Homework: Ideal Flows

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Solution of a the velocity field around a cylinder, and pressure distribution on the cylinder. Stream-function of an asymptotic suction profile. Hill's spherical vortex

Transcript of Homework: Ideal Flows

  • THE UNIVERSITY OF TEXAS AT DALLAS, MECHANICAL ENGINEERING DEPARTMENT

    Incompresible Fluid Mechanics, MECH-6370,HW 3

    Edgardo Javier Garca Cartagena

    October 21, 2015

    PROBLEM 1

    The streamfunction for flow over a circular cylinder is =Ur sin(1 r

    20

    r 2

    )Evaluate the pres-

    sure distribution on the cylinder.

    vr = 1r

    =U(

    1 r20

    r 2

    )cos()

    v =

    r=U

    (1+ r

    20

    r 2

    )sin()

    On the surface of the cylinder (r = r0) vr = 0 and v =2Usin. Using the Bernoulli equationwe obtain:

    P0+ 12U2 = Ps + 1

    2v2s

    Ps = P0+ 12U2 1

    2

    (U(

    1+ r20

    r 2

    )sin()

    )2

    Ps = P0+ 12U2 1

    2U24sin2

    Ps = P0+ 12U2(14sin2)

    1

  • 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

    3.0

    2.5

    2.0

    1.5

    1.0

    0.5

    0.0

    0.5

    1.0

    PsP

    01 2U

    2

    Figure: Pressure distribution on the surface of a circular cylinder

    2

  • PROBLEM 2

    Find the streamfunction for the aymptotic suction pforile, u(y)=U0[1exp(yV0/)], whichoccurs when a streaming flow with outer velocity,U0, flows above a porous wall with suctionvelocity, V0. Use Matlab to plot streamlines with equal increments of .

    v =x

    V0 =x

    (x, y)=V0x+ f (y)

    u = y

    U0[1exp(yV0/)]= y

    (x, y)=U0y +U0 V0

    eyV0 + f (x)

    (x, y)=V0x+U0y +U0 V0

    eyV0

    0.0 0.2 0.4 0.6 0.8 1.0

    X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    1.500

    3.000

    4.500

    6.000

    7.500

    9.000

    10.500

    (a)

    0.0 0.2 0.4 0.6 0.8 1.0

    X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (a) Streamfunction, (b) Velocity Vector Field

    3

  • PROBLEM 3

    The velocity potential for a spiral vortex flow is given by: (r,) = (/2pi) (m/2pi)log (r ), where and m are constants. Show that the angle, , between the velocity vector, u =ur ir +Ui and the radial direction, r , remains constant throughout the flow field.

    ur = r = m2piru = 1r

    = 2pir

    tan()= uur

    = /2pirm/2pir

    = m

    = tan1(m

    )= constant

    PROBLEM 4

    Velocity potential for flow impinging on a flat plate is described by the streamfunction, =Axy . Using MatLab (or some other application), draw streamlines in equal increments of for this flow. The addition of a source of strength, m, at the flow origin, O, creates thepresence of a bump of height, h, protruding in the wall-normal direction. By conservation(i.e. 2 = 0), streamlines will flow around this bump. Add the source to your code, andexplore the relationship between h, A, and m (5 different combinations will be adequate).

    = Axy + m2pi

    = A2r 2sin(2)+ m

    2pi

    Stagnation point is at x = 0, y = h ( = pi2 ,r = h) where h is the height of the bump.

    vr = 1r

    = Arcos(2)+ m

    2pir

    v =

    r= Ar sin(2)

    At this stagnation point the velocity is zero meaning ur = 0 and v = 0, therefore from previ-ous equations the following relation is obtained:

    0= Ahcos(pi)+ m2pih

    or

    h2 = m2piA

    4

  • 1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    -0.5

    00-0.

    250

    0.000

    0.000

    0.25

    0

    0.250

    0.500 0.750

    1.000

    (a)

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (A = 1,m = 1,h = 1/2pi) (a) Stremfunction, (b) Velocity Vector Field

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    -0.7

    50

    -0.5

    00

    -0.2

    50

    0.000

    0.250

    0.500

    0.750

    (a)

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (A = 1,m = 0,h = 0) (a) Stremfunction, (b) Velocity Vector Field

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    0.000

    0.000

    0.250

    0.250

    0.50

    00.

    500

    0.750

    0.750

    1.000

    (a)

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (A = 1,m = 2,h = (1/pi)1/2) (a) Stremfunction, (b) Velocity Vector Field

    5

  • 1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    -1.2

    00

    -0.6

    00

    0.000

    0.00

    0

    0.600

    1.200

    1.800

    (a)

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (A = 2,m = 1,h = (1/(4pi))1/2 (a) Stremfunction, (b) Velocity Vector Field

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    0.080

    0.16

    0

    0.24

    0

    0.320

    0.400

    0.480

    (a)

    1.0 0.5 0.0 0.5 1.0X

    0.0

    0.2

    0.4

    0.6

    0.8

    1.0

    Y

    (b)

    Figure: (A = 0,m = 1,h =) (a) Stremfunction, (b) Velocity Vector Field

    6

  • PROBLEM 5

    Show that the volume of an arbitrary region is given by

    dVARdt

    =niwidS

    the volume of an arbitrary region can be wrriten by the integral

    VAR =R(t )

    1d

    using Leibnitz theorem

    dVARdt

    = ddt

    R(t )

    1d=1

    td+

    niwi 1dS

    1

    td= 0

    dVARdt

    =niwi 1dS

    7

  • PROBLEM 6

    Prob. 6.1 Panton Stokes flow over a sphere has velocity components

    vr =Ucos()[

    1+ 12

    (r0r

    )3 3

    2

    (r0r

    )]

    v =Usin()[1+ 1

    4

    (r0r

    )3+ 3

    2

    (r0r

    )]Compute all components of the viscous stress tensor in r,, coordinates.The viscous stress tensor is defined by

    =23v+2S

    if incompressible= 2S

    r r = 2vrr

    = 2Ucos()[3

    2

    r 30r 4+ 3

    2

    r0r 2

    ]

    = 2[

    1

    r

    v

    + vrr

    ]= 2

    [Ucos()

    r

    [1+ 1

    4

    (r0r

    )3+ 3

    2

    (r0r

    )]+Ucos()

    r

    [1+ 1

    2

    (r0r

    )3 3

    2

    (r0r

    )]]= 3

    2

    Ucos()

    r

    (r0r

    )3 = 2

    [1

    r sin()

    v

    + vrr+ vcot ()

    r

    ]= 2

    [vrr+ vcot ()

    r

    ]= 2

    [Ucos()

    r

    [1+ 1

    2

    (r0r

    )3 3

    2

    (r0r

    )]+Ucos()

    r

    [1+ 1

    4

    (r0r

    )3+ 3

    2

    (r0r

    )]]= 3

    2

    Ucos()

    r

    (r0r

    )3

    8

  • r = r =[r

    r

    (vr

    )+ 1r

    vr

    ]=[r

    r

    [Usin()

    [1r+ 1

    4

    r 30r 4+ 3

    2

    r0r 2

    ]]+ 1r

    [Ucos()

    [1+ 1

    2

    (r0r

    )3 3

    2

    (r0r

    )]]]

    =[Usin()

    (1

    r r

    30

    r 4 6

    2

    r0r 2

    )Usin()

    (1

    r+ 1

    2

    r 30r 4 3

    2

    r0r 2

    )]

    =Usin()[3

    2

    r 30r 4+ 3

    2

    r0r 2

    ]

    = =[sin()

    r

    (v

    sin()

    )+ 1r sin()

    v

    ]= 0

    r = r =[

    1

    r sin()

    vr

    + r r

    (vr

    )]= 0

    9

  • PROBLEM 7

    Prob. 6.3 Panton An ideal "inviscid" flow over a cylinder has the velocity components givenin problem 5.1. Compute all components of the viscous stress tensor. Compute . Why isthis flow called inviscid?

    vr =Ucos()[

    1(r0r

    )2]v =Usin()

    [1+(r0r

    )2]p = 1

    2U2

    [2(r0r

    )2(12sin2())

    (r0r

    )4]the viscous stress tensor is defined as

    =23(v)+2S

    first the divergence of the velocity is computed in cylindrical coordinates

    v= 1r

    r(r vr )+ 1

    r

    v

    + vzz

    expanding and setting vz = 0

    v= 1r

    [rvrr+ vr]+ 1r

    v

    vrr

    = 2Ucos()r20

    r 3

    v

    =Ucos()(

    1+ r20

    r 2

    )

    substituting back into the divergence equation of the velocity

    v= 1r

    [2Ucos()

    r 20r 3Ucos()

    [1(r0r

    )2]+Ucos()

    (1+ r

    20

    r 2

    )]

    = 1rUcos()

    [2r

    20

    r 21+1+ r

    20

    r 2+ r0r 2

    ]= 0

    the viscous stress tensor is reduced to

    10

  • = 2S

    r r = 2vrr

    =4Ucos()r20

    r 3

    = 2(

    1

    r

    v

    + vrr

    )= 2

    (1

    r

    (Ucos()

    [1+ r

    20

    r 2

    ]) 1r

    (Ucos()

    [1 r0

    r 2

    ]))

    = 2rUcos()

    [1+ r

    20

    r 21+ r

    20

    r 2

    ]

    = 4Ucos()r20

    r 3

    zz = 0

    r = r =[r

    r

    (vr

    )+ 1r

    vr

    ]=[vr

    vr+ 1r

    vr

    ]=[2Usin()r

    20

    r 3Usin()

    r

    (1+ r

    20

    r 2

    )+Usin()

    r

    (1 r

    20

    r 2

    )]

    =Usin()[2r

    20

    r 3 1r r

    20

    r 3+ 1r r

    20

    r 3

    ]

    =4Usin()r20

    r 3

    zr = r z = 0z = z = 0

    []r = 1r

    r(rr r )+ 1

    r

    r

    r

    = 1r

    8Ucos()r 20r 3 1r

    4Ucos()r 20r 34Ucos r

    20

    r 4

    = 0

    [] =1

    r 2

    r(r 2r)+

    1

    r

    = 1r 2uUsin()

    r 20r 2+ 1r

    4U (sin())r20

    r 3

    = 0

    11

  • []z = 0

    = 0since this is the term that accounts for viscous forces in the momentum equation , we can saythat the flow is "inviscid"

    12

  • PROBLEM 8

    Prob. 6.7 Panton For a Newtonian fluid, show that the dissipation is given by

    :v=23(v)2+2S : S (0.1)

    the viscous tress tensor is

    =23v+2S (0.2)

    :v=(2

    3v+2S

    ):v

    =(2

    3i j

    vkxk