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<title type="245">Memorandum from Raymond L. Murray and A. C. Menius, Jr.  to Dr. Clifford K. Beck</title>
<title type="gmd">Machine readable transcription</title>
<author>Murray, Raymond L.</author>
<author>Menius, A. C., Jr.</author>
<respStmt>
<resp>Creation of machine-readable version:</resp>
<name>Russell S. Koonts</name>
<resp>Creation of digital images:</resp>
<name>Russell S. Koonts</name>
<resp>Conversion to TEI.2-conformant markup:</resp>
<name>Russell S. Koonts</name>
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<extent>ca. 9 kilobytes</extent>

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<publisher>NCSU Libraries</publisher>
<pubPlace>Raleigh, NC</pubPlace>
<idno type="ETC"> Modern English, MurNBweight080251</idno>
<availability>
<p>Available from: NC State University Archives</p>
<p>Publicly-accessible</p>
<p n="public">URL: http://www.lib.ncsu.edu/archives/etext/engineering/reactor/murray/</p>
</availability>
<date>20 October, 2000</date>
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<seriesStmt>
<p>Nuclear Reactor Digitization Project</p>
<p>Raymond L. Murray Reactor Project Notebook</p>
</seriesStmt>

<notesStmt>
<note>Illustrations have been included from the print version.</note>
<note>Scanned by Russell Koonts with Photoshop 5.0 software.</note>
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<titleStmt>
<title>Memorandum from Raymond L. Murray and A. C. Menius, Jr.  to Dr. Clifford K. Beck</title>
<author>Raymond L. Murray</author>
<author>A. C. Menius, Jr.</author>
<respStmt>
<resp></resp>
<name></name>
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<p></p>
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<extent>2 pp.</extent>
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<pubPlace></pubPlace>
<date></date>
<idno>Manuscript copy consulted UA 105.16</idno>
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<p>Prepared for the North Carolina State University Science and Technology Electronic Text Center</p>
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<p>Keywords in the header are a local Science and Technology Electronic Text Center scheme to aid in establishing analytical groupings.</p>
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<date>August 2, 1951</date>
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<language id="en">English</language>
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<text id="MurNBweight080251T">

<front><div1 type="summary" n="1">
<head><hi rend="bold"><hi rend="center">Memorandum from Raymond L. Murray and A. C. Menius, Jr.  to Dr. Clifford K. Beck</hi><lb/>
<bibl><abbr>Typescript</abbr><lb/> <extent>2 pp.</extent> <lb/><date value="1951-08-02">August 2, 1951</date><lb/> <idno rend="suppress">MurNBweight080251</idno></bibl></hi></head>
<p>

</p>
</div1>
</front>

<body>
<pb n=""/>
<p><seg><xref id="reactorlg/MurNBweight080251a.jpg" rend="new">
<figure entity="MurNBweight080251a"></figure></xref></seg></p>
<div1 type="memorandum" n="1">
<head><hi rend="italics">212 lb/ft<hi rend="sup">3</hi><lb/><date value="1951-08-02">Aug 2, 1951</date><lb/>NCSC-29<lb/><hi rend="underline">CORRECTED</hi><lb/>RM</hi></head>
<opener>TO: Dr. <name type="person">Clifford K. Beck</name><lb/>
FROM: <name type="person">Raymond L. Murray</name> and <name type="person">A. C. Menius, Jr.</name><lb/>
SUBJECT: Weight of Reactor Shield<lb/>
CC: <name type="person">Newton Underwood</name> and <name type="person">A. W. Waltner</name></opener>
<p>For purposes of estimating needs for material in the reactor shield,<lb/>
and for crane load considerations, the volumes and weights are computed. Voids<lb/>
due to various pipes are neglected.
<table>
<row>
<cell><lb/></cell><cell><seg><xref id="reactorlg/MurNBweight080251aa.jpg" rend="new">
<figure entity="MurNBweight080251aa"></figure></xref></seg></cell>
</row>
<row>
<cell><lb/></cell><cell><hi rend="center">Figure 1</hi></cell>
</row>
</table></p>
<p>A. <hi rend="underline">Cap.</hi> Consider one quadrant as in figure 2.
<table>
<row>
<cell><lb/></cell><cell><seg><xref id="reactorlg/MurNBweight080251ab.jpg" rend="new">
<figure entity="MurNBweight080251ab"></figure></xref></seg></cell>
</row>
</table></p>
<p>The system can be further subdivided into fourths.
<table>
<row>
<cell><lb/></cell><cell><seg><xref id="reactorlg/MurNBweight080251ac.jpg" rend="new">
<figure entity="MurNBweight080251ac"></figure></xref></seg></cell>
</row>
</table></p>
<p>If the Level is excluded, the triangular block has a volume
<list>
<item>1/2 (6) (6 tan 22 1/2&#x00B0;) (2.5) = 18.64 ft.<hi rend="sup">3</hi></item>
</list>
The volume beveled off is-- 1/2 (1.5) (1.5) (5.25 tan 22 1/2&#x00B0;) = 2.45 ft.<hi rend="sup">3</hi>,<lb/>
leaving a volume of 16.19 ft.<hi rend="sup">3</hi> or 64.76 ft.<hi rend="sup">3</hi> total. The correction to <orig reg="account">ac-<lb/>
count</orig> for the removable top block 1' x 1' x 1' = 1 ft.<hi rend="sup">3</hi>, and the overhang<lb/>
into the reactor cavity of 2.5' x 2.5' x 4" = 2.08 ft.<hi rend="sup">3</hi> is a net 108 ft.<hi rend="sup">3</hi>.<lb/>
Thus, each quadrant of the cap is 65.84 ft.<hi rend="sup">3</hi>, with a density of 213 lb./ft<hi rend="sup">3</hi>.<lb/>
The weight is 14,024 lb. = 7.01 tons.
</p>
<pb n="2"/>
<p><seg><xref id="reactorlg/MurNBweight080251b.jpg" rend="new">
<figure entity="MurNBweight080251b"></figure></xref></seg></p>
<p><table>
<row>
<cell>B.  The area of this block is<lb/>
= (4) (3.5) - 1/2 (1.5) (1.5) - 2 (1.5)<lb/>
= 14 - 1.125 - 3 = 9.875 ft.<hi rend="sup">2</hi>. Subtract the recess 3" x 6" = .125 ft<hi rend="sup">3</hi>,<lb/>
leaving 9.75 ft.<hi rend="sup">2</hi>, or a volume of (9.75) (6) = <hi rend="strike">48.75</hi> <hi rend="italics">58.50</hi> ft.<hi rend="sup">3</hi>, a weight of<lb/>
<hi rend="strike">10,384 lb.</hi> <hi rend="italics">12,460 lb</hi> or <hi rend="strike">5.19</hi> <hi rend="italics">6.23</hi> tons.</cell><cell><seg><xref id="reactorlg/MurNBweight080251ba.jpg" rend="new">
<figure entity="MurNBweight080251ba"></figure></xref></seg></cell>
</row>
</table></p>
<p><table>
<row>
<cell>C. The area of this block is<lb/>
= (3.5) (3.5) - (1.5) (1.5) (1) (0.25)<lb/>
= 12.25 - 2.25 - 0.25 = 9.75 ft.<hi rend="strike"><hi rend="sup">3</hi></hi><hi rend="italics"><hi rend="sup">2</hi></hi><lb/>
The volume and weight is thus the same as for <hi rend="strike">A.</hi><hi rend="italics">B.</hi><lb/>
<hi rend="italics">58.5 ft<hi rend="sup">3</hi>, 12,460 lb, 6.23 tons</hi>
</cell><cell><seg><xref id="reactorlg/MurNBweight080251bb.jpg" rend="new">
<figure entity="MurNBweight080251bb"></figure></xref></seg></cell>
</row>
</table></p>
<p><table>
<row>
<cell>D. The slab may be considered as being two pieces of total volume<lb/>
5' x 5' x 6" = 12.3 ft.<hi rend="sup">3</hi> plus 5' 6" x 5' 6" x 6" = 15.125 ft.<hi rend="sup">3</hi><lb/>
or 27.625 ft.<hi rend="sup">3</hi>, weight 5,884 lb. or 2.94 tons.</cell><cell><seg><xref id="reactorlg/MurNBweight080251bc.jpg" rend="new">
<figure entity="MurNBweight080251bc"></figure></xref></seg></cell>
</row>
</table>
</p>
<p>
<table>
<row>
<cell>E. Divide the slab into two parts of total volume<lb/>
5 x 5 x 6" = 12.5 ft.<hi rend="sup">3</hi> plus 5' 6" x 5' 6" x 1' = 30.25 ft.<hi rend="sup">3</hi> or 42.75 ft.<hi rend="sup">3</hi><lb/>
The weight is 9,106 lb. or 4.55 tons.
</cell><cell><seg><xref id="reactorlg/MurNBweight080251bd.jpg" rend="new">
<figure entity="MurNBweight080251bd"></figure></xref></seg></cell>
</row>
</table>
</p>

<p>In order to determine the <hi rend="underline">total weight of shield</hi>, including the <orig reg="permanent">per-<lb/>
manent</orig> sides, consider a solid octagon 8' 6", neglecting the bevel.
</p>
<p>Since an octagon of diameter has an area .8284 d<hi rend="sup">2</hi>, the volume of this<lb/>
port is (8.5)(0.8284)(17)<hi rend="sup">2</hi> = 2035.0 ft.<hi rend="sup">3</hi>.
</p>

<p><table>
<row>
<cell>Subtract the following voids:</cell><cell>a. Bevel. Divide the octagonal rim into 16 parts. the<lb/> average length of each is (<hi rend="sup">8.5+7.</hi>/<hi rend="sub">2</hi>) tan 22 1/2&#x00B0; =<lb/>3.21 ft; the area is 1/2 (l.5) 1.5) = 1.125 ft.<hi rend="sup">2</hi><lb/>
the volume is 3.611 ft.<hi rend="sup">3</hi>. For 16 of these, the<lb/>
volume is 57.8 ft.<hi rend="sup">3</hi>.
</cell>
</row>
<row>
<cell></cell><cell>b. Reactor Cavity. 8' 2" x 5' x 5' = 204.2 ft.<hi rend="sup">3</hi>.
</cell>
</row>
<row>
<cell></cell><cell>c. Thermal column cavity. 5 x 5 x 5 = l25 ft.<hi rend="sup">3</hi><lb/>
plus the ledge 3" x 1' x 5' = 1.25 ft.<hi rend="sup">3</hi>, giving<lb/>
126.2 ft.<hi rend="sup">3</hi>.
</cell>
</row>
</table></p> 
<p>The sum of the voids is 385.7 ft.<hi rend="sup">3</hi>. The complete cap has a total volume (now<lb/>
including the top block of 4 ft.<hi rend="sup">3</hi>) of 4(65.84) + 4 = 267.4 ft.<hi rend="sup">3</hi>.
</p>
<p>The final shield volume is<lb/>
2035.0 - 385.7 + 267.4 = 1916.7 ft.<hi rend="sup">3</hi> or 71.0 yds.<hi rend="sup">3</hi>
</p>
<p>The weight is<lb/>
408,200 lbs. or 204.1 tons.
</p>
<p>Since there are a number of unoccupied spaces--exposure tubes, sample box, locks<lb/>
and ducts, this is probably high by a few per cent.
</p>


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