12. Load Sensing

Service Training August 1999

INTRODUCTION TO LOAD SENSING/PRESSURE COMPENSATED HYDRAULIC SYSTEMS

LOAD SENSING / PRESSURE COMPENSATION PUMP CONTROL VALVE

LOAD

ENGINE

VARIABLE DISPLACEMENT PUMP

DOUBLE CHECK VALVE

50 PSI SPRING

PRESSURE REDUCING VALVE

CLOSED CENTER CONTROL VALVE

STUDENT HANDBOOK

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Student Materials

COURSE PLAN/OUTLINE GET ACQUAINTED MODULE: INTRODUCTION TO LOAD SENSING PRESSURE COMPENSATED HYDRAULICS - Objectives - Lesson: Basic LS/PC Hydraulics - Slide presentation on LS/PC hydraulics - Quiz: LS/PC Hydraulics

MODULE: INTRODUCTION TO LOAD SENSING PRESSURE COMPENSATED HYDRAULICS 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 system characteristics of Load Sensing and Pressure Compensation and components used in this type of hydraulic system. 2. On a quiz in the back of the Student Handbook, determine system pressures and pressure differences at various locations on provided quiz schematics.

Materials Needed: LS/PC Student Handout with Quiz Video (optional) "Introduction to LS/PC Hydraulics Systems"

SEVN1804

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Student Materials

1

INTRODUCTION Load sensing/pressure compensated” hydraulic systems are now used in a number of Caterpillar machines. Some of the current Caterpillar built machines which use these systems include the 416-446 Backhoe Loaders, Challenger 65, H-Series Track-Type Tractors, updated G-Series Motor graders, and 916-936 Wheel Loaders (IT18 and IT28). For more detailed information on the hydraulic systems of these machines refer to their respective individual Service Training Meeting Guides and Service Manuals. This presentation will show the evolution from a very basic open center hydraulic system to a closed center pressure compensating hydraulic system and finally a load sensing/pressure compensating hydraulic system. NOTE: This material should be used to explain the basic concepts prior to teaching machine using an LS/PC hydraulic system.

a

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Student Materials

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 hydraulic design principles; they may or may not exist together.

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Student Materials

BASIC SYSTEM CYLINDER

TANK LOAD ENGINE

FIXED DISPLACEMENT PUMP

2

PRESSURE GAUGE

OPEN CENTER CONTROL VALVE

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, whether going straight to tank or to the cylinder. This constant flow of a large volume of oil has the potential of creating large amounts of heat if there are any restrictions in the flow path (i.e., valves). Heat reduces component life. By using large control valves to minimize the restriction or an oil cooler to remove heat, we can reduce the effects of heat; however, this may not always be practical due to extra cost or components too large for the machine. Note: With multiple envelope or multiple position valves, the envelopes in yellow have oil flowing through them while the envelopes in grey do not.

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Student Materials

BASIC SYSTEM RELIEF VALVE ADDED TANK

CYLINDER

LOAD ENGINE

FIXED DISPLACEMENT PUMP

3

PRESSURE GAUGE

2700 PSI SPRING

OPEN CENTER CONTROL VALVE MAIN RELIEF VALVE

Relief Valve With this simple system we can create high system pressures if we stall out the cylinder, whether due to bottoming out the cylinder or due to an extremely large load. To protect the system we should add another component — a main relief valve. One drawback to this is, we unload the system at high pressure, resulting in high heat buildup. The higher 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 flow rate to change).

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Student Materials

FLOW FORCES

4

SUPPLY WORKPORT

FORCE

RESULTANT FORCE

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 tend to keep the spool in the open position as long as there is flow passing through the orifice created by the open spool. These flow forces are directly proportional to the amount of flow and the pressure differential 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 to keep the stem in the open position. In our simple valve, the closer the spool comes to closing off the supply oil (decreasing the orifice size), the greater is the difference in pressure between the supply oil and 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 a restriction to the air flow. The closer you got to shutting the door, the resistance or forces working against you became stronger. What you felt was the effect of flow and pressure across an orifice known as flow forces.

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Student Materials

BASIC SYSTEM CENTERING SPRING ADDED

CYLINDER

LOAD ENGINE

FIXED DISPLACEMENT PUMP

5

2700 PSI SPRING MAIN RELIEF VALVE

OPEN CENTER CONTROL VALVE CENTERING SPRING

Centering Spring With hydraulic control valves it is helpful if the control spool centers itself. We can easily do this by adding 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 “flow forces,” 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. If we can minimize one or both of these factors, we can then minimize the flow forces and less the centering 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 control spool. 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 port load changed, he would have to continually change his lever displacement (vary the orifice size) to maintain the same pressure differential across the control spool. We know from basic hydraulic theory that when a pressure differential is constant across an orifice the flow rate through the orifice stays the same. This is not easy to do because it adds to operator fatigue and requires constant lever movement by the operator to maintain constant implement speed. Couple this constant lever movement with high lever effort and the operator will quickly become fatigued.

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Student Materials

PRESSURE COMPENSATION LOAD ENGINE FIXED DISPLACEMENT PUMP DOUBLE CHECK VALVE

6

50 PSI SPRING

2700 PSI SPRING PRESSURE REDUCING VALVE

CLOSED CENTER CONTROL VALVE

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 pressure reducing 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 or rod end of the cylinder and send this work port pressure to the pressure reducing valve. This pressure is called “signal” pressure and is always equal to the work port pressure. NOTE: This pressure reducing valve is sometimes called a “pressure compensator”, “flow control valve” or “flow compensator”, while the double check valve is sometimes called “shuttle valve”, or “ball resolver”. NOTE: Some LS/PC implement valves communicate the work port pressure by means of three crossed drilled passages and one axial drilled passage in the main directional control stem 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 our discussion of flow forces and their effect on lever efforts, the only way to reduce flow forces is to reduce flow and/or the pressure differential across the control spool. Since the flow is determined by the (fixed-displacement) pump and the work port requirements, which we can cannot change, the only remaining variable is the pressure differential across the spool.

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Student Materials

From the schematic we can see that the pressure reducing valve senses “work port” pressure. This pressure works with the pressure reducing valve spring to control downstream pressure. The downstream pressure is equal to work port pressure plus spring pressure. From the pressure reducing valve, this downstream pressure feeds the directional (main control) spool. If the pressure to the the main control spool equals work port pressure plus the value of the spring in the spring chamber, then it is obvious that the pressure differential (difference between supply pressure to the main control spool and work port pressure) across the main control spool is equal to the value of the spring. If the spring has a force of 50 psi (345 kPa), we see that this 50 psi (345 kPa) controlled maximum pressure differential across the main control spool will minimize flow forces, and allow us to reduce the centering spring size, reducing lever efforts. This same pressure reducing valve also acts to negate the effect of varying engine speed on cylinder speed. As engine speed increases pump flow increases, increasing pump supply pressure. The pressure reducing valve will react to this increase in pump supply pressure and restrict input flow to maintain the same pressure differential across the main control spool. By doing so, we maintain a constant flow to the 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 not affect 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 (345 kPa) 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 our spring. If the work port pressure increases to 1000 psi (6900 kPa). This pressure works once again with the 50 psi (345 kPa) spring in our pressure reducing valve resulting in 1050 psi (7245 kPa) at the main control spool. Our pressure differential across our main control spool once again is 50 psi (345 kPa) which is the value of our spring. Even though the load changed the pressure differential across the main control spool stayed the same which results in constant flow. NOTE: This holds true unless the load is higher than what the system is designed for or the pump can’t produce adequate flow to meet flow requirements.

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Student Materials

DEFINITION OF PRESSURE COMPENSATION A control system which results in 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 in the pressure reducing valve (which we use to control flow). With pressure compensation, as we are describing here, there are actually two pressure differentials occurring. 1) One is the pressure differential across the pressure reducing valve itself. This pressure differential varies depending on the difference between pump supply pressure and work port pressure (plus the spring value). 2) Two is the differential across the main control spool, which is limited or controlled by the spring in the pressure reducing valve. Note: None of the following slides show values for flow rates or pressures. This was done to provide the instructor with increased flexibility in developing his own teaching scenarios.

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Student Materials

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 fixed displacement pump is capable of supplying 30 gpm (114 l/m). The spring in the pressure reducing valve is equal to 50 psi (345 kPa), so the spring plus the work port pressure limits the downstream pressure to the main control stem to 1050 psi (7245 kPa). Since we don’t need full flow to the implement the pump supply pressure increases to 2700 psi (18630 kPa) and the excess flow is dumped to tank across the main relief valve. The first pressure differential across the pressure reducing valve will be 2700 psi (18630 kPa) minus 1050 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 high pressure through the relief valve is going to rapidly build up a lot of heat which can shorten component life. We mentioned earlier that we could add an oil cooler to reduce heat in the system. We also have at least two 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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Student Materials

DEFINITION OF LOAD SENSING A control system that maintains a pump supply pressure at a fixed value above the highest system pressure requirement.

8

LOAD SENSING (CLOSED CENTER SYSTEM) By using a flow control (dump) valve or a variable displacement pump with a pump control valve to regulate 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 the highest 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.

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Student Materials

LOAD SENSING / PRESSURE COMPENSATION LOAD ENGINE

FIXED DISPLACEMENT PUMP

FLOW CONTROL VALVE

9

DOUBLE CHECK VALVE

200 PSI SPRING

50 PSI SPRING

2700 PSI SPRING PRESSURE REDUCING VALVE

CLOSED CENTER CONTROL VALVE

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 fixed displacement pump is capable of supplying 30 gpm (114 l/m). The spring in the flow control valve or dump 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 excess flow not needed by the implement is dumped to tank at a pressure a set value higher than what is needed at 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 center control 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 a pump 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 valve setting. 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 always supplying maximum flow regardless of what the implement needs. Excess flow goes back to tank. This is wasted energy.

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Student Materials

VARIABLE DISPLACEMENT PUMP PUMP OUTPUT LARGE ACTUATOR

YOKE PAD SWASH PLATE

10

DRIVE SHAFT

FLOW COMPENSATOR ( MARGIN SPOOL)

PRESSURE COMPENSATOR (PRESSURE CUT-OFF)

SMALL ACTUATOR & BIAS SPRING

Variable Displacement Pump We can reduce this energy wasted by the fixed displacement pump by replacing it with a variable displacement pump. The pump uses a control valve to control pump flow by changing the angle of the swashplate. The work port or signal pressure will act with the flow compensator spring in the pump control valve to give us a pump 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 just like the flow control valve did with the fixed displacement pump. As flow requirements change because of changing control lever movements, the difference in pressure between work port pressure and pump supply pressure will change in reaction to those lever movements. This will cause the position of the flow compensator spool to change. It will either send more or less flow to the large actuator 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 supply pressure and is set to open at a given maximum pressure. This allows us to destroke the pump (reduce pump flow) to maintain maximum system pressure without the use of a main relief valve. It called a pressure compensator or pressure cut-off spool. By setting the pump and its control valve to give exactly the flow needed to meet system work port pressure demands, the system is going to work much more efficiently (as compared to a system with a fixed displacement pump).

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Student Materials

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, a flow 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 actually using 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 and dumped back to the tank in the form of heat. By using a variable displacement pump how much wasted energy do we have? Practically none! The pump will give exactly the flow needed at a slightly higher pressure than is required. What the pump gives 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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Student Materials

LOAD SENSING / PRESSURE COMPENSATION PUMP CONTROL VALVE

LOAD

ENGINE

VARIABLE DISPLACEMENT PUMP

11

DOUBLE CHECK VALVE

50 PSI SPRING

PRESSURE REDUCING VALVE

CLOSED CENTER CONTROL VALVE

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

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Student Materials

LOAD SENSING TWO VALVE OPERATION MOTOR

LOAD

12 LOAD 200 PSI SPRING

2700 PSI SPRING

A

B

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 has a pressure reducing valve in it’s circuit. There is a double check valve located between the rod and head end of each cylinder. If either one or both implements are operated, another double check valve will send 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 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 pressure of 500 psi (3450 kPa). The higher of the two work port pressures, which is 2000 psi (13800 kPa), will end up at the flow control valve through the “signal network”. This pressure will work along with the 200 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 main control spools.

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Student Materials

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 differential across 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). The pressure differential across the main control spool is 1700 psi (11700 kPa). This higher pressure differential will try to open the orifice at the main control spool even more to allow more flow than what we 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 the main spool control is held in a fixed position). The operator will be required to constantly adjust the main valve spool if he is to maintain constant speed. The higher pressure differential across this type of valve will require heavier centering springs to prevent spool sticking. This will result in higher lever efforts and operator fatigue.

LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION ENGINE LOAD

200 PSI SPRING

13

LOAD 50 PSI SPRING

2700 PSI SPRING

A

50 PSI SPRING

B

Load Sensing with Pressure Compensation (Fixed Displacement Pump) In this schematic we also have two implement control valves. Both of them now have a pressure reducing valve in their circuits. There is a double check valve located between the rod and head end of each cylinder. If either one or both implements are operated, another double check valve will send the higher of the two work port signals from the two valve bodies back to the the flow control valve.

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Student Materials

We will treat these two implement valves as being “load sensing and pressure compensating". We are going 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 pressure of 500 psi (3450 kPa). The higher of the two work port pressures, which is 2000 psi (13800 kPa), will end up at the flow control valve through the “signal network”. This pressure will work along with the 200 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 control spools. 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. This pressure works with the pressure reducing valve spring. The resultant pressure downstream of the pressure reducing valve is equal to work port pressure plus spring pressure. In valve “A” this pressure is 2000 psi (13800)work port pressure plus 50 psi (345 kPa) for the spring, or 2050 psi (14145) at the supply side of the main control spool. We can now calculate the pressure differentials. Pump supply pressure of 2200 psi (15180 kPa) less 2050 psi (14145 kPa) equals 150 psi (1035 kPa) across the pressure 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 spring value of 50 psi (345 kPa) in the pressure reducing valve limits the supply pressure at the main control stem to 550 psi (3795 kPa). We can now calculate the pressure differentials. Pump supply pressure of 2200 psi (15180 kPa) less 550 psi (3795 kPa) equals 1650 psi (11385 kPa) across the pressure reducing valve. 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 pressure reducing valve. We have a maximum controlled pressure differential of 50 psi (345 kPa) across each main control spool due to the 50 psi (345 kPa) spring in the pressure reducing valves (even with varying loads). The pressure reducing valves minimize flow forces at the main control spools, regardless of what the system supply pressure is, and allows us to reduce the centering spring size, which reduces lever efforts and operator fatigue.

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Student Materials

LOAD SENSING TWO VALVE OPERATION PUMP CONTROL VALVE

ENGINE

LOAD

14

LOAD

A

B

Load Sensing Only (Variable Displacement Pump) To get all of the benefits of load sensing we need to use a variable displacement pump, which will reduce the amount of energy wasted. The effect on the individual control valves which are “load sensing” will be the same as per our discussion with the flow control (dump) valve. NOTE : This slide and the next have been included to allow the instructor additional flexibility in using this material.

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Student Materials

LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION PUMP CONTROL VALVE

ENGINE LOAD

LOAD

15

50 PSI SPRING

A

50 PSI SPRING

B

Load Sensing/Pressure Compensation (Variable Displacement Pump) The effect on the individual control valves which are “load sensing and pressure compensating” will be the 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 they relate to Caterpillar implement hydraulics. This presentation should make it easier to understand material presented in other Service Training Meeting Guides on Load Sensing/Pressure Compensated Hydraulic Systems.

STUDENT NOTES

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Student Materials

INTRODUCTION TO LOAD SENSING/PRESSURE COMPENSATING HYDRAULICS QUIZ I. Match the following: 1. ______ A control system that maintains a pump supply pressure at a fixed value above the highest system pressure.

A. Load Sensing Hydraulics

2. ______ Can also be called work port pressure.

C. Margin pressure

3. ______ A control system which results in a constant implement speed for a given distance of lever displacement.

D. Pressure Compensating Hyd.

B. Pressure drop

E. Signal Pressure

4. ______ A pressure reduction between two points. 5. ______ The difference in pressure between pump supply pressure and the highest signal pressure.

II. The following are true or false. Enter T for true and F for false. If false, circle the word or words that 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 controlling force (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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Student Materials

QUIZ LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION ENGINE LOAD

300 PSI SPRING LOAD

90 PSI SPRING 3000 PSI SPRING 60 PSI SPRING

A

B 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 ) is 1800 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 = ∆ P across control spool =

psi -

psi -

psi =

psi =

psi psi

Valve B work port pressure = 700 psi

∆ P across pressure reducing valve = ∆ P across control spool=

psi -

psi psi =

psi = psi

psi

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Student Materials

QUIZ LOAD SENSING / PRESSURE COMPENSATION TWO VALVE OPERATION ENGINE LOAD

300 PSI SPRING LOAD

90 PSI SPRING 3000 PSI SPRING

A

60 PSI SPRING

B 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 ) is 3000 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 = ∆ P across control spool =

psi -

psi -

psi =

psi =

psi psi

Valve B work port pressure = 1000 psi

∆ P across pressure reducing valve = ∆ P across control spool=

psi -

psi psi =

psi = psi

psi

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STUDENT NOTES

Student Materials