Stagnation Point Flow

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    Stagnation Point flow.

    For an ideal fluid the flow against an infinite flat plate in the plane y = 0 is given by

    u = Ux, (122)

    v = -Uy, (123)

    where Uis a constant. When viscosity is included, it still must be true that u is proportional tox,for smallx and for ally.Thus, for smallx, at least, we may take

    u = kxF(y) (124)

    The governing equations for steady flow in terms of dimensional variables are given by

    equations (2.1) to (2.3). If we substitute equation (7.124) into the continuity equation (2.3), weobtain

    (125)

    This suggest that we take

    u = kxf'(Ay), (126)

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    v = -Bf(Ay). (127)

    Now, we substitute equations (7.126) and (7.127) into the governing equations (2.1) to (2.3). We

    obtain from

    (128)

    (129)

    (130)

    This last equation implies thatBA = k, so we can write equations (7.128) and (7.130) in the form

    (131)

    (132)

    We can solve equation (7.131) for the pressure derivative with respect tox to obtain

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    (133)

    Since the the term between brackets in the right side of the equation above is not a function ofx,we set

    (134)

    which implies, according to equation (7.133), that

    (135)

    If we substitute equation (7.135) into equation (7.132), we have the relation

    (136)

    and if we integrate this equation, we obtain that

    (137)

    Next, we substitute the expression for C(y) above into the equation (7.135) for the pressure,which gives

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    (138)

    To simplify equation (7.134), we chose the coefficient off''' equal to one, which implies that

    (139)

    and sinceBA = k, we have that

    (140)

    With equations (7.138) and (7.139), we can write the equation (7.138) as follows

    (141)

    with andp0 is the pressure as . The function , where

    satisfies the ordinary differential equation

    (f')2-f f''-f'''-1 = 0 (142)

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    with boundary conditions:

    No slip condition at the plate surface.

    (143)

    No flux across the plate:

    (144)

    At , we have u = kx, which implies that

    (145)

    In summary, the function is the solution of the boundary value problem given by the equations

    (7.142) to (7.145), which has no closed form solution. The equation ordinary differentialequation (7.142) is non-linear and has to be solved numerically together with the boundary

    conditions (7.143) to (7.145). Figure2below illustarte the result of the numerical evaluation of

    the boundary value problem given by equations (7.142) to (7.145) for in the range

    .

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    Figure: Functions and . The horizontal axis represents the range of values

    of considered, and in the vertical axis we have the values of the functions and

    .

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    Once we know the values of and , we can obtain the velocities u and v

    given, respectivelly, by equations (7.126) and (7.127) withB andA given, respectively, by

    equations (7.139) and (7.140), the pressure field is given by equation (7.141).

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