Introduction

1- Operation

A regenerative circuit that is used to speed up the extending speed of a double-acting hydraulic cylinder. Here the pipelines to both ends of the hydraulic cylinder are connected to pump, one end (A) through the 2 / 3 way DCV and the other end (B) directly. The operation of the cylinder during the retraction stroke is the same as that of a regular double-acting cylinder. Fluid flows through the DCV zeroposition from the actuator A side during retraction. In this position, fluid from the pump directly enters the rod end of the cylinder ( direct connection). Fluid in the blank end drains back to the tank through the DCV as the cylinder retracts. When the DCV is shifted to 1 position due to manual actuation, the cylinder extends. The speed of extension is greater than that for a regular double-acting cylinderbecause flow from the rod end (QR) regenerates with the pump flow (QP) to provide a total flow rate (QT), which is greater than the pump flow rate to the A side of the cylinder. ( Area of blank end is more than rod end, thereby blank end provide least resistance )

2- Cylinder Extending Speed

The total flow rate entering the blank end (A) of the cylinder equals the pump flow rate plus the regenerative flow rate coming from the rod end of the cylinder: QT = QP + QR

Or QP = QT - QR

We know that the total flow rate equals the piston area multiplied by the extending speed of the piston (Vpext). Similarly, the regenerative flow rate equals the difference of the piston and rod areas (Ap - Ar) multiplied by the extending speed of the piston. Substituting these two relationships into the eq (1) yields

QP = ApVpext (Ap Ar)Vpext

Therefore QP = ArVpext

Hence the extending speed of the piston V =

Thus the extending speed equals the pump flow divided by the rod area. Thus, a small rod area (which produces a large regenerative flow) provides a large extending speed. In fact the extending speed can be greater than the retracting speed if the rod area is made small enough.

3- Ratio of Extending and Retracting Speeds

Lets determine under what condition the extending and retracting speeds are equal. We know that the retracting speed (Vpret) equals the pump flow divided by the difference of the piston and rod areas

We see that when the piston area equals two times the rod area, the extension and retraction speeds are equal. In general, the greater the ratio of piston area to rod area, the greater the ratio of extending speed to retracting speed.

4- Load-Carrying Capacity during Extension :

It should be noted that the load-carrying capacity of a regenerativecylinder during extension is less than that obtained from a regular double-acting cylinder. The load-carrying capacity (Fload) for a regenerative cylinder equals the pressure multiplied by piston rod area rather than the pressure multiplied by piston area. This is due to the same system pressure acting on both sides of the piston during the extending stroke of the regenerative cylinder Thus

Fload = P*Ar

Thus, the power obtained from the regenerative cylinder is less because the extending speed is increased at the expense of load-carrying capacity.

System from question

System part Figure shows an application of regenerative circuit in a drilling machine. Here a 3-position, 4-way , regenerative center directional control valve is used. When the DCV is in the spring-centered position, port P is connected to A and B and tank port T is blocked. In this position pump flow goes to A and flow from rod end of the cylinder also joins the pump flow to gives rapid spindle advance ( no work is doneduring this period ) Why does the spring-centered position give rapid extension of the cylinder (drill spindle)? The reason is simple. Oil from the rod end regenerates with the pump flow going to the blank end. This effectively increases pump flow to the blank end of the cylinder during the spring-centered mode of operation. Once again we have a regenerative cylinder. It should be noted that the cylinder used in a regenerative circuit is actually a regular double-acting cylinder. What makes it a regenerative cylinder is the way it is hooked up in the circuit. The blank and rod ends are connected in parallel during the extending stroke of a regenerative center. When the DCV shifts to 1st position, P is connected to A and B to T gives slow feed (extension) when the drill starts to cut into the work piece. Similarly when the DCV shifts to 2nd position, P is connected to B and A is connected to T, since the ring area is less the cylinder will have fast return motion.

Metric data is as follows:

Electric Motor: Power= 25HP * = 18.65 KW

Overall efficiency = 90%

Pump: Discharge pressure = 1000 psi * = 6897(KPa)

Oil: Kin Visc v0001 * = 0.000093Weight Dens = Cylinder : Piston Diameter = 8 in * * = 0.203 m

Rod Diameter = 4 in = 0.102 m

Elbows: K factor = 0.75

Pipes:NO:Length (m)Diameter (m)

10.610.0508

26.100.0445

39.150.0445

49.150.0445

56.100.0445

We have the following useable equations:

V = *

Pump Power = P (kPa) * Q ( = 0.90 18.65 16.79 KW

= = 0.00243 (

= * - * ( )

= = 0.00243

we have the following two equations to solve for the flow rates in lines 3 and 4:

= * = = 0.00972

= * = = 0.00729

Solving for the velocities yields: V =

= = 1.20

= = 1.56

= = 6.24

= = 4.69

We can now calculate the Reynolds numbers.

= 655

= 746

= 2990

= 2240

Also f = and P = *

= 0.74m= 0.74*7850= 5840 Pa

=2.08 m = 16300 Pa

= 10.2 m = 80300 Pa

= 7.43 m = 58400 Pa

Thus we have

F ( KN) = (6897KPa 5.84-16.3-80.3) * * - (6897KPa 5.84-16.3-58.4) * *

F = 220-168 = 52 KN

Extend due regenerative

End of rod

Retrace circuit

End of rod

Extend circuit

End of rod

Application

The hydraulic press depends onPascal's principle: thepressurethroughout a closed system is constant. One part of the system is apistonacting as a pump, with a modest mechanical force acting on a small cross-sectional area; the other part is a piston with a larger area which generates a correspondingly large mechanical force. Only small-diametertubing(which more easily resists pressure) is needed if the pump is separated from the press cylinder.Pascal's law: Pressure on a confined fluid is transmitted undiminished and acts with equal force on equal areas and at 90 degrees to the container wall.A fluid, such asoil, is displaced when either piston is pushed inward. Since the fluid is incompressible, thevolumethat the small piston displaces is equal to the volume displaced by the large piston. This causes a difference in the length of displacement, which is proportional to the ratio of areas of the heads of the pistons given that volume = area X length. Therefore, the small piston must be moved a large distance to get the large piston to move significantly. The distance the large piston will move is the distance that the small piston is moved divided by the ratio of the areas of the heads of the pistons. This is howenergy, in the form ofworkin this case, is conserved and theLaw of Conservation of Energyis satisfied. Work is force applied over a distance, and since the force is increased on the larger piston, the distance the force is applied over must be decreased.

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