A hydraulic cylinder has a piston of cross sectional area 25 cm^2 and a fluid pressure of 2 MPa. If the piston is moved 0.25 m

Question 1

A hydraulic cylinder has a piston of cross sectional area 25 cm2 and a fluid pressure of 2 MPa. If the piston is moved 0.25 m how much work is done?

Question 2

The work is a force with a displacement and force is constant: F = PA

W = ∫ F dx = ∫ PA dx = PA ∆x

= 2000 kPa × 25 × 10-4 m2 × 0.25 m = 1.25 kJ

Units: kPa m2 m = kN m-2 m2 m = kN m = kJ

kratos · Answer 1 · 2020-10-23T00:36:47+0000

The work is a force with a displacement and force is constant: F = PA

W = ∫ F dx = ∫ PA dx = PA ∆x

= 2000 kPa × 25 × 10-4 m2 × 0.25 m = 1.25 kJ

Units: kPa m2 m = kN m-2 m2 m = kN m = kJ

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