\fdO'^ - Old Forge Coal Mines

192 HYDROMECHANICS.
(47 1 ) (a) 36 in. = 3 f t. A column of water 1 in. square
and 1 ft. high weighs .434 lb. .434 X 43 = 18.662 lb. per
sq. in., pressure on the bottom of the cylinder. .434 x 40
— 17.36 lb. per sq. in., pressure on the top of the cylinder.
Area of base of cylinder = 20* X .7854 = 314.16 sq. in.
314.16 X 18.662 = 5,862.85 lb., total pressure on the bottom.
Ans.
(^) 314.16X17.36 = 5,453.82 lb., total pressure on the
top. Ans.
(472) 2 lb. = 32 oz. 32 — 10 = 22 oz. = loss of weight
of the bottle in water. 32 + 16 = 48 = weight of bottle and
sugar in air. 48 — 16 = 32 oz. = loss of weight of bottle
and sugar in water. 32 — 22 = 10 oz. = loss of weight of
sugar in water = weight of a volume of water equal to the
volume of the sugar. Then, by formula 27,
specific gravity = ^^_ ^, = — = 1.6. Ans.
(473) 33 = i/2£7i (see Art. 1005), or
33"
k = —r-^r^ = 16.931 ft. per
64.32
sec. Ans.
(474) (a) Use formula 42, and multiply by 7.48 X 60
X 60 to reduce cu. ft. per sec. to gal. per hour.
a = .41 X ^ X 1/2 X 32.16 X X 7.48 X 60 X 60 =
(^y
(6) By formula 43,
216,551 gal. per hr. Ans.
r, .41x||x|/2x32.16x(j-|y
^ - ^« ^ 1^
Vi±Z_ = 3.676
*" ^^ ^y— ft. per sec.
12 ^ 12 Ans.
(475) /= .0193 for v„, = 12. Therefore, using formula
47,
''=Sfd+ "'''-' =
.0193 X 6,000 X ly ^ ^^33 ^ ^.^. ^ ^_^^^ 3^ ^^ ^^
D.oo X o