Sales Industry Sector Metallurgy
Formulary Hydraulics
Hydraulic Formulary
Author: Houman Hatami Tel.: +49-9352-18-1225 Fax: +49-9352-18-1293
[email protected]
11.04.2012
1
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 11.04.2012
2
Sales Industry Sector Metallurgy
Formulary Hydraulics
HYDRO ACCUMULATOR .................................................................................................................... 40 HEAT EXCHANGER (OIL- WATER) .................................................................................................... 41 LAYOUT OF A VALVE ......................................................................................................................... 43
11.04.2012
3
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
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
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
11.04.2012
5
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
6
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
Equation / Equation Variations
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
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] 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
Sales Industry Sector Metallurgy
Formulary Hydraulics
Continuity Equation Figure
Equation / Equation Variations
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
Equation / Equation Variations
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
Equation / Equation Variations
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]
11.04.2012
8
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
10
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
11
Sales Industry Sector Metallurgy
Formulary Hydraulics
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]
12
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
[s]
13
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
2
2
14
Sales Industry Sector Metallurgy
Formulary Hydraulics
Double Acting Cylinder
pA = pB = QA =
4 • 104
•
π 4 • 104
dK = Piston diameter [mm]
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
11.04.2012
2
2
2
15
Sales Industry Sector Metallurgy
Formulary Hydraulics
Cylinder in Differential Control
dK = Piston diameter [mm]
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
11.04.2012
2
2
16
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
Sales Industry Sector Metallurgy
Formulary Hydraulics
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 =
ω 0 = Angular frequency
400 • A R π
4
mred mölred + mred
ω 01 2π
11.04.2012
18
Sales Industry Sector Metallurgy
Formulary Hydraulics
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 =
ω 0 = Angular frequency
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
11.04.2012
19
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
11.04.2012
20
Sales Industry Sector Metallurgy
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
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
11.04.2012
21
Sales Industry Sector Metallurgy
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 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
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 ] = 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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
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
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 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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
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
Sales Industry Sector Metallurgy
Formulary Hydraulics
Differential Cylinder Retracting with Negative Load
Calculation: Layout:
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
FT = -4450 daN PS = 210 bar PT = 0 bar 2 A1 = 81,3 cm 2 A2 = 61,3 cm ϕ = 1,3 vmax = 25,4 cm/s ==> p1 und p2
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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
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) + 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
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.
11.04.2012
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 • 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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
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•(µ•cosα-sinα)] daN
p1 =
(210 • 106) + 1,22 [−6675 + (0 • 106)] = 131bar 106(1 + 1,43 )
Given Parameters
FT = -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 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 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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
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
Sales Industry Sector Metallurgy
Formulary Hydraulics
Differential Cylinder Retracting at an Inclined Plane with Negative Load
Calculation: Layout:
(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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
29
35 = 34 l/min 210 − 107
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
210 + 0 10 • π • 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 = 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
Selection of a Servo valve 10% larger than the calculated nominal volume flow.
11.04.2012
210 + 0 10 • π • ( −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:
Γ = I • θ ′′
Γ = torque [Nm] Í= moment of inertia [kgm2]
θ ′′ = angular acceleration [rad/s2]
11.04.2012
32
Sales Industry Sector Metallurgy
Formulary Hydraulics
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
Sales Industry Sector Metallurgy
Formulary Hydraulics
1 v = m • l2 • c r 2 = =
2
(vc=r• θ ′ )
1 l2 2 m • 2 • vc 2 r
1 2 M • vc 2
11.04.2012
2
M=m•i
34
and i=l/r
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 • 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.
11.04.2012
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.
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
11.04.2012
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).
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
11.04.2012
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 • θ ′) 2 =
=
1 m • v2 2 v=r• θ ′
1 m • r 2 •θ ′2 2
1 Ie • θ ′2 2
11.04.2012
Ie= m•r
2
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 • 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]
11.04.2012
39
Sales Industry Sector Metallurgy
Formulary Hydraulics
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 =
