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Service Training August 1999 STUDENT HANDBOOK INTRODUCTION TO LOAD SENSING/PRESSURE COMPENSATED HYDRAULIC SYSTEMS ENGINE VARIABLE DISPLACEMENT PUMP CLOSED CENTER CONTROL VALVE DOUBLE CHECK VALVE PRESSURE REDUCING VALVE 50 PSI SPRING PUMP CONTROL VALVE LOAD LOAD SENSING / PRESSURE COMPENSATION

12. Load Sensing

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Page 1: 12. Load Sensing

Service TrainingAugust 1999

STUDENT HANDBOOK

INTRODUCTION TOLOAD SENSING/PRESSURE COMPENSATED HYDRAULIC

SYSTEMS

ENGINE

VARIABLEDISPLACEMENT

PUMP

CLOSED CENTERCONTROL VALVE

DOUBLECHECK VALVE

PRESSUREREDUCING VALVE

50 PSISPRING

PUMPCONTROL

VALVE LOAD

LOAD SENSING / PRESSURE COMPENSATION

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COURSE PLAN/OUTLINE

GET ACQUAINTED

MODULE: INTRODUCTION TO LOAD SENSING PRESSURECOMPENSATED HYDRAULICS

- Objectives- Lesson: Basic LS/PC Hydraulics

- Slide presentation on LS/PC hydraulics- Quiz: LS/PC Hydraulics

MODULE: INTRODUCTION TO LOADSENSING PRESSURE COMPENSATEDHYDRAULICS

This module is a review of basic "load sensing/pressure compensated"hydraulics.

OBJECTIVES

1. On a quiz in the back of the Student Handbook, identify systemcharacteristics of Load Sensing and Pressure Compensation andcomponents used in this type of hydraulic system.

2. On a quiz in the back of the Student Handbook, determine systempressures and pressure differences at various locations on providedquiz schematics.

Materials Needed:

LS/PC Student Handout with Quiz

Video (optional)

"Introduction to LS/PC Hydraulics Systems" SEVN1804

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1

INTRODUCTION

Load sensing/pressure compensated” hydraulic systems are now used in a number of Caterpillarmachines.

Some of the current Caterpillar built machines which use these systems include the 416-446 BackhoeLoaders, Challenger 65, H-Series Track-Type Tractors, updated G-Series Motor graders, and 916-936Wheel Loaders (IT18 and IT28).

For more detailed information on the hydraulic systems of these machines refer to their respectiveindividual Service Training Meeting Guides and Service Manuals.

This presentation will show the evolution from a very basic open center hydraulic system to a closedcenter pressure compensating hydraulic system and finally a load sensing/pressure compensatinghydraulic system.

NOTE: This material should be used to explain the basic concepts prior to teaching amachine using an LS/PC hydraulic system.

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The following colors will be used to represent the various oil pressures throughout this discussion:

Green - Return oil to tank

Blue - Blocked oil

Red - Pump supply pressure

Red and White Stripes - A lower supply pressure

Red Dots - An even lower supply pressure

Pink - The lowest supply pressure

Orange - Signal or pilot pressure

Orange and White Stripes - A lower signal pressure

It is important to remember that pressure compensation and load sensing are two different hydraulicdesign principles; they may or may not exist together.

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ENGINE

LOAD

TANK

FIXEDDISPLACEMENT

PUMP

CYLINDER

OPEN CENTERCONTROL VALVE

PRESSUREGAUGE

BASIC SYSTEM

2

BASIC SYSTEM (OPEN CENTER)

Introduction

We will begin our discussion by building a basic system composed of:

(1) A reservoir/tank,

(2) A fixed displacement pump,

(3) A pressure tap,

(4) An open center, lever actuated control valve, and

(5) A double-acting hydraulic cylinder.

With an open center system full pump flow is going through the control valve at all times, whethergoing straight to tank or to the cylinder. This constant flow of a large volume of oil has the potential ofcreating large amounts of heat if there are any restrictions in the flow path (i.e., valves). Heat reducescomponent life. By using large control valves to minimize the restriction or an oil cooler to removeheat, we can reduce the effects of heat; however, this may not always be practical due to extra cost orcomponents too large for the machine.

Note: With multiple envelope or multiple position valves, the envelopes in yellow have oilflowing through them while the envelopes in grey do not.

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ENGINE

TANK

FIXEDDISPLACEMENT

PUMP

CYLINDER

OPEN CENTERCONTROL VALVE

PRESSUREGAUGE

MAINRELIEF VALVE

2700 PSISPRING

LOAD

BASIC SYSTEMRELIEF VALVE ADDED

3

Relief Valve

With this simple system we can create high system pressures if we stall out the cylinder, whether due tobottoming out the cylinder or due to an extremely large load. To protect the system we should addanother component — a main relief valve.

One drawback to this is, we unload the system at high pressure, resulting in high heat buildup. Thehigher pressure can also reduce component life.

There are two other problems with this hydraulic system:

1) "Sticky" control spool movement, and

2) the cylinder speed varies with engine speed or changes in work port load (this causes the flowrate to change).

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FORCE

RESULTANT

FORCE

FLOW FORCES

SUPPLY

WORKPORT

4

FLOW FORCES

“Sticky" Control Spool

A “sticky” control valve is caused by something commonly known as “flow forces.”

Flow forces are those forces acting on a control spool. In this case, we will assume that they tend tendto keep the spool in the open position as long as there is flow passing through the orifice created by theopen spool. These flow forces are directly proportional to the amount of flow and the pressuredifferential across the spool land. In other words, as the flow and/or the pressure differential increases,the forces trying to keep the spool open increase.

The force vector, acting parallel (vertical) to the centerline of the control valve is the force trying tokeep the stem in the open position. In our simple valve, the closer the spool comes to closing off thesupply oil (decreasing the orifice size), the greater is the difference in pressure between the supply oiland the work port oil, and the greater is the force trying to keep the spool open.

To illustrate, think of closing a door against a strong wind. As you pushed on the door you created arestriction to the air flow. The closer you got to shutting the door, the resistance or forces workingagainst you became stronger. What you felt was the effect of flow and pressure across an orifice knownas flow forces.

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ENGINEFIXED

DISPLACEMENTPUMP

CYLINDER

OPEN CENTERCONTROL VALVEMAIN

RELIEF VALVE

2700 PSISPRING

CENTERINGSPRING

LOAD

BASIC SYSTEMCENTERING SPRING ADDED

5

Centering Spring

With hydraulic control valves it is helpful if the control spool centers itself. We can easily do this byadding a centering spring below the spool to close off the orifice when the operator releases the lever.Remember; however, the greater the flow and/or greater the system pressure, the greater the “flowforces,” therefore the heavier the centering spring becomes. What is the result? Higher lever efforts!This results in quicker operator fatigue!

How do we best solve this problem? Flow forces are related to both flow and pressure differential. Ifwe can minimize one or both of these factors, we can then minimize the flow forces and less thecentering spring force will be required, reducing lever effort.

VARYING CYLINDER SPEED

In a simple circuit like this, the cylinder speed is determined by the amount of flow across the controlspool. This can be affected by engine speed, work port load, lever displacement and pump output.

If the operator tried to maintain a constant cylinder speed as the engine changed speed or the work portload changed, he would have to continually change his lever displacement (vary the orifice size) tomaintain the same pressure differential across the control spool. We know from basic hydraulic theorythat when a pressure differential is constant across an orifice the flow rate through the orifice stays thesame. This is not easy to do because it adds to operator fatigue and requires constant lever movementby the operator to maintain constant implement speed.

Couple this constant lever movement with high lever effort and the operator will quickly becomefatigued.

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ENGINEFIXED

DISPLACEMENTPUMP

CLOSED CENTERCONTROL VALVE

2700 PSISPRING

DOUBLECHECK VALVE

PRESSUREREDUCING VALVE

50 PSISPRING

LOAD

PRESSURE COMPENSATION

6

PRESSURE COMPENSATION (CLOSED CENTER SYSTEM)

Wouldn’t it be great to solve both of these problems at the same time? We can, by adding a pressurereducing valve to control flow. We are also going to use a closed center control valves now.

A “double” check valve has been added which will sense work port pressure in either the head end orrod end of the cylinder and send this work port pressure to the pressure reducing valve. This pressure iscalled “signal” pressure and is always equal to the work port pressure.

NOTE: This pressure reducing valve is sometimes called a “pressure compensator”, “flowcontrol valve” or “flow compensator”, while the double check valveis sometimes called “shuttle valve”, or “ball resolver”.

NOTE: Some LS/PC implement valves communicate the work port pressure by means ofthree crossed drilled passages and one axial drilled passage in the main directional controlstem instead of by means of a double check valve or ball resolver.

How will adding a pressure reducing valve to control flow reduce lever efforts? If you recall from ourdiscussion of flow forces and their effect on lever efforts, the only way to reduce flow forces is toreduce flow and/or the pressure differential across the control spool. Since the flow is determined bythe (fixed-displacement) pump and the work port requirements, which we can cannot change, the onlyremaining variable is the pressure differential across the spool.

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From the schematic we can see that the pressure reducing valve senses “work port” pressure. Thispressure works with the pressure reducing valve spring to control downstream pressure. Thedownstream pressure is equal to work port pressure plus spring pressure. From the pressure reducingvalve, this downstream pressure feeds the directional (main control) spool. If the pressure to the themain control spool equals work port pressure plus the value of the spring in the spring chamber, then itis obvious that the pressure differential (difference between supply pressure to the main control spooland work port pressure) across the main control spool is equal to the value of the spring. If the springhas a force of 50 psi (345 kPa), we see that this 50 psi (345 kPa) controlled maximum pressuredifferential across the main control spool will minimize flow forces, and allow us to reduce thecentering spring size, reducing lever efforts.

This same pressure reducing valve also acts to negate the effect of varying engine speed on cylinderspeed. As engine speed increases pump flow increases, increasing pump supply pressure. The pressurereducing valve will react to this increase in pump supply pressure and restrict input flow to maintain thesame pressure differential across the main control spool. By doing so, we maintain a constant flow tothe cylinder. The opposite action will occur if the engine speed decreases.

This valve will also negate the effect of changing work port loads. Varying work loads will also notaffect implement speed. Implement speed will be constant.

Let's work through some examples:

Let's assume work port pressure equals 500 psi (3450 kPa). This pressure works with the 50 psi (345kPa) spring in our pressure reducing valve resulting in 550 psi (3795 kPa) at the main control spool.Our pressure differential across our main control spool is 50 psi (345 kPa) which is the value of ourspring.

If the work port pressure increases to 1000 psi (6900 kPa). This pressure works once again with the 50psi (345 kPa) spring in our pressure reducing valve resulting in 1050 psi (7245 kPa) at the main controlspool. Our pressure differential across our main control spool once again is 50 psi (345 kPa) which isthe value of our spring. Even though the load changed the pressure differential across the main controlspool stayed the same which results in constant flow.

NOTE: This holds true unless the load is higher than what the system is designed for orthe pump can’t produce adequate flow to meet flow requirements.

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DEFINITION OF

PRESSURE COMPENSATION

A control system which resultsin a constant implement speed for a

given lever displacement.

7

Definition of Pressure Compensation

We are now ready to give you a definition for "Pressure Compensation".

It is a control system which results in a constant implement speed for a given lever displacement.

This is done by maintaining a constant pressure differential across the main control spool by a spring inthe pressure reducing valve (which we use to control flow).

With pressure compensation, as we are describing here, there are actually two pressure differentialsoccurring.

1) One is the pressure differential across the pressure reducing valve itself. This pressuredifferential varies depending on the difference between pump supply pressure and work portpressure (plus the spring value).

2) Two is the differential across the main control spool, which is limited or controlled by the springin the pressure reducing valve.

Note: None of the following slides show values for flow rates or pressures. This was doneto provide the instructor with increased flexibility in developing his own teachingscenarios.

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Let’s work through an example:

An implement requires 5 gpm (19 l/m) and develops 1000 psi (6900 kPa) at the work port. The fixeddisplacement pump is capable of supplying 30 gpm (114 l/m). The spring in the pressure reducingvalve is equal to 50 psi (345 kPa), so the spring plus the work port pressure limits the downstreampressure to the main control stem to 1050 psi (7245 kPa). Since we don’t need full flow to theimplement the pump supply pressure increases to 2700 psi (18630 kPa) and the excess flow is dumpedto tank across the main relief valve.

The first pressure differential across the pressure reducing valve will be 2700 psi (18630 kPa) minus1050 psi (7245 kPa) which equals 1650 psi (11385 kPa) and the second pressure differential is 1050 psi(7245 kPa) minus 1000 psi (6900 kPa)which equals 50 psi (345 kPa) or the value of the spring.

When slow cylinder movement is required, the operator will just crack the closed center control spool,allowing only a small part of total pump flow to go to the cylinder. With a fixed displacement pump,the pump supply pressure increases and excess pump flow returns to tank. This high rate of flow at highpressure through the relief valve is going to rapidly build up a lot of heat which can shorten componentlife.

We mentioned earlier that we could add an oil cooler to reduce heat in the system. We also have at leasttwo other options:

1) We could add a flow control or dump valve to the system, or

2) we could replace the fixed displacement pump with a variable displacement pump.

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8

LOAD SENSING (CLOSED CENTER SYSTEM)

By using a flow control (dump) valve or a variable displacement pump with a pump control valve toregulate the system flow to meet the pressure needs of the system provides us with a feature known as“Load Sensing.”

Definition of Load Sensing

Load Sensing is a control system that maintains a pump supply pressure at a fixed value above thehighest system pressure requirement.

A signal network is used in both systems that sends the highest work port pressure (senses the load)back to either the flow control (dump) valve or to a pump control valve.

Within this signal network are several double check valves that are called “resolvers” or “shuttle”valves.

DEFINITION OF LOAD SENSING

A control system that maintains apump supply pressure at a fixedvalue above the highest system

pressure requirement.

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Flow Control "Dump" Valve

A flow control "dump" valve has now been added to the system.

Let’s work through an example using it:

The implement requires 5 gpm (19 l/m) and develops 1000 psi (6900 kPa) at the work port. The fixeddisplacement pump is capable of supplying 30 gpm (114 l/m). The spring in the flow control valve ordump valve is equal to 200 psi (1380 kPa), so the spring plus the work port pressure equaling 1200 psi(8280 kPa) is acting against pump supply pressure limiting it also to 1200 psi (8280 kPa). The excessflow not needed by the implement is dumped to tank at a pressure a set value higher than what is neededat the work port. This difference is equal to the spring or 200 psi (1380 kPa). This value is the“margin” pressure which ensures good implement response.

What if the implement is in “hold” and we don’t need any flow to it? Since we’re using a closed centercontrol valve you’d expect the system supply pressure to build up and go over the main relief; however,by using the flow control valve, system supply pressure will act against the 200 psi (1380 kPa) spring(which has 0 psi or 0 kPa work port pressure acting with it) and dump all 30 gpm (114 l/m) to tank at apump supply pressure of 200 psi (1380 kPa).

In both cases we are returning supply oil back to tank at a pressure lower than the main relief valvesetting. The flow control valve will minimize heat buildup and increase component life.

Now the biggest concern is wasted hydraulic horsepower. With this system the pump is alwayssupplying maximum flow regardless of what the implement needs. Excess flow goes back to tank. Thisis wasted energy.

9

ENGINE FIXEDDISPLACEMENT

PUMP

CLOSED CENTERCONTROL VALVE

2700 PSISPRING

DOUBLECHECK VALVE

50 PSISPRING

200 PSISPRING

LOAD

LOAD SENSING / PRESSURE COMPENSATION

FLOWCONTROL

VALVE

PRESSUREREDUCING VALVE

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VARIABLE DISPLACEMENT PUMP

PUMP OUTPUTLARGE ACTUATOR

SMALL ACTUATOR& BIAS SPRING

SWASH PLATE

DRIVESHAFT

YOKE PAD

PRESSURECOMPENSATOR

(PRESSURE CUT-OFF)

FLOW COMPENSATOR( MARGIN SPOOL)

10

Variable Displacement Pump

We can reduce this energy wasted by the fixed displacement pump by replacing it with a variabledisplacement pump.

The pump uses a control valve to control pump flow by changing the angle of the swashplate. The workport or signal pressure will act with the flow compensator spring in the pump control valve to give us apump supply pressure at a fixed value called “margin pressure” that is above the work port pressure.

The flow compensator spool will be able to sense pump supply pressure and work port pressure justlike the flow control valve did with the fixed displacement pump. As flow requirements changebecause of changing control lever movements, the difference in pressure between work port pressureand pump supply pressure will change in reaction to those lever movements. This will cause theposition of the flow compensator spool to change. It will either send more or less flow to the largeactuator piston in the pump. This will change the pump swashplate angle, which changes pump output.

We usually have a secondary control stem in the pump control valve that also reacts to pump supplypressure and is set to open at a given maximum pressure. This allows us to destroke the pump (reducepump flow) to maintain maximum system pressure without the use of a main relief valve. It called apressure compensator or pressure cut-off spool.

By setting the pump and its control valve to give exactly the flow needed to meet system work portpressure demands, the system is going to work much more efficiently (as compared to a system with afixed displacement pump).

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For example, the formula for hydraulic horsepower is:

(gpm x psi)/1714 = hp = gpm x psi x .000583

(l/m x kPa)/60240 = hkw = l/m x kPa x .0000166

If we use a 30 gpm (114 l/m) fixed displacement pump and a 1000 psi (6900 kPa) work port pressure, aflow control (dump) valve that senses work port pressure and acts with a 200 psi (1380 kPa) spring(margin), the hydraulic horsepower (hydraulic kilowatts) being pulled from the engine is :

30 gpm x (1000 psi + 200 psi) x .000583 = 21 hp

114 l/m x (6900 kPa + 1380 kPa) x .0000166 = 15.67 hkw

However, if we are metering flow to the cylinder, only using 5 gpm (19l/m) of flow, we are actuallyusing only:

5 gpm x (1000 psi + 200 psi) x .000583 = 3.5 hp

19 l/m x (6900 + 1380) x .0000166 = 2.61 hkw

What happened to that other 17.5 hp (13.06 hkw) being pulled from the engine? It is wasted anddumped back to the tank in the form of heat.

By using a variable displacement pump how much wasted energy do we have? Practically none! Thepump will give exactly the flow needed at a slightly higher pressure than is required. What the pumpgives us is:

5 gpm x 200 psi x .000583 = .58 hp wasted, which was used for the margin pressure.

19 l/m x 1380 x .0000166 = .44 hkw

Two advantages of using a variable displacement pump are:

1. Less heat is generated, giving longer component life, and

2. less horsepower is wasted from the engine, which results in using less fuel.

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ENGINE

VARIABLEDISPLACEMENT

PUMP

CLOSED CENTERCONTROL VALVE

DOUBLECHECK VALVE

PRESSUREREDUCING VALVE

50 PSISPRING

PUMPCONTROL

VALVE LOAD

LOAD SENSING / PRESSURE COMPENSATION

11

Getting back to the basic diagram we've moved the function of the flow control (dump) valve and themain relief valve to the pump control valve. As we said one of the spools in the pump control valve iscalled a "flow compensator or margin spool" which controls flow while the other is our "pressurecompensator or pressure cut-off" spool which limits maximum system pressure.

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MOTOR

2700 PSISPRING

200 PSISPRING

LOAD SENSING TWO VALVE OPERATION

A

B

LOAD

LOAD

12

LS/PC HYDRAULIC SYSTEMS - TWO VALVE OPERATION

Load Sensing Only (Fixed Displacement Pump)

We have already seen how pressure compensation works and have discussed how load sensing works.Now let’s put them together and show why we desire pressure compensation in a load sensing system.

In this schematic we have two implement control valves labeled "A" and "B". Neither one of them hasa pressure reducing valve in it’s circuit. There is a double check valve located between the rod and headend of each cylinder. If either one or both implements are operated, another double check valve willsend the higher of the two work port signals from the two valve bodies back to the the flow control(dump) valve.

These two implement valves we will treat as being “load sensing” only. We are going to operate bothvalves at the same time.

Valve “A” has a work port pressure of 2000 psi (13800 kPa), while valve”B” has a work port pressureof 500 psi (3450 kPa). The higher of the two work port pressures, which is 2000 psi (13800 kPa), willend up at the flow control valve through the “signal network”. This pressure will work along with the200 psi (1380 kPa) spring in the flow control valve. This will limit system supply pressure to 2200 psi(15180 kPa). Both implements will have 2200 psi (15180 kPa) felt on the supply side of their maincontrol spools.

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Let’s now calculate the pressure differential across each control spool.

At valve “A” we need 2000 psi (13800 kPa) at the work port. Pump supply is 2200 psi (15180 kPa).The difference between the two pressures is 200 psi (1380 kPa). This relative small pressure differentialacross the main control spool will not induce operator fatigue due to high lever efforts.

At valve “B” we need 500 psi (3450 kPa) at the work port. Pump supply is 2200 psi (15180 kPa). Thepressure differential across the main control spool is 1700 psi (11700 kPa). This higher pressuredifferential will try to open the orifice at the main control spool even more to allow more flow than whatwe need, due to the fact that this is the path of least resistance in the circuit. Flow across the orifice(created at the main control spool) will not remain constant as the load requirement varies (even if themain spool control is held in a fixed position). The operator will be required to constantly adjust themain valve spool if he is to maintain constant speed. The higher pressure differential across this type ofvalve will require heavier centering springs to prevent spool sticking. This will result in higher leverefforts and operator fatigue.

LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION

ENGINE

2700 PSISPRING

50 PSISPRING

200 PSISPRING

50 PSISPRING

LOAD

LOAD

A

B

13

Load Sensing with Pressure Compensation (Fixed Displacement Pump)

In this schematic we also have two implement control valves. Both of them now have a pressurereducing valve in their circuits. There is a double check valve located between the rod and head end ofeach cylinder. If either one or both implements are operated, another double check valve will send thehigher of the two work port signals from the two valve bodies back to the the flow control valve.

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We will treat these two implement valves as being “load sensing and pressure compensating". We aregoing to operate both valves at the same time.

Valve “A” has a work port pressure of 2000 psi (13800 kPa), while valve”B” has a work port pressureof 500 psi (3450 kPa). The higher of the two work port pressures, which is 2000 psi (13800 kPa), willend up at the flow control valve through the “signal network”. This pressure will work along with the200 psi (1380 kPa) spring in the flow control valve. This will limit system supply pressure to 2200 psi(15180 kPa). Both implements will have 2200 psi (15180) felt on the supply side of their main controlspools. Both pressure reducing valves use a 50 psi (345 kPa) spring.

From the schematic we can see that the pressure reducing valve will sense “work port” pressure. Thispressure works with the pressure reducing valve spring. The resultant pressure downstream of thepressure reducing valve is equal to work port pressure plus spring pressure. In valve “A” this pressureis 2000 psi (13800)work port pressure plus 50 psi (345 kPa) for the spring, or 2050 psi (14145) at thesupply side of the main control spool. We can now calculate the pressure differentials. Pump supplypressure of 2200 psi (15180 kPa) less 2050 psi (14145 kPa) equals 150 psi (1035 kPa) across thepressure reducing valve. The second pressure differential, which occurs across the main control spool,is 2050 psi (14145) less the work port pressure of 2000 psi (13800 kPa) is equal to 50 psi (345 kPa)which also happens to be the value of the spring in the pressure reducing valve.

Now lets see what happens with valve “B”. Work port pressure of 500 psi (3450 kPa) plus the springvalue of 50 psi (345 kPa) in the pressure reducing valve limits the supply pressure at the main controlstem to 550 psi (3795 kPa). We can now calculate the pressure differentials. Pump supply pressure of2200 psi (15180 kPa) less 550 psi (3795 kPa) equals 1650 psi (11385 kPa) across the pressure reducingvalve. The second pressure differential is 550 psi (3795 kPa), less the work port pressure of 500 psi(3450 kPa), is equal to 50 psi (345 kPa), which also happens to be the value of the spring in the pressurereducing valve.

We have a maximum controlled pressure differential of 50 psi (345 kPa) across each main control spooldue to the 50 psi (345 kPa) spring in the pressure reducing valves (even with varying loads). Thepressure reducing valves minimize flow forces at the main control spools, regardless of what the systemsupply pressure is, and allows us to reduce the centering spring size, which reduces lever efforts andoperator fatigue.

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ENGINE

LOAD SENSING TWO VALVE OPERATION

LOAD

LOAD

A

B

PUMPCONTROL

VALVE

14

Load Sensing Only (Variable Displacement Pump)

To get all of the benefits of load sensing we need to use a variable displacement pump, which willreduce the amount of energy wasted.

The effect on the individual control valves which are “load sensing” will be the same as per ourdiscussion with the flow control (dump) valve.

NOTE : This slide and the next have been included to allow the instructor additionalflexibility in using this material.

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LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION

ENGINE

50 PSISPRING

50 PSISPRING

LOAD

LOAD

A

B

PUMPCONTROL

VALVE

15

Load Sensing/Pressure Compensation (Variable Displacement Pump)

The effect on the individual control valves which are “load sensing and pressure compensating” will bethe same as per our discussion with the flow control (dump) valve.

We have now discussed some of the basics of “load sensing” and of “pressure compensation” as theyrelate to Caterpillar implement hydraulics. This presentation should make it easier to understandmaterial presented in other Service Training Meeting Guides on Load Sensing/Pressure CompensatedHydraulic Systems.

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STUDENT NOTES

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INTRODUCTION TOLOAD SENSING/PRESSURE COMPENSATING HYDRAULICS

QUIZ

I. Match the following:

1. ______ A control system that maintains a pump supplypressure at a fixed value above the highestsystem pressure.

2. ______ Can also be called work port pressure.

3. ______ A control system which results in a constantimplement speed for a given distance of leverdisplacement.

4. ______ A pressure reduction between two points.

5. ______ The difference in pressure between pumpsupply pressure and the highest signalpressure.

II. The following are true or false. Enter T for true and F for false. If false, circle the word or wordsthat make the statement false and replace with a word or words that make the statement correct.

A. A relief valve is "normally open."

B. In a pressure reducing valve, the valve spool is "normally open."

C. A pressure reducing valve limits the downstream pressure to a value equal to the controllingforce (either the spring or the spring and load signal).

D. The pressure reducing valve as used with a closed center control valve is sometimes called a"flow compensator spool," a "pressure compensating spool," or a "flow control spool."

E. A resolver valve can also be called a "shuttle valve" or a "relief valve."

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A. Load Sensing Hydraulics

B. Pressure drop

C. Margin pressure

D. Pressure Compensating Hyd.

E. Signal Pressure

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QUIZLOAD SENSING / PRESSURE COMPENSATION

TWO VALVE OPERATION

ENGINE

3000 PSISPRING

300 PSISPRING

LOAD

LOAD

A

B

90 PSISPRING

60 PSISPRING

III. Fill in the correct response.

If two valves are operated at the same time and at valve A the work port pressure (signal pressure ) is1800 psi and at valve B the work port pressure is 700 psi then:

System pressure will be:

1800 psi + psi = psi

CALCULATIONS:

Valve A work port pressure = 1800 psi

∆ P across pressure reducing valve = psi - psi = psi

∆ P across control spool = psi - psi = psi

Valve B work port pressure = 700 psi

∆ P across pressure reducing valve = psi - psi = psi

∆ P across control spool= psi - psi = psi

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QUIZLOAD SENSING / PRESSURE COMPENSATION

TWO VALVE OPERATION

ENGINE

3000 PSISPRING

300 PSISPRING

LOAD

LOAD

A

B

90 PSISPRING

60 PSISPRING

IV. Fill in the correct response.

If two valves are operated at the same time and at valve A the work port pressure (signal pressure ) is3000 psi and at valve B the work port pressure is 1000 psi then:

System pressure will be:

psi + psi = psi

CALCULATIONS:

Valve A work port pressure = 3000 psi

∆ P across pressure reducing valve = psi - psi = psi

∆ P across control spool = psi - psi = psi

Valve B work port pressure = 1000 psi

∆ P across pressure reducing valve = psi - psi = psi

∆ P across control spool= psi - psi = psi

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STUDENT NOTES

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