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Ashutosh Mishra - IE Consultant
BE, PP (APO), SSBB
New Delhi
M - 8882031712
1) DEFINITIONS
1.1) HYDRAULICS
1.2) CLASSIFICATION
1.2.1) HYDROSTATICS
1.2.2) HYDRODYNAMICS
1.3) FORCE , PRESSURE , AREA
1.4) PASCAL’S LAW
2) MULTIPLICATION OF FORCES
2.1) BRAMAH’S PRESS
2.2) LAW OF CONSERVATION OF ENERGY
3) HYDRAULIC POWER TRANSMISSION
3.1) LINEAR ACTUATOR
3.2) ROTARY ACTUATOR
17/02/20192
BASICS OF HYDRAULICS
BASIC
DEFINITIONS
&
FORMULAE
Uses of
hydraulics
BASICS OF HYDRAULICS
ADVANTAGES OF HYDRAULICS
4.1) SPEED CONTROL
4.2) DIRECTION CONTROL
4.3) FORCE CONTROL
4.4) OVERLOAD PROTECTION
4.5) COMPACTNESS
5) PRESSURE IN FLUID COLUMN
5.1) WEIGHT OF OIL
5.2) PRESSURE HEAD
5.3) ATM. PR. CHARGES THE PUMP
6) HOW PRESSURE IS CREATED
17/02/20193
Advantages of
hydraulics
Practical
details in
hydraulics
BASICS OF HYDRAULICS
7) PARALLEL & SERIES FLOW PATHS
8) FLOW THROUGH AN ORIFICE
9) FLOW
9.1) SPEED OF AN ACTUATOR
9.2) FLOW , VELOCITY , PRESSURE DROP
9.3) BERNOULLI’S PRINCIPLE
17/02/20194
Practical
details in
hydraulics
BASICS OF HYDRAULICS
HYDRO AULUS
( meaning Water ) ( meaning Pipe )
17/02/20195
HYDRAULICS
HYDRAULICS :
Work done by fluids in pipes.
HYDROSTATICS HYDRODYNAMICS
Hydraulics is Classifed as
17/02/20196
HYDROSTATICS
Eg.:-
F1 = 1 Kg
A1 = 1 Cm2
P = F1 = 1 Kg
A1 1 Cm2
= 1 Kg / Cm2
A2 = 10 Cm2
F2 = P x A2
= 1 x 10
= 10 Kg
(Same pressure) P
HYDROSTATICS
Force F1
Area A1
Area A2
FORCE
F2
HYDRAULICS BASICS
17/02/20197
Nozzle
Liquid at high velocity
Turbine
HYDRODYNAMICS
PRESSURE
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
F (Kg / Cm2 )
P=
A
P A
FORCE = PRESSURE x AREA
THE ATMOSPHERIC AIR EXERTS UNIFORM PRESSURE
ALL ROUND. THIS PRESSURE IS APPROX. 1 Kg / Cm2 AND
IS DENOTED AS 1 BAR ( BAROMETER )
17/02/20198
F
HYDRAULIC BASICS
17/02/20199
PASCAL’S LAW
FORCE F1
SMALL AREA
A1
LARGE AREA
A2
P = F1
A1
F2 = P x A2
FORCE F2
HYDRAULIC BASICS
17/02/201910
HYDRAULIC LEVERAGE
10
kg
1Cm2
10 Kg ON A
1Cm2 AREA PRESSURE
DEVELOPED
THROUGHOUT
IS 10 Kg / Cm2
THIS PRESSURE
SUPPORTS A WT
OF 100 Kg IF AREA
IS 10 Cm2
100 kg
10 Cm2
BRAMAH’S PRESS
INPUT OUTPUTTHE FORCES ARE PROPORTIONAL TO
THE PISTON AREAS
10 Kg
1 Cm2 =
100 Kg
10 Cm2
100 Kg
10 Kg
A LOAD OF
10 Kg
HERE
WILL BALANCE
A LOAD OF 100
Kg HERE
MECHANICAL LEVERAGE 10 1
HYDRAULIC BASICS
17/02/201911
MOVING THE SMALL PISTON
10 Cm DISPLACES
1 Cm2 x 10 Cm = 10 Cm3 OF LIQUID
10 Cm OF LIQUID WILL
MOVE LARGER PISTON
ONLY 1Cm.
10 Cm2 x 1 Cm = 10 Cm3
10
kg
1Cm2
100 kg
10 Cm2
Q = A x h
10Cm
1Cm
WORK DONE = FORCE x DISTANCE MOVED
W = F x d
= 10 Kg x 10 Cm
= 100 Kg-Cm
W = F x d
= 100 Kg x 1 Cm
= 100 Kg-Cm
W = F x d
 ENERGY CAN NEITHER BE CREATED NOR DESTROYED.
 WHAT IS GAINED BY FORCE IS SACRIFICED IN THE
DISTANCE MOVED.
HYDRAULIC BASICS
17/02/201912
HYDRAULIC BASIC
17/02/201913
HYDRAULIC POWER TRANSMISSION
LINEAR ACTUATOR
PUMP
PISTON & ROD
TO RESERVOIR
LOAD
HYDRO MOTOR
PUMP
ROTARY ACTUATOR
HYDRAULIC BASIC
17/02/201914
REVERSIBLE :
CAN CHANGE DIRECTION IN FULL MOTION SINCE IT
DOES NOT INVOLVE BRAKE OR CHANGE IN GEAR.
Eg. : DUPLEX MILLING MACHINE .
 RACK AND PINION
 HUGE ELECTRIC MOTOR
 FREQUENT REPAIRS.
 HEAVY NOISE.
– CHANGED TO HYDRAULIC CYLINDER
 MINIMUM MAINTANANCE
 SMOOTH FUNCTION WHILE
 OPERATING
 DIRECTION CHANGE
 LESS NOISE
17/02/201915
 HYDRAULIC DRIVES ARE REVERSIBLE
ADVANTAGES OF HYDRAULICS
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.
17/02/201916
ADVANTAGES OF HYDRAULICS
 COMPACTNESS
 SMALL IN SIZE
 COMPARING BASIC MACHINE, AREA OCCUPIED IS
SMALL.
( SOMETIMES THEY ARE KEPT INSIDE THE MACHINE )
 CONSIDERING THE AMOUNT OF FLOW AND PRESSURE
HANDLING CAPABILITIES THE HYDRAULIC COMPONENTS
ARE COMPACT.
 THEY PROVIDE HIGH POWER OUTPUT WITH VERY SMALL
WEIGHT AND SIZE.
17/02/201917
PRESSURE IN A FLUID COLUMN
17/02/201918
1 Cm3 ( 1cc ) OF OIL
WEIGHT ABOUT 0.85
Gms.
PRESSURE AT
THE BOTTOM
IS 0.85 Gms /
Cm2
1 Cm
1 Cm
1 Cm
2 Cm
A 2 Cm COLUMN WEIGHS
= 2 x 0.85 = 1.7 Gms
PRESSURE AT THE BOTTOM IS
= 1.7 Gms / Cm2
AT THE BOTTOM OF 300 Cms OF OIL THE PRESSURE IS = 300 x 0.85 Gms / Cm2
= 255 Gms / Cm2
= 0.255 Kg / Cm2
P =  x h
= 0.85 gms / Cm3 x h ( Cm )
PRESSURE HEAD
PUMP INLET LOCATIONS
OIL LEVEL
ABOVE PUMP
CHARGES INLET
100 Cm
PRESSURE HERE
IS 0.85 x 100 gm /
Cm2
= 0.085 Kg / Cm2
INLET OUTLET
PUMP
INLET OUTLET
PUMP
OIL LEVEL BELOW
PUMP REQUIRES
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.
ATMOSPHERIC PRESSURE CHARGES THE PUMP
17/02/201920
EXTERNAL FORCE
HYDRAULIC FORCE
TO ACTUATOR
Eg. : HYDRAULIC
JACK PUMP
PRESSURE DIFFERENCE PUSHES
OIL INTO PUMP
RECOMMENDED VACUUM AT
INLET 0.2 BAR.
SPEED REDUCTION DUE TO LEAKAGE
17/02/201921
FORCE
4000 Kgs
50 Cm2
80 Kg/Cm2
NO LEAKAGE
PUMP
V = Q
A
10 lpm
PRESSURE = 4000 = 80 Kg/Cm2
50
IF 2 Lpm IS LOST
THROUGH A LEAK
4000 Kgs
50 Cm2
LEAKAGE
PUMP
80 Kg/Cm2
10 lpm
PRESSURE LOSS REQUIRES
FULL LOSS OF PUMP OUTPUT
STILL A 4000 Kg FORCE ON OIL AND
PRESSURE IS 80 Kg/Cm2
HOW PRESSURE IS DEVELOPED
17/02/201922
NO RESTRICTION
WITH RESTRICTION
CLOSING
Set at 100 Kg/Cm2
RELIEF
VALVE
PUMP
PRESSURE BUILDS UP
NO PRESSURE
PRESSURE BUILDS UPTO
RELIEF VALVE SETTING
(100 Kg / Cm2 )
PUMP
PUMP
RELIEF
VALVE
RELIEF
VALVE
Set at 100 Kg/Cm2
Set at 100 Kg/Cm2
PARALLEL FLOW PATHS
17/02/201923
THE OIL CAN CHOOSE
3 PATHS
OIL TAKES THE PATH OF
LEAST RESISTANCE
PUMP
PUMP
10 BAR OPENS VALVE A
20 BAR OPENS VALVE B
30 BAR OPENS VALVE C
A
A
B
C
IF FLOW IS BLOCKED BEYOND “ A”
OIL WILL FLOW THRO
“B” WHEN PRESSURE
REACHES 20 BAR
10
20
SERIES RESISTANCE ADD PRESSURE
17/02/201924
P1 = 0
P2 = ( P1 + 10 )
= 0 + 10
= 10 BAR
P3 = ( P2 + 20 )
= 10 + 20
= 30 BAR
P = ( P3 + 30 )
= 30 + 30
= 60 BAR
A
10 BAR
B
20 BAR
C
30 BAR
0
10
30
60
SERIES RESISTANCE ADD PRESSURE
17/02/201925
ORIFICE
• AN ORIFICE IS A RESTRICTED PASSAGE IN A
HYDRAULIC LINE OR COMPONENT, USED TO
CONTROL FLOW OR CREATE A PRESSURE
DIFFERENTIAL ( PRESSURE DROP )
• AN INCREASE IN PRESSURE DROP ACROSS AN
ORIFICE WILL ALWAYS BE ACCOMPANIED BY AN
INCREASE IN FLOW.
• IF FLOW IS BLOCKED BEYOND AN ORIFICE, THE
PRESSURE WILL IMMEDIATELY EQUALIZE ON
BOTH SIDES OF THE ORIFICE IN ACCORDANCE
WITH PASCAL’S LAW.
17/02/201926
FLOW THROUGH AN ORIFICE AND PRESSURE DIFFERENCE
17/02/201927
PRESSURE
DIFFERENCE
 P = P1 - P2
PUMP
RELIEF VALVE
Orifice
Area - “A”
P1 P2
Set at
100 Kg / Cm2
Q = K x A x  Where
Q  Flow through an orifice. ( l / min )
K  Constant. ( Depends on Coefficient of discharge, viscosity of oil )
A  Area of Orifice ( Cm 2 )
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 PRINCIPLE
FLOW IS THE ACTION IN THE HYDRAULIC SYSTEM THAT GIVES THE
ACTUATOR ITS MOTION.
PRESSURE GIVES THE ACTUATOR ITS FORCE , BUT FLOW IS
ESSENTIAL TO CAUSE MOVEMENT.
FLOW IN THE HYDRAULIC SYSTEM IS CREATED BY THE PUMP
PRESSURE INDICATES WORK LOAD.
17/02/201928
PRINCIPLES OF FLOW
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 ACTUATOR
SIZE AND RATE OF FLOW INTO IT.
Q = A x V
FLOW IN Cm3 / min : AREA IN Cm 2 : VELOCITY IN Cm / min
17/02/201929
17/02/201930
MAX.
PRESSURE
HERE
BECAUSE OF
THE HEAD OF
THE FLUID
SUCEEDINGLY LOWER LEVEL OF
LIQUID SHOWS PRESSURE IS
REDUCED AT POINTS DOWNSTREAM
FROM SOURCE.
PRESSURE GRADIENT
FRICTION IN PIPE DROPS PRESSURE PRESSURE IS
ZERO HERE AS
THE FLUID FLOWS
OUT
UNRESTRICTEDDUE 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
LAMINAR FLOW
17/02/201931
NOR DOES A
GRADUAL
CHANGE IN
DIRECTION.
LAMINAR FLOW
LOW VELOCITY FLOW IN
A STRAIGHT PIPE IS
STREAMLINED. THE
FLUID PARTICLES MOVE
PARALLEL TO FLOW
DIRECTION.
SO DOES AN
ABRUPT CHANGE
IN DIRECTION.
THE FLOW MAY START OUT
STREAMLINED.
AN ABRUPT CHANGE
IN CROSS-SECTION
MAKES IT
TURBULENT. NON PARALLEL PATHS OF PARTICLES
INCREASE RESISTANCE TO FLOW.
TURBULENT FLOW
TURBULENT FLOW
BERNOULLI’S PRINCIPLE
THE SUMS OF THE PRESSURE ENERGY AND KINETIC
ENERGY AT VARIOUS POINTS IN A SYSTEM MUST
BE CONSTANT IF FLOW RATE IS CONSTANT.
KINETIC ENERGY : FLUID’S WEIGHT & VELOCITY
POTENTIAL ENERGY : PRESSURE
WITH A CONSTANT FLOW RATE , ENERGY IS
TRANSFORMED FROM ONE FORM TO THE OTHER
EACH TIME THE PIPE CROSS - SECTION SIZE
CHANGES.
17/02/201932
BERNOULLI’S PRINCIPLE
17/02/201933
VELOCITY DECREASES IN LARGER
PIPE. KINETIC ENERGY LOSS MADE UP
BY INCREASE IN PRESSURE.
IGNORING FRICTION LOSS PRESSURE
AGAIN BECOMES SAME AS THAT OF “
A” WHEN VELOCITY BECOMES THAT
OF “ A”
VELOCITY IS HIGH
PRESSURE IS
LOWER
PUMP
A B C
BERNOULLI’S PRINCIPLE - APPLICATION
17/02/201934
CARBURETOR BARREL
BUTTERFLY VALVE
VENTURI THROAT
AIR STREAM
FUEL BOWL
VENTURI EFFECT IN A GASOLINE ENGINE CARBURETOR IS AN
APPLICATION OF BERNOULLI’S PRINCIPLE
FRICTION AND VELOCITY AFFECT PRESSURE
17/02/201935
FRICTION REDUCES
THE HEAD AT
SUCCEDING POINTS
DOWNSTREAM
EXCEPT WHERE ....
THE LARGER PIPE
REDUCES
VELOCITY OF THE
FLOW
Pressure
P1
P2
P3
* * *
Friction & Pressure Losses
17/02/201936
The size of pressure loss (frictional loss) is dependent on :
- Length of Pipe
- Cross-sectional area of pipe
- Roughness of pipe walls
- Number of pipe bends
- Velocity of flow and
- Viscosity of the fluid
PRINCIPLES OF FLOW
 FLOW IS THE ACTION IN THE HYDRAULIC
SYSTEM THAT GIVES THE ACTUATOR ITS
MOTION.
 PRESSURE GIVES THE ACTUATOR ITS FORCE ,
BUT FLOW IS ESSENTIAL TO CAUSE MOVEMENT.
 FLOW IN THE HYDRAULIC SYSTEM IS CREATED
BY THE PUMP
 PRESSURE INDICATES WORK LOAD.
17/02/201937

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Basic Hydraulics

  • 1. Ashutosh Mishra - IE Consultant BE, PP (APO), SSBB New Delhi M - 8882031712
  • 2. 1) DEFINITIONS 1.1) HYDRAULICS 1.2) CLASSIFICATION 1.2.1) HYDROSTATICS 1.2.2) HYDRODYNAMICS 1.3) FORCE , PRESSURE , AREA 1.4) PASCAL’S LAW 2) MULTIPLICATION OF FORCES 2.1) BRAMAH’S PRESS 2.2) LAW OF CONSERVATION OF ENERGY 3) HYDRAULIC POWER TRANSMISSION 3.1) LINEAR ACTUATOR 3.2) ROTARY ACTUATOR 17/02/20192 BASICS OF HYDRAULICS BASIC DEFINITIONS & FORMULAE Uses of hydraulics
  • 3. BASICS OF HYDRAULICS ADVANTAGES OF HYDRAULICS 4.1) SPEED CONTROL 4.2) DIRECTION CONTROL 4.3) FORCE CONTROL 4.4) OVERLOAD PROTECTION 4.5) COMPACTNESS 5) PRESSURE IN FLUID COLUMN 5.1) WEIGHT OF OIL 5.2) PRESSURE HEAD 5.3) ATM. PR. CHARGES THE PUMP 6) HOW PRESSURE IS CREATED 17/02/20193 Advantages of hydraulics Practical details in hydraulics
  • 4. BASICS OF HYDRAULICS 7) PARALLEL & SERIES FLOW PATHS 8) FLOW THROUGH AN ORIFICE 9) FLOW 9.1) SPEED OF AN ACTUATOR 9.2) FLOW , VELOCITY , PRESSURE DROP 9.3) BERNOULLI’S PRINCIPLE 17/02/20194 Practical details in hydraulics
  • 5. BASICS OF HYDRAULICS HYDRO AULUS ( meaning Water ) ( meaning Pipe ) 17/02/20195 HYDRAULICS HYDRAULICS : Work done by fluids in pipes. HYDROSTATICS HYDRODYNAMICS Hydraulics is Classifed as
  • 6. 17/02/20196 HYDROSTATICS Eg.:- F1 = 1 Kg A1 = 1 Cm2 P = F1 = 1 Kg A1 1 Cm2 = 1 Kg / Cm2 A2 = 10 Cm2 F2 = P x A2 = 1 x 10 = 10 Kg (Same pressure) P HYDROSTATICS Force F1 Area A1 Area A2 FORCE F2
  • 7. HYDRAULICS BASICS 17/02/20197 Nozzle Liquid at high velocity Turbine HYDRODYNAMICS
  • 8. PRESSURE 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 F (Kg / Cm2 ) P= A P A FORCE = PRESSURE x AREA THE ATMOSPHERIC AIR EXERTS UNIFORM PRESSURE ALL ROUND. THIS PRESSURE IS APPROX. 1 Kg / Cm2 AND IS DENOTED AS 1 BAR ( BAROMETER ) 17/02/20198 F
  • 9. HYDRAULIC BASICS 17/02/20199 PASCAL’S LAW FORCE F1 SMALL AREA A1 LARGE AREA A2 P = F1 A1 F2 = P x A2 FORCE F2
  • 10. HYDRAULIC BASICS 17/02/201910 HYDRAULIC LEVERAGE 10 kg 1Cm2 10 Kg ON A 1Cm2 AREA PRESSURE DEVELOPED THROUGHOUT IS 10 Kg / Cm2 THIS PRESSURE SUPPORTS A WT OF 100 Kg IF AREA IS 10 Cm2 100 kg 10 Cm2 BRAMAH’S PRESS INPUT OUTPUTTHE FORCES ARE PROPORTIONAL TO THE PISTON AREAS 10 Kg 1 Cm2 = 100 Kg 10 Cm2 100 Kg 10 Kg A LOAD OF 10 Kg HERE WILL BALANCE A LOAD OF 100 Kg HERE MECHANICAL LEVERAGE 10 1
  • 11. HYDRAULIC BASICS 17/02/201911 MOVING THE SMALL PISTON 10 Cm DISPLACES 1 Cm2 x 10 Cm = 10 Cm3 OF LIQUID 10 Cm OF LIQUID WILL MOVE LARGER PISTON ONLY 1Cm. 10 Cm2 x 1 Cm = 10 Cm3 10 kg 1Cm2 100 kg 10 Cm2 Q = A x h 10Cm 1Cm WORK DONE = FORCE x DISTANCE MOVED W = F x d = 10 Kg x 10 Cm = 100 Kg-Cm W = F x d = 100 Kg x 1 Cm = 100 Kg-Cm W = F x d  ENERGY CAN NEITHER BE CREATED NOR DESTROYED.  WHAT IS GAINED BY FORCE IS SACRIFICED IN THE DISTANCE MOVED.
  • 13. HYDRAULIC BASIC 17/02/201913 HYDRAULIC POWER TRANSMISSION LINEAR ACTUATOR PUMP PISTON & ROD TO RESERVOIR LOAD HYDRO MOTOR PUMP ROTARY ACTUATOR
  • 15. REVERSIBLE : CAN CHANGE DIRECTION IN FULL MOTION SINCE IT DOES NOT INVOLVE BRAKE OR CHANGE IN GEAR. Eg. : DUPLEX MILLING MACHINE .  RACK AND PINION  HUGE ELECTRIC MOTOR  FREQUENT REPAIRS.  HEAVY NOISE. – CHANGED TO HYDRAULIC CYLINDER  MINIMUM MAINTANANCE  SMOOTH FUNCTION WHILE  OPERATING  DIRECTION CHANGE  LESS NOISE 17/02/201915  HYDRAULIC DRIVES ARE REVERSIBLE ADVANTAGES OF HYDRAULICS
  • 16. 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. 17/02/201916
  • 17. ADVANTAGES OF HYDRAULICS  COMPACTNESS  SMALL IN SIZE  COMPARING BASIC MACHINE, AREA OCCUPIED IS SMALL. ( SOMETIMES THEY ARE KEPT INSIDE THE MACHINE )  CONSIDERING THE AMOUNT OF FLOW AND PRESSURE HANDLING CAPABILITIES THE HYDRAULIC COMPONENTS ARE COMPACT.  THEY PROVIDE HIGH POWER OUTPUT WITH VERY SMALL WEIGHT AND SIZE. 17/02/201917
  • 18. PRESSURE IN A FLUID COLUMN 17/02/201918 1 Cm3 ( 1cc ) OF OIL WEIGHT ABOUT 0.85 Gms. PRESSURE AT THE BOTTOM IS 0.85 Gms / Cm2 1 Cm 1 Cm 1 Cm 2 Cm A 2 Cm COLUMN WEIGHS = 2 x 0.85 = 1.7 Gms PRESSURE AT THE BOTTOM IS = 1.7 Gms / Cm2 AT THE BOTTOM OF 300 Cms OF OIL THE PRESSURE IS = 300 x 0.85 Gms / Cm2 = 255 Gms / Cm2 = 0.255 Kg / Cm2 P =  x h = 0.85 gms / Cm3 x h ( Cm )
  • 19. PRESSURE HEAD PUMP INLET LOCATIONS OIL LEVEL ABOVE PUMP CHARGES INLET 100 Cm PRESSURE HERE IS 0.85 x 100 gm / Cm2 = 0.085 Kg / Cm2 INLET OUTLET PUMP INLET OUTLET PUMP OIL LEVEL BELOW PUMP REQUIRES 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.
  • 20. ATMOSPHERIC PRESSURE CHARGES THE PUMP 17/02/201920 EXTERNAL FORCE HYDRAULIC FORCE TO ACTUATOR Eg. : HYDRAULIC JACK PUMP PRESSURE DIFFERENCE PUSHES OIL INTO PUMP RECOMMENDED VACUUM AT INLET 0.2 BAR.
  • 21. SPEED REDUCTION DUE TO LEAKAGE 17/02/201921 FORCE 4000 Kgs 50 Cm2 80 Kg/Cm2 NO LEAKAGE PUMP V = Q A 10 lpm PRESSURE = 4000 = 80 Kg/Cm2 50 IF 2 Lpm IS LOST THROUGH A LEAK 4000 Kgs 50 Cm2 LEAKAGE PUMP 80 Kg/Cm2 10 lpm PRESSURE LOSS REQUIRES FULL LOSS OF PUMP OUTPUT STILL A 4000 Kg FORCE ON OIL AND PRESSURE IS 80 Kg/Cm2
  • 22. HOW PRESSURE IS DEVELOPED 17/02/201922 NO RESTRICTION WITH RESTRICTION CLOSING Set at 100 Kg/Cm2 RELIEF VALVE PUMP PRESSURE BUILDS UP NO PRESSURE PRESSURE BUILDS UPTO RELIEF VALVE SETTING (100 Kg / Cm2 ) PUMP PUMP RELIEF VALVE RELIEF VALVE Set at 100 Kg/Cm2 Set at 100 Kg/Cm2
  • 23. PARALLEL FLOW PATHS 17/02/201923 THE OIL CAN CHOOSE 3 PATHS OIL TAKES THE PATH OF LEAST RESISTANCE PUMP PUMP 10 BAR OPENS VALVE A 20 BAR OPENS VALVE B 30 BAR OPENS VALVE C A A B C IF FLOW IS BLOCKED BEYOND “ A” OIL WILL FLOW THRO “B” WHEN PRESSURE REACHES 20 BAR 10 20
  • 24. SERIES RESISTANCE ADD PRESSURE 17/02/201924 P1 = 0 P2 = ( P1 + 10 ) = 0 + 10 = 10 BAR P3 = ( P2 + 20 ) = 10 + 20 = 30 BAR P = ( P3 + 30 ) = 30 + 30 = 60 BAR A 10 BAR B 20 BAR C 30 BAR 0 10 30 60
  • 25. SERIES RESISTANCE ADD PRESSURE 17/02/201925
  • 26. ORIFICE • AN ORIFICE IS A RESTRICTED PASSAGE IN A HYDRAULIC LINE OR COMPONENT, USED TO CONTROL FLOW OR CREATE A PRESSURE DIFFERENTIAL ( PRESSURE DROP ) • AN INCREASE IN PRESSURE DROP ACROSS AN ORIFICE WILL ALWAYS BE ACCOMPANIED BY AN INCREASE IN FLOW. • IF FLOW IS BLOCKED BEYOND AN ORIFICE, THE PRESSURE WILL IMMEDIATELY EQUALIZE ON BOTH SIDES OF THE ORIFICE IN ACCORDANCE WITH PASCAL’S LAW. 17/02/201926
  • 27. FLOW THROUGH AN ORIFICE AND PRESSURE DIFFERENCE 17/02/201927 PRESSURE DIFFERENCE  P = P1 - P2 PUMP RELIEF VALVE Orifice Area - “A” P1 P2 Set at 100 Kg / Cm2 Q = K x A x  Where Q  Flow through an orifice. ( l / min ) K  Constant. ( Depends on Coefficient of discharge, viscosity of oil ) A  Area of Orifice ( Cm 2 )
  • 28. 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 PRINCIPLE FLOW IS THE ACTION IN THE HYDRAULIC SYSTEM THAT GIVES THE ACTUATOR ITS MOTION. PRESSURE GIVES THE ACTUATOR ITS FORCE , BUT FLOW IS ESSENTIAL TO CAUSE MOVEMENT. FLOW IN THE HYDRAULIC SYSTEM IS CREATED BY THE PUMP PRESSURE INDICATES WORK LOAD. 17/02/201928
  • 29. PRINCIPLES OF FLOW 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 ACTUATOR SIZE AND RATE OF FLOW INTO IT. Q = A x V FLOW IN Cm3 / min : AREA IN Cm 2 : VELOCITY IN Cm / min 17/02/201929
  • 30. 17/02/201930 MAX. PRESSURE HERE BECAUSE OF THE HEAD OF THE FLUID SUCEEDINGLY LOWER LEVEL OF LIQUID SHOWS PRESSURE IS REDUCED AT POINTS DOWNSTREAM FROM SOURCE. PRESSURE GRADIENT FRICTION IN PIPE DROPS PRESSURE PRESSURE IS ZERO HERE AS THE FLUID FLOWS OUT UNRESTRICTEDDUE 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
  • 31. LAMINAR FLOW 17/02/201931 NOR DOES A GRADUAL CHANGE IN DIRECTION. LAMINAR FLOW LOW VELOCITY FLOW IN A STRAIGHT PIPE IS STREAMLINED. THE FLUID PARTICLES MOVE PARALLEL TO FLOW DIRECTION. SO DOES AN ABRUPT CHANGE IN DIRECTION. THE FLOW MAY START OUT STREAMLINED. AN ABRUPT CHANGE IN CROSS-SECTION MAKES IT TURBULENT. NON PARALLEL PATHS OF PARTICLES INCREASE RESISTANCE TO FLOW. TURBULENT FLOW TURBULENT FLOW
  • 32. BERNOULLI’S PRINCIPLE THE SUMS OF THE PRESSURE ENERGY AND KINETIC ENERGY AT VARIOUS POINTS IN A SYSTEM MUST BE CONSTANT IF FLOW RATE IS CONSTANT. KINETIC ENERGY : FLUID’S WEIGHT & VELOCITY POTENTIAL ENERGY : PRESSURE WITH A CONSTANT FLOW RATE , ENERGY IS TRANSFORMED FROM ONE FORM TO THE OTHER EACH TIME THE PIPE CROSS - SECTION SIZE CHANGES. 17/02/201932
  • 33. BERNOULLI’S PRINCIPLE 17/02/201933 VELOCITY DECREASES IN LARGER PIPE. KINETIC ENERGY LOSS MADE UP BY INCREASE IN PRESSURE. IGNORING FRICTION LOSS PRESSURE AGAIN BECOMES SAME AS THAT OF “ A” WHEN VELOCITY BECOMES THAT OF “ A” VELOCITY IS HIGH PRESSURE IS LOWER PUMP A B C
  • 34. BERNOULLI’S PRINCIPLE - APPLICATION 17/02/201934 CARBURETOR BARREL BUTTERFLY VALVE VENTURI THROAT AIR STREAM FUEL BOWL VENTURI EFFECT IN A GASOLINE ENGINE CARBURETOR IS AN APPLICATION OF BERNOULLI’S PRINCIPLE
  • 35. FRICTION AND VELOCITY AFFECT PRESSURE 17/02/201935 FRICTION REDUCES THE HEAD AT SUCCEDING POINTS DOWNSTREAM EXCEPT WHERE .... THE LARGER PIPE REDUCES VELOCITY OF THE FLOW Pressure P1 P2 P3 * * *
  • 36. Friction & Pressure Losses 17/02/201936 The size of pressure loss (frictional loss) is dependent on : - Length of Pipe - Cross-sectional area of pipe - Roughness of pipe walls - Number of pipe bends - Velocity of flow and - Viscosity of the fluid
  • 37. PRINCIPLES OF FLOW  FLOW IS THE ACTION IN THE HYDRAULIC SYSTEM THAT GIVES THE ACTUATOR ITS MOTION.  PRESSURE GIVES THE ACTUATOR ITS FORCE , BUT FLOW IS ESSENTIAL TO CAUSE MOVEMENT.  FLOW IN THE HYDRAULIC SYSTEM IS CREATED BY THE PUMP  PRESSURE INDICATES WORK LOAD. 17/02/201937