# XYLENE POWER LTD.

## FNR MATHEMATICAL MODEL

#### By Charles Rhodes, P.Eng., Ph.D.

OVERVIEW:
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 power control system, including thermal expansion, to respond. The safety case lies in mathematical certainty that the with the chosen materials and dimensions the FNR will always be 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 at the start of a fuel cycle and at the end of a fuel cycle. At the start of a fuel cycle the active fuel bundle control portion only partially overlaps the active fuel bundle surround portion. At the end of a fuel cycle the active fuel bundle control portion fully overlaps the active fuel bundle surround portion. The overlap distance contros the fraction of fission neutrons that remain in the core zone. The core fuel rods must be sufficiently long to meet the requirement of the maximum overlap.

A FNR inherently operates with its core zone chain reaction on the threshold of criticality. Whether or not the core zone is critical depends on the fraction of fission neutrons that diffuse from the core zone into the adjacent blanket zones. For safety at the FNR operating point this fraction should rapidly increase with increasing FNR 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 99.8% of these neutrons are prompt and are rekeased at the instant of fission. The remaining 0.2% of these neutrons are delayed and are released about 3 seconds later by fission products. 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 plutonium concentration. A fractional neutron is lost to non-productive neutron absorption by steel, sodium, fission products and zirconium in the core zone. The rest of the neutrons should diffuse into the adjacent blanket zones where they are absorbed by U-238, steel, sodium and zirconium.

From a reactor stability perspective the key issue is criticality in the core zone. If the core zone goes critical with respect to prompt neutrons the reactor will explode. Precise power control is realized via the delayed neutrons. The rate of core zone reactivity increases must be limited to allow the delayed neutron flux to stay in phase with the prompt neutron flux. Hence FNR fuel geometry changes that potentially increase reactivity should be made very slowly.

If a change in the fuel geometry is made which increases the core zone reactivity 99.8% of the neutron flux will respond almost instantly but the remaining neutron flux will take about 3 seconds to respond. Thus 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 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 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 active fuel bundle control portion fuel has the effect of shifting part of the U-238 absorption mass from a blanket zone into the core zone. Active fuel bundle control portion insertion and withdrawal is used to compensate for the effect of changing core zone composition on the reactor discharge temperature setpoint and for achieving a reactor cold shutdown.

In FNR operation the exact fraction of fission neutrons that diffuse from the core zone into the blanket zone is determined by the reactor material temperature. For safety at the operating point the increase in this fraction with increasing temperature should be as large as possible. 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 rod alloy, core rod dimensions, blanket rod alloy and blanket rod dimensions. The formulae also allow calculation of the amount of blanket rod material that must be inserted into a core zone to lower the reactor discharge temperature setpoint by a specified amount.

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. The effect of rising temperature in the FNR is to increase the fraction of fission neutrons that are absorbed within the blanket zones.

The core zone height 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.

Define:
Lac = distance along the path of a neutron in the core zone to point of absorption;
Lsc = distance along the path of a neutron in the core zone between successive scatters;
Lab = distance along the path of a neutron in the blanket zone to point of absorption;
Lsb = distance along the path of a neutron in the blanket zone between successive scatters;
Lb = length of blanket zone
Lc = length 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 core zone - blanket zone junction
Z = (Lc / 2) + Lb at the outer end of the blanket zone
N(Z) = neutron concentration as a function of Z
N(Z) = relative maximum at Z = 0
d[N(Z)]/ dZ = 0 at Z = 0
N(Z) = minimum 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 number of plutonium atoms per unit volume in core zone (Note that Np decreases by a factor of 2 during the fuel cycle)
Ni = average number of iron in atoms per unit volume both the core zone and the blanket zone
Ncr = average number of chromium in atoms per unit volume both the core zone and the blanket zone
Nsc = average number of sodium atoms per unit volume in the core zone
Nsb = average number of sodium atoms per unit volume in the blanket zone
Nuc = average number of U-238 atoms per unit volume in the core zone
Nub = average number of U-238 atoms per unit volume in the blanket zone
Nzc = average number of zirconium atoms per unit voume in the core zone
Nzb = average number of zirconium atoms per unit voume 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 number of fission product atoms per unit volume 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 sodium = .927 gm / cm^3
Rhoi = mass density of iron = 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
Aws = atomic weight of sodium = 23
Awi = atomic weight of iron = 55.845
Awu = atomic weight of uranium = 238
Awz = atomic weight of zirconium = 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 dimensions 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.

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 decrease across the barrier

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 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 neutron flux through the 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 poaaition in the core zone

- Kdb A (dN / dZ) = net neutron diffusion flux in the Z direction 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)

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

OK TO HERE? Tan(Cd Lc / 2)] = [( Cd Tc / Cb Tb)] [1 - Exp(- Cb Lb)] / {1 + Exp[- Cb Lb]}

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

Lc is proportional to (1 / Nu)^0.3333
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 = 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
__________

This web page last updated February 20, 2017