BASIC HYDRAULICS LEARNING ACTIVITY PACKET PRINCIPLES OF HYDRAULIC PRESSURE AND FLOW TM BB831-XA03XEN
LEARNING ACTIVITY PACKET 3 PRINCIPLES OF HYDRAULIC PRESSURE AND FLOW INTRODUCTION Previous LAPs discussed the construction and operation of basic hydraulic circuits and how they are represented using standard schematic symbols. It is important to understand the principles that allow power to be transmitted by a pressurized liquid. A good understanding of this LAP will help develop skills that are used in analyzing, specifying, and troubleshooting hydraulic components. This LAP will provide the foundation for understanding the theory of fluid power with the principles of pressure and flow. It will explain Pascal s Law and how it is used to multiply and transmit forces, how to determine cylinder forces, and how fluid flows. ITEMS NEEDED Amatrol Supplied 1 85-BH Basic Hydraulic Training System 1 85-HPS Hydraulic Power Unit School Supplied 1 Ruler 1 Adjustable Wrench 1 Flathead Screwdriver FIRST EDITION, LAP 3, REV. A Amatrol, AMNET, CIMSOFT, MCL, MINI-CIM, IST, ITC, VEST, and Technovate are trademarks or registered trademarks of Amatrol, Inc. All other brand and product names are trademarks or registered trademarks of their respective companies. Copyright 2009 by AMATROL, INC. All rights Reserved. No part of this publication may be reproduced, translated, or transmitted in any form or by any means, electronic, optical, mechanical, or magnetic, including but not limited to photographing, photocopying, recording or any information storage and retrieval system, without written permission of the copyright owner. Amatrol,Inc., P.O. Box 2697, Jeffersonville, IN 47131 USA, Ph 812-288-8285, FAX 812-283-1584 www.amatrol.com Copyright 2009 Amatrol, Inc. 2
TABLE OF CONTENTS SEGMENT 1 PRESSURE VS. CYLINDER FORCE... 4 OBJECTIVE 1 Describe how to calculate the force output of an extending cylinder SKILL 1 Calculate the extension force of a cylinder given its size and pressure SKILL 2 Measure the force output of an extending cylinder OBJECTIVE 2 Describe how to calculate the force output of a hydraulic cylinder in retraction (pull) SKILL 3 Calculate the retraction force of a cylinder given its size and pressure SKILL 4 Measure the force output of a retracting cylinder SEGMENT 2 HYDRAULIC LEVERAGE....28 OBJECTIVE 3 State Pascal s Law and explain its significance in hydraulics Activity 1 Verification of Pascal s Law for hydraulics OBJECTIVE 4 Explain how force is multiplied using Pascal s Law Activity 2 Demonstrate how distance is sacrificed to obtain force multiplication SEGMENT 3 FLUID FRICTION....40 OBJECTIVE 5 Describe two types of resistance in a hydraulic system OBJECTIVE 6 Explain how Delta P describes hydraulic resistance SKILL 5 Measure Delta P across a hydraulic component Activity 3 Effect of flow and orifice size on Delta P OBJECTIVE 7 Explain how pressure is distributed in a hydraulic system Activity 4 Characteristics of circuit pressure drops SEGMENT 4 ABSOLUTE VS. GAUGE PRESSURE... 58 OBJECTIVE 8 Describe two methods of representing hydraulic pressure SKILL 6 Convert between absolute pressure and gauge hydraulic pressure OBJECTIVE 9 Describe how oil flows on the suction side of the pump Copyright 2009 Amatrol, Inc. 3
SEGMENT 1 PRESSURE VS. CYLINDER FORCE OBJECTIVE 1 DESCRIBE HOW TO CALCULATE THE FORCE OUTPUT OF AN EXTENDING CYLINDER In a hydraulic cylinder, the pressure will rise to a level that creates a force which is equal and opposite to the force created by the load. If a cylinder is trying to extend a load, as shown in figure 1, the pressure developed in the oil is exerted over all the surfaces inside the cap end of the cylinder. However, only on the area of the piston does this pressure exert a force to move the load. The amount of pressure required to move the load depends on the area of the cylinder s piston, as shown in figure 1. LOAD FORCE LOAD CROSS SECTION OF PISTON PISTON/ROD FORCE D FLUID FORCE FLUID PRESSURE Figure 1. Cylinder Extending a Load Copyright 2009 Amatrol, Inc. 4
The theoretical amount of force created by the cylinder can be calculated by knowing the pressure and the area of the piston. Since the piston is circular, the area can be obtained by using the formula for the area of a circle: πr 2, or 0.7854D 2. Therefore, the formula for determining force output of a cylinder during extension can be stated as follows: FORMULA: FORCE OUTPUT OF CYLINDER DURING EXTENSION U.S. Customary Units: S.I. Units: F = P A F = P A = P D 2 0.7854 F = P A = P D 2 0.07854 Where Note: F = Force output of cylinder rod extending (lbs or Newtons) P = Pressure on piston (psi or kpa) A = Area of piston (in 2 or cm 2 ) D = Diameter of piston (in or cm) The 0.07854 in the S.I. units occurs in converting meters and Pascals to centimeters and kpa. Copyright 2009 Amatrol, Inc. 5
SKILL 1 CALCULATE THE EXTENSION FORCE OF A CYLINDER GIVEN ITS SIZE AND PRESSURE Procedure Overview In this procedure, you will calculate the theoretical force output of the cylinders used on the 850 Series trainer for several pressures. In step 1, you will be given an example to show how to do this. 1. Let s assume you have a cylinder with a bore of 2.0 in/5.08 cm and a pressure of 1000 psi/6900 kpa. The force output of the cylinder would be calculated as follows: U.S. Customary Units: S.I. Units: F = P D 2.7854 = 1000 (2) 2.7854 = 1000 4.7854 = 3,142 lbs. F = P D 2.07854 = 6900 (5.08) 2.07854 = 6900 25.81.07854 = 13,987 N 2. Now calculate the theoretical force of extension of the two cylinders used in the 850 Series hydraulic trainer for each of the pressures shown in the table. Use the formula for the force output of a cylinder during extension. For the 850 Series hydraulic trainer, the piston diameter of the large bore cylinder is 1.5 in (3.81 cm) and the piston diameter of the small bore cylinder is 1.125 in. (2.86 cm). PRESSURE (psi/kpa) THEORETICAL EXTENSION FORCE Large Cylinder (lbs/n) Small Cylinder (lbs/n) 150/1035 / / 200/1380 / / 250/1725 / / 300/2070 / / Copyright 2009 Amatrol, Inc. 6
3. Calculate the minimum bore diameter of a cylinder that will lift the load shown in figure 2. The maximum pressure available is 1000 psi/6900 kpa. The minimum bore diameter required is (in/cm) 1767 LBS / 7,860 N LOAD P BORE DIAMETER Figure 2. Cylinder Under Load NOTE This is a typical application of the F = P A formula. Copyright 2009 Amatrol, Inc. 7
SKILL 2 MEASURE THE FORCE OUTPUT OF AN EXTENDING CYLINDER Procedure Overview In this procedure, you will learn a simple method of measuring the force output of the cylinder. This test will verify that the formula you used in Skill 1 works. 1. Connect the circuit shown in figures 3 and 4. In this circuit, the flow from the pump deadheads against the cylinder s piston. The fluid pressure at the cylinder s piston can then be adjusted by adjusting the pressure setting of the relief valve. In the steps that follow, you will verify that the cylinder force calculation formula actually works by measuring the load on the cylinder at different pressures. To do this, we will use a spring. HYDRAULIC INSTRUMENTATION PANEL GAUGE A GAUGE B GAUGE C FLOW METER SUPPLY MANIFOLD CYLINDER FLOW FLOW CONTROL CONTROL #1 #2 A A B B MOTOR RETURN MANIFOLD CYLINDER HYDRAULIC ACTUATOR MODULE Figure 3. Pictorial of Test Circuit for Measuring Force Output of a Cylinder in Extension Copyright 2009 Amatrol, Inc. 8
RELIEF VALVE PRESSURE GAUGE A ROD CAM COMPRESSED SPRING LENGTH L 1 CYLINDER (LARGE) RELIEF VALVE PRESSURE FORCE OF CYLINDER SPRING FORCE Figure 4. Schematic of Circuit for Measuring Force Output of a Cylinder in Extension 2. If not already installed, perform the following substeps to mount the load spring, as shown in figure 6. A. Remove the clear plastic guard covering the load rod. B. Then unscrew the load rod from the rod cam, as shown in figure 5. The load rod has wrench flats on one end to help unscrew it. C. Place the spring over the load rod and screw the rod cam back into the load rod, as shown in figure 6. D. Replace the clear plastic guard. Figure 5. Rod Cam Unscrewed from Load Rod Copyright 2009 Amatrol, Inc. 9
Figure 6. Load Spring Mounted for Extension 3. Perform the following checkout procedures before starting the power unit. A. Check the oil level. Fill if necessary. B. Press the stop pushbutton on the motor starter to make sure the starter is in the Off position. C. Plug in the power cord to a wall outlet. D. Reduce the relief valve to its minimum pressure setting by turning CCW fully. E. Make sure the shutoff valve is closed. Copyright 2009 Amatrol, Inc. 10
4. Before starting the power unit, use a ruler to measure the uncompressed length, L, of the spring, as shown in figure 7. All lengths should be read to the nearest 1/32 inch (0.5 mm). Record this length below. Uncompressed Length (L) = (in/cm) L 1 2 3 4 RULER Figure 7. Measure Length of Spring WARNING Keep your hands and fingers away from the load spring while the power unit is running. The load spring will compress during operation. This will make it possible for you to pinch your fingers. If the spring needs to be aligned while the power unit is running, use a screwdriver, pencil, or pen to move the spring. Copyright 2009 Amatrol, Inc. 11
The principle you will be using to measure the force output is called Hooke s Law. It says that a spring compresses proportionately to the force placed on it. Here, the force is from the rod of the cylinder, as shown in figure 8. P FROM PUMP SPRING L 1 = F CYL F LS Figure 8. Measuring Cylinder Force Output Using a Spring The load spring reacts to a load by compressing to the distance where the spring force balances the cylinder force. If the cylinder force increases, the spring compresses further until the forces are balanced. The force needed to compress a spring a given distance is constant and is called the spring rate (K). This is also called Hooke s constant. This characteristic of the spring allows us to calculate the actual force output of the cylinder by measuring the distance of spring compression and multiplying it by the spring rate. This is shown in the following formula: FORMULA: LOAD SPRING FORCE F LS = (L - L 1 ) K Where: F LS = Force of load spring (lbs or Newtons) F LS also equals the force of cylinder rod, F CYL L = Uncompressed length of spring (in or cm) L 1 = Compressed length of spring (in or cm) K = Spring rate (lbs / in or Newton / cm) For the 850 trainer, the spring the rate is 294 lbs / in or 515 Newtons / cm Copyright 2009 Amatrol, Inc. 12
5. Start the power unit and open the shutoff valve. 6. Increase the relief valve s setting until the pressure at gauge A reads 150 psi/1035 kpa. As you do this, watch the cylinder s rod and spring. You should observe that the increase in pressure causes the cylinder piston to extend forward until it compresses the spring enough that the spring force balances the cylinder s force. 7. Measure the new compressed spring length, L 1, for the pressure of 150 psi/1035 kpa. PRESSURE (psi/kpa) COMPRESSED SPRING LENGTH (L 1 ) (in/cm) 150/1035 / 200/1380 / 250/1725 / 300/2070 / 8. Repeat steps 6 and 7 for each of the other pressures listed in the chart. Each time you increase the pressure you should observe that the spring compresses a little further. The more the spring is compressed the greater its force output. Since the spring force is equal to the force created by the fluid pressure on the piston this also means the cylinder s force output increases with pressure. 9. After you have completed the chart, experiment with your ability to position the cylinder by changing the pressure with the relief valve. WARNING Do not exceed 400 psi/2760 kpa setting at the relief valve. Above this pressure, the spring becomes completely compressed and can no longer be used to calculate forces. 10. Reduce the relief valve s setting to minimum and perform the following substeps to allow the cylinder to fully retract. NOTE The spring force can only retract the cylinder to a point where its force is balanced by the friction of the cylinder seals. These substeps will completely retract the cylinder to allow easy removal of the load spring in the next step. Copyright 2009 Amatrol, Inc. 13
A. Turn off the power unit. B. Reverse the hose connections to the cylinder. C. Start the power unit. D. Turn the power unit back off and close the shutoff valve. 11. Perform the following substeps to remove the load spring. A. Remove the clear plastic guard. B. Unscrew the load rod from the rod cam. C. Remove and store the load spring. D. Re-attach the load rod to the rod cam. E. Replace the clear plastic guard. 12. Calculate the spring length change and use the load spring force formula to calculate the actual force output for each pressure listed in the chart below. PRESSURE (psi/kpa) SPRING LENGTH CHANGE (L-L1) (in/cm) ACTUAL EXTENSION FORCE (lbs/n) 150/1035 / / 200/1380 / / 250/1725 / / 300/2070 / / 13. Compare these forces with those obtained from the force formula in Skill 1. The actual values obtained should be smaller than those obtained by calculation. The reason for this is that other forces that resist rod movement have been omitted from the calculation. These include friction, which always resists movement, and back pressure from fluid leaving the backside of the cylinder. Frictional forces occur in two areas. One area is between the piston and cylinder body. The other area is between the rod and the rod end cap. NOTE Designers always add at least 10% to their calculation of pressure or cylinder size theoretical to account for these frictional forces. Copyright 2009 Amatrol, Inc. 14
OBJECTIVE 2 DESCRIBE HOW TO CALCULATE THE FORCE OUTPUT OF A HYDRAULIC CYLINDER IN RETRACTION (PULL) When a cylinder retracts or pulls a load, the oil pressure is applied to the rod end side of the piston, as shown in figure 9. In this case, the oil pressure cannot act against the entire surface of the piston because of the rod, it can only act against the donut or annular area, as shown in figure 9. This means that the retracting force capability of a cylinder is less than its extending force capability. PISTON FORCE CROSS SECTION OF PISTON FLUID FORCE FLUID PRESSURE GENERATED TO PULL THE LOAD ANNULAR AREA ROD AREA LOAD FORCE LOAD Figure 9. Cylinder Retracting a Load Copyright 2009 Amatrol, Inc. 15
The area used in the formula for calculating the cylinder force in extension can be modified with the rod area to calculate the force in retraction by using the following formula: FORMULA: FORCE OUTPUT OF CYLINDER RETRACTING U.S. Customary Units: F = P A F = P A a D = P [ 2 2 P DR ] 0.7854 S.I. Units: F = P A a D = P [ 2 2 P DR ] 0.07854 Where F = Force output of cylinder rod retracting (lbs or Newtons) P = Pressure on piston (psi or kpa) A a = Annular area of piston (in 2 or cm 2 ) D P = Diameter of piston (in or cm) D R = Diameter of rod (in or cm) Copyright 2009 Amatrol, Inc. 16
SKILL 3 CALCULATE THE RETRACTION FORCE OF A CYLINDER GIVEN ITS SIZE AND PRESSURE Procedure Overview In this procedure, you will calculate the theoretical retraction force for the two cylinders used in the 850 Series trainer at several different pressures. In step 1, you will be given an example. 1. Let s assume you have a cylinder with a bore of 2.0 in/5.08 cm, a rod diameter of 1.0 in/2.54 cm. and a pressure of 1000 psi/6900 kpa. The force output of the cylinder in retraction would be calculated as follows: U.S. Customary Units: S.I. Units: F = P x [D P 2 - D R 2 ] x 0.7854 = 1000 x [(2.0) 2 - (1.0) 2 ] x 0.7854 = 1000 x [4-1] x 0.7854 = 1000 x [3] x 0.7854 = 2,356 lbs. F = P x [D P 2 - D R 2 ] x 0.07854 = 6900 [ (5.08) 2 - (2.54) 2 ] x 0.07854 = 6900 [25.8-6.5] x 0.07854 = 6900 x [19.3] x 0.07854 = 10,459 N Compare this result with the calculation of the force in extension in step 1 of skill 1. As you can see, the retraction force is less at the same pressure. 2. Now, calculate the theoretical force of retraction of the two cylinders for the pressures shown in the table. For these calculations, you will be using the cylinders of the hydraulic actuator module, the same as used in Skill 1. The large bore cylinder has a piston diameter of 1.5 in. (3.81 cm) and a rod diameter of 0.44 in. (1.12 cm). The small bore cylinder has a piston diameter of 1.125 in. (2.86 cm) and a rod diameter of 0.31 in. (0.79 cm). PRESSURE (psi/kpa) THEORETICAL RETRACTION FORCE Large Cylinder (lbs/n) Small Cylinder (lbs/n) 150/1035 / / 200/1380 / / 250/1725 / / 300/2070 / / Copyright 2009 Amatrol, Inc. 17
3. Calculate the minimum pressure needed to lift the load shown in figure 10, given the following information: Load = 5000 lbs / 22,241 Newtons Cylinder Diameter = 3 in / 7.62 cm Rod Diameter = 1 in / 2.54 cm P LOAD Figure 10. Cylinder Retracting a Load Minimum pressure (psi/kpa) NOTE This is a typical design calculation. Designers must usually calculate the load retracting as well as the load extending. Copyright 2009 Amatrol, Inc. 18
SKILL 4 MEASURE THE FORCE OUTPUT OF A RETRACTING CYLINDER Procedure Overview In this procedure, you will measure the actual force output in retraction of both cylinders on the 850 Series trainer using the load spring. This will also show that a cylinder s actual force output in retraction is less than extension and demonstrate that piston size has an effect on the net force output. 1. Perform the following substeps to mount the load spring on the other side of the rod cam of the large bore cylinder, as shown in figure 11. WARNING Keep the relief valve setting at minimum during this step. Installing the load spring for this procedure requires the hands and fingers to be near pinch points. Use extreme care during this step. Figure 11. Load Spring Mounted for Retraction Copyright 2009 Amatrol, Inc. 19
A. Set up a cylinder reciprocation circuit, as shown in figure 12. IN OUT A B DIRECTIONAL CONTROL VALVE CYLINDER Figure 12. Cylinder Reciprocation Circuit B. Perform the power unit checkout procedures and then turn on the power unit. C. Open the shutoff valve. D. Retract the cylinder with the DCV. The minimum pressure should move the cylinder with no load. If the cylinder will not move, increase the relief valve setting just enough to move the cylinder. E. Reduce the pressure to minimum and shutoff the power unit. F. Remove the clear plastic guard covering the load rod. Copyright 2009 Amatrol, Inc. 20
G. Remove the load rod from the cam and slide it to the right until the threads are flush with the load block, as shown in figure 13. H. Remove the cam from the cylinder, as shown in figure 13, and set it aside. STEP H: REMOVE CYLINDER CAM LOAD BLOCK STEP G: REMOVE LOAD ROD. SLIDE LOAD ROD TO THE RIGHT. STOP WHEN THREADS ARE FLUSH HERE. Figure 13. Removing the Load Rod and Cylinder CAM I. Install the loose rod extender on the end of the cylinder rod and slide the spring over the extender rod, as shown in figure 14. ROD EXTENDER Figure 14. Extender Piece Attached to Rod Copyright 2009 Amatrol, Inc. 21
J. Turn on the power unit and verify that the relief valve is set at minimum. K. While holding the spring against the cylinder, use the DCV to extend the rod until the threads are just beyond the spring, as shown in figure 15. L. Attach the cam to the extender rod, as shown in figure 15. M. Slide the load rod to the left and attach it to the cam, as shown in figure 15. Use the wrench flats at the end of the load rod if necessary. CAM STEP L: ATTACH CAM TO EXTENDER ROD LOAD ROD WRENCH FLAT STEP K: HOLD SPRING AGAINST CYLINDER. EXTEND ROD JUST FAR ENOUGH TO ATTACH CAM. STEP M: SLIDE LOAD ROD TO THE LEFT AND ATTACH TO CAM. Figure 15. Setup Diagram N. Retract the cylinder until the spring is snug and centered. O. Close the shutoff valve and turn off the power unit. P. Replace the clear plastic guard. Copyright 2009 Amatrol, Inc. 22
2. Set up the test circuit using the large bore cylinder shown in figure 16. GAUGE A SPRING CYLINDER ROD CAM L 1 Figure 16. Test Circuit for Measuring the Force Output of a Cylinder in Retraction WARNING Keep your hands and fingers away from the load spring while the power unit is running. The load spring will be compressing during operation. This will make it possible for you to pinch your fingers. If the spring needs to be aligned while the power unit is running, use a screwdriver, pencil, or pen to move the spring. 3. Turn on the power unit and open the shutoff valve. 4. Increase the relief valve setting until the pressure at Gauge A reads 150 psi/1035 kpa. You should observe that the cylinder s rod retracts and compresses the spring slightly. 5. Measure the new compressed spring length, L 1, for a pressure of 150 psi/1035 kpa. Measure the length to the nearest 1/32 inch (0.5 mm). Record it in the following chart. LARGE BORE CYLINDER PRESSURE (psi/kpa) COMPRESSED SPRING LENGTH (L 1 ) (in/cm) 150/1035 / 200/1380 / 250/1725 / 300/2070 / Copyright 2009 Amatrol, Inc. 23
6. Repeat steps 4 and 5 for each of the other pressures listed in the chart. 7. After you have completed the chart, experiment with your ability to change the position of the cylinder by varying the relief valve s pressure setting. WARNING Do not exceed 400 psi/2760 kpa setting at the relief valve. Above this pressure, the spring becomes completely compressed and can no longer be used to calculate forces. 8. Reduce the relief valve s pressure setting to minimum and turn off the power unit. 9. Perform the following substeps to remove the load spring. A. Remove the plastic guard, disconnect the load rod from the cam and slide it to the right. B. Remove the cam from the extender rod and set it aside. C. Remove the load spring and extender rod. D. Re-attach the cam to the cylinder rod. E. Keep your fingers clear of the cylinder and start the power unit. The cylinder should retract with the minimum relief valve setting. F. Turn off the power unit and close the shutoff valve. G. Replace the plastic guard. 10. Calculate the spring length change and use the load spring formula to calculate the actual force output for each pressure in the chart below: LARGE BORE CYLINDER PRESSURE (psi/kpa) SPRING LENGTH CHANGE (L-L 1 ) (in/cm) ACTUAL EXTENSION FORCE (lbs/n) 150/1035 / / 200/1380 / / 250/1725 / / 300/2070 / / 11. Compare these output forces with those obtained from the force formula in Skill 3. The actual values obtained should be smaller than those obtained by calculation but they should be close. 12. Compare these retraction forces with those obtained during extension in Skill 2. Retraction forces should be less at each pressure because the annular area is smaller than the cap area. Copyright 2009 Amatrol, Inc. 24
13. Disconnect the hoses from the large bore cylinder. 14. Now perform the following substeps to mount the load spring to the small bore cylinder. A. Set up a cylinder reciprocation circuit, as shown in figure 12. B. Make sure the relief valve is set for minimum pressure (fully CCW), then turn on the power unit. C. Open the shutoff valve and extend the cylinder completely with the DCV. If the cylinder will not move, increase the relief valve setting just enough to move the cylinder. D. Reduce the pressure to minimum and shut off the power unit. E. Remove the rod cam and slide the spring over the cylinder rod. F. Replace the rod cam. G. Turn on the power unit. H. While centering the spring against the cylinder, use the DCV to retract the cylinder until the spring is snug. I. Close the shutoff valve and turn off the power unit. 15. Connect the small bore cylinder in the test circuit shown in figure 16. This circuit will allow you to measure the actual force output in retraction of the small bore cylinder for comparison with the actual force output in retraction of the large bore cylinder. 16. Turn on the power unit and open the shutoff valve. 17. Increase the relief valve setting until the pressure at Gauge A reads 150 psi/1035 kpa. You should observe that the cylinder s rod retracts and compresses the spring slightly. 18. Measure the new compressed spring length, L 1, for a pressure of 150 psi/1035 kpa. Measure the length to the nearest 1/32 in. (0.5 mm). Record it in the following chart: SMALL BORE CYLINDER PRESSURE (psi/kpa) COMPRESSED SPRING LENGTH (L 1 ) (in/cm) 150/1035 / 200/1380 / 250/1725 / 300/2070 / Copyright 2009 Amatrol, Inc. 25
19. Repeat steps 17 and 18 for each of the other pressures listed in the chart. As you record your measurements, compare them to the measurements in step 5. Notice whether the small bore cylinder compresses the spring more or less at the same pressure. 20. Reduce the relief valve s pressure to minimum and turn off power unit. 21. Perform the following substeps to remove the load spring. A. Remove the rod cam. B. Remove the spring and then replace the rod cam. C. Start the power unit. The cylinder should retract with the minimum relief valve setting. D. Turn off the power unit and close the shutoff valve. 22. Calculate the spring length change and use the load spring formula to calculate the actual force output for each pressure in the chart below. SMALL BORE CYLINDER PRESSURE (psi/kpa) SPRING LENGTH CHANGE (L-L 1 ) (in/cm) ACTUAL EXTENSION FORCE (lbs/n) 150/1035 / / 200/1380 / / 250/1725 / / 300/2070 / / 23. Compare these output forces with those obtained from the large bore cylinder in step 10. The forces obtained should be smaller because of the smaller annular area with the small bore cylinder. 24. Disconnect the circuit and store the components. Copyright 2009 Amatrol, Inc. 26
SEGMENT 1 SELF REVIEW 1. For a double-acting cylinder operating at one pressure, the force of extension will be than the force of retraction. 2. The annular area of a cylinder is less than the cap (piston) area because of the. 3. To move a cylinder rod, you must overcome the load, back pressure, and. 4. Using a load spring to measure forces provides force values. 5. A 4-inch diameter cylinder with a 2-inch diameter rod will provide a theoretical push force of lbs at 500 psi. 6. The theoretical pull (retraction) force that the cylinder in question 5 would provide is lbs. 7. A smaller cylinder rod in the example of question 5 would provide a(n) force. Copyright 2009 Amatrol, Inc. 27
SEGMENT 2 HYDRAULIC LEVERAGE OBJECTIVE 3 STATE PASCAL S LAW AND EXPLAIN ITS SIGNIFICANCE IN HYDRAULICS LAPs 1 and 2 explained that the fluid pressure at an actuator can be changed into mechanical force to perform work. The basis for this comes from a concept known as Pascal s Law. This law is named after the 17th Century discoverer of this concept, Blaise Pascal. Pascal s Law states that fluid pressure in a confined vessel is transmitted undiminished to every portion of the surface of the containing vessel and acts at right angles to the surface. The term confined in this application means the fluid cannot flow anywhere because it is contained on all sides by the vessel. Fluid flowing in a pipe, for example, is not confined because the pipe is not closed on all sides. The concept of Pascal s Law is shown in figure 17. The weight sitting on the stopper cork causes a force to be applied to the fluid. Since the fluid is contained by the bottle, it compresses until the pressure of the fluid exerts an equal force on the cork in the opposite direction to support the weight. This same pressure exists throughout the entire volume of fluid and acts perpendicular to all the surfaces of the bottle. 100 POUNDS LOAD BOTTLE FLUID PRESSURE OIL Figure 17. Fluid Pressure Generated by an External Force Copyright 2009 Amatrol, Inc. 28
One significance of Pascal s Law is that it makes it possible for hydraulic systems to generate very large forces to move heavy loads with a small input force. This concept is called hydraulic leverage. More about the details of hydraulic leverage will be covered later in this LAP. Another significance of Pascal s Law is a cylinder or motor will generate full force output when it first starts to move because the pressure at the actuator is at full pressure. This is a big advantage because it usually takes more force to start a load than it does once the load is moving. Electric motors can provide a high output when started too, but if the load is too great, the motor will burn up. Hydraulic systems can be stalled indefinitely without damage to the components and will produce full force at zero speed. HEAVY LOAD 6900kPa 6900kPa PUMP RESERVOIR Figure 18. Pascal s Law Assures High Force to Move Loads from Rest And finally, understanding Pascal s Law is necessary in order to correctly place pressure gauges in a circuit to measure pressure. This is discussed in the next activity. Copyright 2009 Amatrol, Inc. 29
Activity 1. Verification of Pascal s Law for Hydraulics Procedure Overview In this activity, you will verify Pascal s Law by showing that the pressure at two different points in a volume of fluid is the same if the fluid is not flowing. This will be accomplished by measuring the pressure at different points in a hydraulic system. First, you will use a simple circuit with pressure gauges and hoses. Then you will demonstrate the same principle with a hydraulic cylinder. 1. Set up the hydraulic circuit shown in figures 19 and 20. This circuit has two gauges (A and B) connected at the same point in the circuit and another gauge (C) connected downstream. A needle valve is used to stop and start the flow. NOTE Be sure to connect the loose tee directly onto the gauge block A tee, as shown in figures 19 and 20. HYDRAULIC INSTRUMENTATION PANEL GAUGE A GAUGE B GAUGE C FLOW METER TEE FITTING SUPPLY MANIFOLD RELIEF \ SEQUENCE VALVE PRESSURE REDUCING VALVE 3 1 2 3 1 2 SHUTOFF VALVE D.C.V. #1 NEEDLE VALVE A B IN CHECK VALVE #1 A B B A CHECK VALVE #2 OUT B A BASIC HYDRAULIC VALVE MODULE RETURN MANIFOLD Figure 19. Pictorial of a Circuit for Demonstrating Pascal s Law Copyright 2009 Amatrol, Inc. 30
GAUGE A GAUGE B GAUGE C TEE NEEDLE VALVE GAUGE GAUGE BLOCK TEE TEE Figure 20. Schematic of a Circuit for Demonstrating Pascal s Law 2. Perform the power unit checkout procedures. 3. Close the needle valve fully. 4. Verify that the shutoff valve is closed. 5. Turn on the power unit. 6. Adjust the relief valve setting until the pressure at Gauge S reads 500 psi/3447 kpa. Copyright 2009 Amatrol, Inc. 31
7. Open the shutoff valve and observe the pressure gauge readings. Gauge A (psi/kpa) Gauge B (psi/kpa) Gauge C (psi/kpa) You should notice that Gauges A, B, and C read the same (or very close). This is because the closed needle valve creates a confined fluid in the hoses. This is Pascal s Law. NOTE You may observe that gauge readings differ as much as 40 psi from each other. This is not because the pressure isn t the same but because the gauges are not set accurately. To set the gauges accurately, a procedure called calibration would be required. 8. Now open the needle valve one full turn to allow the oil to flow through the hoses. Observe the new pressure gauge readings. Gauge A (psi/kpa) Gauge B (psi/kpa) Gauge C (psi/kpa) You should observe that the pressure reading of Gauges A and B are about the same but Gauge C is different. The reason is the fluid between Gauges A and C is no longer confined. It is flowing. Since pressure drops as it flows through the hose, the pressure is lower at Gauge C than Gauge A. However, the fluid in the hose connected between Gauges A and B is confined. It is not moving and, therefore, the pressure is the same at both Gauges A and B. This is Pascal s Law. This demonstrates an important point about the connection of pressure gauges. The hose length from the pressure line to the pressure gauge can be as long as you want it to be because the flow is deadheaded at the gauge. This means you can locate a pressure gauge on a panel that may not be close to the machine itself. 9. Reduce the relief valve setting to minimum. 10. Turn off the power unit and close the shutoff valve. Copyright 2009 Amatrol, Inc. 32
11. Set up the hydraulic circuit shown in figure 21. This circuit uses a directional control valve to reciprocate a cylinder. In the next few steps, you will demonstrate that Pascal s law affects the operation of any hydraulic circuit. GAUGE C DIRECTIONAL CONTROL VALVE GAUGE A IN OUT A B GAUGE B LARGE BORE CYLINDER Figure 21. Schematic of a Cylinder Circuit for Demonstrating Pascal s Law 12. Turn on the power unit. 13. Increase the relief valve setting until the pressure at Gauge S reads 500 psi/3447 kpa. 14. Open the shutoff valve. 15. Now push in on the lever of the directional control valve (DCV) to extend the cylinder and observe the pressure at Gauges A and C while the cylinder is extending. Gauge A (psi/kpa) Gauge C (psi/kpa) When the cylinder is extending, the pressure will be lower in both gauges because there is no load on the cylinder. Copyright 2009 Amatrol, Inc. 33
16. When the cylinder fully extends, continue to hold the lever of the DCV shifted and observe the readings of Gauges A and C. Gauge A (psi/kpa) Gauge C (psi/kpa) Since fluid is no longer flowing between Gauges A and C, they should be at the same pressure according to Pascal s Law. This means that you have the full force output of the cylinder generated while the cylinder is stopped. This fact enables hydraulic systems to move heavy loads from rest. 17. Now pull out on the lever to retract the cylinder and observe the pressure at Gauges B and C while the cylinder is retracting. Gauge B (psi/kpa) Gauge C (psi/kpa) You should observe that the pressure is lower when the cylinder is retracting because there is no load on the cylinder. The only pressure created is caused by the frictional resistance in the lines and at the cylinder. 18. When the cylinder fully retracts, continue to hold the lever of the DCV shifted and observe the readings of Gauges B and C. Gauge B (psi/kpa) Gauge C (psi/kpa) Fluid no longer flows between Gauges B and C, so they should be at the same pressure. 19. Release the lever. 20. Repeat steps 15 through 19 to verify your readings. 21. Reduce the relief valve setting to minimum, turn off the power unit, cycle the DCV to relieve any pressure remaining in the gauges, and close the shutoff valve. Copyright 2009 Amatrol, Inc. 34
OBJECTIVE 4 EXPLAIN HOW FORCE IS MULTIPLIED USING PASCAL S LAW One of the most important features of a hydraulic system is its ability to generate a very large force output with a very small force input. This principle is called force multiplication or hydraulic leverage. The principle of force multiplication is based on the force/pressure relationship in a fluid system and Pascal s Law. An example is shown in figure 22. In this example, an outside force of 10 pounds is applied over a small piston with a 1-in 2 area. This creates 10 psi pressure in the confined fluid. Pascal s Law says this 10 psi is the same throughout the fluid. This means that 10 psi acts against the larger 10-square-inch piston. This 10 psi causes an output force of 100 pounds because of the piston s larger area. So, for the 10-pound input force in the example, we increased the force from 10 to 100 pounds. This is hydraulic leverage. 10 LBS/4.5Kg 2 AREA = 10 IN 2 2 AREA = 1 IN (64.51 CM ) 2 (6.45 CM ) 100 LBS/45.4Kg 10 PSI 10 PSI CONFINED FLUID Figure 22. Force Multiplication or Hydraulic Leverage Copyright 2009 Amatrol, Inc. 35
A practical example of hydraulic leverage or force multiplication, is the automotive brake system. A small input force applied across a small cylinder (master cylinder) creates a high output force at the larger cylinder (brake cylinder). BRAKE PEDAL OPERATOR'S FOOT BRAKE CYLINDER SPRING OIL BRAKE DRUM PISTON BRAKE LINE BRAKE SHOE Figure 23. Hydraulic Brake System In an industrial-type hydraulic system, a pump and electric motor replace the piston in the valve system, but the principle of hydraulic leverage still applies. In this case, it means that a very small electric motor and pump could generate a very high force output at the cylinder by using a cylinder with a very large area. This ability to multiply the force output is a major advantage of hydraulics and is the reason why hydraulics is used in many other applications. Unfortunately, hydraulic leverage or force multiplication is not free. Just like the mechanical lever, distance is sacrificed. In our example of figure 22, to move the 100-pound load (force) 1 inch, the 10-pound input load would need to be moved 10 inches. In industrial-type hydraulic systems where a small pump is used with a large area cylinder to multiply force, the tradeoff, therefore, is the speed of the cylinder. It would be slower with a larger piston area. Copyright 2009 Amatrol, Inc. 36
Activity 2. Demonstrate How Distance is Sacrificed to Obtain Force Multiplication Procedure Overview In this activity, you will connect two cylinders of different sizes together and measure the distance each cylinder moves for a certain amount of oil volume input. 1. Before connecting the two cylinders together, position them about mid-stroke using the following substeps: A. Set up the large bore cylinder in the cylinder reciprocation circuit shown in figure 24. IN OUT A B DIRECTIONAL CONTROL VALVE LARGE BORE CYLINDER Figure 24. Cylinder Reciprocation Circuit B. Perform the power unit checkout procedures. C. Turn on the power unit. D. Increase the relief valve setting until the pressure at Gauge S reads 100 psi / 690 kpa. Open the shutoff valve. E. Extend and retract the cylinder by moving the lever on the directional control valve back and forth to remove any air from the system. F. Use the lever of the DCV to position the cylinder rod approximately mid-stroke (half-way). G. Reduce the relief valve setting to minimum, turn off the power unit, and close the shutoff valve. H. Cycle the DCV by pushing in and pulling out on the handle. This will remove any remaining pressure in the circuit. Copyright 2009 Amatrol, Inc. 37
I. Disconnect the hoses at the large bore cylinder and reconnect them to the small bore cylinder located at the top of the hydraulic actuator module. J. Repeat steps C through H for the small bore cylinder. 2. Connect the cylinders together, as shown in figure 25. SMALL BORE CYLINDER RETURN MANIFOLD LARGE BORE CYLINDER Figure 25. Schematic of Cylinders Connected Together 3. Set a ruler next to each cylinder rod and mark the starting points so that the movement of each rod can be measured. 4. Now, use your hand to push the rod of the small bore cylinder in about 1.25 inches and hold for about 30 seconds before releasing. 5. Measure the final distance each rod has moved. Small bore cylinder rod movement (in/cm) Large bore cylinder rod movement (in/cm) The large bore cylinder rod should have moved about half the distance of the small bore cylinder rod showing that distance is sacrificed to obtain force multiplication. 6. Disconnect the circuit and store the components. Copyright 2009 Amatrol, Inc. 38
SEGMENT 2 SELF REVIEW 1. Pascal s Law states that pressure on a(n) fluid is transmitted undiminished to every portion of a container. 2. When using hydraulic leverage, is multiplied. 3. To increase the force out of an actuator, increase the pressure over the. 4. When using hydraulic leverage, is lost. 5. 1000 psi acting across an area of five square inches produces lbs of force. Copyright 2009 Amatrol, Inc. 39
SEGMENT 3 FLUID FRICTION OBJECTIVE 5 DESCRIBE TWO TYPES OF RESISTANCE IN A HYDRAULIC SYSTEM As you know, hydraulic pumps do not produce pressure, they produce flow. The pressure at the pump is caused by the resistance in the system to the flow produced by the pump. In fact, if there were no resistance, a hydraulic system could pump oil into the system at zero pressure, as shown in figure 26. Unfortunately, we always have some type of resistance caused by either: Fluid friction Load resistance The pressure at the pump is equal to the load resistance plus the fluid friction resistance. 0 kpa PUMP RESERVOIR Figure 26. Hydraulic Pump Produces Flow, Not Pressure Copyright 2009 Amatrol, Inc. 40
Fluid Friction As fluid flows to the actuator, the internal surfaces of the components in the hydraulic system create a frictional resistance to flow. No matter how smooth the surfaces, there is always some frictional resistance. This resistance causes the fluid upstream of each component to have a higher pressure in order to push its way through the component. Even the conductors create a resistance, as shown in figure 27. PUMP PRESSURE 200 kpa PUMP RESERVOIR Figure 27. Fluid Pressure Drop Due to Frictional Resistance in a Pipe The amount of resistance of each component depends on the characteristics of the component such as the size, shape, and roughness of the inside surfaces. The resistance is also affected by characteristics of the fluid including fluid viscosity (thickness) and flow rate. Normally, the frictional resistance should be as low as possible because the energy used to overcome this resistance is lost as heat. Friction can be minimized by sizing the components properly. The larger the component, the less friction it will create. Copyright 2009 Amatrol, Inc. 41
Load Resistance The load on the actuator also contributes to the pump pressure. If the actuator is a cylinder, this pressure is equal to the load divided by the area of the cylinder (P = F Α). The load pressure, as shown in figure 28, is 7000 kpa. The frictional pressure is 60 kpa. This is the sum of the difference between A and B plus the difference between C and D. This makes the total pump pressure 7060 kpa. [(7060-7020) + (20-0) = (40 + 20) = 60 kpa] B = 7020 kpa A = 7060 kpa LOAD 7000 kpa C = 20 kpa PUMP D = 0 kpa RESERVOIR Figure 28. Pump Pressure Created by a Load and Frictional Resistance Copyright 2009 Amatrol, Inc. 42
OBJECTIVE 6 EXPLAIN HOW DELTA P DESCRIBES HYDRAULIC RESISTANCE The term delta P ( P) is often used to describe resistance in a hydraulic system or component. Delta P is the difference in pressure between one point and another point in the system. For example, in figure 29, the P between Gauge A (pump outlet) and Gauge B (needle valve inlet) is 20 kpa. This is the difference between the pressure at Gauges A and B (540-520=20). For a component, the P is the difference in pressure between the inlet and the outlet. For example, in figure 29, the P across the needle valve is 500 kpa, the pressure difference between Gauges B and C (520-20= 500). The P can be caused by either load resistance (i.e. a cylinder) or frictional resistance. For the needle valve and the conductors shown in figure 29, the P is caused by frictional resistance. 520 kpa B C 20kPa P A-B = 20kPa 540 kpa A P B-C = 500kPa P C-D = 20kPa PUMP D 0 kpa RESERVOIR Figure 29. Delta P Is Caused by Various Components Delta P is used to describe each individual resistance because the pressure downstream of the component may not be zero. The P allows you to describe the component resistance no matter what the downstream pressure is. As an example, figure 29 shows that the downstream pressure of the needle valve is 20 kpa. This pressure is caused only by components that are further downstream from the needle valve, which in this case is the hose resistance. It is not affected by the needle valve. Instead, the needle valve causes a pressure resistance upstream that adds to the total pump pressure and reduces the pressure downstream. This difference between upstream and downstream pressure is the P. In figure 29, the P of the needle valve is 500 kpa. Copyright 2009 Amatrol, Inc. 43
Understanding P is important when designing a hydraulic system, because components are sized by looking at curves that show how the P varies with flow rate. Another situation where you need to understand P is when you troubleshoot a system. If a component is partially clogged, its P will be higher than normal. SKILL 5 MEASURE DELTA P ACROSS A HYDRAULIC COMPONENT Procedure Overview In this procedure, you will demonstrate that components in a hydraulic system exert a frictional resistance to fluid flow. To do this, you will measure the pressure drop across the following hydraulic components: hoses with fittings, needle valve, and directional control valve. 1. Set up the test circuit shown in figure 30. This circuit will be used to measure the pressure drop caused by two hoses and a tee fitting. In this circuit, the oil flows between the supply and return manifolds through two hoses and a tee. Gauge A measures the pressure at the supply manifold and Gauge B measures the pressure at the return manifold. The difference between these two readings is the pressure drop ( P) caused by the resistance to the pump s flow of the two hoses and fittings. GAUGE A INSERT T FITTING EXACTLY AS SHOWN GAUGE B DIRECTION OF FLOW Figure 30. Schematic of Circuit for Measuring Pressure Drop Across Two Hoses and a Tee Fitting Copyright 2009 Amatrol, Inc. 44
2. Perform the power unit checkout procedures. 3. Verify that the shutoff valve is closed. 4. Turn on the power unit and adjust the relief valve s setting until Gauge S reads 500 psi/3447 kpa. 5. Open the shutoff valve and record below the readings of Gauge A and Gauge B. Also, calculate the pressure drop ( P). Full pump flow should now be flowing through the hoses and tee connected between the two manifolds. The frictional resistance of these components will cause a pressure difference between Gauges A and B. Two Hoses and Tee Fitting: Gauge A (psi/kpa) Gauge B (psi/kpa) Delta P (psi/kpa) (Delta P = Gauge A - Gauge B) A typical pressure drop for this measurement is 70 psi/483 kpa. However, your data will vary according to the oil you are using and other factors. As you can see, the pressure is fairly high. The reason for this is the valves inside the quick-connect fittings and the small cross-sectional areas of the fittings restrict flow. For this reason, you should use quick-connect fittings in industry only when absolutely necessary. 6. Now turn off the hydraulic power unit. Copyright 2009 Amatrol, Inc. 45
7. Replace the tee with a needle valve, as shown in figure 31. GAUGE A NEEDLE VALVE A B GAUGE B Figure 31. Schematic of Circuit for Measuring Drop Across Two Hoses and a Needle Valve 8. Open the needle valve completely (turn CCW fully). 9. Turn on the power unit. 10. Record below the readings of Gauges A and B. Also, calculate the pressure drop. The oil is now flowing through the hoses and the wide-open needle valve. You should observe that the two hoses and needle valve will cause a resistance which will make the pressure drop greater than with just two hoses and a tee in the circuit. Two Hoses and Needle Valve: Gauge A (psi/kpa) Gauge B (psi/kpa) Delta P (psi/kpa) (Delta P = Gauge A - Gauge B) A typical pressure drop for this circuit is 115 psi/794 kpa. Your data will vary. However, it should be higher than the drop measured in step 5 because the needle valve has a higher resistance than the tee. 11. Turn off the power unit. Copyright 2009 Amatrol, Inc. 46
12. Replace the needle valve with a directional control valve, as shown in figure 32. GAUGE A IN OUT DIRECTIONAL CONTROL VALVE A B GAUGE B Figure 32. Schematic of a Circuit to Measure the Delta P Across Two Hoses and a Directional Control Valve 13. Push and hold the lever of the directional control valve in so that the oil will flow through the valve from the P port to the A port. 14. Turn on the power unit. 15. With the lever held pushed in, record below the readings of gauge A and B. Then calculate the pressure drop. Two hoses and directional control valve: Gauge A (psi/kpa) Gauge B (psi/kpa) Delta P (psi/kpa) (Delta P = Gauge A - Gauge B) A typical pressure drop for this circuit is 75 psi / 518 kpa. Your data will vary but the drop through the DCV should be less than the needle valve because this particular DCV s internal flow path is larger. This is not always the case because DCVs and needle valves are available in many different sizes. The key point to understand is that all components create a pressure drop no matter what their size. 16. Release the lever. 17. Reduce the relief valve s pressure setting to minimum (turn CCW fully). 18. Close the shutoff valve and turn off the power unit. Copyright 2009 Amatrol, Inc. 47
Activity 3. Effect of Flow and Orifice Size on Delta P Procedure Overview In this activity, you will demonstrate that the pressure drop across a component increases when the flow is increased or the orifice size is decreased. For this demonstration, you will use a needle valve because it has a variable orifice. The pressure drop across the valve will be measured at a number of flow rates for a given orifice size. The pressure drop will then be measured at different orifice sizes for a given flow. 1. Set up the circuit shown in figure 33. In this circuit, oil flows through the needle valve. Gauges A and B will show the pressure drop across the needle valve. You will use the shutoff valve to change the flow rate to the needle valve. The flowmeter will indicate this flow. GAUGE A GAUGE B IN FLOW METER OUT A B NEEDLE VALVE Figure 33. Schematic of Circuit for Measuring Pressure Drop Copyright 2009 Amatrol, Inc. 48
2. Perform the power unit checkout procedures. 3. Verify that the shutoff valve is closed. 4. Turn on the power unit. Then increase the setting of the relief valve until Gauge S reads 500 psi/3447 kpa. 5. Close the needle valve completely. Then open it one complete turn. 6. Now open the shutoff valve until the flowmeter reads 1.0 gpm/3.8 lpm. Fluid will now be flowing through the needle valve and there should be a pressure difference between Gauges A and B. This is caused by the frictional resistance of the needle valve. Observe the readings of Gauges A and B. Then calculate and record the pressure drop. FLOW (gpm/lpm) GAUGE A (psi/kpa) GAUGE B (psi/kpa) PRESSURE DROP [DELTA P (P A - P B )] (psi/kpa) 1.0/3.8 / / / 1.25/4.8 / / / 1.5/5.7 / / / 2.0/7.6 / / / 7. Repeat step 6 for each of the other flow rates listed in the chart above. You should observe that the P increases as flow through the needle valve increases. This is why components must be sized for the flow rate they will handle. If the component is too small for a given flow rate, you will have a high pressure drop and a lot of wasted energy. 8. Now open the shutoff and needle valves completely. In the remaining steps of this activity, you will measure the effect orifice size has on pressure drop. Copyright 2009 Amatrol, Inc. 49