EnergyA hydraulic system is not a source of energy The energy source isA hydraulic system is not a source of energy. The energy source isthe prime mover, which drives the pump. Thus, in reality, a hydraulicsystem is merely an energy transfer system.

The above figure provides a block diagram illustrating how energy istransferred throughout a hydraulic system (contained within thedashed lines). As shown, the prime mover (such as an electric motoror an internal combustion engine) delivers input energy to a pump ofthe hydraulic system via a rotating shaft. 1

The pump converts this mechanical energy into hydraulic energy byincreasing the fluid's pressure and velocity. The fluid flows to anactuator via a hydraulic circuit that consists of pipelines containingvalves and other control components. The hydraulic circuit controlsthe pressures and flow rates throughout the hydraulic system.

The actuator (either a hydraulic cylinder or motor) converts thehydraulic energy from the fluid into mechanical energy to drive theexternal load via a force or torque in an output shaft. Some of theq phydraulic energy of the fluid is lost due to friction as the fluid flowsthrough pipes, valves, fittings, and other control components. Thesefrictional energy losses show up as heat energy which is transferredfrictional energy losses show up as heat energy which is transferredto the environment and/or to the fluid as evident by an increase influid temperature. 2

Mechanical power

Power is defined as the rate of doing work or expending energy.Thus the rate at which the prime mover adds energy to the pumpequals the power input to the hydraulic system.

Likewise, the rate at which the actuator delivers energy to theexternal load equals the power output of the hydraulic system.q p p y y

The power output is determined by the requirements of the externalload. The greater the force or torque required to move the externalg q qload and the faster this work must be done, the greater must be thepower output of the hydraulic system.

In case of linear motion

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Mechanical power in linear motion

Power (in the English system of units) is usually measured in unitsof horsepower (HP).

By definition, 1 HP equals 550 ft.lb/s or 33,000 ft.lb/min. Thus, wehave

The unit of horsepower was created by James Watt at the end of thep ynineteenth century, when he attempted to compare the rate of doingwork by a horse in comparison with a steam engine.

During an experiment he showed that a horse could raise a 150-lbweight (using a block-and-tackle) at an average velocity of.3.67 ft/s.

This rate of doing work by an average horse equals 150 lb x 3.67 ft/sor 550 ft.lb/s which he defined as 1 horsepower. 4

Mechanical power in angular motion

The amount of horsepower transmitted in rotational motion can befound from

It is commonly called brake horsepower (BHP) because a pronybrake (a mechanical device) is used to measure the amount ofhorsepower transmitted by a torque-driven rotating shaft. The pronyhorsepower transmitted by a torque driven rotating shaft. The pronybrake system uses a disk with a flat belt wrapped around itsperiphery to measure the torque in a rotating shaft. The value of63 000 for the constant in the denominator is valid only when the63,000 for the constant in the denominator is valid only when thetorque T has units of in . lb and the rotational speed N has units ofrpm. 5

HYDRAULIC POWER

Consider a hydraulic cylinder as shown

The horsepower delivered by the fluid to the cylinder is calledp y yhydraulic horsepower (HHP).

The horsepower delivered by the cylinder to the load is called outputp y y phorsepower.

Output horse-power is always less than hydraulic horsepower due toOutput horse power is always less than hydraulic horsepower due tofriction and leakage losses.

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In calculations related to hydraulic system there are three questionsto answer:

1. How do we determine how large a piston diameter is required forthe cylinder?2. What is the pump flow rate required to drive the cylinder throughits stroke in a specified time?3. How much hydraulic horsepower does the fluid deliver to thecylinder?

Answer to Question 1. A pump receives fluid on its inlet side atabout atmospheric pressure (0 psig) and discharges the fluid on theoutlet side at some elevated pressure p sufficiently high to overcomep p y gthe load. This pressure p acts on the area of the piston A to producethe force required to overcome the load:

pA= Fload7

Solving for the piston area A, we obtain

The load is known from the application, and the maximum allowablepressure is established based on the pump design. Thus, the aboveequation allows us to calculate the required piston area if the frictionbetween the piston and cylinder bore is negligibly small.

Answer to Question 2. What is the pump flow rate required to drivethe cylinder through its stroke in a specified time?

The volumetric displacement VD of the hydraulic cylinder equals thefluid volume swept out by the piston traveling through its stroke S:

VD(ft3) = A(ft2) x S(ft)D( ) ( ) ( )If there is negligibly small leakage between the piston and cylinder

bore, the required pump volume flow rate Q isQ(ft3/s) =VD(ft3) / t(s) = A(ft2) x S(ft) / t(s) = A(ft2) x v(ft/s)Q(ft /s) VD(ft ) / t(s) A(ft ) x S(ft) / t(s) A(ft ) x v(ft/s)

where v = piston velocity. Note that the larger the piston area andvelocity, the greater must be the pump flow rate. 8

Answer to Question 3. How much hydraulic horsepower does thefluid deliver to the cylinder?

Energy equals force times distance:energy = (F)(S) = (pA)(S)

Since power is the rate of doing work, we havePower = (pA)(S)/t = p(Av)

Since Q = Av, the final result ishydraulic power (ft .lb/s) = p(lb/ft2) x Q(ft3/s)

Recalling that 1 hp = 550 ft . lb/s, we obtainhydraulic horse power = HHP = p(lb/ft2) x Q(ft3/s) /550 (ft. lb/s)

Pressure in psi (lb/in2) and flow rate in gallons per minute (gpm) arethe most common English units used for hydraulic systems. Thus ahydraulic horsepower equation using these most common Englishhydraulic horsepower equation using these most common Englishunits is developed as follows:

HHP = p(psi) x Q(gpm) /1714 9

All three types of power (mechanical, electrical and hydraulic) are typically involved in hydraulic systems as illustrated in the Figure.

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