE

Engineering.

Deqiang Pei “Definition and Expression of Thermal Neutron Effective Proliferation Coefficient in the Uniformity

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Dispersion

Nuclear

Reactor

with

S.Glasstone, M.C.Edlund :“Atomic Nuclear Reactor Theory Compendium”;