Suppose a flow of fluid is delivered to a hydraulic cylinder, causing it to extend. The cylinder has a cross-sectional area A and delivers a force F while moving a distance x (Fig. 1.2). The distance moved is related to the fluid volume delivered to the cylinder.

x = V/A (1.5)

where

x = distance (in)

V = volume (in^{3} )

A = area (in^{2})

The force is related to the pressure developed at the cap end.

F = PA (1.6)

where

F = force (lbf)

P = pressure (lbf/in^{2})

A = area (in^{2})

(Throughout this text, pressure in lbf/in^{2} is expressed as psi.)

Work done is given by

Power is work per unit time,

Power = PV/t (1.8)

Flow is defined as volume per unit time, Q = V/t ; therefore,

Power = PQ (1.9)

Mechanical power is the product of T and N and hydraulic power is the product of P and Q.

The relationship in Eq. (1.9) is the fundamental concept for our study of fluid power. The units used for pressure are typically lbf/in^{2}, or psi, and the units for flow are gal/min, or GPM. To obtain hydraulic power with units of lbf-ft/min, the following conversions are needed.

To obtain hydraulic horsepower,

It is useful to memorize this formula, as it will give a quick frame of reference when beginning a design. For example, if a pump delivers 5 GPM at 2,000 psi pressure drop, how much power is required?

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