In a two-piston hydraulic arrangement, the displacement relation is described by which equation?

In a two-piston hydraulic arrangement, the displacement relation is described by which equation?

• A. Displacement equals pressure times velocity
• B. Flow rate constant.
• C. A_s x x_s = A_L x x_L
• D. Displacements independent of area.

Answer: (C)

In a two-piston hydraulic system, the key idea is conservation of fluid volume for an incompressible fluid. The volume moved by the small piston is the cross-sectional area of that piston times its displacement, A_s × x_s. The volume moved by the large piston is A_L × x_L. Because the fluid is incompressible and the system is closed, these two volumes must be equal: A_s × x_s = A_L × x_L. This equation shows how displacements scale with piston area: a larger area piston moves less for the same volume, and a smaller area piston moves more. The other options don’t express this volume- conservation relationship: displacement isn’t set by pressure times velocity, flow rate continuity isn’t describing the displacement- area link, and displacements do depend on area.

In a two-piston hydraulic system, the key idea is conservation of fluid volume for an incompressible fluid. The volume moved by the small piston is the cross-sectional area of that piston times its displacement, A_s × x_s. The volume moved by the large piston is A_L × x_L. Because the fluid is incompressible and the system is closed, these two volumes must be equal: A_s × x_s = A_L × x_L. This equation shows how displacements scale with piston area: a larger area piston moves less for the same volume, and a smaller area piston moves more. The other options don’t express this volume- conservation relationship: displacement isn’t set by pressure times velocity, flow rate continuity isn’t describing the displacement- area link, and displacements do depend on area.