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

To determine the displacement and piston motion needed for each hydraulic cylinder to deliver 1 kJ of work, we apply the relationship W = P·A·H. For a pressure of 1200 kPa and cross-sectional areas of 0.01 m² and 0.03 m², the respective displacements needed are 8.3 cm and 2.8 cm.

Step-by-step explanation:

The student's question involves calculating the displacement and piston motion required in two hydraulic cylinders to deliver a specific amount of work, given their pressure and cross-sectional areas. To solve for displacement (V) and piston motion (H) for each cylinder, we first need to recognize that work (W) is the product of pressure (P) and the volume change (ΔV), which can also be calculated from the cross-sectional area (A) and displacement (H), since ΔV = A·H.

Given that the pressure (P) is 1200 kPa for both cylinders and the work (W) desired is 1 kJ (1000 J), we can set up the following equation for both cylinders:

W = P·ΔV = P·A·H

Solving for H, we get:

H = W / (P·A)

Calculate this for each cylinder:

• Cylinder 1 (A = 0.01 m2):
• H1 = 1000 J / (1200· 103 Pa · 0.01 m2)
• H1 = 0.083 m or 8.3 cm
• Cylinder 2 (A = 0.03 m2):
• H2 = 1000 J / (1200· 103 Pa · 0.03 m2)
• H2 = 0.028 m or 2.8 cm

The required displacements for the cylinders are 8.3 cm and 2.8 cm, respectively.

Answer 2

Answered

Step-by-step 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 = (0.000833)/(0.01) = 0.0833`

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

`H_2 = (0.000833)/(0.03) = 0.0278`