空气动力学习题
,
2k8315m1.1解: R,,,259.84232(s,k)m
p,,RT
6P5,10kg ,,,63.506, 3mRT2.5984,303
气瓶中氧气的重量为
G,,vg,63.506,0.15,9.8,93.3541.2解:建立坐标系
根据两圆盘之间的液体速度分布量呈线性分布
则离圆盘中心r,距底面为h处的速度为 u,kn,u 0
当n=0时 u=0推出u,0 0
wr当n=h时 u=wr推出 k,h
则摩擦应力为 ,
duwr,,u,u dnh
上圆盘半径为r处的微元对中心的转矩为
3wrwrd,,,dA,r,urdrd,,r,udrd, hh
D332,ruD,,,2则udrd,,,, ,,00h321.4解:在高为10000米处
T=288.15-0.0065,10000=288.15-65=223.15
PT,,5.2588 压强为,,,PaTa,,
5.2588T,,KN ,,ppa26.43,,MTa,,
5.2588,T,,密度为 ,,,,aTa,,
5.2588T,,kg ?,,,,a0.4127,,mTa,,
PKG1-7解: p,,RT ?,,,24.4642MRT
空气的质量为 m,,v,662.98kg
22-3解:将y+2xy=常数两边微分
2ydy+2xdx+2ydx=0
整理得ydx+(x+y)dy=0 (1)
dydy 将曲线的微分方程代入上式得 ,VVxy
yVx+(x+y)V=0 y
22 由V,x,2xy,2y得
(22 22 V+V=x+2xy+y(2) xy
由(1)(2)得,,v,,x,y,v,,y xy
dxdy习题二2-2解流线的微分方程为 ,vvxy
dxdy 将v,,xdx,ydy和v的
达式代入得 xy222xy2xy
22 将上式积分得y-x=c,将(1,7)点代入得c=7
22 因此过点(1,7)的流线方程为y-x=48 2-5解:直角坐标系与柱坐标系的转换关系如图所示
v,vcos,,vsin,,xr 速度之间的转换关系为 ,v,vsin,,v,cos,yr
,r,r,,sin,,cos,x,rcos,,,,y,x 由, ,,,v1,v1y,rsin,,,,,sin,,cos,,xr,yr,
,,,,,,,vvrvv1,,xxx,,,,,,,,Vcos,Vsin,cos,,Vcos,,Vsin,,sin,,,r,r,,x,r,x,,,x,r,,r,,
,,,V,VV1V,,,,r,r,,cos,,sin,cos,,cos,,Vsin,,sin,,Vcos,sin,,,,,r,,r,rr,,,,,,,,
,V,V1,V11,V1222rr,,,cos,,sin,cos,,sin,cos,,Vsin,,sin,,Vsin,cos,r,,r,rr,,rr,,r
,V,V,V,r,v,,1yyy,,,,,,,,,Vsin,,Vcos,sin,,Vsin,,Vcos,cos,r,r,,V,V,y,v,y,r,,rxx
,,,V,VVV1,,,,r,r,,sin,,cos,sin,,sin,,Vcos,,cos,,Vsin,cos, ,,,,r,,r,r,,,,r,,,,,V,V1,V11,V1222rr,,,sin,,sin,cos,,sin,cos,,Vcos,,cos,,Vsin,cos,r,,r,rr,vrr,,r
,V,V,V,V1,V,Vy,,xzr,z, ?div,,,,,V,,,,r,x,y,z,rr,,,z,,
,V,V,V,Vyy22xx2-6解:(1) ,,3xsiny,3xsiny,,0,x,y,x,y
?此流动满足质量守恒定律
,V,V,V,Vyy222xx (2) ,3xsiny,3xsiny,,6xsiny,0,x,y,x,y
?此流动不满足质量守恒定律
22y2xy,,,2 (3)V=2rsin, V=-2rsin,xyrr
2323,V,V,V4xy4xy,2y,Vx 2yyyx ,,,,0,,,333,xr,x,yr,yr
?此流动不满足质量守恒方程
dxdy22 (4)对方程x+y=常数取微分,得 ,,dyx
2kkdxdy22由流线方程v,得v,v,(2) (1) 由 ,xy4rrvvxy
kxky由(1)(2)得方程v,, v,,xy33rr
,V,V,V3kxy,V3kxyyyxx ,,0,?,,552,x,y,y2,xrr
?此流动满足质量守恒方程
,V,V,V,V3yz3yz,V,Vyyxzxz2—7解: ,,0,,,,,,,0同样,,07722,x,y,y,z22,z,xrr
?该流场无旋
222xdx,ydy,zdz1dx,y,z,, d,,vdx,vdy,vdz,,, xyz33 222222222,,,,x,y,zx,y,z
1 ?,,,,c222x,y,z
,V,V,Vyxz2—8解:(1) ,,,a,,,,a,,,axzy,x,z,y
,V,,,,1,V1,V,V1,V,V,,yzxzxx ,,,, v,,,0;v,,,0;v,,,0,,xyz,,,,2,x,z2,z,x2,x,y,,,,,,
,V,V,,,,1,V1,V,V1,V,,yyzxzx (2),,,, ,,,,0;,,,,0;,,,,0,,xyz,,,,2,y,z2,z,x2,x,y,,,,,,
?该流线无旋,存在速度位
(3)d,,vdx,vdy,vdz,axdx,aydy,2azdz xyz
11222 ?,,ax,ay,az,c22
222—9解:曲线x,,fx,y,xy,4,0y=-4,
2fyfxx2xy 切向单位向量t,i,j,i, 2222422422x,4xyx,4xyf,ff,fxyxy
v,,切向速度分量,v,v,t,,,,tt
,,,,2 把x=2,y=-1代入得,,v,,,i,j,,,x,2x,yi,,x,2xj, ,x,y
2,,x2xy11,, t,i,j,i,j422422,,22x,4xyx,4xy,,
331133,, vvt,,,, v,vt,,i,j,,i,j,,ttt222222,,
mkm2—14解:v=180=50 sh
1122 根据伯努利方程p,pap,,V,,,,V ,,,22
1122 驻点处v=0,表示为p,pa,,V,,1.225,50,1531.25pa ,22
m处得表 s
111222 示为 p,pa,,V,,V,1531.25,,1.225,60,,637.75, 相对流速为60222
Qy3—1解:根据叠加原理,流动的流函数为,, ,x,y,Vy,arctg,2,x,,Qx,,Qy速度分量是V,,V,,V,,,,; x,y2222,y2,x,y,x2,x,y
Q驻点A的位置由V=0 V=0求得 x,,;y,0AXAyAA2V,,
y,yQQA过驻点的流线方程为yarctgyarctg ,,,,VVAy,2,x2,x2A
,,,Qy,,,, 即y,,arctg或r,,,,,2Vx2,Vsin,,,,,
2,,QQysinvsin,在半无限体上,垂直方向的速度为,,, vy22,,2xy,2r,-,
2dv2vsincosvsin,,,y,,线面求极值 ,,,02d-,,,,,-,,
tg,当 ,,2 v,v,0v,vsin,,0yyyymaxmin,,-
tg,用迭代法求解,,2得 ,,-
, ,,1.9760315,113.2183时,v取最小值1y
, ,,4.3071538,246.7817时,v取最大值2y
2,,QQysinvsin,由,,, vy22,,2xy,2r,-,
xcos,sin,cos,QQvvvv,,,,,, x,,,222,xy2,r,-,,
可计算出当,,,时,v,0.724611v,v,0.6891574v 1y,x,
,,,时,v,,0.724611v,v,0.6891514 2y,x
22V,v,v,v ,xy
3—3解:设点源强度为Q,根据叠加原理,流动的函数为
合速度
,y,y,y-3a ,arctgarctgarctg,,, 2,xa2,xa2,x,,
,,,x,ax,ax两个速度分量为 x,,,,,222222,2,,,,x,a,yx,a,y,,,,x,y-3a,,
,,,yyy-3a v,,,,,y222222,2,,,,x,a,yx,a,y,,,,x,y-3a,,
3对于驻点,v,v,0x,0,y,a,解得 xyAA3
3—4解:设点源的强度为Q,点涡的强度为T,根据叠加原理得合成流动的位函数为
,, Q ,,lnr,2,2,
,,,,1,1,1 V,,;V,,r,,rr2,r,,r2,
V,, 速度与极半径的夹角为,,, arctgarctgVQr
,,yy3—5根据叠加原理得合成流动的流函数为 ,,,Vaarctgaarctgy,,,,,,yaya,,,,
,,,,,,,,,,axaaxa 两个速度分量为 v1,,V,,,,,x2222,,,,,yx,a,yx,a,y,,
,,,,yy v,,,Va,,,y,2222,y,,,,x,a,yx,a,y,,
由驻点 ,,v,v,0得驻点位置为,3a,0xy
yy 零流线方程为Vy,Vxaarctg,aarctg,0 ,,y,ay,a
2ay222 对上式进行改变,得 x,y,a,,y,,tan,,a,,
当时,数值求解得y,,1.03065a x,0
3—9解:根据叠加原理,得合成流动的流函数为
yyQQ,,vy,arctg,arctg, 2,y,a2,y,a
x,ax,aQQ 速度分量为v,vy,, ,x2222,,22,,,,x,a,yx,a,y
x,ax,aQQv,,, y2222,,22,,,,x,a,yx,a,y
,,Qa2由,,v,v,0a,0得驻点位置为,, xy,,,v,,,
QyQyvy,,arctg,arctg,0,过驻点的流线方程为 ,,2y,a2y,a
2vyy,,上面的流线方程可改写为y,arctg,arctg Qy,ay,a,,,,,2vyy2ay, ,,,,?tany,tanarctg,arctg,222,,,,Qy,ay,ax,y,a,,,,
容易看出y=0满足上面方程
2ay222当时,包含驻点的流线方程可写为 y,0xya,,,,,2vy,,,tan,,Q,,
Q2y22当,,,,,时,包含驻点的流线方程为 a,v,,1xy1,tany2,
3—10解:偶极子位于原点,正指向和负x轴夹角为,,其流函数为
ycos,xsinM,,, ,,, 当时 ,,45222x,y,
M2y,x ,,,222,2x,y
,3—11解:圆柱表面上的速度为v,,2vsin,, ,2,a
222,,,,,,v2222 ,,v4vsin,,,,4sin,,4sin,, ,22222,,2,a4,av2,av4,av,,,,,
22,,,,v,2压强分布函数为,,,,C,1,,1,4sin,1, p,,,,v4,asin,v,,,,,,
kg,5u1.7810 ,,n
,VL1.225,30,0.66, R,,,1.23876,104—1解:查表得
大气的粘性系数为el,5u1.78,10
平板上下两面所受的总得摩擦阻力为
0.664122F,2,,,VS,0.789N ,2ReL
4—2解:沿边阶层的外边界,伯努利方程成立
12,cp,v,,2
,v,p2mm,12m,1,,,,vxvxvx,,v,,,,m,000 xx,,
pp,,?当m,0时,0;当m,0时,0xx,,
?m,0代表顺压梯度,m,0代表逆压梯度
2v3y1y,,,,4—4解:(a)将x带入(4—90)中的第二式得 ,,,,,,v22,,,,,,,
,,,vv39,,xx ,, ,,,,,1dy,,,0vv280,,,,
,,u,v3x,由牛顿粘性定律,,下面求动量积分关系式,因为是平板附面层 ,,,uuw,,,,y2,,y,0
,,dv,,2d,w?,0积分关系式可表示为 ,v,dxdx,
13dx将上述关系式代入积分关系式,得,,边界条件为x=0时, ,,0d,u140,v,
积分上式,得平板边界层的厚度沿板长的变化规律
4.64,,
Rxl ,,39,,,?Rx,0.646,,4.64lx280
,,,v3,x,,,,1,dy,,,,,0v8(b),,,
,3,,,?Rx,,4.64,1.74lx8
, ,,Rx,4.64lx
,34.64x(c)由(a)知;,,,u,w,2Rxl
,0.646w(d) 由(—)得432C,,f12Rxl,v,2
?CRx,0.646fl
l12,,X,C,vbdx假设版宽为bfF,,02
X1.292(e)单面平板的摩擦阻力为F 摩阻系数为C,,f12Rxl,vs,2
?CRx,1.292fl
4—6解:全部为层流时的附面层流厚度由式(4—92)得
L ,,,L,5.48,0.01918RLe
全部为湍流时的附面层流厚度由式(4—10)得
1,5,, ,L,0.37LRL,0.0817e
5—3证明(1)将r()表示为下列三角级数 ,
,cos,,, 将其代入(5—35)得 ,,r,2vAAnsinn,,,,,,,0sin,n,1,,
,,,1dy2dydyfff ,,,,A,A,,可得 A,,,d,;A,cosn,d,ncosn,01n1101,,00,dx,dx,dxn1
dycos,f对于平板,A,,,,,故有, ,0A,A,?A,0?r,,2v,012n,dxsin,
当,,r,,0时,,不满足后缘条件 ,,,
,1,cos,,,(2)将将其带入(5—35)积分得 ,,r,2vAAnsins,,,,,,,0sin,n,1,,
,,,Ansinn,sin,,,,,,,,11,cos,d,dy,fn1 ,, ,,,,,,vd,22,,00,cos,,cos,cos,,cos,dx,,
,,,,
,dy f,,,,,A,A, ncosn,01,sin,n11
,1dy2dyff A,,,d,A,cosn,d,01n11,,0,dx,dx
dyf对于平板?A,,;A,A,?A,0, ,0012n dx
1,cos,,, ?r,,2v,,sin,
当,,时,r,,0,满足后缘条件 ,,,
315—2解:设在弦线处布涡的强度为,则该涡在弦线处产生的诱导速度为 ,44
,, ,,vyic,c2,2
若取3弦点为控制点,在改点满足边界条件4
dydy,,,,,dy,,2fffvcv,,,?,,,,,L,,,,,,,,, 因此开力为 vcv,,,,,,,,,,cdxdxdx,,,,,,,
Ldydy,,ff开力系数为对于平板 C,,,0,,2,,,L12dxdx,,,vc,2
,?C,2,,;C,2, LL
dy1f,0.8,2x;0,x,0.4,,,dx85—4解对于薄翼型,C,2,对于2412翼型, Ldyf,,,0.05550.8,2x;0.4,x,1dx
1 令,,,,arccos0.2x,1,cos,,则当x=0.4时, 112
dy1f,0.8,0.2;0,x,arccos0.2,,dx8 dyf,,,0.05550.8,0.2;arccos0.2,x,,dx
,arccos0.21dy11f ,,,,,,?,,1,cos,d,,cos,,0.21,cos,d, 011111,,00,dx,8
,1 ,,,,,0.0555cos,,0.21,cos,d, 111,arccos0.2,
,2dyf A,cosn,d, n11,0,dx
arccos0.2,212,,,,,,,,,,,,,,,,Acos,0.21cos,d,0.0555cos,0.21cos, 111111,,,,0arccos0.2,8,,,
arccos0.2,21,,,,,,,,,,Acos,0.2cos2,d,0.0555cos,0.2cos2,d, 2111111,,,,0arccos0.2,8,,
4,, C,A,Amp12,4
2225—5解:根据余弦定理 c,a,b,2abcosc,0.9849?c,0.9924
2222222a,c,ba,a,b,2abcosc,ba,bcosc cosB,,,,0.9962 2ac2acc
00 ?,B,4.9878,5
,1dy,,,,C,,,,,,L,f,2,;,1,cosd,0011dx0
dy20f,tan5;0,x,dx3
dy2折算后的迎角为00f, ,tan170;,x,110dx3
121,,令,,x,1,cos,当x,时,,arccos,,1.9106弧度 ,,11233,,
1.9106,1100,,,, ?,,tan51,cos,d,,tan1701,cos,d,01111,,01.9106,,
001.91061.9106tan5tan170,,,,,1,cos,d,,1,cos,d,,,0.1253 1111,,00,,
10,,,,,,?C,2,,2,0.1253,1.8837,,,, L0,,180,,
325—7解:,,,,,,y,kxx-1x,2,kx,3x,2x f
dydy32ff ,,x,1,正号舍去,,,k3x,6x,2 令,0得 3dxdx
22dydy3ff ,,,k6x,6,0将代入,得 x,1,22dxdx3
3因此yf,2在处取得极大值,% x,1,f3
3代入得k=0.052 yx,1,f3
将dy3311,,2f令,cos,,cos,,k代入(1)得 ,,x,1,cos,,,111dx4242,,
,1dyf,, ?,,1,cos,d,011,0,dx
,,,,?C,2,,,,,2,0.0524,0.1105,1.0235 L0
,2dyf A,cos,d,,0.07794111,0,dx
,1dyf A,,,d,,0.0458701,0,dx
,2dy,,fA,cos2,d,,0.0186 ,,211,0,dx,,
,1,,C,2A,A,,0.533,, C,A,A,C,,0.1798L01L21L44
L1d,26—5解:根据开力线理论,,v,d,, yi,L,,,4,,,d,,2
112222,,,,2,d,,12,,2,,,,,0已知 ,,,,,,1,;,1,,,,,,,,,02Ld,LL,,,,,,,,,,,,
122,,,2,,1,,,,,LL,,3,LLL,,2,,0,,?v,,,d,;令,,,cos,;,,,cos,;d,,sin,d,yi1112,L,,,,,222L2
2,,,33sin,cos,sin,3,,0110 则 ,,,,,vd,1,,yi12,0,,Lcos,cos,8Lsin,,,1
,0L23,,,,时,,,,yiv438L当 ,3L0,时,,,,,,,vyi24L
6—6解(1)有叠加原理可知,a处的下洗速度为
2,,,,,,L,,2LL,,,,,,,a,,2,,a,,,,,,,,,22v,,,,21,,,1,yi,,,,,,222L4,a,aLLLL,,,,,,4,222,,,,,,,a,a,a,,,,,,2,,,,,,222,,,,,,,,,,,,
2,,L,,2,,,a,,v2,L2,,,,,,yi,,,,,1,;C,,a处的下洗角为 L,,12VLVVL,a,,,V,,,,2,,,,
2,,L2,,,,,a,,C2L,,,,,,1,,,,,2aCVL因此L,,,代入下洗角中得 ,,,,2a,,
,C,,,22L,,,,,,,,,,,,,,C00L,,2,,2CL,1,1,(2)对于椭圆翼,, ,,
22,,,,LL,,,,22,,,,,a,a,,,,22C1,,,,,,,,L ,,d,,1,,1,,, i0,,,,2,,a,,2a,,,,,,,,,,,,
2,,L2,,,,,a,,i2 dd1,,,,当时 ,,8,a,0.4?,,1,,,,d,2a,,
,,,,
ddi ,0.26 d,
,LC,,,1L6—9解:C,,0.274;C,,4.68rad LL,121,C,L,Vs,2
C00 L,,,,,3.354,;,3.354,1.2,2.13 0,CL
2CL,,0.00385 CDi,,
2LCL6—11解:,, ,,0.846;,1,,,0.09985CCLDi12,,,sV,2
x12i % x,C,Vs,1017N;,4.71iDi,2L
7—1解状态方程 p,,RT
P,506.62KPa;P,506.62KPa;P,1019.25KPa123
1;,;,,,,,,12132 2
v;v,w;v,v12132
T,300K;T;T123
(1)由状态1等压膨胀到2的过程中,根据质量守恒方程
1v,2v所以 ,,,21212
,T21等压变化,,T,T?,,2;T,2T,600K 112221,T12
由,,,等容变化,根据质量方程 2,332
PTP323等容变化,?,2;T,2T 32TTT322
(2)介质只在过程中膨胀做功 w,p,v,21.53KJ1,2
(3),,,Q,CT,CTm,182.996 pv
(4) ,q,du,pdv?du,,q-pdv,161.466KJ
r,,,,,Pkj21(5),,,, ,,?,,,sCln0.298v,,kP,,,12,,,,
7—3解根据质量守恒小截面与A截面的流量相等即 2
PAPAA,q,,010211,,,,,,c,q,c,q?,q,,0.388122 ATT 200
?,,0.252
0, ,,11.91
00总的外折角度, ,,,,15,26.917—4解:气流从Ma=1加速到Ma1=1.5需要的外折角度为
,,,,PPPPP22002查表得Ma2=2.02,,,, ,,,,0.456,,,,PPPPP01,,,,101
7—5解:经过正激波时绝热,总温度T不变 0
Tr,1T,220根据总静温之比,1,Ma?, T2Tr,10
TRT22r,00T;CRT?,,,r,,r,1r,1
vv22波后的速度系数为,,, ,2C2rRT0
r,1
2rRT10根据波前波后的速度关系,,,1 ,?,121vr,12根据马赫数与速度系数的关系,得得波德马赫数
22,12r,1 Ma,1r,12,1,1r,1
总压损失系数为 ,
11,2r,1r,1,,,,2rr,1r,1Ma,,21,,Ma,,,,, 12,,,,r1r1r,1Ma,2,,1,,