NCSC-29

CORRECTED

RM

TO: Dr.

FROM:

SUBJECT: Weight of Reactor Shield

CC:

For purposes of estimating needs for material in the reactor shield,

and for crane load considerations, the volumes and weights are computed. Voids

due to various pipes are neglected.

Figure 1 |

A. Cap. Consider one quadrant as in figure 2.

The system can be further subdivided into fourths.

If the Level is excluded, the triangular block has a volume

- 1/2 (6) (6 tan 22 1/2°) (2.5) = 18.64 ft.³

leaving a volume of 16.19 ft.³ or 64.76 ft.³ total. The correction to

count

into the reactor cavity of 2.5' x 2.5' x 4" = 2.08 ft.³ is a net 108 ft.³.

Thus, each quadrant of the cap is 65.84 ft.³, with a density of 213 lb./ft³.

The weight is 14,024 lb. = 7.01 tons.

C. The area of this block is = (3.5) (3.5) - (1.5) (1.5) (1) (0.25) = 12.25 - 2.25 - 0.25 = 9.75 ft.³² The volume and weight is thus the same as for A.B. 58.5 ft³, 12,460 lb, 6.23 tons |

D. The slab may be considered as being two pieces of total volume 5' x 5' x 6" = 12.3 ft.³ plus 5' 6" x 5' 6" x 6" = 15.125 ft.³ or 27.625 ft.³, weight 5,884 lb. or 2.94 tons. |

E. Divide the slab into two parts of total volume 5 x 5 x 6" = 12.5 ft.³ plus 5' 6" x 5' 6" x 1' = 30.25 ft.³ or 42.75 ft.³ The weight is 9,106 lb. or 4.55 tons. |

In order to determine the total weight of shield, including the

manent

Since an octagon of diameter has an area .8284 d², the volume of this

port is (8.5)(0.8284)(17)² = 2035.0 ft.³.

Subtract the following voids: | a. Bevel. Divide the octagonal rim into 16 parts. the average length of each is (8.5+7./2) tan 22 1/2° = 3.21 ft; the area is 1/2 (l.5) 1.5) = 1.125 ft.² the volume is 3.611 ft.³. For 16 of these, the volume is 57.8 ft.³. |

b. Reactor Cavity. 8' 2" x 5' x 5' = 204.2 ft.³. | |

c. Thermal column cavity. 5 x 5 x 5 = l25 ft.³ plus the ledge 3" x 1' x 5' = 1.25 ft.³, giving 126.2 ft.³. |

The sum of the voids is 385.7 ft.³. The complete cap has a total volume (now

including the top block of 4 ft.³) of 4(65.84) + 4 = 267.4 ft.³.

The final shield volume is

2035.0 - 385.7 + 267.4 = 1916.7 ft.³ or 71.0 yds.³

The weight is

408,200 lbs. or 204.1 tons.

Since there are a number of unoccupied spaces--exposure tubes, sample box, locks

and ducts, this is probably high by a few per cent.