\fdO'^ - Old Forge Coal Mines

PNEUMATICS. 203
Therefore, 7"== —————- = 439.35°. Since this is less
I.411I00
than 460°, the temperature is 460 — 439.35 = 20.65° below
zero, or — 20.65°. Ans.
(516) A hollow space from which all air or other gas
(or gaseous vapor) has been removed. An example would
be the space above the mercury in a barometer.
(517) One inch of mercury corresponds to a pressure of
.49 lb. per sq. in.
1 . .49
-— inch of mercury corresponds to a pressure of '——
lb. per
40 40
sq.
49
in. '-— X 144 = 1.764 lb.
40
per sq. ft. Ans.
(518) (a) 325 X .14= 45.5 lb. = force necessary to
overcome the friction. 6 X 12 = 72" = length of cylinder.
72 — 40 = 32 = distance which the piston must move. Since
the area of the cylinder remains the same, any variation in
the volume will be proportional to the variation in the
length between the head and piston. By formula 53,
rru c ^ A^, 14.7X40 _,2 ,,
pv=pv^. Therefore, / =:'^-^—^= — = 8.1- lb. per
sq. in. = pressure when piston is at the end of the cylinder.
Since there is the atmospheric pressure of 14. 7 lb. on one
2
side of the piston and only 8.1 -lb. on the other side, the
o
2
force required to pull it out of the cylinder is 14.7 — 8.1- =
O
6.5- lb. per sq. in. Area of piston = 40' X .7854 = 1,256.64
o
sq. in. Total force = 1,256.64 x 6.5^- = 8,210.05. Adding
o
the friction, 8,210.05 + 45.5 = 8,255.55 lb. Ans.
{Jti) Proceeding likewise in the second case, pv =. p^ z/„ or
/ ='^J-^ = i^t^iA^ ^ ^g ^^ ,^g - 14.7 = 83.3 lb. per sq. in.
1,266.64 X 83.3 -f 45.5 = 104,723.612 lb. Ans.