当前位置:文档之家› 《流体力学》课后习题答案.pdf

《流体力学》课后习题答案.pdf


得:T1 = t1 + 273 = 50 + 273 = 323K ,T2 = t2 + 273 = 78 + 273 = 351K
根据
p
=
mRT V
,有:
p1
=
mRT1 V1

p2
=
mRT2 V2
得: V2 V1
=
p1 p2
T2 T1
=
9.8067 104 5.8840 105
351 323
=
0.18
设管段长度 l,管段表面积: A = dl
单位长度管壁上粘滞力: = A u = dl u − 0 = 3.14 0.025 0.03
l y l
0.001
1-8 解: A = 0.8 0.2 = 0.16m2 ,u=1m/s, = 10mm , = 1.15Pa s
T = A u = A u − 0 = 1.15 0.16 1 = 18.4N
1
=
T1 b
=
A b
u
−0 −h
=
0.7 0.06b b
15 − 0 0.04 − 0.01
=
21N
/m,方向水平向左
下表面单位宽度受到的内摩擦力:
2
=
T2 b
=
Au−0 b h−0
=
0.7 0.06b 15 − 0
b
0.01− 0
= 63N
/m,方向水平向左
平板单位宽度上受到的阻力:
= 1 + 2 = 21+ 63 = 84N ,方向水平向左。
h1 = 5.6m
2.4 解:如图 1-2 是等压面,3-4 是等压面,5-6 段充的是空气,因此 p6 = p5 ,6-7 是等压面,
建立各点的压强:
p1 = pA − g ( x + H ) = gy
p2 = pB − gx − gH
y
p1 = p2
联立有: pA − pB = ( − ) H p3 = g ( y + H )
T2 T1
p1
=
323 395 = 293
435kPa
1-2 解:受到的质量力有两个,一个是重力,一个是惯性力。 重力方向竖直向下,大小为 mg;惯性力方向和重力加速度方向相反为竖直向上,大小
为 mg,其合力为 0,受到的单位质量力为 0 惯性力: F=-mg
自由落体: 加速度 a=g
重力: G=mg
1-2 是等压面
p1 = pM + g ( H + h1 + h2 )
p2 = oil gh1 + Hg gh2
1
2
p1 = p2
有: pM + g ( H + h1 + h2 ) = oil gh1 + Hg gh2
−980 + 9.81000(1.5 + h1 + 0.2) = 9.8800h1 +136009.8 0.2
1-3 解:已知:V=10m3, T = 50 ℃,V = 0.0005 ℃-1
根据 V
=1 V
V T
,得: V
= V
V
T
= 0.00051050 = 0.25m3
1-4 解:已知: p1 = 9.8067104 Pa , p2 = 5.8840105 Pa , t1 = 50 ℃, t2 = 78 ℃
hB
=
pB g
=
3.0mH2O
pB = ghB = 9.81000 3.0 = 29400Pa
pA = pB − ghAB = 2.94104 -9.810003.5=-4900Pa
hA
=
pA g
=
−0.5mH2O
2-2 解: z = 1m , h = 2m , p0 = 0Pa
管中为空气部分近似的为各点压强相等。
sin
sin =
d
2
=
H
2
+
d 2
2
0.3 = 0.514 0.52 + 0.32
圆锥体所受到的合力矩:
d 4
M
=
dM
=
d
02
r
r
2 r
dr sin
= 1 2 2 sin
= 0.1 3.1415 0.34 = 37.1N m 2 0.001 0.514
习题【2】
2-1 解: hB = 3.0m , hAB = 3.5m
学无 止 境
习题【1】
1-1 解:已知: t1 = 20 ℃, p1 = 395kPa , t2 = 50 ℃
T1 = 20 + 273 = 293K ,T2 = 50 + 273 = 323K
据 p = RT ,有: p1 = RT1 , p2 = RT2
得:
p2 p1
=
T2 T1
,则
p2=Leabharlann τ1 τ21-6 解: = 0.5mm , = 2Pa ,u=0.25m/s
根据 = u ,有: = y = = 2 0.510−3 = 0.004Pa s
y
u u − 0
0.25 − 0
1-7 解: t = 20 ℃,d=2.5cm=0.025m, = 1mm =0.001m,u=3cm/s=0.03m/s
y
0.01
1-9 解: =15rad/s , = 1mm , = 0.1Pa s , d = 0.6m , H = 0.5m
学无 止 境
根据牛顿内摩擦定律,切应力: = u = r y
小微元表面积: dA = 2 r dr sin
小微元受到的粘滞力: dT = dA
小微元粘滞力的力矩: dM = r dT = r r 2 r dr
pA

pB
=
(

)
H
=
a b
H
2-5 解: pM = 4900Pa , h 1= 0.4m , h 2 = 1.5m
p1 = p0 − gh = − gh p2 = pA − gz
p1 = p2
hAB 2
hB 1
学无 止 境
有: pA = g(z − h) = 9.81000 (1− 2) = −9800Pa
2-3 解: H = 1.5m , h2 = 0.2m , oil =800kg/m3
根据真空表读数,可知 M 点相对压强 pM = −980Pa
76
1
2
5
z
x
3
4
p4 = p5 + gz
p6 = p5
p6 = p7
p7 = g ( y + H − b) + g (b − a − z)
p4 = p3
联立有: g ( y + H ) = g ( y + H − b) + g (b − a − z) + gz
学无 止 境
= b−a b
有:
,即V2
= 0.18V1
体积减小了 (1− 0.18)100% = 82%
学无 止 境
1-5 解:已知: = 40mm , = 0.7Pa s ,a=60mm,u=15m/s,h=10mm
根据牛顿内摩擦力定律:T = A u y
设平板宽度为 b,则平板面积 A = a b = 0.06b
上表面单位宽度受到的内摩擦力:
相关主题