PH-LecNotes-HH3 (AY2013-V1.0)

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Power Hydrau l i cs Module PH-LecNotes -HH3 (AY2013-V1 .0)

Hydraulic actuators convert hydraulic power into mechanical power. In cylinders, mechanical powerproduces force and linear motion while hydraulic motors produce torque and rotational motion.

1. CYLINDERS (LINEAR ACTUATORS)

They are similar to pneumatic cylinders but can withstand higher pressures. Figure 1 shows a typicalhydraulic cylinder.

Bore Area, A

Rod Area, a Annulus Area, ( A - a )

Figure 1

CHAPTERH3

HYDRAULIC ACTUATORS

C linders & Motors



NGEE ANN POLYTECHNIC Chapter H3 Hydraulic Actuators

P a g e 2

1.1 Cylinder Calculations

The output force and the speed developed by a cylinder piston during its extension and retractionstrokes can be calculated as follows :

Extension force, F E is computed by

f b E F a AP APF )(

Where F E = Extension force [ N ]

P = System pressure [ N/m2 ] ,

Pb = Back pressure [ N/m2 ],

A = Piston bore area [ m2 ], or Area on which pressure P is acting,

a = Piston rod area [ m2 ],

A - a = Annulus area [ m2 ], or Area on which back pressure Pb is acting, and

F f = Friction force within the cylinder [ N ]

In many instances, when the cylinder extends,

the bore-end pressure is the system pressure

the rod-end pressure, which is the back pressure, is zero, and

the frictional force is negligible

Hence,

APF force Extension E , ,

A

Qvspeed Extension E , , and

a Avq flowrateOutput E , .

Pb a

Rod Area A

Bore Area

Q q

P

v E Extension

Speed

During Extension




P a g e 3

Retraction Force, F R is computed by

f b R F APa APF

Where F R = Retraction force [ N ]

P = System pressure [ N/m2 ] ,

Pb = Back pressure [ N/m2 ],

A = Piston bore area [ m2 ], or Area on which pressure P is acting,

a = Piston rod area [ m2 ],

A - a = Annulus area [ m2 ], or Area on which back pressure Pb is acting, and

F f = Friction force within the cylinder [ N ]

Hence,

a APF forcetraction R ,Re ,

)(,Re

a A

Qvspeed traction R

, and

Avq flowrateOutput R , .

1.2 Speed Control of Cylinders

Piston speed varies with the flowrate of fluid entering the cylinder. This may be achieved by

varying the displacement of the pump

adjusting flow control valves (refer to Chapter H 4)

using the "Regenerative" circuit.

During Retraction

Pb a

Rod Area A

Bore Area

q Q

P

v R

Retraction

Speed




P a g e 4

2. REGENERATIVE CIRCUIT

In some applications, when the extension or retraction speed of the cylinder has to be increased, ahydraulic circuit in the regenerative form can be connected.

The “regenerative” circuit for increasing the extension speed, as shown in figure 2, has the return flow ofthe extending cylinder added to the pump flow to increase the rate of flow entering the cylinder. In thisway, the extension speed becomes faster than in “normal” connection; the extension force, however,becomes smaller.

During retraction, the speed and force remain the same as in “normal” connection.

2.1 Principle of Regenerative Circuit

Consider the extension of the cylinder piston as shown in figure 2.

Extension Speed

Rate of input flow entering the cylinder at bore end

= Pump flowrate + output flowrate of cylinder from rod end

a

Qv

Qav

av AvQ Av

av AvQ Av

a AvQ Av

qQQ

E

E

E E E

E E E

E E

2

2

2

2

2

2

)(

v E : Regenerative Extension Velocity,

Q2 : Output (Actual) Flowrate of Pump

Figure 2

The Regenerative

Extension Speed is




P a g e 5

Extension Force

Since pressures are equal at both ends,i.e.

aP

aP AP AP

aP AP AP

a AP APF E

)(

aPF E

Retraction Force & Speed

Comparing the extension speed and force obtained from section 3.1.1 using the “normal” circuitconfiguration,

Speed,)(

2

a A

Qv R

Force, )( a APF R

It shows that regenerative speed is increased but regenerative force is reduced.

Also, note that the regenerative action occurs only during the extension stroke. Hence, the speedand force during the retraction stroke are the same as in normal connection.




P a g e 6

2.2 Comparison of Regenerative Circuit with Normal Circuit

Worked Example 1

A hydraulic cylinder which has bore diameter of 40 mm and rod diameter of 15 mm is driven by a

pump of output delivery 15 /min. The system pressure is set at 100 bar.

(a) If the cylinder and pump is connected in the configuration of Figure 3(a), determine at thecylinder piston,

i) the extension speed in m/min,

ii) the retraction speed, and

iii) the extension force in N and

iv) the retraction force

Solution : (for normal configuration)

(a)

23

242

232

10080.1)(

10767.1)015.0(4

10257.1)040.0(4

ma A

ma

m A

(i)

min93.11

10257.1

10153

33

mm

A

Qv E

(ii)

min89.1310080.1

1015

)( 3

33

mm

a A

Qv R

(iii)

N

m N

APF E

12570

10257.110100325

(iv)

N

m N

a APF R

10800

10080.110100

)(

325

Figure 3(a)




P a g e 7

(b) If the cylinder and pump are connected in the regenerative configuration of Figure 3(b),determine at the cylinder piston,

(i) the extension speed in m/min,

(ii) the retraction speed, and(iii) the extension force in N and

(iv) the retraction force

Solution : (for regenerative configuration)

(i) The regenerative extension speed

min89.8410767.1

10154

33

mm

a

Qv E

which is faster than the normal extension speed.

(ii) The retraction is at normal retraction speed

min89.13 mv R

(iii) The regenerative extension output force

N

m N

APF E

1767

10767.110100425

(iv) The retraction formal is same as that of normal configuration

N F R 10800

Figure 3(b)




P a g e 8

3. HYDRAULIC MOTORS

Hydraulic motors convert hydraulic power into torque and rotation, i.e. mechanical power. Hydraulicmotors resemble hydraulic pumps in construction and many pumps can in fact be used as motors.

3.1 Operating Principle

A pump pushes fluid into the hydraulic system when a torque is applied to its input drive shaft,whereas a hydraulic motor is pushed by fluid, thereby developing torque and continuous rotarymotion at its output drive shaft. Thus, input power of the motor is in the form of hydraulic powerand the output power is in the form of mechanical power.

3.2 Motor Displacement

Motor displacement is the theoretical amount of fluid required to push the motor through onerevolution. Hydraulic motors are available as either fixed or variable-displacement type.

mm nQ

where m = motor displacement (cm3/rev)

Q = flowrate (cm3/min)

nm = pump rotational speed (rev/min)

3.3 Basic Motor Calculations

The symbol of a simple non-reversible hydraulic motor is shown in Figure 4. Points (3) and (4)are the pump's inlet and outlet respectively. Since the motor converts Hydraulic Energy intoMechanical Work, the input power to the motor will be hydraulic and the output powermechanical.

Fixed Displacement

Motor

Variable Displacement

Motor

Figure 4

(3)

(4)

n p, T p n p, T p,th

(d)(c)




P a g e 9

m : the motor displacement [cm3/rev]

nm : the motor speed or the rotational speed of the load [rpm]

Q3 : the flowrate entering the motor at point (3) [litre/min ]

Q4 : the flowrate leaving the motor at point (4) [litre/min ]

T m : the actual output torque produced at the motor shaft [Nm]T m,th : the theoretical output torque develop at motor shaft [Nm]

Pm : the pressure drop across the motor [bar]

Pm= P3 - P4

Flowrate

The actual flowrate driving the motor will be Q4 ,

Here Q4 = motor displacement x motor speed

mm nQ 4

Because of leakage and slippage problem at the motor, input flowrate Q3 output flowrate Q4.

Hence, not all fluid flowing into the motor will drive it.

Hence, volumetric efficiency is defined as

3

4

Q

Qvm

Loss of flow as a result of leakages or leakage flow through case drains of motor is computed as,

43 QQQ L

Power Generated by the Motor

(3)

(4)

Q3

Q4




P a g e 1 0

Because of friction, slippage and windage losses across the couplings, the actual torque T m

generated by the hydraulic motor at (d) is therefore smaller than the ideal torque T m,th available at

(c) at the motor shaft,

ie T m < T m,th

Now, the actual Mechanical Output Power at (d) is

mm T n 2

and Mechanical Power available at (c) is

thmm T n ,2

Therefore, mechanical efficiency of a hydraulic motor mm , which is defined as ratio of the Actual

Output Mechanical Power at (d) to the Mechanical Power At (c) is given as,

thm

mmm

thmm

mm

mm

T

T

T n

T n

cat Power Mechanical

d at Power MechanicalOutput

,

,2

2

)(

)(

The Input Hydraulic Power At (3) with the flowrate Q3 3QPm

and the Hydraulic Power Available At (4)

4QPm

Logically, the Hydraulic Power available at outlet at (4) should be converted to the MechanicalPower at the ideal torque at (c)

(c) (d)

T mnm

T m,th

nm

(3)

(4)




P a g e 1 1

mm

mmm

mmm

mm

m

mm

mm

P

T

nP

T n

QP

T n

at Power Hydraulic

d at Power Mechanical


d at Power Mechanical

2

2

2

)4(

)(

)(

)(

4

Now, overall efficiency of a motor mo , is defined as the ratio of Output Mechanical Power at (d) to

the Input Hydraulic Power at (3).

Therefore,

Alternatively.

)(

)(

)3(

)4(

)3(

)(



at Power Hydraulic Input

at Power Hydraulic

at Power Hydraulic Input


om

om

mmvmom

Important considerations when using hydraulic motors:There is a tendency for a load to 'run-away' when lowering it with a hoist, or when stopping ahigh inertia load by cutting off the input flowrate. When such situations arise, the motor is runningtemporarily as a pump and cavitations will damage the motor just as it will damage a hydraulicpump.

Volumetric Efficiency Mechanical Efficiency

3

2

)3(

)(

QP

T n

at Power Input Hydraulic

d at Power Output Mechanical

m

mmom

om




P a g e 1 2

Worked Example 2

An fixed-displacement hydraulic motor of displacement 110 cm3/rev operates at a speed of 150 rpm

through a pressure drop of 100 bar. Given that the volumetric and mechanical efficiencies are 95%

and 90% respectively. Determine

(a) the inlet flowrate Q3 supplied to the motor in /min,

(b) the leakage flow (case drain) in /min,

(c) the output mechanical power available at the motor shaft in kW

(d) the output torque T m in Nm, and

Solution :

(a)

min5.16

min150101103

4

revrev

nvQmm

min368.1795.0

min5.16

43

mv

QQ

(b) min868.043 QQQ L

(c)

855.09.095.0 efficiencyOverall

kW QPPower MechnicalOutput

Power Hydraulic Input Power MechnicalOutput

Power Hydraulic Input

Power MechnicalOutput

mom

mo

mo

475.2855.060

10368.1710100

35

3

(d)

m N

n

Power MechanicalOutput T

T nPower MechanicalOutput

m

m

mm

56.157601502

2475

2

2

(3)

(4)

n p, T p

Q3

Q4

Q L




P a g e 1 3

4. TYPES OF MOTORS

Positive-displacement motors are available in a variety of designs, e.g. gear motor, vane motor and in-line piston motor. These motors can be of fixed displacement as well as variable displacement. Sincethey are the exact replicas of pumps, only a brief description of the piston pump shall be given below.

Gear Motors

The gear motor consists of a pair of matched gears enclosed in one housing. Both gears have thesame tooth form and are driven by fluid under pressure. Fluid pressure enters the housing on oneside at a point where the gear mesh and forces the gear to rotate. The fluid exits at low pressure,at the opposite side of the motor.

The torque developed is a function of the hydraulic imbalance of only one tooth of gear at a time;the other of the teeth are hydraulic balanced.

Figure 5




P a g e 1 4

Vane Motor

In vane motors, torque is developed by pressure on exposed surfaces of the rectangular vanes,which slide in and out the slots in a rotor splined to the driveshaft. As the rotor turns, the vanefollow the surface of a cam ring, forming sealed chambers which carry the fluid from the inlet to theoutlet.

Figure 6




P a g e 1 5

Swash Plate Axial Piston Motors

Piston motors generate torque through pressure on the ends of reciprocating pistons operating in acylinder block. The swash plate axial-piston motor resembles the pump of the same name. Its driveshaft and cylinder block are centred on the same axis (Figure 7).

In-line piston motors are built in both fixed and variable-displacement models. The swash-plate angledetermines the displacement. In the variable-displacement model, the swash plate is mounted in aswinging yoke and the angle can be changed by various means, e.g. lever, handwheel, pressurecompensator, or servo control.

1. Yoke return spring

initially moves yoke to a

minimum displacement

position for maximum

speed and minimum torque

2. Adjustment spring sets

initial compensating

pressure

3. Compensator spool is

forced open against spring

by system pressure and

ports oil to yoke piston at

its pressure setting

4. Yoke actuating piston

responds to pressure to

increase displacement and

reduce speed and increase

torque

ADJUSTMENT

SCREW

OUT

IN

(Sys Pressure)

YOKE (non-rotating)

1. Oil under pressure

at inlet2. exerts a force on pistons,

forcing them out of the

cylinder block

3. The piston thrust is

transmitted to the angled

swash plate causing rotation

4. The pistons, shoes plate, and

cylinder block rotate together.

The drive shaft is splined to

the cylinder block

5. As the piston passes the

inlets, it begins to return

into its bore because of its

swash plate angle. Exhaust

fluid is pushed int the outlet port.

Figure 7




P a g e 1 6

3.6 Case Drain

Leakage which occurs continuously within the vane and piston pumps must be drained externally (i.e., anadditional pipe line added) from the housing chamber to the tank so that pressure does not build up inthe motor casing and damage the shaft seal. Gear motors do not require external case drain as the leakoccurs from the high-pressure side to the return line only.

Figure 7




P a g e 1 7

Reference :

SELECTION OF CYLINDER AND MOTOR

A. Selection of CylindersCylinders are selected based on :

1. Applied load

2. Stroke length

3. Speed

4. Buckling load

5. Pressure

6. Types of mounting

B. Selection of MotorsThe characteristics of various types of motors are given in Table 1.

Table 1 : Motor characteristics

Motor

Type

Output

Torque and Speed

Max.

operating

pressure

Other characteristics

Gear Low torque (15-200 Nm) Smooth speed to reach 400 rpm Not reversible

Medium/high(up to 200 bar)

Simple design Dirt tolerant Lower efficiency Some can reach 6000

rev/min but noisy

Vane

Medium/high torque (to 16 kNm) Medium/high speed (500 -

4000rpm) Not reversible

Medium

(up to 175 bar)

Cannot run below 500

rev/min as vanes needcentrifugal force

Piston

High torque (300 Nm to 190 kNm) Medium/high speed (to 8000 rpm) Reversible

Very high(up to 450 bar)

High efficiency, highpower-to-weight ratio

Most costly good oil filter

High service andmaintenance standardrequired

*** Always refers to the manufacturers' catalogues

C. Considerations in Motor Selection

1. Torque requirement

2. Reversible or not

3. Speed requirement

4. Displacement

5. Pressure requirement

6. Efficiencies

maximum torque and shaft speed must match the load.

displacement is determined by the speed required and the pump's flow rate.

pressure rating must exceed the maximum operating pressure.

operating speed should be near the maximum efficiency condition.

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PH-LecNotes-HH3 (AY2013-V1.0)