Carlos Nunes Hw5 Ae515

10
AE 515 SPRING 2015 HW#5 CARLOS ALBERTO NUNES UIUC

description

wing theory hw

Transcript of Carlos Nunes Hw5 Ae515

  • AE 515 SPRING 2015

    HW#5

    CARLOS ALBERTO NUNES

    UIUC

  • 2

    21

    pCM

    =

    (1.1)

    And:

    uu

    ll

    yxyx

    =

    = +

    (1.2)

    Using the given equations for the upper and lower surfaces:

    max max

    max

    max

    1 12 1 2

    12 1

    2 1 2

    u

    u

    u

    y t tx xcx c c c c c c

    y t x xx c c c

    y t xx c c

    =

    =

    =

    (1.3)

  • Since l uy yx x

    =

    then we have:

    max2 1 2ly t xx c c

    = (1.4)

    Plug back eq. 1.3 and 1.4 in eq. 1.1 to find the equation for the pressure distribution:

    max2

    max2

    2 2 1 21

    2 2 1 21

    u

    l

    p

    p

    t xCc cM

    t xCc cM

    =

    = +

    (1.5)

    For = 0 we found the following distribution, where both up

    C and lp

    C are the same

    When = 6:

  • We know that:

    2

    41

    LCM

    =

    (1.6)

    The drag coefficient is given by:

    ( )2

    2 2

    2 2

    4 21 1

    D u lCM M

    = + +

    (1.7)

    Where:

    2

    2

    0

    1 c uu

    y dxc x

    =

    (1.8)

    For this geometry 2 2u l = , then substituting eq. 1.4 in eq. 1.8 and integrating we get:

    2

    2 max43u

    tc

    =

    (1.9)

    Therefore, the eq. 1.7 may be re-written as:

    22

    max2 2

    4 2 831 1

    DtCcM M

    = + (1.10)

    Using eq. 1.6 and 1.10 we found the following values for the two given angle of attack:

    = 0 = 6 CL 0 0.3700 CD 2.33E-02 6.25E-02

    To calculate the center of pressure location we did the following:

    I. Calculate the lift per unit length of each element of upper and lower surface:

  • '

    ' 4

    cos

    cos

    upper upper

    lower upper

    upperi p

    loweri p

    yL C q x

    x

    yL C q xx

    =

    =

    (1.11)

    II. Compute the total lift per unit length of each element:

    ( )' ' 'upper loweri i itotal

    L L L= + (1.12)

    III. Determine the total lift and moment ( about the leading edge) of the airfoil:

    ' '

    1

    ' '

    1

    wing i

    wing i

    n

    in

    ii

    L L

    M L x

    =

    =

    =

    =

    (1.13)

    IV. And finally find the pressure center location:

    '

    'wing

    cpwing

    Mx

    L= (1.14)

    L'/q_inf 0.3690 M'/q_inf -0.1845

    xcp 0.500

    The result is in agreement with the linear theory for supersonic airfoils, since it is at 50% of the

    airfoil chord.

  • As we can see, the 2nd order theory shows that we have less suction in the upper surface and more positive

    pressure in the lower surface for positive angles of attack when compared with the linear theory.

  • The wave drag is given by:

    2

    24

    64W

    VD ul

    = (1.15)

    Using the following values for the given altitude and speed:

    We found:

    123.30[ ]wD lbs=

  • The previous charts and table shows how the wave drag changes with the length for a fixed volume body.

    We can see that the drag drastically decays as we increase the body length, but to calculate the body total

    drag we have also to consider the skin friction drag, that is proportional to the wet surface area, and then to

    the body length. Therefore, we must find an optimum length that minimizes the total drag, which for this

    case must be something between 10 and 20 ft.

    The wave drag coefficient for a von Karman body is given by:

    2 21 0.04

    5wb

    DdCl

    = = =

    (1.16)

    2

    2

    78.53

    6142.77[ / ]BS ft

    q lbs ft

    =

    =

  • von Karman body wave drag:

    19298.09[ ]ww B D

    D S q C lbs= = (1.17)

    The conical forebody wave drag:

    20262.99[ ]wD lbs= (1.18)

    The radial distribution in the von Karman body is given by (as function of ):

    ( ) ( ) ( )2sin 2

    2s l

    r

    = +

    (1.19)

    And for the conical body:

    ( )2

    bdxr xl

    = (1.20)

    Then we have the following radius distribution along the body length: