Pressure Limitation

2-3

Exercise 2-1

Pressure Limitation

EXERCISE OBJECTIVE

C To introduce the operation of a pressure relief valve.C To establish the oil flow path in a circuit using a pressure relief valve.C To connect and operate a circuit using a pressure relief valve.

DISCUSSION

Pressure limitation

Pressure is the amount of force exerted against a given surface. Flow is themovement of fluid caused by a difference in pressure between two points. Fluidalways flows from a higher pressure point to a lower pressure point. The citywaterworks, for example, builds up a pressure greater than the atmospheric pressurein our water pipes. As a result, when we turn on a water tap, the water is forced out.

When two parallel paths of flow are available, fluid will always take the path of leastresistance. An example of this in the everyday life would be a garden hose branchinginto two sections, as Figure 2-2 shows. One section is blocked, while the othersection allows water to move freely in it. All the water will flow through the unblockedsection since it is less restrictive than the blocked section. The input pressure willrise just enough for the water to flow through the unblocked section. The pressurein the blocked section will not build up beyond the level required to make the waterflow in the unblocked section. The pressure gauges in Figure 2-2, therefore, willindicate low, equal pressures.

Figure 2-2. Unrestricted flow path.

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Now what happens if we squeeze the unblocked section so that water is restrictedbut not completely confined, as Figure 2-3 shows? All the water will flow through thesqueezed section since it is still less restrictive than the blocked section. The inputpressure will rise to the level necessary to flow through the restricted path. Thepressure in the blocked section will not build up beyond the needs of the squeezedsection. The pressure gauges in Figure 2-3, therefore, will indicate high, equalpressures.

Figure 2-3. Restricted flow path.

So we see that the pressure in the blocked section can never be higher than thepressure in the unblocked section. In fact, these pressures will always be equal. Ifthe restricted section were closed completely, confining water instead of merelyrestricting it, the pressure in both sections would equal the maximum pressureavailable at the input.

In a hydraulic circuit, flow is produced by the action of a pump, which continuouslydischarges the oil at a certain flow rate. Pressure is not created by the pump itselfbut by resistance to the oil flow. When the oil is allowed to flow with no resistancethrough a hydraulic circuit, the pressure in that circuit is theoretically zero. When theflow is resisted, however, the circuit pressure increases to the amount necessary totake the easier path.

Relief valves

Figure 2-4 shows a hydraulic circuit consisting of two parallel paths of flow. The oilfrom the pump can pass through a relief valve or through a hydraulic circuitconsisting of a directional control valve and a cylinder.

The relief valve can be compared to the hand in the hose example previouslydescribed. It limits the maximum pressure in the system by providing an alternateflow path to the reservoir whenever the oil flow to the circuit is blocked, as when thedirectional valve is in the blocked center position or when the cylinder is fullyextended or retracted.

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The relief valve is connected between the pump pressure line and reservoir. It isnormally non-passing. It is adjusted to open at a pressure slightly higher than thecircuit requirement and divert the pumped oil to the reservoir when this pressure isreached.

In Figure 2-4, for example, all the oil from the pump flows through the circuit as longas the cylinder is not fully extended because the circuit provides an easier path thanthe relief valve. While the cylinder is extending, the pressure rises only to the amountnecessary to force oil on the rod side of the cylinder into the reservoir (here 700 kPa,or 100 psi).

Figure 2-4. Oil flows through the circuit.

Once the cylinder is fully extended, the cylinder circuit becomes blocked and thepumped oil can no longer flow through it. The system pressure climbs to 3450 kPa(500 psi), then the relief valve opens and the oil is dumped back to the reservoir atthe relief valve pressure setting of 3450 kPa (500 psi), as Figure 2-5 shows.Thereafter, no flow occurs throughout the circuit and the pressure is equalthroughout. The circuit pressure, therefore, cannot build up beyond the relief valvepressure setting.

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Figure 2-5. Oil flows through the relief valve.

Hydraulics Trainer relief valves

Your Hydraulics Trainer contains two relief valves. One of these valves, called mainrelief valve, is located inside the Power Unit. The other valve, called secondaryrelief valve, is supplied with your kit of hydraulic components. The two valves areidentical. However, you will operate the secondary valve only. The main valve isfactory-set at a higher pressure than the secondary valve. It is used as an additionalsafety device for backing up the secondary valve. It should not be readjusted ortampered with.

Figure 2-6 illustrates the relief valve supplied with your kit of hydraulic components.This valve is of pilot-operated type. The valve body has three ports: a pressure (P)port, which is to be connected to the pump pressure line, a tank (T) port, which is tobe connected to the reservoir, and a vent (V) port, which is used for control of thevalve from a remote point by external valve. The use of the vent port will bediscussed in detail in Exercise 4-4. When not used, this port should be leftunconnected.

By sensing the upstream pressure on the P port of the valve, an internal spoolcontrols the flow of oil through the valve by acting on a large spring. The pressurelevel where the spool is wide open and all the pumped oil passes through the valveis called relieving pressure, or full-open pressure.

The relieving pressure can be set by using the adjustment knob on the valve body.Turning the knob clockwise increases the compression of a small spring locatedabove the valve spool, which increases the relieving pressure and allows higherpressures to build up in the circuit. Notice that the knob must first be pulled beforeit can be turned. When the knob is released, a spring forces the knob to engage afixed spline. This prevents vibrations and shocks from changing the adjustment.

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2-7

The pressure at which the relief valve begins to open is called cracking pressure.This pressure is below the valve relieving pressure. At cracking pressure, the valveopens just enough to let the first few drops of oil through. Pressure override is thepressure difference between the cracking pressure and the relieving pressure.

Before turning on the Power Unit, the valve should always be completely open(adjustment knob turned fully counterclockwise) to allow the pump to start under thelightest load and to prevent the system components from being subjected topressure surges. Once the Power Unit is running, the relief valve can be closedgradually until the desired pressure is reached.

REFERENCE MATERIAL

For detailed information on pilot-operated relief valves, refer to the chapter entitledPilot Operated Pressure Control Valve in the Parker-Hannifin’s manual IndustrialHydraulic Technology.

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2-8

Figure 2-6. The trainer Relief Valve.

Procedure summary

In the first part of the exercise, you will measure the cracking pressure of the ReliefValve supplied with your kit of hydraulic components. You will adjust the valverelieving pressure by modifying the compression of its spring.

In the second part of the exercise, you will test the effect of pressure limitation on abasic hydraulic circuit.

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2-9

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of this manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

Relief valve operation

G 1. Connect the circuit shown in Figure 2-7. Refer to the connection diagramshown in Figure 2-8 to make your connections.

Note: As Figure 2-7 shows, the vent (V) port of the Relief Valveis unused in this circuit. Therefore, leave this port unconnected.

Figure 2-7. Schematic diagram of the circuit for adjusting the Relief Valve.

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2-10

Figure 2-8. Connection diagram of the circuit for adjusting the Relief Valve.

G 2. Before starting the Power Unit, perform the following start-up procedure:

a. Make sure the hoses are firmly connected.b. Check the level of the oil in the reservoir. Add oil if required.c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF

position.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve completely. To do so, pull the valve adjustment

knob and turn it fully counterclockwise.

G 3. Turn on the Power Unit by setting its power switch to ON. Since the oil flowis blocked at gauge A, all the pumped oil is now being forced through theRelief Valve.

The pressure reading of gauge A is the minimum pressure required todevelop an oil flow through the valve (cracking pressure). It corresponds tothe pressure required to counteract the resistance of the spring inside thevalve. Record below the pressure reading of gauge A.

Cracking pressure = kPa or psi

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2-11

Note: The Trainer Pressure Gauges provide “bar” and “psi”readings. Since bar is a metric unit of measurement forpressures, students working with S.I. units must multiply themeasured pressure in bars by 100 to obtain the equivalentpressure in kilopascals (kPa).

G 4. Now, compress the spring of the Relief Valve by turning its adjustment knobclockwise 2 turns. Use the vernier scale on the knob for the adjustment.What is the reading of gauge A?

Pressure = kPa or psi

G 5. Why does the pressure reading increase as the spring compression isincreased?

G 6. Turn the Relief Valve adjustment knob fully clockwise while watching thereading of gauge A. Can the pressure level be increased beyond 6200 kPa(900 psi)? Why?

G 7. Turn off the Power Unit.

G 8. Based on the cracking pressure recorded in step 3, at which pressure willthe Relief Valve start to open if the relieving pressure is set to 3450 kPa(500 psi)?

Limiting system pressure

G 9. Modify the existing circuit in order to obtain the circuit shown in Figures 2-9and 2-10. Make sure to mount the 3.81-cm (1.5-in) bore cylinder in aposition where its rod can extend freely.

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2-12

Note: For ease of connection, the Directional Control Valvesupplied with your Hydraulics Trainer is bolted to a subplate towhich the hoses can be connected. The arrangement of ports P,T, A, and B on the valve subplate does not follow the symbol forthe directional valve appearing in Figure 2-9 and on themanufacturer nameplate on top of the valve. Thus, port P actuallyfaces port B on the subplate, while port T faces port A. Therefore,always refer to the letters stamped on the valve subplate whenconnecting the valve into a circuit.

Figure 2-9. Schematic diagram of the cylinder actuation circuit.

G 10. Make sure the hoses are firmly connected. Open the Relief Valvecompletely by turning its adjustment knob fully counterclockwise.

G 11. Turn on the Power Unit.

G 12. Turn the Relief Valve adjustment knob clockwise until gauge A reads1400 kPa (200 psi).

G 13. Stay clear of the cylinder rod. Move the lever of the Directional ControlValve toward the valve body, which should extend the cylinder rod. Then,move the lever outward from the valve body, which should retract the rod.

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Figure 2-10. Connection diagram of the cylinder actuation circuit.

G 14. While watching the reading of gauge A, move the lever of the directionalvalve toward the valve body to extend the cylinder rod. What is the pressureat gauge A during the extension stroke of the rod?


G 15. What is the pressure at gauge A when the cylinder is fully extended?


G 16. Move the lever of the directional valve outward from the valve body toretract the cylinder rod.

G 17. Turn the Relief Valve adjustment knob clockwise until gauge A reads2100 kPa (300 psi).

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G 18. While watching the reading of gauge A, move the lever of the directionalvalve toward the valve body to extend the cylinder rod. What is the pressureat gauge A during the extension stroke of the cylinder rod?


G 19. What is the pressure at gauge A when the cylinder rod is fully extended?


G 20. Move the lever of the directional valve outward from the valve body toretract the cylinder rod.

G 21. Turn off the Power Unit. Open the Relief Valve completely by turning itsadjustment knob fully counterclockwise.

G 22. Explain the reason for the nearly identical pressures registered duringcylinder extension at the two relief valve pressure settings.

G 23. Why does the circuit pressure increase when the cylinder rod is fullyextended?

G 24. Disconnect all hoses. It may be necessary to move the directional valvelever back and forth to relieve static pressure; the quick connects can thenbe removed. Wipe off any hydraulic oil residue.

G 25. Remove all components from the work surface and wipe off any hydraulicoil residue. Return all components to their storage location.

G 26. Clean up any hydraulic oil from the floor and the trainer. Properly disposeof any paper towels and rags used to clean up oil.

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2-15

CONCLUSION

In the first part of the exercise, you measured the minimum pressure setting of arelief valve by connecting the valve between the pump pressure line and reservoirand by opening the valve completely.

You then modified the valve relieving pressure by increasing the compression of itsinternal spring, which increased the circuit pressure.

In the second part of the exercise, you tested the effect of pressure limitation on abasic hydraulic circuit. You learned that pressure changes depend on the movementof oil through the circuit. When the cylinder rod extends or retracts, the circuitpressure rises only to the amount required to force oil out of the cylinder back intothe reservoir. When the cylinder rod becomes fully extended or retracted, however,the circuit pressure rises to the relief valve pressure setting.

Up to that point, we have seen that pilot-operated relief valves provide pressurecontrol by sensing pressure upstream on their input line. Pilot-operated relief valvescan also sense pressure in another part of the system or even in a remote systemby means of a vent line. This type of operation is identified as remote control and isachieved through the use of the relief valve vent port. Remote control of a relief valvewill be described in detail in Exercise 4-4.

REVIEW QUESTIONS

1. What is the purpose of a relief valve?

2. Explain the difference between the main relief valve in the Power Unit and theRelief Valve supplied with your kit of hydraulic components (secondary reliefvalve)?

3. What type of relief valve is used in your Hydraulics Trainer?

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2-16

4. What might happen to a hydraulic system if the tank port of the relief valve is notconnected to the power unit return line port?

5. Define the term cracking pressure.

6. In the circuit of Figure 2-11, what will be the pressure reading of gauge A duringcylinder extension and when the cylinder is fully extended if the relief valvepressure setting is changed from 3400 kPa (500 psi) to 6900 kPa (1000 psi)?

Note: The pressure required to extend the cylinder rod is 600 kPa(85 psi).

Figure 2-11. Circuit for review question 6.

2-17

Exercise 2-2

Pressure and Force

EXERCISE OBJECTIVE

C To verify the formula F = P x A using a cylinder and a load spring;C To discover what happens to a cylinder when equal pressure is applied to each

side of its piston;C To explain the concept of pressure distribution in a cylinder in equilibrium of

forces;C To determine the weight of the Power Unit given the pressure required to lift it.

DISCUSSION

Pascal’s Law

Pascal’s Law states that pressure applied on a confined fluid is transmittedundiminished in all directions, and acts with equal force on equal areas, andat right angles to them.

Figure 2-12 illustrates this basic properties of fluids. The bottle in this example iscompletely filled with a non-compressible fluid. When a stopper is placed in the topof the bottle, and a force is applied to the top of the stopper, the fluid inside the bottleresists compression by pushing with an equal pressure in all directions.

Figure 2-12. Force applied to a confined fluid.

The generated pressure is equal to the force applied to the top of the stopper dividedby the area of the stopper. In equation form:

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2-18

S.I. units:

English units:

As you can see, pressure is measured in “Newtons per 10 square centimeters (kPa)”in S.I. units, or in “pounds per square inch” (psi) in English units. In physics formula,the word “per” can be rewritten as a division sign.

Memorizing the pyramid in Figure 2-13 will make rearranging of formula P = F/Aeasier. In the pyramid, the letter on the top row equals the product of the bottom twoletters. A letter on the bottom row equals the top letter divided by the other bottomletter.

Figure 2-13. Rearranging formulas.

Hydraulic pressure versus cylinder force

In a hydraulic circuit, the force that pushes the oil, attempting to make it flow, comesfrom a mechanical pump. When the oil pushed on by the pump is confined withina restricted area, as in the body of a cylinder, there is a pressure build-up, and thispressure can be used to do useful work.

As you can see, pressure is not created by the pump but by resistance to theoil flow. The amount of pressure created in a circuit will only be as high as requiredto counteract the least resistance to flow in the circuit. Resistance to flow mainlycomes from three sources: resistance to motion of the load attached to the cylinder,frictional resistance of the cylinder seals, and frictional resistance of the inner wallof the hoses.

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In Figure 2-14, the oil from the pump is confined in the cap end of the cylinder. Asa result, pressure develops in the cap end of the cylinder. This pressure is exertedevenly over the entire surface of the cap end of the cylinder. It acts on the piston,resulting in a mechanical force to push the load.

Figure 2-14. Cylinder pushing a load.

To find the amount of force generated by the piston during its extension, we canrewrite the formula P = F/A as F = P x A. Therefore, the generated force is equal tothe pressure in the cap end of the cylinder times the piston area being acted upon.This area is called full area, or “face” area.

In Figure 2-15, the oil from the pump is confined in the rod end of the cylinder. Asa result, pressure develops in the rod end of the cylinder. This pressure is exertedevenly over the entire surface of the rod end of the cylinder. It acts on the piston,resulting in a mechanical force to pull the load.

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Figure 2-15. Cylinder pulling a load.

This time, however, the generated force is lower because the piston area availablefor the pressure to act on is reduced by the fact that the cylinder rod covers a portionof the piston. This area is called annular area, or “donut” area. Therefore, the systemmust generate more pressure to pull than to push the load.

Conversion factors

Table 2-1 shows the conversion factors used to convert measurements of force,pressure, and area from S.I. units to English units and vice versa.

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2-21

Force

Newtons (N) x 0.225 = Pounds-force(lb; lbf)

x 4.448 = Newtons (N)

Pressure

Kilopascals (kPa) x 0.145 = Pounds-forceper square inch(psi; lb/in2; lbf/in2)

x 6.895 = Kilopascals (kPa)

Area

Square centimeters(cm2)

x 0.155 = Square inches(in2)

x 6.45 = Square centimeters(cm2)

Table 2-1. Conversion factors.

For example, the pressure generated by the fluid in Figure 2-12 is 10 kPa, inS.I. units, or 1.45 psi, in English units, as demonstrated below:

S.I. units:

English units:

REFERENCE MATERIAL

For additional information on the relationship between force and pressure, refer tothe chapters entitled Hydraulic Transmission of Force and Energy and HydraulicActuators in the Parker-Hannifin’s manual Industrial Hydraulic Technology.

Procedure summary

In the first part of the exercise, you will verify the formula F = P x A by measuring thecompression force of a cylinder on a loading device.

In the second part of the exercise, you will predict and demonstrate what happenswhen equal pressure is applied to both sides of a piston.

In the third part of the exercise, you will determine how much pressure there is ineach side of a cylinder in equilibrium of forces.

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2-22

In the fourth part of the exercise, you will measure the pressure required to lift thePower Unit in order to determine its weight.

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of this manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

Conversion of pressure to force

G 1. What is the formula for determining the force in a hydraulic system?

G 2. If the bore diameter, D, of a cylinder piston equals 3.81 cm (1.5 in), calculatethe full area, Af, of this piston. Use the formula shown in Figure 2-14. It isgiven below for your convenience.

G 3. Using this area and the formula from step 1, calculate the theoretical forceof the cylinder for the pressure levels in Table 2-2. Record your calculationsin Table 2-2 under “THEORETICAL”.

PRESSURE APPLIED ONFULL PISTON AREA

THEORETICAL CYLINDERFORCE

ACTUAL CYLINDER FORCE

3500 kPa (500 psi)

2800 kPa (400 psi)

2100 kPa (300 psi)

Table 2-2. Cylinder force versus pressure.

G 4. Remove the 3.81-cm (1.5-in) bore cylinder from its adapter by unscrewingits retaining ring. Make sure the cylinder tip (bullet) is removed from thecylinder rod end.

G 5. As Figure 2-16 (a) shows, screw the cylinder into the Loading Device untilthe load piston inside the Loading Device begins to push on the spring andthe cylinder fittings point upwards. Do not use a tool to turn the cylinder!

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Figure 2-16. Loading device assembly.

Note: If the 3.81-cm (1.5-in) bore cylinder is not fully retracted,do not try to screw the cylinder into the Loading Device. Insteadconnect the cylinder actuation circuit of Figure 2-10. Turn theknob of the relief valve fully counterclockwise, then turn on thepower unit. Actuate the lever of the directional control valve toretract the cylinder rod fully, then turn off the power unit.Disconnect the circuit. Now screw the cylinder into the LoadingDevice as shown in Figure 2-16 (a).

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G 6. Clip the NEWTON/LBF-graduated ruler to the Loading Device, and align the“0 Newton” or “0 lbf” mark with the colored line on the load piston.Figure 2-16 (b) shows ruler installation for measurement of forces inNewtons (N). The ruler must be installed on the other side of the LoadingDevice in order to measure forces in pounds (lbf or lb).

G 7. Connect the circuit shown in Figures 2-17 and 2-18.

Figure 2-17. Schematic diagram of the circuit for measuring the output force of a cylinder.

G 8. Before starting the Power Unit, perform the following start-up procedure:

a. Make sure the hoses are firmly connected.b. Check the level of the oil in the reservoir. Add oil if required.c. Put on safety glasses.d. Make sure the power switch on the Power Unit is set to the OFF pos-

ition.e. Plug the Power Unit line cord into an ac outlet.f. Open the Relief Valve by turning its adjustment knob fully

counterclockwise.

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Figure 2-18. Connection diagram of the circuit for measuring the output force of a cylinder.


G 10. Move the lever of the directional valve toward the valve body to directthe pumped oil toward the cap end of the cylinder. While keeping the valvelever shifted, turn the Relief Valve adjustment knob clockwise until thepressure at gauge A equals 4100 kPa (600 psi). Observe that the appliedpressure caused the cylinder to compress the spring in the Loading Device.

G 11. With the lever of the directional valve still shifted toward the valve body, turnthe Relief Valve adjustment knob counterclockwise to decrease thepressure at gauge A at 3500 kPa (500 psi). Note the force reading on theLoading Device, and record this value in Table 2-2 under “ACTUAL”.

Note: To counteract hysteresis of the spring load device andobtain a more accurate force reading at 3500 kPa (500 psi), thepressure applied to the cylinder piston must first be set at4100 kPa (600 psi) and then decreased at 3500 kPa (500 psi).

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G 12. By shifting the directional valve lever and adjusting the knob on the ReliefValve to decrease the pressure at gauge A in steps, measure the forcereading for the pressure levels in Table 2-2. Record your results inTable 2-2.

G 13. When you have finished, move the directional valve lever outward from thevalve body to retract the rod, then turn off the Power Unit. Open the ReliefValve completely (turn knob fully counterclockwise).

G 14. Compare the actual forces you obtained in the experiment with thetheoretical forces in Table 2-2. Are these values within 10% of each other?

G Yes G No

G 15. Does force increase or decrease as pressure increases?

Applying equal pressure on both sides of a piston

G 16. Connect the circuit shown in Figure 2-19. Use the 2.54-cm (1-in) borecylinder.

Figure 2-19. Applying equal pressure on both sides of a piston.

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Note: If the 2.54-cm (1-in) bore cylinder is not fully retracted, donot connect the circuit of Figure 2-19. Instead connect thecylinder actuation circuit shown in Figure 2-10. Open the reliefvalve completely, then turn on the power unit. Actuate the leverof the directional valve to retract the cylinder rod fully, then turnoff the power unit. Disconnect the circuit. Now, connect the circuitof Figure 2-19.

G 17. Examine the circuit of Figure 2-19. This circuit applies equal pressure to thefull and annular sides of the piston. However, the piston area available forthe pressure to act on is less on the annular side because the cylinder rodcovers a portion of the piston. Given that force is equal to pressuremultiplied by area, predict which side of the piston will develop the mostforce.

G 18. What do you think will happen to the cylinder rod?


G 20. Turn the Relief Valve adjustment knob clockwise until the circuit pressureat gauge A equals 2100 kPa (300 psi).

G 21. While observing the cylinder rod, move the lever of the directional valvetoward the valve body so that the pumped oil is directed toward both sidesof the cylinder piston. In which direction does the rod move? Why?

G 22. Turn off the Power Unit. Open the relief valve completely (turn knob fullycounterclockwise).

Pressure distribution in a cylinder in equilibrium of forces

G 23. Disconnect the 2.54-cm (1-in) bore cylinder from the circuit. Connect the twoports of this cylinder to the return manifold. Slowly push the piston rod inuntil it is retracted half way. Disconnect the cylinder from the returnmanifold, then connect the circuit shown in Figure 2-20 (a).

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Figure 2-20. Determining pressure distribution in a cylinder.

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G 24. Examine the circuit in Figure 2-20 (a). The oil in the rod side of the cylinderis captured because there is no return path to the reservoir. The pistoncannot move because the oil cannot be compressed. The pressures in thecap and rod sides build until the forces exerted on both sides of the pistonare exactly equal.

– The force on the full area of the piston is:

– The force on the annular area of the piston is:

– Since these forces are equal:

Based on the above formulas, predict which gauge in Figure 2-20 (a) willread the most pressure. Explain.

G 25. Turn on the Power Unit. Turn the relief valve adjustment knob clockwiseuntil the input pressure at gauge A (Pf) is 1400 kPa (200 psi). Then, recordthe output pressure at gauge B (Pa) in Table 2-3.

PART A

INPUT PRESSUREAPPLIED ONFULL PISTON

AREA

INPUT PRESSUREAT GAUGE A

(Pf)

OUTPUT PRESSURE ATGAUGE B

(Pa)

INPUT/OUTPUTPRESSURE RATIO

(Pf/Pa)

RECIPROCAL OFAREA RATIO (Aa/Af)

1400 kPa(200 psi)

2100 kPa(300 psi)

PART B

INPUTPRESSURE

APPLIED ONANNULAR

AREA

INPUT PRESSUREAT GAUGE A (Pa)

OUTPUT PRESSURE ATGAUGE B (Pf)

INPUT/OUTPUTPRESSURE RATIO

(Pa/Pf)

RECIPROCAL OFAREA RATIO (Af/Aa)

1400 kPa(200 psi)

2100 kPa(300 psi)

Table 2-3. Pressure distribution in the cylinder of Figure 2-20.

G 26. Increase the Relief Valve pressure setting until the input pressure atgauge A is 2100 kPa (300 psi) and again record the output pressure atgauge B in Table 2-3.

G 27. Turn off the Power Unit. Open the Relief Valve completely (turn knob fullycounterclockwise).

G 28. Switch the two hoses connected to the ports of the cylinder with each otherso that the rod end is connected to gauge A and the cap end is connectedto gauge B, as Figure 2-20 (b) shows.

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G 29. Examine the circuit in Figure 2-20 (b). Predict which gauge will read themost pressure. Explain why.

G 30. Turn on the Power Unit. Turn the Relief Valve adjustment knob clockwiseuntil the input pressure at gauge A (Pa) is 1400 kPa (200 psi). Record theoutput pressure at gauge B (Pf) in Table 2-3.

G 31. Increase the Relief Valve pressure setting until the input pressure atgauge A is 2100 kPa (300 psi) and again record the output pressure atgauge B in Table 2-3.


G 33. Complete the fourth column of Table 2-3, “INPUT/OUTPUT PRESSURERATIO”, using the input and output pressures registered in parts A and B ofthe experiment.

G 34. Complete the fifth column of Table 2-3, “RECIPROCAL OF AREA RATIO”,given that the piston diameter, D, is 2.54 cm (1 in) and the rod diameter, d,is 1.59 cm (0.625 in). Use the formulas below to determine Af and Aa.

Af = D² x 0.7854

Aa = (D² ! d²) x 0.7854

G 35. In a cylinder in equilibrium of forces, the ratio of input to output pressure istheoretically equal to the reciprocal (inverse) of the area ratio.

Compare the input/output pressure ratios Pf/Pa in Table 2-3, part A, with thereciprocal of area ratio, Aa/Af. Also, compare the input/output pressure ratiosPa/Pf in Table 2-3, part B, with the reciprocal of area ratio, Af/Aa. Are thepressure ratios approximately equal to the reciprocal of area ratio?

G Yes G No

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Measuring the weight of the Power Unit

G 36. Disconnect the Power Unit line cord from the wall outlet.

G 37. Disconnect the 2.54-cm (1-in) bore cylinder from the circuit, then remove thecylinder from its adapter.

G 38. Insert the cylinder rod into the cylinder hole in the Power Unit lifting frame,then fasten the cylinder to the lifting frame by tightening its retaining ringsecurely. Position the lifting frame over the Power Unit, with its open side atthe rear of the Power Unit.

G 39. Connect the two cylinder ports together using a hose full of oil, then pull thepiston rod out until it touches the lifting attachment on the Power Unit.Fasten the cylinder to the Power Unit by screwing the lifting attachment ontothe threaded end of the cylinder rod. Then, disconnect the hose from thecylinder.

G 40. Connect the circuit shown in Figure 2-21.

CAUTION!

Make sure the hoses and Power Unit line cord will notbecome wedged between rigid parts of the trainer when thePower Unit is lifted.

G 41. Plug the Power Unit line cord into the wall outlet, then turn on the PowerUnit.

G 42. Move the lever of the directional valve outward from the valve body andslowly turn the Relief Valve adjustment knob clockwise until the Power Unitbegins to rise. Then, release the valve lever.

G 43. According to gauge A, how much pressure is currently applied on theannular area of the cylinder piston?


G 44. Move the lever of the directional valve toward the valve body to return thePower Unit to the ground.

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Figure 2-21. Circuit used to lift the Power Unit.


G 46. Based on the annular pressure recorded in step 43, determine the weight(mass) of the Power Unit in both S.I. and English units.

Note: 1 Newton is equal to 0.1019 kilogram.

G 47. Disconnect the Power Unit line cord from the wall outlet, then disconnect allhoses. Wipe off any hydraulic oil residue.

G 48. Unscrew the cylinder from the Power Unit lifting attachment. Unscrew thering retaining the cylinder to the lifting frame. Remove the cylinder from thelifting frame. Reinstall the cylinder on its adapter by fastening its retainingring securely.

G 49. Remove all components from the work surface and wipe off any hydraulicoil residue. Return all components to their storage location.

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G 50. Clean up any hydraulic oil from the floor and from the trainer. Properlydispose of any paper towels and rags used to clean up oil.

CONCLUSION

In this exercise, you learned that the force exerted on a given surface is directlyproportional to the pressure applied on this surface. Since the relationship betweenforce and pressure is linear, it is possible to predict the force exerted by the cylinderat any pressure setting.

You also learned what happens when equal pressure is applied to both sides of apiston. Since the working area on the rod side of the cylinder is less than the workingarea on the cap side of the cylinder, the piston has a tendency to extend when equalpressure is applied to each end.

You then determined the pressure distribution in a cylinder in equilibrium of forces.The cylinder was blocked and the oil was captured in the cylinder, so the pressuresin the cap and rod sides had to build until the forces exerted on both sides of thepiston were exactly equal. The rod side of a cylinder in equilibrium of forces mustbuild more pressure than the cap side because the working area on the rod side(annular area) is less than the working area on the cap side (full area).

Finally, you measured the pressure required to lift the Power Unit using the 2.54-cm(1-in) bore cylinder. You then calculated the force exerted on the annular area of thepiston using the measured pressure and the formula F = P x A. This force wascorresponding to the weight (mass) of the Power Unit.

REVIEW QUESTIONS

1. What is the formula for calculating the force in a hydraulic system? How can yourewrite this formula to calculate pressure?

2. What is the formula for calculating the surface area of a piston?

3. How much pressure must be applied to the cap end of a 2.54-cm (1-in) borecylinder in order to compress a spring 5.08 cm (2 in), if the spring rate is728 N/cm (416 lb/in)?

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4. In the circuit of Figure 2-20 (a), what will be the pressure at gauge B if thepressure at gauge A pressure is raised to 3500 kPa (500 psi)?

5. In the circuit of Figure 2-20 (b), what will be the pressure at gauge B if thepressure at gauge A pressure is raised to 3500 kPa (500 psi)?

Documents

Pressure Limitation