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**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 FNR can operate at full rated power (1000 MWt) at a primary liquid sodium discharge temperature of 445 C and a primary liquid sodium inlet temperature of 350 C in the presence of more than 15% linear fuel tube swelling.

**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 active fuel bundle chimneys, over the top surface of the primary sodium pool toward the side of the pool where it cools as it flows down between the intermediate heat exchange tubes, contracts and sinks. The higher density cooled liquid sodium flows along the bottom of the pool back to the pool bottom center where it again rises due to heating by the 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 fuel bundles is normally about 440 degrees C. The hot sodium discharge from the active fuel bundle chimneys keeps the primary sodium from the sodium top surface down to the tops of the active fuel bundle chinmeys at about 440 degrees C, except very close to the pool walls where the sodium temperature is reduced by the intermediate heat exchanger. The balance of the liquid sodium is at a variable temperature that varies from 440 degrees C at no load to 330 degrees C at full load.

The height of the column of hot sodium in the active fuel bundles that drives the natural circulation is:

[(6 m - 1.6 m - (0.5)(.35 m)) + 3 m] = 7.2 m

The temperature difference between the hot liquid sodium in the active fuel bundles and the cooler sodium in the lower portion of the primary liquid sodium pool 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 1.6 m of the fuel tube:

(T(Z) - To)) = 0

For the next 0.40 m of the fuel tube:

(T(Z) - To)) = 47.5 C

For the next 4.4 m of the fuel tube:

(T(Z) - To)) = 95 C

For the 3 m high chimney:

(T(Z) - To)) = 95 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)(1.6 m) + (47.5 C)(0.4 m) + (95 C)(4.4 m)] + (95 C) (3.0 m)

= g [271 X 10^-6 / C] Rhoo [95 C (.2 m) + 95 C (4.4 m) + 95 C (3 m)]

= g [271 X 10^-6 / C] Rhoo [722] C-m

= **g (0.195622 m) Rhoo**

**Pg** = g (0.195622 m) Rhoo

= (9.8 m / s^2) (0.195622 m)(927 kg / m^3)

= **1777.51 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 (0.118427 m) Rhoo

= 2 g (0.195622 m)

= 2 (9.8 m / s^2)(0.195622 m)

= 3.83419 (m / s)^2

or

**Vco = 1.958 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 1.958 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** = **1777.51**

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

FIX FROM HERE ONWARDS- fix Pg

[(Rhos Pg) / 2]^0.25 = [(927 kg / m^3) (1075.86 kg /m-s^2) / 2]^0.25

= [498,661 kg^2 / m^4-s^2]^0.25

= 26.5736 (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 26.5736 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5

= **1.9990**

**(N + 4) = 3.9990**

= {[Pi X 1075.86 kg / s^2-m X (4.87532 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.9990)(3.9990) channel]}

= 5687.275 X 10^-8 m^3 / s-channel

=

Total reactor flow = [532 active bundles X 544 channels / bundle X 0.56873 X 10^-4 m^3 / s-channel]

= **16.4594 m^3 / s**

Assume a 95 deg C temperature differential.

The corresponding maximum potential reactor heat output is:

16.4594 m^3 / s X 927 kg / m^3 X 95 deg K X 1.26 kJ / kg deg K

= 1,826,369.73 kJ / s

= **1,826 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 350 C to 445 C. In reality the thermal power will be less because of the fuel tube heat transfer limitation in the core zone.

**AFTER 10% LINEAR FUEL TUBE DIAMETER SWELLING:**

**Ro** = **4.24032 X 10^-3 m**

= 4.24032 X 10^-3 m X 26.5736 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5

=

**(N + 4) = 3.7387**

= {[Pi X 1075.86 kg /m-s^2 X (4.24032 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.7837)(3.7837) channel]}

= 3854.89 X 10^-8 m^3 / s-channel

Total reactor flow = [532 core bundles X 544 channels / bundle X 0.385489 X 10^-4 m^3 / s-channel]

= **11.1563 m^3 / s**

The corresponding maximum potential reactor heat output is:

11.1563 m^3 / s X 927 kg / m^3 X 95 deg K X 1.26 kJ / kg deg K

= 1,237,930 kJ / s

= **1,238 MWt**

**AFTER 15% LINEAR FUEL TUBE DIAMETER SWELLING:**

**Ro** = **3.92282 X 10^-3 m**

= 3.92282 X 10^-3 m X 26.5736 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5

=

**(N + 4) = 3.6085**

**Fv = {Pi Pg Ro^4 / [Muv Zo (N + 2)(N + 4)]}**

= {[Pi X 1075.86 kg /m-s^2 X (3.92282 X 10^-3 m)^4] / [( 7 X 10^-4 kg / m-s)(6.0 m)(1.6085)(3.6085) channel]}

= 3283.23 X 10^-8 m^3 / s-channel

Total reactor flow = [532 core bundles X 544 channels / bundle X 0.328323 X 10^-4 m^3 / s-channel]

= **9.5019 m^3 / s**

The corresponding maximum potential reactor heat output is:

9.5019 m^3 / s X 927 kg / m^3 X 95 deg K X 1.26 kJ / kg deg K

= 1,054,353 kJ / s

= **1,054.4 MWt**

**CLEARLY LIMITING THE LINEAR FUEL TUBE DIAMETER SWELLING TO 15% OF THE INITIAL FUEL TUBE OD IS AN IMPORTANT ISSUE IN FNR ECONOMICS.** When the fuel tube linear expansion reaches about 17% it will become a reactor power limiting factor. At lesser values of fuel tube linear swelling the limited thermal power transfer capacity of the fuel tubes in the core region will be the reactor power limiting factor.

This web page last updated February 21, 2017.

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