\fdO'^ - Old Forge Coal Mines

(527)
PNEUMATICS. 205
See Art. 1043.
30 23
Pressure in condenser = —— x 14.7 = 3.43 lb. per sq.
in. Ans.
(528) 144 X 14.7 = 2,116.8 lb. per sq. ft. Ans.
(529) .27 -7- 3 = .09 = weight of 1 cu. ft. Using formula
56,
/ W, =/, W, or 30 PT, = 65 X .09. IV^ = .195 lb. Ans.
(530) Using formula 61,
p V= .37052 fT r, or 30 X 1 = .37052 X .09 X T.
^ = o^A-o nn = 899. 6°. 899. 6° - 460° = 439. 6°. Ans.
.370o2x.09
(531 ) 460°+ 32° = 492° ; 460°+ 212° = 672° ; 460°+62°
= 522°, and 460° + ( - 40°) = 420°.
(532) Using formula 61,/ F=. 37052 W T, and
substituting,
(14.7X10) X4=. 37052X3.5 X r, or r=imf|^* =
453.417°. 453.417 - 460° = - 6.583°. Ans.
(533) Using formula 63, V P=vp + v^p^, we find
P= o 15X63+19X14.7X3
—r-
25
^, oi.. lu
=71.316 lb.
a
Ans.
(534) Using formula 60,/F=. 37052 T, or P =
.37052X540 . . .
.
,,
,
-r = 20 lb. per sq. m., nearly, Ans.
(535) One inch of mercury represents a pressure of
.49 lb. Therefore, the height of the mercury column is
12.5 -i- .49 = 25.51 in. Ans.
(536) Thirty inches of mercury corresponds
of water. (See Art. 1043.) Therefore,
30' : 34 ft. ::27' : x ft., or x = 30.6 ft. Ans.
A more accurate way is (27 X .49) -^ .434 = 30.5 ft.
to 34 ft.