液压元件设计库

Training
HCD:
28
ˆ ‹ /ˆ
x= ß> / 1m ¯ ¨ ˙7(,Xdead volume hA„1 b 10cm3
7(,X '/ˆ
x=0 ˚ ¨ 10 + 100*(pi/4) = 88.54cm3
Training
©IMAGINE SA 1998-2005
HCD
29
) ı + ,X ˆ ‹8V# •
HCD
33
,X" $ª ) ı +
:
3 π D .C R .∆ P Q = 12 µ L
F viscous
=
πµ L ∆ P
C
R
.V
: (e = ,¨ X ,X #
Q =
3 R
)[ )
2
π D .C e .∆ P . 1 + 1 . 5 CR 12 µ L
©IMAGINE SA 1998-2005
HCD:
16
ˆ ‹ /ˆ
:
Q0 = Ap*V*ρ(P)/ρ(P0)
0bar>< _ ,X# G£¨ A„# G£ B ' ! rL _ oEfl> Z ´!7
Q = Qin(0)*ρ(P0)/ρ(P)
©IMAGINE SA 1998-2005
Training
HCD: !
17
ˆ ‹ /ˆ
B = −V
Training
HCD
36
h*
4-E K
Supply
Return
Loads
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Training
HCD
37
h*
©IMAGINE SA 1998-2005
Training
HCD
38
h*
BOSCH
©IMAGINE SA 1998-2005
Training
HCD
©IMAGINE SA 1998-2005
Training
HCD
12
,X ‹G£
HCD g
ı +0ˆ • ‹G£
F2 = F3 - P*A Q1 = -A*V2 V3 = V2
©IMAGINE SA 1998-2005
Training
HCD:
13
ˆ ‹ /ˆ
Example1.ame
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A=
Dp =
π
=
©IMAGINE SA 1998-2005
Training
HCD:
15
ˆ ‹ /ˆ
:
:
- 4 ,X _ o 283.3bar -‘ G * &# G£$d ı +Au1k k ,X _ o281bar, !¤ , Aˆ ´ ‰
©IMAGINE SA 1998-2005
Training
HCD ,X ) ı + ,X ) ı +
?U5 <% ˆ ‹8V# •
©IMAGINE SA 1998-2005
Training
HCD
30
) ı + ,X ˆ ‹8V# •
/ !/ˇ R # _ · p?
BAP025
BAP026
©IMAGINE SA 1998-2005
Training
HCD
31
©IMAGINE SA 1998-2005
Training
HCD:
23
ˆ ‹ /ˆ
©IMAGINE SA 1998-2005
Training
HCD: !
24
ˆ ‹ /ˆ
chamber length at zero displacement 0 Œ5 B ,X '/ˆM2 G¡?U‰ ?
X=0 X=
Training
34
K
/ ,X
/# / !Ł ' '
D )
(n
liquid
gas
(KH=B/Vol)
©IMAGINE SA 1998-2005
Training
35
1. 2. 3. 4. 5.
HCD g1T h* r _ A’Au ) K CY ı + „ K A’Au EK
? V ) .
©IMAGINE SA 1998-2005
A K1 Pc Vc
39
h*
Constant pressure line, Ps x Ks
£ _K
Qc PR, Vt QL Hydraulic load
©IMAGINE SA 1998-2005
Training
HCD
40
h*
REXROTH
©IMAGINE SA 1998-2005
Training
HCD
Training
HCD:
20
ˆ ‹ /ˆ
HCD g # _ /ˆ, G,XG¡?U length at zero displacement
D
Chamber
©IMAGINE SA 1998-2005
Training
HCD:
21
ˆ ‹ /ˆ
zero displacement
or … or … or …
Training
50
:#
_ g ,X1T ) ı _
(M ß: _ o = 10bar # G£ = 100 L/min @ 30bar
©IMAGINE SA 1998-2005
Training
51
:
#
_ g ,X œOE ‘,X1T ) ı _
(M ß: _ o = 10bar # G£ = 100 L/min @ 30bar OE ‘ = 100 L/min @ 30 bar
z
ˆ ‹# E M6/ˆ
©IMAGINE SA 1998-2005
Training
54
˛0 ) K ı _,X Æ ª
: (1) * HCD g ,X ı + X ˚,X |
D=8mm
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Training
7
...
©IMAGINE SA 1998-2005
Training
8
©IMAGINE SA 1998-2005
Training
9
HCD g X ˛,X1T ),X#
Qb Qa Pa Va Vb Aa Ab Pb
_5 ı _
load force FL
δP δV
©IMAGINE SA 1998-2005
Training
HCD:
18
ˆ ‹ /ˆ
/ˆ ) ı + 9
•#
_g
HCD g ,X ' /ˆ ı + ˜
BHC11
(4
©IMAGINE SA 1998-2005
0ˆ •)
Training
HCD:
19
ˆ ‹ /ˆ
Example8a.ame
©IMAGINE SA 1998-2005
Example9.ame
©IMAGINE SA 1998-2005
Training
HCD:
26
ˆ ‹ /ˆ
0.1cm3OE ‘ Z ¨
AMESim FS ! 0 Œ5 B ,X '/ˆ ˛) V0/100 ¨ J V0 = # _ 7( n ,X
?
¨ OE ‘A’ n Dead volume
©IMAGINE SA 1998-2005
AME_HYD2
Chapter 2
Hydraulic Component Design (HCD)
"© zK0 @ #K •><
©IMAGINE SA 1998-2005 Updated: April 2005
Training
HCD
2
¯
1. 2. 3. 4. 5.
HCD g1T h* r _ A’Au ) K CY ı + „ K A’Au EK
©IMAGINE SA 1998-2005
Training
?
5
!
©IMAGINE SA 1998-2005
Training
L ?U# _ ˚A’Au g
?
6
Q = 0 if ∆P < Pcrack Q = coef ⋅ ∆P otherwise
)„ ü È å À å5×<%8V# ·,X,ó r,X ´ ) 6(Š È KÔ8ƒ,X Ã û È | o … #†
) ı + ,X ˆ ‹8V# •
G£ n
/ G¡ ·
Dead volumes
©IMAGINE SA 1998-2005
Training
HCD
32
,X" $ª ) ı +
(
5×<% J
ª ´ p?˜ í,XA± ÈÃ*ü $ õ _ DG£?U5Ï á) Training
©IMAGINE SA 1998-2005
41
h*
/¯ 2 + K¨,X" $ª
©IMAGINE SA 1998-2005
Training
HCD
42
h*
©IMAGINE SA 1998-2005
Training
HCD
43
h*
Caterpillar
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Training
HCD
44
h*
©IMAGINE SA 1998-2005
ß> /
X=0 n 10cm3
#
2
b
” !

˜
˛ !5B
¯ dead volume
X=
ß> /
dead volume
10cm3
©IMAGINE SA 1998-2005
Training
HCD:
25
ˆ ‹ /ˆ
:
Stroke = 1m
X=0 X=stroke
Piston diameter = 10mm Rod diameter ( Dead volume = 10cm3 Ports' diameter = 5mm Displacement: 10s Y ¢ 0-1m 4› ”7( ) = 5mm
< _
E⁄ |
: 1uCˆ | (M ß ?
Q=f(v) F=f(P, k, x)
©IMAGINE SA 1998-2005
Training
HCD
11
4§
HCD gG * „/ 4§ X ) 4 ,X "© S k* ˆ „E E ˆ6 ,X4§ X ) ı + X ˛ ˆ6 ,X „/ 2ˇ4‡ ı _ ‰
HCD g
!
/#
2
,XB G£
Training
©IMAGINE SA 1998-2005
HCD
8: K 8 Q=f(x) F=f(∆P, Jet) ∆ : X”=f(m, damp, stops) # : 2
10
_$
: |+
’/ˆ :
M
P=f(β, V, ΣQ, fluid β props)
> ^ ^#
©IMAGINE SA 1998-2005
Training
48
©IMAGINE SA 1998-2005
Training
49
P
X=0
: _ o = 10bar # G£ = 100 L/min @ 30bar , : _ o$d = 50 bar ˜ 10 Y _ o+ 0 7˙
©IMAGINE SA 1998-2005
Training
HCD:
27
ˆ ‹ /ˆ
dead volumes ,X '# A„ 7( = -,X '/ˆ ˜
2E⁄ | W,X4 0ˆ
ß> / *# 2M6/ˆ zero displacement
dead volume + ´!8LÔ?U4- nchamber length at
©IMAGINE SA 1998-2005
Training
HCD
VVA
45
h*
©IMAGINE SA 1998-2005
Training
HCD
46
h*
©IMAGINE SA 1998-2005
Training
47
1. 2. 3. 4. 5.
HCD g1T h* r _ A’Au ) K CY ı + „ K A’Au EK
? V ) .
?…
? 1 fl4{ ı _
©IMAGINE SA 1998-2005
Training
52
:
!¤EW!7 8V# •,XOE ‘ ˘4
(M ß: # G£ = 100 L/min @ 30bar
©IMAGINE SA 1998-2005
Training
53
:
*
ˆ ‹8V# • ˛0
)
K
ı_
/ _o
? V ) .
©IMAGINE SA 1998-2005
Training
AMESim
3
,X # _g
©IMAGINE SA 1998-2005
Training
L ?U# _ ˚A’Au g
?
/¡2O _,X# _5
4
: Œ+
!8 A’# _5 ,X5 ' ˛ n,X ˜ V p# _5 ,X5 ' ˆ |,XA– ¨ Fw # _5 2O _,X D,´ ?U5ˇ Ɖ ˚˝ /¡2O _E‹L ?U5 <%0ˆ • Æ ,X · p? (C Œ R)
©IMAGINE SA 1998-2005
Training
HCD:
22
ˆ ‹ /ˆ
chamber length [Vadd= L*AP] Dead Volume 1000cm3 ˜# 2,XE z 0.1m/s ¨ t=10s ˚ ¨ 4 ,X !/ˇ 1m ¨ ·!8 4 7(,X '/ˆ 1000-16.67 = 983.33cm3. chamber length at zero displacement 1m ¨ æ ,X 7(,X '/ˆ ‹ V0=1000 + 16.67 = 1016.67cm3 ¨ 5 4 ,X '/ˆ ‹ 1000cm3
Example8.ame Training
HCD: ˛ {*
14
ˆ ‹ /ˆ
0.1m/s ¨ 0.1L/min,X# G£L ?U,X#
Q 0. 1 1 1 100 = . = m2 = mm 2 6 V 60000 0.1 60000
4A 20 mm ≈ 4.607 mm 6π
ˆ „Au1k 2M6/ˆ