火炮单筒身管设计
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单筒身管设计教师:徐亚栋
孙天云
0901510425
2012年5月19日星期六
(1)口径:121.92d m m = (2)炮弹断面积:21.192S dm = (3)弹丸行程长:lg 35.21dm = (4)药室容积:307.5W dm = (5)药室长: 4.74l dm =ys (6)弹丸重:21.76q kg = (7)装药重: 4.75kg ω= (8)铜柱测压:()280M P m p T = (9)膛压与时间表:
一. 高低温压力曲线:
1.考虑弹丸旋转和摩擦力的次要功系数:1 1.02ϕ=
3002
7.5dm
2=
=6.29195dm 1.192dm
W l S
=
.药室自由容积缩颈长:
0 6.29195dm =
=1.327424.74dm
ys
l l χ=
3.药室扩大系数:
3
3
4.75kg kg =
=0.63333
dm
7.5dm
W ω
∆=
4.装填密度:
35.21dm =
=5.596046.29195dm
g g l l Λ=
5.弹丸相对行程长:
()
1
1
5.59604 1.32742
6.0.48130
2(1 5.59604)
21g g χ
λΛ+
+
=
=
=++Λ1系数:
()
1
1
5.59604 1.327427.0.32087
3(1 5.59604)
31g g χ
λΛ+
+
=
=
=++Λ2系数:
8. 4.7435.2139.9539.952632.7740
121.921.05
sg ys g sg a l l l dm dm dm l dm d m m
a =+=+=<
=
=<=系数:
则:
2
4.759 1.050.32087 1.12004
21.76kg a q
kg
ωϕλ=+=+⨯
=:次要功系数:
10.数据放大: 序号为
26,则
()28026306m T p M P a M P a M P a
=+=
()1.12 1.12306342.72tm m T p p M Pa M Pa
==⨯=
1
1
1 4.75 1.02(1)
(10.48130) 1.00451.0221.76 1.12004
tm m m m
kg p p p p q kg
ϕω
λϕϕ
=+=+⨯
⨯
=⨯342.72341.184671.0045
1.0045
tm m p M P a p M P a =
=
=
老师给的数据最大平均压力为:257.89999MPa
则放大系数为341.18467MPa/257.89999MPa =
1.322934014 11.弹底压力:
111
1
0.93341333111 4.75 1.071336745
1133 1.0221.76d p p p p p
kg q
kg
ωϕ=
=
=
=+
+
⨯
⨯
0.933413331弹底压力相对于膛压的放大系数:
12.时间与行程关系:
2
-6
2
1537.0510
p d
d d l S p p d t
q
ϕ=
=⨯
d p 是时间的函数,根据数据分成三段拟合:
1) 第一段:直线,从t=0~0.002252s :
68980.1925436.30442079
d p kt b t =+=+代入二阶微分方程并一次积分得:2
11537.05(68980.1925436.30442079)2
p
dl v t t C dt =
=⨯⨯++ 二次积分得:3
2
121
1537.05(68980.1925436.30442079)6
2
p l t t C t C =⨯⨯+
⨯++
边界条件:100,C =0t v ==时,则
20 4.74=0.474m ,C =0.474p ys t l l dm m ===时,则
则:3
2
1
1537.05(68980.1925436.30442079)0.474
6
2
p l t t =⨯⨯+
⨯+
2)第二段:抛物线,从t=0.002252s~0.00485s :
用matlab 拟合得:22-38229200 t + 316399.5t-335.8905d p at bt c =++= 拟合曲线与实际点对比图:
代入二阶微分方程并一次积分得: 32
311537.0538229200316399.5335.8905t 32p
dl v t t C dt ⎛⎫=
=⨯-⨯+⨯-+ ⎪⎝⎭
二次积分:
432
34111537.0538229200316399.5335.8905t
12
62p l t t C t C ⎛⎫
=⨯-⨯+⨯-⨯++ ⎪⎝⎭
边界条件:
2
32
3
0.0022521537.05(68980.192540.00225236.304420790.002252)
2
11537.0538*******.002252316399.50.002252335.89050.00225232t s v C ==⨯⨯⨯+⨯⎛⎫=⨯-⨯⨯+⨯⨯-⨯+ ⎪⎝⎭
时,
则:3191.3658806C =
4324
32
0.002252111537.0538*******.002252316399.50.002252335.89050.002252191.36588060.002252126211537.05(68980.192540.00225236.304420790.002252)0.474=0.59395716
62
p t s l C =⎛⎫
=⨯-⨯⨯+⨯⨯-⨯⨯+⨯+ ⎪⎝⎭=⨯⨯⨯+⨯⨯+时,
则:4C =0.340982542
432111537.0538229200316399.5335.8905191.36588060.340982542
1262p l t t t t ⎛⎫
=⨯-⨯+⨯-⨯++ ⎪⎝
⎭