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# Basics of hydraulics

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### Basics of hydraulics

1. 1. BASICS OF HYDRAULICS1) DEFINITIONS 1.1) HYDRAULICS BASIC DEFINITIONS 1.2) CLASSIFICATION & 1.2.1) HYDROSTATICS FORMULAE 1.2.2) HYDRODYNAMICS 1.3) FORCE , PRESSURE , AREA 1.4) PASCAL’S LAW2) MULTIPLICATION OF FORCES 2.1) BRAMAH’S PRESS 2.2) LAW OF CONSERVATION OF ENERGY3) HYDRAULIC POWER TRANSMISSION USES OF 3.1) LINEAR ACTUATOR HYDRAULICS 3.2) ROTARY ACTUATOR
2. 2. BASICS OF HYDRAULICS4) ADVANTAGES OF HYDRAULICS 4.1) SPEED CONTROL 4.2) DIRECTION CONTROL 4.3) FORCE CONTROL 4.4) OVERLOAD PROTECTION ADVANTAGES OF 4.5) COMPACTNESS HYDRAULICS5) HOW PRESSURE IS CREATED PRACTICAL DETAILS IN HYDRAULICS
3. 3.  HYDRAULICS HYDRO AULUS( meaning Water ) ( meaning Pipe )HYDRAULICS :Work done by fluids in pipes. H y d r a u lic s is C la s s ife d a sH Y D R O S T A T IC S H Y D R O D Y N A M IC S
4. 4. HYDROSTATICSHYDROSTATICS Eg.:- FORCE F1 F1 = 1 Kg A1 = 1 Cm2P = F1 = 1 Kg A1 1 Cm2 AREA A1 = 1 Kg / Cm2( Same Pressure P ) A2 = 10 Cm2 F2 = P x A2 = 1 x 10 FORCE AREA F2 A2 = 10 Kg
5. 5. HYDRODYNAMICS LIQUID AT HIGH VELOCITYNOZZLE TURBINE
6. 6.  IN ORDER TO DETERMINE THE TOTAL FORCE EXERTED ON A SURFACE WE NEED TO KNOW THE PRESSURE OR FORCE PER UNIT AREA. PRESSURE = FORCE FORCE IN → KILOGRAMS ( Kg ) AREA AREA IN → SQ. CM ( Cm2 ) PRESSURE IN → KILOGRAM / SQ.CM (Kg / Cm2 ) P = F A FORCE = PRESSURE x AREA F THE ATMOSPHERIC AIR EXERTS UNIFORM PRESSURE ALL ROUND. THIS PRESSURE IS APPROX. 1 Kg / Cm2 AND IS DENOTED AS 1 BAR ( BAROMETER ) P A
7. 7. PRESSURE APPLIED ON A CONFINED FLUID IS TRANSMITTED UNDIMINISHED IN ALL DIRECTIONS AND ACTS WITH EQUAL FORCE ON EQUAL AREAS AND AT RIGHT ANGLES TO THEM. PRESSURE APPLIED ON A  FRENCH SCIENTIST PASCAL DISCOVERED CONFINED FLUID THIS LAW IN THE IS TRANSMITTED 17th CENTURY.  UNDIMINISHED  IN ALL DIRECTIONS • RELATES TO USE OF CONFINED FLUID  ACTS WITH EQUAL IN FORCE ON EQUAL AREAS ♦ TRANSMITTING AND POWER ♦ MODIFYING  AT RIGHT ANGLES TO MOTION THEM ♦ MULTIPLYING FORCE.
8. 8. PASCAL’S LAW FORCE F1 SMALL AREA A1 PRESSUREP = F1 P A1F2 = P x A2 LARGE AREA A2 FORCE F2
9. 9. BRAMAH’S PRESS THISHYDRAULIC LEVERAGE PRESSURE SUPPORTS A WT OF 100 Kg 10 Kg ON A IF AREA IS 1Cm2 100 kg 10 Cm2 10 AREA PRESSURE kg DEVELOPED 1Cm2 THROUGHOUT 10 Cm2 IS 10 Kg / Cm2 INPUT THE FORCES ARE PROPORTIONAL OUTPUT TO THE PISTON AREAS 10 Kg = 100 Kg 1 Cm2 10 Cm2MECHANICAL LEVERAGE WILL BALANCE A LOAD 100 Kg A LOAD OF 10 Kg OF 10 Kg 100 Kg HERE HERE 10 1
10. 10. LAW OF CONSERVATION OF ENERGY MOVING THE SMALL PISTON 10 Cm OF LIQUID WILL 10 Cm DISPLACES MOVE LARGER PISTON 1 Cm2 x 10 Cm = 10 Cm3 OF LIQUID ONLY 1Cm. 10 Cm2 x 1 Cm = 10 Cm3 10 Q=Axh 100 kg 1 Cm kg 1Cm2 10 Cm210 Cm WORK DONE = FORCE x DISTANCE MOVED W=Fxd W=Fxd W=Fxd = 10 Kg x 10 Cm = 100 Kg x 1 Cm = 100 Kg-Cm = 100 Kg-Cm  ENERGY CAN NEITHER BE CREATED NOR DESTROYED.  WHAT IS GAINED BY FORCE IS SACRIFICED IN THE DISTANCE MOVED.
11. 11. HYDRAULIC POWER TRANSMISSIONLINEAR ACTUATOR PUMP LOAD PISTON & ROD TO RESERVOIRROTARY ACTUATOR HYDRO MOTOR PUMP
12. 12. ADVANTAGES OF HYDRAULICS  SPEED CONTROL PISTON MOVESMAXIMUM SPEED “X” Cm IN 1 min. THIS VOL.(No speed control ) 10 lpm IS 10 Lts.Q=AxV PUMPQ  Flow (Cm3/min)A  Area FLOW ( Cm2 ) CONTROL VALVE 10 lpmV  Velocity(Speed control ) PUMP 5 lpm (Cm/ min ) RELIEF VALVE ACTUATOR GETS ONLY 5 LPM AND TRAVELS “X/2” Cm IN ONE MIN.
13. 13. ADVANTAGES OF HYDRAULICS  HYDRAULIC DRIVES ARE REVERSIBLEDIRECTION CONTROL PUMP RELIEF VALVE DIRECTIONAL THE CYLINDER ROD EXTENDS VALVE RELIEF PUMP VALVE DIRECTIONAL THE CYLINDER VALVE ROD RETRACTS
14. 14. ADVANTAGES OF HYDRAULICS  OVER LOAD PROTECTION RELIEF VALVE PROTECTS THE SYSTEM BY MAINTAINING THE SYSTEM SET PRESSURE. ANY INCREASE IN PRESSURE IN SYSTEM IS RELEAVED TO TANK . ( MOMENTARILY DIVERTING FLOW TO THE TANK. ) THUS OVERLOAD PROTECTION IS ACHIEVED.
15. 15. PRESSURE HEAD PUMP INLET LOCATIONS PRESSURE HERE IS OIL LEVEL ABOVE PUMP 0.85 x 100 gm / Cm2 CHARGES INLET 100 Cm = 0.085 Kg / Cm2 PUMP INLET OUTLET INLET OUTLET PUMPOIL LEVEL BELOW PUMPREQUIRES VACUUM TO“LIFT “ OIL 100 Cm THERE MUST BE A VACUUM EQUIVALENT TO 0.085 Kg / Cm2 TO LIFT THE OIL PUMP MECHANISM CREATES THE LOWER PRESSURE CONDITION.
16. 16. HOW PRESSURE IS DEVELOPED NO PRESSURE PUMPNO RESTRICTION RELIEF VALVE Set at 100 Kg/Cm2 PRESSURE BUILDS UPWITH RESTRICTION PUMP RELIEF VALVE Set at 100 Kg/Cm2 PRESSURE BUILDS UPTO RELIEF VALVE SETTING (100 Kg / Cm2) PUMPCLOSING RELIEF VALVE Set at 100 Kg/ Cm2
17. 17. PARALLEL FLOW PATHS 10THE OIL CAN CHOOSE A 10 BAR OPENS VALVE A3 PATHS PUMP B 20 BAR OPENS VALVE B C 30 BAR OPENS VALVE C IF FLOW IS BLOCKED OIL TAKES THE BEYOND “ A” PATH OF LEAST 20 RESISTANCE OIL WILL FLOW THRO “B” WHEN PUMP PRESSURE REACHES 20 BAR
18. 18. SERIES RESISTANCE ADD PRESSURE P1 = 0 A 0 10 BAR P2 = ( P1 + 10 ) 10 B = 0 + 10 20 BAR = 10 BAR P3 = ( P2 + 20 ) C 30 = 10 + 20 30 BAR = 30 BAR P = ( P3 + 30 ) PUMP 60 = 30 + 30 = 60 BAR
19. 19. PRINCIPLES OF FLOW♦ HOW FLOW IS MEASURED ?  VELOCITY  FLOW ( FLOW RATE )♦ FLOW RATE AND SPEED♦ FLOW AND PRESSURE DROP♦ LAMINAR AND TURBULENT FLOW♦ BERNOULLI’S PRINCIPLEFLOW IS THE ACTION IN THE HYDRAULIC SYSTEMTHAT GIVES THE ACTUATOR ITS MOTION.PRESSURE GIVES THE ACTUATOR ITS FORCE , BUTFLOW IS ESSENTIAL TO CAUSE MOVEMENT.FLOW IN THE HYDRAULIC SYSTEM IS CREATED BY THEPUMPPRESSURE INDICATES WORK LOAD.
20. 20. VELOCITY : IS THE AVERAGE SPEED OF THE FLUID’S PARTICLES PAST A GIVEN POINT OR THE AVERAGE DISTANCE THE PARTICLES TRAVEL PER UNIT OF TIME. Unit :m/Sec or m / min ( Metres / Sec or Metres/min )FLOW RATE : IS THE VOLUME OF FLUID PASSING A POINT IN A GIVEN TIME. Unit: Cm3 / min or l / min ( cc / minute or litres / min )SPEED OF AN ACTUATOR DEPENDES ON THE ACTUATORSIZE AND RATE OF FLOW INTO IT. Q=AxVFLOW IN Cm3 / min : AREA IN Cm 2 : VELOCITY IN Cm / min
21. 21. FLOW AND PRESSURE DROPMAX. PRESSURE SUCEEDINGLY LOWER LEVEL OF LIQUIDHERE BECAUSE SHOWS PRESSURE IS REDUCED AT POINTSOF THE HEAD OF DOWNSTREAM FROM SOURCE.THE FLUID PRESSURE GRADIENT FRICTION IN PIPE PRESSURE IS DROPS PRESSURE ZERO HERE AS THE FLUID FLOWS OUT UNRESTRICTED DUE TO EFFECT OF FRICTION RECOMMENDED VELOCITY RANGES ARE : 1.) PUMP INLET LINE : 0.6 ~ 1.2 metres / Second 2.) WORKING LINE ( PR. LINES) : 2~6 metres / Second
22. 22. NOR DOES A GRADUAL CHANGE INLAMINAR FLOW DIRECTION.LOW VELOCITY FLOW IN ASTRAIGHT PIPE IS STREAMLINED.THE FLUID PARTICLES MOVEPARALLEL TO FLOW DIRECTION. SO DOES AN ABRUPTTURBULENT FLOW CHANGE IN DIRECTION. AN ABRUPTTHE FLOW MAY START OUT CHANGE IN CROSS-STREAMLINED. SECTION MAKES IT NON PARALLEL PATHS OF PARTICLES TURBULENT. INCREASE RESISTANCE TO FLOW.