Formulas Rexroth

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Formulary Hydraulics

Hydraulic Formulary

Author: Houman Hatami Tel.: +49-9352-18-1225 Fax: +49-9352-18-1293 [email protected]

11.04.2012

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CONTENTS RELATIONS BETWEEN UNITS ............................................................................................................. 4 IMPORTANT CHARACTERISTIC VALUES OF HYDRAULIC FLUIDS ................................................ 6 GENERAL HYDRAULIC RELATIONS ................................................................................................... 7 PISTON PRESSURE FORCE ...................................................................................................................... 7 PISTON FORCES ..................................................................................................................................... 7 HYDRAULIC PRESS ................................................................................................................................. 7 CONTINUITY EQUATION ........................................................................................................................... 8 PISTION SPEED....................................................................................................................................... 8 PRESSURE INTENSIFIER .......................................................................................................................... 8 HYDRAULIC SYSTEM COMPONENT ................................................................................................... 9 HYDRO PUMP ......................................................................................................................................... 9 HYDRO MOTOR ....................................................................................................................................... 9 Hydro motor variable ...................................................................................................................... 10 Hydro motor fixed ........................................................................................................................... 11 Hydro motor intrinsic frequency ..................................................................................................... 12 HYDRO PISTON ..................................................................................................................................... 13 Differential piston ............................................................................................................................ 14 Double acting cylinder .................................................................................................................... 15 Cylinder in differential control ......................................................................................................... 16 Cylinder intrinsic frequency at differential cylinder ......................................................................... 17 Cylinder intrinsic frequency at double acting cylinder .................................................................... 18 Cylinder intrinsic frequency at plunger cylinder.............................................................................. 19 PIPING................................................................................................................................................... 20 APPLICATION EXAMPLES FOR SPECIFICATION OF THE CYLINDER PRESSURES AND VOLUME FLOWS UNDER POSITIVE AND NEGATIVE LOADS ........................................................ 21 DIFFERENTIAL CYLINDER EXTENDING WITH POSITIVE LOAD ..................................................................... 22 DIFFERENTIAL CYLINDER RETRACTING WITH POSITIVE LOAD ................................................................... 23 DIFFERENTIAL CYLINDER EXTENDING WITH NETAGIVE LOAD.................................................................... 24 DIFFERENTIAL CYLINDER RETRACTING WITH NEGATIVE LOAD .................................................................. 25 DIFFERENTIAL CYLINDER EXTENDING AT AN INCLINED PLANE WITH POSITIVE LOAD .................................. 26 DIFFERENTIAL CYLINDER RETRACTING AT AN INCLINED PLANE WITH POSITIVE LOAD................................. 27 DIFFERENTIAL CYLINDER EXTENDING AT AN INCLINED PLANE WITH NEGATIVE LOAD ................................ 28 DIFFERENTIAL CYLINDER RETRACTING AT AN LINCLINED PBLANE WITH NEGATIVE LOAD ............................ 29 HYDRAULIC MOTOR WITH POSITIVE LOAD ............................................................................................... 30 HYDRAULIC MOTOR WITH NEGATIVE LOAD.............................................................................................. 31 IDENTIFICATION OF THE REDUCED MASSES OF DIFFERENT SYSTEMS ................................... 32 LINEARE DRIVES .................................................................................................................................. 33 Primary applications (Energy method) ........................................................................................... 33 Concentrated mass at linear movements....................................................................................... 35 Distributed mass at linear movements ........................................................................................... 36 ROTATION ............................................................................................................................................ 37 COMBINATION OF LINEAR AND ROTATIONAL MOVEMENT .......................................................................... 38 HYDRAULIC RESISTANCES………………………………………………………………………………...39 ORIFICE EQUATION ............................................................................................................................... 39 TROTTLE EQUATION .............................................................................................................................. 39 11.04.2012

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HYDRO ACCUMULATOR .................................................................................................................... 40 HEAT EXCHANGER (OIL- WATER) .................................................................................................... 41 LAYOUT OF A VALVE ......................................................................................................................... 43

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Relation between Units Size Lengths

Surfaces

Volume

Density

Unit

Symbol

Relation

Micrometer

µm

1µm = 0,001mm

Millimeter

mm

1mm = 0,1cm = 0,01dm = 0,001m

Centimeter

cm

1cm = 10mm = 10.000µm

Decimeter

dm

1dm = 10cm = 100mm = 100.000µm

Meter

m

1m = 10dm = 100cm = 1.000mm = 1.000.000µm

Kilometer

km

1km = 1.000m = 100.000cm = 1.000.000mm

Square centimeter

cm2

1cm2 = 100mm2

Square decimeter

dm2

1dm2 = 100cm2 = 10.000mm2

Square meter

m2

1m2 = 100dm2 = 10.000cm2 = 1.000.000mm2

Are

a

1a = 100m2

Hectare

ha

1ha = 100a = 10.000m2

Square kilometer

km2

1km2 = 100ha = 10.000a = 1.000.000m2

Cubic centimeter

cm3

1cm3 = 1.000mm3 = 1ml = 0,001l

Cubic decimeter

dm3

1dm3 = 1.000cm3 = 1.000.000mm3

Cubic meter

m3

1m3 = 1.000dm3 = 1.000.000cm3

Milliliter

ml

1ml = 0,001l = 1cm3

Liter

l

1l = 1.000 ml = 1dm3

Hectoliter

hl

1hl = 100l = 100dm3

Gram/

g cm3

1

g kg t g =1 3 =1 3 =1 cm3 dm m ml

Cubic centimeter

Force

Newton

N

1N = 1

Weight

kg • m J =1 s2 m

1daN = 10N

Torque

Newton meter

Nm

1Nm = 1J

Pressure

Pascal

Pa

Bar

Bar

1Pa = 1N/m2 = 0,01mbar = 1kg m • s2

psi =

pound inch 2

1bar = 10 Psi

1psi = 0,06895 bar

kp cm 2

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N N = 100.000 2 = 10 5 Pa 2 cm m

1

4

kp = 0,981bar cm 2



Mass

Acceleration

Milligram

mg

1mg = 0,001g

Gram

g

1g = 1.000mg

Kilogram

kg

1kg = 1000g = 1.000.000 mg

Ton

t

1t = 1000kg = 1.000.000g

Mega gram

Mg

1Mg = 1t

Meter/

m s2

1

per square second

Angular speed

One/ Second Radiant/ Second

Power

m N =1 s2 kg

1g = 9,81 m/s2 ω = 2•π•n

1 s

n in 1/s

rad s

Nm J kg • m m =1 =1 2 • s s s s

Watt

W

Newton meter/ second

Nm/s

Joule/ second

J/s

Work/ Energy

Watt second

Ws

Heat volume

Newton meter

Nm

Joule

J

Kilowatt hour

kWh

1kWh = 1.000 Wh = 1000•3600Ws = 3,6•106Ws

Kilo joule

kJ

= 3,6•103kJ = 3600kJ = 3,6MJ

Mega joule

MJ

Mechanic

Newton/ square

N mm2

1

tension

millimeter

Plane angle

Second

´´

1´´ = 1´/60

Minute

´

1´ = 60´´

Degree

°

Radiant

rad

1° = 60´ = 3600 ´´= π rad 180°

1W = 1

1Ws = 1Nm = 1

kg • m • m = 1J s2

N = 10bar = 1MPa mm2

1rad = 1m/m = 57,2957° 1rad = 180°/π

Speed

One/second

1/s

One/minute

1/min

1 −1 = s = 60 min −1 s

1 1 = min −1 = 60s min

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Important Characteristic Values of Hydraulic Fluids

HLP

HFC

HFA (3%)

HFD

0,00087

0,00105-0,00108

0,0001

0,000115

10-100

36-50

0,7

15-70

12000-14000

20400-23800

1500017500

1800021000

[Bar] Specific Heat at 20°C

2,1

3,3

4,2

1,3-1,5

[kJ/kgK] Thermal Conductivity at 20°C

0,14

0,4

0,6

0,11

[W/mK] Optimal Temperatures

40-50

35-50

35-50

35-50

[°C] Water Content

0

40-50

80-97

0

[%] Cavitation Tendency

low

high

very high

low

Density at 20°C 3

[kg/cm ] Kinematic Viscosity at 40°C 2

[mm /s] Compressions Module E at 50°C

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General Hydraulic Relations Piston Pressure Force Figure

Equation / Equation Variations

Symbols / Units

F = 10 • p • A

F = p • A • η • 10 A= d=

F = piston pressure force[N]

d2 • π 4

p = fluid pressure[bar] A = piston surface[cm2] d = piston diameter[cm] η = efficiency cylinder

4 • F • 0,1 π•p

p = 0,1 •

4• F π •d2

Piston Forces Figure


Symbols / Units

F = pe • A • 10 F = pe • A • η • 10 A=

F = piston pressure force[N] pe = excess pressure on the piston[bar]

d2 • π 4

A = effective piston surface[cm2] d = piston diameter[cm] η = efficiency cylinder

A For annulus surface:

A=

(D2 − d 2 ) • π 4

Hydraulic Press Figure


Symbols / Units F1 = Force at the pump piston[N]

F1 F = 2 A1 A 2

F2 = Force at the operating piston[N] A1 = Surface of the pump piston [cm2]

F1 • s1 = F2 • s2

A2 = Surface of the operating piston [cm2] s1 = Stroke of the pump piston [cm] s2 = Stroke of the operating piston [cm]

ϕ=

11.04.2012

F1 A1 s 2 = = F2 A2 s1

ϕ = Gear ratio

7



Continuity Equation Figure


Q1 = Q 2

Symbols / Units

Q1,2 = Volume flows [cm3/s, dm3/s, m3/s]

Q1 = A 1 • v 1

A1,2 = Area surfaces [cm2, dm2, m2] v1,2 = Velocities

Q2 = A 2 • v2

[cm/s, dm/s, m/s]

A 1 • v1 = A 2 • v 2

Piston Speed Figure


v1 =

Q1 A1

v1,2 = Piston speed [cm/s] Q1,2 = Volume flow [cm3/s]

Q v2 = 2 A2

A1 =

Symbols / Units

A1 = Effective piston surface (circle) [cm2] A2 = Effective piston surface (ring) [cm2]

d2 •π 4

(D2 − d 2 ) • π A2 = 4 Pressure Intensifier Figure


Symbols / Units

p1 = Pressure in the small cylinder [bar]

p1 • A 1 = p 2 • A 2

A1 = Piston surface [cm2] p2 = Pressure at the large cylinder [bar] A2 = Piston surface [cm2]

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Hydraulic System Components Hydro Pump Q=

V • n • η vol 1000

p•Q Pan = 600 • η ges

Q = Volume flow [l/min] [l/min]

V = Nominal volume [cm3] n = Drive speed of the pump [min-1]

[kW]

Pan = Drive power [kW] p = Service pressure [bar]

1,59 • V • ∆p [Nm] M= 100 • η mh

M = Drive torque [Nm] ηges = Total efficiency (0,8-0,85)

η ges = η vol • η mh

ηvol = Volumetric efficiency (0,9-0,95) ηmh = Hydro-mechanic efficiency(0,9-0,95)

Hydro Motor

Q = Volume flow [l/min] V = Nominal volume [cm3]

Q=

V• n 1000 • η vol

n = Drive speed of the pump [min-1] ηges = Total efficiency (0,8-0,85) ηvol = Volumetric efficiency (0,9-0,95)

Q • η vol • 1000 n= V M ab

ηmh = Hydro-mechanic efficiency

(0,9-0,95)

∆p • V •η mh = = 1,59 • V • ∆p •η mh • 10 −2 20 • π

Pab =

∆p = Pressure difference between motor inlet and

outlet (bar)

∆p • Q • η ges

11.04.2012

Pab = Output power of the motor [kW] Mab = Output torque [Nm]

600

9



Hydro Motor Variable

Md = P=

n=

Md = Torque [Nm]

π

n = Speed [min-1]

• Md • n

30000

30000

π

Md = n=

30000 P • n π

P • Md

M d max i • η Getr

i = Gear ratio ηGetr = Gear efficiency ηmh = Mech./hydraulic efficiency ηvol = Vol. efficiency

Vg = Flow volume [cm3]

Md Vg • η mh

Vg • n 1000 • η vol

QP = P=

Mdmax = Max torque [Nm]

n max i

∆p = 20π • Q=

P = Power [kW]

Vg • n • η vol 1000

Q • ∆p 600 • η ges

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Hydro Motor Fixed

Md = P=

n=

Md = Torque [Nm]

π

n = Speed [min-1]

• Md • n

30000

30000

π

Md = n=

30000 P • n π

P • Md

M d max i • η Getr

i = Gear ratio ηGetr = Gear efficiency ηmh = Mech./hydraulic efficiency ηvol = Vol. efficiency

Vg = Flow volume [cm3]

Md Vg • η mh

Vg • n 1000 • η vol

QP = P=

Mdmax = Max torque [Nm]

n max i

∆p = 20π • Q=

P = Power [kW]

Vg • n • η vol 1000

Q • ∆p 600 • η ges

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Hydro Motor Intrinsic Frequency

V ( G )2 2• E 2π ω0 = • VG J red + VR ) ( 2

ω f0 = 0 2π

11.04.2012

VG = Displacement [cm3] ω0 = Intrinsic angular frequency [1/s]

f0 = Intrinsic frequency [Hz] Jred = Moment of inertia red. [kgm2] Eöl = 1400 N/mm2 VR = Volume of the line [cm3]

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Hydro Cylinder d • π d 1 • 0,785 2 [cm ] A= 1 = 400 100 2

d1 = Piston diameter [mm]

2

d2 = Piston rod diameter [mm] p = Service pressure [bar]

d 2 • 0,785 2 [cm ] 100 2

A st =

v = Stroke speed [m/s] V = Stroke volume [l]

(d − d 2 ) • 0,785 2 [cm ] AR = 1 100 2

2

Q = Volume flow, considering the leakages (l/min)

p • d1 • 0,785 [kN] FD = 10000

Qth = Volume flow, without considering the

p • (d 1 − d 2 ) • 0,785 [kN] Fz = 10000

ηvol = Volumetric efficiency (approx. 0,95)

h Q [m/s] = v= t • 1000 A • 6

t = Stroke time [s]

2

2

Qth = 6 • A • v =

Q=

V= t=

leakages (l/min)

2

V • 60 t

h = Stroke [mm]

FD

[l/min] FZ

Q th

η vol.

FS

A•h [l] 10000

A• h•6 Q • 1000

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[s]

13



Differential Cylinder

dK = Piston diameter [mm]

4 • FD π • pK

d K = 100 •

dst = Rod diameter [mm] FD = Pressure force [kN]

pK =

4 • 10 • FD π • d K2

p St =

4 • 104 • FZ π • (d K 2 − d St 2 )

ϕ=

4

Fz = Traction force [kN] pK = Pressure at the piston side [bar] ϕ = Aspect ratio

QK = Volume flow piston side [l/min] QSt = Volume flow rod side [l/min] va = Extending speed [m/s]

2

dK 2 2 (d K − d St )

ve = Retracting speed [m/s] Volp = Working volume [l]

6•π 2 • va • d K QK = 400

QSt = ve =

va =

VolF = Fill-up volume [l] h = Stroke [mm]

6•π 2 2 • v e • (d K − d St ) 400 QSt

6π 2 2 • (d K − d St ) 400

QK

6π 2 • dK 400

Vol p =

Vol F =

π 4 • 10

6

• d St • h

6

• h • (d K − d St )

2

π 4 • 10

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2

2

14



Double Acting Cylinder

pA = pB = QA =

4 • 104

•

π 4 • 104


FA 2 2 (d K − d StA )

dstA = Rod diameter A-side [mm] dstB = Rod diameter B-side [mm]

FB • 2 2 (d K − d StB )

π

FA = Force A [kN] FB = Force B [kN]

6•π 2 2 • v a • (d K − d StA ) 400

pA = Pressure at the A-side [bar] pB = Pressure at the B-side [bar] QA = Volume flow A-side [l/min]

6•π 2 2 QB = • v b • (d K − d StB ) 400

ve =

va =

QB = Volume flow B-side [l/min] va = Speed a [m/s]

QSt

vb = Speed b [m/s]

6π 2 2 • (d K − d St ) 400

Volp = Compensating volume [l] VolFA = Fill-up volume A [l]

QK

VolFB = Fill-up volume B [l]

6π 2 • dK 400

Vol p =

π 4 • 10

Vol FA = Vol FB =

• d St • h 2

6

π 6

• h • (d K − d StA )

6

• h • (d K − d StB )

2

4 • 10

π 4 • 10

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2

2

2

15



Cylinder in Differential Control


4 • FD d st = 100 • π • p St

dst = Rod diameter [mm] FD = Pressure force [kN]

4 • 10 • FD π • d St 2 4

pK =

Fz = Traction force [kN] pK = Pressure at the piston side [bar] pSt = Pressure at the rod side [bar]

4 • 104 • FZ p St = π • (d K 2 − d St 2 ) Q=

h = Stroke [mm] QK = Volume flow piston side [l/min]

6•π 2 • v a • d St 400

QSt = Volume flow rod side [l/min]

Extension:

va =

QP = Pump flow [l/min] va = Extending speed [m/s]

QP

6π 2 • d St 400

Q •d QK = P 2 K d St

ve = Retracting speed [m/s] Volp = Working volume [l] VolF = Fill-up volume [l]

2

Q P • (d K − d St ) 2 d St 2

QSt =

2

Retraction:

ve =

QP

6π 2 2 • (d K − d St ) 400

QSt=QP

QP • d K QK = 2 2 (d K − d St ) 2

Vol p = Vol F =

π 4 • 10

6

• d St • h

6

• h • (d K − d St )

2

π 4 • 10

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2

2

16



Cylinder Intrinsic Frequency at Differential Cylinders

AK = Piston surface [cm2]

d π AK = K 4 100 2

AR = Piston ring surface [cm2] dK = Piston diameter [mm] dSt = Piston rod diameter [mm]

(d − d St )π AR = K 4 100 2

2

dRK = NW- piston side [mm] LK = Length of piston side [mm] dRSt = NW-rod side [mm]

d π L = RK • K 4 1000 2

VRK

LSt = Length of rod side [mm] h = Stroke [cm]

d π L = RSt • St 4 1000

VRK = Volume of the line piston side [cm3]

2

VRSt

mRK =

VRSt = Volume of the line rod side [cm3] mRK = Mass of the oil in the line piston side [kg]

VRK • ρ Öl 1000

mRSt = Mass of the oil in the line rod side [kg] hK = Position at min intrinsic frequency [cm]

V • ρ öl = RSt 1000

mRSt

f0 = Intrinsic frequency [Hz]

 A •h V V  R + RSt + RK 3 3 3  AR AR AK  hk = 1 1 ( ) + AR AK

ω 01 = ω 0 • f 01 =

A • E ÖL AR • E Öl 1 ) •( K + AR • h − h K m AK • h K + V RK + V RSt 10 10 2

ω0 =

f0 =

ω 0 = Angular frequency

   

2

ω0 2π 4

mölred

 1 d  = mRK  K  + mRSt   d RK   d RSt

11.04.2012

400 • A R   π 

17

ω 01 2π

mred mölred + mred



Cylinder Intrinsic Frequency at Double Acting Cylinders

AR = Piston ring surface [cm2]

(d − d St )π AR = K 4 100 2

2

dK = Piston diameter [mm] dSt = Piston rod diameter [mm] dR = NW [mm]

d π L VR = RK • K 4 1000

LK = Length of the piston side [mm]

V • ρ öl mR = R 1000

VR = Volume of the line [cm3]

2

h = Stroke [mm]

mR = Mass of the oil in the line [kg] f0 = Intrinsic frequency

AR 2 • E öl ω 0 = 100 • •( ) AR • h m red + VRSt 10 2

f0 =

ω0 2π

mölred

 1 = 2 • mRK   dR

ω 01 = ω 0 • f 01 =


400 • A R   π 

4

mred mölred + mred

ω 01 2π

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Cylinder Intrinsic Frequency for Plunger Cylinders

AK = Piston surface [cm2]

d π AK = K 4 100 2

dK = Piston diameter [mm] dR = Diameter of the piping [mm] LK = Length piston side [mm]

d π L VR = K • K 4 1000 2

LR = Length of the line [mm] h = Stroke [mm]

V • ρ öl mR = R 1000

VR = Volume of the line [cm3] MR = Mass of the oil in the line [kg] 2

E AK ω 0 = 100 • öl • ( ) mred AK • h + VRSt f0 =


ω0 2π

m ölred

d  = 2 • mR  K   dR 

ω 01 = ω 0 • f 01 =

f0 = Intrinsic frequency

4

mred mölred + mred

ω 01 2π

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Piping

∆p = λ •

l • ρ • v 2 • 10 d•2

λ lam. λturb.

0,316 = 4 Re v•d

υ

λturb. = Pipe friction coefficient for turbulent flow

l = Length of the line [m] v = Velocity in the line [m/s]

• 103

6• d2 •

d =

3

λlam. = Pipe friction coefficient for laminar flow

Q

v=

ρ = Density [kg/dm ] (0,89) λ = Pipe friction coefficient

64 = Re

Re =

∆p = Pressure loss at direct piping [bar]

π

• 102

d = Internal diameter of the piping [mm] ν = Kinematic viscosity [mm /s] 2

Q = Volume flow in the piping [l/min]

4

400 Q • 6•π v

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20



Application Examples for Specification of the Cylinder Pressures and Volume Flows under Positive and Negative Load Nomenclature

Parameters

Symbols

Units

Acceleration / deceleration

A

m/s2

Cylinder surface

A1

cm2

Ring surface

A2

cm2

Aspect ratio

ϕ=A1/A2

-

FT

daN

Fa=0,1•m•a

daN

External forces

FE

daN

Friction forces (coulomb friction)

FC

daN

Sealing friction force

FR

daN

Weight force

G

daN

Total force Acceleration force

Mass

G + mK g

kg

Piston mass

mK

kg

Volume flow

Q=0,06• A•vmax

l/min

vmax

cm/s

T=α•J+ TL

Nm

m=

Torque Load torque

TL

Nm

Angular acceleration

α

rad/s2

Inertia moment

J

kgm2

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21



Differential Cylinder Extending with Positive Load

Calculation: Layout: FT = Fa+FR+FC+FE

p1 =

[daN]

Given Parameters

p 2 = 5,25 +

FT = 4450 daN PS = 210 bar PT = 5,25 bar A1 = 53,50 cm2 A2 = 38,10 cm2 ϕ = 1,40 vmax = 30,00 cm/s ==> p1 und p2

Q N = 96

p S A2 + R [ FT + ( pT A2 )] bar A2 (1 + ϕ 3 ) p −p p2 = pT + S 2 1 bar

ϕ

Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1. Q= 0,06•A1•vmax

QN = Q

l/min

35 p S − p1

l/min

Selection of a Servo valve 10% larger than the calculated nominal volume flow.

11.04.2012

210 − 120 = 52bar 1,4 2

Q= 0,06•53,5•30=96 l/min

2

p1 =

210 • 38,1 + 1,4 2 [4450 + (5,25 • 38,1)] = 120bar 38,1(1 + 1,4 3 )

22

35 = 60l / min 210 − 120



Differential Cylinder Retracting with Positive Load

Layout:

Calculation:

FT = Fa+FR+FC+FE

[daN]

p2 =

Given Parameters

p 1 = 5,25 + [(210 − 187)1,4 2 ] = 52bar

FT = 4450 daN PS = 210 bar PT = 5,25 bar A1 = 53,50 cm2 A2 = 38,10 cm2 ϕ = 1,40 vmax = 30,00 cm/s ==> p1 und p2

Q= 0,06•38,1•30=69 l/min

( p A ϕ 3 ) + FT + ( pT A2ϕ )] bar p2 = S 2 A2 (1 + ϕ 3 )

Q N = 96

p1 = pT + [( p S − p2 )ϕ 2 ] bar Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1.

Q= 0,06•A2•vmax

l/min

35 pS − p 2

l/min

QN = Q


11.04.2012

(210 • 38,1 • 1,42 ) + 4450 + (5,25 • 38,1 • 1,4)] = 187bar 38,1(1 + 1,43 )

23

35 = 84l / min 210 − 187



Differential Cylinder Extending with Negative Load

Calculation: Layout:

FT = Fa+FR-G

p1 =

[daN]

Given Parameters

p2 = 0 +

FT = -2225 daN PS = 175 bar PT = 0 bar 2 A1 = 81,3 cm 2 A2 = 61,3 cm ϕ = 1,3 vmax = 12,7 cm/s ==> p1 und p2

p S A2 + ϕ 2 [ FT + ( pT A2 )] bar A2 (1 + ϕ 3 ) p −p p2 = pT + S 2 1 bar

Q N = 62

ϕ

Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1.

Q= 0,06•A1•vmax

l/min

35 p S − p1

l/min


11.04.2012

175 − 36 = 82bar 1,32

Q= 0,06•81,3•12,7=62 l/min

p1 =

QN = Q

175 • 61,3 + 1,32 [−2225 + (0 • 61,3)] = 36bar 61,3(1 + 1,33 )

24

35 = 31l / min 175 − 36



Differential Cylinder Retracting with Negative Load


FT = Fa+FR-G

p2 =

[daN]

(210 • 61,3 + 1,32 ) − 4450 + (0 • 61,3 • 1,3)] = 122bar 61,3(1 + 1,33 )

Given Parameters

p 1 = 0 + [(210 − 122)] = 149 bar


Q= 0,06•61,3•25,4=93 l/min

p2 =

Q N = 93

( p S A2ϕ ) + FT + ( pT A2ϕ )] bar A2 (1 + ϕ 3 ) 3

p1 = pT + [( p S − p2 )ϕ 2 ] bar Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1.

Q= 0,06•A2•vmax

l/min

35 pS − p 2

l/min

QN = Q


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25

35 = 59l / min 210 − 122



Differential Cylinder Retracting at an Inclined Plane with Positive Load


(140 • 19,9) + 1,62 [2225 + (3,5 • 19,9)] = 85bar 19,9(1 + 1,63 )

FT = Fa+FE+FS+[G•(µ•cosα+sinα)] daN

p1 =

Given Parameters

p 2 = 35 +

FT = 2225 daN PS = 140 bar PT = 3,5 bar 2 A1 = 31,6 cm 2 A2 = 19,9 cm R = 1,6 vmax = 12,7 cm/s ==> p1 and p2

p1 =

Q= 0,06•31,6•12,7=24 l/min

Q N = 24

p S A2 + ϕ 2 [ F + ( pT A2 )]

p2 = pT +

A2 (1 + ϕ 3 ) p S − p1

ϕ2

bar

bar


Q= 0,06•A1•vmax

l/min

35 p S − p1

l/min

QN = Q


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140 − 85 = 25bar 1,6 2

26

35 = 19 l/min 140 − 85



Differential Cylinder Retracting at an Inclined Plane with Positive Load


(140 • 19,9 • 1,63 ) + 1780 + [3,5 • 19,9 • 1,6)] = 131bar 19,9(1 + 1,63 )

FT =Fa+FE+FS+[G•(µ•cosα+sinα)] daN

p2 =

Given Parameters

p 1 = 3,5 + [(140 − 131) • 1,6 2 = 26bar

FT = 1780 daN PS = 140 bar PT = 3,5 bar 2 A1 = 31,6 cm 2 A2 = 19,9 cm ϕ = 1,6 vmax = 12,7 cm/s ==> p1 and p2

p2 =

Q= 0,06•19,9•12,7=15 l/min

Q N = 15

( p S A2ϕ 3 ) + F + ( pT A2ϕ )] A2 (1 + ϕ 3 )

bar

p1 = pT + [( p S − p2 )ϕ 2 ] bar Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1. Q= 0,06•A2•vmax

QN = Q

l/min

35 pS − p 2

l/min


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27

35 = 30 l/min 140 − 131



Differential Cylinder Extending at an Inclined Plane with Negative Load

Calculation: Layout: FT = Fa+FE+FR+[G•(µ•cosα-sinα)] daN

p1 =

(210 • 106) + 1,22 [−6675 + (0 • 106)] = 131bar 106(1 + 1,43 )

Given Parameters


p1 =

Caution!!!

Negative load is leading to cylinder cavitation. Specified parameters to be changed by means of using a larger cylinder size, increasing the system pressure or reducing the necessary total force. A1 = 126 cm

p S A2 + ϕ 2 [ F + ( pT A2 )]

p2 = pT +

A2 (1 + ϕ 3 ) p S − p1

bar

bar

ϕ2

Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1. Q= 0,06•A1•vmax

QN = Q

p2 =

l/min


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A2 = 106 cm

2

210 − 44 = 116bar 1,2 2

Q= 0,06•126•25,4=192 l/min 35 Q N = 192 = 88 l/min 210 − 44

l/min

35 p S − p1

2

28

R=1,2



Differential Cylinder Retracting at an Inclined Plane with Negative Load


(210 • 38,1 • 1,43 ) + [ −6675 + (0 • 38,1 • 1,4)] = 107 bar 38,1(1 + 1,43 )

F = Fa+FE+FR+[G•(µ•cosα-sinα)] daN

p2 =

Given Parameters

p 1 = 0 + [(210 − 107) • 1,4 2 ] = 202 bar

F = -6675 daN PS = 210 bar PT = 0 bar 2 A1 = 53,5 cm 2 A2 = 38,1 cm ϕ = 1,4 vmax = 25,4 cm/s ==> p1 and p2

Q= 0,06•38,1•25,4=58 l/min

Q N = 58

( p S A2ϕ 3 ) + F + ( pT A2ϕ )] bar p2 = A2 (1 + ϕ 3 )

p1 = pT + [( p S − p2 )ϕ 2 ] bar Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p2. Q= 0,06•A2•vmax

QN = Q

l/min

35 pS − p 2

l/min


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29

35 = 34 l/min 210 − 107



Hydraulic Motor with a Positive Load

Calculation: Layout: T = α•J+TL

p1 =

[Nm]

p 2 = 210 − 127 + 0 = 83bar

Given Parameters

T = 56,5 Nm PS = 210 bar PT = 0 bar 3 DM = 82 cm /rad ωM = 10 rad/s

QM= 0,01•10•82=8,2 l/min

Q N = 8,2

==> p1 and p2

p S + p T 10πT bar + 2 DM p 2 = p S − p1 + p T bar p1 =

Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1. QM= 0,01•ωM•DM

QN = QM

l/min

35 p S − p1

l/min


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210 + 0 10 • π • 56,5 + = 127bar 2 82

30

35 = 5,3 l/min 210 − 127



Hydraulic Motor with a Negative Load

Calculation: Layout: T = α•J-TL

p1 =

[Nm]

p 2 = 210 − 40 + 0 = 170bar

Given Parameters

T = -170 Nm PS = 210 bar PT = 0 bar 3 DM = 82 cm /rad ωM = 10 rad/s

QM= 0,01•10•82=8,2 l/min

Q N = 8,2

==> p1 and p2

p S + p T 10πT bar + 2 DM p 2 = p S − p1 + p T bar p1 =

Verification of the cylinder dimensioning and calculation of the nominal volume flow QN, depending on the load pressure p1. QM= 0,01•ωM•DM

QN = QM

l/min

35 p S − p1

l/min


11.04.2012

210 + 0 10 • π • ( −170) + = 40bar 2 82

31

35 = 3,6 l/min 210 − 40



Identification of the Reduced Masses of Different Systems The different components (cylinder / motors …) have to be dimensioned for the layout of the necessary forces of a hydraulic system, so that the acceleration and the deceleration of a mass is correct and targeted. The mechanics of the system are defining the stroke of the cylinders and motors. Speed- and force calculations have to be carried out. Statements with view to acceleration and its effects on the system can be made by fixing the reduced mass of a system. The reduced mass (M) is a concentrated mass, exerting the same force – and acceleration components as the regular mass at the correct system. The reduced moment of inertia (Ie) has to be considered for rotational systems. The reduced mass has to be fixed in a first step for considerations with stroke measuring systems or for applications with deceleration of a mass! Newton’s second axiom is used for the specification of the acceleration forces.

F = m• a

F= force [N] m= mass [kg] a= acceleration [m/s2]

The following equation is applied for rotational movements:

Γ = I • θ ′′

Γ = torque [Nm] Í= moment of inertia [kgm2]

θ ′′ = angular acceleration [rad/s2]

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32



Linear Drives Primary Applications (Energy Method)

The mass is a concentrated mass and the rod l is weightless. The cylinder axis is positioned rectangular to the rod l. Relation between cylinder and rod:

θ′ =

vc v m = r l

θ ′′ =

ac am = r l

Needed torque for acceleration of the mass:

Γ = IXθ ′′ = F • r

= m • l 2 Xθ ′′ = m • l2 X

I = m • l2

am l

θ ′′ =

am l

= m • lXa m ==>

F=

m• l• am = m• i• am r

i=

l r

m•i can be considered as mass movement m.

F = m• i• am = m• i•

l • ac = m • i2 • a c = M • a c r

mit

ac am = r l

F= cylinder force M= reduced mass ac= acceleration of the cylinder rod General validity: M = m•i The same result can be obtained by the aid of the energy method (kinetic energy of the mass m). The dependence of the mass movement with the cylinder movement can be specified with the help of the geometry of the system. 2

Energy of the mass:

1 1 KE = I • θ ′ 2 = m • l 2 • θ ′ 2 2 2

11.04.2012

(I=m•i2)

33



1 v  = m • l2 •  c   r 2 = =

2

(vc=r• θ ′ )

1 l2 2 m • 2 • vc 2 r

1 2 M • vc 2

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2

M=m•i

34

and i=l/r



Concentrated Mass with Linear Movements

v is the horizontal component of v´. v´ is positioned rectangular to rod l. Energy method:

1 1 KE = I • θ ′ 2 = m • l 2 • θ ′ 2 2 2  v′  1 = m • l2 •   2  r

2

( θ ′ =v´/r)

l2 1 2 = m • 2 • v′ r 2 1 2 = m • i2 • v′ 2 with v=v´•cosα

==> KE =

=

1 m • i2 • v′ 2 2

1 m • i2 1 • v2 = M • v2 2 2 (cosα ) 2

with M = m If:

i2 ==> M is position-depending (cosα ) 2 α= 0 then, α=1

and M=mi2

α=90° then, cosα=0

and M=∝

α=30° then, cosα=±0,866

and Mα = m

i2 0,75

If a cylinder is moving a mass, as shown in the preceding figure, and the movement is situated between -30° and +30°, the acceleration- and deceleration forces have to be calculated in the center of motion with a reduced mass, twice as large as the one in the neutral center.

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35



Distributed Mass at Linear Movements

When considering the same rod l with the mass m, you can here also calculate the reduced mass of the rod.

1 1 1 KE = I • θ ′ 2 = X • m • l 2 • θ ′ 2 2 2 3  v′  1 1 = X • m • l2 •   2 3  r

1 • m • l2 3

2

( θ ′ =v´/r)

l2 1 1 2 = X • m • 2 • v′ 2 3 r 1 1 2 = X • m • i2 • v′ 2 3 with v=v´•cosα

=

M=

1 1 m • i2 1 X• • • v2 = • M • v2 2 2 3 (cos a ) 3

1 m • i2 • 2 (cos a ) 2

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36



Rotation

If considering now a rotating mass with a moment of inertia I, driven by a motor (ratio D/d).

1 1 d KE = I • θ ′ 2 m = I • (θ ′ • ) 2 2 2 D

I= moment of inertia [kgm2]

2

1  d = I •   •θ ′2 2  D

θ ′ = angular acceleration [rad/s2]

1 = I • i2 •θ ′ 2 2

1 = Ie •θ ′2 2

Ie= I•i2 i=d/D

If a gearbox has to be used, i has to be considered. 2

If i=D/d, then Ie=I/i

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37



Combination of Linear and Rotational Movement

A mass m is here moved by a wheel with radius r. The wheel is weightless.

KE = =

1 2 m • ( r • θ ′) 2 =

=

1 m • v2 2 v=r• θ ′

1 m • r 2 •θ ′2 2

1 Ie • θ ′2 2

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Ie= m•r

2

38



Hydraulic Resistances The resistance of an area reduction is the change of the applied pressure difference corresponding volume flow change.

R=

∆p to the

d (∆p) dQ

Orifice Equation

QBlende = 0,6 • α K •

dB •π 2 • ∆p • ρ 4 2

αK = flow coefficient (0,6-0,8) ρ = 0,88 [kg/dm3]

dB = orifice diameter [mm]

∆ p = pressure difference [bar] Qorifice= [l/min]

Throttle Equation

Q Drossel

π • r4 • ( p1 − p 2 ) = 8•η • l

η=ρ•ν

Qthrottle= [m3/s] η = dynamic viscosity [kg/ms]

l = throttle length [m] r = radius [m] ν = kinematic viscosity [m2/s] ρ = 880 [kg/m3]

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39



Hydro Accumulator

1 1    p 0  κ   p1  κ  ∆V = V0   • 1 −    p1    p 2    

p2 =

V0 =

p1    ∆V 1 − 1   p0  κ  V0   p    1

      

κ

∆V 1 1    p 0  κ   p1  κ    • 1 −     p1   p2   

11.04.2012

κ = 1,4 (adiabatic compression) ∆V = effective volume [l]

V0 = accumulator size [l] p0 = gas filling pressure [bar] p1 = service pressure min [bar] (pressure loss at the valve) p2 = service pressure max [bar]

p0 = <0,9*P1 Provide an accumulator in the pressure circuit for pressure-controlled pumps! Swivel time of pump tSA of the pump catalog.

∆V = Q • t SA

40



Heat Exchanger (Oil - Water) ETD = t öl − t K

VÖl = oil flow [l/min] PV = dissipation power [kW]

P p 01 = V ETD

tÖl = inlet temperature oil [°C]

14 • PV ∆t K = VK

tK = inlet temperature cooling water [°C]

∆tÖl = cooling of the oil [K] ∆tK = heating of the cooling water [K]

VK = cooling water flow [l/min] Calculation of ∆tÖl is different, depending on the

ETD = inlet temperature difference [K]

respective hydraulic fluid.

p01 = spec. cooling capacity [kW/h]

HFA

∆t öl =

HLP/HFD

14,7 • PV Völ

∆t öl =

HFC

36 • PV Völ

∆t öl =

17,2 • PV Völ

The size of the heat exchangers can be defined by the calculated value p01 of the diagrams of the different manufacturers.

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41



Example AB-Standards:

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42



Layout of a Valve

The necessary volume flow can be calculated based on the cylinder data as well as on the extending – and retracting speeds.

Q= 0,06•A2•vmax

l/min

X pS − p2

l/min

QN = Q

P= PS system pr.-PLload pr.-P Treturn pressure

(Load pressure ≈

2 *System pressure) 3

X= 35 (Servo valve) pressure loss via a leading edge X= 35 (Prop valve) pressure loss via a leading edge (Prop valve with shell)

At optimal efficiency FT = load force [daN] PS = system pressure [bar] PT = return pressure [bar] A1 = piston surface cm2 A2 = ring surface cm2 ϕ = aspect ratio cylinder vmax = extending speed of the cylinder cm/s

X= 5 (Prop valve) pressure loss via a leading edge (Prop valve without shell)

Selection of a valve 10% larger than the calculated nominal volume flow

 p1 and p2

p2 =

( p S A2ϕ 3 ) + FT + ( pT A2ϕ )] bar A2 (1 + ϕ 3 )

p1 = pT + [( p S − p2 )ϕ 2 ] bar


Translation: Harm/March 16, 2011 11.04.2012

43

Formulas Rexroth

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