北航空气动力学
5-1 一架低速飞机的平直机翼采用NACA2415翼型,问此翼型的
xfc,和各是多少, f
f解:此翼型的最大弯度=2%
x=40% 最大弯度位置f
c 最大厚度=15%
1-2 有一个小α下的平板翼型,作为近似,将其上的涡集中在弦 54
,,13Crad点上,见图。试证明若取弦点处满足边界条件,则=2π l4
31解:点涡在处,在处满足边界条件,即 44
,
v',v',,,3 ,42,b,
4
dyfv',v,v,,, 代入边界条件表达式 中, dx
,,,,v, ,b,
,,,bv,? ,
?,,,v, 升力 ,
,,v,,bv, ,,
2,,v,,b, ,
,
C,,2,,y 12vb,,,2
dCy,,1?C,,2,rady d,
,(,)5-3 小迎角下平板翼型的绕流问题,试证明可以有以下两种形式
的解:
,cos,(,),,2v,1) ,sin,
,1,cos,(,),,2v,2) ,sin,
而解1)满足边界条件,解2)不满足边界条件。
解:迎角弯度问题的涡强方程为
bdy1d,,v,(,,) (*) ,,x(,)0,dx2,
置换变量后,上面方程化为
dy,,,,,()sindf,,v(,,) ,,02(cos,cos)dx,,,1
,cos,(,),,2v,对1) ,sin,
带入方程(*)
,cos,,,2v,sind,,,sin,,左 ,02,(cos,,cos,)1
,,,,2vcosd,,, ,02,(cos,,cos,)1
,,,,vcosd,,,, ,0,cos,,cos,1
,,sinv1,,,,, sin,,1
,,v, ,
,v(,,),,v,右 故方程满足 ,,
,1,cos,(,),,2v,对于2), ,sin,
代入方程(*)
,1,cos,,,,2vsind,,,sin,,左 ,02,(cos,,cos,)1
,,,,2v(1,cos)d,,, ,02,(cos,,cos,)1
,(1,cos,)d,,,, 0cos,,cos,1
d,cos,d,v,,,,,(,,) ,,00cos,,cos,cos,,cos,,11
,,sinv01,,,,(,,) ,,,sinsin11,,v,, 右 故方程满足 ,
后缘条件:
,cos,(,),,2v,? ,sin,
cos,
,,,2v,,,,,0,,,当后缘处 ,sin,
,,0 故不满足后缘处的条件
,1,cos
,(,),,2v,,? sin,
,1,cos0
,,,2v,,2v,,,, 后缘处, ,,sin0,
,1,cos
lim,,, 当时取极限 sin,
,0,sin,lim cos,
,0,sin
, cos,
0
,,0 ,1
, 故=0 ,,,
满足后缘条件
5-4 NACA2412翼型中弧线方程是
12y,[0.80x,x]0,x,0.4 f前8
2y,0.0555[0.20,0.80x,x]0.4,x,1.0 f后
,,Cmx见图。试根据薄翼型理论求,,和并与表5-1中实0yZF0
,,,1x,0.25m,,0.05309,,,2.095C,2,rad验数据相比较。[, ,,] Z0Fy0
,C,2,/rad解: y
dy,1f,,(1,cos,)d, 0,0dx,
bx,(1,cos,)b,1 由变量置换 取 2
x,0.4 知时
cos,,0.2,,78.463,1.369rad,0.44, ff
1,dy[0.82]0.10.25,x,,x,f,8, 又 dx,0.0555[0.82]0.04440.111,x,,x,
,,f1
?,,[(0.1,0.25x)(1,cos,)d,,(0.0444,0.111x)(1,cos,)d,]0,,0,f,
,,111f,{[0.1,0.25,(1,cos,)](1,cos,),d,[0.0444,0.111,(1,cos,)](1,cos,)d,},,0,f22,
,xx,,2.095 (注意:是焦点,是最大弯度位置) fF
dy,1fm,(cos2,,cos,)d, Z,002dx,1f,(0.1,0.25x)(cos2,,cos,)d, ,02
,1,(0.0444,0.111x)(cos2,,cos,)d, ,,f2
,,0.053
,C,0.985,2,实验值为 y
,,,1.90 0
x,0.243 F
m,,0.05 Z0
,155-5 一个翼型前段是一平板,后段为下偏的平板襟翼,见图。
,C,,5 试求当时的值。 y
22,AB,AC,BC,2AC,BC,cos165,0.99246,1解:
,BC,sin15,AB,sin, 1
,,,,4.98,5,0.087rad 1
,,,,15,,,10 21
dy,,,tan,,0.087,, 1dx,,AC
dy,,,,tan,,,0.1745,, 2dx,,BC
2b,xb?x,(1,cos,) h32
?,,1.916 h
dy,f,,(1,cos,)d, 0 ,0dx
1.91,11,0.087(1,cos,)d,,(,0.1745)(1,cos,)d, ,,0911.,,11.916,,,,0.087(,,sin,),(,0.1745)(,,sin,) 01.916,
,,,0.0939rd,,5.38
C,2,(,,,),,/180,1.69 y0
y,kx(x,1)(x,2)f,2%b,15-7 一个弯板翼型,,,k为常数。。 f
,mC,,3试求:时的和。 yZ
dy,1f,,(1,cos,)d, 解: 011,0dx,
,12,k(3x,6x,2)(1,cos,)d, 11,0,
,11112,k[3(,cos,),6,(1,cos,),2](1,cos,)d, 1111,0222,
,13312,k[cos,,cos,,](1,cos,)d, 1111,0424,
5,,k 8
dy,1fm,(cos2,,cos,)d, 111Z,002dx
9,,k, 32
2k3y,y,,0.02x,1,当时, max333
?k,0.033,0.052
5,,,C,23,,,0.052,0.533,,, y1808,,
191m,m,C,,,0.052,,,0.533,,0.179 ZZy04324
,V5-10 低速气流以小流过一个薄对称翼型, ,
C
y,4()x(1,x),试用迎角问题和厚度问题,求 C2
xC? 表面与的函数关系表达式。 P
1C(x,)? 的值 P2
解:应用薄翼理论,将该问题分解为迎角问题和厚度问题。
,迎角问题:攻角流过平板
A,0A,,, n0
,()2cot,V,,,故 ,2
x,,1,
C,,,,,,,2cot,2P ,V2x,
厚度问题:攻角0度,流过对称翼型
dyc,,d
1,d2
C,,P,c 0,x,,
1,,c,d22(12),, ,0,x,,
4c1,,,,[2,,2x,1ln,,x] 0,
,,4c1,x,,2,(2x,1)ln ,,,x,,
C,C,CPPP ,c
1,x4c1,x, ,,2,[2,(2x,1)ln]
,xx
8c1,C,,2,x,当时, P,2
2b,1.5m,,4S,35m6-1 有一平直梯形翼,,, 1
, 求该机翼的值。
b,1.5?,,4 解: 1
?b,,b,6 01
(b,b)l10?S,,,2 22
35,2?l,,9.33 1.5,6
22l9.33?,,,,2.49 S35
,tan,,,6-2 试从几何关系证明三角翼的 0
证明:
2l
,, S
cl0tan,,S,c, 而 00l22
2cl0tan?,,,,0 lS
2
2lc0,,4
l
c,02
2,,C6-8(旧
) 使用三角级数法计算无扭转矩形翼的环量分布,,y
,,,,,(,),沿展向取,,三个位置(n=3),试求出的表达式。 623
解:根据升力线理论的三角级数解法,可知
,
,(,),2lVAsin(n,) ? n,,n1,
A系数可用下式确定 n
,
,,sin,,Asin(n,)(,n,sin,) ? an,
n1,
,C()b(),,y,, ,4l
b(,),const对该题,
,C,, y,
,,const,,,const 0a
b,?,,0.25, 4l
,,,,,将,,代入?得(?取三项) 632
,(,sin,)Asin,,(,3,sin,)Asin3,,(5,,sin,)Asin5,,,,sin,135a
,,,,,,,35,,,,,,(,sin)sin,(3,sin)sin,(5,sin)sin,sinAAA135a,6666666,,,,,,,5,,,,,,,(,sin)sin,(3,sin)sin,(5,sin)sin,sinAAA,135a333333,
35,,,,,,,,(,sin)Asin,(3,sin)Asin,(5,sin)Asin,sin,,,,,135a,2222222,
,0.375A,1.25A,0.875A,0.125,135a,,0.96651A,1.83253A,0.21651,15a即 ,1.25A,1.75A,2.25A,0.25,135a,
A,0.0277,A,0.0038,A,0.232,解得 3a5a1a
?,,(),2lV(Asin,,Asin3,,Asin5,) ,135
,lV(0.464sin,0.0554sin3,0.0076sin5),,,, ,a
,,,,,4C,2,rad6-8一个有弯度的翼型,,, y0,
,,,5,,8若将此翼型放到一个无扭转的椭圆翼上,试求此机翼在
C时的。 y
,C,(,,,)C 解: y0y
由于是无扭转机翼
,?,,,4,, 00,
,C,2y,,C,,,4.486/rady,,2 Cy,1,1,,5,,
?C,0.078,(8,4),0.94 y
V,300km/hG,14700Nh,3000m6-9一架重量的飞机,在以巡,
2NACA,,6.2Y,GS,17m航平飞(),机翼面积,,23012翼型,
,,,C(,C)(,,1.2,C,0.108/), 无扭转椭圆形平面形状。求:,,Ly0L,
C(,C) DXVi
YG,,,,0.274Cy 解:1130022,,0.90913,(),1.7VS,223.6
,?,,,,,1.2因是无扭转椭圆翼 00,
,57.3,C0.108,57.3y,,,C,,,4.69/rad,0.082/y,0.108,57.3Cy,1,1,,,6.2,,
,?C,C(,,,) yy0
Cy?,,,,,2.140 ,Cy
22C0.274y,,,0.00385C Xi,,,6.2
4G,7.38,10N6-10 有一架重量的单翼飞机,机翼为椭圆形平面形l,15.23m90m/s状,,现以的速度在海平面直线飞行,是计算其X,涡阻及根部剖面处的值。 i0
4,,G,7.38,10 解:平飞
G.S.qX, X,iI
2GG= XI,,
427.38x10()X,故, i122x1.225x90,,2
X,代入,得1507 i
,,,0Y=2 .,,,4
,=55.99 0
2,,6l,12mG/s,900N/m6-11 矩形机翼,,,翼载荷 。试计
v,150km/h算飞机在海平面以平飞时的诱导阻力以及诱导与总升
力之比。
,,0.049解:矩形机翼
2Cy,(1,,)CX故 i,,
,900C,,,0.846y 11232,vS,,1.225,(150,10/3600),22
20.864?C,(1,0.049),0.0399 Xi3.14,6
21l2,,,,,0.0399,1063.83,,1018.7XCvS iX,i,2
CX0.0399Xii,,,0.047 C,0.846y
,,2.56-12一个A=9,无扭转值机翼在某雷诺数下实验所得的
,,,C,1.22,,,1.5C,0.084/C,,曲线见图。,,,若其L0LLmax
,,C他参数不变,只是A减小为5,求此时和,并画出A=5时机翼0L
,C的曲线。 L
,,2.5 解:无扭转直机翼
,C,1.22,,,1.5C,0.084A=9时,, L0Lmax
,,,,1.5当A=5时,不变 00
,CL,,,?CL, CL,1,(1,,)
,,
,CL,?0.084,, CL,1,(1,,)
9,
,假定为0,则
,C0.084,,LC,,,0.1012/L,, 0.084,57.3CL1,1,,9,,
故
,C0.1012L,,,C,,,0.074/,4.24/radLA,5,0.1012,57.3CL,1,,1,(1,),5,5
8-4 二维翼型在气流中这样放置,使它的最低压强点出现在下表面。当远前方来流马赫数为0.3时,这点的压强系数为-0.782。试用普朗特—葛劳渥法则,求出翼型的临界马赫数。
1
C,C,,0.782M,0.3P解:时,,应用普—葛法则,即,minP,min,
C,0 PM,
,0.782
?C,Pmin ? 21,M,
CPM,,0,0.782,,C,,0.746P或用 M,,021,0.3
,0.746C,P则 min21,M,
又应用等熵关系
,,,,,,1P1,,,20,,,1,1,1M,,,,,,2min1M,,2Pmin,,,,minP,2min,,,,,1,, ,,1P2,,,,1,M,P1,,,,202,,,1,,M,,,,2P,,,,
M,1临界马赫数时 min
,
,,1P,,21,,,,2min1?,,M,,, ,,12,P,,,,,,
,,,,,1,,,,,2P221,,,,,2,,,,,,,C11M1,,,,P,,,22? ,,min,P12MM,,,,,,,,,,,,,,,,
C,0.987M,0.654联立??得, P,min
M8-6 某翼型在增大到0.8时,翼型上最大速度点的速度已达音速。,
问此翼型在低速时最大速度点的压强系数是多少,假设普朗特—葛
涝渥法则可用。
M,0.8C,?解: 求 PminM,,0,临
C,0PM,C,P 21,M,
,,P2,,C,,1 P2,,P,M,,,,
,,,,,1,,,221,,,,,2,,,1M1,,,,, ,,2,12M,,,,,,,,,,,
,,0.4346
2?C,,0.4346,1,0.8,,0.26076PminM,0 ,
M,0.6,8-9 一展弦比为10的矩形机翼,以马赫数作等速水平飞,
,C行,试求该机翼的升力线斜率的,并将此结果与相同机翼在不可y
,C压缩流中的进行比较 y
,C为: 解: 相同翼型在不可压流中的y
,Cy,,C,,5.09y2C y,,1,(1,)
,,
M,0.6时,根据普朗特-葛劳渥法则,对应不可压机翼后掠角还,
,C',',,,,8是0,仍为矩形翼,展弦比变小为,其不可压为 y
,Cy,,C',,4.86y,C y,,1,(1,)
',,
14.86,,C,C',,6.075yy而 ,0.8
,M,2109-3 二维平板在2千米高度,以飞行,迎角为。试分别,
用激-膨理论和线化理论,计算上下表面间的压强差。
解:如图,用激-膨理论
P2下,,1.69M,2,,10,下,查图7-20, 1P1
P1,,0.1278M,2,',26.38上表面膨胀波,查表5,,对应, ,P0
P2,,,0.07,',10,36.38故从M=1外折后, P0
P0.072上,,0.55所以, P0.12781
P22C,(,1),(1.69,1),0.2464P下故 22,MP,,2,,
1
C,(0.55,1),,0.1607 P上2.8
,C,C,C,0.2464,0.1607,0.407 PP下P上
2,C,,线化理论 PB
2,10/1802,3.14159/18,C,,,,,,0.2015 P上21.7322,1
C,0.2015 P下
,C,C,C,2,0.2015,0.403 PP下P上
9-6 有一机翼,平面形状如图所示。试求超音速前缘和亚音速
后缘的马赫数范围
c2c3,arctan,arctan,前解: b3b
2
1
sin,, M,
2tan,M,1, ,
,,,故 时 超音速前缘 前
22cc2,,2M,,1,M,1,, 即 ,,b33b,,
2
c4c3,,,arctan,,arctan后b 3b
2
1M,Mcos,M,,,n,,后24c,, 1,,,
3b,,
M,1 要亚音速后缘,即 ,n
24c,,M,1,,, 故 ,3b,,
224c16c
,1,M,1, 所以,当时满足要求。 ,229b9b
,,V,450m/s459-7 有一三角形机翼,前缘后掠角为,现以0,速度飞行。试考虑飞行高度分别为海平面、5500米和11000米时,该机翼前缘性质作何变化。
tan,,1解: 前
V,M,,1.324a,340海平面时, , ,a
111tan,,,1, 20.870.87M1,,
2450,,tan,,,1,1.0012,,h=5500 a=318 时 318,,
2
450,,tan,,,1,1.1519,,h=11000 a=295 时 295,,
故前缘分别是亚音速、音速、超音速。