\fdO'^ - Old Forge Coal Mines

HYDROMECHANICS.
(QUESTIONS 454-503.)
(454) The area of the surface of the sphere
X 3.1416= 1,256.64 sq. in. (See rule, Art. 817.)
is 20 X 20
The specific gravity of sea water is 1.026. (See tables
of Specific Gravity, )
The pressure on the sphere per square
in. in cross-
inch is the weight of a column of wat^r 1 sq.
section and 2 miles long. The total pressure is, therefore,
lb. Ans.
1,256.64 X 5,280 X 2 X .434 X 1.026 =-5,908,971
(455) 125 — 83.5 = 41.5 lb. = loss of weight in water
= weight of a volume of water equal to the volume of the
sphere. (See Art. 987.) 1 cu. in. of water weighs .03617
lb.; hence, 41.5 lb. of water must contain 41.5 h- .03617 =
1,147.4 cu. in. = volume of the sphere. Ans.
Note.— It is evident that the specific gravity of the sphere need not
be taken into account.
/^=r^\ n 225,000 ^c, k c.
(456) Q = gQ^gQ = 62.5 cu. ft. per min.
Substituting the values given in formula 51,
^ = 1.229 / ^'^»°^"^-^' = 10.38'+.
Substituting this value of d in formula 49,
^»'=
24.51X62.5
16.38-
^ ^--_
^^•^^^^-
The value of f (from the table) corresponding to v^ =
5.7095 is .0216, using but four decimal places. Hence, applying
formula 52,
j^ 2.57 < /^^^ X 2,800 + i X 16.38)62.5- ^ ^^, ^^^_
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