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液压与气压传动-王守城-习题答案


2-2 解:由公式ν =

得: ρ
6.31 E

= νρ = 32 ×106 × 900 = 0.0288 N i s / m 2
∵ν = 7.31 E ∴ E = 4.6
2-3 解:

∵ν = 7.31 E ∴ν 20 ∵ν =

6.31 E =

72.469mm2 / s,ν 80 = 23.785mm2 / s

ν 80 ν 20 ∴ λ = 0.0112 ν 60 20 (1 λ 40) = = 1 40λ = 1 40 × 0.0112 = 0.552 ν 20 20 ν 60 = 0.552 ×ν 20 = 0.552 × 72.469 = 40mm 2 / s
2-8 解:由连续性方程可得

, t = 0 (1 λt ) ρ (1 λ 60) = 20 = 1 60λ = 0.328 20

v1 A1 = v2 A2 = q
q1 = v1[π (
从而可得

D1 2 d ) π ( 1 )2 ] 2 2

v1 = 0.1m / s
又因为

v2 =

v1 A1 A2

= 0.036m / s D2 2 d ) π ( 2 ) 2 ] = 23.78 L / min 2 2

q2 = v2 [π (

2-9 解:由题可知油液在钢管中处于层流状态 临界

Re = 2320 Re = vd

因为

ν

v=

Re v 2320 × 40 = = 1856mm / s = 1.856m / s d 50

q = vA =

π

4

× 250 × 106 × 1.856 = 21.85 L / min

2-10 解:因为不计油液流动的能量损失 由图可得

p1 +

1 2 1 2 ρ v1 = p2 + ρ v2 2 2

v1 A1 = v2 A2 v2 = q 30 L / min 4 × 30 × 103 = = m / s = 1.59m / s A2 π × 20 2 mm 2 π × 60 × 20 2 × 106 4 q 30 × 103 × 4 = = 25.5m / s A1 60 × π × 52 × 106

v1 =

p1 p2 =

ρ
2

2 (v12 v2 ) =

900 (25.52 1.59 2 ) Pa = 291474 Pa ≈ 0.29 MPa 2

2-11 解:对油箱液面与泵入口处列伯努利方程,以油箱液面为基准面

p1 α1v12 p α v2 + + h1 = 2 + 2 1 + h2 ρ g 2g ρ g 2g v1 = 0, h1 = 0 Q 4 × 25 × 103 v2 = = = 85cm / s A d 3.14 × 2.52 × 60 v2 85 × 2.5 Re = = = 1063 < 2320层流 γ 20 ×102 α2 = 2

λ=

75 75 = = 0.07 R e 1063

p1 p2 α 2 v 2 2 = +H 2g ρg 2 真空度: pv = p1 p2 = ρ g α 2 v 2 + H 2g
2 × 0.852 + 0.4) 2 × 9.8 = 4.18 × 103 pa = 900 × 9.8(
2-12 解:对油箱液面与泵入口处列伯努利方程,以油箱液面为基准面

p1

ρ

+

α1v12
2

=

p2

ρ

+

2 α 2 v2

2

+ gh +

pw

ρ

4q 4 × 103 (m3 / s ) = = 2.04(m / s ) π d 2 π × 0.025(m) vd 2.04 × 0.025 Re = 2 = = 359 < 2320 γ 1.42 ×104 (m2 / s ) α =2 其中:v1 = 0, v2 =

l ρv2 75 2 900 × 2.042 pw = p + pλ = p + λ × = p + × × d 2 359 0.025 2 = 10000 + 31298 = 41298( pa ) p1 40000 2 × 2.042 41298 代入方程有: = + + 9.8h + 2 ρ ρ ρ hmax = 1.696(m)
2-13 解: v1 =

4Q 4 × 18 × 103 = = 382cm / s π d12 3.14 ×12 × 60

v2 =

4Q 4 × 18 × 103 = = 1062cm / s 2 π d 0 2 3.14 × 0.6)× 60 (

判断流态:

v1d 382 × 1 = = 1910 < 2320 层流 v 0.2 v d 1062 × 0.6 Re 2 = 0 = = 3186 > 2320 紊流 v 0.2 Re1 =
阻力系数:

λ1 =

75 75 = = 0.0393 R e1 1910
1 4

λ2 = 0.3164 Re ξ = 0.35
压力损失:

= 0.3164

1 = 0.0421 31860.25

pλ1 = λ1 pλ 2 = λ0

l ρv1 3 900(3.82) 2 = 0.0393 = 77420 Pa d1 2 0.01 2
2

l ρ v0 2 3 900(10.62) 2 = 0.0421 = 1.06835MPa d0 2 0.006 2
2

l ρv1 900(3.82) 2 pξ = ξ = 0.035 = 2298 Pa d1 2 2
总压力损失:

pw = p λ1 + p λ 0 + pξ = 77420 × 106 + 1.06835 + 2298 ×10 6 = 1.148Mpa

对进出口端面列伯努利方程,取过中心线的水平面为基准面

h1 = h2 = 0, hw = pw / ρ g, p 2 = 0 v1 = 3.82m / s, v2 = v0 = 10.62m / s p1 α1v12 α 2 v2 p w + = + 2g ρ g 2g ρg 层流α1 = 2,紊流α 2 = 1 v 2 2v12 pw + p1 = ρ g 2 2g ρg
2 2 900 1× (10.62 ) 2 × ( 3.82 ) + 1.148 = 2 = 37620 ×10-6 + 1.148 = 1.186Mpa

2-14 解: (1)由公式:pAt =

∑ mv

pAt = ρ Av 2t ∴ p = ρ v 2 = 900 × 36 = 32400 Pa F = pA = p ×

π d12
4

= 32400 ×

25π ×10 6 = 0.6 N 4

(2) F1 = F sin 60 = 0.6 × 2-15 解: q =

π d 4 P
128ul

= =

π d 4 P 128V ρ l

3 = 0.52 N 2

l=

π d 4 p

3.14 (1× 103 ) ×1× 106 × 60
4

128vq = 27.3cm

128 × 20 × 10 6 × 900 × 0.3 × 103

= 0.273M

2-17 解:由q = KAT P 得
m

其中薄壁 m=0.5 细长孔 m=1

对于薄壁孔

q1 = q2

p1 q2 502 p2 = 22 p1 = 2 × 0.05MPa = 0.2MPa 25 p2 q1

对于细长孔
2-18

q1 p1 q 50 = p2 = 2 × p1 = × 0.05MPa = 0.1MPa q2 p2 q1 25

解: (1)根据柱塞运动状态和式(2-64)得

q=

π dhu0 V π dh3 = p t 12 l 2

式中

v=

π
4

d 2H
是柱塞下降 0.1m 排出的液体体积

p =

F H u0 = 2 u0 π d / 4 ; 是柱塞下降的运动速度, t ;

将上述参数代入上式并整理可得

1 d 2 H + π dδ H 3π ld 2 H 2 = t= 4 (d + 2δ ) π d 2δ 3p 4 Fh3 12 l = 3 × π × 50 × 10 3 × 70 × 103 × 20 2 × 106 × 0.1 × (20 × 10 3 + 0.1× 10 3 ) 5 3 4 × 100 × (5 × 10 )

π

= 530s
(2)完全偏心 则 ε = 1

π dhu0 V π dh3 q= = p × 2.5 2 t 12 l
t=

3π ld 2 H (d + 2δ ) 10 Fh3 3 × π × 50 × 10 3 × 70 × 103 × 20 2 × 106 × 0.1 = × (20 × 10 3 + 0.1× 10 3 ) 5 3 1000 × (5 × 10 ) = 212 s

3-5

3-6 解: (1)求偏心量 以单作用式叶片泵的理论排量公式计算(忽略叶片厚度)

q = 2π DeB, 则e =
(2)

q 16 ×103 = = 0.95mm 2π DB 2π × 89 × 30

emax = qmax

D d 0.5 = 2.5mm 2 2 = 2π Demax B = 2π × 89 × 2.5 × 30 = 4.194 × 10 4 mm 3 / r = 41.94ml / r

3-7 解:忽略泄露流量的影响

q = nv v= q 50 L / min = = 0.0167 L / r n 3000r / min

∵v =

π

4

d 2 DZ tan r 4v 4 × 0.0167 × 103 = = 0.074 π d 2 DZ π ×162 ×106 ×125 ×103 × 9

∴ tan r =
3-8 解: (1)

液压马达的理论流量qtm为:
qtm = qmηv = 22 × 0.92 L / min = 20.24 L / min

(2)液压马达的实际转速为 q 20.24 L / min nM = tM = = 80.96r / min VM 250mL / r
(3)液压马达的输出转速为 TM = PM VM η M × 2π η MV 2π × 0.92

(10.5 1) ×106 × 250 ×106 × 0.9 N im = 369.96 N im =
4-3 解: (1)∵ P3 为大气压

∴ P3 = 0

P1 A1 0.9 × 10 6 × 100 × 10 4 P2 = = = 0.5(MPa ) A1 + A2 100 × 10 4 + 80 × 10 4
∴ F1 = F2 = P2 A1 P3 A2 = 0.5 × 10 6 × 100 × 10 4 = 5000( N ) V1 = q 12 × 10 3 / 60 = = 0.02(m / s ) A1 100 × 10 4

q 80 × 10 4 × 0.02 = = 0.016(m / s ) A2 100 × 10 4 P (2) F1 = P A1 P2 A2 = P A1 1 A2 = 0.9 × 106 × 100 × 10 4 0.9 × 106 / 2 × 80 × 10 4 = 5400 ( N ) 1 1 2 P F2 = P2 A1 P3 A2 = 1 A1 = 0.9 × 106 / 2 ×100 × 10 4 = 4500 ( N ) 2 V2 =
(3) F1 = P1 A1 P2 A2 = 0

P2 =

P1 A1 0.9 × 10 6 × 100 × 10 4 = = 1.125(MPa ) A2 80 × 10 4

4-4 解:对于摆动式液压缸

(1)ω =2π n=

2q b( R 2 r 2 )

∴b =

2q 2 × 25 L / min = = 0.142m 2 2 ω ( R r ) 0.7rad / s × (1002 402 )mm 2
b 2 2 ( R r )( p1 p2 ) 2

(2) T =

=

2q q ( R 2 r 2 )( p1 p2 ) = ( p1 p2 ) 2 2 2ω ( R r ) ω 25 L / min × (10 0.5) MPa = 5654.76 N m 0.7 rad / s

∴T =

4-5 (1)快进时为差动连接 则

v3 =

4q πd2

∴d =

4q 4 × 25 L / min = = 0.073m π v3 π × 6m / min
∵ v2 = v3 ∴

快退时

4q v2 = π D2 - d 2 ) (
∴D =

4q 4q = 2 π d π (D2 - d 2 )

∴ D = 2d
2

2

2d = 0.103m

取标 准直 径 d = 70 mm , D = 100 mm (2)工作进给时液压缸无杆腔的压力

p1

π D2
4

= F + p2

π (D2 d 2 )
4

4F D2 d 2 4 × 25000 0.12 0.07 2 5 p1 = + p2 =( + 2 × 10 × ) Pa = 32.85 × 105 Pa < 16 MPa 2 2 2 2 πD D π × 0.1 0.1

故取试验压力p y = 1.5 p1 = 4.93MPa 缸筒材料是45钢,其材料的抗拉强度σ b = 610 MPa, 取安全系数n = 5, 许用应力[σ ] = σ b / n = 122 MPa
δ≥

Py D 2[σ ]

=

49.3 × 105 × 0.1 m = 0.00202m = 2.02mm 2 × 122 × 106

故取缸筒壁厚 3mm. (3)活塞杆截面最小回转半径

rk =

J π d 4 / 64 = = d / 4 = 17.5mm A πd4 / 4
l = 15 / 0.0175 = 85.7 rk
8

活塞杆细长比

根据已知条件以及查表可得ψ 1 = 85,ψ 2 = 2, f = 4.9 × 10 Nm, a = 1/ 5000

ψ 1 ψ 2 = 85 × 2 = 120 > l / rk
故活塞杆保持稳定工作的临界值

Fcr =

fA 4.9 × 108 × π × 7 2 × 104 = N = 1.087 × 106 N a l 2 1 2 × 85.7 ] 1+ ( ) 4[1 + 5000 × 2 ψ 2 rk Fcr 1.087 × 106 = = 43.5 >> 2 4 F 25000

安全系数n =

所以该活塞稳定. 4-6 解:(a)图:

π 1 Fa = 2[ ( D 2 d 2 ) p ] = π ( D 2 d 2 ) p 4 2 2q va = π (D2 d 2 )

运动方向向左 (b)图:

4 4d 2 4q1 4qd 2 vb = = π D2 π D4
运动方向向右 4-8 解:由图可知为工进状态,则:

Fb =

π

D 2 p1 =

π D4

p

(1) F = p1

π
4

D 2 p2

π
4

(D2 d 2 )

4 4 = 43155.375N 4q (2)v = πd2 4q 4 × 10 L / min ∴ vmax = max = = 0.815m / min 2 πD π × 1252 mm 2

= 4 ×106 ×

π

× 1252 × 10 6 1× 106 ×

π

× (1252 90 2 ) ×10 6

∴ vmin =
4-9 解

4qmin 4 × 0.05 L / min = = 0.004m / min π D2 π × 1252 mm 2

(1)在整个缓冲行程 l1 中,缓冲室要吸收的能量包括:活塞连同负载等运动部件在刚进入 缓冲过程一瞬间的动能 Ek ; 整个缓冲过程压力腔压力液体所做的功 E p ; 整个缓冲过程消耗 在摩擦上的功 E f . 它们分别为:

Ek =

1 2 1 mv = × 2000kg × 0.32 (m / s )2 = 90 J 2 2

E p = p2 A2l2 = 70 × 105 ×

π

4

× 25 × 10 3 m × 104 = 54524 J

E f = F f l = 950 × 25 × 103 = 23.75 J
则整个缓冲过程的平均缓冲压力为: pc =

E p + Ek + E f A1l1 E + 2E p + E f Ek + pc = k 1.75MPa A1l1 A1l1

而最大缓冲压力,即冲击压力为: pmax =

(2)如缸筒强度不够,需采取必要的改进措施,可供选择的方法有: 1.允许的话,可减小缸的额定工作压力;2.加大缓冲行程;3.在缓冲柱塞上开轴向三角 槽,或采用锥角为 5°—15°的锥面缓冲柱塞,一缓和最大缓冲压力;4.在油路上安装行程 减速阀,使在进入缓冲之前先减速降低动能,从而减小最大冲击压力. 4-10 解: (1)快进时 v3 =

4q 4 × 25 L / min 4q ,∴ d = = = 79.8mm 2 πd π v3 π × 5L / min
4q 4q 4q ,∵ v2 = v3 ∴ = 2 2 2 π (D d ) πd π ( D2 d 2 )

快退时 v2 =

∴ D 2 = 2d 2 , D = 2d = 112.85mm
(2) F3 = p1 ( A1 A2 ) = p1

π
4

d2

∴ p1 =

π

F3 d2

=

4

25 ×103 = 5Mpa π D2 × 4 2

所以溢流阀的调定压力应等于 5Mpa 4-11 解:对于单杆活塞缸 工进

(1) p1

π
4

D 2 = F + Ff

D=

4( F + Ff )

π p1

=

4 × (2 × 104 + 12 × 102 ) m = 0.0735m = 73.5mm 3.14 × 5 × 106

选择缸的内径为 74mm.

(2)泵的流量q =

vA

ηV

=

π vD 2 0.04 × 3.14 × 742 ×106 3 = m / s = 191.05 × 106 m3 / s 4ηV 4 × (1 0.1)
qp

(3)电动机驱动功率P =

η

=

5 × 191.05 W = 1123.8W = 1.12 KW 0.85
减压阀阀芯全开

5-11 解: (1)夹紧缸作空载快速运动时, p A = 0 , p B = 0

(2)工件夹紧时,夹紧缸压力即为减压阀调整压力, p A = 5Mpa, pC = 2.5Mpa . 减压阀开口很小这时仍有一部分油通过减压阀阀芯的小开口(或三角槽) ,将先导阀打开而 流出,减压阀阀口始终处在工作状态. 5-13 解: (1) p A = 4 Mpa, pb = 4 Mpa (2) p A = 1Mpa, pb = 3Mpa (3) p A = 5Mpa, pb = 5Mpa 7-2 解: (1) p A = pB = 5MPa

pc = 3MPa
(2) p A = 4 Mpa, pB =

F 35000 = = 3.5MPa A1 100 × 104

pc = 3MPa
终端: p A = pB = 5MPa

pc = 3MPa
(3)

p A = p B = p c = 0 MPa

固定时: p A = pB = 5MPa

pc = 3MPa

7-3 解:

p1 A1 p2 A2 = F , p2 = 0 p1 = F 7000 = = 1.4 ( MPa ) A1 50 × 104

p = p y p1 = 2.4 1.4 = 1( MPa )
Q1 = kAT p m = Cq 2

= 0.584 × 104 ( m3 / s ) = 58.4 ( cm3 / s )
速度 v =

ρ

AT p 0.5 = 0.62 × 0.02 × 104

2 × 1× 106 900

Q1 58.4 = = 1.168 ( cm / s ) A1 50

7-4 解: (1)η =

F v 40000 × 18 × 102 = = 0.053 pY q 5.4 × 106 × 2.5 × 103

( 2 ) p1 A1 = p2 A2 + F
p2 = p1 A1 F 5.4 × 106 × 80 × 104 + 40000 = = 0.8 ( MPa ) = p A2 40 × 104

q2 = A2 v = 40 × 18 × 103 = 0.72 L q2 = KAT p

(

min

)



K=

q2 AT p

若F = 0
p1 A1 = p2 A2 p2 = 80 A1 p1 = × 5.4 = 10.8 ( MPa ) = p′ A2 40 q2 10.8 × AT p′ = 0.72 = 2.65 L min 0.8 AT p

q2′ = KAT p′ = v′ =

(

)

q2′ 2.65 × 103 = = 66 cm min 40 A2

(

)

7-7 解: (1) 泵理论流量q p = V p n = 40 ×10 3 × 1500 = 60L/ min

泵实际流量qt = q p ×ηVP = 60 × 0.95 = 57L/ min 马达最高转速nmax = qt ×ηVM 57 × 0.98 = = 1117 r / min Vm 50 ×103 nmin = 0r / min

( 2 ) 马达进口压力8MPa,出口压力0,p = 8MPa

p Vm 8 × 106 × 50 × 106 T= ηmM = × 0.8 = 50.96 N m 2π 2π

( 3) P0 = 2π nT = 2π ×1117 × 50.96 = 357472W = 357.472KW

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