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KIM [24]
2 years ago
11

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
Physics
1 answer:
fredd [130]2 years ago
5 0

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

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