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**INTRODUCTION:**

This web page develops a closed form solution for the natural convection FNR liquid sodium flow rate and hence thermal power as a function of sodium differential temperature. It is assumed that the return sodium to the fuel assemby is well mixed. It is shown that the contemplated FNR can operate at 1000 MWt at a liquid sodium discharge temperature of 460 C and a liquid sodium return temperature to the fuel bundles of about 410 C.______ However, the FNR power should be regulated by controlling the NaK flow along with the nitrate salt flow. From time to time the low load sodium discharge temperature setpoint should be adjusted by adjusting the insertion of the movable fuel bundles into the matrix of fixed fuel bundles.

**SODIUM NATURAL CIRCULATION:**

Natural circulation of the 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. This thermal expansion causes a decrease in sodium density which drives natural sodium circulation. The hot sodium flows out of the tops of the active fuel bundles, over the top surface of the sodium pool flowing toward the perimeter of the pool where it encounters cooler intermediate heat exchange tubes. This sodium cools and contracts as it flows down beside and between the intermediate heat exchange bundle tubes. The higher density cooled liquid sodium flows along the bottom of the liquid sodium pool to the pool bottom center where it mixes with warmer sodium and then again rises due to heating by the active reactor fuel tubes.

In order to naturally circulate there must be a temperature difference between the top and bottom of the fuel tubes. At full load the sodium contained in the active cooling channels normally rises from about 410 degrees C to about 460 degrees C, but might reach as much as 510 degrees C adjacent to a dual isolated cooling channel blockage. The hot sodium discharge from the active fuel bundles keeps the sodium region from the tops of the fuel tubes up to the sodium pool top surface at about 460 degrees C, except close to the pool walls where the liquid sodium temperature is reduced by the intermediate heat exchangers. The balance of the liquid sodium is at a variable temperature that varies from 460 degrees C after prolonged operation at no load down to a theoretical minimum of 410 degrees C at full rated load.

However, the cold sodium discharge temperature from the intermediate heat exchange bundles may be a cold as 340 degrees C. The FNR relies on mising of recirculated 460 degree C sodium with 340 degree C sodium to get 410 degree C sodium at the bottom of the fuel tubes.

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 - 0.30 m] = 3.9 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 natural sodium circulation. Note that the cool sodium discharge from the intermediate heat exchange bundles is level with the top of the fuel tubes.

The fission thermal power output of the FNR is set by the product of the liquid sodium mass flow and the liquid sodium differential temperature across the fuel assembly. The liquid sodium discharge temperature is typically set at 460 degrees C in order to keep the fuel center line temperature under 600 degrees C, even with two adjacent sodium cooling channels blocked. The rational for these temperature related decisions is set out at FNR Temperature Setpoint.

The chosen steam pressure of about 10 MPa sets the saturated steam temperature at 310 degrees C which in turn sets the coldest full load nitrate salt temperature at about 320 degrees C. The corresponding coldest full load NaK temperature is about 330 degrees C and the corresponding full load sodium return temperature is about 340 degrees C. This 340 deg C sodium mixes with 460 deg C sodium in the FNR sodium pool to give 410 deg C sodium at the bottom input to the FNR fuel assembly.

As shown on the web page: FNR Temperature Setpoint under full load conditions the FNR fuel assembly inlet temperature should be about 410 degrees C.

Note that with a fixed pressure in the steam generators of 10 MPa at full load the coldest nitrate salt temperature is about 320 degrees C and the coldest NaK temperature is about 330 degrees C which is sufficient to prevent NaOH deposits in the NaK loop that can occur below 318 degrees C. However, this temperature is not high enough to prevent KOH deposits in the NaK loop which can occur at NaK temperatures below 360 degrees C. Fortunately, such deposits can be cleared by operating the FNR for a short time at part load with the NaK return temperature setpoint raised from 340 dgrees C to above 360 degrees C.

As set out at FNR Temperature Setpoint the maximum full load sodium temperature differential temperature across the fuel assembly is about:

460 degrees C - 410 degrees C = 50 degrees C

The fission thermal power is:

(sodium differential temperature) X (sodium volumetric flow rate) X (sodium density) X (sodium heat capacity)

The sodium volumetric flow rate = flow cross section X average sodium axial flow velocity

From the web page titled: FNR Temperature Setpoint the maximum permisible temperature rise across the fuel assembly is 60 deg C.

As shown at FNR Fuel Bundle the total open area for sodium flow is **42.918 m^2**. In order to achieve a thermal output of 1000 MWt when the sodium temperatue rise is 40 degrees C the minimum required sodium average axial flow velocity in the active zone is given by:

10^6 kWt = (Minimum average axial velocity) X 40 deg C X 42.918 m^2 X 840 kg / m^3 X 1.26 kJ / Kg deg C

or

(Minimum average axial sodium flow velocity)

= 10^6 kWt / [40 deg C X 42.918 m^2 X 840 kg / m^3 X 1.26 kJ / kg deg C X 1 kWt-s / kJ]

= **0.55036 m / s**

The actual axial sodium flow velocity near the slot walls will be reduced by the sodium viscosity. However, as a first cut we can easily find the average axial sodium velocity when the sodium viscosity is zero.

A simple conservation of energy argument gives:

(Delta Rhos) H g = Rhos V^2 / 2

where:

Rhos = .840 kg / m^3 = density of liquid sodium at 460 degrees C

(Delta Rhos) = 40 degrees C X 210 / 10^6 deg C X Rhos

H = 3.9 m = height of the warmed sodium column in the fuel assembly

g = 9.8 m / s^2 = acceleration of gravity

V = axial sodium velocity

Rearranging this equation gives:

V = [2 H g (Delta Rhos) / Rhos]^0.5

= [2 X 3.9 m X 9.8 m / s^2 X 40 X 290 / 10^6]^0.5

= **0.9416 m / s**

Thus even when the viscosity of sodium is taken into account the FNR as designed should have more than sufficient fuel assembly power capacity.

**SLOT APPROXIMATION:**

The geometry of a fuel bundle from the viscosity perspective is complicated. To simplify the calculations we can replace the fuel assembly by an equivalent vertical slot. The slot open area is set equal to the fuel assembly open area for sodium flow. The slot height is:

Zo = 6 m,

same as the fuel bundle.

The slot perimeter length is set equal to the total metal perimeter length seen by the flowing sodium. The change in pressure driving the sodium flow is set by the temperature profile of the rising sodium column.

Slot horizontal length = L

and

Slot width = (2 Ro).

The slot open area (2 Ro L) should be set equal to the open area Ab of the active fuel bundles.
Then the slot width (2 Ro) is given by:

2 Ro = (2 Ro L) / L

= (open area) / L

= Ab / [(metal inside perimeter length) / 2]

= 2 Ab / (metal inside perimeter length)

or

**Ro = Ab / (metal perimeter length)**

As shown at FNR Fuel Bundle the equivalent slot open area for sodium flow is:

Ab = **42.918 m^2**

and the slot length is:

5641.8 m

giving a slot width of:

2 Ro = 42.918 m^2 / (5641.8 m)

= **0.007607 m
= 7.607 mm**

and the corresponding value of Ro is given by:

=

=

The vertical slot height is 6 m

Consider a vertical column of 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 tubes. Since the fuel bundle is immersed in the cool portion of the liquid sodium pool the density of liquid sodium outside the fuel tube assembly 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 top of this sodium column is:

Fg = Integral from Z = 0 to Z = Zo of

[Ac g (Rhoo - Rhos) dZ]

where:

Rhos = sodium density at height Z.

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 Z in the column but is independent of radial position R in the column.

For liquid sodium the fractional change in density with temperature is about:

[2.9 X 10^-4 / deg C],

which is exceptionally large for a metal.

Let To be the temperature of the cool sodium at the bottom of the fuel tubes.

Hence:

(Rhoo -Rhos) = [Rhoo - (Rhoo - [290 X 10^-6 / C] Rhoo (T(Z) - To))]

= [290 X 10^-6 / C] Rhoo (T(Z) - To)

**Assume a 40 degree C sodium temperature rise.**

For the bottom 1.8 m of the fuel tube:

(T(Z) - To)) = 0

For the next 0.60 m of the fuel tube:

(T(Z) - To)) = 20 deg C

For the next 3.6 m of the fuel tube:

(T(Z) - To)) = 40 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 (290 X 10^-6 / C) Rhoo (T(Z) - To) dZ

= g [290 X 10^-6 / C] Rhoo [(0 C)(1.8 m) + (20 C)(0.6 m) + (40 C)(3.6 m)]

= g [290 X 10^-6 / C] Rhoo [20 C (7.8 m)]

= g [290 X 10^-6 / C] Rhoo [156] C-m

= **g (0.04524 m) Rhoo**

**Pg** = g (0.04524 m) Rhoo

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 R.

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

**IDEAL CASE OF NO VISCOUS FORCE:**

For the **ideal case** of zero viscous force in the center of the fluid column:

Vc = Vco

and

(Kinetic power) = (Rhoo Vco Acc)(Vco^2 / 2)

Power applied to a column of fluid

= (Pg Ac) Vco = Pg Fv

where

Fv = volumetric flow rate

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 (0.04524 m) Rhoo]

= 2 g (0.04524 m)

= 2 (9.8 m / s^2)(0.04524 m)

= 0.886704 (m / s)^2

or

**Vco = 0.94165 m / s**

Hence due to viscous force Vc(R) will be always be less than 0.94165 m / s.

**VISCOUS FORCE WITHIN A SLOT:**

Consider a slot shaped vertical flow cooling passage.

Zo = slot height

L = slot length

2 Ro = slot width

The component of 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 slot shaped pipe:

F = viscous force

Muv = dynamic viscosity of liquid sodium = 7 X 10^-4 kg / m-s

Ap = pipe inside surface area

Vc = column vertical velocity

R = distance from slot center line

Ro = distance from slot line to slot wall

dPv = pressure difference over fluid column length dZ due to viscous force

L = slot length

Zo = slot height

Seek a half slot solution:

For a half slot:

Ap = L Zo

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

At the bottom of each extended sheet of fluid of wall thickness dR, length L and height Zo:

Pg = Pk + Pv

where:

Pk = kinetic pressure

and

Pv = viscous pressure

**BOUNDARY CONDITIONS:**

**Assume that the slot is wide enough and the viscosity is low enough that at the slot center line:
Pv = 0**

Then at the slot center line:

R = 0;

Pg = Pk

Vc = Vco

At the slot walls:

R = Ro,

Vc = 0

so that:

Pk = 0

and

Pv = Pg

**SOLUTION:**

We will attempt to find a consistent solution of the form:

Pv = Pg (|R / Ro|^N

which satisfies both boundary conditions.

(dPv / dR) = [N Pg / Ro] [R / Ro]^(N - 1)

From definition of viscosity:

Consider two sheets with differential velocity dVc and center to center separation dR. The viscosity equation is:

F = Muv Ap (-dVc / dR)|R = R

= (kg / m-s) m^2 (m / s-m)

= kg m / s^2 = force

dF = Pv(R) L dR

dF / dR = Pv(R) L

dF / dR = - Muv Ap d^2(Vc) / dR^2

Equate the two expressions for (dF / dR) to get:

Pv(R) L = - Muv L Zo d^2(Vc) / dR^2

Pv(R) = - Muv Zo d^2(Vc) / dR^2

Apply Boundary Conditions:

At R = Ro:

Vc = 0

Pv = Pg

Pg = - Muv Zo d2(Vc) / dR^2|R = Ro

At R = 0:

Pv = 0

Recall that:

Pv(R) = - Muv Zo d^2(Vc) / dR^2

or

{d^2(Vc) / dR^2|R = 0} = 0

Try solution:

Vc = Vco [1 - (R / Ro)^N]

dVc / dR = - [Vco N / Ro] (R / Ro)^(N-1)

d^2 (Vc) / dR^2 = - [Vco N / Ro^2](N - 1)(R / Ro)^(N-2)

Check for boundary condition compliance:

At R = Ro:

Pg = - Muv Zo d2(Vc) / dR^2|R = Ro

which can be rearranged to give N.

Then:

**Volumetric Flow through the entire cooling channel**

= 2 Integral from R = 0 to R = Ro

Vc(R) L dR

= 2 Integral from R = 0 to R = Ro

Vco [1 - (R / Ro)^N] L dR

= 2 Vco L Ro

- Integral from R = 0 to R = Ro of

2 Vco L (R / Ro)^N

= 2 Vco L Ro

- 2 Vco L [Ro / (N + 1)](R / Ro)^(N + 1)|R = Ro

= 2 Vco L Ro - 2 Vco L [Ro / (N + 1)]

= 2 Vco L Ro [1 - 1 / (N + 1)]

= **2 Vco L Ro [N / (N + 1)]**

Thus we have a closed form solution for the liquid sodium flow through a slot.

**DETERMINATION OF N:**

Recall that at R = 0:

Pk = Pg

and

Pk = Muv Zo [Vco N / Ro^2](N - 1)

or

[Muv Zo Vco / Ro^2] N^2 - [Muv Zo Vco / Ro^2] N - Pg = 0

or

N^2 - N - [(Pg Ro^2) / (Muv Zo Vco)] = 0

Recall that at R = 0 it is assumed that:

Pg = Pk = Rhoo Vco^2 / 2

or

Vco^2 = 2 Pg / Rhoo

or

Vco =[2 Pg / Rhoo]^0.5

Thus:

[(Pg Ro^2) / (Muv Zo Vco)]

= [(Pg Ro^2) / (Muv Zo)][Rhoo / 2 Pg]^0.5

= [(Ro^2) / (Muv Zo)] [Rhoo Pg /2]^0.5

Hence:

N^2 - N - [Pg Rhoo / 2]^0.5 [Ro^2 / Muv Zo] = 0

Let:

C = - [Pg Rhoo / 2]^0.5 [Ro^2 / Muv Zo]

Then:

N^2 - N + C = 0

or

N = {1 +/- [1 - 4(1)C]^0.5} / 2

Recall that with a 40 degree C sodium temperature rise across the fuel bundle: Pg = g (0.04524 m) Rhoo

Hence:

- C = [Pg Rhoo / 2]^0.5 [Ro^2 / Muv Zo]

= [g (0.04524 m / 2)]^0.5 [Ro^2 Rhoo / Muv Zo]

= [(9.8 m / s^2)(0.04524 m / 2)]^0.5 [((.0038036 m)^2 840 kg / m3)) / (6.0 m X 7 X 10^-4 kg / m-s)]

= [9.8(0.04524)/ 2]^0.5 [14.467 X 10^-6 X 840 / (42 X 10^-4)](m / s) (kg / m)(m-s / m kg)

= [9.8(0.04524)/ 2]^0.5 [0.14467 X 840 / (42)]

= 0.47082 X 2.8935

= 1.3623

N = {1 +/- [1 - 4(1)C]^0.5} / 2

= {1 + [6.4493]^0.5} / 2

= 1.7698

Volumetric Flow = 2 Vco L Ro [N / (N + 1)]

= 2 Vco L Ro [1.7698 / 2.7698]

= 0.6389 (open area) Vco

Recall that with a 40 degree C temperature rise:

Vco = 0.94165 m / s

Volumetric Flow = (open area) (0.6389)(.94165 m / s)

= 0.601667 m / s (open area)

**REACTOR POWER:**

The open area is 42.918 m^2.

At a temperature rise across the fuel bundles of 40 degrees C the corresponding sodium volumetric flow is:

0.601667 m / s X 42.918 m^2 = 25.8224 m^3 / s

The temperature drop across the NaK flow loop is 120 degrees C. Hence th maximum NaK flow should be about:

25.8224 m^3 / s / 3 = 8.607 m^3 / s

The corresponding reactor thermal power is:

= 25.8224 m3 / s X 840 kg / m^3 X Cp X 40 deg C

= 25.8224 m3 / s X 840 kg / m^3 X 1.26 kJ / Kg deg C X 40 deg C

= 1,093,215 kJ / s

= 1,093,215 kWt

The heat transport system limits the reactor power to about 10^6 kWt. Hence, at full rated power this FNR will operate with about a 40 degree C sodium temperature rise. However, this temperature rise will increase a bit over time due to fuel aging and fuel tube swelling.

If we can solve the issue of the NaK gaskets there is a possibility of raising the reactor setpoint from 460 degees C to about 490 ______degrees C.

In practice the temperature difference between the top and bottom of the sodium pool will self adjust to make the heat generation rate equal 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 NaK and nitrate salt flows. Thus the actual temperature rise across the fuel assembly will only be about 40 degrees C.

Even if the heat exchangers are ideal the system power is limited by:

(reactor setpoint temperature - water temperature in the steam generator) X (NaK mass flow) X Cp.

These calculations suggest that the core fuel rods should be ~ 0.60 m long.

**SODIUM LOOP RESIDENCE TIME:**

At a sodium temperature rise of 40 degrees C the sodium volumetric flow rate is:

25.8224 m^3 / s

The corresponding reactor thermal power is:

1,093,215 kWt

The sodium pool volume is about:

Pi (10 m)^2 15 m = Pi 1500 m^3

Hence the sodium loop residence time is about:

Pi (1500 m^3) / (25.8224 m^3 / s) = 182.5 seconds.

This is the time available for warming 340 degree C sodium discharged from the intermediate heat exchange bundles with warmer sodium to bring the mixed sodium temperature up to 400 degrees C. This temperature mixing is achieved by distributed thermal conduction within the liquid sodium pool. This warming is essential to prevent the reactor thermal power spiking due to low sodium inlet temperature. If the sodium mixing is insufficient then a circulation pump or like device is required to achieve the required liquid sodium mixing.

**NUMERICAL VALUATION OF OTHER PARAMETERS:**

Length over which viscous drag occurs:

Zo = 6.0 m

Muv|hot liquid sodium = 7 X 10^-4 kg / m-s

This web page last updated February 29, 2024.

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