automative lif rasise a 3500lb car 7ft above floor level if the hydraulic cylinder contain 8in diameter piston?

An automotive LIFOtraisese a 3500lb car 7ft above floor level if the hydraulic cylinder contain 8in diameter piston and a 4in diameter rod,determine the
A. work necessary to lift the car
B. required pressure
C. power if the lift raises the car 10 s
D. Descending speed of the lift for a 10-gpm flow rate
E. flow rate for the auto too descend in 10 s

Chosen Answer:

All these answers based on 8 inch lifting cylinder. You haven’t specified anything about the 4 inch rod so haven’t used that in these answers

a). work = force x distance
work = 3500 lb x 7 ft = 24,500 ft-lbs of work
————————————————-
b) 8 inch diameter piston = area of pi * r^2 = 50.26 sq inches
3500 lb / 50.26 in^2 = 69.6 psi pressure
——————————————–
c) power = work / time = 24,500 ft-lbs / 10s = 2450 ft-lb/ second
(one foot-lb /second = 0.738 watt so –> 2450 * .738 = 1,808 watts of power in SI system)
—————————————————————-
d) volume of a cylinder = pi * radius squared * height
one gallon = 231 cubic inches
10 gallons = 2,310 cubic inches
V =2,310 cubic inches = 3.14 * 4in ^2 * height
height = 2310 / (pi * 16) = 45.96 inches

So 10 gallons a minute will lift car or descend car 45.96 inches or 3.8 feet a minute
————————————————————————————————–

e) Flow rate to descend in 10 seconds instead of a minute is 6 times faster (60 sec a minute / 10 sec) = 6.
6 * 10gpm = 60 gpm
.
by: biire2u
on: 14th April 10

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