Formula Book
for
Hydraulics and Pneumatics
Fluid and Mechanical Engineering Systems
Department of Management and Engineering
Linköping University
Revised 2008-10-27
Contents
1 Elementary Equations
1.1 Nomenclature . . . . . .
1.2 The continuity equation
1.3 Reynolds number . . . .
1.4 Flow equations . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
1
1
1
2 Pipe flow
2.1 Nomenclature . . . . . . . . .
2.2 Bernoullis extended equation
2.3 The flow loss, ∆pf . . . . . .
2.4 The friction factor, λ . . . . .
2.5 Disturbance source factor, ζs
2.6 Single loss factor, ζ . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2
2
2
2
2
2
3
3 Orifices
3.1 Nomenclature . . . . . . . . . . . . . . .
3.2 The flow coefficient, Cq . . . . . . . . .
3.3 Series connection of turbulent orifices . .
3.4 Parallel connection of turbulent orifices
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6
6
6
6
7
4 Flow forces
4.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Spool . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
8
8
5 Rotational transmissions
5.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Efficiency models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
9
6 Accumulators
6.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Calculating of the accumulator volume, V0 . . . . . . . . . . . . . . . . . . . . . . .
10
10
10
7 Gap theory
7.1 Nomenclature . . . . . . . . . . . .
7.2 Plane parallel gap . . . . . . . . .
7.3 Radial gap . . . . . . . . . . . . . .
7.4 Gap between cylindrical piston and
7.5 Axial, annular gap . . . . . . . . .
.
.
.
.
.
11
11
11
11
11
12
8 Hydrostatic bearings
8.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Circled block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 Rectangular block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
13
13
14
9 Hydrodynamic bearing
9.1 Nomenclature . . . .
9.2 Cartesian coordinate
9.3 Polar coordinates . .
15
15
15
15
.
.
.
.
.
.
.
.
. . . . .
. . . . .
. . . . .
cylinder
. . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
theory
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Non-stationary flow
10.1 Nomenclature . . . . . . . . . . . . . . .
10.2 Joukowskis equation . . . . . . . . . . .
10.3 Retardation of cylinder with inertia load
10.4 Concentrated hydraulic inductance . . .
10.5 Concentrated hydraulic capacitance . .
10.6 Concentrated hydraulic resistance . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
17
17
17
17
17
17
18
10.7 Basic differential equations on flow systems with parameter distribution in space .
10.8 Speed of waves in pipes filled with liquid . . . . . . . . . . . . . . . . . . . . . . . .
18
18
11 Pump pulsations
11.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 System with closed end . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Systems with low end impedance (e.g. volume) . . . . . . . . . . . . . . . . . . . .
20
20
20
21
12 Hydraulic servo systems
12.1 Nomenclature . . . . . . . . . . . . . . . . . . . .
12.2 Introduction . . . . . . . . . . . . . . . . . . . . .
12.3 The servo valve’s transfer function . . . . . . . .
12.4 Servo valve . . . . . . . . . . . . . . . . . . . . .
12.5 The hydraulic system’s transfer function . . . . .
12.6 The servo stability . . . . . . . . . . . . . . . . .
12.7 The servo’s response – bandwidth . . . . . . . . .
12.8 The hydraulic system’s and the servo’s sensitivity
12.9 The servo’s steady state error . . . . . . . . . . .
12.10Control technical resources . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
23
23
24
24
26
27
28
30
31
33
33
13 Hydraulic fluids
13.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
38
14 Pneumatic
14.1 Nomenclature . . . . . . . . . . . . . . . . . . . . . .
14.2 Stream through nozzle . . . . . . . . . . . . . . . . .
14.3 Series connection of pneumatic components . . . . .
14.4 Parallel connected pneumatic components . . . . . .
14.5 Parallel- and series connected pneumatic components
14.6 Filling and emptying of volumes . . . . . . . . . . .
42
42
42
43
44
44
44
Appendix A Symbols of Hydraulic Components
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
to loading – stiffness
. . . . . . . . . . . .
. . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
47
1
1.1
Re
V
d
dh
p
q
qin
t
1.2
Elementary Equations
Nomenclature
:Reynold’s number
:volume
:diameter
:hydraulic diameter
:pressure
:flow (volume flow)
:flow in into volume
:time
v
βe
[-]
[m3 ]
[m]
[m]
[Pa]
[m3 /s]
[m3 /s]
[s]
∆p
η
ν
ρ
:(mean-) flow velocity
:effective bulk modulus
(reservoir and fluid)
:pressure variation, upstream
to downstream
:dynamic viscosity
:kinematic viscosity (= ηρ )
:density
[m/s]
[Pa]
[Pa]
[Ns/m2 ]
[m2 /s]
[kg/m3 ]
The continuity equation
dV
qin
X
p, V
qin =
dV
V dp
+
dt
βe dt
Flow out from the volume is counted negative.
1.3
Reynolds number
vdh ρ
(dh
hydraulic diameter)
η
4 × cross section area
dh =
(dh = d at cirkular crosssection)
circumference
Re =
1.4
Flow equations
For flow through fix orifices applies (see also section 3, Orifices):
Laminar flow
Turbulent flow
q ∝ ∆p
√
q ∝ ∆p
1
2
2.1
Re
d
d1
d2
h
h1
h2
g
ℓ
ℓx
ℓs
2.2
Pipe flow
Nomenclature
:Reynolds number
:diameter
:upstream diameter
:downstream diameter
:height
:upstream height
:downstream height
:gravitation
:length
:distance after disturbance source
:distance after disturbance source
when the flow profile is
completely developed
p
p1
p2
r
v
v1
v2
α
∆pf
[-]
[m]
[m]
[m]
[m]
[m]
[m]
[m/s2 ]
[m]
[m]
ϕ
λ
ρ
ζ
ζs
[m]
:pressure
:upstream pressure
:downstream pressure
:radius
:(mean-) flow velocity
:upstream (mean-) flow velocity
:downstream (mean-) flow velocity
:correction factor
:pressure loss, upstream
to downstream
:angle
:friction factor
:density
:single loss factor
:disturbance source factor
Bernoullis extended equation
At stationary incompressible flow
p1 +
2.3
2.4
The flow loss, ∆pf

ℓ ρv 2


λ



d 2





ρv 2
∆pf = ζs

2






2


 ζ ρv
2
at a straight distance
after a disturbance source
at a single disturbance source
The friction factor, λ
λ=
2.5
ρv 2
ρv12
+ ρgh1 = p2 + 2 + ρgh2 + ∆pf
2
2

64



 Re
Re < 2300 laminar flow

0,316


 √
4
Re
2300 < Re < 105
turbulent flow in smooth pips
Disturbance source factor, ζs
The laminar flow is completely developed at the distance ℓs after a disturbance source.
ζs ≈ 1,21
ℓs ≥ 0,06dRe
(Re < 2300),
For ℓ < ℓs applies
ζs = 1,28 tanh(6,28x0,44 )
where x =
ℓx < 0,06dRe, see diagram in figure 1
ℓx
dRe
2
[Pa]
[Pa]
[Pa]
[m]
[m/s]
[m/s]
[m/s]
[-]
[Pa]
[◦ ]
[-]
[kg/m3 ]
[-]
[-]
ζs
1,30
1,00
0,50
0,00
lx / d Re
-5
10
-4
-3
10
10
-2
-1
10
0
10
10
0,06
Figur 1: The disturbance source factor ζs as function of
ℓx
.
dRe
The turbulant flow is completely developed at the distance ℓs after a disturbance source.
ζs ≈ 0,09
ℓs ≥ 40d
(Re > 2300),
For ℓ < ℓs applies
ζs = 0,09
2.6
r
ℓx
40d
ℓx < 40d
Single loss factor, ζ
Pipe connections to reservoir
tank
v
The pipe starts at a given distance inside the reservoir
(
1
Sharp edge
ζ=
0,5 Slightly rounded edge
tank
v
Pipes in the reservoir wall
ζ = 0,5
tank
Pipe in the reservoir wall with rounded edge
r
d
v
r/d
0,1
0,15
0,25
0,6
ζ
0,12
0,08
0,05
0,04
3
The pipe starts at a given distance inside the reservoir
with entrance cone: For given ℓ, optimal cone angle and
resistance factor is stated.
tank
ϕ
v
d
l
ℓ/d
0,1
0,15
0,25
0,6
1,0
ϕ[◦ ]
60–90
60–80
60–70
50–60
50–60
ζ
0,40
0,26
0,17
0,13
0,10
Pipe in reservoir wall with entrance cone: For given ℓ,
optimal cone angle and resistance factor is stated.
tank
ϕ
v
d
l
ℓ/d
0,1
0,15
0,25
0,6
ϕ[◦ ]
50–60
50–60
45–55
40–50
ζ
0,18
0,14
0,12
0,10
Area variations in the pipe
d1
ϕ
d2
v
ζ 0,9
Increase of area: The loss factor is found in the figure 2
ϕ = 180°, dashed
ϕ = 60°
0,8
0,7
40°
0,6
0,5
30°
0,4
20°
0,3
15°
0,2
10°
0,1
5°
0,0
1,50
2,00
2,50
3,00
3,50
4,00
d2/d1
Figur 2: The loss coefficient ζ as function of the area relationship with the angle ϕ as parameter at increase of area.
Decrease of area: The loss factor is described by ζ =
ζ0 α, where ζ0 for different geometries is received from
section above about Pipe connection to reservoir.
d1
d2 v
According to von Mises applies when
d2 /d1
0,90
0,80
0,70
0,50
0,30
0,10
α
0,19
0,37
0,51
0,76
0,91
0,99
Disturbance source factor ζs applies in a similar way
"
3 #
d2
ζs = ζs0 1 −
d1
The uncompensated disturbance source factor ζs0 is received from section 2.5, Disturbance source
factor, ζs .
4
Pipe bend
ϕ
For pipe bend relates: ζ = ζ90
90
where ζ90 is
ϕ
r
d
d/r
0,20
0,40
0,60
0,80
1,00
ζ90
0,13
0,14
0,16
0,21
0,29
For pipe angle gives ζ directly by
ϕ[◦ ]
10
20
30
40
50
ζ
0,04
0,10
0,17
0,27
0,40
ϕ[◦ ]
60
70
80
90
ζ
0,55
0,70
0,90
1,20
ϕ
Special geometries
ζ = 0,10
ζ = 0,50
ζ = 1,20
45˚
ζ = 0,15
45˚
ζ = 2,5 à 3
ζ = 0,50
ζ = 0,06
5
Banjonipple
ζ = 2,5 à 4
3
3.1
A
A1
Cq
Re
d
K
Orifices
Nomenclature
[m2 ]
[m2 ]
[-]
[-]
[m]
√
[m3 /s P a]
:area
:upstream area
:flow coefficient
:Reynold’s number
:diameter
p
:constant Cq A 2/ρ
p1
ℓ
p
p1
p2
q
ρ
:length
:pressure
:upstream pressure
:downstream pressure
:flow (volume flow)
:density
p2
r
2
q = Cq A
(p1 − p2 )
ρ
q
3.2
[m]
[Pa]
[Pa]
[Pa]
[m3 /s]
[kg/m3 ]
The flow coefficient, Cq
Hole orifice (sharp edged)
p2
p1
A1
Cq = Cq (Re,
A
A
)
A1
If nothing else is stated Cq = 0,67 can be used.
Pipe orifice (sharp edged)
p1
p2
d
l
Laminar flow in the orifice (Re < 2300)
1
Cq = p
1,5 + 1,28 tanh(6,28x0,44 ) + 64x
där x =
ℓ
dRe
The term 1,28 tanh(6,28x0,44 ) agrees with ζs and can be received from the diagram in
figure 1 in section Pipe flow. When x ≥ 0,06, the value 1,21 is accepted.
Turbulent flow in the orifice with 2 ≤
Cq =
ℓ
≤ 20
d

1


s



ℓ
0,316 ℓ


1,46 + 0,088 + √

4


d
Re d



1


r



ℓ

 1,46 + 0,115
d
3.3
2300 ≤ Re ≤ 2 · 104
2 · 104 ≤ Re
Series connection of turbulent orifices
p0
p1
p2
pi
pn
q
K1
K2
Ki
6
Kn
The sum of the orifices applies:
√
q = K p0 − pn
där
1
K= v
u n
uX 1
t
Ki2
i=1
Ki = Cqi Ai
r
The pressure after the j:th orifice is given by
pj = p0 − (p0 − pn )K 2
3.4
j
X
1
2
K
i
i=1
Parallel connection of turbulent orifices
p0
q
q1
q2
qi
qn
K1
K2
Ki
Kn
p1
The sum of the orifices applies:
√
q = K p0 − p1
där K =
n
X
i=1
The flow through the i:th orifice is given by
qi =
Ki
q
K
7
Ki
Ki = Cqi Ai
r
2
ρ
2
ρ
4
4.1
Fs
d
ℓ
p
p1
p2
4.2
Flow forces
Nomenclature
:flow force
:spool diameter
:length
:pressure
:upstream pressure
:downstream pressure
q
v
w
x
δ
ρ
[N]
[m]
[m]
[Pa]
[Pa]
[Pa]
:flow (volume flow)
:(mean-)flow velocity
:area gradient
:spool opening
:jet angle
:density
[m3 /s]
[m/s]
[m]
[m]
[◦ ]
[kg/m3 ]
Spool
Fluidelement
Fs
d
δ
For this valve, without pressure
relief grooves, is the area gradient:
w = πd
l
v
p
p1
p2
x
Fs = |2Cq wx(p1 − p2 ) cos(δ)| + ρℓq̇
The term with absolute value is the static part of the flow force and has a closing effect.
If the spool and bushing have sharp and right angle edges and if the gap between the spool and
the bushing is small and also x ≪ d, then are:
0,62 ≤ Cq ≤ 0,67
and
8
δ = 69◦
5
Rotational transmissions
5.1
Nomenclature
Cv
D
Min
Mut
kp
kv
kε
n
:laminar leakage losses
:displacement
:driving torque pump
:output torque motor
:Coulomb friction
:viscous friction losses
:displacement coefficient
:revs
[-]
[m3 /rev]
[Nm]
[Nm]
[-]
[-]
[-]
[rev/s]
qe
:effective flow
∆p
:pressure difference
ε
:displacement setting
ηhm
:hydraulic mechanical efficiency
ηvol
:volumetric efficiency
Sub index
p
pump
m
motor
[m3 /s]
[Pa]
[-]
[-]
[-]
Pump
Effective flow
np, Min
εp
Dp
qep
qep = εp Dp np ηvolp
Torque
Min =
∆p
1
εp Dp
∆p
2π
ηhmp
Motor
Effective flow
qem
nm, Mut
εm
Dm
qem = εm Dm nm
Torque
∆p
5.2
Mout =
1
ηvolm
εm Dm
∆pηhmm
2π
Efficiency models
Pump
Volumetric efficiency
ηvolp = 1 − Cv
Hydraulic mechanical efficiency
∆p
|εp |np η
ηhmp =
1
np η (kε (1 − |εp |))
1 + (kp + kv
)e
∆p
Motor
Hydraulic mechanical efficiency
Volumetric efficiency
1
ηvolm =
1 + Cv
ηhmm = 1 − (kp + kv
∆p
|εm |nm η
9
nm η (kε (1 − |εm |))
)e
∆p
6
6.1
V0
n
p0
6.2
Accumulators
Nomenclature
[m3 ]
[-]
:accumulator volume
:polytrophic exponent
:pre-charged pressure (absolute pressure,
normally ≈ 90 % av p1 )
p1
p2
∆V
:minimum working pressure (absolute)
:maximum working pressure (absolute)
:working volume
[Pa]
Calculating of the accumulator volume, V0
A. Both the charging and discharging is either adiabatic or isotherm process
p1
p0
V0 =
1
p1 n
1−
p2
∆V
n=
(
1
isotherm process
1,4(1,5) adiabatic process
B. Isotherm charging and adiabatic discharging
∆V
V0 =
p2
p1
p2
p0
1
n
n = 1,4(1,5)
−1
10
[Pa]
[Pa]
[m3 ]
7
7.1
Ff
Mf
Pf
b
Gap theory
Nomenclature
ℓ
:friction force
:friction torque
:power loss
:gap width perpendicular
to flow direction
:eccentricity
:gap height
h1 + h2
mean gap height (
)
2
:length
7.2
Plane parallel gap
e
h
h0
p
qℓ
r
r1
r2
v
∆p
[N]
[Nm]
[W]
[m]
[m]
[m]
γ
η
ω
[m]
[m]
:pressure
:leakage flow
:radius
:inner radius
:outer radius
:relative velocity
:pressure difference through
the gap (p1 − p0 )
:angle
:dynamic viscosity
:angular speed
[Pa]
[m3 /s]
[m]
[m]
[m]
[m/s]
[Pa]
[rad]
[Ns/m2 ]
[rad/s]
Leakage flow relative to the fix wall
l
p1
qℓ =
v
ql
h
p0
vbh bh3 ∆p
+
2
12η ℓ
Friction force
Ff =
ηvbℓ bh
− ∆p
h
2
Effect losses (flow and frictional losses)
Pf =
7.3
bh3 ∆p2
ηv 2 bℓ
+
12η ℓ
h
Radial gap
p0 ql
r
γ
When h ≪ r can the equations for plane parallel
gap be used with following substitution:
ω
h
v = rω
ℓ = γr
p1
7.4
Gap between cylindrical piston and cylinder
r
Leakage flow relative to the fix gap wall
"
2 #
e
πrh30 ∆p
1 + 1,5
qℓ = πvrh0 +
6η ℓ
h0
l
h2
h1
Frictional force
v
e
p1
p0
ql
Ff =
11
2πrηvℓ
s
2 − πrh0 ∆p
e
h0 1 −
h0
Effect losses (flow and frictional losses)
"
2 #
e
πrh30 ∆p2
1 + 1,5
+
Pf =
6η
ℓ
h0
7.5
2πrηv 2 ℓ
s
2
e
h0 1 −
h0
Axial, annular gap
Leakage flow
qℓ =
p0 ql
r2
ω
p1
∆pπh3
r2
6η ln
r1
Pressure as function of radius
r
ln
r1
pr = p1 − (p1 − p0 ) r2
ln
r1
r1
h
Friction moment
Mf =
Effect losses (flow and frictional losses)
Pf =
πh3
6η
∆p2
π ηω 2 4
+
(r − r14 )
r2
2 h 2
ln
r1
12
π ηω 4
(r − r14 )
2 h 2
8
8.1
Ae
B
F
K1
K2
L
ae
f
h
ℓ
8.2
Hydrostatic bearings
Nomenclature
[m2 ]
[m]
[N]
[Ns]
[Nm2 s]
[m]
[m]
[N/m]
[m]
[m]
:effective area
:bearing chamber length
:load
:constant
:constant
:bearing surface length
:effective area/width
:load/width
:gap height
:length
kb
k2
p
pb
qs
qsB
r1
r2
η
:constant
:constant
:pressure
:pressure in the bearing chamber
:flow through the bearing
:flow through
the bearing/width
:inner radius
:outer radius
:dynamic viscosity
[Ns/m2 ]
[Nms]
[Pa]
[Pa]
[m3 /s]
[m2 /s]
[m]
[m]
[Ns/m2 ]
Circled block
h = constant
Flow
qs =
p=0
r2
Load
F = Ae pb
r1
pb
h3
pb
kb
where
6η
ln
kb =
π
and
h
r2
r1
π r22 − r12
Ae =
r2
2
ln
r1
Squeeze
Pressure
p=0
r2
pb
pb = −
K1
ḣ
h3
F =−
K2
ḣ
h3
Load
r1
where
F
.
K1 =
h
h
and
p=0
r2
pb
3ηr22
"
1−
r1
r2
2 #
"
4 #
r1
3π 4
ηr 1 −
K2 =
2 2
r2
If pb = 0
Load
r1
F =−
F
K2
ḣ
h3
where
.
h
h
"
3 4 #
3π 4
r1
r1
r1
K2 =
ηr2 1 − 2 + 2
−
2
r2
r2
r2
13
8.3
Rectangular block
h = constant
Flow
p=0
qsB =
L
pb
h3
pb
kb
Load
f = ae p b
f
B
where
kb = 6ηL
h
and
ae = B + L
L
Squeeze
Pressure
p=0
L
pb = −
K1
ḣ
h3
f =−
k2
ḣ
h3
Load
pb
f
B
.
h
h
where
K1 = 6ηL(B + L)
and
4 2
2
k2 = 6ηL B + 2BL + L
3
L
p=0
L
pb
f
.
B
h
h
If pb = 0
Load
f =−
k2
ḣ
h3
where
k2 = 2ηL3
L
14
9
Hydrodynamic bearing theory
9.1
U1
U2
V1
V2
T
c
h
p
qx
qz
qr
Nomenclature
:velocity in x-axis surface 1
:velocity in x-axis surface 2
:velocity in y-axis surface 1
:velocity in y-axis surface 2
:temperature
:specific heat
:gap height
:pressure
:flow/width unit in x-axis
:flow/width unit in z-axis
:flow/width unit in r-axis
9.2
qθ
r
t
u
w
η
τx
τθ
ρ
ω1
ω2
θ
[m/s]
[m/s]
[m/s]
[m/s]
[K]
[kJ/kg K]
[m]
[Pa]
[m2 /s]
[m2 /s]
[m2 /s]
:flow/width unit in θ-axis
:radius
:time
:flow velocity in x-axis
:flow velocity in z-axis
:dynamic viscosity
:shear stress in x-axis
:shear stress in θ-axis
:density
:velocity in θ-axis surface 1
:velocity in θ-axis surface 2
:angle
[m2 /s]
[m]
[s]
[m/s]
[m/s]
[Ns/m2 ]
[N/m2 ]
[N/m2 ]
[kg/m3 ]
[rad/s]
[rad/s]
[rad]
Cartesian coordinate
y
V2
U2
z
x
V1
U1
Speeeds
u=
y
y
1 ∂p
+ U2
[y(y − h)] + U1 1 −
2η ∂x
h
h
Flows
qx = −
h3 ∂p
h
+ (U1 + U2 )
12η ∂x
2
1 ∂p
[y(y − h)]
2η ∂z
w=
qz = −
h3 ∂p
12η ∂z
Shear stresses
τx|y=0 = −
τx|y=h =
y
h ∂p
U2 − U1
+η
2 ∂x
h
p
z
U2 − U1
h ∂p
+η
2 ∂x
h
τx|y=h
x
τx|y=0
Reynolds equation
∂ ρh3 ∂p
∂
∂ ρh3 ∂p
+
= 6(U1 − U2 ) (ρh) + 12ρ(V2 − V1 )
∂x
η ∂x
∂z
η ∂z
∂x
Adiabatic energy equation
" 2 #
2
∂p
∂p
∂T
∂T
∂T
h3
η
2
+ qz
+h
+
ρc qx
= (U1 − U2 ) +
∂x
∂z
∂t
h
12η
∂x
∂z
9.3
Polar coordinates
y ω2
ω1
V2
w
x
V1
15
y
u
θ
r
Velocities
u=
1 ∂p
[y(y − h)]
2η ∂r
Flows
qr = −
w=
h3 ∂p
12η ∂r
y
y
1 ∂p
+ rω2
[y(y − h)] + rω1 1 −
2η r∂θ
h
h
qθ = −
h3 ∂p
rh
+ (ω1 + ω2 )
12η r∂θ
2
Shear stresses
τθ|y=0 = −
τθ|y=h =
y
r(ω2 − ω1 )
h ∂p
+η
2 r∂θ
h
h ∂p
r(ω2 − ω1 )
+η
2 r∂θ
h
p
τθ|y=h
τθ|y=0
θ
Reynolds equation
3 ∂
∂
ρh ∂p
∂ ρh3 r ∂p
+
= 6r(ω1 − ω2 ) (ρh) + 12ρr(V2 − V1 )
∂r
η ∂r
r∂θ
η ∂θ
∂θ
Adiabatic energy equation
" #
2
2
r2 η
∂p
1 ∂p
p ∂p
∂T
∂T
∂T
h3
2
=
+ 2
+
ρc qr
+ qθ
+h
(ω1 − ω2 ) +
∂r
∂θ
∂t
h
12η
∂r
r
∂θ
r ∂r
16
10
10.1
A
B
CH
C0
F
LH
L0
P
Q
RHℓ
RHt
R0ℓ
R0t
V1
V2
Z0
a
d
ℓ
m
10.2
Non-stationary flow
Nomenclature
:line sectional area
:constant (L0 a)
:conc. hydr. capacitance
:conc. hydr. capacitance/l.enh.
:force
:conc. hydr. inductance
:conc. hydr. inductance/l.enh.
:pressure (frequency dependent)
:flow (volume flow) (frequency dependent)
:conc. hydr. resistance (lam.)
:conc. hydr. resistance (turb.)
:conc. hydr. res./l.unit. (lam.)
:conc. hydr. res./l.unit. (turb.)
:volume
:volume
:impedance
:speed of sound
:diameter
:length
:mass
p
[m2 ]
p
[kg/s m40 ]
p
[m5 /N] 1
p
4
[m /N] 2
ps
[N]
q
4
[kg/m ]
q
5
[kg/m ]0
s
[Pa]
t
3
[m /s]
t
[Ns/m5 v]
v
[Ns/m5 ]0
6
[Ns/m ]
α
[Ns/m6 ]
βe
3
[m ]
∆p
3
[m ]
η
[Ns/m5 ]
λ
[m/s]
[m]
ρ
[m]
ζ
[kg]
:pressure
:pressure in point of operation
:upstream pressure
:downstream pressure
:supply pressure
:flow (volume flow)
:flow in point of operation
:Laplace operator (iω)
:time
:valve closing time
:flow velocity
:velocity of cylinder
:dimensionless area
:effective bulk modulus
:change in pressure due to pressure peek
:dynamic viscosity
:friction coefficient
:parameter
:density
:single resistant loss
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[m3 /s]
[m3 /s]
[1/s]
[s]
[s]
[m/s]
[m/s]
[-]
[Pa]
[Pa]
[Ns/m2 ]
[-]
[1/m]
[kg/m3 ]
[-]
Joukowskis equation
∆p = ρav0
Reduction due to valve closing time.
∆pred = ∆p
10.3
2ℓ
atv
for
tv >
2ℓ
a
Retardation of cylinder with inertia load
V2
V1
p2
p1
v0
m
FL
Pressure difference for retardation (v0 → 0)
p1max − p2min = v0
ps = const
10.4
pT = 0
Concentrated hydraulic inductance
p1 − p2 = L H
10.5
dq
dt
were LH =
ρℓ
A
Concentrated hydraulic capacitance
q1 − q2 = CH
dp
dt
were CH =
17
Aℓ
βe
r
2βe m
V1
10.6
Concentrated hydraulic resistance
p1 − p2 = RH q
10.7
were RH =

128ηℓ


RHℓ =


πd4




ℓ ρq0


 RHt = λ d A2

For laminar flow
For turbulent flow,
with linearization around the working
point with the flow q0
Basic differential equations on flow systems with parameter distribution in space
∂q
∂p
+ L0
+ R0 q|q|m = 0
∂x
∂t
∂p
∂q
+ C0
=0
∂x
∂t
Parameter values (per length unit)
independent of flow regime
C0 =
A
βe
L0 =
ρ
A
with laminar flow
R0ℓ =
128η
πd4
m=0
with turbulent flow
R0t =
10.8
0,1582 η 0,25 ρ0,75
d1,25 A1,75
m = 0,75
Speed of waves in pipes filled with liquid
1
a= √
=
L0 C0
s
βe
ρ
Graphical solution
F-wave
f-wave
F
f
∆p = B∆q
∆p = −B∆q
were B = L0 a
18
Boundary conditions:
At a valve
r
q
p
=α
q0
p0
were α = dimensionless area and p0 , q0 is stationary state.
At a pressure source with concentrated friction loss
ℓ ρq 2
p = p0 − ζ + λ
d 2A2
Solution with impedance method
Transfer matrices
cosh(λℓ)
P (s, ℓ)
=
− Z10 sinh(λℓ)
Q(s, ℓ)
P (s, 0)
Q(s, 0)
P (s, x)
Q(s, x)
P (s, x)
Q(s, x)
were
−Z0 sinh(λℓ)
cosh(λℓ)
=
cosh(λℓ)
1
Z0 sinh(λℓ)
=
cosh(λx)
− Z10 sinh(λx)
=
cosh(λ(ℓ − x))
1
Z0 sinh(λ(ℓ − x))
λ=
Z0 sinh(λℓ)
cosh(λℓ)
−Z0 sinh(λx)
cosh(λx)
p
(L0 s + R0 )C0 s
19
P (s, 0)
Q(s, 0)
P (s, ℓ)
Q(s, ℓ)
Z0 =
P (s, 0)
Q(s, 0)
Z0 sinh(λ(ℓ − x))
cosh(λ(ℓ − x))
r
P (s, ℓ)
Q(s, ℓ)
L0 s + R0
C0 s
11
11.1
A
T
D
L
P
V
a
d
fp
n
11.2
Pump pulsations
Nomenclature
[m2 ]
[s]
[m3 /varv]
[m]
[N/m2 ]
[m3 ]
[m/s]
[m]
[-]
[rev/s]
:the pipe’s cross-sectional area
:wave propagation time
:pump displacement
:pipe length
:pulsation amplitude
:volume
:wave propagation speed
:pipe diameter
:dim. free flow spectrum
:pump speed
ps
z
η
αp
β
ε
γ
ρ
τ
ω
:static pressure level
:the pump’s piston number
:dynamic viscosity
:dim.free cylinder volume
:bulk modulus
:the pump’s displacement
:dim.free dead volume
:density
:dim.free charging time
:angular frequency
[Pa]
[-]
[Ns/m2 ]
[-]
[Pa]
[-]
[-]
[kg/m3 ]
[-]
[rad/s]
System with closed end
Resonances in a pipe system with closed end are obtained at following frequency/-ies:
ω=
πk
T
k = 1, 2, 3, . . .
where
L
T =
a
a=
s
β
ρ
Figur 3: Amplitude for the resonances k = 1, 2, 3 and 4 in a system with closed end.
Flow disturbance from the pump rises at following frequencies:
ω = 2πnzj
j = 1, 2, 3, . . .
Under condition that the pumps flow disturbance frequencies coincide with the pipe system’s
resonances, can the resulting pressure amplitude at this frequency be calculated with following
equation:
r
P 2fp αp D
ρn
=
ps 3
dL
π ηzj
max
where
1+ε
2
αp =
+γ
3
1 + (τ j)
2
The dimensionless charging time τ is the relationship between the time you, due to the oil’s
compressibility, receive a back flow into a cylinder, and the total time period, i.e. the time
between two volumes in succession are charged. This parameter is very difficult to decide, a
typical value of τ is between 0,05 till 0,3. The higher values are referred to pumps designed for
low flow pulsations for example pumps with pressure relief grooves.
The dead volume γ has often the magnitude of 0,2, that is 20% of the effective cylinder volume.
Anti-resonances are received in a pipe system with closed end at following frequencies:
fp =
π(k + 21 )
k = 0, 1, 2, . . .
T
Under condition that the pump flow disturbance frequencies coincide with the pipe system’s
anti-resonances, can the resulting maximum pressure amplitude at this frequency be calculated
with following equation:
P 4fp αp Dn
=
ps πd2 a
max
ω=
20
Example: The above equations are used for following example. A constant pressure pump
presumed work against a closed valve. Following data is obtained:
d = 38 · 10−3
n = 25
z=9
D = 220 · 10−6
βe = 1,5 · 109
γ = 0,2
ε=0
η = 0,02
ρ = 900
τ = 0,25
[m]
[rev/s]
[m3 /rev]
[Pa]
[Ns/m2 ]
[kg/m3 ]
Four different pipe lengths between pump and valve is analysed:
L
L
L
L
= 1,45
= 1,91
= 2,15
= 2,90
m
m
m
m
Gives
Gives
Gives
Gives
resonances for j = 2, 4, 6, . . .
resonance for j = 3, 6, . . .
resonance for j = 4, . . .
resonance for j = 1, 2, 3, . . .
anti-resonances for j = 1, 3, 5, . . .
no anti-resonance
anti-resonances for j = 2, 6, . . .
no anti-resonance
In the table below shows the obtained relationship between pulsation’s pressure amplitude and the
system’s pressure level for respective disturbance harmonic. Note, for L = 1,91 m and L = 2,15 m
can some disturbance harmonics not be analysed with the equations above, since they don’t
coincide with any of the pipe’s resonances or anti-resonances. However, the amplitudes at these
frequencies are relative small because they don’t coincide with any of the pipe’s resonances. In
the table below, these values are in parenthesis.
Relative pulsation amplitude |P/ps |
11.3
j
f [Hz]
L = 1,45 m
L = 1,91 m
L = 2,15 m
L = 2,90 m
1
225
0,01
(0,01)
(0,01)
0,35
2
450
0,45
(0,01)
0,00
0,22
3
675
0,00
0,22
(0,01)
0,14
4
900
0,18
(0,00)
0,12
0,09
5
1125
0,00
(0,00)
(0,01)
0,05
6
1350
0,07
0,05
0,00
0,03
Systems with low end impedance (e.g. volume)
Resonances in a pipe system with low end impedance is obtained at following frequencies:
ω=
π(k + 21 )
T
k = 0, 1, 2, . . .
Figur 4: Amplitude for the resonances k = 0, 1, 2 and 3 for a system with low end impedance.
Anti-resonances in a pipe system with low end impedance is obtained at following frequencies:
ω=
πk
T
k = 1, 2, 3, . . .
The same equations as in previous section can be used here for calculation of maximum pressure
amplitude for the pulsations.
21
If a volume, which size is not infinite, is connected to the pipe system is a dislocation of the
line’s resonances from the values in above equations obtained. This dislocation can be calculated
according to following equation
Vω
1 π
− arctan
∆ω =
T 2
Aa
I.e. if a finite volume is used the resonance frequency is increased.
As example on this section, a high pressure filter is placed before the valve in the example with
the constant pressure pump. The valve is closed in this example too. The volume of the filter is
2 liters; other parameters are the same as the previous example except for the length of the line.
The following line lengths are analysed:
L = 1,47 m
L = 1,87 m
L = 3,00 m
Gives resonance for j = 3, 5, . . .
Gives resonance for j = 1, 4, . . .
Gives no resonance
anti-resonance for j = 2, 4, 6, . . .
anti-resonance for j = 3, 6, . . .
anti-resonance for j = 1, 2, 3, 4, 5, 6, . . .
Note, the volume is relative small and therefore the dislocation equation has to be used. When
the new resonance frequency is calculated, ”‘passningsräkning”’ has to be used. The method is
shown bellow for L = 1,47 m and k = 1.
∆ω = 430 rad/s
ω = 1810 rad/s
∆ω
=
340
rad/s
ω
= 1720 rad/s
π k + 21
∆ω
=
350
rad/s
ω
=
1730 rad/s
= 1350rad/s ⇒
ω=
T
The size of the pulsation amplitude in relation to the static system pressure is shown in the table
below. The values in parenthesis show, as in previous example, a more correct analyze of the
disturbance harmonic which can not be calculated with the equation given in this handbook.
Relative pulsation amplitude |P/ps |
j
f [Hz]
L = 1,47 m
L = 1,87 m
L = 3,00 m
1
225
(0,01)
(0,54)
0,01
2
450
0,00
(0,01)
0,00
3
675
0,28
0,00
0,00
4
900
0,00
0,14
0,00
5
1125
0,11
(0,00)
0,00
6
1350
0,00
0,00
0,00
22
12
12.1
A
Ah
Am
Ar
Au
Bm
Bp
Ce
Ci
Ct
Cq
D
FL
Gc
Go
Greg
Jt
Kc
Kp
Kq
Kreg
Kv
Mt
Np
S
TL
U
V
Hydraulic servo systems
Nomenclature
:piston area
:control piston area (3-port valve)
:amplitude margin
:piston area, rod side
(3-port valve)
:the open system’s transfer function
:viscous friction coeff. (motor)
:viscous friction coeff. (cylinder)
:external leakage flow coeff.
(cylinder/motor/pump)
:internal leakage flow coeff.
(cylinder/motor/pump)
:total leakage flow coeff.
:leakage flow coeff.
:displacement (motor/pump)
:external (load-) force on cylinder
:the closed system’s transfer function
:the open system’s transfer function
:controller transfer function
:total moment of inertia
(motor and load)
:flowpressure coeff. (servo valve)
:pressure gain (servo valve)
:flow gain (servo valve)
:control gain
:loop gain
:total mass (cylinder piston
and load)
:pump speed
:stiffness
:external (load-)moment on motor
:under lap
:volume
Vh
Vt
e
2
[m ]
kp
[dB]
p
pc
[m2 ]
ps
q
[Nms/rad]
qc
[Ns/m]
s
t
[m5 /Ns]
xv
xp
[m5 /Ns]
w
5
[m /Ns]
βe
[-]
δh
[m3 /rad]
ε0
[N]
ρ
θm
φp
ϕm
ω
2
[kg m ]
ωb
5
[m /Ns]
ωc
[Pa/m]
ωh
[m2 /s]
Re
Im
[m2 ]
[kg]
[rad/s]
[Nm]
[m]
[m3 ]
23
:control volume (3-port valve)
:total volume
:control error
:displacement gradient (pump)
:pressure
:control pressure (3-port valve)
:supply pressure
:flow (volume flow)
:centre flow
:Laplace operator (iω)
:time
:position (servo valve)
:position (cylinder piston)
:area gradient
:effective bulk modulus
:hydraulic damping
:control error (stationary)
:density
:angular position (motor)
:displacement angle (pump)
:phase margin
:angular frequency
:bandwidth
:crossing-out frequency
:hydraulic eigen frequency
:real part
:imaginary part
Tilläggsindex
0
e
m
p
v
t
L
:working point
:effective
:motor
:cylinder (piston), pump
:valve
:total
:load
[m3 ]
[m3 ]
[m3 /rad2 ]
[Pa]
[Pa]
[Pa]
[m3 /s]
[m3 /s]
[rad/s]
[s]
[m]
[m]
[m]
[Pa]
[-]
[kg/m3 ]
[rad]
[rad]
[◦ ]
[rad/s]
[rad/s]
[rad/s]
[rad/s]
12.2
Introduction
The servo technical section discusses following system:
• valve controlled cylinder
• valve controlled motor
• pump controlled cylinder
• pump controlled motor
As example in this section will a position servo of the type valve controlled cylinder in a constant
pressure system be used. In this example is the servo valve a 4-port valve with negligible dynamic
and the cylinder is symmetric. See figure 5.
xp
xp ref
+
Greg
V1 A1
A2 V2
p1
p2
FL
Mt
Bp
xv
pT = 0
ps = constant
Figur 5: Lägesservo: ventilstyrd cylinder i konstanttryckssystem.
A position servo looks, in general, out as follow:
∆FL (∆TL)
∆Xp ref (∆Θm ref)
∆E
+
-
Controller
∆X v (∆Φp)
∆Xp (∆Θm)
Hydraulic system
The special case when the hydraulic system consists of a valve controlled cylinder becomes the
block diagram as:
∆FL
Kce
A2p
∆Xp ref
+
∆E
-
12.3
Controller
∆X v
(
1+
Kq
Ap
Vt
4βeKce
+
s
(
s
Hydraulic
system
)
∆Xp
1
s2
ωh2
)
+ 2δh s + 1
ωh
The servo valve’s transfer function
All the transfer functions in this section are applied when the load spring constant K is negligible
and its viscous friction Bp and Bm respectively is small (can often be set to 0). The direction
24
dependent friction coefficient Cf is neglected also in the motor case.
Valve controlled systems
The dynamic of the servo valve in the valve controlled systems is assumed to be negligible compared
to the system. Following is valid for the servo valve (see also section 12.4):
4-ports servo valve
∂qL
Kq =
∂xv
∂qL
Kc = −
∂pL
∂pL
Kq
=
Kp =
Kc
∂xv
3-ports servo valve
∂qL
Kq =
∂xv
∂qL
Kc = −
∂pc
∂pc
Kq
=
Kp =
Kc
∂xv
the servo valve’s flow gain
the servo valve’s flowpressure coefficient
the servo valve’s pressure gain
The load flow and load pressure through a 4-port valve controlled symmetric cylinder/motor is
defined according to following expression:
qL = (qL1 + qL2 )/2
pL = p1 − p2
Pump controlled systems
The dynamic of the pump in the pump controlled systems is assumed to be negligible compared
to the system. The pump’s ideal flow is:
qp ideal = εp Dp Np = kp φp Np
Linearised and Laplace transformed equations which describes the dynamic of the following hydraulic systems are presented in section 12.5.
25
xp
V1 A1
A2 V2
p1
p2
q1
xp
FL
Mt
Vh Ah
Mt
pc
Bp
q2
FL
Ar
Bp
xv
xv
ps = const
pT = 0
ps = const
a. Valve controlled symmetric cylinder (4-port valve)
pT = 0
b. Valve controlled asymmetric cylinder (3-port valve)
TL Bm
TL B m
Jt
θm
Dm
p1
Jt
xp
V1 A1
p2
V1
q1
Bp
φ
xv
ps = const
pT = 0
Dm
Mt
p1
V2
q2
θm
FL
A2
p1
V1
p2 = const
φ
M
c. Valve controlled motor
M
d. Pump controlled cylinder
e. pump controlled motor (transmission)
Figur 6: Different types of servo systems.
12.4
Servo valve
4-port zero lapped valve
Load flow:
s 1
xv
ps −
pL
qL = Cq wxv
ρ
|xv |
Centre flow:
qc = 0
Zero coefficients:
Kq0
ideal
r
ps
= Cq w
ρ
qL0 = 0
pL0 = 0
p2 =
const
Kc0 = 0 ideal
xv0 = 0
26
Kp0 = ∞ ideal
4-port under lapped valve
Load flow:
Centre flow:
Zero coefficients:
r
r
ps − pL
ps + pL
qL = Cq w (U + xv )
for |xv | ≤ U
− (U − xv )
ρ
ρ
r
ps
qc = 2Cq wU
ρ
r
r
2ps
ps
1
Kc0 = Cq wU
Kp0 =
Kq0 = 2Cq w
ρ
ps ρ
U
qL0 = 0
pL0 = 0
xv0 = 0
3-port zero-lapped valve

r
2


(ps − pc ) då xv ≥ 0
Cq wxv



ρ
Load flow:
qL =
r


2


 Cq wxv
pc
då xv ≤ 0
ρ
Centre flow:
qc = 0
ideal
r
ps
= Cq w
ρ
Zero coefficients:
Kq0
qL0 = 0
pc0 =
Kc0 = 0 ideal
ps
2
Kp0 = ∞ ideal
xv0 = 0
3-port under lapped valve
Load flow:
Centre flow:
Zero coefficients:
12.5
r
r
2(ps − pc )
2pc
qL = Cq w (U + xv )
for |xv | ≤ U
− (U − xv )
ρ
ρ
r
ps
qc = Cq wU
ρ
r
r
ps
ps
1
Kc0 = 2Cq wU
Kp0 =
Kq0 = 2Cq w
ρ
ps ρ
U
ps
qL0 = 0
pL0 =
xv0 = 0
2
The hydraulic system’s transfer function
Valve controlled symmetric cylinder with mass load (4-port valve)
Vt
Kce
Kq
∆Xv − 2 1 +
s ∆FL
Ap
A
4βe Kce
2p
∆Xp =
s
s
s
+1
+ 2δh
2
ωh
ωh
where
s

4βe A2p



if V1 ≈ V2


Vt Mt
ωh = s

βe A2p 1

1


if V1 6= V2
+


Mt
V1
V2
Cep
Kce = Kc + Cip +
Vt = V1 + V2
2
27
Kce
δh =
Ap
r
Bp
βe M t
+
Vt
4Ap
Ap = A1 = A2
r
Vt
βe M t
Valve controlled asymmetric cylinder with mass load (3-port valve)
where
ωh =
s
βe A2h
Vh Mt
Vh
Kq
Kce
∆Xv − 2 1 +
s ∆FL
Ah
A
βe Kce
2h
∆Xp =
s
s
+1
+ 2δh
s
ωh2
ωh
r
r
Kce βe Mt
Bp
Vh
δh =
+
2Ah
Vh
2Ah βe Mt
Kce = Kc + Cip
Valve controlled motor with moment of inertia
Kq
Vt
Kce
∆Xv − 2 1 +
s ∆TL
Dm
D
4βe Kce
2m
∆Θm =
s
s
+1
+ 2δh
s
ωh2
ωh
s
r
r
2
Kce βe Jt
Bm
4βe Dm
Vt
where ωh =
δh =
+
Vt Jt
Dm
Vt
4Dm βe Jt
Cem
Kce = Kc + Cim +
Vt = V1 + V2 , V1 = V2
2
Pump controlled cylinder with mass load
where
ωh =
s
βe A2p
V0 Mt
V0 = V1
Ct = Cit + Cet
kp Np
V0
Ct
∆φp − 2 1 +
s ∆FL
Ap
A
βe Ct
2 p
∆Xp =
s
s
s
+1
+ 2δh
ωh2
ωh
r
r
βe M t
V0
Ct
Bp
δh =
+
2Ap
V0
2Ap βe Mt
Ap = A1 = A2
= Cip(iston) + Cip(ump) + Cep(iston) + Cep(ump)
Pump controlled motor with moment of inertia (transmission)
∆Θm
where
ωh =
s
V0 = V1
12.6
2
βe D m
V0 Jt
V0
Ct
kp Np
∆φp − 2 1 +
s ∆TL
Dm
D
βe Ct
2 m
=
s
s
s
+
1
+
2δ
h
ωh2
ωh
r
r
βe Jt
V0
Ct
Bm
δh =
+
2Dm
V0
2Dm βe Jt
Ct = Cit + Cet = Cip + Cim + Cep + Cem
The servo stability
Feedback systems can become instable if the feedback is incorrect dimensioned. In this case we
study a position servo with proportional feedback Greg = Kreg . The open loop transfer function
become:
Kv
Au = Greg Go = 2
s
s
+
1
+
2δ
s
h
ωh2
ωh
28
where Go is the transfer function which describes the hydraulic system’s output signal (cylinder
position) as function of the hydraulic system’s input signal (valve position) when the disturbance
signal (∆FL ) is zero. The steady state loop gain Kv (also called the velocity coefficient) is:
Kv =
Kv =
Kv =
Kv =
Kv =
Kq
Kreg
Ap
Kq
Kreg
Ah
Kq
Kreg
Dm
kp Np
Kreg
Ap
kp Np
Kreg
Dm
valve controlled symmetric cylinder (4-port valve)
valve controlled asymmetric cylinder (3-port valve)
valve controlled motor
pump controlled cylinder
pump controlled motor
Then, the open system’s Bode-diagram for a position servo become as figure 7.
1
270
ωc ≈ Kv
180
|Au| = 1
0
|Au| [-], (solid line)
10
Am
90
Kv / 2δhωh
-1
10
0
ωh
-90
ϕm
-2
10
-180
-3
10
-1
10
0
1
10
10
Angular frequency [rad/s]
phase(Au) [degrees], (dashed line)
10
-270
2
10
Figur 7: The open system’s transfer function for a position servo. Amplitude, (solid line) and phase, (dashed).
Stability condition
A stable system is obtained when
• the amplitude margin Am > 0 dB at −180◦ phase shift. If the phase intersects −180◦ more
than one time the Nyquist diagram is needed.
• the phase margin ϕm > 0◦ at 0 dB amplitude. If the amplitude curve intersects 0 dB more
than one time the Nyquist diagram is needed.
For a proportional position servo with a feedback is, for the hydraulic, the amplitude margin the
critical stability margin. It means that a stable system needs the open loop gain |Au | to be <= 1
(0 dB) when the phase shift is <= −180◦.
Figure 7 shows that the stability condition becomes
Kv
<1
2δh ωh
Stability margins
Position servo (with Bode-diagram according to figure 7) amplitude margin can be written as:
Kv 10
Am = −20 log [dB]
−2δh ωh The following margins should be used when the control parameters shall be dimensioned in a
hydraulic system with a feedback.
29
amplitude margin: Am ≈ 10 dB
phase margin: ϕm ≥ 45◦
The system’s critical working condition
Since hydraulic systems are non-linear systems, the stability margin will become different in
different working condition. From the figure for the open system’s transfer function Au the
stability margin become worst when both ωh and δh are low and steady state loop gain Kv is big.
This happened for a valve controlled symmetric cylinder when
• Cylinder piston is centered (xp = 0), i.e. V1 = V2 =
Vt
2 .
ωh is minimised.
• The servo valve is closed (xv = qL = 0), i.e. when Kc = Kc0 . Kce and consequently δh is
minimised.
√
• Cylinder piston is out balanced, i.e. when pL = 0. Kq , which is proportional to ∆ps − ∆pL ,
is maximised and consequently also Kv (proportional to Kq ).
With similar discussion, the critical working condition can be decided for other systems.
In practical dimensioning, the hydraulic damping is often set to δh ≈ 0, 1.
12.7
The servo’s response – bandwidth
The bandwidth (ωb ) for the closed loop system specifies how high frequency the servo’s output
signal can follow a sinusoidal input signal, when the disturbance ∆FL = 0, without:
• the gain lowers more than 3 dB (29,3 %)
• or the phase shift become more than −90◦
If the feedback is −1, the closed loop system’s transfer function for a position servo with any of
the earlier described hydraulic systems becomes:
Gc =
1
1
2δ
1
h
s2 +
s+1
s3 +
2
K v ωh
K v ωh
Kv
∆Xp
for a position servo with a valve controlled cylinder.
∆Xp ref
1
10
270
ωb,1
|Gc| = 0.71 ⇔ -3 dB
|Gc| [-], (solid line)
180
|Gc| = 1
0
10
90
0
-1
10
-90
-180
ωb,2
-2
10
-1
10
0
1
10
10
Angular frequency [rad/s]
phase(Gc) [degrees], (dashed line)
where Gc =
Au
=
1 + Au
-270
2
10
Figur 8: The closed loop system’s transfer function for a position servo. Amplitude (solid) and phase (dashed).
30
The closed loop system’s transfer function can also be written as
Gc = s
+1
ωb
1
s
s2
+
1
+
2δ
nc
2
ωnc
ωnc
If δh and Kv /ωh is small
ωnc ≈ ωh
ωb ≈ K v
Kv
2δnc ≈ 2δh −
ωh
12.8
The hydraulic system’s and the servo’s sensitivity to loading –
stiffness
Cylinder - respective motor position sensitivity to disturbance force ∆FL or a disturbance torque
∆TL is described with its stiffness S. When the stiffness is studied is all other input signals
assumed to be constant (∆Insignal = 0). The stiffness is defined as
S=
∆FL
∆Xp
or
S=
∆TL
∆Θm
The hydraulic system’s transfer function can be found in section 12.5.
As example on a system without respective with a feedback, the stiffness for the valve controlled
(4-port valve) symmetric cylinder is decided. The same approach is used when the stiffness is
decided for the other hydraulic.
System without feedback
For a hydraulic system without feedback, the stiffness for the valve controlled cylinder is calculated
as
2
s
s
+1
+ 2δh
s
ωh2
ωh
1
∆FL
=
S=
=−
∆Xp
s
Kce
∆Xp
1
+
∆FL
A2p
ωs
where
ωs =
4βe Kce
= {if Bp isneglected} = 2δh ωh
Vt
The transfer function for the stiffness is shown in figure 9.
10
10
9
|S| [N/m]
10
8
10
qh!h Ap!h 2
2
Kce
7
10
ωs
2
1+ (! )
s
ωh
6
10
-1
10
0
1
10
10
Angular frequency [rad/s]
2
10
Figur 9: The transfer function for the stiffness for the non-feedback valve controlled cylinder.
31
System with feedback
For a system with feedback with the feedback −1 and proportional gain of the control error is
following stiffness obtained for a valve controlled cylinder(∆Xp ref are set to 0).
1
2δh 2
1
s +
s+1
s3 +
2
K v ωh
K v ωh
K
1
∆FL
v
=
S=
= −Kv
∆Xp
s
Kce
∆Xp
1+
∆FL
A2p
ωs
where
ωs =
4βe Kce
= {if Bp isneglected} = 2δh ωh
Vt
The transfer function for the stiffness in the position servo is shown in figure 10.
10
10
9
|S| [N/m]
10
8
10
2
KvAp / Kce
7
10
ωs
ωh
6
10
-1
10
0
1
10
10
Angular frequency [rad/s]
2
10
Figur 10: The transfer function for the stiffness for the valve controlled cylinder with proportional feedback.
32
12.9
The servo’s steady state error
The control error e(t) in a servo system is defined as the difference between the output value and
the input value when ∆Disturbance signal = 0.
According to end value theorem, the steady state control error will be:
ε0 = lim e(t) = lim s∆E(s)
t→∞
s→0
where the error gets the following expression if the feedback is −1:
∆E(s) = ∆Xpref − ∆Xp = ∆Xpref
1
1 + Au
The end value theorem is only usable on an asymptotic stable system, i.e. if the output signal
has a finite limit value. For all systems can the transient be studied in the time domain (inverse
transformation, see section 12.10).
The input signal is a step
If the input signal ∆Xpref respective ∆Θmref is a step with the amplitude A becomes the input
signal A/s in frequency domain. The end value theorem becomes
ε0 = lim s
s→0
1
A
A
A
=
→
=0
s 1 + Au
1 + lim Au
∞
s→0
Practical, the steady state error does never become 0, because of the components which are
included in the control loop do not have ideal characteristic.
The input signal is a ramp
If the input signal ∆Xpref respective ∆Θmref is a ramp A · t (i.e. the speed A is desired), becomes
the input signal A/s2 . The steady state position error becomes
ε0 = lim s
s→0
12.10
1
A
A
=→
2
s 1 + Au
Kv
Control technical resources
Linearize
Non-linear differential equations are linearized around a working point with the first term of the
Taylor series according to
∆f
= ∆f (x10 , x20 , . . . , xn0 ) =
0
n
X
∆xj
j=1
∂f (x1 , x2 , . . . , xn )
∂xj
(x10 ,x20 ,...,xn0 )
where the working point is assumed to have stationary working conditions.
33
Laplace transformation
The following Laplace transformation table translates equations in the time domain to equations
in frequency domain (rad/s) and vice versa.
Z ∞
Definition
L{f (t)} = F (s) =
e−st f (t)dt
0
Function in time domain
Transform to frequency domain
f (t)
F (s)
af (t) + bg(t)
a, b constants
aF (s) + bG(s)
f (t − a)
a≥0
e−as F (s)
f (at)
a>0
1
F
a
tn f (t)
n = 0, 1, 2, . . .
(−1)n F (n)(s)
f (t)
t
f (t)
lim
exists
t→0+ t
Z
e−at f (t)
a constant
F (s + a)
f (t − a)H(t − a)
a > 0, H(t)
s
a
∞
F (u)du
s
0
then t < 0
1
otherwise
f ′ (t)
e−as F (s)
sF (s) − f (0)
f (n) (t)
sn F (s) −
n = 0, 1, 2, . . .
n
X
sn−k f (k−1) (0)
k=1
Rt
0
F (s)
s
f (u)du
f (t) ∗ g(t) =
Z
t
f (u)g(t − u)du =
0
Z
t
g(u)f (t − u)du
F (s)G(s)
0
Function in time domain
Transform to frequency domain
tn
n = 0, 1, 2, . . .
eat
a constant
tn eat
n = 0, 1, 2, . . .
n!
sn+1
1
s−a
n!
(s − a)n+1
s
s2 + a 2
a
s2 + a 2
s−a
(s − a)2 + b2
b
(s − a)2 + b2
s2 − a 2
(s2 + a2 )2
2s3
2
(s + a2 )2
2as2
2
(s + a2 )2
2a3
2
(s + a2 )2
cos at
sin at
eat cos bt
eat sin bt
t cos at
2 cos at − at sin at
sin at + at cos at
sin at − at cos at
34
Re(s) > 0
Re(s) > Re(a)
Re(s) > Re(a)
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Re(s) > |Im(a)|
Block diagram reduction
Reduction
System
u
Equivalent system
+
±
1. Feedback
y
G
u
H
u
2. Series connection
G
H
u
3. Parallel connection
G
+
H
+
u
4. Move branch point
after a block
G
y
u
y
u
y
u
u
5. Move branch point
in front of a block
y
G
u
7. Move summation point
in front of a block
u
−
y
+
G
+
1
G
−
+
u
G
G
−
G
u
+
−
1
G
−
y
+
+
−
z
35
G
z
y
z
u
y
+
z
y
z
u + y −z
+
8. Shift summation order
y
G
z
y
u
y
G
y
G
y
+
y
G+H
u
y
6. Move summation point
after a block
y
GH
u
u
y
G
±
1 GH
+
u
u + y −z
+
+
−
z
u + y −z
Bode diagram
For the transfer functions
s
s
s
1+
··· 1+
1+
K
z1
z2
z
m
1.
G(s) = p s
s
s 1+ s
1+
··· 1+
p1
p2
pn
2.
G(s) =
(real numbers)
1
s
s2
+ 2δ0
+1
ω02
ω0
δ0 < 1 (complex numbers)
the amplitude curve becomes
1.
2.
log |G(iω)| =
iω iω log K − p log |ω| + log 1 + + log 1 + + · · ·
z1
z2
iω iω iω iω − log 1 + − log 1 + − · · · − log 1 + · · · + log 1 +
zm p1
p2
pn
1
|G(iω)| = s
2
2
ω2
ω
1− 2
+ 2δ0
ω0
ω0
and the phase becomes
1.
arg G(iω) =
−p90◦ + arctan
ω
z1
+ arctan
ω
z2
+ · · · + arctan
ω
zm
−
ω
ω
ω
− arctan
− · · · − arctan
p1
p2
pn



ω



 2δ0 ω0 





−
arctan
då 0 ≤ ω ≤ ω0



ω2 



1
−


ω02

◦
arg G(iω) = −90
då ω = ω0





ω



 2δ0 ω0 


◦
 då ω > ω0

−180 + arctan 

 ω2





−
1
ω02
arctan
2.
Nyquist diagram
If the transfer function is plotted direct in the complex domain, the Nyquist diagram is obtained
which is more usable than the Bode diagram. From the Bode diagram can Nyquist diagram be
constructed in following way:
36
1
270
180
0
|Au| [-], (solid line)
10
90
|Au(iωi)|
-1
10
0
-90
arg(Au(iωi))
-2
10
-180
ωi
-3
10
-1
10
0
1
2
10
10
Angular frequency [rad/s]
phase(Au) [degrees], (dashed line)
10
-270
10
0,50
Im(Au)
0,25
0,00
|Au(iωi)|
arg(Au(iωi))
-0,25
-0,50
-1,50
-1,00
-0,50
Re(Au)
0,00
0,50
Linear systems - linearized system
Under presumption that Au does not have poles in the right side are following valid:
• For a feedback system shall be stable must not the open loop system’s transfer function
enclose Re(Au ) = −1 in the Nyquist diagram.
• Mnemonic rule: Pull the ”rope” downward. If Re(Au ) = −1 follows - the system is instable!
Non-linear systems
With non-linear systems can a host of phenomenon occur due to occurrence of play, hysteresis in
the system etc despite the system is seemingly stable. A analyse method for investigation of the
stability is the descriptive functions where the open loop system’s transfer function is divided in
a linear part and a non-linear part according to
Au = Glinear Gnon−linear
which results in the stability condition
Glinear Gnon−linear = −1
By plotting Glinear and (−1/Gnon−linear) in the Nyquist diagram can the stability be investigated.
The intersection point gives in many cases the frequency and the amplitude for the self oscillation.
See other literature for determination of the non-linear transfer function.
37
13
13.1
V
V1
V2
Vc
Vg
Vℓ
Vt
p
p0
p1
p2
u
n
ys
yt
Hydraulic fluids
Nomenclature
:volume
:start volume (secant)
:end volume (secant)
:volume (reservoir)
:volume (gas)
:volume (fluid)
:total volume
:absolute pressure
:absolute pressure at NTP (= 0,1 MPa)
:start pressure (secant)
:end pressure (secant)
:flow velocity in x-led
:polytrophic exponent
:correction coefficient (secant)
:correction coefficient (tangent)
[m3 ]
[m3 ]
[m3 ]
[m3 ]
[m3 ]
[m3 ]
[m3 ]
[Pa]
[Pa]
[Pa]
[Pa]
[m/s]
[-]
[-]
[-]
x0
ν
ρ
βe
βc
βg
βℓ
βt
βs
βbt
βbs
τ
η
:amount of air in the oil
(gas volume/total volume at
normal state, NTP)
:kinematic viscosity
:density
:effective bulk modulus
:bulk modulus (reservoir)
:bulk modulus (gas)
:bulk modulus (fluid)
:bulk modulus with no air in the
:bulk modulus with no air in the
:bulk modulus with air in the oil
:bulk modulus with air in the oil
:skjuvspänning
:dynamic viscosity
u(y)
y
Definition, η
dynamic viscosity
for Newton fluid
τ =η
oil (tangent)
olja (secant)
(tangent)
(secant)
du
dy
x
η
ρ
ν=
Kinematic viscosity
Bulk modulus for oil with no air
Tangent value, is used at small changes
βt = −V
dp
dV
βt is shown in figure 11. Most of the normal hydraulic oils have bulk modulus between the Naften
based oil and Paraffin based oil.
Bulk modulus, tangent [MPa]
2500
ASME-36 Naften based oil
ASME-31 Paraffin based oil
T = 0°
25°
2000
50°
75°
100°
1500
1000
0
10
20
30
Pressure [MPa]
40
Figur 11: Tangent value of the bulk modulus for oil with no air.
38
50
[-]
[m2 /s]
[kg/m3 ]
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[N/m2 ]
[Ns/m2 ]
Secant value, is used at big changes from normal pressure
(p1 = 0, V1 ) till (p2 , V2 )
p2
βs = −V1
V2 − V1
βs is shown in figure 12. Most of the normal hydraulic oils have bulk modulus between the Naften
base oil and Paraffin base oil.
Bulk modulus, secant [MPa]
2500
ASME-36 Naften based oil
ASME-31 Paraffin based oil
T = 0°
2000
25°
50°
75°
1500
100°
1000
0
10
20
30
Pressure [MPa]
40
50
Figur 12: Secant value of bulk modulus for oil with no air.
Bulk modulus for oil with air
Tangent value
Simplified model and can be used when x0 ≤ 0,1.
βbt = yt βt
där
1
yt =
x0
βt
1+
np (1 − x0 )
p0
p
1
n
yt is shown in figure 13 for different amount of air in the oil, x0 , at the special case:
polytrophic exponent n = 1,4
βt = 1500 + 7,5∆p [MPa]
1,0
x0 = 0,1
Compensator factor yt
0,8
x0 = 0,05
x0 = 0,02
0,6
x0 = 0,01
x0 = 0,005
0,4
p0 = 0,1 MPa
n = 1,4
0,2
βt = 1500 + 7,5p MPa
0,0
0
10
20
30
Pressure [MPa]
40
Figur 13: Correction for the air included in oil, tangent value.
39
50
Secant value
Simplified model and can be used when x0 ≤ 0,1.
βbs = ys βs
där
1
ys =
1 − x0 +

βs x0 
1−
(p − p0 )
p0
p
1
n


ys is shown in figure 14 for different amount of air in the oil, x0 , at the special case:
polytropexponent n = 1,4
βs = 1500 + 3,7∆p [MPa]
Compensator factor ys
1,0
βs = 1500 + 3,7p MPa
x0 = 0,005
p0 = 0,1 MPa
0,8
n = 1,4
x0 = 0,01
0,6
x0 = 0,02
x0 = 0,05
0,4
0,2
x0 = 0,1
0,0
0
10
20
30
Pressure [MPa]
40
50
Figur 14: Correction for the air included in oil, secant value.
Effective bulk modulus, βe
The effective bulk modulus, βe ,
is defined as
∆Vt
1
=
βe
Vt ∆p
Total initial volume:
At compression:
Gas
Vg
Vt = Vℓ + Vg
∆Vt = −∆Vℓ − ∆Vg + ∆Vc
Fluid
Vl
where ℓ, g och c refer to fluid, gas respective reservoir.
In general, the bulk modulus is calculated as (secant
value)
1
Vg 1
Vℓ 1
1
=
+
+
βe
Vt βg
Vt βℓ
βc
Gas
−∆Vg
For oil with no air:
Fluid
1
1
1
=
+
βe
βℓ
βc
∆Vc
40
∆Vt
Specific heat capacity
2500
3
Specific heat [J/kg K]
ρ20 = 840 kg/m
2300
860
880
900
920
940
960
2100
1900
1700
0
50
100
Temperature [degrees C]
150
Thermal conductivity ability
0,14
3
Thermal conductivity [W/m K]
ρ20 = 840 kg/m
860
880
0,13
900
920
940
0,12
960
0,11
0
50
100
Temperature [degrees C]
41
150
14
14.1
Pneumatic
Nomenclature
A0
A12
A23
Ae
Cd
Ci
Cs
K
:min. cross-section area of the orifice
:effective entrance area
:effective exit area
:effective orifice area (Cd A0 )
:flow coefficient
:C-value for component i
:C-value for system
:constant
Kt
N
R
T0
T1
T3
Tv
:temperature correction ( T0 /T1 )
:parameter
:gas constant (287 for air)
:reference temperature (NTP)
:upstream total temperature
:downstream total temperature
:total temperature i volume
14.2
[m2 ]
[m2 ]
[m2 ]
[m2 ]
[-]
[-]
[-]
p
[ kgK/J]
p
[-]
[-]
[J/kg K]
[K]
[K]
[K]
[K]
b
bi
bs
ṁ
p1
p2
p3
pv
q
t
α
κ
ω
τ
:critical pressure ratio
:b-value for component i
:b-value for system
:mass flow
:upstream total absolute pressure
(= static + dynamic pressure)
:downstream static absolute pressure
:atmospheric pressure (0,1 MPa)
:atmospheric pressure in volume
:volume flow
:time
:parameter
:isentropic exponent
:parameter
√
At
)
:dimension free time (= RT
V
Stream through nozzle
According to the thermodynamic, the mass flow ṁ through a nozzle can be written as
v
u
u κ + 1
u
p1 Cd A0 KN
2
tκ
κ−1
√
ṁ =
where
K=
R κ+1
T1



1





v


u


2 κ + 1
u
u p2 κ
p2
κ
N= u
−
u

p
p1
1

u

u


κ + 1

u
t
2
κ
−
1

κ−1



2
κ+1
critical pressure ratio:
b=
p2
p1
∗
=
p2
≤
for
p1
p2
p1
∗
p2
>
p1
p2
p1
∗
for
κ
2
κ−1
κ+1
With b- and C-value hold for volume flow q following expression at NTP
q = p1 Kt Cω


1




v

u
2
p
2
−b
ω= u
u


 p1
u


 t1 −  1 − b 



42
for
p2
≤b
p1
for
p2
>b
p1
[-]
[-]
[-]
[kg/s]
[Pa]
[Pa]
[Pa]
[Pa]
[m3 /s]
[s]
[-]
[-]
[-]
[-]
14.3
Series connection of pneumatic components
p1
b1
p2 b2
p3
bi
bn
pn+1
q1
T1
C1 q2 C2 q3
Ci
Cn
qn+1
bs ,Cs
Figur 15: Series connection of pneumatic components with b- and C-values
On condition that
• every component can be described with q = p1 Kt Cω
• b- and C-value is known for every component
• absolute static pressure after one component is equal to the absolute total pressure before
the next component
• the entrance temperature holds for all system
• every component have the same mass flow (q1 = q2 = · · · = qn )
qs = p1 Kt Cs ω
med Kt =


1




v
p
2
u
n+1
− bs
ω= u
u


u 1 −  p1




t


1
−
b
s


r
T0
T1
för
pn+1
≤ bs
p1
för
pn+1
> bs
p1
Case A If b- and C-value is about the same
n
X 1
1
=
Cs3
Ci3
i=1
bs = 1 − Cs2
n
X
1 − bi
i=1
Ci2
Case B In cases of, the components’ characteristics show large divergence,
the components sequence have to be considered (gradual reduction)
Calculation sequence:
α12 =
C1
C2 b1
α12 < 1
Critical pressure drop first over component 1, and because of
the decrease of pressure ratio as well in component 2.
α12 = 1
Both components is critical at the same time
α12 > 1
Critical pressure drop only in component 2

C1


s

2



1 − b1

2
α12 b1 + (1 − b1 ) α12 +
−1
C12 =
b1


2
 C2 α12

1 − b1

2 +

α

12
b1
43
for α12 ≤ 1
for α12 > 1
α13
14.4
1 − b2
1 − b1
+
C12
C22
C12
=
osv . . .
C3 b12
2
b12 = 1 − C12
Parallel connected pneumatic components
p1
q1
T1
b1
b2
bn
C1
C2
Cn
bs ,Cs
p2
Figur 16: Parallel connected pneumatic components with b- and C-values
qs = p1 Kt Cs ω
med Kt =


1





v
p
2
u
2
− bs
ω= u
u
p1


u1 − 




t


1
−
b
s


T0
T1
p2
≤ bs
p1
forr
for
r
p2
> bs
p1
For systems with the same further line (see figure 16) relates to
Cs =
n
X
n
Ci
i=1
14.5
X Ci
Cs
√
√
=
1 − bs
1 − bi
i=1
Parallel- and series connected pneumatic components
For system with separate further lines are dealt as series links, after which the total flow is obtained
as the sum of all the partial flows in every series links.
14.6
Filling and emptying of volumes
Assumptions
• isotherm process (T = T1 = Tv = T3 )
dAe
dV
dp1
dp3
dT
= 0,
= 0,
= 0,
= 0,
=0
• stationary conditions
dt
dt
dt
dt
dt
• p3 = atmospheric pressure
• p1 and pv are absolute pressure
• A12 and A23 are effective areas
44
Charging a volume
The diagrams below shows pv as a function of dimensionless time τ =
A23
as parameter.
relation of
A12
A12
p1
T1
A23
pv
Tv
p3
T3
V
Pressure pv [MPa absolute]
0,6
√
RT A12 t
with the area
V
A23/A12 = 0,0
0,5
0,5
1,0
0,4
1,5
0,3
2,0
0,2
0,1
(a) The downstream pressure p3 = 0,1 MPa absolute.
0,0
0,0
0,5
1,0
1,5
2,0
2,5
tao [-]
(b) The upstream pressure p1 = 0,6 MPa (absolute)
1,1
0,5
0,9
0,8
1,0
0,7
1,5
0,6
2,0
0,5
0,4
0,3
0,2
0,1
0,0
0,0
A23/A12 = 0,0
2,0
Pressure pv [MPa absolute]
Pressure pv [MPa absolute]
1,0
2,2
A23/A12 = 0,0
1,8
0,5
1,6
1,0
1,4
1,5
1,2
2,0
1,0
0,8
0,6
0,4
0,2
0,5
1,0
1,5
2,0
0,0
0,0
2,5
0,5
tao [-]
1,0
1,5
2,0
2,5
tao [-]
(d) The upstream pressure p1 = 2,1 MPa (absolute)
(c) The upstream pressure p1 = 1,1 MPa (absolute)
Figur 17: Charging of the volume V .
45
Discharging a volume
The diagrams below shows pv as a function of dimensionless time τ =
A12
as parameter.
relation of
A23
√
RT A23 t
with the area
V
A12
p1
T1
A23
pv
Tv
p3
T3
V
Pressure pv [MPa absolute]
0,6
A12/A23 = 1,0
0,5
0,8
0,4
0,6
0,3
0,4
0,2
0,2
0,0
0,1
(a) The downstream pressure p3 = 0,1 MPa absolute.
0,0
0,0
1,0
2,0
3,0
4,0
5,0
tao [-]
(b) The upstream pressure p1 = 0,6 MPa (absolute)
1,1
2,2
2,0
A12/A23 = 1,0
0,9
Pressure pv [MPA absolute]
Pressure pv [MPa absolute]
1,0
0,8
0,8
0,6
0,7
0,6
0,4
0,5
0,4
0,2
0,3
0,2
0,0
0,1
0,0
0,0
1,0
2,0
3,0
4,0
A12/A23 = 1,0
1,8
1,6
0,8
1,4
0,6
1,2
1,0
0,4
0,8
0,2
0,6
0,4
0,0
0,2
0,0
0,0
5,0
1,0
tao [-]
2,0
3,0
4,0
5,0
tao [-]
(c) The upstream pressure p1 = 1,1 MPa (absolute)
(d) The upstream pressure p1 = 2,1 MPa (absolute)
Figur 18: Discharging the volume V .
For both charging and discharging a volume
If p1 is between the assumed levels in the diagrams can the time be calculated with linear interpolation of the two diagram according to the equation below ((p1 in MPa)).

 t5 + p1 − 0,6 (t10 − t5 )
for 0,6 ≤ p1 ≤ 1,1 MPa
0,5
t=

t10 + (p1 − 1,1)(t20 − t10 ) for 1,1 ≤ p1 ≤ 2,1 MPa
46
Appendix A
Symbols for hydraulic diagrams
Correspond to the international standard CETOP RP3 and the Swedish SMS 712. It is specified
when the two standards differ.
General symbols...................................................................
Mechanical elements.............................................................
Pipes and connections..........................................................
Control systems....................................................................
Pumps and Motors...............................................................
Cylinders..............................................................................
Directional control valves.....................................................
Check valves or non-return valves........................................
Pressure control valves................................´.......................
Flow control valves...............................................................
Components for cooling, filtering, energy storage etc..........
Energy sources.....................................................................
Measurement equipments.....................................................
47
48
48
48
49
49
51
52
53
54
55
56
57
57
General symbols
1)
2)
Flow direction
1)Hydraulic
2)Pneumatic
Variability
Joined
nents
compo-
Components belonging to one assembly or
functional group
Mechanical elements
D
D<5E
Shaft, lever arm,
bar, rod piston
Rod piston can be drawn with a single line
Rotational shaft
One respective two rotational directions
Spring
Pipes and connections
Solid line = main pipe
Dotted line with L>10E = control pipe
Dotted line with l<5E = drain pipe
(E = line width)
E
L<10E
Line
E
L<10E
Flexible
hose
pipe,
The symbol is used mainly for movable parts
d
1)
1)
3)
2)
2)
1)
2)
3)
4)
Pipe connection
d = 5E (E = line width)
Air-drain
air outlet
1) Air-drain for hydraulic pipe
2) Air outlet without possibility of connection
3) Air outlet with possibility of connection
and
Connection
Quick connect
fittings
Rotatable connection
1) Plugged connection
2) Connection with connected pipe.
mally is the connection plugged.)
(Nor-
Enable connection of pipes without tools.
1) Quick connect fitting without valve (interconnected)
2) Quick connect fitting with valve (interconnected)
3) Half of a quick connect fitting without valve
4) Half of a quick connect fitting with valve
Can rotate during operation, with one line
48
Control systems
Manual controls
1)
2)
3)
4)
Mechanical controls
1) Plunge
2) Roll
3) Spring
Electric controls
1) Electromagnetic with one winding
2) Electromagnetic with two winding (active
in two directions)
3) Control with electric motor
Pressure control
Pressure control through pressure rise respective pressure reduction
Pressure control
Direct control through differential pressure
Indirect
pressure control
Simplified symbol for pre-controlled valve.
Control at pressure rise respective pressure reduction.
Internal control
line
Control line is inside the valve.
Combined controls
1) Control with electromagnetic controlled
pre-control valve.
2) Control with electromagnetic or pre-control
valve.
2)
Pump
with
constant
displacement
1) One flow direction
2) Two flow directions
2)
Pump
with
variable
displacement
1) One flow direction
2) Two flow directions
1)
2)
3)
4)
1)
1)
1)
2)
2)
3)
3) M
2)
General symbol
Control with push button
Control with lever arm
Control with pedal
Pumps and Motors
1)
1)
49
1)
2)
Example of controls for variable
pump
1) Manual control
2) Pressure control via control valve
Compressor
with one flow
direction
The two distorted lines is not included in SMS
Vacuum pump
1)
1)
1)
2)
2)
2)
Motor
with
constant
displacement
1) One flow direction
2) Two flow directions
Pneumatic
motor
with
constant
displacement
1) One flow direction
2) Two flow directions
Motor
with
variable
displacement
1) One flow direction
2) Two flow directions
Motor
with
limited angle of
twist
1)
2)
3)
Pump/motor
with constant
displacement
Component working as pump/motor at:
1) Shifted flow direction, maintained pressure
side.
2) Maintained flow direction, shifted pressure
side.
3) Shifted flow direction and pressure side.
Hydrostatic
gear
Pump and motor together as one unit without
external pipe system.
50
Cylinders
Single-acting
cylinder
The fluid pressure exercises a force in one direction only.
Single-acting
cylinder
The fluid pressure exercises a force in one direction only and the return stroke by return
spring.
Double-acting
cylinder
The fluid operates alternatively in both directions.
Double-acting
symmetric
cylinder
Double-acting cylinder where the area different between the both sides are essential for the
function.
Differential
cylinder
1)
Cylinder
cushion
2)
1)
2)
1)
x
y
x
y
2)
x
y
x
y
with
1) Single acting cushion.
2) Double acting cushion.
Cylinder with
variable cushion
1) Single acting cushion.
2) Double acting cushion.
Pressure intensifiers
Unit converting a pressure X into a higher
pressure Y.
1) For one medium (air).
2) For two mediums (air and oil).
Air-oil actuators
Unit converting a pneumatic pressure into an
equal hydraulic pressure.
51
Directional control valves
Opening and closing of one or more flow paths. Symbols with
several squares. The external flow lines are normally situated
at the square which indicates the neutral or normal position.
Other positions can be shown by displacement of the squares
until the external flow lines are situated at the corresponding
square.
1)
2)
3)
4)
5)
6)
7)
Example
squares
different
paths.
of
with
flow
1)
2)
3)
Non-throttling
directional
control valve
1)
2)
Several service positions each shown by a
square;
1) Valve with two distinct positions.
2) Valve with three distinct positions.
3) Valve with two distinct positions (outer
squares), but between the distinct positions
the valve passes a central position with
essential function.
2/2 directional
control valve
First number in the description denotes the
number of ports, the second the number of
positions. Pilot ports are not included. Control e.g.
1) manual
2)by pressure against return spring (unloading
valve).
3/2 directional
control valve
1) Control by pressure from both ends.
2) Control by solenoid against return spring,
with transient intermediate position.
Electro
hydraulic
servo
valve with pilot
and mechanical
feedback.
4/2 directional
control valve
Combined with solenoid operated pilot valve
with return spring. Representation
1) detailed
2) simplified.
5/2 directional
valve
Control by pressure in both directions.
1)
2)
1) Two pipe connections and free throughflow.
2) Two connections that are closed.
3) 4) Four connections and free throughflow.
5) Four connections where all are tied to each
other.
6) Four connections were two are closed and
two are tied together.
7) Five connections were one is closed.
52
1)
2)
Throttling directional control
valve
Two end positions and intermediate throttling
positions.
1) Shows only the end positions.
2) Shows the end positions and the centre
(neutral) position.
All valve symbols have parallel lines outside
the envelope.
One throttling
orifice (2 ports)
Tracer valve; plunger operated against return
spring.
Two throttling
orifices (3 ports)
Pressure controlled against return spring in
two directions.
Four throttling
orifices (4 ports)
Tracer valve; plunger operated against return
spring.
Single
stage
electrohydraulic servo
valve
Amplification of infinitely variable electrical
input signals transformed onto hydraulic output; without pilot operation.
Two
stage
electrohydraulic
servo
valve
with mechanical
feedback
Two
stage
electrohydraulic
servo
valve
with hydraulic
feedback
Check valves or non-return valves
1)
1)
2)
Check valve
1) without, 2) with back pressure. Opening
if inlet pressure is 1) higher than the outlet
pressure, 2) higher than the outlet pressure
plus spring pressure.
2)
Pilot controlled
check valve
Pilot controlled
1) Opening can be prevented
2) Closing can be prevented
Shuttle valve
The inlet under pressure is automatically connected to the common outlet and the other
inlet is closed.
53
One way restrictor
Valve which allows free flow in one direction
and restricted flow in the other direction.
Quick
valve
When pressure falls at the inlet connection,
the outlet is automatically opened to exhaust.
exhaust
Pressure control valves
Pressure control
valves
Automatic control of pressure;
1) One throttling orifice, normally closed.
2) One throttling orifice, normally open.
3) Two throttling orifices. Arrows with or
without tails.
Pictures in the right column are not according
to SMS standard.
Pressure
valve
Inlet pressure is controlled by opening the exhaust port to the reservoir or the atmosphere
against an opposing force. The left valve has
fixed preloaded spring force, the right has variable.
1)
2)
3)
relief
Pilot controlled
pressure relief
valve
Inlet pressure is controlled by spring force or
by a value determined by the pressure in a
outer pilot port. In the left figure pilot pressure acts against the spring force. In the right
figure it acts together with the spring force.
Proportioning
pressure relief
valve
Inlet pressure is limited to a value proportional to the pilot pressure.
Sequence valve
When the inlet pressure exceeds the opposing
force of the spring, the valve opens permitting
flow through the outlet port.
Pressure regulator
With varying inlet pressure the outlet pressure
remains substantially constant. Inlet pressure
must however remain higher than the selected
outlet pressure.
1) without,
2) with unloading device.
1)
2)
3)
54
Pilot controlled
pressure regulator
Outlet port pressure proportional to pilot
pressure.
Differential
pressure regulator
The outlet pressure is reduced by a fixed
amount with regard to the inlet pressure.
Proportioning
pressure regulator
The outlet pressure is reduced by a fixed ratio
with regard to the inlet pressure.
Variable
flow
control valve
Manual controlled throttle valve.
1) Detailed
2) Simplified
Mechanical controlled throttle valve.
3) Mechanical controlled against spring (braking valve).
Series flow control valve
Regulator with fixed setting, without exceeding oil bleed off.
1) Detailed
2) Simplified
By-pass
flow
control valve
Regulator with fixed setting, with exceeding
oil bleed off.
1) Detailed
2) Simplified
Flow divider
The flow is divided into two fixed flows substantially independent of pressure variations.
Shut-off valve
Simplified symbol.
Flow control valves
1)
2)
3)
1)
2)
1)
2)
55
Components for cooling, filtering, energy storage etc.
1)
1)
2)
Orifices
Reservoirs
1),2), and 4) Reservoirs with atmospheric
pressure.
3) Pressured reservoir. 1),3) The flow line flow
into the reservoir above the the fluid level. 2)
The flow line flow into the reservoir below the
the fluid level. 4) The flow line flow connected
under the reservoir.
Accumulators
1) Hydraulic, the fluid is subjected to pressure
from a spring, weight or gas (air, nitrogen, etc)
2) pneumatic (receiver)
3)
2)
1)
Acc. CETOP:
1) Viscosity dependent (pipe orifice)
2) 1) Viscosity dependent (sharp edged)
Acc. SMS:
1) General symbol
3) Viscosity dependent (pipe orifice)
3)
2)
4)
Filter; strainer
1)
2)
1)
2)
Water trap
1) Manual control of draining, 2) automatic
draining.
Filter with water trap
The apparatus is a combination of filter and
water trap..
Desiccator
Air drying by chemicals.
Lubricator
For lubrication of apparatus small quantities
of oil are added to the air which is flowing
through the lubricator.
Maintenance
unit
Apparatus comprising filter, pressure regulator and lubricator assembled as a unit, 1) detailed 2) simplified.
1)
2)
56
1)
2)
Temperature
controller
The fluid temperature is controlled between
two predetermined values. The arrows indicate both heat introduction and dissipation.
Cooler
1) Without 2) with indication of the flow lines
of the coolant. The arrows in the square indicate the heat dissipation.
Heater
The arrows indicate introduction of heat.
Silencer
Energy sources
Pressure source
Μ
1)
Electric motor
2)
Μ
Combustion engine
1) According to CETOP
2) According to SMS
Measurement equipments
Manometer
(Pressure transducer)
Σ
Thermometer
The symbol can be placed arbitrarily.
Flow rate meter
Measure flow (volume per time unit)
Integrated flow
meter
Measure the total volume which is passed.
Pressure switch
The symbol shows a switching contact.
57