电液控制习题答案
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流量增益:
)/(4.1870
10
7010814.362.025
3s m p W
C K s
d q =⨯⨯⨯⨯⨯==-ρ 流量——压力系数:
(
)
)/(1008.7107.83210814.310514.3323
123
32
62s pa m W
r K c c ⋅⨯=⨯⨯⨯⨯⨯⨯⨯==----μπ
)(107.8101087036s pa ⋅⨯=⨯⨯==--ρυμ
压力增益:
)/(1097.1107.84.1113
m pa K K K c q
p ⨯=⨯==-
P39 习题2
)/(67.1601005.01052
3
30s m U q K c q =⨯⨯⨯==-- )/(1095.5107060210523
122
30s pa m p q K s c c ⋅⨯=⨯⨯⨯⨯==--- )/(108.210
95.567.111
12
000m pa K K K c q p ⨯=⨯==-
P65 习题1
⎪⎪⎭
⎫ ⎝⎛++=
1222
s s s D K X h h
h m
q
V
m
ωζωθ
)/(107.610
645
6
s m rad D K m q
⋅⨯=⨯=-
(
)
)/(98.402
.010310
6107444
2
68
2
s rad J V D t t m
e h =⨯⨯⨯⨯⨯⨯==--βω
t t
e m ce h V J D K βζ=
tm c ce C K K +=
(
)
s m q q n t /10327.61066.695.0%95344--⨯=⨯⨯==
()(
)
s
m q q q q n t n /1034.0%95327.6667.634-⨯=-=-=∆
()
pa s m p q C n tm ⋅⨯=⨯⨯=∆=--/1043.210
1401034.03
125
4
()
pa
s m K ce ⋅⨯=⨯+⨯=---/1043.21043.2105.13121216
28.01032.010********.24
8612=⨯⨯⨯⨯⨯==
---t
t
e m
ce
h V J D K βζ
⎪
⎪⎭
⎫ ⎝⎛++⨯=14156
.01618107.625
s s s X V
m
θ P66 习题4
t X x m p ωsin =
t X x m p ωωcos =∙
t X x m p ωωsin 2-=∙
∙ t m X f t m ωωsin 2-=
t m X t
t m X N t m t m
ωωωωω2sin 2
1cos sin 3
23
2⋅⋅⋅=⋅⋅⋅⋅=
当2
2π
ω=
t ,
4
π
ω=
t 时
t m m X N ⋅⋅=3
221max ω
2
2
'ωm p X x =∙ 222'
ωm p X x -=∙∙ 2
22
't m p m X f ω-= 当s L p p 3
2
=
2
2322
t m s m X Ap p ω=
()3
3225
21064.13003010810140423423m
m X p A t m s p --⨯=⨯⨯⨯⨯⨯⨯==ω
()s m A X A x q p m p p
m /1082.41064.122
30108322333332'0---∙⨯=⨯⨯⨯⨯⨯⨯===
ω
35max max
01082.410140870
162.01-⨯=⨯⨯⨯⨯==V s V d m WX p WX C q ρ ()25
max 10
12.6m WX V -⨯=
P80 习题1
P I V X L L X L L L X 2
1
221-+=
221L L L K I +=
2
1
L L K F = 开环传递函数:
⎪⎪⎭
⎫ ⎝⎛++=⎪⎪⎭⎫ ⎝⎛++=12122222s s s K s s s A K K X X h h h V h h
h P F
q V P ωζωωζω P F q V A K K K =
62lg
20=-h
h V
K ωζ
643202.0=⨯==h h V K ωζ
)/(59.0870102010262.025
2
s m p W C K s d q =⨯⨯⨯⨯==-ρ
21.059
.01020644
=⨯⨯==-q P V F K A K K
P139 习题1
开环传递函数:
⎪⎪⎭
⎫
⎝⎛++=⎪⎪⎭
⎫
⎝⎛++=12122222s s s K s s s D nK K K U X h h h V
h h h m f
q a r P
ωζωω
ζω m
f
q a V D nK K K K =