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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.

Respuesta :

Answer:

a)[tex]A=0.021\ ft^2[/tex]

b))[tex]A=2.04\times 10^{-3}\ m^2[/tex]

c)[tex]Q=5.1\times 10^{-4}\ m^3/s[/tex]

d)Power = 14.14 hp

e)Power =13.85 hp

Explanation:

Given that

t =30 s

S= 25 ft

F= 9500 lb

P= 3000 psi  

1 psi = 6894.68 Pa

[tex]P=2.07\times 10^7\ Pa[/tex]

psi = pound force per inch square

a)

We know that

F = P A

9500 = 3000 A

[tex]A=3.166\ in^2[/tex]

We know that

1 inch = 0.0833 ft

[tex]1\ in^2=6.8\times 10^{-3}\ ft^2[/tex]

[tex]3.166\ in^2=0.021\ ft^2[/tex]

[tex]A=0.021\ ft^2[/tex]

b)

[tex]1\ in^2=6.4\times 10^{-4}\ m^2[/tex]

[tex]3.166\ in^2=2.04\times 10^{-3}\ m^2[/tex]

C)

Flow rate Q

[tex]Q=\dfrac{A\times S}{t}\ m^3/s[/tex]

[tex]Q=\dfrac{2.04\times 10^{-3}\times 7.62}{30}\ m^3/s[/tex]

[tex]Q=5.1\times 10^{-4}\ m^3/s[/tex]

d)

Power = P Q

[tex]Power=2.07\times 10^7\times 5.1\times 10^{-4}\ W[/tex]

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

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