Hyd Formelsammlung En

Sales Industry Sector Metallurgy

Formulary Hydraulics

Hydraulic Formulary

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

30.08.2011

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Sales Industry Sector Metallurgy

Formulary Hydraulics

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 30.08.2011

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Formulary Hydraulics

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

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Formulary Hydraulics

Relation between Units Size Lengths

Surfaces

Volume

Density

Unit

Symbol

Relation

Micrometer

m

1m = 0,001mm

Millimeter

mm

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

Centimeter

cm

1cm = 10mm = 10.000m

Decimeter

dm

1dm = 10cm = 100mm = 100.000m

Meter

m

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

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

Sales Industry Sector Metallurgy

Formulary Hydraulics

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

m N 1 s2 kg

1g = 9,81 m/s2  = 2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 ´´=  rad 180

Power

1W  1

1Ws  1Nm  1

kg  m  m  1J s2

N  10bar  1MPa mm2

1rad = 1m/m = 57,2957° 1rad = 180°/

Speed

One/second

1/s

One/minute

1/min

1 1  s  60 min 1 s

1 1  min 1  min 60s

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Important Characteristic Values of Hydraulic Fluids

HLP

HFC

HFA (3%)

HFD

Density at 20°C

0,00087

0,00105-0,00108

0,0001

0,000115

[kg/cm3] Kinematic Viscosity at 40°C

10-100

36-50

0,7

15-70

[mm2/s] Compressions Module E at 50°C

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

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General Hydraulic Relations Piston Pressure Force Figure

Equation / Equation Variations

Symbols / Units

F  10  p  A F  p  A    10 A

d

F = piston pressure force[N]

d2   4

p = fluid pressure[bar] 2

A = piston surface[cm ] d = piston diameter[cm]  = efficiency cylinder

4  F  0,1 p

p  0,1 

4 F  d2

Piston Forces Figure

Equation / Equation Variations

Symbols / Units

F  pe  A  10 F  pe  A    10

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

d2   A 4

2

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

A For annulus surface:

A

(D2  d 2 )   4

Hydraulic Press Figure

Equation / Equation Variations

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

F1 F  2 A1 A 2

F2 = Force at the operating piston[N]

F1  s1  F2  s2

A2 = Surface of the operating piston [cm2]

A1 = Surface of the pump piston [cm2]

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



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F1 A1 s 2   F2 A2 s1

 = Gear ratio

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Continuity Equation Figure

Equation / Equation Variations

Q1  Q 2

Symbols / Units

3

3

3

Q1,2 = Volume flows [cm /s, dm /s, m /s]

Q1  A 1  v 1

2

2

2

A1,2 = Area surfaces [cm , dm , m ] v1,2 = Velocities

Q2  A 2  v2

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

A 1  v1  A 2  v 2 Piston Speed Figure

Equation / Equation Variations

v1 

Q1 A1

v1,2 = Piston speed [cm/s] 3

Q v2  2 A2 A1 

Symbols / Units

Q1,2 = Volume flow [cm /s] 2

A1 = Effective piston surface (circle) [cm ] 2

A2 = Effective piston surface (ring) [cm ]

d  4 2

(D2  d 2 )   A2  4 Pressure Intensifier Figure

Equation / Equation Variations

Symbols / Units

p1 = Pressure in the small cylinder [bar]

p1  A 1  p 2  A 2

2

A1 = Piston surface [cm ] p2 = Pressure at the large cylinder [bar] 2

A2 = Piston surface [cm ]

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Formulary Hydraulics

Hydraulic System Components Hydro Pump

Q

V  n   vol 1000

pQ Pan  600   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 M [Nm] 100   mh

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

 ges   vol   mh

vol = Volumetric efficiency (0,9-0,95) mh = Hydro-mechanic efficiency(0,9-0,95)

Hydro Motor

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

Q

V n 1000   vol

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

Q   vol  1000 n V

M ab

mh = Hydro-mechanic efficiency

(0,9-0,95)

p  V  mh   1,59  V  p  mh  10 2 20  

Pab 

p = Pressure difference between motor inlet and

outlet (bar)

p  Q   ges

30.08.2011

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

600

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Formulary Hydraulics

Hydro Motor Variable

Md  P n



 Md  n

30000 30000



Md  n

30000 P  n 

P  Md

M d max i   Getr

n = Speed [min-1] Mdmax = Max torque [Nm] i = Gear ratio Getr = Gear efficiency mh = Mech./hydraulic efficiency vol = Vol. efficiency

Vg = Flow volume [cm3]

Md Vg   mh

Vg  n 1000   vol

QP  P

P = Power [kW]

n max i

p  20 

Q

Md = Torque [Nm]

Vg  n   vol 1000

Q  p 600   ges

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Formulary Hydraulics

Hydro Motor Fixed

Md  P n



 Md  n

30000 30000



Md  n

30000 P  n 

P  Md

M d max i   Getr

n = Speed [min-1] Mdmax = Max torque [Nm] i = Gear ratio Getr = Gear efficiency mh = Mech./hydraulic efficiency vol = Vol. efficiency

Vg = Flow volume [cm3]

Md Vg   mh

Vg  n 1000   vol

QP  P

P = Power [kW]

n max i

p  20 

Q

Md = Torque [Nm]

Vg  n   vol 1000

Q  p 600   ges

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Hydro Motor Intrinsic Frequency

V ( G )2 2 E 2  0  VG J red  VR ) ( 2

 f0  0 2

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VG = Displacement [cm3] 0 = 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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Formulary Hydraulics

Hydro Cylinder

d 12   d 12  0,785 2 A  [cm ] 400 100 A st 

d1 = Piston diameter [mm] d2 = Piston rod diameter [mm] p = Service pressure [bar]

d 2 2  0,785 2 [cm ] 100

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

(d 12  d 2 2 )  0,785 2 AR  [cm ] 100

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

p  d1  0,785 [kN] FD  10000

Qth = Volume flow, without considering the

p  (d 12  d 2 2 )  0,785 Fz  [kN] 10000

vol = Volumetric efficiency (approx. 0,95)

2

h = Stroke [mm]

h Q [m/s]  t  1000 A  6

v

Qth  6  A  v 

Q V t

leakages (l/min)

V  60 t

t = Stroke time [s] FD

[l/min] FZ

Q th

 vol.

FS

Ah [l] 10000

A h6 Q  1000

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[s]

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Differential Cylinder

dK = Piston diameter [mm]

4  FD   pK

d K  100 

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

pK 

4  10  FD   d K2

p St 

4  104  FZ   (d K 2  d St 2 )



4

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

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

d K2 (d K 2  d St 2 )

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

6 2 QK   va  d K 400 QSt 

ve 

va 

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

6 2 2  v e  (d K  d St ) 400

QSt

6  (d K 2  d St 2 ) 400 QK

6 2  dK 400

Vol p  Vol F 

 4  10

6

 d St  h

6

 h  (d K  d St )

2

 4  10

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2

2

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Double Acting Cylinder

pA  pB  QA 

4  104



 4  104

dK = Piston diameter [mm]

FA 2 (d K  d StA 2 )

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

FB  2 (d K  d StB 2 )



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

6 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 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]

6 2 2  (d K  d St ) 400

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

QK

VolFB = Fill-up volume B [l]

6  d K2 400

Vol p 

 4  10

Vol FA  Vol FB 

 d St  h 2

6

 6

 h  (d K  d StA )

6

 h  (d K  d StB )

2

4  10

 4  10

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2

2

2

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Cylinder in Differential Control

dK = Piston diameter [mm]

4  FD d st  100    p St

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

4  104  FD pK    d St 2

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

4  104  FZ p St    (d K 2  d St 2 ) Q

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

6 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

6  d St 2 400

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

Q d 2 QK  P 2 K d St QSt 

Q P  (d K 2  d St 2 ) d St 2

Retraction:

ve 

QP

6 2 2  (d K  d St ) 400

QSt=QP

QP  d K2 QK  (d K 2  d St 2 ) Vol p  Vol F 

 4  10

6

 d St  h

6

 h  (d K  d St )

2

 4  10

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2

2

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Cylinder Intrinsic Frequency at Differential Cylinders

d K 2 AK  4 100

AK = Piston surface [cm2]

(d 2  d St 2 ) AR  K 4 100

dSt = Piston rod diameter [mm]

VRK

VRSt

AR = Piston ring surface [cm2] dK = Piston diameter [mm] dRK = NW- piston side [mm] LK = Length of piston side [mm] dRSt = NW-rod side [mm]

d 2 L  RK  K 4 1000

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

d 2 L  RSt  St 4 1000

mRK 

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

VRK   Öl 1000

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

V   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

 01   0  f 01 

A  EÖL AR  EÖl 1 ( K  ) AR  h  hK m AK  hK VRK VRSt 10 10 2

0 

f0 

 0 = Angular frequency

   

2

0 2 4

mölred

 1 d   mRK  K   mRSt   d RK   d RSt

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400  A R    

17

 01 2

mred mölred  mred

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Cylinder Intrinsic Frequency at Double Acting Cylinders

AR = Piston ring surface [cm2]

(d K 2  d St 2 ) AR  4 100

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

d 2 L VR  RK  K 4 1000

LK = Length of the piston side [mm]

V   öl mR  R 1000

VR = Volume of the line [cm3]

h = Stroke [mm]

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

AR2 2 Eöl  0  100 ( ) AR  h mred  VRSt 10 f0 

0 2

mölred

 1  2  mRK   dR

 01   0  f 01 

 0 = Angular frequency

400  A R    

4

mred mölred  mred

 01 2

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Cylinder Intrinsic Frequency for Plunger Cylinders

d K 2 AK  4 100

AK = Piston surface [cm2]

d 2 L VR  K  K 4 1000

LK = Length piston side [mm]

dK = Piston diameter [mm] dR = Diameter of the piping [mm] LR = Length of the line [mm] h = Stroke [mm]

V   öl mR  R 1000

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

E AK 0  100 öl  ( ) mred AK  h VRSt f0 

f0 = Intrinsic frequency

 0 = Angular frequency

0 2

m ölred

d   2  mR  K   dR 

 01   0  f 01 

4

mred mölred  mred

 01 2

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Formulary Hydraulics

Piping

p   

l    v 2  10 d2

 lam. turb.

0,316  4 Re vd



lam. = Pipe friction coefficient for laminar flow turb. = Pipe friction coefficient for turbulent flow

6 d2  d 

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

 103

Q

v

 = Density [kg/dm3] (0,89)  = Pipe friction coefficient

64  Re

Re 

p = Pressure loss at direct piping [bar]



 102

d = Internal diameter of the piping [mm]  = Kinematic viscosity [mm2/s]

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

4

400 Q  6 v

30.08.2011

20

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Formulary Hydraulics

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

=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=J+ TL

Nm

m

Torque Load torque

TL

Nm

Angular acceleration



rad/s2

Inertia moment

J

kgm2

30.08.2011

21

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Formulary Hydraulics

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  = 1,40 vmax = 30,00 cm/s ==> p1 und p2

p S A2  R [ FT  ( pT A2 )] bar A2 (1   3 ) p p p2  pT  S 2 1 bar

Q N  96



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.

30.08.2011

210  120  52 bar 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

Sales Industry Sector Metallurgy

Formulary Hydraulics

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 ]  52 bar

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

Q= 0,0638,130=69 l/min

( p A  3 )  FT  ( pT A2 )] p2  S 2 bar A2 (1   3 )

Q N  96

p1  pT  [( p S  p2 ) 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

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

30.08.2011

(210 381 , 14 , 2 )  4450  (525 ,  381 , 14 , )]  187bar 3 381 , (114 , )

23

35  84l / min 210  187

Sales Industry Sector Metallurgy

Formulary Hydraulics

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 A1 = 81,3 cm2 A2 = 61,3 cm2  = 1,3 vmax = 12,7 cm/s ==> p1 und p2

p S A2   2 [ FT  ( pT A2 )] bar A2 (1   3 ) p p p2  pT  S 2 1 bar

Q N  62



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

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

30.08.2011

175  36  82 bar 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

Sales Industry Sector Metallurgy

Formulary Hydraulics

Differential Cylinder Retracting with Negative Load

Calculation: Layout:

FT = Fa+FR-G

p2 

[daN]

(210 613 , 13 , 2)  4450  (0 613 , 13 , )]  122bar 613 , (113 , 3)

Given Parameters

p 1  0  [(210  122)]  149 bar

FT = -4450 daN PS = 210 bar PT = 0 bar A1 = 81,3 cm2 A2 = 61,3 cm2  = 1,3 vmax = 25,4 cm/s ==> p1 und p2

Q= 0,0661,325,4=93 l/min

p2 

( p S A2 )  FT  ( pT A2 )] bar A2 (1   3 ) 3

Q N  93

p1  pT  [( p S  p2 ) 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

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

30.08.2011

25

35  59l / min 210  122

Sales Industry Sector Metallurgy

Formulary Hydraulics

Differential Cylinder Retracting at an Inclined Plane with Positive Load

Calculation: Layout:

(14019,9) 16 , 2[2225 (35 , 19,9)]  85bar 19,9(116 , 3)

FT = Fa+FE+FS+[G(cos+sin)] daN

p1 

Given Parameters

p 2  35 

FT = 2225 daN PS = 140 bar PT = 3,5 bar A1 = 31,6 cm2 A2 = 19,9 cm2 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   2 [ F  ( pT A2 )]

p2  pT 

A2 (1   3 ) p S  p1

2

bar

bar

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

QN  Q

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

30.08.2011

140  85  25bar 1,6 2

26

35  19 l/min 140  85

Sales Industry Sector Metallurgy

Formulary Hydraulics

Differential Cylinder Retracting at an Inclined Plane with Positive Load

Calculation: Layout:

(14019,9 16 , 3) 1780 [35 , 19,9 16 , )]  131bar 19,9(116 , 3)

FT =Fa+FE+FS+[G(cos+sin)] daN

p2 

Given Parameters

p 1  3,5  [(140  131)  1,6 2  26bar

FT = 1780 daN PS = 140 bar PT = 3,5 bar A1 = 31,6 cm2 A2 = 19,9 cm2  = 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 A2 3 )  F  ( pT A2 )] A2 (1   3 )

bar

p1  pT  [( p S  p2 ) 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

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

30.08.2011

27

35  30 l/min 140  131

Sales Industry Sector Metallurgy

Formulary Hydraulics

Differential Cylinder Extending at an Inclined Plane with Negative Load

Calculation: Layout: FT = Fa+FE+FR+[G(cos-sin)] daN

p1 

(210106) 12 , 2[6675 (0106)]  131bar 106(114 , 3)

Given Parameters

FT = -6675 daN PS = 210 bar PT = 0 bar A1 = 53,5 cm2 A2 = 38,1 cm2  = 1,4 vmax = 25,4 cm/s ==> p1 und p2

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 cm2 A2 = 106 cm2

p S A2   2 [ F  ( pT A2 )]

p2  pT 

A2 (1   3 ) p S  p1

bar

bar

2

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 

Q= 0,0612625,4=192 l/min 35 Q N  192  88 l/min 210  44

l/min

35 p S  p1

l/min

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

30.08.2011

210  44  116bar 1,2 2

28

R=1,2

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Formulary Hydraulics

Differential Cylinder Retracting at an Inclined Plane with Negative Load

Calculation: Layout:

(210 381 , 14 , 3) [6675 (0 381 , 14 , )]  107bar 3 3811 , ( 14 , )

F = Fa+FE+FR+[G(cos-sin)] daN

p2 

Given Parameters

p 1  0  [(210  107)  1,4 2 ]  202 bar

F = -6675 daN PS = 210 bar PT = 0 bar A1 = 53,5 cm2 A2 = 38,1 cm2  = 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 A2 3 )  F  ( pT A2 )] bar p2  A2 (1   3 ) p1  pT  [( p S  p2 ) 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

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

30.08.2011

29

35  34 l/min 210  107

Sales Industry Sector Metallurgy

Formulary Hydraulics

Hydraulic Motor with a Positive Load

Calculation: Layout: T = J+TL

p1 

[Nm]

p 2  210  127  0  83bar

Given Parameters

T = 56,5 Nm PS = 210 bar PT = 0 bar DM = 82 cm3/rad M = 10 rad/s

QM= 0,011082=8,2 l/min

Q N  8,2

==> p1 and p2

p S  p T 10T 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MDM

QN  QM

l/min

35 p S  p1

l/min

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

30.08.2011

210  0 10    56,5   127bar 2 82

30

35  5,3 l/min 210  127

Sales Industry Sector Metallurgy

Formulary Hydraulics

Hydraulic Motor with a Negative Load

Calculation: Layout: T = J-TL

p1 

[Nm]

p 2  210  40  0  170bar

Given Parameters

T = -170 Nm PS = 210 bar PT = 0 bar DM = 82 cm3/rad M = 10 rad/s

QM= 0,011082=8,2 l/min

Q N  8,2

==> p1 and p2

p S  p T 10T  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MDM

QN  QM

l/min

35 p S  p1

l/min

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

30.08.2011

210  0 10    ( 170)   40bar 2 82

31

35  3,6 l/min 210  40

Sales Industry Sector Metallurgy

Formulary Hydraulics

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:

  I   

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

  = angular acceleration [rad/s2]

30.08.2011

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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:

 

vc v m  r l

  

ac am  r l

Needed torque for acceleration of the mass:

  IX   F  r

 m  l 2 X   m  l2 X

I  m  l2

am l

  

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

M  m i General validity: 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    2  m  l 2    2 2 2

30.08.2011

(I=mi2)

33

Sales Industry Sector Metallurgy

Formulary Hydraulics

1 v   m  l2   c   r 2  =

2

(vc=r   )

l2 1 m  2  vc 2 r 2

1 2 M  vc 2

30.08.2011

M=mi2 and i=l/r

34

Sales Industry Sector Metallurgy

Formulary Hydraulics

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    2  m  l 2    2 2 2  v  1  m  l2    2  r 

2

(   =v´/r)

1 l2  m  2  v 2 2 r 1 2 = m  i2  v 2 with v=v´cos

==> KE 



1 m  i2  v 2 2 1 m  i2 1  v2  M  v2 2 2 (cos ) 2

with M  m If:

i2 ==> M is position-depending (cos ) 2 = 0 then, =1

and M=mi2

=90° then, cos=0

and M=

=30° then, cos=0,7

and M=0

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.

30.08.2011

35

Sales Industry Sector Metallurgy

Formulary Hydraulics

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.

KE 

1 1 1 I  2  X  m  l2  2 2 2 3

 v  1 1  X  m  l2    2 3  r

1  m  l2 3

2

(   =v´/r)

1 1 l2  X  m  2  v 2 2 3 r 1 1 2 = X  m  i2  v 2 3 with v=v´cos



M

1 1 m  i2 1 X   v2   M  v2 2 2 3 (cos a ) 3

1 m  i2  2 (cos a ) 2

30.08.2011

36

Sales Industry Sector Metallurgy

Formulary Hydraulics

Rotation

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

KE 

1 1 d I    2 m  I  (   ) 2 2 2 D

I= moment of inertia [kgm2]

2

1  d  I     2 2  D

  = angular acceleration [rad/s2]

1  I  i2    2 2 1 = Ie  2 2

Ie= Ii2 i=d/D

If a gearbox has to be used, i has to be considered. If i=D/d, then Ie=I/i2

30.08.2011

37

Sales Industry Sector Metallurgy

Formulary Hydraulics

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    2 

=

1 m  v2 2 v=r  

1 m  r 2  2 2

1 Ie   2 2

30.08.2011

Ie= mr2

38

Sales Industry Sector Metallurgy

Formulary Hydraulics

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   K 

dB  2  p  4  2

K = flow coefficient (0,6-0,8)  = 0,88 [kg/dm ] 3

dB = orifice diameter [mm]

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

Throttle Equation Q Drossel

  r4   ( p1  p 2 ) 8  l

=

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

l = throttle length [m] r = radius [m]  = kinematic viscosity [m /s] 2

 = 880 [kg/m ] 3

30.08.2011

39

Sales Industry Sector Metallurgy

Formulary Hydraulics

Hydro Accumulator

1 1    p 0     p1    V  V0    1     p1    p 2    

p2 

V0 

p1    V 1  1   p0    V0   p    1

      



V 1 1    p 0     p1       1      p1   p2   

30.08.2011

 = 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 =