This problem is to show the additional air consumption needed when the air over oil tank is low with oil that is used in car garages for the car hoist.
Solving with a full tank of oil:
Area of cylinder: 10² x .7854 = 78.54 sq. in
Stroke is 8’ 8’ x 12” = 96”
Volume of cylinder: 78.54 in² x 96 in = 7,539.84 in³
Cubic feet = 7,539.84 / 1728 (cu. in. / cu. ft.) = 4.36 cu. ft.
Compression ratio (C.R.): 100 + 1 = 7.8 C.R.
14.7
SCF: 4.36 CF x 7.8 = 34 SCF when the tank of oil is full when starting to lift
Air/Oil Tank is only half full of oil when starting. We now have to pressurize the air/oil tank before the hoist will start to lift
Volume of cylinder is 4.36 CF
Volume of air/oil tank is 4.36 x 1.5 (size of air/oil tank) = 6.54 CF
Half full of oil means we need an additional volume of air that equals
6.54 /2 = 3.27 CF
Volume of air needed is: 4.36 (cyl.) + 3.27 (tank) = 7.63 CF
SCF = 7.63 x 7.8 C.R. = 59.51 SCF
Note: If the tank of oil is only half full, 59.51 / 34 = 1.75 times as much air is needed every time the hoist is fully raised.
By Ernie Parker, CFPAI, CFPSD, CFPS, CFPMM, CFPMT, CFPMIP, CFPMMH, CFPMIH
