2. BASICS OF HYDRAULICS
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
BASIC
DEFINITIONS
&
FORMULAE
USES OF
HYDRAULIC
S
3. BASICS OF HYDRAULICS
4) ADVANTAGES OF HYDRAULICS
4.1) SPEED CONTROL
4.2) DIRECTION CONTROL
4.3) FORCE CONTROL
4.4) OVERLOAD PROTECTION
4.5) COMPACTNESS
5) HOW PRESSURE IS CREATED
ADVANTAGES
OF
HYDRAULICS
PRACTICAL
DETAILS IN
HYDRAULIC
S
4. HYDRAULICS
HYDRO AULUS
( meaning Water ) ( meaning Pipe )
HYDRAULICS :
Work done by fluids in pipes.
H Y D R O S T A T I C S H Y D R O D Y N A M I C S
H y d r a u lic s is C la s s ife d a s
7. • 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
• THE ATMOSPHERIC AIR EXERTS UNIFORM PRESSURE
ALL ROUND. THIS PRESSURE IS APPROX. 1 Kg / Cm2
AND
IS DENOTED AS 1 BAR ( BAROMETER )
PRESSURE
F
P A
8. PASCAL’S LAW
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
CONFINED FLUID
IS TRANSMITTED
UNDIMINISHED
IN ALL DIRECTIONS
ACTS WITH EQUAL
FORCE ON EQUAL AREAS
AND
AT RIGHT ANGLES TO
THEM
• FRENCH SCIENTIST
PASCAL DISCOVERED
THIS LAW IN THE
17th
CENTURY.
• RELATES TO
USE OF CONFINED FLUID
IN
♦ TRANSMITTING
POWER
♦ MODIFYING
MOTION
♦ MULTIPLYING
FORCE.
10. BRAMAH’S PRESS
HYDRAULIC LEVERAGE
MECHANICAL LEVERAGE
10
kg
1Cm2
10 Cm2
100 kg
INPUT OUTPUT
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
THE FORCES ARE PROPORTIONAL
TO THE PISTON AREAS
10 Kg
1 Cm2
= 100 Kg
10 Cm2
A LOAD
OF 10 Kg
HERE
WILL
BALANCE A
LOAD OF
100 Kg
HERE
10 Kg
100 Kg
10
1
11. 1Cm
LAW OF CONSERVATION OF ENERGY
1Cm2
10 Cm2
100 kg
10
kg
10Cm
ENERGY CAN NEITHER BE CREATED NOR DESTROYED.
WHAT IS GAINED BY FORCE IS SACRIFICED IN THE
DISTANCE MOVED.
WORK DONE = FORCE x DISTANCE MOVED
W = F x d
W = F x d
= 10 Kg x 10 Cm
= 100 Kg-Cm
W = F x d
= 100 Kg x 1 Cm
= 100 Kg-Cm
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
Q = A x h
13. ADVANTAGES OF HYDRAULICS
PUMP
10 lpm
10 lpm
PISTON MOVES
“X” Cm IN 1 min.THIS VOL.
IS 10 Lts.
MAXIMUM SPEED
(No speed control )
(Speed control )
RELIEF
VALVE
FLOW
CONTROL
VALVE
Q = A x V
Q Flow
(Cm3
/min)
A Area
( Cm2
)
V Velocity
(Cm/ min )
5 lpm
ACTUATOR GETS ONLY
5 LPM AND TRAVELS
“X/2” Cm IN ONE MIN.
SPEED CONTROL
PUMP
15. ADVANTAGES OF HYDRAULICS
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.
OVER LOAD PROTECTION
16. PRESSURE HEAD
PUMP INLET LOCATIONS
OIL LEVEL ABOVE PUMP
CHARGES INLET 100 Cm
PUMP
PUMP
OIL LEVEL BELOW PUMP
REQUIRES VACUUM TO
“LIFT “ OIL
INLET OUTLET
PRESSURE HERE IS
0.85 x 100 gm / Cm2
= 0.085 Kg / Cm2
THERE MUST BE A VACUUM
EQUIVALENT TO
0.085 Kg / Cm2
TO LIFT THE
OIL
PUMP MECHANISM CREATES
THE LOWER PRESSURE
CONDITION.
100 Cm
INLET OUTLET
17. HOW PRESSURE IS DEVELOPED
NO PRESSURE
NO RESTRICTION
WITH RESTRICTION
CLOSING
RELIEF
VALVE
RELIEF
VALVE
PRESSURE
BUILDS UP
PRESSURE BUILDS
UPTO RELIEF VALVE
SETTING (100 Kg / Cm2
)
PUMP
PUMP
Set at 100
Kg/Cm2
Set at 100
Kg/Cm2
RELIEF
VALVE
PUMP
Set at 100
Kg/Cm2
18. A
B
C
10 BAR OPENS VALVE
A
20 BAR OPENS VALVE
B
30 BAR OPENS VALVE
C
PARALLEL FLOW PATHS
10
PUMP
PUMP
THE OIL CAN CHOOSE
3 PATHS
20
IF FLOW IS
BLOCKED BEYOND “
A”
OIL WILL FLOW
THRO “B” WHEN
PRESSURE
REACHES 20
BAR
OIL TAKES THE
PATH OF LEAST
RESISTANCE
19. SERIES RESISTANCE ADD PRESSURE
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
PUMP
0
10
30
60
20. 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.
21. 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
22. FLOW AND PRESSURE DROPFLOW AND PRESSURE DROP
MAX. PRESSURE
HERE BECAUSE
OF THE HEAD OF
THE FLUID
FRICTION IN PIPE
DROPS PRESSURE
PRESSURE IS
ZERO HERE AS
THE FLUID
FLOWS OUT
UNRESTRICTED
PRESSURE
GRADIENT
SUCEEDINGLY LOWER LEVEL OF LIQUID
SHOWS PRESSURE IS REDUCED AT POINTS
DOWNSTREAM FROM SOURCE.
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
23. LAMINAR FLOWLAMINAR FLOW
TURBULENT FLOWTURBULENT FLOW
LOW VELOCITY FLOW IN A
STRAIGHT PIPE IS STREAMLINED.
THE FLUID PARTICLES MOVE
PARALLEL TO FLOW DIRECTION.
NOR DOES A
GRADUAL CHANGE
IN DIRECTION.
THE FLOW MAY START OUT
STREAMLINED.
AN ABRUPT
CHANGE IN CROSS-
SECTION MAKES IT
TURBULENT.
SO DOES AN ABRUPT
CHANGE IN DIRECTION.
NON PARALLEL PATHS OF PARTICLES
INCREASE RESISTANCE TO FLOW.