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Nuclear Reactor Digitization Project

Raymond L. Murray Reactor Project Notebook

Prepared for the North Carolina State University Science and Technology Electronic Text Center

The lineation of the manuscript has been maintained and all end-of-line hyphens have been preserved.

Keywords in the header are a local Science and Technology Electronic Text Center scheme to aid in establishing analytical groupings.

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

The volume beveled off is-- 1/2 (1.5) (1.5) (5.25 tan 22 1/2°) = 2.45 ft.33,
leaving a volume of 16.19 ft.

3 or 64.76 ft.3 total. The correction to
count

3, and the overhang
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.

= (4) (3.5) - 1/2 (1.5) (1.5) - 2 (1.5)

= 14 - 1.125 - 3 = 9.875 ft.

2. Subtract the recess 3" x 6" = .125 ft3,
leaving 9.75 ft.

2, or a volume of (9.75) (6) = 48.75 58.50 ft.3, a weight of

10,384 lb. 12,460 lb or 5.19 6.23 tons.

= (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 ft 3, 12,460 lb, 6.23 tons

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.

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

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

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.

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.
3.
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

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