Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (9)

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  • Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

    Exercise 9.1. a) Write out the three components of (9.1) in x-y-z Cartesian coordinates. b) Set u = (u(y), 0, 0), and show that the x- and y-momentum equations reduce to:

    0 = 1 px

    +d 2udy2

    , and 0 = 1 py

    .

    Solution 9.1. a) Equation (9.1) is the constant-viscosity Navier-Stokes' momentum equation for incompressible flow:

    DuDt

    = 1p+2u ,

    where is the kinematic viscosity of the flow. Using u = (u, v, w) and

    = x , y , z( ) , the three components of this equation become:

    x:

    ut

    + uux

    + v uy

    + w uz

    = 1px

    + 2ux 2

    + 2uy 2

    + 2uz2

    &

    ' (

    )

    * + ,

    y:

    vt

    + uvx

    + v vy

    + w vz

    = 1py

    + 2vx 2

    + 2vy 2

    + 2vz2

    &

    ' (

    )

    * + , and

    z:

    wt

    + uwx

    + v wy

    + w wz

    = 1pz

    + 2wx 2

    + 2wy 2

    + 2wz2

    &

    ' (

    )

    * + .

    b) When u = (u(y), 0, 0), all the terms involving v and w disappear, so the part a) equations simplify to:

    x:

    ut

    + uux

    + 0 + 0 = 1px

    + 2ux 2

    + 2uy 2

    + 2uz2

    &

    ' (

    )

    * + ,

    y: 0+ 0+ 0+ 0 = 1 py

    + 0+ 0+ 0( ) , and

    z:

    0 + 0 + 0 + 0 = 1pz

    + 0 + 0 + 0( ).

    And, when u depends only on y, then u/t = u/x = u/z = 0 so the part a) equations simplify further:

    x:

    0 = 1px

    + 2uy 2&

    ' (

    )

    * + ,

    y:

    0 = 1py

    , and

    z:

    0 = 1pz

    .

    These x- and y-direction equations match those in the problem statement.

  • Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

    Exercise 9.2. For steady pressure driven flow between parallel plates (see Figure 8.3), there are 7 parameters: u(y), U, y, h, , , and dp/dx. Determine a dimensionless scaling law for u(y), and rewrite the flow-field solution (8.5) in dimensionless form. Solution 9.2. The parameters are: u(y), U, y, h, , , and dp/dx. First, create the parameter matrix: u U y h dp/dx Mass: 0 0 0 0 1 1 1 Length: 1 1 1 1 -3 -1 -2 Time: -1 -1 0 0 0 -1 -2 Next, determine the number of dimensionless groups. This rank of this matrix is three so 7 parameters - 3 dimensions = 4 groups, and construct the groups:

    1 = u U ,

    2 = y h ,

    3 = Uh , and

    4 = h2(dp /dx) U . Now write a dimensionless law:

    uU

    = f yh, Uh

    , h

    2

    Udpdx

    #

    $ %

    &

    ' (

    where f is an unknown function. When rewritten in dimensionless form, (8.5) is:

    uU

    =yhh2

    2Udpdx

    yh

    #

    $ %

    &

    ' ( 1

    yh

    #

    $ %

    &

    ' ( or

    1 =2 1242 12( ).

    In this flow, there is no fluid acceleration so the Reynolds number, 3, does not appear.

  • Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

    Exercise 9.3. An incompressible viscous liquid with density fills the gap between two large smooth parallel walls that are both stationary. The upper and lower walls are located at x2 = h, respectively. An additive in the liquid causes its viscosity to vary in the x2 direction. Here the flow is driven by a constant non-zero pressure gradient:

    p x1 = const. a) Assume steady flow, ignore the body force, set

    u = u1(x2),0,0( ) and use

    t

    +xi

    ui( ) = 0 ,

    u jt

    + uiu jxi

    = px j

    + g j +xi

    uix j

    +u jxi

    %

    & ' '

    (

    ) * *

    +

    , - -

    .

    / 0 0

    +x j

    v 23

    %

    & '

    (

    ) * uixi

    +

    , -

    .

    / 0

    to determine u1(x2) when

    = o 1+ x2 h( )2( ) .

    b) What shear stress is felt on the lower wall? c) What is the volume flow rate (per unit depth into the page) in the gap when = 0? d) If 1 < < 0, will the volume flux be higher or lower than the case when = 0?

    Solution 9.3. a) The continuity equation is satisfied by the form of the velocity field. The j =1-component of momentum equation simplifies to:

    0 = p x1( ) + x2( ) u1 x2( )[ ]. Integrate once with

    p x1 = const. to find:

    u1 x2( ) = p x1( )x2 + C . Divide by and integrate again:

    u1 =1

    px1

    x2 + C#

    $ %

    &

    ' ( dx2 =

    p x1( )x2 + Co 1+ x2 h( )

    2( )dx2

    = h2

    2opx1

    ln 1+ x2h

    #

    $ %

    &

    ' (

    2#

    $ % %

    &

    ' ( ( +

    Cho

    tan1 x2 h

    #

    $ %

    &

    ' ( + D.

    The boundary conditions,

    u1(h) = 0, determine the values of the constants: C = 0, and

    D = h2 2o( ) p x1( ) ln 1+ ( ) , thus:

    u1(x2) = h2

    2opx1ln1+ x2 h( )

    2

    1+

    %

    & ' '

    (

    ) * * .

    b) From the solution of part a) with C = 0:

    w = u1 x2( )y=h = h p x1( )

    c) When = 0, the flow profile is parabolic:

    q = u1(x2)dx2h

    +h

    = h2

    2opx1

    1 x22

    h2%

    & '

    (

    ) * dx2

    h

    +h

    = 2h3

    3opx1

    d) The volume flux will be higher because the viscosity will be reduced at the wall. Manipulation of the near-wall viscosity with additives is sometimes used in long piping systems to reduce pumping power requirements.

  • Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

    Exercise 9.4. An incompressible viscous liquid with density fills the gap between two large smooth parallel plates. The upper plate at x2 = h moves in the positive x1-direction at speed U. The lower plate at x2 = 0 is stationary. An additive in the liquid causes its viscosity to vary in the x2 direction. a) Assume steady flow, ignore the body force, set

    u = u1(x2),0,0( ) and

    p x1 = 0, and use the equations specified in Exercise 8.3 to determine u1(x2) when

    = o 1+ x2 h( ) . b) What shear stress is felt on the lower plate? c) Are there any physical limits on ? If, so specify them.

    Solution 9.4. a) For

    u = u1(x2),0,0( ) , no body force, and

    p x1 = 0 in steady incompressible flow, the continuity equation is automatically satisfied, and the momentum equation for j = 1 simplifies to:

    0 = + x2( ) u1 x2( )[ ], or, after integrating once:

    C = u1 x2( ) , where C is a constant. Now use the specified relationship for the viscosity and integrate to find:

    u1(x2) =Cdx2 =

    Cdx2o 1+ x2 h( )

    =Cho

    ln 1+ x2 h( ) + D

    where D is another constant. The boundary conditions u1(0) = 0 and u1(h) = U allow

    C =Uo h ln(1+ )( ) and D = 0 to be determined yielding:

    u1(x2) =U ln 1+ x2 h( ) ln 1+ ( ). b) From part a), the shear stress is constant:

    w = u1 x2( ) = C =Uo h ln(1+ )( ) . c) Negative viscosities violate the second law of thermodynamics, thus > 1 is required.

  • Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

    Exercise 9.5. Planar Couette flow is generated by placing a viscous fluid between two infinite parallel plates and moving one plate (say, the upper one) at a velocity U with respect to the other one. The plates are a distance h apart. Two immiscible viscous liquids are placed between the plates as shown in the diagram. The lower fluid layer has thickness d. Solve for the velocity distributions in the two fluids.

    Solution 9.5. For steady viscous flow between infinite parallel plates, the fluid velocity will be unidirectional: u = (u, 0, 0). For this problem, no pressure gradient is specified so assume it to be zero. Thus, the horizontal (x1-direction) momentum equation reduces to:

    21 x2 = x2( ) u x2( )= 2u y2 = 0 , where the last equality follows when the viscosity is constant and x2 = y. Here, the viscosity is assumed constant within each fluid. This means that the flow profile in each fluid will be piece-wise linear:

    u(y) =A1 + B1y for 0 y dA2 + B2y for d y h

    # $ %

    & ' (

    ,

    where the As & Bs are constants and 1 implies the upper fluid layer with viscosity 1, and 2 implies the lower fluid layer with viscosity 2. The four constants can be determined from the four boundary conditions: i) u(0) = 0 (match the speed of the lower boundary) ii) u(h) = U (match the speed of the upper boundary) iii) u(d) = u(d+), and (match flow speeds at the internal fluid-fluid interface) iv) (d) = (d+) (match shear stress at the internal fluid-fluid interface) where is the shear stress in the fluid. These four boundary conditions imply:

    A2 = 0,

    A1 + B1h =U ,

    A2 + B2d = A1 + B1d , and

    2B2 = 1B1 Use the first two equations to eliminate A1 and A2 from the second two equations to find:

    B2d =U B1(h d) , and

    2B2 = 1B1. Eliminate B2 and solve for B1:

    1 2( )B1d =U B1(h d) >

    B1 = 2U 2h + 1 2( )d[ ]. So,

    B2 = 1U 2h + 1 2( )d[ ], and

    A1 =U 1 2( )d 2h + 1 2( )d[ ] with A2 = 0. Thus:

    u(y) = U2h + 1 2( )d

    1y for 0 y d1 2( )d + 2y for d y h

    $ % &

    ' ( )

  • Fluid Mechanics, 6th Ed.