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Unformatted text preview: An Application Using Complex Numbers 62 Example of Programming with Complex Numbers Conformal Mapping of a Circle into an Airfoil 2D Flow: ()5 is velocity potential, «ﬂ is stream function. _3¢_3¢ [email protected]_ 9E “‘63:"631 U_By__623 Complex Numbers d<I>
z=$+iy (I):¢+i¢ —=u——iv
, dz ‘
Simple Example
V U (I
X u‘zUcosa véUsina
¢=chosa+Uysina ¢=UycosaUmsina <1) 2 ¢+i111 = chosoz+Uysina+incosa—an:sina 8(1)
——=Ucosoz—2Usma=u—w (9(1) 1 8(1) .04) ~—— ~2— 3(iy) i 5y 8y H
II —iUsina + Ucosa = u ——‘iv 63 CVIQ. Lu; J)
Now we map a circle in the z—plane to an airfoil in the (—plane. Streamlines in z—plane map into streamlines in Cplane. The circle is a streamline in the z—plane and the airfoil is a streamline in
the {—plane. , _ d<I> HdQ/dz (u —iv)z
(“ _ WAC " (Tc: dC/dz dg/dz The KarmanJI‘refftz mapping function is: (z+a)’A + (z —a)’A C: Aa (z+a)A— —(z—a)A A and a are real numbers and A > 1. g_ 4A22<z—a)*1<z+a>“
dz .{<z+a)i— (z—aw For large 2‘, C— An; (zA +aAzA 1+.. .)+(zA —aAzA 1+...)
(2A + aAzA 1 +.. .)— (2A — aAzA 1 + ..)
AaZz’A +
C — M—ZAzA‘la + — z +
Far ﬁeld ﬂow in z—plane is equal to far ﬁeld ﬂow in Cplane. dC / dz = O at z = a and at z = ~a. "If either of these points are in the ﬂow
ﬁeld, u — iv must equal zero there to avoid inﬁnite velocity in C—plane. Approach Locate circle so that z _= —a is inside it.
Locate circle so that z = a is on circle and u — iv there is zero. z = (1
maps into the trailing edge of the airfoil and since dC / dz = 0 there it can be sharp. _ 64 bVlL mi
lllll
% ___ l r" r LN“
§§l\
{/1 ) S , N“
B
«.3 minm
l/ﬁ
r. _
§‘\§ , uh“ Ill];
I'll“!!!
sir?
gm ii I ll Ill! ‘ . Flow around a circle with zero circulation. The center of the circle is located
at a: = —.3,y = 0.4. The circle passes through a: : a = 1.0. The flow angle of attack is
10 degrees. ‘ The inﬂow angle is oz 2’ 10 degrees, the circle radius is rC : V1232 + 0.42 =
1.3602 and the ﬂow is: 7'c 2 .
u=Ucosa—U( )c0s(20~—a) 7' 2
u : Usina — U (as) sin(20 — a) This ﬂow is not zero at z = a. 65 cvlo/ To make the ﬂow zero at z 2 a add circulation F P : 47rch sin(—ﬂ — a) ,8 : sin—1 EC
‘  TC
Then:
To 2 P .
u = Ucosa — U (—) cos(2t9 — a) — ——sm6
r 27rr
. re 2 . F
v = Usma —~ U (—) sm(20 —— a) + ——cos€
rt ' 27rr g '1
A 7/7)”) 'Illll/l'llllll ‘ m:
lbym‘ V‘ﬂf . mm
mm Flow around a circle with circulation. The center of the circle is located at
a: = —.3,y = 0.4. The circle passes through ac = a = 1.0. Note that the rear stagnation
point has moved to :L‘ = a. 66 CU’e s
e
m
9
e
d
5
2
t 3 5
2 _l!% I II... II... 'qMrial '4'5574'4 'u'i‘.‘ ‘
W... ..I l
I %% 0 n _IIEEBsw‘EBsg§§§=§l _I—§§§§§§ﬁu _§§§§;§§T
_§§§i!§§F
_£§§§ll%%
“Wm.“ﬁhhﬁﬁﬁrlilé waggllgge
maggﬁlggﬁ Hg? '1' I’ll. al'nflur. aida: 1 4i
_a%
2%? The circle mans into an airfoil shape. The included angle , T (in degrees) at the tail is: T = 180(2 — A) q2=u2+v2 1 1
§pU2 — ipq2 P—Pooz The Pressure Distribution 67 '64”: Procedure to Compute Pressure Coefﬁcient 1. Make a sequence of points on the circle. 2. Determine value of z for each point. 3. Use complex number programming to determine the value of z and
dC/dz for each point. 4. (u — iv); = (u — iv)z/%§.
5. q2 = (u ~ iv)c (u + iv)c.
6. Cp = 1 — (q/U)2. 68 «FD ‘
M1 [email protected]{
’/ cpl
% cpl 1n mat1ab a=1.0;
a1pha=0.1745;
1ambda=1.8611;
xc = 0.3;
yc =0.4;
UU=1.0;
amma= 7. 779695;
pr=180. /p1;
rc = rt( ((1: 0— —xc) .A2 + yc .AZ);
f1d= op en( 'cpm. dat' 'w ');
degv = (1:1: 360);
angv=degv ./dpr;
xv xc + ( rc .* cos(angv));
yv yc + ( rc .* s1n(angv));
zv xv + 1*yv;
zetav=1ambda*a*((zv + a) .A 1ambda + (zv—a) .A1ambda) ./ ...
((zv+a) .A 1ambda — (zva) .A 1ambda);
1m = 1ambda  1.0: dzetadzv = 4.0 * 1ambda A2 * a A2 * (zv—a) .A 1m .*(zv+a) .A1m ./ ... _(((zv + a) * 1ambda — (zv a) .A 1ambda) .A 2);
(UU*cos(a1pha))  UU*cos(2. 0 .* angv — a1pha) —
:—(gamma / (2. p0*ﬁ1*rc)) * s1n(angv);
(UU * s1n(a1p a)) — UU*s1n(2. 0*angv — a1pha) + ...
(gamma/(2. 0*p1*rc)) .* cos(angv);
wz = uv 1*vv;
wzeta = wz ./(dzetadzv + eps);
q = wzeta. *(conj (wzeta));
Cp: 1' o _ q / (UU .AZ);
Cpm = ‘Cp !
for m = 1: 360
fpr1ntf(f1d, '%7. 3f %7. 3f %7. 3f %7. 3f %7. 3f %7. 3f %7. 3f\n' rea1(zetav(m)), 1mag(zetav(m)), cpm(m), rea1(zv(m)),1mag(zv(m)5,... d rea1(w2(m)) imagcwzcmD);
en
fc1ose(f1d) , Page 1 69 1.5 0.5 2 1.5 70 0.5 \XT F’ R QQL 1.5 2 20 15 10 2.5 2 71 0.5 1.5 W? ané ...
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 Spring '07
 HenrikSchmidt

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