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XYLENE POWER LTD.

FAST NEUTRON REACTOR (FNR) EARTHQUAKE PROTECTION

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

INTRODUCTION:
Elsewhere on this website Fast Neutron Reactors (FNRs) have been identified as the primary source of energy for meeting mankind's future energy needs. This web page focuses on FNR features which prevent earthquake damage.
 

EARTHQUAKE OVERVIEW:
Severe earthquakes can cause short term horizontal oscillating ground accelerations of up to 3 g. In principle a FNR fuel assembly could be fabricated to be sufficiently robust to tolerate that acceleration. However, the amount of steel required to achieve that level of robustness would seriously reduce the FNR's fuel breeding performance. Instead the design approach taken herein is to design the FNR fuel bundles to resist normal handling stresses and to mount the fuel bundles on an inertial low friction fuel assembly such that if a severe earthquake occurs the fuel assembly stays almost stationary while the ground shakes underneath it.

The primary sodium pool walls and the intermediate heat exchangers are firmly connected to the ground and are intended to follow any earthquake vibration. The fuel assembly is inerial. During an earthquake the liquid sodium between the pool walls and the fuel assembly will slosh around to relieve stress.

The fuel assembly weighs almost 4000 tons. It rests on a layer of one inch diameter ball bearings that cover the entire primary sodium pool floor. That floor is made slightly bowled. The bottom plate of the open steel lattice which supports the fuel assembly has a matching bowl shape. That bottom plate must have sufficient flexibility to conform to the ball bearing surface contour when under load.

If 1 inch diameter ball bearings are used the required number of bearings is:
Pi (10 m)^2 / (.0254 m / inch)^2 = 486,947 inch^2
implying that about 486,947 one inch diameter steel ball bearings would be required. The cost of these bearing balls is close to $1 million.

During a severe earthquake the fuel assembly will likely slide off center in the sodium pool. The purpose of slightly bowling the bottom of the sodium pool is to cause gravity to naturally return the fuel assembly to pool center after the earthquake subsides.
 

BEARING FORCE
The bottom of the fuel assembly is a disk 14 m in diameter. The area of this disk is:
Pi (7 m)^2 / (0.0254 m / inch)^2 = 238,604 inch^2.

The weight of the fuel assembly is about:
4000 tonnes X 2200 lb / tonne = 8,800,000 lbs.

Hence on average the force born by each one inch diameter ball is:
8,800,000 lbs / 238,604 balls = 36.88 lb / ball

Thus the bottom plate of the open steel lattice is likely just a sheet of (3 / 8) inch steel. This steel sheet should be edge welded to slightly flex to distribute the load and to match the primary sodium pool bottom contour. The welds must be ground flush with the sheet steel surface. Around the bottom perimeter of the open steel lattice support plate there must be a (1 / 2) inch projecting rim to contain the ball bearings.

The approximate cost of the required ball bearings is:
238,604 balls X $2.00 / ball = $477,208
 

FIND SYSTEM NATURAL FREQUENCY:
Let X denote distance off pool center.
Assume an earthquake vibration of the form:
X = A sin(wt)
dX / dt = w A cos(wt)
d^2(X) / dt^2 = w^2 A [- sin(wt)]
where:
w^2 A = 3 g
= 3 (9.8 m / s^2)
= 29.4 m / s^2

The viscous damping by the primary liquid sodium should be sufficient for critical damping. The earthquake excited oscillation of the fuel assembly must not be permitted to grow.

Experimental measurements suggest that:
w A = 1.2 m / s

The fuel assembly moving on the ball bearings will have the same natural oscillation frequency as does a pendulum that describes the same arc as does the fuel assembly CM motion. The radius of arc curvature Rp is very long (~ 300 m) which makes the pendulum frequency low.

Rp = pendulum length

Theta = pendulum angular deviation from vertical.

Pendulum KE = M V^2 / 2 = (M / 2) (Rp d(Theta) / dt)^2

H = fuel assembly center of mass height above its minium height.

Pendulum PE = M g H = M g Rp Theta^2

TE = KE + PE
= (M / 2) [Rp d(Theta) / dt]^2 + M g Rp Theta^2
= (M Rp / 2) [Rp (d(Theta) / dt)^2) + 2 g Theta^2]

Theta = B sin(w t)
Theta^2 = B^2 sin^2(w t)
dTheta / dt = B w cos (w t)
(d(Theta) / dt)^2 = B^2 w^2 cos^2(w t)
giving:
TE = (M Rp / 2) [Rp (d(Theta) / dt)^2) + 2 g Theta^2]
= (M Rp / 2) [Rp B^2 w^2 cos^2(w t) + 2 g B^2 sin^2(w t)
= (M Rp / 2) B^2 2 g [(Rp w^2 / 2 g) cos^2(w t) + sin^2(w t)
]
= constant

Recall identity that:
cos^2(wt) + sin^2(wt) = 1

Hence:
Rp w^2 / 2 g = 1
or
w = [2 g / Rp]^0.5

H / (Rp Theta) = sin(Theta)
R / (Rp Theta) = cos(Theta)
[H / (Rp Theta)]^2 + [R / (Rp Theta)]^2 = 1
or
H^2 + R^2 = [Rp Theta]^2

(Theta) = (H / R)

At small Theta angles:
tan(Theta) ~ Theta
giving:
H^2 + R^2 = [Rp Theta]^2
= [Rp H / R]^2
or
Rp^2 = R^2 + (R^4 / H^2)
or for R >> H:
Rp = R^2 / H

Hence:
w = [2 g / Rp]^0.5
= [2 g H / R^2]^0.5
or
f = (1 / 2 Pi R)[2 g H]^0.5
where typically:
R = 3 m
H ~ 3 cm
giving:
f = (1 / 2 Pi R)[2 g H]^0.5
= [1 / (2 Pi 3 m)] [(2 9.8 m 0.03 m)/ s^2]^0.5 =[1 / 6 Pi m][0.7668 m / s]
= 0.0406 Hz

Rp = R^2 / H
= 9 m^2 / 0.03 m
= 300 m

Thus with these parameters after an earthquake impulse it will take several minutes for the fuel assembly to precisely re-center itself. The returning velocity generates relatively little viscous drag as compared to the earthquake. The natural frequency of the fuel assembly suspension is about 20 fold less than the typical earthquake wave frequency.

This web page last updated November 12, 2019.

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