Question

A hydraulic cylinder is to compress a car body down to bale size in 30 sec. The operation requires a 25 ft stroke and a 9500-lb force. If a 3000 psi pump has been selected and assuming the cylinder is 98% efficient find:

a) The required piston area in ft^2.

b)The required piston area in m^2.

c)The necessary pump flow rate.

d)The hydraulic horsepower delivered to the cylinder.

e)The output horsepower delivered by the cylinder to the load.

Answer 1

a)
`A=0.021\ ft^2`

b))
`A=2.04* 10^(-3)\ m^2`

c)
`Q=5.1* 10^(-4)\ m^3/s`

d)Power = 14.14 hp

e)Power =13.85 hp

Step-by-step explanation:

Given that

t =30 s

S= 25 ft

F= 9500 lb

P= 3000 psi

1 psi = 6894.68 Pa

`P=2.07* 10^7\ Pa`

psi = pound force per inch square

a)

We know that

F = P A

9500 = 3000 A

`A=3.166\ in^2`

We know that

1 inch = 0.0833 ft

`1\ in^2=6.8* 10^(-3)\ ft^2`

`3.166\ in^2=0.021\ ft^2`

`A=0.021\ ft^2`

b)

`1\ in^2=6.4* 10^(-4)\ m^2`

`3.166\ in^2=2.04* 10^(-3)\ m^2`

C)

Flow rate Q

`Q=(A* S)/(t)\ m^3/s`

`Q=(2.04* 10^(-3)* 7.62)/(30)\ m^3/s`

`Q=5.1* 10^(-4)\ m^3/s`

d)

Power = P Q

`Power=2.07* 10^7* 5.1* 10^(-4)\ W`

Power = 10.55 KW

1 KW = 1.3 hp

Power = 14.14 hp

e)

Horsepower delivered by the cylinder to the load.= 0.98 x 14.14 =13.85 hp