Formula Book for Hydraulics and Pneumatics Fluid and Mechanical Engineering Systems Department of Management and Engineering Link¨oping 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 . . . . .

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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 . . . . . .

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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

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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 . . . . . . . . .

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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

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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 . . . .

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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 . . . . . . . . . . . .

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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

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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

d¨ ar 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

d¨ ar

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

d¨ ar 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)| + ρℓq˙ 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 ˙ h h3

F =−

K2 ˙ h 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 ˙ h 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 ˙ h h3

f =−

k2 ˙ h 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 ˙ h 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, ”‘passningsr¨ akning”’ 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

Till¨ aggsindex 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: L¨ agesservo: 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 ) d˚ a xv ≥ 0 Cq wxv    ρ Load flow: qL = r   2    Cq wxv pc d˚ a 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 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 d˚ a 0 ≤ ω ≤ ω0    ω2     1 −   ω02  ◦ arg G(iω) = −90 d˚ a ω = ω0      ω     2δ0 ω0    ◦  d˚  −180 + arctan  a ω > ω0   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 :skjuvsp¨ anning :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

d¨ ar

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

d¨ ar

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 m ˙ 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 m ˙ through a nozzle can be written as v u u  κ + 1 u p1 Cd A0 KN 2 tκ κ−1 √ m ˙ = 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

f¨or

pn+1 ≤ bs p1

f¨or

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 D10E = control pipe Dotted line with l