complex_num_appn

# complex_num_appn - An Application Using Complex Numbers 62...

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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 ¢=Uycosa-Umsina <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 C-plane. 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 C-plane. 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 min-m 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... 'q-Mrial '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 — (zv-a) .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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