Handbook of Machine Design P50

Valve controls make use of one of many types of directional control valves to regulate the distribution of energy throughout the circuit.. Meter-in The flow control valve is in the suppl

CHAPTER 43FLUID POWER SYSTEMS AND CIRCUIT DESIGN

Russ Henke, RE.

Russ Henke Associates Elm Grove, Wisconsin

43.6 OPEN-LOOP AND CLOSED-LOOP CIRCUITS / 43.5

43.7 CONSTANT-FLOW VERSUS DEMAND-FLOW CIRCUITS—OPEN LOOP / 43.12 43.8 DEMAND-FLOW CIRCUITS / 43.20

43.9 HYDRAULIC VERSUS PNEUMATIC SYSTEMS / 43.28

43.10 PNEUMATIC CIRCUITS / 43.28

43.11 EFFECT OF FLUID CHARACTERISTICS ON ACTUATOR PERFORMANCE / 43.28 43.12 EFFECT OF FLUID CHARACTERISTICS ON CONTROL-VALVE

PERFORMANCE / 43.31

43.13 BASIC PNEUMATIC POWER CIRCUIT / 43.32

43.14 FLUID LOGIC SYSTEMS / 43.38

43.7 PRESSUREPLOTS

The pressure plots are the same as the load plots discussed in Sec 42.4 except for theintroduction of a constant reflecting the interface area over which the load is dis-tributed The equation is

Note that the constant \IA P is different for each size of cylinder

Once the load and pressure plots have been completed, the level of energy

trans-fer occurring throughout the machine cycle has been defined

43.2 FLOWPLOTS

The next step is to define the rate of energy transfer in the machine This is a

func-tion of the velocities of the various piston rods (or motor shafts, if rotary motors areused)

To determine the required rod velocities, the designer can go back to the sequencechart The time scale along the bottom of the chart indicates how much time is avail-able to complete each part of the machine cycle Time (Ref [43.1], pp 14,15) has twobasic implications in the design of a cycle:

1 It determines the flow rate requirement relative to the actuator motion pattern

2 It determines the horsepower requirement of the circuit or branch

After establishing time increments, the engineer must turn to the element layouts to determine the length of stroke necessary to complete each

machine-motion Next, the steady-state velocity required of the piston can be calculated,

allow-ing for acceleration and deceleration At this point the designer must make a choice

of velocity patterns (Ref [43.1], pp 16,19,22,26,29,347,355,358)

The superimposition of the flow plots for individual actuators has an important

implication If study of the sequence diagram indicates simultaneous operation of

two or more cylinders or motors, their flow plots must be superimposed, as shown in

Fig 43.1 Such superimposition of the plots will give the designer an insight into the

maximum flow rate required This affects the selection of the pump or pumps and

influences separation of circuit branches

MAXIMUM PUMP DISCHARGE REQUIRED^

LOAD DISPLACEMENT OF CYLINDER 1, %

LOAD DISPLACEMENT OF CYLINDER 2, %

FIGURE 43.1 Flow plot of two actuators superimposed to indicate total flow

rates (From Ref [43.1].)

43.3 POWERPLOTS

With the information developed for the pressure and flow plots, the designer can now

make power plots This is a necessary preliminary to selection of the prime mover It

is particularly useful in pointing out power peaks which might otherwise be hidden in

averaged calculations Such peak power demands, occurring when unexpected, could

be great enough to stall an undersized prime mover The fluid horsepower can be

cal-culated from

where p is in pounds per square inch (psi) and Q is in gallons per minute (gpm).The

input horsepower of the prime mover is then

e 0 where e = overall efficiency of the pump.

43.4 CYCLEPROFILE

A complete cycle profile for a single actuator might look something like that shown

in Fig 43.2 Note that the cycle must be plotted for both directions of motion

If the circuit designer has intelligently followed the cycle-profile procedure, acomplete graphic portrayal of what should happen at any point in the cycle of opera-tion of the machine is the result The designer should be able to communicate anyinformation necessary to an understanding of the operational capabilities and limita-tions of the equipment Even more important, the designer should be able to spot anymalfunctions of the machine much more quickly and surely than if she or he had toguess what combinations of events were supposed to transpire and compare themwith what was observed (Ref [43.1], Chap 26)

FIGURE 43.2 Cycle profile for a single actuator (From Ref [43.1].)

43.5 CIRCUITDESlGN

Fluid power circuits can be thought of as consisting of four sections, as shown in Fig.43.3 Section I represents energy output, where energy is transferred to the loadacross the hydromechanical interface Section II is the control area; fluid switching

EXTEND STROKECYLINDER

FIGURE 43.3 The four sections of a fluid power circuit (From Ref [43.1].)

and energy modulation are effected in this section Section III is the energy input tion; this is where the pumps are involved Section IV is the auxiliaries area; this con-sists of piping and fittings and all the other components and equipment necessary tomake a circuit work, including the fluid

sec-The pattern of Fig 43.3 suggests a logical method for solving a circuit design lem A format like that of Fig 43.4 is helpful Divide the sketch sheet into three areas

prob-by drawing vertical lines The left-hand column is reserved for energy input devices,i.e., pumps The right-hand column is for energy output devices, i.e., motors or actua-tors The middle area is for control devices

Next, sketch the symbols (Ref [43.5]) for the output components in the right-handcolumn, in vertical array, as shown in Fig 43.4 Then divide the page into rows bydrawing horizontal lines which separate each actuator from its neighbors We nowhave a matrix of sorts, with the columns representing circuit functions (energy output,energy control, and input), and the rows representing machine functions, as typified

by the actuators

The circuit designer can now select functions to match the requirements of themachine functions

43.6 OPEN-LOOP AND CLOSED-LOOP CIRCUITS

Fluid power systems can be divided into two major groups: open-loop and loop

closed-In a closed-loop system, a feedback mechanism continually monitors system

out-put, generating a signal proportional to this output and comparing it to an input orcommand signal If the two match, there is no adjustment, and the system continues

to operate as programmed If there is a difference between the input command signaland the feedback signal, the output is adjusted automatically to match commandrequirements

There is no feedback mechanism in an open-loop system The performance

char-acteristics of the circuit are determined entirely by the charchar-acteristics of the ual components and their interaction in the circuit A typical open-loop circuit isillustrated in Fig 43.5« Most industrial circuits fall into this category

individ-LOAD

FIGURE 43.4 Graphical layout of a circuit design problem (From Ref [43.1].)

An electrohydraulic servo system is a feedback system in which the output is a

mechanical position or function thereof; see Fig 43.55

Open-loop circuits can be grouped by the functions performed or by the controlmethods

FIGURE 43.5 (a) Typical open-loop circuit; (b) typical closed-loop circuit.

(From Ref [43.1].)

1 Directional controls regulate the distribution of energy (Ref [43.4], Chap 12, pp.

79-91,151-164)

2 Flow controls regulate the rate at which energy is transferred by adjusting the flow

rate in a circuit or branch of circuit (Ref [43.4], Chaps 10,11, pp 65-75,164-168)

3 Pressure controls regulate energy transfer by adjusting the pressure level or by

using a specific pressure level as a signal to initiate a secondary action (Ref [43.4],Chaps 8,9, pp 47-60,143-151)

43.6.2 Control Methods

Directional Control Valve controls make use of one of many types of directional

control valves to regulate the distribution of energy throughout the circuit Thesevalves switch flow streams entering and leaving the valve

Pump control is limited to reversal of the direction of flow from a displacement reversible pump Fluid motor control is similar to pump control; it uses

variable-reversible, variable-displacement motors

TRANSDUCER

Flow Control Valve controls use one of several types of pressure-compensated or

noncompensated flow control valves (Ref [43.3], Chap 9, pp 91-98) The position ofthe flow control valve in the circuit determines the appropriate type to use These are

as follows:

1 Meter-in The flow control valve is in the supply line to the actuator and controls

the energy transfer by limiting the rate of flow out of that actuator; see Fig 43.60

2 Meter-out The flow control valve is in the return line from the actuator and

con-trols the energy transfer by limiting the rate of flow out of that actuator; see Fig.43.66

FIGURE 43.6 Valve controls for open-loop circuits, (a) Meter-in; (b)

meter-out; (c) bleed-off (From Ref [43.1].)

LOAD

LOAD

LOAD

3 Bleed-off The flow control valve is in parallel with the actuator It limits the rate

of energy transfer to the actuator by controlling the amount of fluid bypassedthrough the parallel circuit; see Fig 43.6c

Pump control involves the use of one of two methods, depending on the type of pump used Multiple pumps provide a step variation in flow (Fig 43.ld)\ variable- displacement pumps deliver infinitely (from zero to maximum) variable flows (Fig 43.Ib).

Fluid motor controls use techniques similar to pump controls, and this involves the

use of multiple motors as in Fig 43.80 for step variation or variable-displacement

motors as in Fig 43.Sb for infinite variation in output speeds.

Pressure Control Valve controls use one or more of six types of pressure control

valves:

1 Relief valves limit the maximum energy level of the system by limiting the

maxi-mum operating pressure; see Fig 43.9

2 Unloading valves regulate the pressure level by bypassing the supply fluid to the

tank at a low energy level Unloading valves shift when the system pressurereaches a preset level; see Fig 43.10

FIGURE 43.9 Pressure relief valve regulates system output fluid pressure.

(From Ref [43.1].)

FIGURE 43.7 Pump controls for

open-loop circuits, (a) Multiple pumps;

(b) variable displacement (From Ref.

[43.1].)

FIGURE 43.8 Actuator controls for

open-loop circuits, (a) Multiple-fluid motors; (b) variable displacement (From Ref [43.1].)

TO SYSTEM

TO TANKDRAIN LINE

FIGURE 43.12 Pressure-reducing valve allows one branch of a circuit to

operate at a lower pressure than the main system (From Ref [43.1].)

FIGURE 43.11 Sequence valve prevents fluid from entering one branch of a

circuit before a preset pressure is reached in the main circuit (From Ref [43.1].)

FIGURE 43.10 Pressure unloading valve unloads pump output to the tank at

low pressure when high-pressure flow is not required (From Ref [43.1].)

3 Sequence valves react to a pressure signal to divert energy from a primary circuit

to a secondary circuit; see Fig 43.11

4 Reducing valves react to a pressure signal to throttle flow to a secondary circuit,

thus delivering energy at a lower level to the secondary than to the primary circuit;see Fig 43.12

TO TANK

TANK

PILOT PUMPFROM PUMP

FIGURE 43.13 Counterbalance valve holds fluid pressure in part of a circuit

to counterbalance weight on the external force (From Ref [43.1].)

5 Counterbalance valves control the potential energy differential across an actuator

by maintaining a preset backpressure in the return line; see Fig 43.13 Their pose is to prevent a load from drifting

pur-6 Decompression valves provide

con-trolled release of energy stored inhigh-pressure systems, because of theelasticity in the system; see Fig 43.14

Pump control of pressure fluid in

open-loop circuits is generally achievedwith pressure-compensated variable-displacement pumps Energy transfer iscontrolled by varying the flow from thepump in response to a pressure-level sig-nal across the compensator; see Fig 43.15

Rotary actuator control of fluid

pres-sure is not generally used

FIGURE 43.15 Pressure-compensated variable-displacement pump

Gover-nor spring loads pump toward full-displacement position As output pressure

rises, it supplies the required force to stroke the cam ring toward deadhead

position (From Ref [43.1].)

FIGURE 43.14 Decompression valve releases

fluid at controlled rate to release energy stored

in high-pressure system (From Ref [43.1].)

OUTLET

PILOT LINE INLET

43.7 CONSTANT-FLOW VERSUS DEMAND-FLOW

CIRCUITS—OPEN LOOP

The next step in implementing the circuit shown in Rg 43.4 is to decide which basictype of circuit to use An understanding of the characteristics of each is required.Fluid power circuits are broadly categorized as open-loop or closed-loop, as we have

seen Open-loop circuits are further subdivided into constant-flow and demand-flow

circuits Figure 43.16 illustrates the constant-flow principle in a simple circuit that hasonly one directional control valve Figure 43.17 illustrates this for a multiple- or stack-

valve installation where pump flow is returned directly to the tank only when both

valves are in the neutral position If either valve is shifted, normal four-way valvedirectional control will start

FIGURE 43.16 Typical constant-flow system.

When directional control valve is in neutral position, pump output bypasses to tank through tandem cen-

ter (From Ref [43.1].)

43.7.1 Pump Discharge Pressure

In constant-flow circuits, the pressure at which the pump discharges fluid is a function

of the load resistance encountered by and reflected across the actuator; see Fig 43.4.The system operating pressure generated by the load is a function of the actuatorgeometry If the prime mover can satisfy the energy demand, it will do so If not, theprime mover will stall; or, as is more likely to happen in actual practice, the reliefvalve will open to bypass fluid to tank—this wastes energy

FIGURE 43.17 Constant-flow multiple-valve system Pump output bypasses to

tank only when both directional valves are in neutral position (From Ref [43J].)

43.7.2 Relation of Pump Discharge to Actuator Speed

In constant-flow circuits, the pump output is not determined by the actuator's

instan-taneous speed requirements The discharge rate is a function of pump displacementand its speed of rotation Pump output and actuator displacement jointly determine

a steady-state speed, according to the equations

V = Qf- (for a cylinder) N 0 = ^- (for a motor) (43.4)

V m = motor displacement per revolution

Load inertia may preclude rapid acceleration to this steady-state speed; if it does,excess flow from the pump must return to the tank through the relief valve; see Fig.43.18 At time ^0 the control valve shifts, porting pressure fluid to the actuator There

is a slight time lag caused by such factors as the compressibility of the oil in the

sys-tem and throttling while the valve spool is shifting Actuator flow Q 0 then increases to

full rated output Q = Q ; see the vertical dotted line in Fig 43.18.

FIGURE 43.18 Should load inertia prevent acceleration to steady-state

speed, excess flow from the pump returns to the tank through the relief

valve (From Ref [43.1].)

At time r0 the actuator velocity is zero At time J0 + dt Q (the time that corresponds

to full buildup of pump output), the actuator has not yet started to move Therefore,

actuator flow Qa is zero at that instant It could be demonstrated mathematically that for these conditions of finite pump output Q p and zero actuator flow Q a to coexist, theinstantaneous acceleration of the actuator would have to be infinite An infinitelypowerful driving force would be required for this

If we examine the pressure plot in Fig 43.18a and compare it with the flow plot ofFig 43.18Z?, we see that the fluid pressure rises rapidly and peaks at some level abovethe relief valve setting This level depends on the response time of the relief valve; inaddition, it depends on the internal slip in the pump, valve leakage, and actuator slip.Once the relief valve opens, fluid pressure in the system levels out at the relief valvesetting

Now consider the plot of the actuator flow rate Q a In any well-designed system

consisting of one pump and one actuator, the pump output just matches the actuatorinput requirement at design speed Thus, under steady-state conditions,

Qp = Qa (43.5)

At the time dt , however, Q is equal to rated flow and Q is zero

(dumped overrelief valve)

Rated pump flowPump flow Qp

Relief valve setting

The actuator and load must accelerate from zero to design velocity This takes afinite interval of time, from ^0 to J1 During this time interval, Q a increases until Q 0 =

QP at time J1, which is when the actuator reaches design speed Note that at that time,the system pressure drops to the steady-state design level, and the relief valve closes.The shaded area between the two flow curves (Fig 43.18/?) represents the volume ofoil returned to the tank through the relief valve during the acceleration period.Because this complex sequence of events takes place in a fraction of a second, it isdifficult to observe under normal operating conditions And in most constant-flowcircuit applications, it is not even a matter of concern The designer would analyze thissequence only when dealing with applications that have high performance require-ments or when an operating malfunction cannot otherwise be explained Such a mal-function might occur if the pump's output flow rate and the actuator flow rate werebadly matched

For this reason, the designer must make sure that these two quantities are erly matched, especially when designing multibranch circuits If a pump must be sizedfor multibranch circuit operation, as is frequently the case, the designer must choose

prop-a pump with prop-a cprop-approp-acity thprop-at equprop-als peprop-ak flow requirements Note thprop-at the cprop-approp-acity ofsuch a pump exceeds the fluid needs of a single actuator

Sizing the Actuator In constant-flow circuits, the designer tries to size actuators to

meet speed requirements as a function of pump output For example, a cylinder might

QP = pump flow rate

In some instances this formula may call for a cylinder with a capacity larger than thatrequired for force output alone

The designer would ordinarily select a fluid motor with a capacity (at desiredoperating speed) equal to the rated pump output

Qp = VJ* (43.7)

Unloading the Pump In a constant-flow circuit, the directional control valve

unloads the pump when the valve is in its neutral position This is an advantage in thatauxiliary controls are not required By unloading the pump, the designer reducesunnecessary energy dissipation during dwell or passive intervals in the cycle, thusminimizing the generation of heat Care must be taken that the directional control

valve selected has enough capacity to bypass the full pump output without causing

excessive pressure drop

Output Speed Control One way to control actuator speed in a constant-flow

cir-cuit is to restrict flow with a metering or flow control device The most commonmetering approach uses one of the many types of flow control valves in combina-tion with one of the basic methods of flow control described previously Anotherapproach takes advantage of the throttling characteristics of the directional control

FIGURE 43.19 Example of a typical resistive load system (From Ref [43.1].)

W = weight of load, actuator elements, machine-tool carriage, etc.

Since at startup VI = O, we have

states a relationship among the resistive force F R required to overcome the resistive

load, the cylinder piston area A p , and an initial system pressure/?/ However, the

equa-tion is not complete because we must consider several other factors: the fricequa-tional

component of the resistive load F fo the breakaway friction F^ and the running tion F fr

fric-The frictional component of the resistive load is given by

oper-Application Problem Let us consider a typical example of constant-flow analysis.

As previously discussed, a thorough analysis of system objectives is fundamental togood circuit design The cycle-profile technique was suggested as one approach toorderly design Remember that this approach divides the circuit into sections; see Fig.43.19 Note that the load is primarily a resistive one Therefore, under steady-stateconditions,

RESISTIVE LOAD

where W=weight of the load, actuator elements, etc., and L N = normal component of

any applied force (such as cable tension or cutter reaction), we may write

F f =\i(W+L N )

We have noted that breakaway and running friction also enter into the ship We must distinguish between these two quantities because the coefficient of fric-tion varies between the static condition Ji5 and the dynamic condition u^, so that

relation-14 > \i d The breakaway friction is

Ffl, = ^(W + L N )

and the running friction is

Ffr = \UW + L N ) Note that breakaway friction force F^ may vary during the cycle because of a vari- able normal component force L N (There is another frequently neglected component

of the total resistance energy requirement which, in some cases, cannot be looked—the energy needed to accelerate the mass of oil within the system We do notdiscuss this component here, since it is beyond the scope of this presentation See Ref.[43.1], Chap 27.)

over-Thus the total resistive force at breakaway is

represents the short interval during which breakaway from zero velocity takes place.This is a transient state, and it would be difficult to plot without use of an analyticalinstrument such as an oscilloscope

Qualitatively, however, load components are functions of static friction and the

resistive load itself In the interval dt < t < Ar (also of short duration), the load and

actuator masses are accelerated Again, this is a transient state The components aredynamic friction, resistive load, and the load due to the acceleration of a mass

FIGURE 43.20 Typical load-cycle plot for the system illustrated in Fig 43.19.

Note the nonlinear time scale of the abscissa (From Ref [43.1].)

Beyond At < t, acceleration of the load is essentially a steady-state quantity; at least

we usually assume this, even if it is not quite true in actual practice In this time val, the important components are resistive load and dynamic friction

inter-Figure 43.21 shows the shape of a typical system pressure plot determined fromthe load-cycle plot shown in Fig 43.20 Note that the highest pressures appear when

O < dti < At These are the familiar transients, frequently seen on oscilloscopes, caused

by the breakaway phenomenon and the superimposition of acceleration forces on

normal load resistance Figure 43.21 also shows that a steady-state pressure p ss isachieved when the load speed corresponds to the actuator design speed

TIME

FIGURE 43.21 Shape of typical pressure plot determined from the load-cycle

plot shown in Fig 43.20 Note that the highest pressures appear in the interval O

The relief valve setting at this point is ordinarily between p ss and p a + p 0 If Ar1 isbrief in comparison with J2, the relief valve setting can be close to p ss , because the

pressure transient will be so short that the relief valve cannot respond—or even if itcould, the relatively small quantity of oil bypassed would not affect circuit operationsignificantly If, however, A^ is large in comparison with fc, the relief valve will have to

be set higher

When interpreting such pressure plots, you should bear in mind that the system

will develop a fluid pressure p 0 caused by the acceleration of the oil column in the linethat connects the pump and the actuator This pressure is superimposed on the other

pressures reflected by the load at the actuator; this relationship develops because p 0

is not load-reflective and occurs only in the oil in the line to the actuator.

Characteristics of Constant Flow In light of this discussion, remember these

char-acteristics of constant-flow circuits:

1 The pump discharge pressure is a function of the load resistance and must buildfrom zero

2 The pump output is not determined by the actuator speed requirements

3 Actuators are sized to meet speed requirements as a function of the pump output.Statements 2 and 3 make sense only if we assume that the pump in a multibranchcircuit is sized to accommodate maximum system demand, no matter when or where

it occurs During parts of the cycle when demand is not at peak, the pump output willexceed that required by one actuator In the simple example of Fig 43.19, the pumpand actuator displacements would have to be matched Let us analyze the basic cir-cuit of Fig 43.22 required for the application

FIGURE 43.22 Basic circuit for application illustrated in Fig 43.19 (From

Ref.[43.1J.)

Assume that the load resistance of the circuit calls for a piston area A p for a design

pressure/?/ and that the load must be moved through a stroke S The cylinder will

dis-place a volume

If the job to be done requires that the load be moved in t seconds, then the necessary

flow rate to the cylinder is

where Q is in cubic inches per second.

(At this point in the analysis, the designer must check the columnar strength of thepiston rod, which may turn out to be the critical factor If a larger rod is needed, thecylinder bore may have to be increased accordingly Such a change would, in turn,require adjustment of the pump-displacement calculation.)

Assuming that we have satisfactorily calculated the required pump output Q p , we

can complete the input segment of the circuit we are using as an example; see Fig.43.22 By the very nature of this circuit, we must use the tandem-center four-wayvalve shown Also, a constant-flow circuit requires a relief valve with a fixed-displacement pump

The main functional element still missing from the simple circuit is a method forspeed control Since it was indicated that speed control could be accomplished only

by throttling, the manually operated directional control or proportional valve can beused to throttle flow In all circuits, the pressure drop across the valve has the effect

of reducing the pressure available at the actuator Consequently, the force available

to accelerate the load and overcome friction is reduced

A flow control valve used in a meter-in circuit (Fig 43.23) would have essentiallythe same effect, unless it were used in a bleed-off circuit In this case, the flow to thecylinder would actually be reduced One could use a bleed-off circuit (Fig 43.24) forminor speed adjustment, but a valve in a meter-in circuit would be preferred foradjustment over a wide flow range A meter-out circuit could be used, but since theload in this example is a resistive one, this alternative would have little advantageover a meter-in circuit

43.8 DEMAND-FLOW CIRCUITS

A closed-center circuit is one in which the port from the pump to the directional trol valve is blocked when the valve is in its neutral position (see Fig 43.25) Typically,demand-flow circuits are equipped with a fixed-displacement pump, an unloadingvalve, and an accumulator, as shown in Fig 43.26, or a variable-displacement,pressure-compensated pump, as shown in Fig 43.27 Closed-center circuits are moreaccurately characterized as demand-flow circuits

con-43.8.1 Fixed-Displacement Pump Circuits

In demand-flow circuits that use a fixed-displacement pump, unloading valve, andaccumulator, fluid pressure from the pump is not directly determined by actuatorforce requirements As Fig 43.26 illustrates, the pump charges the accumulator todesign pressure when the directional control valve is centered

FIGURE 43.23 Flow control valve in meter-in circuit provides speed control

for circuit in Fig 43.22 (From Ref [43.1].)

Maximum design pressure in the circuits is controlled by the spring setting

of an unloading valve When this setting is reached, the valve opens and bypassesoil to the tank, at low pressure Note that the pilot signal to the relief valve issensed downstream of a check valve placed between the pump and the accumula-tor The check valve prevents the unloading of the accumulator as well as thepump

When the directional control valve is shifted so that it ports oil to the actuator, thefull design pressure (as stored in the accumulator) is immediately available to the sys-tem As the cylinder moves, oil is forced from the accumulator by the compressed gascharge behind the oil After a time interval, system pressure drops because of theexpansion of the gas charge in the accumulator

At some pressure level for which it has been designed, the unloading valve closesand causes output from the pump to reenter the system rather than bypass to thetank At this time, the pump will do one of two things:

1 Add its output to that from the accumulator at the lower pressure level

2 Recharge the accumulator to a higher pressure

Which event occurs is a function of many other factors

Some accumulator circuits are designed so that the accumulator supplies all the oilused during the active part of the cycle It cannot do so at constant pressure, becausethe pressure of the gas charge drops as the gas expands when the oil flows out of theaccumulator The load cycle must be designed so that the system can still function at

FIGURE 43.24 Bleed-off circuit could be used to control flow to cap end of

cylinder, but meter-in circuit is preferred for fine adjustments (From Ref [43.1].)

the lowest pressure level delivered by the accumulator This design feature is usedwhere the active, or work, segment of the cycle is rather short and is followed by a rel-atively long passive, or dwell, segment during which the pump recharges the accumu-lator In such circuits, the pump is sized to charge the accumulator during thework-cycle dwell segment; see Fig 43.28

Example L Assume a circuit similar to that in Fig 43.26, in which an accumulator

supplies 924 cubic inches (in3) of oil to the circuit in 10 seconds (s) What is therequired pump output rate if the dwell time between work periods is 50 s?

Solution The required pump discharge, in gallons per minute, is

accumulator which it charges during the passive part of the work cycle When the

operator shifts the directional control valve, the accumulator output flow is added to

the pump flow Note that the combined flows may exceed several times the output ofthe pump alone However, this condition will exist for only a very short time Indesigns where a peak flow of short duration may be desirable, this configuration may

be much more economical than one that relies on one large pump; see Fig 43.29

Example 2 Assume that a pump is used instead of an accumulator to supply the

required oil in Example 1 If the operating pressure is 1000 psi, what is the difference

in horsepower required to drive the pump in these two examples?

Solution The total flow to the system is

FIGURE 43.25 In a simple demand-flow

circuit, the line from the fixed-displacement

pump to the valve is blocked when the

direc-tional control valve is in the neutral position.

(From Ref [43.1].)

FIGURE 43.26 In a simple demand-flow cuit powered by a fixed-displacement pump,

cir-an accumulator is added to supply full design

pressure immediately (From Ref [43.1].)

Therefore, using a pump instead of anaccumulator to supply the requiredshort-duration, high-volume flows in-creases the horsepower requirement by

600 percent

In these types of circuits, actuatorsare sized to meet the force require-ments based on load-cycle analysis Fre-quently actuators, particularly cylinders,can be sized smaller than in compara-ble, constant-flow circuits This is truebecause in constant-flow circuits, thecylinders must be sized to provide therequired speed based on availablepump output In demand circuits, how-ever, a given force is available thataccelerates the load at a rate propor-tional to the mass Thus, the cylinderdemands oil from the accumulator inproportion to its instantaneous velocity.The accumulator delivers only on de-mand because, unlike a pump, it is not apositive-displacement device

43.8.2 Pressure-Compensated Pumps

We discussed the simplest form of pressure-compensated pump in a demand-flowcircuit in Fig 43.27 If the demand-flow circuit uses a pressure-compensated pump

FIGURE 43.27 A variable-displacement,

pressure-compensated pump supplies pressure

fluid to this demand-flow circuit Note the

ab-sence of a relief valve (From Ref [43.1].)

TIME

FIGURE 43.28 In systems with cycles that have short work segments and long

dwell segments, the pump is sized to charge the accumulator during dwell (From

Pump

charges

accumulatorActive load cycle

Passive cycleAccumulator

pressure Accumulator flow

to actuator Pump flowto circuit

DesignpressureOperatingpressure bandMinimum accumulatoroperating pressureUnloadedpump pressure