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**OVERVIEW:**

The FNR Mathematical Model is used to calculate the thicknesses of the FNR core and blanket regions and to calculate the temperature dependence of the FNR core zone bulk reactivity.

The FNR Mathematical Model underlies FNR performance and safety. If any portion of a FNR goes critical with respect to prompt neutrons thermal power growth occurs on a sub-microsecond time scale. There is no time for any external power control system to respond. The safety case lies in mathematical certainty that the with the chosen materials and dimensions thermal expansion will always keep the FNR stable. To realize such mathematical certainty the FNR requires a mathematical model that is a good representation of physical reality.

The mathematical model needs to be solved both with all new fuel and with steady state used fuel. At the start of a fuel cycle the movable fuel bundles only partially overlap the fixed fuel bundles. At the end of a fuel cycle the movable fuel bundles fully overlap fixed fuel bundles. The overlap distance controls the fraction of fission neutrons that diffuse out of the core zone and into the blanket zone. The total core fuel rod length must be greater than the required core zone thickness when the fuel burnup fraction is 15%.

The FNR has a pancake shaped core zone that is surrouned by six layers of blanket material that total 1.8 m thick. Each active fuel bundle has an average blanket cross sectional area of about 13 inch X 13 inch.

As neutrons propagate across the blanket zone the neutron flux decreases and the individal neutron K.E. decreases. The neutron path length is enlarged by neutron scattering.

A FNR inherently operates with its core zone on the threshold of criticality. Whether or not the core zone is critical depends on the fraction of fission neutrons that diffuse out of the core zone into the adjacent blanket zones. As discussed on the web page titled: FNR Criticality this fraction should be a constant independent of temperature. Typically this fraction is about (1 / 2) at the beginning of a fuel cycle and about (1 / 3) at the end of a fuel cycle. With plutonium based fuel every fission produces 3.1 neutrons. About 99.7% ____of these neutrons are prompt and are released at the instant of fission. The remaining 0.3%____ of these neutrons are delayed and are released by fission products after an average 3 second delay. One neutron must be absorbed by Pu-239 in the core zone to sustain the chain reaction. Between (1 / 2) and 1 neutron is absorbed by U-238 in the core zone to help sustain the core zone TRU concentration. A fractional neutron is lost to non-productive neutron absorption by zirconium, iron, chromium, sodium, and fission products in the core zone. The rest of the neutrons diffuse into the adjacent blanket zones where most of them are absorbed by U-238, zirconium, iron, chromium and sodium.

Neutrons that escape from the outer surface of the blanket zone face an ~ 2 m wide guard band before impacting the sodium pool side wall.

From a reactor stability perspective the key issue is criticality in the core zone. The fuel atomic concentration of plutonium, Np and the fuel atomic concentration of uranium Nu respond almost instantly to changes in reactor power. However, the other atomic concentrations response is delayed by thermal conduction times. However, the FNR thermal behaviour is dominated by its fuel temperature. A step increase in fuel temperature causes an instantaneous decrease in reactor reactivity.

However, this control range, which relies on thermal expansion of the fuel, is limited. If there is a sudden change in fuel geometry which forces the core zone critical with respect to prompt neutrons then the reactor can potentially blow up.

Recall from the web page titled: FNR Criticality that the effect of the core zone non-fuel elements is to make the reactor slightly less stable. The concentrations of these non-fuel elements respond to changes in reactor power after a thermal conduction time delay.

Further reactor safety is realized via the delayed neutrons. The rate of core zone reactivity increases should be limited to allow the delayed neutron flux to stay in phase with the prompt neutron flux. The core zone should be kept subcritical with respect to the prompt neutrons. The power control should be realized via delayed neutrons. Hence FNR fuel geometry changes that potentially increase FNR reactivity should be made very slowly.

If a change in the fuel geometry is made which increases the core zone reactivity the prompt neutron flux will respond almost instantly but the remaining delayed neutron flux will take about 3 seconds to respond. Thus for safety any reactor geometry change controlled by the neutron flux or the corresponding gamma flux must not change the reactor from being subcritical with respect to prompt neutrons to being above critical with respect to prompt neutrons in a 3 second time period.

In a solid fuel FNR the rate of reactor fuel geometry changes is easily controlled. However, in a Molten Salt Reactor (MSR) the flowing motion of the fuel through the core zone together with liquid waves and vorticies can easily lead to unintended transient criticality with respect to prompt neutrons. Thus the mathematical model developed herein is only valid for solid fuel reactors. This author has severe reservations about the safety of liquid fuel power MSRs, especially those operating with fast neutrons.

Over time there will be formation of Pu-239 in the portions of the blanket zone close to the core zone. Hence towards the end of a fuel cycle there will be some fission reactions in the portions of the blanket zones close to the core zone. However, the concentration of Pu-239 in the blanket zone is not sufficient to sustain the chain reaction. The Pu-239 concentration in the core zone will gradually decrease and the fission product concentration will gradually increase until the chain reaction can no longer be maintianed.

Withdrawal of the movable fuel bundles has the effect of shifting part of the U-238 absorption mass from a blanket zone into the core zone. Movable fuel bundle insertion and withdrawal is used to compensate for the effect of the changing core zone composition on the reactor fuel temperature setpoint and for achieving a reactor cool shutdown.

In FNR operation the exact fraction of fission neutrons that diffuse out of the core zone into the blanket zone is determined by the reactor fuel geometry. For safety at the operating point this fraction must be constant or must increase with increasing reactor material temperature. Establishing the best FNR operating point requires a good mathematical model of the FNR.

About 99.8% of the fast fission neutrons are prompt and the remaining 0.2% of the fast neutrons are delayed due to being emitted from fission products instead of being emitted directly from Pu-239 fission. Provided that the FNR is structurally uniform these delayed neutrons perform the important function of limiting the peak amplitude of the nuclear power spike that is emitted in response to an increment of active fuel bundle control portion insertion. Even so, the active fuel bundle control portion positioning apparatus should not have any hysterisis or other position control instabiity.

From an operational perspective it is important that the FNR primary sodium operating temperature remain stable while meeting a rapidly changing external thermal load, without any change in active fuel bundle control portion position.

**The mathematical model developed on this web page for a FNR gives explicit formulae relating the core fuel rod alloy, core thickness, blanket fuel rod alloy and blanket thickness.**

The contemplated core rod material at the beginning of a fuel cycle is:

20% Pu, 70% U-238, 10% Zr by weight.

The contemplated core rod material at the end of a fuel cycle is:

10% Pu____, 65% U-238____, 10% Zr and 15% fission products by weight.

The contemplated blanket rod material at the beginning of a fuel cycle is:

90% U-238, 10% Zr by weight.

**MATHEMATICAL MODEL:**

The fast neutron concentration in a FNR decreases monotonically from the middle of the core zone to the far ends of the adjacent blanket zones. The fast neutrons go through a series of scatters. Some of the fast neutrons are absorbed within the core zone. Other fast neutrons are absorbed within the blanket zone. A few neutrons escape into the surrounding sodium guard band. The effect of rising temperature in the FNR is to decrease the fraction of fission neutrons that react within the core zone which causes an increase in the fraction of fission neutrons that are absorbed by the blanket zones.

The core zone thickness Lc used in the EBR-2 was Lc = 14.22 inches. The core zone height Lc used for preliminary FNR design on this web site is:

0.35 m < Lc < 0.40 m (13.78 inch to 15.74 inch).

For calculation purposes use Lc = 0.375 m = 14.76 inches. One of the purposes of this web page is to determine the optimum value of Lc in terms of other FNR parameters.

**DEFINITIONS:**

Lac = average distance along the path of a neutron in the core zone to point of absorption;

Lsc = average distance along the path of a neutron in the core zone between successive scatters;

Lab = average distance along the path of a neutron in the blanket zone to point of absorption;

Lsb = average distance along the path of a neutron in the blanket zone between successive scatters;

Lb = thickness of blanket zone

Lc = thickness of core zone

Z = distance along the vertical axis of the FNR where:

Z = 0 at the center of the core zone.

Z = (Lc / 2) at the upper core zone - blanket zone junction

Z = (Lc / 2) + Lb at the upper end of the blanket zone

N(Z) = neutron concentration as a function of Z

N(Z) = has a relative maximum at Z = 0

d[N(Z)]/ dZ = 0 at Z = 0

N(Z) = minimum value at Z = (Lc / 2) + Lb

Nj = N(Z)|Z = (Lc / 2)

Vn = neutron velocity (assume constant to simplify the problem)

A = cross sectional area of a FNR

A dZ = an element of volume of the active region of a FNR

Np = average concentration of plutonium atoms in core zone (Note that Np decreases by a factor of about 2 during the fuel cycle)

Ni = average concentration of Fe atoms in both the core zone and the blanket zone

Nc = average concentration of Cr atoms both the core zone and the blanket zone

Nsc = average concentration of Na atoms in the core zone

Nsb = average concentration of Na atoms in the blanket zone

Nuc = average concentration of U-238 atoms in the core zone

Nub = average concentration of U-238 atoms in the blanket zone

Nzc = average concentration of Zr atoms in the core zone

Nzb = average concentration of Zr atoms in the blanket zone

Nuco = average number of U-238 atoms per unit volume in the core zone at the beginning of fuel bundle life when mobile fuel bundles are fully inserted.

Nuca = additional average number of U-238 atoms per unit volume in core zone due to full withdrawal of the active fuel bundle control portion

Nf = average concentration of fission product atoms in core zone at the end of fuel ccycle

Gp = number of neutrons released per Pu-239 fission = 3.1

Gu = number of neutrons released per U-235 fission = 2.6

Tc = neutron lifetime in core zone

Tb = neutron lifetime in blanket zone

Kdc = neutron diffusion rate constant in the core zone

Kdb = neutron diffusion rate constant in the blanket zone

Rhos = mass density of liquid Na = .927 gm / cm^3

Rhoi = mass density of Fe = 7.874 gm / cm^3

Rhou = mass density of U-238 = 19.1 gm / cm^3

Rhop = mass density of Pu-239 = 19.8 gm / cm^3

Rhoc = mass density of Cr = ______

Rhoz = mass density of Zr = _____

Aws = atomic weight of Na = 23

Awi = atomic weight of Fe = 55.845

Awu = atomic weight of U = 238

Awz = atomic weight of Zr = 91.22

Av = Avogadro's Number

Vf = effective neutron diffusion velocity along the Z axis through the neutron scattering barrier

?????_______
Lg = 1 / [Gp Sigmafp Np + Gu Sigmafu Nu]

????????

Data from Kaye & Laby for a FNR core:

Sigmaas = fast neutron absorption cross section of sodium = 0.0014 b

Sigmass = fast neutron scatter cross section of sodium = 2.62 b

Sigmaai = fast neutron absorption cross section of iron = .0086 b

Sigmasi = fast neutron scatter cross section of iron = 4.6 b

Sigmaaf = fast neutron absorption cross section of fission products = ____ b

Sigmaau = fast neutron absorption cross section of U-238 = 0.25 b

Sigmasu = fast neutron scatter cross section of U-238 = 9.4 b

Sigmaap = fast neutron absorption cross section of plutonium-239 = 0.040 b

Sigmaaz = fast neutron absorption cross section of zirconium = 0.0066 b

Sigmafp = fast neutron fission cross section of plutonium-239 = 1.70 b

Sigmafu = fast neutron fission cross section of uranium-238 = 0.041 b

In a practical power FNR the core zone and blanket zone thicknesses are large compared to the average distance between successive neutron scatters, which is typically about 0.06 m. Hence diffusion theory can be used to find Kdb and Kdc in terms of fast neutron scattering parameters.

**SCATTERING BARRIER:**

Introduce the concept of scattering atoms acting as a barrier to neutron absorption.
Each neutron impinges on this scattering barrier many times before encountering a neutron absorbing atom.

**DERIVATION OF DIFFUSION CONSTANTS Kdc and Kdb:**

Consider diffusion of neutrons through a barrier composed of scattering centers where:

Ls = mean distance between successive scatters

La = mean distance travelled along the neutron path to cross the barrier

Vn = neutron velocity which is assumed to be almost constant because the mass of the individual scattering atoms is much much greater than a neutron mass.

N = neutron concentration

Neutron diffusion flux is:

Flux = - Kd A (dN / dZ)

The number of scatters required for a single neutron to cross the barrier is:

(La / Ls)

From random walk theory the barrier thickness is:

[(number of scatters) / 3]^0.5 Ls

= [(La / 3 Ls]^0.5 Ls

= **[La Ls / 3]^0.5**

The change in neutron concentration across the barrier is:

dN / dZ = - N / (barrier thickness)

= {- N / [La Ls / 3]^0.5}

Hence the neutron flux through the barrier is:

**Flux** = - Kd A (dN / dZ)

= - Kd A {- N / [La Ls / 3]^0.5}

= **{Kd A N / [La Ls / 3]^0.5}**

The transit time required for a neutron to cross the scattering barrier is:

(La / Vn)

The effective neutron velocity Vf through the barrier along the Z axis is:

Vf = (barrier thickness) / (transit time)

= {[La Ls / 3]^0.5} / (La / Vn)

The neutron flux through the barrier is:

**Flux** = N Vf A

= **N A {[La Ls / 3]^0.5} / (La / Vn)**

Equating the two expressions for the neutron flux through the scattering barrier gives:

{Kd A N / [La Ls / 3]^0.5} = N A {[La Ls / 3]^0.5} / (La / Vn)

or

{Kd / [La Ls / 3]^0.5} = {[La Ls / 3]^0.5} / (La / Vn)

or

Kd = {[La Ls / 3]} / (La / Vn)

or

**Kd = [Ls Vn / 3]**

Hence in the core zone:

**Kdc = Lsc Vn / 3**

and in the blanket zone:

**Kdb = Lsb Vn / 3**

The scattering lengths are given by:

**Lsc = 1 / [Ns Sigmass + Ni Sigmasi + Nu Sigmasu + Nz Sigmasz]_____**

and

**Lsb = 1 / [Nsb Sigmass + Nib Sigmasi + Nub Sigmasu + Nzb Sigmasz]_____**

**NEUTRON FLUX:**

**- Kdc A (dN / dZ)** = net neutron diffusion flux in the Z direction at any position in the core zone

**- Kdb A (dN / dZ)** = net neutron diffusion flux in the Z direction at any position in the blanket zone

Consider an element of volume A dZ at Z = Z:

The diffusion flux into the element of volume at Z = Z is:

- Kd A (dN / dZ)|Z = Z

The diffusion flux out of the element of volume at Z = Z+ dZ is:

- Kd A (dN / dZ)|Z = Z + dZ

The net diffusion flux of neutrons into the element of volume is:

[- Kd A (dN / dZ)|Z = Z] - [- Kd A (dN / dZ)|Z = Z + dZ]

= - Kd A {[(dN / dZ)|Z = Z] - [(dN / dZ)|Z = Z + dZ}

= + Kd A {[(dN / dZ)|Z = Z + dZ] - [(dN / dZ)|Z = Z]}

= **+ Kd A d[(dN / dZ)]**

NEUTRON GENERATION:

The rate of neutron generation within an element of volume (A dZ) in the core zone is:

**(Gp Sigmafp Np + Gu Sigmafu Nu) N Vn A dZ**

It is convenient to define the pseudo length Lg by:

**Lg = [1 / (Gp Sigmafp Np + Gu Sigmafu Nu)]**

Then the rate of neutron generation within an element of volume (A dZ) in the core zone is:

**N Vn A dZ / Lg**

NEUTRON ABSORPTION:

The rate of neutron absorption within an element of volume (A dZ) within the core zone is:

**(N / Tc) A dZ**

The lifetime Tc of a free neutron in the core zone is given by:

**Tc = [Lac / Vn]
= 1 / {Vn [Ns Sigmaas + Nuc Sigmaau + Ni Sigmaai + Ncr Sigmaacr + Np Sigmaap + Np Sigmafp]}**

Similarly the rate of neutron absorption within an element of volume within the blanket zone is:

**(N / Tb) A dZ**

The lifetime Tb of a free neutron in the blanket zone is given by:

**Tb = [Lab / Vn]
= 1 / {Vn [Nsb Sigmaas + Nub Sigmaau + Ni Sigmaai + Ncr Sigmaacr + Npb Sigmaap + Npb Sigmapf]}**

At steady state conditions the neutron profile N(Z) in the core zone is constant in time giving:

N (Vn A / Lg) dz - (N / Tc) A dZ + Kdc A d[(dN / dZ)] = 0

or

N (Vn / Lg) - (N / Tc) + Kdc d[(dN / dZ)] / dZ = 0

or

**[(Vn / Lg) - (1 / Tc)] N + Kdc d[(dN / dZ)] / dZ = 0**

This equation has a boundary condition that at Z = 0:

dN(Z) / dZ = 0

This equation has a second boundary condition that at Z = (Lc / 2) the value of N = Nco is the same as the value of N = Nco for the blanket zone for Z = (Lc / 2).

**BLANKET ZONE:**

At steady state conditions the neutron profile N(Z) in the blanket zone is constant in time giving:

- (N / Tb) A dZ + Kdb A d[(dN / dZ)] = 0

or

** - (N / Tb) + Kdb d[(dN / dZ)] / dZ = 0**

This equation has a boundary condition that at Z = (Lc / 2) + Lb the neutron diffusion flux must be close to zero. Hence at

Z = (Lc / 2) + Lb :

N ~ 0

and

dN / dZ ~ 0.

For Z > (Lc / 2) try solution in the blanket zone of the form:

N(Z) = Nj exp(- [Z - (Lc / 2)] Cb)

where Nco = the value of N at Z = (Lc / 2)

d[(dN / dZ)] / dZ = Cb^2 Nj exp(- [Z - (Lc / 2)] Cx)

Substitution into the differential equation for the blanket zone gives:

- (N / Tb) + Kdb d[(dN / dZ)] / dZ = 0

or

- (Nj / Tb) exp(- [Z - (Lc / 2)] Cb) + Kdb Cb^2 Nj exp(- [Z - (Lc / 2)] Cb) = 0

or

- (1 / Tb) + Kdb Cb^2 = 0

or

**Cb^2 = Tb / Kdb**

Thus in the blanket zone:

**N(Z) = Nj exp{- [Z - (Lc / 2)] (Tb / Kdb)^0.5}**

and

dN / dZ = - Nj (Tb / Kdb)^0.5) exp{- [Z - (Lc / 2)] (Tb / Kdb)^0.5}

= - Nj Cb exp{- [Z - (Lc / 2)] Cb}

The neutron diffusion flux of neutrons flowing from the core zone into the blanket zone is:

**{- Kdb A [(dN / dZ)]|Z = (Lc / 2)}**
= -Kdb A [-Cb Ca Exp(-Cb Z) + Cb Ca Exp(- Cb (Lc + Lb - Z))]|Z = (Lc / 2)

= - Kdb A [-Cb Ca Exp(-Cb Lc / 2) + Cb Ca Exp(- Cb ((Lc / 2) + Lb))]

= Kdb A [Cb Ca Exp(-Cb Lc / 2)][1 - Exp(- Cb Lb)]

**CORE ZONE SOLUTION:**

Recall that in the core zone:

[(Vn / Lg) - (1 / Tc)] N + Kdc d[(dN / dZ)] / dZ = 0

Try for a solution of the form:

N(Z) = Cc Cos(Cd Z)

or

dN(Z) / dZ = - Cd Cc Sin(Cd Z)

or

d[dN(Z) / dZ] / dZ = - Cd^2 Cc Cos(Cd Z)

Note that this solution satisfies the boundary condition that:

dN(Z) / dZ = 0

at Z = 0

Hence:

[(Vn / Lg) - (1 / Tc)] N + Kdc d[(dN / dZ)] / dZ = 0

or

[(Vn / Lg) - (1 / Tc)] Cc Cos(Cd Z) + Kdc [- Cd^2 Cc Cos(Cd Z)] = 0

or

[(Vn / Lg) - (1 / Tc)] + Kdc [- Cd^2] = 0

or

**Cd = {[(Vn / Lg) - (1 / Tc)] / Kdc}^0.5**

This equation potentially allows quantification of Cd.

Hence:

N(Z) = Cc Cos(Cd Z)

= Cc Cos({[(Vn / Lg) - (1 / Tc)] / Kdc}^0.5 Z)

At Z = (Lc / 2):

**[N(Z)|z = (Lc / 2)] = Cc Cos({[(Vn / Lg) - (1 / Tc)] / Kdc}^0.5 (Lc / 2))**

The diffusion flux at Z = (Lc / 2) is given by:

- Kdc A [dN(Z) / dZ]|Z = (Lc / 2)

= - Kdc A [- Cd Cc Sin(Cd Lc / 2))]

**SHARED BOUNDARY CONDITIONS AT Z = (Lc / 2):**

At the junction between the core zone and the blanket zone N(Z)|Z = (Lc / 2) is the same for both zones and the neutron diffusion flux out of the core zone equals the neutron diffusion flux into the blanket zone.

At the junction between the two zones:

[N(Z)|Z = (Lc / 2)]

= **Cc Cos[Cd (Lc / 2)] = Ca Exp[-Cb (Lc / 2)] + Ca Exp[- Cb(Lb + (Lc / 2))]**

or

**(Cc / Ca) = {Exp[-Cb (Lc / 2)] + Exp[- Cb(Lb + (Lc / 2))]} / Cos[Cd (Lc / 2)]**

At the junction between the two zones the diffusion flux is equal giving:

-Kdc A [- Cd Cc Sin(Cd (Lc / 2))] = Kdb A [Cb Ca Exp(-Cb Lc / 2)][1 - Exp(- Cb Lb)]
or

**Kdc [Cd (Cc / Ca) Sin(Cd (Lc / 2))] = Kdb [Cb Exp(-Cb Lc / 2)][1 - Exp(- Cb Lb)]**

Combining these two boundary condition equations gives:

Kdc[Cd Sin(Cd (Lc / 2))]{Exp[-Cb (Lc / 2)] + Exp[- Cb(Lb + (Lc / 2))]} / Cos[Cd (Lc / 2)] = Kdb [Cb Exp(-Cb Lc / 2)][1 - Exp(- Cb Lb)]

or

Kdc Cd [Tan(Cd Lc / 2)][1 + Exp(- Cb Lb)] = Kdb Cb [1 - Exp(- Cb Lb)]

or

[Tan(Cd Lc / 2)] = [Kdb Cb / Kdc Cd] {[1 - Exp(- Cb Lb)] / [1 + Exp(- Cb Lb)]}

or

**Lc = (2 / Cd) Arc Tan{[Kdb Cb / Kdc Cd] [1 - Exp(- Cb Lb)] / [1 + Exp(- Cb Lb)]}**

This equation potentially allows quantification of Lc

**FNR DESIGN:**

The last important relationship is that the total neutron absorption in the half core is about equal to the corresponding neutron absorption in the adjacent blanket + neutron loss out end of blanket.

In the half core the total neutron absorption is:

Integral from Z = 0 to Z = (Lc / 2) of:

(Cc / Tc) Cos(Cd Z) dZ

= **[(Cc / Cd Tc) Sin(Cd Lc / 2)]**

In the adjacent blanket zone the corresponding rate of neutron absorption is:

Integral from Z = (Lc / 2) to Z = (Lc / 2)+ Lb of:

(N / Tb) A dZ

= Integral from Z = (Lc / 2) to Z = (Lc / 2)+ Lb) of:

(Nco A / Tb) exp{- [Z - (Lc / 2)] (Tb / Kdb)^0.5} dZ

= - (Kdb / Tb)^0.5 (Nco A / Tb) exp{- [Z - (Lc / 2)] (Tb / Kdb)^0.5}|Z = (Lc / 2)+ Lb

+ (Kdb / Tb)^0.5 (Nco A / Tb) exp{- [Z - (Lc / 2)] (Tb / Kdb)^0.5}|Z = (Lc / 2)

= - (Kdb / Tb)^0.5 (Nco A / Tb) exp{- [Lb] (Tb / Kdb)^0.5}

+ (Kdb / Tb)^0.5 (Nco A / Tb)

= **(Kdb / Tb)^0.5 (Nco A / Tb) [1 - exp{- [Lb] (Tb / Kdb)^0.5}]**

Generally the object is to make the blanket sufficiently long that:

Lb (Tb / Kdb)^0.5 > 4
so that the rate of neutron absorption in one blanket zone is approximately:

**(Kdb / Tb)^0.5 (Nco A / Tb)**

Recall that from the core zone solution for N(Z): Nco = Cc Cos[(Cd Lc) / 2]

Hence the total neutron absorption by the blanket zone is:

(Kdb / Tb)^0.5 (Nco A / Tb) [1 - exp{- [Lb] (Tb / Kdb)^0.5}]
= (Kdb / Tb)^0.5 (Cc Cos[(Cd Lc) / 2]) (A / Tb) [1 - exp{- [Lb] (Tb / Kdb)^0.5}]

= **Cb (Cc Cos[(Cd Lc) / 2]) (A / Tb) [1 - exp(- Cb Lb)]**

The total neutron absorption by both the blanket and out its end is:

**Cb Cc Cos[(Cd Lc) / 2] (A / Tb)**

**CRITICALITY MAINTENANCE:**

In order to maintain core zone criticality with the chosen core zone fuel at the end of the fuel cycle choose to operate with the neutron absorption in the half core approximately equal to 2X the neutron absorption in one blanket. Then:

(neutron absorption in half core) = 2X(neutron absorption in one blanket + end losses)

or

[(Cc / Cd Tc) Sin(Cd Lc / 2)] ~ 2X Cb (Cc Cos[(Cd Lc) / 2] A / Tb)

or

tan(Cd Lc / 2) = 2X[Cd Tc Cb A / Tb]

or

**Lc = (2 / Cd) Arc Tan[2 Cd Tc Cb A / Tb]**

Recall that:

Tc = Lac / Vn

and

Tb = Lab / Vn

Hence:

** Tan(Cd Lc / 2)] = [( Cd Lac / Cb Lab)] [1 - Exp(- Cb Lb)] / [1 + Exp(- Cb Lb)]**

Note that Lc goes down if Cb Lb is not >> 1.

**This is a key equation for FNR design.**

This equation has several important features:

1) (Cb Lb) ~ 3 to make:

Exp[- Cb Lb] < < 1

This condition imposes important requirements on the blanket rod length Lb, the blanket rod thickness and the blanket rod material composition. If these requirements are not met the FNR thermal safety shutdown characteristic is compromised.

and

2) [Tan(Cd Lc / 2) has a negative temperature coefficient. A FNR relies on this negative temperature coefficient to cause a reactivity decrease as its temperature increases.

Recall that:

Cd^2 = {[(Vn / Lg) - (1 / Tc)] / Kdc}

= {[(Vn / Lg) - (Vn / Lac)] / (Lsc Vn / 3)}

= {(3 / Lsc)[(1 / Lg) - (1 / Lac)]}

Hence:

Cd Lc = {(3 / Lsc)[(1 / Lg) - (1 / Lac)]}^0.5 Lc

Lg, Lac, Lsc are all proportional to (1 / Nu) making Cd proportional to Nu.

Hence:

Cd Lc is proportional to Nu^0.6666

As the temperature increases Nu decreases reducing neutron absorption in the core zone, which causes a drop in reactivity. Note that the temperature dependence of Cd Lac is the same as the temperature dependence of Cb Lab.

**EVALUATION OF Lac, Lab, Lsc, Lsb, Lg:**

The above equations allow calculation of Lc and FNR fuel constraints. The practical calculation procedure is to assume particular fuel mixes in the core and blanket zones to calculate:

Lac = 1 / [Ns Sigmaas + Nu Sigmaau + Ni Sigmaai + Np Sigmaap + Np Sigmafp]

Lab = 1 / [Nsb Sigmaas + Nub Sigmaau + Ni Sigmaai + Npb Sigmaap + Npb Sigmapf]

Lsc = 1 / [Ns Sigmass + Ni Sigmasi + Nu Sigmasu + Nz Sigmasz]

Lsb = 1 / [Nsb Sigmass + Nib Sigmasi + Nub Sigmasu + Nzb Sigmasz]

Lg = [1 / (Gp Sigmafp Np + Gu Sigmafu Nu)]

**EVALUATION OF Lc:**

Recall that:

Lc = (2 / Cd) Arc Tan[(Lsb / Lsc) (Cb / Cd)]

where:

**(Cb / Cd)** = {{[(Vn / Lg) - (1 / Tc)] / Kdc} Kdb Tb}^0.5

= {[(Vn / Lg) - (Vn / Lac)] (Kdb / Kdc)(Lab / Vn)}^0.5

= {[(1 / Lg) - (1 / Lac)] (Lsb / Lsc)(Lab)}^0.5

= **{[(Lab / Lg) - (Lab / Lac)] (Lsb / Lsc)}^0.5**

Then:

[(Lsb / Lsc) (Cb / Cd)] = (Lsb / Lsc){[(Lab / Lg) - (Lab / Lac)] (Lsb / Lsc)}^0.5

= (Lsb / Lsc)^1.5 [(Lab / Lg) - (Lab / Lac)]^0.5

= (Lsb / Lsc)^1.5 (Lab / Lac)^0.5 [(Lac / Lg) - (Lac / Lac)]^0.5

Choose the parameters of the blanket rods so that:

**(Lsb / Lsc)^1.5 [(Lab / Lg) - (Lab / Lac)]^0.5 = 1**

Note that Lab > Lac but Lsb < Lsc.

Note that: Lg ~ Lac / 2

Then:

**Arc Tan[(Lsb / Lsc) (Cb / Cd)] = Arc Tan[1] = Pi / 4**

To achieve this objective make Nub > Nu by increasing blanket rod thickness and by making the blanket rods of pure metallic uranium. Possibly the fuel tube wall thickness must be slightly reduced.

Recall that:

**Cd** = {[(Vn / Lg) - (1 / Tc)] / Kdc}^0.5

= {[(Vn / Lg) - (Vn / Lac)][ 3 / Lsc Vn]}^0.5

= **{[(1 / Lg) - (1 / Lac)][ 3 / Lsc]}^0.5**

Numerical evaluation gives:

Lg - 1.497 m

Lac = 2.901 m

Lsc = 0.0607 m

Thus:

Cd = {[(1 / Lg) - (1 / Lac)][ 3 / Lsc]}^0.5

= {[(1 / 1.497 m) - (1 / 2.901 m)][ 3 / 0.0607 m]}^0.5

= {[(0.6680 / m) - (0.3447 / m)][ 49.423 / m]}^0.5

= **3.997 / m**

Hence:

**Lc** = (2 / Cd) Arc Tan[(Lsb / Lsc) (Cb / Cd)]

= (2 m / 3.997) (Pi / 4)

= **0.393 m** = 15.47 inches

______________
which is close to the 14.22 inches used for the EBR-2.

**EVALUATION OF Lb:**

Recall that:

Cb Lb ~ 3

Recall that:

Cb^2 = 1 / (Kdb Tb)

= 1 / [(Lsb Vn / 3) ((Lab / Vn)]

= 3 / [Lsb Lab]

or

Cb = [3 / (Lsb Lab)]^0.5

Hence:

**Lb** = [3 / Cb]

= {3 / [3 / (Lsb Lab)]^0.5}

= **[3 Lsb Lab]^0.5**

Typically:

Lsb ~ Lsc

and

Lab ~ 3 Lac

giving:

Lb ~ 3 (Lsc Lac)^0.5

= 3 (0.0607 m X 2.901 m)^0.5

= **1.26 m**

__________

**NEUTRON DIFFUSION:**

In order to sustain the nuclear chain reaction in the reactor core while controlling the reaction by thermal expansion the mean fast neutron travel along the Z axis between successive Pu-239 fissions must be greater than the core zone height.
On the web page_________it was determined that a practical core zone height is about 0.35 m._______

The neutron path between successive fissions is inversely proportional to the average fissionable fuel density in the reactor. Hence a major issue in FNR design is maximizing the ratio of metallic core fuel volume to total reactor core volume while maintaining adequate cooling and while minimizing unwanted neutron absorption.

**REACTIVITY MODULATION VIA TEMPERATURE DEPENDENT DIFFUSION:**

**CONSIDER THE DISTANCE Lsc TRAVELLED BY A NEUTRON IN A CORE ZONE BETWEEN SUCCESSIVE SCATTERS:**

**Lsc** = 1 / [Ns Sigmass + Ni Sigmasi + Nu Sigmasu]

= 1 / {Nu [(Ns / Nu) Sigmass + (Ni / Nu) Sigmasi + (Nu / Nu) Sigmasu]}

= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (2.62 b) + (3.9815)(4.6 b) + (1) (9.4 b)]

= 100 m / {41.435 [10.12 + 18.3149 + 9.4]}

= 100 m / {41.435 [37.835]}

= **0.063787 m**

Note that as **Lsc** increases when **Ns, Ni, Nu** decrease. Hence Lsc increases with increasing temperature.

**CONSIDER THE DISTANCE Lac TRAVELLED BY A NEUTRON IN THE REACTOR CORE ZONE BEFORE ABSORPTION:**

**Lac** = 1 / [Ns Sigmas + Ni Sigmaai + Nu Sigmaau + Nu Sigmafu + Np Sigmaap + Np Sigmafp]

= 1 / {Nu [(Ns / Nu) Sigmaas + (Ni / Nu) Sigmaai + (Nu / Nu) Sigmaau + (Nu / Nu) Sigmafu + (Np / Nu) (Sigmaap + Sigmafp)]}

= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (.0014 b) + (3.9815)(.0086 b) + .25 b + .041 b + (0.2857)(.040 + 1.70)]

= 100 m / {41.435 [.00541 + .03424 + .25 + .041 + .497118]}

= 100 m / {41.435 [.82776]}

= **2.915 m**

Note that **Lac** increases when **Ns, Ni, Nu, Np** decrease. Hence Lac increases with increasing temperature.

Hence the number of neutron scatters before absorption is:

**(Lac / Lsc)** = 2.915 m / 0.063787 m

= **45.71**

Since neutron scattering takes place in a 3 dimensional random walk the distance travelled by a neutron along a single axis before absorption is:

[(La / Ls)^0.5 X (Ls / 3^0.5)]

= [(La Ls) / 3]^0.5
= [(2.915 m X 0.063787 m) / 3]^0.5

= **.2489 m**

Note that diffusion distance [(La Ls) / 3]^0.5 increases with increasing temperature.

Note that by comparison the core zone height is 0.35 m and the adjacent blanket thickness is 1.20 m. The diffusion distance is more than half the core zone height. Hence about half of the fission neutrons diffuse out of the core zone into the blanket zone and do not contribute to maintenance of the chain reaction. An increase in temperature decreases **Ns, Ni, Nu, Np** which increases:

**[(La Ls) / 3]^0.5** which increases the fraction Fp of neutrons diffusing out of the core zone and hence not supporting the chain reaction. Hence the chain reaction will stop at a sufficiently high temperature.

The change in sodium temperature only has a limited effect of the net neutron diffusion rate out of the core overlap zone. That change in diffusion rate in combination with the change in fuel atom concentration must be sufficient for stable reactor power control.

USE DEDICATED WEB PAGE FOR NEXT SECTIONAs indicated at FNR FUEL TUBE WEAR at the end of the fuel tube's useful life when the number of plutonium atom fissions equals 0.59 of the number of plutonium atoms initially present:

(Nf / Nu) = 2 (0.59 Np / Np) (Np / Nu)

= 2 (0.59) (.2857) = 0.3371.

**DEFINE Fp:**

Fp = fraction of plutonium fission neutrons that react in the core zone. Then the fraction of fission neutrons that diffuse into the blanket zones is: (1 - Fp)

**NUMERICAL EVALUATION:**

Recall that:

**(Np / Nuc)** = [(Sigmaas Ns / Nuc) + (Sigmaai Ni / Nuc) + (Sigmaac Nc / Nuc) + (Sigmaau) + (Sigmaaz Nz / Nuc) + (Sigmaaf Nf / Nuc) - (Ln(Gu) Sigmafu)] / [(Ln(Gp Fp) Sigmafp) - (Sigmaap)]

= [0.0014 b (3.38037) + 0.0086 b (4.8123) + (0.25 b) + 0.0066 b (0.3727) + (Sigmaaf Nf / Nu) - (0.9555) 0.041 b] / [(Ln(3.1 Fp) 1.70 b) - (0.040 b)]

= [.0047325 + .041386 + 0.25 + 0.0024598 + (Sigmaaf Nf / Nuc-b) - 0.03917] / [1.70 b Ln(3.1 Fp) - 0.040]

= **[0.2594 + (Sigmaaf Nf / Nuc-b)] / [1.70 Ln(3.1 Fp) - 0.040]**

Define:

At start of fuel cycle:

Np = Npi

Nuc = Nuci

Npi / Nuci = 0.2 / 0.7 = 0.2857

Assume 15% burnup. Then at the end of the fuel cycle the cumulative number of fissions is 0.75 Npi.

In an element of time dT when (Sigmafp Np) Pu atoms fission (Sigmaau Nuc) U-238 atoms capture neutrons and form new Pu atoms. Thus:dNuc = - Sigmaau Nuc dT

and

dNp = - Sigmafp Np dT + Sigmaau Nuc dT

or

dNp / dT = - Sigmafp Np + Sigmaau Nuc

or

dNp / (dNuc / - Sigmaau Nuc) = - Sigmafp Np + Sigmaau Nuc

or

dNp / dNuc = (- Sigmafp Np + Sigmaau Nuc) / (- Sigmaau Nuc)

or

dNuc = dNp (- Sigmaau Nuc) / (- Sigmafp Np + Sigmaau Nuc)

The number of fissions in time dT is:

Sigmafp Np dT = Sigmafp Np dNuc / (- Sigmaau Nuc)

Hence:

0.75 Npi = Integral from Np = Npi to Np = Npf of:

{Sigmafp Np dNuc / (- Sigmaau Nuc)}

or

0.75 Npi = Integral from Np = Npi to Np = Npf of:

{Sigmafp Np / (- Sigmaau Nuc)}{dNp (- Sigmaau Nuc) / (- Sigmafp Np + Sigmaau Nuc)

or

0.75 Npi = Integral from Np = Npi to Np = Npf of:

{Sigmafp Np}{dNp / (- Sigmafp Np + Sigmaau Nuc)

or

0.75 Npi = Integral from Np = Npi to Np = Npf of:

{dNp / (- 1 + [(Sigmaau Nuc) /(Sigmafp Np)] )

OR

0.75 Npi = Integral from Np = Npi to Np = Npf of:

{- dNp / ( 1 - [(Sigmaau Nuc) /(Sigmafp Np)] )

To a first approximation:

(- dNp) = (Npi - Npf)

Solution is about: Npi - Npf ~ 0.375 Npi

**PRACTICAL REACTOR:**

Try (Np / Nuc) = (2 / 7) = 0.2857 as in EBR-2. Then:

Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (Np / Nuc)} + 0.040} / 1.70

= Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (0.2857)} + 0.040} / 1.70

= {[0.9479 + 3.50 (Sigmaaf Nf / Nuc-b)} / 1.70

At the start of a fuel cycle Nf = 0 giving:

Ln(3.1 Fp) = 0.9479 / 1.70 = 0.5576

or

3.1 Fp = Exp (0.5576) = 1.746

or

**Fp** = 1.746 / 3.1 = **0.5633**

At the end of the fuel cycle with 15% burnup:

(Nf / Nuc) ~ (0.3 / 0.65) = 0.4615

and

Np / Nuc = (0.125 / 0.65) = 0.1923

giving:

Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (Np / Nuc)} + 0.040} / 1.70

= {{[0.2594 + 0.4615 (Sigmaaf / b)] / 0.1923} + 0.040} / 1.70

= {1.3889 + 2.4 (Sigmaaf / b)} / 1.70

= 0.817 + 1.4117 (Sigmaaf / b)

Hence:

3.1 Fp = Exp[ 0.817 + 1.4117 (Sigmaaf / b)]

or

Fp = Exp[ 0.817 + 1.4117 (Sigmaaf / b)] / 3.1

which for small values of Sigmaaf gives:

**Fp** = 2.263 / 3.1 = **0.7302**

Thus to realize a long fuel cycle time the reactor must be capable of achieving Fp values as high as **~ 0.75**

Hence the core rod length will likely need to be extended. The full core rod length is only used near the end of the fuel cycle. Early in the fuel cycle the active core length is only about 60% of the core rod length.

If the core rods are made too long there will be a problem of too much spontaneous reactivity in the active fuel bundles.

The major issue is ensuring that the various shutdown systems work as intended.

In order to control the nuclear reaction full withdrawal of the mobile fuel bundles must render the reactor subcritical under all conditions. Fully inserting the mobile fuel bundles should make the reactor critical.

This web page last updated February 20, 2017

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