Two hydraulic cylinders maintain a pressure of 1200 kPa . One has a cross-sectional area of 0.01

step 1 Here, work done is equal to force into displacement. Hence, force is pressure into volume, E, d

Two hydraulic cylinders maintain a pressure of 1200 kPa ....

Two hydraulic cylinders maintain a pressure of $1200 \mathrm{kPa} .$ One has a cross-sectional area of 0.01 $\mathrm{m}^{2},$ the other one of $0.03 \mathrm{m}^{2} .$ To deliver $1 \mathrm{kJ}$ of work to the piston, how large a displacement ( $V$ ) and piston motion $H$ are needed for each cylinder? Neglect $P_{\text {atn" }}$

Key Concepts

Hydraulic Pressure

Hydraulic pressure is a fundamental concept in fluid mechanics and engineering that describes the force per unit area exerted by a fluid. In systems such as hydraulic cylinders, maintaining a constant pressure allows for predictable force outputs, which is crucial for controlled movement and work. This concept is rooted in Pascal's Principle, which states that pressure applied at any point in an incompressible fluid is transmitted undiminished throughout the fluid.

Work in Hydraulic Systems

In hydraulic systems, work is performed when a pressurized fluid does work on a piston by moving it. The work done is the product of the pressure of the fluid and the volume of the fluid displaced. This relationship (Work = Pressure × Displaced Volume) highlights how hydraulic systems can transform force and displacement to perform mechanical work efficiently.

Relationship Between Displacement Volume and Piston Motion

The volume displaced by a moving piston is directly related to the piston’s cross-sectional area and the distance (or travel) it moves. This relationship is given by the equation Volume = Area × Displacement. Understanding this concept is key when designing and analyzing hydraulic systems since it allows for the calculation of the piston’s motion needed to achieve a desired work output based on the system’s pressure and piston area.

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