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**INTER ATOMIC SPACING CONTROL**

An important FNR physics issue relating to Fast Neutron Reactors (FNRs) is the mechanism by which the interatomic spacing increases with increasing temperature. The interatomic spacing controls the rate of neuton diffusion out of the core zone of a FNR. When the temperature of the core zone material increases the interatomic spacing increases which increases the rate of neutron diffusion out of the core zone.

In a FNR when the neutron emission rate from fissions is precisely equal to the sum of the neutron absorption rate plus the neutron diffusion rate out of the core zone the reactor is in a steady state condition at its equilibrium temperature. Lowering the material temperature reduces the interatomic spacing which lowers the neutron loss rate by diffusion. Hence the neutron concentration in the core zone increases causing increased fission reactions which release heat. That heat tends to restore the core zone to its equilibrium temperature. If external heat is added to the core zone while the core zone is at or above its equilibrium temperature fission reactions decrease until ultimately no fission reactions can occur.

An increased interatomic spacing at equilibrium corresponds to an increased reactor operating temperature. However, during a reactor fuel geometry change which increases the equilibrium temperature much of the heat associated with an increase in interatomic spacing does not propagate by normal thermal conduction. Instead the heat pulse is transmitted by a pulse of prompt neutrons which momentarily increase the rate of fission reactions throughout the FNR core zone. This effect is possible because in a FNR fission fuel is almost uniformly distributed throughout the core zone. Almost all the liquid sodium atoms are located within about 0.1 inches of Pu atoms, so when a Pu nucleus fissions liberating about 210 MeV of kinetic energy the temperature of more than 300,000 nearby atoms almost instantly increases.

**TIME T BETWEEN SUCCESSIVE FISSIONS:**

A typical fission neutron has a kinetic energy of 2 MeV.

M V^2 / 2 = 2 X 10^6 eV X 1.602 X 10^-19 J / eV

or

V^2 = 4 X (1.602 / M) X 10^-13 J

M = 1.67 X 10^-27 kg

V^2 = 4 X (1.602 / 1.67 X 10^-27 kg) X 10^-13 J

= 4 X 10^14 (m / s)^2

Hence:

V = 2 X 10^7 m / s

The density of Pu-239 is about:

19.86 gm / cm^3

In FNR fuel containing 20% Pu the number of Pu atoms / unit volume is:

6.023 X 10^23 atoms / mole X 1 mole / 239 gm X 19.86 gm / cm^3 X 0.2

= 0.500 X 10^23 atoms / cm^3 X 0.2

= 0.500 atoms X 10^29 / m^3 X 0.2

= 10^28 atoms / m^3 = N atoms / m^3

The neutron fission cross section for Pu in a FNR is:

1.7 barns X 10^-28 m^2 / barn

= 1.7 X 10^-28 m^2 = SigmaPu m^2

N X SigmaPu X V = 1 / T

where T = time between successive fissions

= 1 / [10^28 atoms / m^3 X 1.7 X 10^-28 m^2 X 2 X 10^7 m / s]

= 10^-7 s / [1.7 X 2] atom

= 100 X 10^-9 s / 3.4

= 29.4 ns / fission

However, in Pu there are 3 neutrons, only one of which needs to fission to sustain the nuclear chain reaction, so the average time per fission step is:

29.4 ns / 3 = **9.8 ns / fission step**

**REACTOR OPERATION**

For a fast neutron reactor (FNR) to not become prompt critical on increasing power the net neutron gain must be less than (1 / .9965)/ fission step

Net neutron gain / fission step

= (neutron multiplication/fission step) X (neutron retention fraction)

For plutonium:

(neutron multiplication / fission step) = 3

Typically a reactor is designed so that at core criticality:

(neutron retention fraction) = (1 / 3)

In a FNR out of every 3 neutrons released by a Pu nuclear fission:

1 neutron is required to sustain the nuclear fission chain reaction;

1 neutron is absorbed by U-238 in the core zone;

1 neutron diffuses into the blanket zone

Thus in equilibrium FNR operation the net neutron gain per fission step is:

3 X (1 / 3) = 1.0000, as required for normal reactor criticality.

A normal control change may temporarily change the net neutron gain / fission step to 0.999 to reduce the thermal power or to 1.001 to increase the thermal power.

In a FNR the fraction of delayed neutrons is 0.0035.

In a CANDU reactor the fraction of delayed neutrons is 0.005.

In a light wate reactor the fraction of delayed neutrona is 0.006

In a FNR provided that the net neutron gain / fission step is less than:

1.0035

= a prompt neutron energy pulse will spontaneously decay. Typically the net neutron gain on a normal contol operation is < 1.001 so a prompt neutron pulse decays by about (1 / 2)
(1.001 / 1.0035)^N = 0.5280 at N = 256
in 256 fission steps.

As shown above the time per fission step is about 10 ns. Hence subject to the net neutron gain constraint the propagating prompt neutron pulse will decay by about (1 / 2) in about:

256 fission steps X 10 ns / fission step = 2560 ns = 2.5 us.

**PROMPT ENERGY PULSE**

Assume that at time t = 0 a control action causes a 1 ns injection of 1000 extra neutrons into a FNR via a fission process. Hence at t = 0 there are 996.5 excess prompt neutrons and 3.5 delayed neutrons (that have not yet been released). The cumulative energy release at t = 0 is about:

1000 X 200 MeV / fission

At t = 10 ns as a result of the initial 1000 neutron injection the core contains:

1000 X (.9965)^2 = 993 prompt neutrons + 7 delayed neutrons (that have not yet been released)

The cumulative energy release at t = 10 ns is about:

(1000 + 996.5) X 200 MeV / fission

At t = 20 ns as a result of the initial 1000 neutron injection the core contains:

1000 X (.9965)^3 = 989.5 prompt neutrons + 10.5 delayed neutrons (that have not yet been released).

The cumulative energy release at t = 20 ns is about:

(1000 + 996.5 + 993) X 200 MeV / fission

At t = 30 ns as a result of the initial 1000 neutron injection the core contains:

1000 X (.9965)^4 = 986.07 prompt neutrons + 13.93 delayed neutrons (that have not yet been released).

The cumulative energy release at t = 30 ns is about:

(1000 + 996.5 + 993 + 989.5) X 200 MeV / fission

After 256 X 10 ns = 2.56 us more than half of the neutron population resulting from the initial 1000 neutron injection consists of delayed neutrons that have not yet been released. The cumulative prompt energy release is about:

1000 (1 + a + a^2 + a^3 + a^4 + ... + a^256) X 200 MeV / fission

where a = .9965

(a + a^2 + a^3 + a^4 + ... + a^256)

= a(1 + a + a^2 + a^3 + a^4 + ... + a^255)

= a Sa

Sa = (1 + a + a^2 + a^3 + a^4 + ... + a^255)

(1 + a + a^2 + a^3 + a^4 + ... + a^256) = 1 + a Sa

= Sa + a^256

Hence:

Sa - a Sa = 1 - a^256

or

Sa = (1 - a^256) / (1 - a)

(1 + a + a^2 + a^3 + a^4 + ... + a^256)
1 + a Sa = 1 + [a / (1 - a)](1 - a^256)

= [(1 - a) / (1 - a)] + [a / (1 - a)] - [a^257 / (1 - a)]

= [1 / (1- a)] - [a^257 / (1 - a)]

= [(1 - a^257) / (1 - a)]

=[(1 - a a^256) / (1 - a)]

= [(1 - .9965 (.40755)) / (1 - .9965)]

= 0.59386 / 0.0035

= 169.677

The cumulative prompt energy release is about:

1000 (1 + a + a^2 + a^3 + a^4 + ... + a^256) X 200 MeV / fission

= 1000 (169.677) X 200 MeV / fission

**Note that provided that the net neutron gain / fission is less than 1.0035 this cumulative prompt energy release does not grow geometrically.** As long as the net neutron gain is less than 1.0035 the energy in the prompt energy pulse is finite and is equally distributed over the reactor core fuel atoms. There is no issue of a thermal propagation delay. The energy pulse is evenly distributed through the core fuel by 256 successive fission steps. That energy distribution happens in a time frame of about 2.56 us, long before the delayed neutrons are released. Thus as far as the ~ 600 delayed neutrons are concerned the interatomic spacing has already increased.

There is scatter in the individual neutron delay times so what started out as 1000 neutrons within a one nanosecond long pulse becomes spread over several seconds. This bunch of delayed neutrons will in turn cause another prompt energy release about 3 seconds later. However, by then thermal expansion of the material has taken place which reduces the net neutron gain which reduces the size of subsequent prompt energy releases until they converge to zero.

**SAFE OPERATION:**

The key issue is that the prompt neutron population must spontaneously decay. The energy released by the prompt neutron pulse must be safely absorbed by the core zone thermal mass. If the prompt neutron population grows instead if decaying there will be an explosion.

As long as the net neutron gain is less than 1.0035 the prompt neutron population will drop to zero long before it can release sufficient energy to create an instability or explosion.

Since the change in net neutron gain is dominated by changes in fuel and coolant geometry the fuel and coolant geometry in the core region must be physically stable over a time of about:

1000 X 10 ns = 10 us

Similarly any intended movement of the fuel assembly toward a more critical geometry should be very slow.

The danger lies in a step increase in the net neutron gain larger than 1.0035 / 10 us which means that the prompt neutron population does not rapidly decay. In a FNR it will take at least 256 X 10 ns for the prompt neutron energy pulse to propagate through the core region thus increasing its temperature. More conservatively we should think in terms of:

1000 X 10 ns = 10 us

for that energy pulse to fully reach all the fuel atoms in the core. By comparison the time for acoustic energy propagation is greater than 100 microseconds.

**SYSTEM SAFETY**

In a real system it takes a finite time for the net neutron gain to transition from a safe 1.001 to a dangerous 1.003. For safety we must design the reactor such that it physically can never reach a net neutron gain of 1.0035 / 10 us.

At a net neutron gain of > 1.0035 / 2.7 us an initial prompt neutron injection can grow without limit. This prompt neutron power increase will continue until the reactor changes its geometry by blowing itself apart.

The change in net neutron gain is a result of a change in the physical geometry of the fuel and coolant in the core region. That change in physical geometry takes a finite time to occur. A protective mechanism to prevent it occurring has to work faster. The biggest challenge is a bomb like attack on a FNR that might in a 1 ms time frame change its physical geometry more than can be compensated for by sodium thermal expansion. The neutron growth might be stopped by changing the sodium to a vapor. However, if that much prompt energy is dumped into the liquid sodium it will blow up just like the super heated water at Chernobyl.

**MATHEMATICAL MODEL**

N = N|t = total number of free neutrons in the core region at time t

G = neutron flux gain / fission step ~ 3

Gp = prompt neutron gain which for a FNR = G (.9965)

Gd = delayed neutron gain which for an FNR is G(.0035)

Fr = neutron flux retention fraction per fission step ~ (1 / 3)

At any instant in time the total number of neutrons N is the sum of the prompt neutrons from recent fissions plus neutrons from delayed fissions less neutrons absorbed or emitted from the core zone.

Nf = fission fuel atom concentration

Sigma = fission cross section

V = neutron velocity

**Change in Neutron Population:**

dN / dt = N|t Nf Sigma V Gp + N|(t - Td) Nf Sigma V Gd - N|t K

where:

Td = a typical delayed neutron time delay

= 3 seconds

**At steady state:**
N|t = N|(t - Td)

or

dN / dt = 0

= N|t Nf Sigma V (Gp + Gd) - N|t K = 0

or

Nf Sigma V (Gp + Gd) - K = 0

or

(Gp + Gd) = K / (Nf Sigma V)

Hence at steady state:

Fr = 1 / (Gp + Gd) = (Nf Sigma V) / K

Gd / G = Gd / (Gp + Gd) = .0035

dN / dt = N|t Nf Sigma V Gp + N|(t - Td) Nf Sigma V Gd - N|t K

At steady state dN / dt = 0 giving:

N|(t - Td) Nf Sigma V Gd] = [N|(t = 0) K] - [N|(t = 0)] Nf Sigma V Gp

= [N|(t = 0)] [K - Nf Sigma V Gp]

Hence if the system is initially at steady state:

dN / dt = N|t Nf Sigma V Gp + N|(t - Td) Nf Sigma V Gd - N|t K

= N|t Nf Sigma V Gp +[N|(t = 0)] [K - Nf Sigma V Gp] - N|t K

= N|t [Nf Sigma V Gp - K] + [N|(t = 0)] [K - Nf Sigma V Gp]

= [N|t - N|(t = 0)] [Nf Sigma V Gp - K]

**If the initial state is at equilibrium** then:

K = Nf Sigma V (Gp + Gd)

giving:

dN / dt = [Nt - N|(t = 0)] [Nf Sigma V Gp - K]

= [Nt - N|(t = 0)] [- Nf Sigma Gd]

Define:

Z = {N|t - [N|(t = 0)]}

Then:

dZ / dt = dN / dt

Hence:

dZ / dt = Z [- Nf Sigma V Gd]

or

dZ / Z = [- Nf Sigma V Gd] dt

or

Ln[Zb / Za] = [- Nf Sigma V Gd] (tb - ta)

or

Zb = Za Exp{[- Nf Sigma V Gd] (tb - ta)}

Then:

{N|(t = tb) - [N|(t = 0)]}

= {N|(t = ta) - [N|(t = 0)]} Exp{[- Nf Sigma V Gd] (tb - ta)}

Gd = .0035 G

and

G = 3

Recall that:

1 / (Nf Sigma V) = 30 ns

Hence:

{[- Nf Sigma V Gd] (tb - ta)} = (-.0035 (3) / 30 ns)(tb - ta)

= - .00035 (tb - ta) / 1 ns

At {[- Nf Sigma V Gd] (tb - ta)} = - 1

or

(tb - ta) = 1 /[Nf Sigma V Gd]

= 30 ns / [3 (.0035)]

= (10 ns / .0035)

= 2857 ns

= 2.857 us

Hence in a FNR initially at equilibrium the decay time constant for a small step transient is about 3 us.

The danger lies in a non-equilibrium initial condition where G F > 1. In this case the number of neutrons N can rapidly grow.

Hence with a FNR we must change the fuel geometry very slowly so that the fuel temperature remains close to its equilibrium value.

The rate of fission power related to heating of the FNR thermal mass must not exceed the heat transfer capacity of the FNR fuel tubes.

This web page last updated March 28, 2019

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