steps to designing a hydraulic 2 8. piston area to annulus … steps to designing a hydraulic system...

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1 STEPS TO DESIGNING A HYDRAULIC SYSTEM 1. Cylinder area = ( ) 1000 9.81 Mpa working pressure Tons : : Where: Area = mm 2 Mpa = Working pressure 2. Piston diameter = Area 4 : r Where: Diameter = mm Area = mm 2 π = 3.142 3. Pump flow rate: Q A V 10 6 : : = - Where: Q = Litres/min A = Piston area V = Cylinder / piston speed in mm/min 4. Motor kilowatt: / kW Mpa Litres Min 60 : = Where: kW = Kilowatt Mpa = Relief valve pressure Litres/min = Pump flow rate 5. Suction line diameter : . DIA V Q 21 22 : = Where: DIA = mm (D) Q = Pump flow rate V = 1m/sec 6. Pressure line diameter: . DIA V Q 21 22 : = Where: DIA = mm (D) Q = Pump flow rate V = Recommended fluid velocity starting at 4.5 m/sec (check Reynolds number refer 7) 7. Check Reynolds number (Re) : Re cSt V D 1000 : : = Where: V = Fluid velocity in m/sec D = Pipe I.D. in mm cSt = Centistoke, use 46 cSt Reynolds number must be <2500 8. Piston area to annulus area ratio = D d D 2 2 2 - Where: D= Piston diameter d = Rod diameter 9. Return line flow rate: Q Ratio Pump flow rate : = 10. Return line diameter: . DIA V Q 21 22 : = Where: DIA = mm Q = Area ratio x pump flow rate V = Fluid velocity (Starting at 3m/sec) 11. Directional control valve flow rate: / Litres min Cylinder area pump flow rate : = 12. Return line filter size: Cylinder area ratio pump flow rate : = 13. Reservoir (tank) capacity: Volume in litres pump flow rate 4 : = 14. Relief valve = Pump flow rate

Transcript of steps to designing a hydraulic 2 8. piston area to annulus … steps to designing a hydraulic system...

Page 1: steps to designing a hydraulic 2 8. piston area to annulus … steps to designing a hydraulic system 1. cylinder area = () 1000 9.81 Mpaw orking pressure Tons:: Where: Area = mm2 Mpa

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steps to designing a hydraulic system

1. cylinder area = ( )1000 9.81

Mpa working pressureTons : :

Where:Area = mm2

Mpa = Working pressure

2. piston diameter = Area 4:r

Where:Diameter = mmArea = mm2

π = 3.142

3. pump flow rate: Q A V 10 6: := -

Where:Q = Litres/minA = Piston areaV = Cylinder / piston speed in mm/min

4. motor kilowatt: /kW

Mpa Litres Min60:

=

Where:kW = KilowattMpa = Relief valve pressureLitres/min = Pump flow rate

5. suction line diameter : .DIA V

Q 21 22:=

Where:DIA = mm (D)Q = Pump flow rateV = 1m/sec

6. pressure line diameter: .DIA V

Q 21 22:=

Where:DIA = mm (D)Q = Pump flow rateV = Recommended fluid velocity starting at 4.5 m/sec (check Reynolds number refer 7)

7. check reynolds number (re) : Re cStV D 1000: :=

Where:V = Fluid velocity in m/secD = Pipe I.D. in mmcSt = Centistoke, use 46 cStReynolds number must be <2500

8. piston area to annulus area ratio = D d

D2 2

2

-Where:D= Piston diameterd = Rod diameter

9. return line flow rate: Q Ratio Pump flow rate:=

10. return line diameter: .

DIA VQ 21 22:

=

Where:DIA = mmQ = Area ratio x pump flow rateV = Fluid velocity (Starting at 3m/sec)

11. directional control valve flow rate:

/Litres min Cylinder area pump flow rate:=

12. return line filter size:

Cylinder area ratio pump flow rate:=

13. reservoir (tank) capacity:

Volume in litres pump flow rate4 :=

14. relief valve = Pump flow rate

Page 2: steps to designing a hydraulic 2 8. piston area to annulus … steps to designing a hydraulic system 1. cylinder area = () 1000 9.81 Mpaw orking pressure Tons:: Where: Area = mm2 Mpa

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Force is exerted on the piston of a hand pump (Figure 1).This force divided by the piston area results in the attainable pressure (p = F/A). The more the piston is pressed on, i.e. the greater the force on the piston is, the higher the pressure rises. However, the pressure only rises with respect to the cylinder area until it is in a position to overcome the load (F = p • A). If the load remains constant, pressure does not increase any further. Consequently, it acts according to the resistance, which is opposed to the flow of the fluid.

The load can therefore be moved if the necessary pressure can be built up. The speed at which the load moves is dependent on the flow which is fed to the cylinder. With reference to Figure 1, this means that the faster the piston of the hand pump is lowered, the more fluid per time unit is supplied to the cylinder, and the faster the load will lift.

A simple hydraulic system is to be set up as a second example. Further elements will be added to the hydraulic system step by step. These elements will:

• Control the di rection of movement of the cylinder(directional control valve).

• Affect the speed of the cyl inder (flow control valve).• Limit the load of the cylinder (pressure relief valve).• Prevent the system at standstill from being completely

drained via the hydraulic pump (check valve).• Prevent the loaded cylinder from retracting when the hydraulic

pump is at a standstill (check valve).

A hydraulic cylinder (5) is loaded by force F. The hydraulic cylinder is to extend and retract with this load. In contrast to Figure 1, the hydraulic pump (1) is driven by a motor (electric motor or internal-combustion engine). This basic design as shown in Figure 1, is demonstrated as a principle in Figure. 2. The hydraulic pump (1) is driven when motor M is running. It sucks fluid out of the tank (2) and pushes it into the lines of the hydraulic circuit up to the hydraulic cylinder (5). As long as there is no resistance to flow, the fluid is merely pushed further.

The hydraulic cylinder (5) at the end of line, which is loaded with force F, represents a resistance to the flow of the fluid. The

design of a simple hydraulic system5

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1

2

Force

Motor

Figure 1. Principle of a hydraulic system

Figure 2.

pressure increases until it can overcome this resistance, i.e. until the piston of the hydraulic cylinder (5) starts to move. However, when the motor is switched off, the force F pushes the hydraulic cylinder (5) back to its end position (the piston retracts). Here, the hydraulic pump (1) acts as a hydraulic motor.

5

3

1

2

F

M

By installing a check valve (3) at the pressure output of the pump (1), the circuit is prevented from being drained and, as in the example, the hydraulic cylinder (5) is prevented from retracting (see Figure 3). Having extended the set-up of the hydraulic system in this manner, it is now possible to stop the hydraulic cylinder (5) at any desired position by switching the motor off. If, however, the piston should fully extend, so to speak reach its mechanical end position, then the pressure would increase until the hydraulic system was destroyed. Therefore a pressure relief valve (4) is added to the hydraulic system corresponding to Figure 4 below.

Figure 3.

1 M2

3

5

4

FFigure 4.

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In order to protect a hydraulic system from excess pressure and hence from overloading, the maximum pressure allowed must basically be limited by a pressure relief valve.

A spring serving as a mechanical force presses a poppet onto the seat of the valve. The pressure in the line acts on the surface of the seat. In accordance with the equation, F = p • A, the poppet is lifted from its seat when the force from pressure • area exceeds the spring force (see Figure 5).

Pressure now no longer rises. The flow still delivered by the hydraulic pump (1) flows via pressure relief valve (4) directly back to the tank (2). Now our set-up is at the point where we can safely extend the hydraulic cylinder. By installing a directional control valve (6), it becomes possible to retract the hydraulic cylinder again.

The directional control valve (6) has been depicted in switching position ‘b’ in Figure 6. There is no change in function compared to Figure 5 in this switching position Let us switch the directional control valve (6) into its three possible switching positions in our minds:

1. Switching position a: Hydraulic cylinder retracts2. Switching position 0: Hydraulic cylinder remains motionless3. Switching position b: Hydraulic cyl inder extends.

In order to change the speed of movement of the piston in the hydraul ic cylinder (5), the amount of flow to the cylinder must be controlled. This may be achieved using a flow control valve (7} as illustrated in Figure 6.

The cross-sectional area of a pipe may be changed using a flow control valve (7). If the area is decreased, less fluid per time unit reaches cylinder (5). The piston in cylinder (5) hence moves slower. The excess fluid , which is now delivered by pump (1), is drained to tank (2) via pressure relief valve (4).

The following pressures are set in the hydraulic system when the hydraulic cylinder extends (5):

• pressure set at pressure relief valve (4) acts between hydraulic pump (1) and flow control valve (7) and;

• pressure dependent on load F acts between flow control valve (7) and cylinder (5).

However, di rectional control valves are always pictured in the initial position in the symbolic depiction of hydraulic systems.

The finally completed hydraulic system for extending and retracting the hydraulic cylinder (5) loaded by force F is depicted symbolically in Figure 7.

1 M2

3

5

4

a o b

6

FFigure 5.

1 M2

3

5

4

a o b

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F

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1 M2

3

5

4

a o b

6

F

7

Figure 6.

Figure 7.