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1
HYDRAULIC CIRCUITS
Welcome to the Session on :
2
HYDRAULIC
POWER
UNIT
HYDRAULIC CIRCUITS
MOTOR &
PUMP
PRESSURE
CONTROL
VALVES
FLOW
CONTROL
VALVES
DIRECTION
CONTROL
VALVES
ACCESSORIES
ACTUATORS
3
HYDRAULIC CIRCUITS
A GOOD HYDRAULIC SYSTEM REQUIREMENT -
SATISFY THE SPECIFICATIONS OF THE OPERATION WITH
SAFETY
PERFORM SMOOTH OPERATION
LOW ENERGY CONSUMPTION – LOW HEAT GENERATION
REDUCE INITIAL COST & RUNNING COST
MAKE MAINTENANCE EASY
HYDRAULIC CIRCUITS ARE GRAPHICAL DIAGRAMS
OF THE HYDRAULIC SYSTEMS.
IT ALSO INDICATES EACH OPERATION OF THE
COMPONENTS.
4
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Variable displacement pump circuit
M
5
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Meter – in Circuit
M
6
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Meter – out Circuit
M
7
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Bleed – off Circuit
M
8
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Deceleration Circuit
M
9
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Feed speed varying circuit
M
10
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Multi Speed Circuit
M
Q1 : High Flow
Q2 : Low Flow
Q1
Q2
11
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Multi Speed Circuit
1
2 3
1 : Rapid Advance
2 : Medium Advance
3 : Slow Advance
M
UCF2-04
12
HYDRAULIC CIRCUITS
SPEED CONTROL CIRCUIT Multi Speed Circuit
Sol. 1 ON Low speed forward
Sol. 3 ON High speed forward
Sol. 3 OFF Speed decrease
Sol. 1 OFF Stop.
Sol. 2 ON Low speed reverse
Sol. 4 ON High speed reverse
Sol. 4 OFF Speed decrease
Sol. 2 OFF Stop.
Sol. 1 Sol. 2 Sol. 3 Sol. 4
M
13
HYDRAULIC CIRCUITS
PRESSURE CONTROL CIRCUIT 2 Operating Pressure Circuit
1
2
14
HYDRAULIC CIRCUITS
PRESSURE CONTROL CIRCUIT Low Pressure Return Circuit
1
2
Main Relief
Valve
Pilot Relief
Valve
15
M
HYDRAULIC CIRCUITS
PRESSURE CONTROL CIRCUIT Decompression Circuit
16
HYDRAULIC CIRCUITS
UNLOADING CIRCUIT
Manual Unloading
M
To Circuit
17
HYDRAULIC CIRCUITS
UNLOADING CIRCUIT
Circuit using Accumulator
Detection of
Pressure by
Pressure Switch
18
HYDRAULIC CIRCUITS
UNLOADING CIRCUIT
Circuit using Accumulator
M
Detection of
Pressure by
Pilot Op. Relief
Valve
19
HYDRAULIC CIRCUITS
UNLOADING CIRCUIT ( Hi-Low Circuit ) Low Pressure Operation
20
HYDRAULIC CIRCUITS
UNLOADING CIRCUIT ( Hi-Low Circuit ) High Pressure Operation
21
HYDRAULIC CIRCUITS
SYNCHRONIZING CIRCUIT Series coupling circuit
M
22
HYDRAULIC CIRCUITS
SYNCHRONIZING CIRCUIT Mechanical Coupling
M
23
HYDRAULIC CIRCUITS
REGENERATIVE CIRCUIT - I Idle Condition
24
HYDRAULIC CIRCUITS
REGENERATIVE CIRCUIT - I Regenerative Advance
25
HYDRAULIC CIRCUITS
REGENERATIVE CIRCUIT - I Retraction
26
HYDRAULIC CIRCUITS
SEQUENCE CIRCUITS Electrically controlled circuit
Seq. Operation Signal Movement
1 Push – ON Sol a Cyl. 1
2 LS - 2 ON Sol c Cyl. 2
3 LS - 3 ON Sol b Cyl. 1
4 LS - 1 ON Sol d Cyl. 2
Cylinder 1 Cylinder 2
a b c d
LS-1 LS-2 LS-3
1 2
3 4
M
27
HYDRAULIC CIRCUITS
SEQUENCE CIRCUITS Automatic control circuit
M
Small Load
Large Load
28
HYDRAULIC CIRCUITS
CLAMPING & SEQUENCING CIRCUIT Extending Clamp Cylinder
29
HYDRAULIC CIRCUITS
CLAMPING & SEQUENCING CIRCUIT Extending Work Cylinder
30
HYDRAULIC CIRCUITS
CLAMPING & SEQUENCING CIRCUIT Limiting Max. Clamping Pr.
31
HYDRAULIC CIRCUITS
CLAMPING & SEQUENCING CIRCUIT Retracting Work Cylinder
32
HYDRAULIC CIRCUITS
CLAMPING & SEQUENCING CIRCUIT Retracting Clamp Cylinder
33
HYDRAULIC CIRCUITS
ACCUMULATOR UNLOADING CIRCUIT Charging
34
HYDRAULIC CIRCUITS
ACCUMULATOR UNLOADING CIRCUIT Unloading
35
HYDRAULIC CIRCUITS
ACCUMULATOR UNLOADING CIRCUIT Supply from Accumulator
36
HYDRAULIC CIRCUITS
ACCUMULATOR CIRCUITS Power saving circuit
M
Starter
motor
Starting
circuit for a
diesel
engine.
37
HYDRAULIC CIRCUITS
ACCUMULATOR CIRCUITS Pressure holding
( leakage compensation )
Vice
M
38
M
HYDRAULIC CIRCUITS
ACCUMULATOR CIRCUITS Safety Device
Safety
device in
a Rolling
Mill
39
HYDRAULIC CIRCUITS
ACCUMULATOR CIRCUITS Surge pressure reducing circuit
M
40
M
HYDRAULIC CIRCUITS
ACCUMULATOR CIRCUITS Pump capacity reducing circuit
Low
Pressure
Pump
High Pressure
Pump
41
HYDRAULIC CIRCUITS
COLLECTION OF DATA FOR CIRCUIT DESIGN
 CYLINDER DETAILS
SINGLE ACTING OR DOUBLE ACTING ?
HOW MANY CYLINDERS ?
SEQUENCE OF CYLINDER MOVEMENT
( ONE AFTER OTHER OR ALMOST TOGETHER )
FUNCTION OF THE CYLINDER ( Eg., Clamping, Drilling )
MACHINE TO WHICH THESE CYLINDERS GO ( Eg., Grinding M/c. )
BORE SIZE & ROD SIZE OF THE CYLINDER
STROKE LENGTH OF THE CYLINDER
MANUAL OR SOLENOID OPERATED MOVEMENT ?
FORCE ACTING ON THE CYLINDER
SPEED OF MOVEMENT REQUIRED
SINGLE SPEED / DOUBLE SPEED / MULTI SPEED ?
LOAD REQUIREMENTS
42
HYDRAULIC CIRCUITS
COLLECTION OF DATA FOR CIRCUIT DESIGN
 OTHER DETAILS
LOCATION OF SYSTEM / EQUIPMENT / ACTUATOR
( Eg. Distance between Power Unit to the Actuator )
LIMITATIONS OF OPERATION ( Eg. Medium, Environment, Space )
AVAILABILITY OF POWER SOURCE & DETAILS ( Eg. AC / DC )
TYPE OF COOLING REQUIRED
SAFETY MEASURES NEEDED
43
HYDRAULIC CIRCUITS
UNDERSTANDING
HYDRAULIC CIRCUITS & HYDRAULIC POWER PACKS
BEGIN WITH THE END
HYDRAULIC CYLINDERS
( LINEAR ACTUATORS )
HYDRAULIC MOTORS
( ROTARY ACTUATORS )
ACTUATORS
44
HYDRAULIC CIRCUITS
A good hydraulic circuit design can be made only when the
parameters influencing the feed drive are clearly understood.
TYPES OF SLIDE
• VERTICAL
• HORIZONTAL
• INCLINED
TYPES OF MACHINING
• ROUGH MACHINING
• FINE MACHINING
45
HYDRAULIC CIRCUITS
FIELD OF
APPLICATION
Pressure ( Kg / Cm2 )
RANGE AVERAGE
MOBILE 70 ~ 300 150
SHIPS ( MARINE ) 40 ~ 250 90
MACHINE TOOL 20 ~ 70 33
FORGES 140 ~ 250 195
INJECTION MOULDING M/c 70 ~ 210 130
INDUSTRIAL ROBOT 5 ~ 140 64
NORMAL WORKING PRESSURES FOR VARIOUS SYSTEMS
46
HYDRAULIC CYLINDERS
 D
 d
AREA A1 AREA A2
F1
F2
PRESSURE = OUTPUT FORCE
EFFECTIVE PISTON AREA
P = F Kg
A Cm2
SELECTION OF AN ACTUATOR
47
HYDRAULIC CIRCUITS
SELECTION OF AN ACTUATOR
Eg. : Pressure = 50 Kg / Cm2
Force required = 4000 Kgs. ( 4 Ton )
P = F
A
Or
A = F
P
= 4000 Kg = 80 Cm2
50 Kg / Cm2
A =  x D2
4
( In this Example A = 80 Cm2 )
80 =  x D2
4 D = 100 mm
Use  100 mm Bore Cylinder
48
HYDRAULIC CIRCUITS
SELECTION OF AN ACTUATOR
STANDARD BORE SIZES OF CYLINDERS ( mm )
32 40 50 63 80 100 125 140 150 160 180 200 220 250 300
Q = A x V
TO CALCULATE THE FLOW “ Q”
Q = Flow in Cm3 / min. ( Divide by 1000 to get flow in LPM )
A = Area in Cm2
V = Velocity in Cm / min
49
HYDRAULIC CIRCUITS
TO CALCULATE THE MOTOR POWER
MOTOR POWER (KW) = P x Q
612 x O
P = Pressure in Kg / Cm2
Q = Flow in LPM
O = Pumps Overall Efficiency ( Eg. 85 %  0.85 )
50
HYDRAULIC CIRCUITS
Heat Generation in a Hydraulic System
SOURCE : Oil Pump
H1 = Li x ( 100 - O ) x 860
100
H1 = Heat generated from the Pump ( Kcal / Hr )
Li = Pump input power ( KW )
O = Pump overall efficiency ( % )
Oil pumps exhaust a large portion of its shaft-input power to perform
an effective task ( Pump output pressure, pump output flow ), while
the rest turns into heat without doing any work.
51
H2 = 10 x 60 x P x Q
427
HYDRAULIC CIRCUITS
Heat Generation in a Hydraulic System
SOURCE : Orifices
H2 = Heat generated ( Kcal / Hr )
P = Differential pressure across an orifice. ( Kg / Cm2 ).
In case of relief valves the set pressure shall be the
differential pressure.
Q = Flow through the orifice ( Lpm )
When pressurised fluid flows through throttle parts at a certain
pressure, the pressure drop is converted into heat ( H2) . Especially
considerable heat will be produced when the pressurised fluid is
released to tank through the Relief valve.
52
M
Load
Pr. 20 Kg/Cm2
HYDRAULIC CIRCUITS
HEAT GENERATION
40 LPM
Set
Pr. 100 Kg/Cm2
60 LPM
Pump
100 LPM
1 Kw = 860 Kcal / Hr
PQ Kw =
612
PQ X 860 Kcal / Hr
612
= 100 x 60 x 860
612
= 8431 Kcal / Hr
Normal Heat Rise
53
HYDRAULIC CIRCUITS
HEAT GENERATION
M
Load
Pr. 20 Kg/Cm2
40 LPM
Set
Pr. 100 Kg/Cm2
60 LPM
Pump
100 LPM
1 Kw = 860 Kcal / Hr
PQ Kw =
612
PQ X 860 Kcal / Hr
612
= 20 x 60 x 860
612
= 1686 Kcal / Hr
With Load
Sensing Heat Rise
54
M
HYDRAULIC CIRCUITS
HEAT GENERATION
Load
Pr. 20 Kg/Cm2
40 LPM
Set
Pr. 100 Kg/Cm2
60 LPM
Pump
100 LPM
With Load
Sensing
55
HYDRAULIC CIRCUITS
Heat dissipation from a Hydraulic System
SOURCES : Reservoir, Tubings, Components
H3 = K x A x ( t1 – t2 )
K = Coefficient of heat dissipation. ( 7 – 9 Kcal / Hr.c.m2 )
A = Effective Area of the Reservoir. ( m2 )
t1 = Oil Temperature (  C )
t2 = Room Temperature (  C )
The dissipated heat ( H3 ) from the surface of the reservoir -
( Kcal / Hr )
In very well ventilated circumstances we can estimate the value of
the Heat transfer coefficient “ K” around 15 Kcal / Hr. C.m2
56
HYDRAULIC CIRCUITS
Heat dissipation from a Hydraulic System
H3 = K x A x ( t1 – t2 )
A = Effective Area of the Reservoir. ( m2 )
( Kcal / Hr )
L = 1000 B = 700
H = 450
A = L x B x H
= 2 [ L x H ] + 2 [ B x H ] + [ L x B ]
= 2 [ 1000 x 450 ] + 2 [ 700 x 450 ] + [ 1000 x 700 ]
= 2 [ 1 x 0.45 ] + 2 [ 0.7 x 0.45 ] + [ 1 x 0.7 ]
= 0.9 + 0.63 + 0.7
= 2.23 m2
57
HYDRAULIC CIRCUITS
Oil Temperature
The oil temperature accelerates the heat transfer as it rises, and
reaches an equilibrium state of thermal relationship H1 + H2 = H3
The equilibrium oil temperature -
t1 = H1 + H2 + t2
KA
At Equilibrium Condition
58
HYDRAULIC CIRCUITS
Oil Temperature
When the temperature is rising
The thermal relationship H1 + H2 > H3 then the oil temperature “t”
at a time “ T” is given by –
- KA T
t = H1 + H2 C + t2
KA 1 – e
C = Heat capacity of the Reservoir ( Cm3 )
C = v x r x S
Where
v = Reservoir capacity. ( cm3 )
r = Specific gravity of oil ( 0.86 x 10 –3 Kgf / Cm3 )
S = Specific heat of oil ( 0.45 Kcal / Kg C )
T = Time ( Hr. )
59
MODULAR VALVES
Features
STACKABLE UNITS
– MAINTENANCE AND SYSTEM CHECK UP MADE EASY.
INSTALLATION AND MOUNTING SPACE MINIMISED.
PIPING ELIMINATED
- OIL LEAKS, VIBRATION AND NOISE CAUSED BY PIPING
MINIMISED.
NO SPECIAL SKILL REQUIRED FOR ASSEMBLY AND ANY
ADDITION OR ALTERATION OF THE HYDRAULIC CIRCUIT
CAN BE MADE QUICKLY AND EASILY.
HYDRAULIC CIRCUITS
60
HYDRAULIC CIRCUITS
Solenoid Operated
Directional Valve
P T B A P T B A
Caution in the Selection of Valves and Circuit designing
Reducing Modular
valve
( for “B” line )
Pilot Operated Check
Modular valve
(for “A” & “B” Lines)
Solenoid Operated
Directional valve
( CORRECT )
( INCORRECT )
61
HYDRAULIC CIRCUITS
Caution in the Selection of Valves and Circuit designing
Solenoid Operated
Directional valve
Throttle and Check
Modular valve
(for “A” & “B” Lines
Meter-out )
Pilot Operated Check
Modular valve
(for “A” & “B” Lines)
Solenoid Operated
Directional Valve
P T B A P T B A
( CORRECT )
( INCORRECT )
62
HYDRAULIC CIRCUITS
Caution in the Selection of Valves and Circuit designing
Solenoid Operated
Directional valve
Throttle and Check
Modular valve
(for “A” & “B” Lines
Meter-out )
( CORRECT )
( INCORRECT )
Brake Modular
valve
Solenoid Operated
Directional Valve
P T B A P T B A
63
HYDRAULIC CIRCUITS
64
HYDRAULIC CIRCUITS
65
HYDRAULIC CIRCUITS
Hyd. Power unit for
Multi Spindle Drilling M/c.
66
HYDRAULIC CIRCUITS
Logic Valves
67
HYDRAULIC CIRCUITS
Logic Valves - Features
MULTIFUNCTION PERFORMANCE IN TERMS OF DIRECTION,
FLOW AND PRESSURE CAN BE OBTAINED BY COMBINING
ELEMENTS AND COVERS.
POPPET TYPE ELEMENTS VIRTUALLY ELIMINATE INTERNAL
LEAKAGE AND HYDRAULIC LOCKING. BECAUSE THERE ARE
NO OVERLAPS, RESPONSE TIMES ARE VERY HIGH,
PERMITTING HIGH-SPEED SHIFTING.
FOR HIGH PRESSURE, LARGE CAPACITY SYSTEMS,
OPTIMUM PERFORMANCE IS ACHIEVED WITH LOW PRESSURE
LOSSES.
68
HYDRAULIC CIRCUITS
Logic Valves - Features
SINCE THE LOGIC VALVES ARE DIRECTLY INCORPORATED IN
CAVITIES PROVIDED IN BLOCKS, THE SYSTEM IS FREE FROM
PROBLEMS RELATED TO PIPING SUCH AS OIL LEAKAGE,
VIBRATION AND NOISE, AND HIGHER RELIABILITY IS
ACHIEVED.
MULTI-FUNCTION LOGIC VALVES PERMIT COMPACT
INTEGRATED HYDRAULIC SYSTEMS WHICH REDUCE MANIFOLD
DIMENSIONS AND MASS AND ACHIEVE LOWER COST
CONVENTIONAL TYPES.
69
HYDRAULIC CIRCUITS
Logic Valves - Features
A
X
B
A
A
X
B
A
X Y
B
Function Graphic Symbols
Working area ratio
(A : A )
A B
Features
Poppet shape
No leakage between port A and B
Flow A to B and B to A are possible
Response time and shock can be adjusted by orifice selection.
Poppet shape
With cushion (LD-*-*-S-1/2/3): flow control.
No leakage between port A and B
Flow A to B only is possible.
Response time and shock can be adjusted by orifice selection.
Remote andunloadingcontrol is possiblewithventcircuit(LB-*-*).
Twoor threepressurecontrols arepossiblein combination ofsolenoid
operated directional valve and pilot relief valve (LBS-*-*).
(2 : 1)
(24 : 1)
Direction
Direction
and Flow
Relief
Without cushion (LD/LDS-*-*): high-speed shift
With cushion (LD/LDS-*-*-S): Shockless shift
/
Functions, working area ratios and features
70
Selection of accumulator capacity.
There are many chances to use accumulator as a source of
energy.
To select the capacity of accumulator, We must know:
(1) Required oil discharge amount: liters.
(2) Max. operating pressure: P3 Kgf / Cm2.
(3) Min. operating pressure : P2 Kgf / Cm2.
(4) Gas charge pressure : P1 Kgf / Cm2.
P1  P2 (0.85 ~ 0.9)
(5) Charging time, discharge time
Especially for discharge time
Incase T > 1 min: use isothermal change.
T < 1 min: use adiabatic change.
From these required specification we can calculate the required
vol. of accumulator.
HYDRAULIC CIRCUITS
71
HYDRAULIC CIRCUITS
CASE STUDY - I : CNC Drilling Machine
Data Available :
1. CYLINDER SPECIFICATION :
Clamping -  125 x  50 x 20 Stroke
Drilling -  63 x  35 x 100 Stroke
2. LOAD OR FORCE ACTING ON THE CYLINDER
Clamping - 400 Kgf
Drilling - 250 Kgf
3. SPEED OF ACTUATORS
Clamping - 1.5 M / min.
Drilling - 0.1 M / min.
72
HYDRAULIC CIRCUITS
CASE STUDY – I : CNC Drilling Machine
1 ) Pressure required for clamping “ P1 ”
P1 = F
A
= 400
122.7
A =  x D2
4
A =  x 12.5 x 12.5
4
= 122.7 Cm2
= 3.3 Kg / Cm2
2 ) Pressure required for drilling “ P2 ”
P2 = F
A
= 250
31.2
A =  x D2
4
A =  x 6.3 x 6.3
4
= 31.2 Cm2
= 8 Kg / Cm2
73
HYDRAULIC CIRCUITS
CASE STUDY – I : CNC Drilling Machine
3 ) Flow required for clamping “ Q1 ”
4 ) Flow required for drilling “ Q2 ”
A =  x D2
4
A =  x 12.5 x 12.5
4
= 122.7 Cm2
Q 1 = A x V
= 122.7 x 1.5 x 100 Cm3 / min
= 18405 Cm3 / min
= 18.4 LPM
A =  x D2
4
A =  x 6.3 x 6.3
4
= 31.2 Cm2
Q 2 = A x V
= 31.2 x 0.1 x 100 Cm3 / min
= 312 Cm3 / min
= 0.31 LPM
74
HYDRAULIC CIRCUITS
CASE STUDY – I : CNC Drilling Machine
5 ) Electric Motor Power
MOTOR POWER (KW) = P x Q
612 x O
8 x 18.4 = 0.28 KW
612 x 0.85 (0.38 HP)
1 hp = 0.746 KW
6 ) Tank Size - ( General Thumb Rule )
For Vane & Gear Pumps = 4 ~ 5 times of System Flow
For Piston Pumps = 2 ~ 3 times of System Flow
7 ) Maximum Pressure to be considered = 8 Kg / Cm2
Maximum Flow to be considered = 18.4 LPM
Electric Motor Power = 0.38 say 0.5 HP
Tank Capacity = 18.4 x 4 = 73.6 say 75 ltrs
75
HYDRAULIC CIRCUITS
SEQUENCE CIRCUITS
Circuit using
Sequence
Valve
Clamp
Drill
1
4
2
3
1 - CLAMPING
2 - DRILLING
3 - DRILL RETURN
4 - DE CLAMP
Sequence of Operation
1
2
3
4
Sequence of Flow
1 2 3 4
M
SEQUENCE
VALVE
SEQUENCE VALVE
76
77
78

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12-ch_hy_circuits..ppt

  • 3. 3 HYDRAULIC CIRCUITS A GOOD HYDRAULIC SYSTEM REQUIREMENT - SATISFY THE SPECIFICATIONS OF THE OPERATION WITH SAFETY PERFORM SMOOTH OPERATION LOW ENERGY CONSUMPTION – LOW HEAT GENERATION REDUCE INITIAL COST & RUNNING COST MAKE MAINTENANCE EASY HYDRAULIC CIRCUITS ARE GRAPHICAL DIAGRAMS OF THE HYDRAULIC SYSTEMS. IT ALSO INDICATES EACH OPERATION OF THE COMPONENTS.
  • 4. 4 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Variable displacement pump circuit M
  • 5. 5 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Meter – in Circuit M
  • 6. 6 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Meter – out Circuit M
  • 7. 7 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Bleed – off Circuit M
  • 8. 8 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Deceleration Circuit M
  • 9. 9 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Feed speed varying circuit M
  • 10. 10 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Multi Speed Circuit M Q1 : High Flow Q2 : Low Flow Q1 Q2
  • 11. 11 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Multi Speed Circuit 1 2 3 1 : Rapid Advance 2 : Medium Advance 3 : Slow Advance M UCF2-04
  • 12. 12 HYDRAULIC CIRCUITS SPEED CONTROL CIRCUIT Multi Speed Circuit Sol. 1 ON Low speed forward Sol. 3 ON High speed forward Sol. 3 OFF Speed decrease Sol. 1 OFF Stop. Sol. 2 ON Low speed reverse Sol. 4 ON High speed reverse Sol. 4 OFF Speed decrease Sol. 2 OFF Stop. Sol. 1 Sol. 2 Sol. 3 Sol. 4 M
  • 13. 13 HYDRAULIC CIRCUITS PRESSURE CONTROL CIRCUIT 2 Operating Pressure Circuit 1 2
  • 14. 14 HYDRAULIC CIRCUITS PRESSURE CONTROL CIRCUIT Low Pressure Return Circuit 1 2 Main Relief Valve Pilot Relief Valve
  • 15. 15 M HYDRAULIC CIRCUITS PRESSURE CONTROL CIRCUIT Decompression Circuit
  • 17. 17 HYDRAULIC CIRCUITS UNLOADING CIRCUIT Circuit using Accumulator Detection of Pressure by Pressure Switch
  • 18. 18 HYDRAULIC CIRCUITS UNLOADING CIRCUIT Circuit using Accumulator M Detection of Pressure by Pilot Op. Relief Valve
  • 19. 19 HYDRAULIC CIRCUITS UNLOADING CIRCUIT ( Hi-Low Circuit ) Low Pressure Operation
  • 20. 20 HYDRAULIC CIRCUITS UNLOADING CIRCUIT ( Hi-Low Circuit ) High Pressure Operation
  • 21. 21 HYDRAULIC CIRCUITS SYNCHRONIZING CIRCUIT Series coupling circuit M
  • 24. 24 HYDRAULIC CIRCUITS REGENERATIVE CIRCUIT - I Regenerative Advance
  • 26. 26 HYDRAULIC CIRCUITS SEQUENCE CIRCUITS Electrically controlled circuit Seq. Operation Signal Movement 1 Push – ON Sol a Cyl. 1 2 LS - 2 ON Sol c Cyl. 2 3 LS - 3 ON Sol b Cyl. 1 4 LS - 1 ON Sol d Cyl. 2 Cylinder 1 Cylinder 2 a b c d LS-1 LS-2 LS-3 1 2 3 4 M
  • 27. 27 HYDRAULIC CIRCUITS SEQUENCE CIRCUITS Automatic control circuit M Small Load Large Load
  • 28. 28 HYDRAULIC CIRCUITS CLAMPING & SEQUENCING CIRCUIT Extending Clamp Cylinder
  • 29. 29 HYDRAULIC CIRCUITS CLAMPING & SEQUENCING CIRCUIT Extending Work Cylinder
  • 30. 30 HYDRAULIC CIRCUITS CLAMPING & SEQUENCING CIRCUIT Limiting Max. Clamping Pr.
  • 31. 31 HYDRAULIC CIRCUITS CLAMPING & SEQUENCING CIRCUIT Retracting Work Cylinder
  • 32. 32 HYDRAULIC CIRCUITS CLAMPING & SEQUENCING CIRCUIT Retracting Clamp Cylinder
  • 35. 35 HYDRAULIC CIRCUITS ACCUMULATOR UNLOADING CIRCUIT Supply from Accumulator
  • 36. 36 HYDRAULIC CIRCUITS ACCUMULATOR CIRCUITS Power saving circuit M Starter motor Starting circuit for a diesel engine.
  • 37. 37 HYDRAULIC CIRCUITS ACCUMULATOR CIRCUITS Pressure holding ( leakage compensation ) Vice M
  • 38. 38 M HYDRAULIC CIRCUITS ACCUMULATOR CIRCUITS Safety Device Safety device in a Rolling Mill
  • 39. 39 HYDRAULIC CIRCUITS ACCUMULATOR CIRCUITS Surge pressure reducing circuit M
  • 40. 40 M HYDRAULIC CIRCUITS ACCUMULATOR CIRCUITS Pump capacity reducing circuit Low Pressure Pump High Pressure Pump
  • 41. 41 HYDRAULIC CIRCUITS COLLECTION OF DATA FOR CIRCUIT DESIGN  CYLINDER DETAILS SINGLE ACTING OR DOUBLE ACTING ? HOW MANY CYLINDERS ? SEQUENCE OF CYLINDER MOVEMENT ( ONE AFTER OTHER OR ALMOST TOGETHER ) FUNCTION OF THE CYLINDER ( Eg., Clamping, Drilling ) MACHINE TO WHICH THESE CYLINDERS GO ( Eg., Grinding M/c. ) BORE SIZE & ROD SIZE OF THE CYLINDER STROKE LENGTH OF THE CYLINDER MANUAL OR SOLENOID OPERATED MOVEMENT ? FORCE ACTING ON THE CYLINDER SPEED OF MOVEMENT REQUIRED SINGLE SPEED / DOUBLE SPEED / MULTI SPEED ? LOAD REQUIREMENTS
  • 42. 42 HYDRAULIC CIRCUITS COLLECTION OF DATA FOR CIRCUIT DESIGN  OTHER DETAILS LOCATION OF SYSTEM / EQUIPMENT / ACTUATOR ( Eg. Distance between Power Unit to the Actuator ) LIMITATIONS OF OPERATION ( Eg. Medium, Environment, Space ) AVAILABILITY OF POWER SOURCE & DETAILS ( Eg. AC / DC ) TYPE OF COOLING REQUIRED SAFETY MEASURES NEEDED
  • 43. 43 HYDRAULIC CIRCUITS UNDERSTANDING HYDRAULIC CIRCUITS & HYDRAULIC POWER PACKS BEGIN WITH THE END HYDRAULIC CYLINDERS ( LINEAR ACTUATORS ) HYDRAULIC MOTORS ( ROTARY ACTUATORS ) ACTUATORS
  • 44. 44 HYDRAULIC CIRCUITS A good hydraulic circuit design can be made only when the parameters influencing the feed drive are clearly understood. TYPES OF SLIDE • VERTICAL • HORIZONTAL • INCLINED TYPES OF MACHINING • ROUGH MACHINING • FINE MACHINING
  • 45. 45 HYDRAULIC CIRCUITS FIELD OF APPLICATION Pressure ( Kg / Cm2 ) RANGE AVERAGE MOBILE 70 ~ 300 150 SHIPS ( MARINE ) 40 ~ 250 90 MACHINE TOOL 20 ~ 70 33 FORGES 140 ~ 250 195 INJECTION MOULDING M/c 70 ~ 210 130 INDUSTRIAL ROBOT 5 ~ 140 64 NORMAL WORKING PRESSURES FOR VARIOUS SYSTEMS
  • 46. 46 HYDRAULIC CYLINDERS  D  d AREA A1 AREA A2 F1 F2 PRESSURE = OUTPUT FORCE EFFECTIVE PISTON AREA P = F Kg A Cm2 SELECTION OF AN ACTUATOR
  • 47. 47 HYDRAULIC CIRCUITS SELECTION OF AN ACTUATOR Eg. : Pressure = 50 Kg / Cm2 Force required = 4000 Kgs. ( 4 Ton ) P = F A Or A = F P = 4000 Kg = 80 Cm2 50 Kg / Cm2 A =  x D2 4 ( In this Example A = 80 Cm2 ) 80 =  x D2 4 D = 100 mm Use  100 mm Bore Cylinder
  • 48. 48 HYDRAULIC CIRCUITS SELECTION OF AN ACTUATOR STANDARD BORE SIZES OF CYLINDERS ( mm ) 32 40 50 63 80 100 125 140 150 160 180 200 220 250 300 Q = A x V TO CALCULATE THE FLOW “ Q” Q = Flow in Cm3 / min. ( Divide by 1000 to get flow in LPM ) A = Area in Cm2 V = Velocity in Cm / min
  • 49. 49 HYDRAULIC CIRCUITS TO CALCULATE THE MOTOR POWER MOTOR POWER (KW) = P x Q 612 x O P = Pressure in Kg / Cm2 Q = Flow in LPM O = Pumps Overall Efficiency ( Eg. 85 %  0.85 )
  • 50. 50 HYDRAULIC CIRCUITS Heat Generation in a Hydraulic System SOURCE : Oil Pump H1 = Li x ( 100 - O ) x 860 100 H1 = Heat generated from the Pump ( Kcal / Hr ) Li = Pump input power ( KW ) O = Pump overall efficiency ( % ) Oil pumps exhaust a large portion of its shaft-input power to perform an effective task ( Pump output pressure, pump output flow ), while the rest turns into heat without doing any work.
  • 51. 51 H2 = 10 x 60 x P x Q 427 HYDRAULIC CIRCUITS Heat Generation in a Hydraulic System SOURCE : Orifices H2 = Heat generated ( Kcal / Hr ) P = Differential pressure across an orifice. ( Kg / Cm2 ). In case of relief valves the set pressure shall be the differential pressure. Q = Flow through the orifice ( Lpm ) When pressurised fluid flows through throttle parts at a certain pressure, the pressure drop is converted into heat ( H2) . Especially considerable heat will be produced when the pressurised fluid is released to tank through the Relief valve.
  • 52. 52 M Load Pr. 20 Kg/Cm2 HYDRAULIC CIRCUITS HEAT GENERATION 40 LPM Set Pr. 100 Kg/Cm2 60 LPM Pump 100 LPM 1 Kw = 860 Kcal / Hr PQ Kw = 612 PQ X 860 Kcal / Hr 612 = 100 x 60 x 860 612 = 8431 Kcal / Hr Normal Heat Rise
  • 53. 53 HYDRAULIC CIRCUITS HEAT GENERATION M Load Pr. 20 Kg/Cm2 40 LPM Set Pr. 100 Kg/Cm2 60 LPM Pump 100 LPM 1 Kw = 860 Kcal / Hr PQ Kw = 612 PQ X 860 Kcal / Hr 612 = 20 x 60 x 860 612 = 1686 Kcal / Hr With Load Sensing Heat Rise
  • 54. 54 M HYDRAULIC CIRCUITS HEAT GENERATION Load Pr. 20 Kg/Cm2 40 LPM Set Pr. 100 Kg/Cm2 60 LPM Pump 100 LPM With Load Sensing
  • 55. 55 HYDRAULIC CIRCUITS Heat dissipation from a Hydraulic System SOURCES : Reservoir, Tubings, Components H3 = K x A x ( t1 – t2 ) K = Coefficient of heat dissipation. ( 7 – 9 Kcal / Hr.c.m2 ) A = Effective Area of the Reservoir. ( m2 ) t1 = Oil Temperature (  C ) t2 = Room Temperature (  C ) The dissipated heat ( H3 ) from the surface of the reservoir - ( Kcal / Hr ) In very well ventilated circumstances we can estimate the value of the Heat transfer coefficient “ K” around 15 Kcal / Hr. C.m2
  • 56. 56 HYDRAULIC CIRCUITS Heat dissipation from a Hydraulic System H3 = K x A x ( t1 – t2 ) A = Effective Area of the Reservoir. ( m2 ) ( Kcal / Hr ) L = 1000 B = 700 H = 450 A = L x B x H = 2 [ L x H ] + 2 [ B x H ] + [ L x B ] = 2 [ 1000 x 450 ] + 2 [ 700 x 450 ] + [ 1000 x 700 ] = 2 [ 1 x 0.45 ] + 2 [ 0.7 x 0.45 ] + [ 1 x 0.7 ] = 0.9 + 0.63 + 0.7 = 2.23 m2
  • 57. 57 HYDRAULIC CIRCUITS Oil Temperature The oil temperature accelerates the heat transfer as it rises, and reaches an equilibrium state of thermal relationship H1 + H2 = H3 The equilibrium oil temperature - t1 = H1 + H2 + t2 KA At Equilibrium Condition
  • 58. 58 HYDRAULIC CIRCUITS Oil Temperature When the temperature is rising The thermal relationship H1 + H2 > H3 then the oil temperature “t” at a time “ T” is given by – - KA T t = H1 + H2 C + t2 KA 1 – e C = Heat capacity of the Reservoir ( Cm3 ) C = v x r x S Where v = Reservoir capacity. ( cm3 ) r = Specific gravity of oil ( 0.86 x 10 –3 Kgf / Cm3 ) S = Specific heat of oil ( 0.45 Kcal / Kg C ) T = Time ( Hr. )
  • 59. 59 MODULAR VALVES Features STACKABLE UNITS – MAINTENANCE AND SYSTEM CHECK UP MADE EASY. INSTALLATION AND MOUNTING SPACE MINIMISED. PIPING ELIMINATED - OIL LEAKS, VIBRATION AND NOISE CAUSED BY PIPING MINIMISED. NO SPECIAL SKILL REQUIRED FOR ASSEMBLY AND ANY ADDITION OR ALTERATION OF THE HYDRAULIC CIRCUIT CAN BE MADE QUICKLY AND EASILY. HYDRAULIC CIRCUITS
  • 60. 60 HYDRAULIC CIRCUITS Solenoid Operated Directional Valve P T B A P T B A Caution in the Selection of Valves and Circuit designing Reducing Modular valve ( for “B” line ) Pilot Operated Check Modular valve (for “A” & “B” Lines) Solenoid Operated Directional valve ( CORRECT ) ( INCORRECT )
  • 61. 61 HYDRAULIC CIRCUITS Caution in the Selection of Valves and Circuit designing Solenoid Operated Directional valve Throttle and Check Modular valve (for “A” & “B” Lines Meter-out ) Pilot Operated Check Modular valve (for “A” & “B” Lines) Solenoid Operated Directional Valve P T B A P T B A ( CORRECT ) ( INCORRECT )
  • 62. 62 HYDRAULIC CIRCUITS Caution in the Selection of Valves and Circuit designing Solenoid Operated Directional valve Throttle and Check Modular valve (for “A” & “B” Lines Meter-out ) ( CORRECT ) ( INCORRECT ) Brake Modular valve Solenoid Operated Directional Valve P T B A P T B A
  • 65. 65 HYDRAULIC CIRCUITS Hyd. Power unit for Multi Spindle Drilling M/c.
  • 67. 67 HYDRAULIC CIRCUITS Logic Valves - Features MULTIFUNCTION PERFORMANCE IN TERMS OF DIRECTION, FLOW AND PRESSURE CAN BE OBTAINED BY COMBINING ELEMENTS AND COVERS. POPPET TYPE ELEMENTS VIRTUALLY ELIMINATE INTERNAL LEAKAGE AND HYDRAULIC LOCKING. BECAUSE THERE ARE NO OVERLAPS, RESPONSE TIMES ARE VERY HIGH, PERMITTING HIGH-SPEED SHIFTING. FOR HIGH PRESSURE, LARGE CAPACITY SYSTEMS, OPTIMUM PERFORMANCE IS ACHIEVED WITH LOW PRESSURE LOSSES.
  • 68. 68 HYDRAULIC CIRCUITS Logic Valves - Features SINCE THE LOGIC VALVES ARE DIRECTLY INCORPORATED IN CAVITIES PROVIDED IN BLOCKS, THE SYSTEM IS FREE FROM PROBLEMS RELATED TO PIPING SUCH AS OIL LEAKAGE, VIBRATION AND NOISE, AND HIGHER RELIABILITY IS ACHIEVED. MULTI-FUNCTION LOGIC VALVES PERMIT COMPACT INTEGRATED HYDRAULIC SYSTEMS WHICH REDUCE MANIFOLD DIMENSIONS AND MASS AND ACHIEVE LOWER COST CONVENTIONAL TYPES.
  • 69. 69 HYDRAULIC CIRCUITS Logic Valves - Features A X B A A X B A X Y B Function Graphic Symbols Working area ratio (A : A ) A B Features Poppet shape No leakage between port A and B Flow A to B and B to A are possible Response time and shock can be adjusted by orifice selection. Poppet shape With cushion (LD-*-*-S-1/2/3): flow control. No leakage between port A and B Flow A to B only is possible. Response time and shock can be adjusted by orifice selection. Remote andunloadingcontrol is possiblewithventcircuit(LB-*-*). Twoor threepressurecontrols arepossiblein combination ofsolenoid operated directional valve and pilot relief valve (LBS-*-*). (2 : 1) (24 : 1) Direction Direction and Flow Relief Without cushion (LD/LDS-*-*): high-speed shift With cushion (LD/LDS-*-*-S): Shockless shift / Functions, working area ratios and features
  • 70. 70 Selection of accumulator capacity. There are many chances to use accumulator as a source of energy. To select the capacity of accumulator, We must know: (1) Required oil discharge amount: liters. (2) Max. operating pressure: P3 Kgf / Cm2. (3) Min. operating pressure : P2 Kgf / Cm2. (4) Gas charge pressure : P1 Kgf / Cm2. P1  P2 (0.85 ~ 0.9) (5) Charging time, discharge time Especially for discharge time Incase T > 1 min: use isothermal change. T < 1 min: use adiabatic change. From these required specification we can calculate the required vol. of accumulator. HYDRAULIC CIRCUITS
  • 71. 71 HYDRAULIC CIRCUITS CASE STUDY - I : CNC Drilling Machine Data Available : 1. CYLINDER SPECIFICATION : Clamping -  125 x  50 x 20 Stroke Drilling -  63 x  35 x 100 Stroke 2. LOAD OR FORCE ACTING ON THE CYLINDER Clamping - 400 Kgf Drilling - 250 Kgf 3. SPEED OF ACTUATORS Clamping - 1.5 M / min. Drilling - 0.1 M / min.
  • 72. 72 HYDRAULIC CIRCUITS CASE STUDY – I : CNC Drilling Machine 1 ) Pressure required for clamping “ P1 ” P1 = F A = 400 122.7 A =  x D2 4 A =  x 12.5 x 12.5 4 = 122.7 Cm2 = 3.3 Kg / Cm2 2 ) Pressure required for drilling “ P2 ” P2 = F A = 250 31.2 A =  x D2 4 A =  x 6.3 x 6.3 4 = 31.2 Cm2 = 8 Kg / Cm2
  • 73. 73 HYDRAULIC CIRCUITS CASE STUDY – I : CNC Drilling Machine 3 ) Flow required for clamping “ Q1 ” 4 ) Flow required for drilling “ Q2 ” A =  x D2 4 A =  x 12.5 x 12.5 4 = 122.7 Cm2 Q 1 = A x V = 122.7 x 1.5 x 100 Cm3 / min = 18405 Cm3 / min = 18.4 LPM A =  x D2 4 A =  x 6.3 x 6.3 4 = 31.2 Cm2 Q 2 = A x V = 31.2 x 0.1 x 100 Cm3 / min = 312 Cm3 / min = 0.31 LPM
  • 74. 74 HYDRAULIC CIRCUITS CASE STUDY – I : CNC Drilling Machine 5 ) Electric Motor Power MOTOR POWER (KW) = P x Q 612 x O 8 x 18.4 = 0.28 KW 612 x 0.85 (0.38 HP) 1 hp = 0.746 KW 6 ) Tank Size - ( General Thumb Rule ) For Vane & Gear Pumps = 4 ~ 5 times of System Flow For Piston Pumps = 2 ~ 3 times of System Flow 7 ) Maximum Pressure to be considered = 8 Kg / Cm2 Maximum Flow to be considered = 18.4 LPM Electric Motor Power = 0.38 say 0.5 HP Tank Capacity = 18.4 x 4 = 73.6 say 75 ltrs
  • 75. 75 HYDRAULIC CIRCUITS SEQUENCE CIRCUITS Circuit using Sequence Valve Clamp Drill 1 4 2 3 1 - CLAMPING 2 - DRILLING 3 - DRILL RETURN 4 - DE CLAMP Sequence of Operation 1 2 3 4 Sequence of Flow 1 2 3 4 M SEQUENCE VALVE SEQUENCE VALVE
  • 76. 76
  • 77. 77
  • 78. 78