Question

Two hydraulic cylinders maintain a pressure of 1200 kPa. One has a cross sectional area of 0.01 mAA2EEAA the other 0.03 mAA2EEAA. To deliver a work of 1 kJ to the piston how large a displacement (V) and piston motion H is needed for each cylinder? g

Answer 1

Answer:

Answered

Explanation:

W = ∫ F dx = ∫ P dV = ∫ PA dx = PA* H = P∆V

∆V = W/P

= 1 kJ / 1200 kPa = 0.0 833 m3

Both cases the height is H = ∆V/A

IN case 1 ( sectional area = 0.01 m^2)

$H_1 = \frac{0.000833}{0.01} = 0.0833$

in case 2 (sectional area = 0.03 m^2)

$H_2 = \frac{0.000833}{0.03} = 0.0278$