4/23/2013. Understanding Mobile Hydraulics. Basic Hydraulics

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1 Understanding Mobile Hydraulics. Basic Hydraulics 1

2 is the capacity to do work or cause physical change If you apply a force over a given distance you will have accomplished work. Work = x Distance 10 lbs x 10 feet = 100 pound feet of work. Work Another example If the cylinder were to move 10 lbs of papers 10 inches it would have accomplished 100 lbs inches of work. 10 lbs x 10 inches = 100 in lbs of work Question 3 How much work would you accomplish if the force is 20 lbs and the distance is 20 inches? Pascal s Law Why We Use Hydraulics Pascal s law states Pressure applied to a confined liquid is transmitted undiminished in all directions, and acts with equal force on all equal areas, and at right angles to those areas. Air is compressible 10 lbs of 100 lbs of Pressure in all directions 1 in 2 10 in 2 Question 2 How many pounds can the cylinder on the right lift if the area is 20 square inches? Nothing on the left cylinder changed. 2

3 This formula triangle is used to find any of one of the missing parts of the equation providing you have the other two. = Pressure x Area Pressure = Area Area = Pressure Example, 1. What is force in lbs if you have an area of 10 square inches a pressure of 20 psi?. equation: 10 sq in x 20 psi = 200 lbs 2. What is the pressure if you have a force of 300 pounds and an area of 10 square inches? equation: 300 lbs 10 sq in. = 30 psi 3. What is the area if you have a force of 200 pounds and a pressure of 10 psi? equation: 200 lbs 10 psi = 20 square inches Basic Hydraulic Formulas P (Pressure) F () A (Area) Area of a circle in square inches can be calculated with this formula; Diameter x Diameter x.7854 = Square inches of area For example: The area of a 5 inch diameter circle = 5 x 5 x.7854 = in 2 Question 1. How many square inches of area are in a 4 inch diameter? Area in square inches 12 x 12 = 144 sq inches Area Pressure is the resistance to flow. This cylinder has no load so the pressure developed will only be created by the friction it takes to move the piston inside the cylinder. The pressure will increase if the load increases. Pressure Pressure = in pounds Area in square inches 10 in 2 10 in 2 10 lbs 20 lbs Pressure Continued Pressure will increase as resistance increases. In the top example; 10 lbs (force) 10in 2 (area)= 1 psi Middle example; 20 lbs (force) 10in 2 (area) = 2 psi Pressure = in pounds Area in square inches 10 in 2 30 lbs Bottom example; 30 lb (force) 10 in 2 (area)= 3 psi Note: the key point is that pressure is affected if the area or force change. 3

4 CYLINDER CAPACITY Cylinder Speed Volume Diameter Area 1000 PSI 1500 PSI 2000 PSI Measured in cubic inch s 2 inch Area Measured in square inch s Length Area 3 inch ,603 14,137 4 inch ,566 18,849 25,132 Length The stroke length of the cylinder 5 inch ,635 29,452 39,270 6 inch ,274 42,411 56,548 Volume = Area x Length POUNDS OF FORCE FORCE America Inc. FORCE America Inc. V Q A Q = Flow V = Velocity A = Area 231 = how many cubic inches of fluid are in 1 gallon You can find any 1 of these if you have the other two. What is the flow rate of this cylinder in gpm? Flow Required for a Cylinder 3 Sec to fully stroke 9 inch Stroke Flow rate = Displacement x Speed / is the number of cubic inches of fluid there are in one gallon. Flow rate is equal to speed so the more flow the system see s the faster the actuator will move. In this example one gallon goes into the cylinder in one minute. Flow Rate Flow (GPM) = Velocity in/min x Area/ sec x 60 sec = 180 in per minute 180 in/min x 3 in = 2.33 gpm 3 in 2 Pump 1 GPM 4

5 In this example three gallons goes into the cylinder in one minute causing the cylinder to mover faster. Key point: Flow is speed so the more flow you have the faster the operation will be. If you lose flow speed will go down. Flow Rate Cylinder Velocity Chart This illustrates the velocity in inches per minute a cylinder will travel based on the flow rates listed. This chart can be found in the Fluid Power Data book on page 17 or also in the Lightning Reference. Diameter 1 gpm 3 gpm 5 gpm 8 gpm 2 74 in/min 221 in/min 368 in/min 588 in/min 3 33 in/min 98 in/min 163 in/min 261 in/min 1 GallonPump 1 Gallon 1 Gallon 3 gallons in a minute 4 18 in/min 55 in/min 92 in/min 147 in/min 5 12 in/min 35 in/min 59 in/min 94 in/min in/min 25 in/min 41 in/min 65 in/min Torque is defined as a turning or twisting force. Torque = Radius x Weight Top example; If the radius is 10 inches and the weight is 10 lbs the torque would be 100 in lbs. Torque = 10 inch (radius) x 10 lbs = 100 in lbs Bottom example; If the radius is 5 inches and the weight is 10 lbs the torque would be 50 in lbs. Torque = 5 inch (radius) x 10 lbs = 50 in lbs 10 inches 10 lbs 5 inches Torque Horsepower James Watt's took an average size horse to a well and determined that the horse could lift 100 lbs 220 feet in one minute. 100 lbs x 220 ft/minute = 22,000 ft-lbs per minute. He decided to add 50% for good measure so 1 horsepower equals 33,000 ft lbs per minute. He did this so he could compare the work of an average horse to the work of his steam engine. Question 5. What is the torque in inch pounds if the radius is 7 inches and the force is 20 lbs? 10 lbs 5

6 2 4/23/2013 W W Power = The Rate of Doing Work = Work per Unit of Time 330 ft. 330 ft. 100 lbs. 100 lbs. 100 min. 1 min. Work = 100 lbs. x 330 ft. = 33,000 ft-lb Power = 33,000 ft-lb 100 min. = 330 ft-lb/min =.01 hp Work = 100 lbs. x 330 ft. = 33,000 ft-lb Power = 33,000 ft-lb 1 min. =33,000 ft-lb/min = 1 hp Horse power = Flow in GPM x Pressure in PSI divided by HP = Gallons per Minute x PSI 1714 Horsepower Formula This means that the horse power requirement will go up if your flow rate goes up, if your pressure goes up or both flow and pressure go up. If you do not have enough HP to meet the requirement the engine can stall. The formula above would be used if a hydraulic system were 100 % efficient but all hydraulic systems have some internal leakage. To account for that leakage we use the formula below as a general guideline to determine input HP. 83 % = GPM x PSI x.0007 Question 6. What is the input horse power at 83 % efficiency if the flow is 20 gpm and the pressure is 1000 psi? A column of air one square inch in cross-section and as high as the atmosphere... Area = 1 in Atmospheric Pressure Works For Us Reduced pressure here allows weighs 14.7 pounds at sea level. Atmospheric pressure is therefore 14.7 psia. Atmospheric pressure here to force fluid up the straw. Atmospheric Pressure Pressure differential causes fluid to flow from the high pressure area to the low pressure area. 6

7 Atmospheric Pressure Provides Fluid To The Pump TERMINOLOGY Reduced pressure here allows... Atmospheric pressure here to force fluid into the pump inlet. Oil Virtually uncompressible Pressure Resistance of work Working Pressure Pressure required operating a system under load FORCE America Inc. Relief Pressure Maximum operating pressures for the system Flow Created by the pump Cavitation Gaseous & vapors Aeration Entrained in the oil Velocity Rate of fluid 4ft/sec suction 20ft/sec working Laminar flow Turbulent flow generates heat Open center TERMINOLGY Closed center Cracking pressure Full flow pressure Pressure override TERMINOLOGY OF HYDRAULIC SYSTEMS Open center systems Pressure compensated closed center systems Pressure and flow compensated closed center systems (Load sensing systems) FORCE America Inc. 7

8 TYPICAL OPEN CENTER SYSTEM Return filter Reservoir Main control valve TYPICAL CLOSED CENTER PRESSURE COMPENSATED SYSTEM Reservoir Return filter Main control valve Pump A hydraulic system in which the control valve allows pump flow to return to the reservoir through the open center core in the control valve Pressure compensated pump with case drain The variable pressure compensated pump uses a closed center control valve. In neutral, flow is blocked making the pump destroke and maintain MAX system pressure until control valve is activated again TYPICAL LOAD SENSING SYSTEM Reservoir Return filter Main control valve Load sense pump with case drain and sensing line The load sensing system is noted for its high efficiency and controllability. The pump senses and compensates to the pressure and flow requirements of the system In neutral the pump only requires PSI standby pressure. Its low standby pressure eliminates excessive horsepower drain and heat build up 8

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