The document provides an overview of hydraulic circuits and components. It discusses key considerations in designing hydraulic circuits such as satisfying operational specifications safely, performing smooth operations, and reducing costs and heat generation. Hydraulic circuits are graphical diagrams that indicate the operation of components in hydraulic systems. The document also covers various types of circuits like speed control, pressure control, unloading, sequencing and accumulator circuits. It emphasizes understanding the application and selecting components appropriately based on factors like required forces, speeds, flows and pressures.
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.
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
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
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
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
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
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