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A double-acting hydraulic cylinder has an inside diameter = 75 mm. The piston rod has a diameter = 14 mm.The hydraulic power source can generate up to 5.0 MPa of pressure at a flow rate of 200,000 mm3/sec todrive the piston.
(a) What are the maximum possible velocity of the piston and the maximum force that canbe applied in the forward stroke?
(b) What are the maximum possible velocity of the piston and themaximum force that can be applied in the reverse stroke?

Respuesta :

Answer:

Forward stroke area A = 0.25π(75)2 = 4418 mm2

Reverse stroke area A = 4418 – 0.25π(14)2 = 4264 mm2

(a) Forward stroke v = 200,000 / 4418 = 45.3 mm/sec

F = 5(4418) = 22,090 N

(b) Reverse stroke v = 200,000 / 4264 = 46.9 mm/sec

F = 5(4264) = 21,320 N

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