液压元件设计库电子版

24
X=0 X=
X=0 10cm3 X=
dead volume
dead volume
10cm3
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Training
HCD:
:
Stroke = 1m Piston diameter = 10mm
X=0 X=stroke
25
Rod diameter ( Dead volume = 10cm3 Ports' diameter = 5mm Displacement: 10s 0-1m
Training
HCD: HCD
BHC11
18
(4
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)
Training
HCD:
19
Example8a.ame
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Training
HCD:
HCD length at zero displacement Chamber
20
HCD
3
: Step 1: 0 to 30 N in 5s Step 2: 30 N during 5s Step 3: 30 to 0 N in 5s
Tank
78
Drod = 5 mm
Pressure source
Dspool = 10 mm
:
P
Forc e
Spool mass: 2 g Viscous friction: 10 N/(m/s) Displacement: 0 to 1.5 mm Hole diameter: 3.5 mm X=0
11
HCD
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Training
HCD HCD
12
F2 = F3 - P*A Q1 = -A*V2 V3 = V2
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Training
HCD:
13
Example1.ame
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Example8.ame
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Training
51
:
: = 10bar = 100 L/min @ 30bar = 100 L/min @ 30 bar
?…
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Training
52
:
: = 100 L/min @ 30bar
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Training
56
: (3)
HCD
89.15 N/mm …
:
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Training
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: (4)
HCD
X=0
x=1.1276mm
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Training
58
: (5)
HCD
HCD ? 0 to 50 bar in 10 s
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Training
HCD:
zero displacement
21
or … or … or …
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Training
HCD: chamber length [Vadd= L*AP] Volume t=10s 1000cm3 1m 1000-16.67 = 983.33cm3. Dead 0.1m/s
22
chamber length at zero displacement 1m V0=1000 + 16.67 = 1016.67cm3 1000cm3
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Training
HCD:
23
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Training
HCD: !
chamber length at zero displacement 0 ?
HCD HCD
29
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Training
HCD
30
/ R
BAP025
BAP026
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Training
HCD
31
/
Dead volumes
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Training
HCD
32
(
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281bar
283.3bar
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Training
HCD: :
16
Q0 = Ap*V*ρ(P)/ρ(P0)
0bar
Q = Qin(0)*ρ(P0)/ρ(P)
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Training
HCD: !
17
B = −V
δP δV
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AME_HYD2
Chapter 2
Hydraulic Component Design (HCD)
©IMAGINE SA 1998-2005 Updated: April 2005
Training
HCD
2
1. HCD 2. 3. 4. 5.
?
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Training
65
AMESim
AMESim
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Training
66
(1)
New Comp Icon
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Training
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(2)
AMESim
AMESim
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Training
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(3)
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Training
53
:
/
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Training
54
: (1)
D=8mm
HCD
(1N/mm) :
P
Mass=10g
X=0
: = 10 bar Area = 8*8*π/4 = 50.27 mm2 Force = P*A = 50.27 N
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Training
48
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Training
49
P
X=0
: = 10bar = 100 L/min @ 30bar : = 50 bar 10 0
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Training
50
:
: = 10bar = 100 L/min @ 30bar
) = 5mm
Example9.ame
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Training
HCD:
26
0.1cm3
AMESim 0 V0 = V0/100
?
Dead volume
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Training
HCD:
dead volumes
27
* zero displacement
AMESim
3
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Training
?
:
4
(C
R)
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Training
?
5
!
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Training
?
6
Q = 0 if ∆P < Pcrack Q = coef ⋅ ∆P otherwise
HCD
37
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Training
HCD
38
BOSCH
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Training
HCD
Constant pressure line, Ps A K1 Pc Vc Qc PR, Vt
QL
39
x Ks
Hydraulic load
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copy to
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Training
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.
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Training
63
( ) AMESim
…
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Training
64
1.
2. 3.
4.
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Training
/ /
liquid gas
(KH=B/Vol)
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Training
35
1. HCD 2. 3. 4. 5.
?
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Training
HCD
36
4-
Supply
Return
Loads
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Training
Training
HCD
44
©IMAGINE SA 1998-2005
Training
HCD
VVA
45
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Training
HCD
46
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Training
47
1. HCD 2. 3. 4. 5.
?
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AMECustom
2
75
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Training
76
1. HCD 2. 3. 4. 5.
?
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Training
HCD
3
77
Hydraulic Component Design
3
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Training
…
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Training
7
...
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Training
8
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Twk.baidu.comaining
9
HCD
Qb Qa Pa Va Vb Aa Ab Pb load force FL
HCD
!
/
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72
AMECustom AMESim :
AMESim
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Training
AMECustom
AMECustom CV50
73
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Training
AMECustom
74
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Training
) Training
HCD :
3 π D .C R Q = .∆ P 12 µ L
33
F viscous
=
πµ L ∆ P
C
R
.V
: (e =
Q =
3 R
)
2
e π D .C .∆ P . 1 + 1 . 5 12 µ L CR
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Training
34
( )
Training
HCD: 0.1m/s 0.1L/min
A= Q 0.1 1 1 100 = = . m2 = mm 2 V 60000 0.1 60000 6
4A 20 mm ≈ 4.607 mm 6π
14
Dp =
π
=
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Training
HCD: :
15
:
Training
HCD
:
: Q=f(x) F=f(∆P, Jet) : X”=f(m, damp, stops) : Q=f(v) F=f(P, k, x) : ? :
10
M
P=f(β, V, ΣQ, fluid props)
> ^ ^
<
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Training
HCD
dead volume + chamber length at
©IMAGINE SA 1998-2005
Training
HCD: x= 10cm3 1m dead volume
28
x=0 10 + 100*(pi/4) = 88.54cm3
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Training
0 to 50 bar in 10 ms
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Training
59
1. HCD 2. 3. 4. 5.
?
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Training
60
AMESim
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Training
61
Edit supercomponent
Training
55
: (2)
HCD
x
X=0
= π.D.x :
Q = A.Cq
2 ∆P
ρ=850 kg/m3, 100 L/min @ 30 bar, Cq = 0.7: = 28.24 mm2 x=1.1276mm
ρ
30 bar ⇔ 100.53 N,
K = F/x = 89.15 N/mm
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Training
69
AMESim
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Training
70
AMESim …
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Training
71
‘Edit constituents’
( )
'Save as'
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Training
AMECustom
Training
HCD
40
REXROTH
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Training
HCD
41
/
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Training
HCD
42
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Training
HCD
43
Caterpillar
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