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XYLENE POWER LTD.

FNR PRIMARY SODIUM FLOW

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

INTRODUCTION:
The natural circulation primary liquid sodium flow limits the FNR thermal power output. This web page develops a closed form solution for the FNR primary liquid sodium flow at full power. It is shown that the contemplated FNR can potentially operate at 1000 MWt at a primary liquid sodium discharge temperature of 440 C and a primary liquid sodium inlet temperature of 340 C in the presence of 15% linear fuel tube swelling. However, this power may have to be reduced early in the fuel life due to a limited amount of fuel bundle overlap which causes only a fraction of the core rods to be active. As the core rod Pu concentration decreases the fuel bundle overlap will be increased to maintain reactor reactivity. That overlap inserts more core rod length into the middle core zone which increases the fuel tube surface area where heat dissipation can occur.
 

PRIMARY SODIUM NATURAL CIRCULATION:
The natural circulation of the primary liquid sodium occurs due to a decrease in liquid sodium density with increasing temperature. Nuclear heating of the sodium in the active fuel bundles causes the sodium to locally expand and rise. The hot sodium flows out of the tops of the active fuel bundles, over the top surface of the primary sodium pool toward the side of the pool where it cools and contracts as it flows down between the intermediate heat exchange tubes. The higher density cooled liquid sodium flows along the bottom of the primary liquid sodium pool to the pool bottom center where it again rises due to heating by the active reactor fuel tubes.

In order to naturally circulate there must be a large temperature difference between the top and bottom of the liquid sodium pool. The sodium contained in the active cooling channels is normally about 440 degrees C, but may be as much as 473 degrees C adjacent to an isolated coolant channel blockage. The hot sodium discharge from the active fuel bundles keeps the primary sodium from the sodium top surface down to the tops of the active fuel bundles at about 440 degrees C, except close to the pool walls where the primary liquid sodium temperature is reduced by the intermediate heat exchangers. The balance of the liquid sodium is at a variable temperature that varies from 450 degrees C at no load to a theoretical minimum of 340 degrees C at full load.

The maximum height of the column of hot sodium in the active fuel bundles that drives the natural circulation is:
[6 m - 1.8 m] = 4.2 m

The temperature difference between the hot liquid sodium in the active fuel bundles and the cooler sodium surrounding the assembly of fuel bundles drives the primary sodium natural circulation.

Consider a column of primary liquid sodium coolant of uniform cross sectional area Ac and height Zo. The density of the liquid sodium in the column is Rhos(Z) where Z indicates the vertical position in the column with respect to the bottom of the fuel tube bundle. Since the fuel tube bundle is immersed in the cool portion of the primary liquid sodium pool the density of primary liquid sodium outside the fuel tube bundle is uniform at Rhoo. Let g = 9.8 m / s^2 be gravitational acceleration. The net buoyancy force Fg causing discharge of liquid sodium from the coolant column is:
Fg = Integral from Z = 0 to Z = Zo of [Ac g (Rhoo - Rhos) dZ]

The corresponding net gravitational buoyancy pressure Pg measured at the bottom of the column is:
Pg = Fg / Ac
= Integral from Z = 0 to Z = Zo of [g (Rhoo - Rhos) dZ]

Note that for a tall thin column of a thermally conductive material such as sodium to a good approximation the sodium temperature depends on vertical position in the column but is independent of radial position in the column.

For liquid sodium the fractional change in density with temperature is about [2.71 X 10^-4 / deg C], which is exceptionally large for a metal.
 

Hence:
(Rhoo -Rhos) = [Rhoo - (Rhoo - [271 X 10^-6 / C] Rhoo (T(Z) - To))]
= [271 X 10^-6 / C] Rhoo (T(Z) - To)

For the bottom 1.8 m of the fuel tube:
(T(Z) - To)) = 0

For the next 0.50 m of the fuel tube:
(T(Z) - To)) = 50 deg C

For the next 3.7 m of the fuel tube:
(T(Z) - To)) = 100 deg C

Hence:
Pg = Integral from Z = 0 to Z = Zo of [g (Rhoo - Rhos) dZ]
= Integral from Z = 0 to Z = Zo of:
g (271 X 10^-6 / C) Rhoo (T(Z) - To) dZ
= g [271 X 10^-6 / C] Rhoo [(0 C)(1.8 m) + (50 C)(0.5 m) + (100 C)(3.7 m)]
= g [271 X 10^-6 / C] Rhoo [50 C (7.9 m)]
= g [271 X 10^-6 / C] Rhoo [395] C-m
= g (0.107045 m) Rhoo

Pg = g (0.107045 m) Rhoo
= (9.8 m / s^2) (0.107045 m)(927 kg / m^3)
= 972.461 kg / s^2-m

This pressure is balanced by the change in liquid sodium fluid kinetic energy and by the viscous force.

Let Vc be the liquid sodium vertical flow velocity which in general is a function of radial position.

Let Vco = liquid fluid flow velocity at the center of the fluid column where the viscous effect is zero.

For the ideal case of zero viscous force in the center of the fluid column:
Vc = Vco and an element of kinetic fluid power = (Rhoo Vco Acc)(Vc^2 / 2)

Power applied to element of fluid = (Pg Ac) Vc = Pg dFv
Hence assuming conservation of energy and no viscous force:
(Pg Ac) Vco = Rhoo Vco Ac (Vco^2 / 2)
or
Pg = Rhoo Vco^2 / 2
or
Vco^2 = 2 Pg / Rhoo
= (2 / Rhoo) g (0.107045 m) Rhoo
= 2 g (0.107045 m)
= 2 (9.8 m / s^2)(0.107045 m)
= 2.0980 (m / s)^2
or
Vco = 1.4484 m / s

Hence due to viscous force except at R = 0 (at the center of the fluid column) Vc(R) will be always be less than 1.4484 m / s.
 

VISCOUS FORCE:
Pressure Pg creating liquid sodium kinetic energy in the fluid column is effectively reduced by the viscous pressure drop Pv. Find the viscous force for a cylindrical pipe:
F = viscous force
Muv = dynamic viscosity of liquid sodium = 7 X 10^-4 kg / m-s
Ap = pipe inside surface area
Vc = column velocity
R = radius from center of fluid column
Ro = outside radius of the liquid sodium fluid column
dPv = pressure difference over fluid column length dZ due to viscous force

Assume laminar flow. The driving pressure is Pg which is independent of radius R.

For each extended cylinder of fluid of wall thickness dR: Pg = Pk + Pv
where:
Pk - kinetic pressure
and
Pv = viscous pressure

In the center of the fluid column:
Pv = 0
and
Pg = Pk

At the pipe wall Vc = 0 so that Pk = 0. Hence at R = Ro: Pg = Pv

Thus the boundary conditions are:
Pv = Pg at R = Ro
and
Pv = 0 at R = 0

We will attempt to find a consistent solution of the form:
Pv = Pg [R / Ro]^N
which satisfies both boundary conditions.

From definition of viscosity:
F = Muv Ap (-dVc / dR)|R = R
= (kg / m-s) m^2 (m / s-m)
= kg m / s^2 = force

For a straight round pipe:
Ap = 2 Pi R Zo
and
F = Integral from R = 0 to R = Ro of:
Pv(R) 2 Pi R dR

Hence:
Integral from R= 0 to R = Ro Pv(R) 2 Pi R dR
= Muv (2 Pi R Zo) (-dVc / dR)|R = R
or
Pv(R) 2 Pi R = Muv (2 Pi Zo) d {R (-dVc / dR)] / dR
or
Pv = - Muv Zo (1 / R) d[R (dVc / dR)] / dR

Recall that:
Pv = Pg [R / Ro]^ N

Thus:
Pg [R / Ro]^ N = - Muv Zo (1 / R) d[R (dVc / dR)] / dR

Try:
Vc = Vco - K [R / Ro]^M
or
(dVc / dR) = - (M K / Ro) [R / Ro]^(M-1)
or
[R (dVc / dR)] = - (M K) [R / Ro]^M or
d[R (dVc / dR)] / dR = - (M^2 K / Ro) [R / Ro]^(M - 1)
or
Pg [R / Ro]^ N = - Muv Zo (1 / R) d[R (dVc / dR)] / dR
= Muv Zo (1 / R) (M^2 K / Ro) [R / Ro]^(M - 1)
= Muv Zo (Ro / R) (M^2 K / Ro^2) [R / Ro]^(M - 1)
= Muv Zo (M^2 K / Ro^2) [R / Ro]^(M - 2)

For this equation to be valid for all R values:
M - 2 = N
and
Pg = Muv Zo (M^2 K / Ro^2)
= Muv Zo ((N + 2)^2 K / Ro^2) or
K = Pg Ro^2 / [Muv Zo (N + 2)^2]

Hence:
Vc = Vco - K [R / Ro]^M
= Vco - {Pg Ro^2 / [Muv Zo (N + 2)^2]} [R / Ro]^(N + 2)

Apply boundary condition that:
[Vc|R = Ro] = 0
to get:
Vco = {Pg Ro^2 / [Muv Zo (N + 2)^2]}

Hence:
Vc = {Pg Ro^2 / [Muv Zo (N + 2)^2]} - {Pg Ro^2 / [Muv Zo (N + 2)^2]} [R / Ro]^(N + 2)
= {Pg Ro^2 / [Muv Zo (N + 2)^2]} {1 - [R / Ro]^(N + 2)}

The volumetric flow rate Fv through the pipe is given by:
Fv = Integral from R = 0 to R = Ro of:
2 Pi R Vc dR
= Integral from R = 0 to R = Ro of:
{2 Pi R Pg Ro^2 / [Muv Zo (N + 2)^2]} {1 - [R / Ro]^(N + 2)} dR
= {2 Pi Pg Ro^2 / [Muv Zo (N + 2)^2]}[Ro^2 / 2] - {2 Pi Pg Ro^2 / [Muv Zo (N + 2)^2]}[1 / Ro]^(N + 2) [Ro^(N + 4)] / (N + 4)]
= {2 Pi Pg Ro^2 / [Muv Zo (N + 2)^2]} {[Ro^2 / 2] - [Ro^2 / (N + 4)]}
= {2 Pi Pg Ro^4 / [Muv Zo (N + 2)^2]} {[1 / 2] - [1 / (N + 4)]}
= {2 Pi Pg Ro^4 / [Muv Zo (N + 2)^2]} {[N + 4 - 2] / [2 (N + 4)]}
= {2 Pi Pg Ro^4 / [Muv Zo (N + 2)]} {1 / [2 (N + 4)]}
= {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}

This equation can be used to find the natural circulation volumetric fluid flow Fv per round coolant flow channel.
Fv = {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}
 

DETERMINATION OF N:
At R = 0:
Pg = (Rhos / 2) Vco^2
= (Rhos / 2){Pg Ro^2 / [Muv Zo (N + 2)^2]}^2
or
(N + 2)^4 = (Rhos / 2)(1 / Pg){Pg Ro^2 / [Muv Zo]}^2
= (Rhos / 2)(Pg) Ro^4 / [Muv Zo]^2
or
(N + 2) = Ro {(Rhos Pg) / (2 [Muv Zo]^2)}^0.25
= Ro [(Rhos Pg) / 2]^0.25 [1 / [Muv Zo]^0.5]
 

FUEL TUBE ARRAY:
The fuel tubes are initially 0.500 inch OD and are in a rectangular array 0.625 inch center to center. Between each group of 4 fuel tubes is an approximately circular liquid sodium flow channel of diameter:
[(.625 inch)^2 + (0.625 inch)^2 ]^0.5 - 0.5 inch = 0.3838834765 inch

After 10% linear fuel tube swelling the flow channel diameter is:
[(.625 inch)^2 + (0.625 inch)^2 ]^0.5 - 0.55 inch = 0.3338834765 inch

The maximum fuel tube swelling that the present fuel tube plenum will support is 15% linear. At 15% linear fuel tube swelling the sodium flow channel diameter is:
[(.625 inch)^2 + (0.625 inch)^2 ]^0.5 - 0.5(1.15) inch = 0.3088834765 inch

PRIOR TO FUEL TUBE SWELLING:
Ro = (0.3838834765 inch / 2) X .0254 m / inch
= 4.87532 X 10^-3 m
 

AFTER 10% LINEAR FUEL TUBE SWELLING:
Ro = (0.3338834765 inch / 2) X .0254 m / inch
= 4.24032 X 10^-3 m
 

AFTER 15% LINEAR FUEL TUBE SWELLING:
Ro = (0.3088834765 inch / 2) X 0.0254 m / inch
= 3.92282 X 10^-3 m
 

NUMERICAL EVALUATION OF OTHER PARAMETERS:
Length over which viscous drag occurs:
Zo = 6.0 m

Pg = 972.461 kg / s^2-m

Muv = 7 X 10^-4 kg / m-s

[(Rhos Pg) / 2]^0.25 = [(927 kg / m^3) (972.461 kg /m-s^2) / 2]^0.25
= [450,735.6 kg^2 / m^4-s^2]^0.25
= [671.368 (kg / m^2-s)]^0.5
= 25.911 (kg / m^2-s)^0.5

[1 / [Muv Zo]^0.5] = [1 / [( 7 X 10^-4 kg / m-s) 6.0 m]^0.5]
= (100 / 6.4807) (s / kg)^0.5
= 15.4304 (s / kg)^0.5
 

PRIOR TO FUEL TUBE SWELLING:
(N + 2) = Ro [(Rhos Pg) / 2]^0.25 [1 / [Muv Zo]^0.5]
= 4.87532 X 10^-3 m X 25.911 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5
= 1.949

(N + 4) = 3.949

Fv = {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}
= {[Pi X 972.461 kg / s^2-m X (4.87532 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.949)(3.949) channel]}
= 5339.32 X 10^-8 m^3 / s-channel
= 0.5339 X 10^-4 m^3 / s-channel

Total reactor flow = [330,016 channels X 0.5339 X 10^-4 m^3 / s-channel]
= 17.621 m^3 / s

For the assumed 100 deg C temperature differential the corresponding maximum potential reactor heat output is:
17.621 m^3 / s X 927 kg / m^3 X 100 deg K X 1.26 kJ / kg deg K
= 2,058,124 kJ / s
= 2,058 MWt

This is the theoretical maximum possible FNR thermal output capacity prior to any fuel tube swelling assuming a primary sodium temperature that varies from 390 C to 490 C. In reality the thermal power will be less because of the fuel tube heat transfer limitation in the core zone. However, this calculation suggests that subject to neutron diffusion length considerations we should consider making the core overlap zone wider so as to increase the active core zone fuel tube area and thus increase the reactor power.

What will happen in practice is that the temperature difference between the top and bottom of the primary sodium pool will fall until the heat generation rate equals the heat extraction rate. The heat extraction rate is limited by the thermal flows through the steam generators, which in turn are limited by the secondary sodium flow.

Even if the heat exchangers are ideal the system power is limited by:
(reactor setpoint temperature - water temperature in the steam generator) X (secondary sodium flow) X Cp.

The bottom of primary sodium pool temperature will rise to make the rate of heat extraction from the fuel assembly equal the rate of heat extraction from the primary sodium pool. Thus the actual temperature rise across the fuel assembly may only be about 40 degrees C.


 

AFTER 10% LINEAR FUEL TUBE DIAMETER SWELLING:
Ro = 4.24032 X 10^-3 m

(N + 2) = Ro [(Rhos Pg) / 2]^0.25 [1 / [Muv Zo]^0.5]
= 4.24032 X 10^-3 m X 25.911 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5
= 1.695

(N + 4) = 3.695

Fv = {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}
= {[Pi X 972.461 kg /m-s^2 X (4.24032 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.695)(3.695) channel]}
= 3754.768 X 10^-8 m^3 / s-channel

Total reactor flow = [330,016 X 0.3754 X 10^-4 m^3 / s-channel]
= 12.391 m^3 / s

The corresponding maximum potential reactor heat output is:
12.391 m^3 / s X 927 kg / m^3 X 100 deg K X 1.26 kJ / kg deg K
= 1,447,293 kJ / s
= 1,447 MWt
 

AFTER 15% LINEAR FUEL TUBE DIAMETER SWELLING:
Ro = 3.92282 X 10^-3 m

(N + 2) = Ro [(Rhos Pg) / 2]^0.25 [1 / [Muv Zo]^0.5]
= 3.92282 X 10^-3 m X 25.911 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5
= 1.568

(N + 4) = 3.568

Fv = {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}
= {[Pi X 972.461 kg /m-s^2 X (3.92282 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.568)(3.568) channel]}
= 3079 X 10^-8 m^3 / s-channel

Total reactor sodium flow = [330,016 X 0.3079 X 10^-4 m^3 / s-channel]
= 10.160 m^3 / s

The corresponding maximum potential reactor heat output is:
10.160 m^3 / s X 927 kg / m^3 X 100 deg K X 1.26 kJ / kg deg K
= 1,186,796 kJ / s
= 1,187 MWt
 

This calculation indicates that the reactor power is not limited by the natural circulation flow rate. Subject to other constraints it appears that the reactor power is limited by the active area of the core fuel tubes and subject to reactor power stability can potentially be increased by increasing the core fuel rod length. With the contemplated reactor geometry the limited thermal power transfer capacity of the fuel tubes in the core region will be the reactor power limiting factor. As the fuel tubes age the temperature rise across the fuel assembly will approach 100 degrees C.
 

This calculation suggests that the core fuel rods should perhaps be made ~ 2 X 0.35 m long.

This web page last updated March 12, 2021.

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