Date post: | 02-Jul-2015 |
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Design and analyze basic hydraulic circuit both for single-acting and double-acting cylinder
A Hydraulic circuit is a group of components such as pumps, actuators, and control valves so arranged that they will perform a useful task.
When analyzing or designing a hydraulic circuit, the following three important considerations must be taken into account:
1. Safety of operation
2. Performance of desired function
3. Efficiency of operation
…Pressure and Temperature ratings
…Interlocks for sequential operations
…Emergency shutdown features
…Power failure locks
…Operation speed
…Environment conditions
…Meet required performance specification
…Life expectancy same as machine
…Facilitate good maintenance practice
…Compatibility with electrical and mechanical components
…Withstand operational hazards
…Keep system Simple, Safe and Functional
…Access to parts need repair or adjustment
…Design to keep min operational cost
…Design to prevent and remove contamination.
Each design must have following section
1. Power supply section – pump, elec motor, engine, etc
2. Power control section – valve, magnetic valve, plc, controller, etc
3. Drive section – cylinders, motors
Usually, the user specifies the final result of design ◦ Eg: Customer need a hydraulic power pack to lift 3
tons load
Engineer needs to get several answers before offer for hydraulic power pack: ◦ For what application
◦ How many cylinders
◦ Nature of the work (lift/clamp/push etc)
Bore size of cylinder
Rod size of cylinder
Stroke length
Speed of movement required
Expected load to take
Q: For what application ◦ A: Special purpose of drilling
Q: How many cylinders ◦ A: two double acting cylinders (1 for clamping &
1 for drilling)
Q: Nature of the work (lift/clamp/push etc) ◦ A: Clamping cylinder acting first, followed by
drilling
Bore size of cylinder (clamping = 80mm, drilling = 63mm)
Rod size of cylinder (standard)
Stroke length (clamping= 20mm, drilling = 120mm)
Speed of movement (clamping = 1.5 m/min, drilling = 200mm/min)
Expected load to take (clamping = 600kg, drilling = 500 kg)
Calculate pump capacity for hydraulic power unit (Q=n.V)
Capacity (cm3/min) = Area of cylinder (cm2) X Speed of movement (cm/min)
For drilling, by using similar approach - pump req = 0.623 lit/min; select 7.5 lit/min
)11000(min/5.77536
min/15024.50
24.50
8);(4
3
2
2
1
2
1
litrecclitcm
cmxcmrequiredPump
cm
cmdcmdAclamping
2
2
2
/05.1615.31
500
/94.1124.50
600
)(
)(
Pr
cmkgkg
presureDrilling
cmkg
cmeaclampingar
kgrceclampingfopresureClamping
AreaForceessure
Max. working pressure = 16.05 kg/cm2
Therefore we can choose the next standard size of electric motor; i.e. 0.5 hp, run at 1440 rpm
hpkW
hp
kWlcmkg
kWinPower
litflowrateQ
cmkgpressureworkingP
PQkWPower
764.0;26.0
2.0600
min)/(5.7)2/(05.16
min)/(
)/(
;600
)(
2
Thumb rule: Reservoir should be 4 times of flow rate of the pump
Here, pump flow rate = 7.5 l/min, therefore, the reservoir should be at least 30 litres
Manufacturer standard size = 50, 75, 100, 125 litres, etc. So, 50 litres reservoir can be chosen
Reservoir capacity = 50 litres
Pump capacity = 8 lit/min (in lieu of 7.5 lit/min)
Motor = 0.5 hp, 1440 rpm
Working pressure = 20 kg/cm2
Components ◦ Single acting cylinder ◦ 3/2-way valve ◦ Pressure relief valve ◦ Hydraulic pump
Initial position ◦ fluid flow goes to the tank via PRV ◦ Piston oil from the blank end drains back into the
tank
When actuated ◦ Fluid goes to the blank end and extends the
cylinder ◦ At full extension, pump flow goes through PRV
Power supply
section
Control section
Drive section
Components ◦ Double-acting cylinder
◦ 4/3 way valve
Centered position ◦ Cylinder is hydraulically locked
◦ Fluid from pump goes to tank
Left position ◦ Cylinder is extended against the load force
◦ Oil in the rod-end flow back to tank via 4/3 way valve
Right position ◦ Cylinder retracts as oil flows into rod-end side
◦ Oil in blank end returned to tank
The output force ( F ) and piston velocity of DAC are not the same for extension and retraction strokes.
During the extension stroke, fluid enters A through the entire circular area of the piston (AP).
retraction stroke, fluid enters the rod end through the smaller area ( AP – AR ), ◦ AP = piston area ◦ AR = rod area.
Since AP > ( AP – AR ), the retraction velocity > extension velocity since the pump flow rate is constant.
The power developed by a hydraulic cylinder for either the extension or retraction stroke, can be found out by
HYDRAULIC AND PNEUMATIC
Regenerative Cylinder Circuit
Regenerative circuit is used to
speed up the extending speed of
a double acting cylinder.
During the extension, flow from
the rod end regenerates with the
pump flow to provide greater
flow rate.
The operation of the cylinder
during the retraction stroke is
the same as that of a regular
double acting cylinder.
HYDRAULIC AND PNEUMATIC
Regenerative Cylinder Circuit
The total flow rate (QT) entering the blank
end of the cylinder equal the pump flow
rate (QP) plus the regenerative flow rate
(QR) coming from the rod end of the
cylinder.
QT= QP + QR
Solving for the pump flow rate,
QP= QT - QR
extrpextPP vAAvAQ )( QR
QT=QP+QR
QP
vext
AP Ar
r
Pext
A
Qv
Hence,
Retracting speed :
rp
Pret
AA
Qv
1
r
p
r
rp
ret
ext
A
A
A
AA
v
v
Ratio of Extending and Retracting Speeds
HYDRAULIC AND PNEUMATIC
Regenerative Cylinder Circuit
1
r
p
r
rp
ret
ext
A
A
A
AA
v
v
Ratio of Extending and Retracting Speeds
Load carrying capacity during extension
This is because system pressure acts on both sides of the piston during the
extending stroke of the regenerative cylinder.
Load carrying capacity during retraction
rload pAFext (Less than regular double acting cylinder, Fload=pAp)
)( rpload AApFret