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