A) Show That an Elliptical Cylinder Is Obtained When a Circle of Radius a in the Z Plane

Problem 1: a) Show that an elliptical cylinder is obtained when a circle of radius “a” in the “z” plane is transformed conformally into the “” plane using b2 b2   z  (b  a)  0.739 z a2 Assuming irrotational flow, obtain a general expression for the velocity of flow crossing the vertical axis of symmetry. The circle is centered at (0, 0). b) Show that the velocity on this line is everywhere greater than U and that a maximum value of 1,15 U occurs on the elliptical boundary at the point of intersection with vertical axis. Find also the distance from this intersection to the point at which the velocity is q()=1,1 U .

Solution: a) The circle in “z” plane has the equation: z  aei ( 0) By applying the Joukowski’s conformal mapping, we will have b2 b2 b2   z   aei   aei  ei ( 0) z aei a Since ei  cos   i sin  and ei  cos   i sin  ( 0) the equation (2) becomes  b2   b2        a   cos   ia   sin  ( 0)  a   a  By separating the real and imaginary parts, we get  b2      a   cos   a  ( 0)  b2      a   sin   a  It is easy to see that 2 2 2  2  1  b2   b2  ( 0)     a   a    a   a  which is the equation of an ellipse. This can be written also as 2 2 2  2  1  b2   b2  ( 0) a21  a21   2   2   a   a  if we take a = 1 as scale factor and if we replace b2/a2 = 0,739, the ellipse will have the final equation: 2 2 2 2   1    1 ( 0) 1 0,7392 1 0,7392 1,7392 0,2612 The velocity is given by formula

W (z) w  d ( 0)

dz zz() The velocity field around the circle of radius “a” is  a2  W (z)  U 1    2  ( 0)  z  and from the Joukowski’s mapping one yields d d  b2  b2  z    1   2 ( 0) dz dz  z  z Introducing in (9), we will get a2 a2 a2 1 1 1 z2 z2 z2 w()  U  U  U ( 0) b2 b2 a2 a2 1 2 1 2 2 1 0,739 2 z a z z zz() By inverting the Joukowski’s mapping, we have b2 1   z   z2  z  b2  0  z    2  4b2  ( 0) z 2   (the formula with “-“ doesn’t make sense, since we need that    z  ) On the vertical axis, we have   i and, replacing in (13), we get 1 1 z(i)  i   2  4b2   i   2  4b2  ( 0) 2   2   2  2 2  2 2 z2     4b    2 b2    2                0,739  ( 0) 2 2  2a 2 2   2a 2  a 4a  4a a   4a  Taking a = 1 and replacing in (12), we will have 1 1 2 2   2    2     0,739     0,739  1  2 4   2 4      w(i)  U  U ( 0)  0,739  2 1   2  2    0,739   0,739   2   2 4     0,739     2 4   

      0,261  w(i)  U 1   w(i)  U ( 0)   2     2      0,739   0,739   2 4       b) From (5) we get the value of () on the top of ellipse: b2    a(1 ) sin  1 0,739  0,261 a2 2 and replacing this in (17) we will find the velocity on the top of ellipse:       0,261 w(0,261i)  U 1   1,15U ( 0)   2    0,261 0,2612      0,739   0,739   2 4       which is the maximum velocity on () axis, since we cannot talk about velocity “inside” the ellipse. The point where velocity is 1,1U can be found from equation (17):       0,261 U 1   1,1U ( 0)   2     2      0,739   0,739   2 4        By putting  x , we get the equation: 2 0,261 0,261 1  1,1   0,1 2 2  x  x2  0,739   0,739  x  x2  0,739   0,739     2 0,261   x  x2  0,739   0,739   2,61   0,1 2   x  x2  0,739   2,61 0,739  1,871    x  x2  0,739  1,871  1,3678  x2  0,739  1,3678  x By solving this equation, we get:  x  0,4138     0,8276 2 The needed distance will be   0,8276  0,261  0,5666 (at a scale factor a = 1)

Problem 2: A solar car design calls for a body shape which is symmetrical airfoil of length 6m and maximum thickness 0,35m. The driver is to be positioned 1,5m back from leading edge. The maximum thickness to chord ratio is given by 3 3be r b whilst the ratio  1 e  e cos  and  1 e  e cos  4b b r a) Confirm the derivation that the real and imaginary components of the shape are given by   2b cos  and   2be1 cos sin  respectively.  b) Show that maximum thickness occurs at   3 c) Determine the flow velocity on the airfoil surface at the site proposed for the driver when the vehicle is travelling at 100 km/h on a horizontal road in still air.

It is assumed that the airfoil performance is not influenced by its proximity to the road and that the edge effects associated with its width are negligible.

Solution: a) The symmetrical airfoil can be obtained by applying Joukowski’s transform on a circle of radius “b” whose center is displaced on “x” axis by amount “m” as it is shown in the figure. a2   z  ( 0) z y η z ζ a2   z  z

-a -m● O a x O 2a ξ ζ (-a-2m)

The displaced circle will have the equation: z  m  a  mei ( 0) which we replace in (20): a2 a2   m  a  mei   m  a  mei  ( 0)  m  a  mei aei  m(1 ei )

aei   m  a  mei   m ( 0) 1 ei 1 ei  a For an aerodynamic body we have m m  a  ei 1 ei   1 a In this case, we can expand the fraction is a geometric series and we keep the first 2 terms: 1 m  1 ei 1 ei  m i i a ( 0) 1 e  1 e   a

i i  m i  (23)    m  a  me  ae 1 e  1 ( 0)  a 

   m  mei  aei  ei  mei ei 1 ( 0)    mei 1 2acos  mei 2 ei 1    mei  11 ei 2  2acos  mei ei 1ei  ei  2acos    m1 ei  2i sin  2acos ( 0) By separating the real and imaginary parts, we will have   Re()  2acos  2m sin2   2acos ( 0)   Im()  2m1 cossin If we denote b  4a (the chord) m ( 0)   2a then   4asin1 cos  bsin1 cos ( 0) The maximum value of () can be found by differentiating the above expression: d  0  cos1 cos  sin2   0  2cos2   cos 1  0 d 1 1 8  1  2  cos   1 ;      0 ;  ( 0) 4  2  3  Since the value 0 is not convenient, we will have maximum for  = 2/3 By replacing in (30), we get 3  1 3 3 max  b 1   b ( 0) 2  2  4

In our problem, we have: b  6m , 2max  0,35m 4 2  0,35 0,7  4a  6  a  1,5    max    0,022 3 3b 3 3  6 18 3 b 6  m  2a     0.022  0,066 2 2 c) The velocity field is given by: W z dF d   2  w where W (z) U z 0           ( 0) d / dz dz dz   z  z0  z0 = the center of displaced circle, 0 = its radius, z0 = -m , 0 = a+m

 a  m2  W(z) U 1      2  ( 0)  z  m  From (20) we have d a2  1 ( 0) dz z2 On the circle, we can write

i i z  m  0e  z  m  a  me ( 0)

W (z)  U 1 ei 2  U ei  2i sin  W (z)  2U sin  C     C  ( 0)

d a2 a2 ei 2 (35)   1  1  1 dz i 2 i i 2 2 C  m  a  me  me 1 ae   m i  1 1 e   a  2 d i 2  m i  i 2  2m i    1 e 1 1 e   1 e 1 1 e  dz C  a   a 

d  2m  2m   1 1 ei 2  ei 3 ( 0) dz C  a  a Since the position of the driver is at 1,5m from leading edge, that means  = -a which is at maximum thickness  = 2/3 (from (28) we have  = -2a cos) 2 (37)  W (z)  2U sin  U 3 ( 0)  3 

4 d  2m  i 2m  2  0,066  1 i 3  2  0,066 1 1 e 3 ei 2 1 1   (38)               dz  a  a  1,5  2  1,5 d   1  0,456  i0,789  0,088  1,368  i0,789  1,3682  0,7892  1,579 dz

U 3 (33) + (39)  w(z )    1,096U ( 0) D 1,579 

Since U = 100 km/h  wzD   1,096 100  109,6 km/h

Problem 3:

Due to an error in construction, the airfoil shape in Question 5 is inclined at an angle of 6 with respect to the horizontal road surface. This nose up condition generates lift which is opposed by 18 bolts which connect the body to the chassis. Calculate the average loading which these bolts must carry. Data provided in Question 5 applies identically in Question 6 but the Kutta-Joukowski condition should be applied. Since the lift generated is the same in the  and z plane, the answer should be structured in the z plane only. Assume air = 1,2 kg/m3. Solution:

An inclined symmetrical airfoil by angle () with respect to the velocity will produce lift which can be computed assuming the hypothesis of Kutta-Joukowski:

WTE = 0. ( 0) That means the velocity on airfoil at trailing edge is null. This condition is equivalent to presence of a circulation around the airfoil which will interact with the main velocity and produces lift according to Kutta-Joukowski’s theorem. The flow potential in (z) plane is given in this case by formula  2  i F(z) U zei 0 ei  ln z z         0  ( 0)  z  z0  2 By differentiating, we will get the velocities field:

dF  2ei  i W (z)   U ei  0     a  m , z  m ( 0) dz  2 2 z  z 0 0  z  z0    0  On the circle in plane (z), we will get  2ei  i i W (z)  U ei  0   U ei ei   ei   ei C   2 i 2  i     0e  20e 2a  m

i     W (z)  ie 2U sin     ( 0) C  2a  m Since in the trailing edge we have  = 0, from above equation we get the value of the circulation: W  W  0    4 a  m U sin TE 0    ( 0) and from here, by using Kutta-Joukowski’s theorem, we can compute the lift:

2 L  U  4a  mU sin ( 0) Assuming that the width of the “aerodynamic” car is 1m and by replacing with numerical values, the lift will be:

U  100 km/h  100 / 3,6  27,78 m/s L  4   1,2  1,5  0,066 27,782  sin 6  1903 N