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**INTRODUCTION:**

The fission thermal power output of the FNR is set by the product of the primary liquid sodium flow and the primary sodium differential temperature. The liquid sodium primary discharge temperature is typically set at 460 degrees C in order to keep the hottest fuel centerline temperature under 560 degrees C, even with two adjacent sodium cooling channels blocked.

The steam generator water temperature is typically set at 310 degrees C which sets the coldest full load nitrate salt temperature at about 320 degrees C. The corresponding coldest full load secondary sodium temperature is about 330 degrees C and the corresponding full load primary sodium return temperature is about 340 degrees C.

Note that with a fixed pressure in the stem generators at 50% load the coldest nitrate salt temperature is about 315 degrees C and the coldest secondary sodium temperature is about 320 degrees C which prevents NaOH deposits in the secondary sodium loop that can occur below 318 degrees C.

Thus the maximum possible full load primary sodium temperature differential is about:

460 degrees C - 340 degrees C = 120 degrees C.

However, operation at this power level will damage the FNR fuel tubes.

The fission thermal power is:

(primary sodium differential temperature) X (primary sodium volumetric flow rate)

X (primary sodium density) X (primary sodium heat capacity)

Due to natural primary sodium circulation, the primary sodium volumetric flow rate is approximately proportional to the differential temperature. Hence:

(fission power) is approximately proportional to

(primary sodium differential temperature)^2

Conservation of energy gives:

(fission thermal power) = (extracted power) + d(primary sodium enthalpy) / dt

(Extracted power) = (secondary sodium differential temperature) X (secondary sodium volumetric flow rate) X (sodium density) X (sodium heat capacity)

This web page develops a closed form solution for the FNR primary liquid sodium flow rate as a function of primary sodium differential temperature. It is shown that the contemplated FNR can potentially operate at over 1000 MWt at a primary liquid sodium discharge temperature of 460 C and a primary liquid sodium return temperature of 410 C. However, this power may have to be reduced early in the fuel life due to a limited amount of movable fuel bundle insertion into the matrix of fixed fuel bundles causing only a fraction of the core fuel rod length to be active. As the core fuel rod Pu concentration decreases over time the movable fuel bundle insertion will be increased to maintain reactor reactivity at the desired primary sodium average fuel temperature setpoint of 460 degrees C. The increased insertion of movable fuel bundles into the fixed fuel bundle matrix increases the active core rod length which increases the fuel tube surface area through which heat can flow.

**PRIMARY SODIUM NATURAL CIRCULATION:**

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. This thermal expansion causes a decrease in sodium density which drives natural circulation. The hot sodium flows out of the tops of the active fuel bundles, over the top surface of the primary sodium pool toward the perimeter of the pool where it encounters cooler intermediate heat exchange tubes. This primary sodium 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. 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 primary sodium from the tops of the fuel tubes up to the pool top surface of the active fuel bundles at about 460 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 460 degrees C after prolonged operation at no load down to a theoretical minimum of 410 degrees C at full rated 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 - 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 primary sodium natural circulation. Note that the cold sodium discharge from the intermediate heat exchange bundles is level with the top of the fuel tubes.

Consider a vertical 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 tubes. Since the fuel bundle is immersed in the cool portion of the primary liquid sodium pool the density of primary 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:

[2.71 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 - [271 X 10^-6 / C] Rhoo (T(Z) - To))]

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

For the bottom 2.1 m of the fuel tube:

(T(Z) - To)) = 0

For the next 0.30 m of the fuel tube:

(T(Z) - To)) = 25 deg C

For the next 3.6 m of the fuel tube:

(T(Z) - To)) = 50 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) + (25 C)(0.3 m) + (50 C)(3.6 m)]

= g [271 X 10^-6 / C] Rhoo [25 C (7.5 m)]

= g [271 X 10^-6 / C] Rhoo [187.5] C-m

= **g (0.050812 m) Rhoo**

**Pg** = g (0.050812 m) Rhoo

= (9.8 m / s^2) (0.050812 m)(850 kg / m^3)

= **423.27 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 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 (0.050812 m) Rhoo]

= 2 g (0.050812 m)

= 2 (9.8 m / s^2)(0.050812 m)

= 0.99591 (m / s)^2

or

**Vco = 0.99795 m / s**

Hence due to viscous force Vc(R) will be always be less than 0.99795 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) tp 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.

**SLOT APPROXIMATION:**

From a viscosity perspective the geometry of the fuel bundles is quite complex. However, this complex geometry can be replaced by a slot
of height Zo, of horizontal length L and of width (2 Ro).
The slot open area (2 Ro L) should be set equal to the open area Ab of a fuel bundle.
Then the slot width (2 Ro) is given by:

2 Ro = (2 Ro L) / L

= (open area) / L

= Ab / [(Viscous perimeter length) / 2]

= 2 Ab / (Viscous perimeter length)

or

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

**MOVABLE FUEL BUNDLE:**

From FNR Geometry each FNR contains 464 movable active fuel bundles, each with 248 fuel tubes.

The open cross sectional area of each movable active fuel bundle is:

Ab = [17 X (5 / 8) inch]^2

- 12 X [(5 / 8) inch]^2

- 2 X (1 / 4) inch X 15 X (5 / 8) inch X 2^0.5

- 248 X Pi [(1 / 4) inch]^2

= 112.890625 inch^2

- 4.6975 inch^2

- 6.629126074 inch^2

- 48.69468608 inch^2

= 52.86931285 inch^2

= 52.86931285 inch^2 X [.0254 m / inch]^2

= **0.0341091659 m^2**

The viscous perimeter length of each movable fuel bundle is:

4 X 13 X (5 / 8) inch

+ 4 X 4 X (5 / 8) inch

+ 4 X 15 X (5 / 8) inch X 2^0.5

+ 248 X Pi X (1 / 2) inch

= 68 X (5 / 8) inch

+ 60 X (5 / 8) inch X 2^0.5
+ 124 Pi inch

= 42.5 inch

+ 53.03300859 inch

+ 389.5574 inch

= 485.0904972 inch

= 485.0904972 inch

X(0.0254 m / inch)

= **12.32129863 m**

Hence to a good approximation for a movable fuel bundle:

Ro = 0.0341091659 m^2 /12.32129863 m

= **.002768 m **

**FIXED FUEL BUNDLE:**

From FNR Geometry each FNR contains 481 fixed active fuel bundles, each with 384 fuel tubes.

The open area of each active fixed fuel bundle is:

Ab = [21 X (5 / 8) inch]^2

- 4 X 5.5 X [(5 / 8) inch]^2

- 2 X (1 / 4) inch X 18 X (5 / 8) inch X 2^0.5

- 384 X Pi [(1 / 4) inch]^2

= 172.265625 inch^2

- 8.59375 inch^2

-7.954951288 inch^2

- 75.3982236 inch^2

= 80.31870011 inch^2

= 80.31870011 inch^2 X (.0254 m / inch)^2

= **0.0518184126 m^2**

The viscous perimeter of each active fixed fuel bundle is:

4 X 15 X (5 / 8) inch

+ 4 X (4 + 2^0.5) X (5 / 8) inch

+ 4 X 18 X (5 / 8) inch X 2^0.5

+ 384 X Pi (1 / 2) inch

76 X (5 / 8) inch

+ 4 X 2^0.5 X (5 / 8) inch

+ 72 X 2^0.5 X (5 / 8) inch

+ 192 Pi inch

= 47.5 inch

+ 67.17514421 inch

+ 603.1857888inch

= 717.6.160649314 m<8609 inch

= 717.8609 inch X 0.0254 m / inch

= **18.2336677 m**

Hence for a fixed fuel bundle:

Ro = Ab / (viscous perimeter length)

= (0.0518184126 m^2) / (18.2336677 m)

= **.0028419 m**

**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 50 degree C primary sodium temperature drop: Pg = g (0.050812 m) Rhoo

Hence:

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

= [g (0.050812 m / 2)]^0.5 [Ro^2 Rhoo / Muv Zo]

= [(9.8 m / s^2)(0.050812 m / 2)]^0.5 [((.0028 m)^2 850 kg / m3)) / (6.0 m X 7 X 10^-4 kg / m-s)]

= [9.8(0.050812)/ 2]^0.5 [7.84 X 10^-6 X 850 / (42 X 10^-4)](m / s) (kg / m)(m-s / m kg)

= [9.8(0.050812)/ 2]^0.5 [.0784 X 850 / (42)]

= 0.4989777 X 1.5867

= 0.791728

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

= 1.52065

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

= 2 Vco L Ro [1.52065 / 2.52065]

= 0.603277 (open area) Vco

Recall that with a 50 degree C temperature rise:

Vco = 0.99795 m / s

Volumetric Flow = (open area) (0.603277)(0.99795 m / s)

= 0.6020405 m / s (open area)

**EQUIVALENT REACTOR:**

Thus from a viscous flow perspective the reactor can be modelled as:

481 slots each 6.0 m high with open area 0.0518184126 m^2 and Ro = .0028419 m

plus

464 slots each 6.0 m high with open area 0.0341091659 m^2 and Ro = .002768 m

Total Open area = [481 X 0.0518184126 m^2] + [464 X 0.0341091659 m^2]

= 24.924 m^2 + 15.826 m^2

= 40.75 m^2

Volumetric Flow = [0.6020405 m / s] (40.75 m^2) = 24.533 m^3 / s

Reactor thermal power = 24.533 m3 / s X 850 kg / m^3 X Cp X 50 deg C

= 24.533 m3 / s X 850 kg / m^3 X 1.26 kJ / Kg deg C X 50 deg C

= 1,313,750 kJ / s

= 1,313,750 kWt

This figure suggests that the initial guess of a 50 degree C reactor primary sodium temperature rise is slightly too high. The tolerable fuel tube heat flux will limit the reactor power to about 10^6 kWt. Hence typically this FNR will operate at full rated power with a ~ 45 degree C primary sodium temperature differential. However, this tempeature differential will rise due to fuel tube swelling, bottom grating obstruction, etc.

CHECK FROM HERE**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 340 C to 460 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**

= 4.24032 X 10^-3 m X 25.911 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5

=

**(N + 4) = 3.695**

= {[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**

= 3.92282 X 10^-3 m X 25.911 (kg / m^2-s)^0.5 X 15.4304 (s / kg)^0.5

=

**(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 ~ 0.60 m long.

This web page last updated March 12, 2021.

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