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. Kundu, Cohen, and Dowling

    Exercise 9.6. Consider plane Poiseuille flow of a non-Newtonian power-law fluid using the coordinate system and geometry shown in Figure 9.3. Here the fluid's constitutive relationship is given by (4.37): xy =m u y( )

    n, where m is the power law coefficient, and n is the power law

    exponent. a) Determine the velocity profile u(y) in the lower half of the channel, 0 y h/2, using the boundary condition u(0) = 0. b) Given that the maximum velocity occurs at y = h/2 and that the flow profile is symmetric about this location, plot u(y) u(h/2) vs. y (h/2) for n = 2 (a shear thickening fluid), n = 1 (a Newtonian fluid), and n = 0.4 (a shear thinning fluid). c) Explain in physical terms why the shear-thinning velocity profile is the bluntest. Solution 9.6. In plane Poiseuille flow, the upper channel wall does not move. In the lower half of the channel, the u/y > 0, so fractional powers are readily managed. a) For fully developed constant density flow (so that p/x is independent of x) the horizontal momentum equation reduces to:

    0 = px+ xyy

    .

    The vertical momentum equation is simply, p/y = 0, so p/x does not depend on x or y. Thus, this horizontal momentum equation can be integrated to find:

    y px

    = xy +C ,

    where C is a constant. However, when y = h/2, xy = 0 so C = (h/2)p/x, therefore: h2 y

    "

    #$

    %

    &'

    px

    "

    #$

    %

    &'= xy =m

    uy

    "

    #$

    %

    &'

    n

    , or uy

    = 1m

    px

    #

    $%

    &

    '(1 n h2 y

    #

    $%

    &

    '(1 n

    ,

    and both equations are written this way because p/x is positive. The second equation can be integrated to find:

    u(y) = 1m

    px

    #

    $%

    &

    '(1 n 11+1 n

    h2 y

    #

    $%

    &

    '(1+1 n

    +D ,

    where D is another constant of integration. The horizontal velocity must go to zero on the lower channel wall at y = 0, so

    D = 1m

    px

    "

    #$

    %

    &'1 n 11+1 n

    h2

    "

    #$

    %

    &'1+1 n

    .

    Thus, the velocity profile is:

    u(y) = nn+1

    1m

    px

    "

    #$

    %

    &'1 n h

    2"

    #$

    %

    &'1+1 n

    h2 y

    "

    #$

    %

    &'1+1 n)

    *++

    ,

    -..=

    nn+1

    "

    #$

    %

    &'h2

    h2m

    px

    "

    #$

    %

    &'1 n

    1 1 yh 2

    "

    #$

    %

    &'

    1+1 n)

    *++

    ,

    -..

    .

    b) Evaluate the result of part a) at y = h/2, and form the ratio:

    u(y)u(h / 2)

    =

    nn+1

    !

    "#

    $

    %&h2

    h2m

    px

    !

    "#

    $

    %&1 n

    1 1 yh 2

    !

    "#

    $

    %&

    1+1 n)

    *++

    ,

    -..

    nn+1

    !

    "#

    $

    %&h2

    h2m

    px

    !

    "#

    $

    %&1 n = 1 1( )

    1+1 n)*

    ,- ,

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

    where = y (h/2) . The three curves for n = 2, 1, and 0.4 are plotted here:

    u()/u( = 1)

    c) From the part a) solution, the shear stress is:

    xy = px

    h2 y

    #

    $%

    &

    '( .

    The highest magnitude shear stress occurs at the upper and lower wall of the channel. In a shear thinning fluid, the local viscosity decreases when the shear stress increases. Thus, to maintain this shear stress profile, the velocity gradient magnitude must increase in regions where the shear stress magnitude is also large. This leads to a blunter profile for a shear thinning fluid compared to a Newtonian (or a shear thickening) fluid.

    0.00#0.20#0.40#0.60#0.80#1.00#1.20#1.40#1.60#1.80#2.00#

    0.000# 0.200# 0.400# 0.600# 0.800# 1.000#

    n=2#

    n=1#

    n=0.4#

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

    Exercise 9.7. Consider the laminar flow of a fluid layer falling down a plane inclined at an angle with respect to the horizontal. If h is the thickness of the layer in the fully developed stage, show that the velocity distribution is

    u(y) = g 2( ) h2 y 2( )sin , where the x-axis points in the direction of flow along the free surface, and the y-axis points toward the plane. Show that the volume flow rate per unit width is

    Q = gh3 3( )sin , and that and the frictional stress on the wall is w = ghsin. Solution 9.7. In the given coordinates, there is a component of gravity, gsin, acting in the x-direction, but the presence of atmospheric pressure on the liquid surface ensures that p/x = 0. Thus, the steady flow x-direction momentum equation is:

    0 = +gsin + 2u

    y 2.

    Integrating twice produces:

    u(y) = gsin2

    y 2 + Ay + B ,

    where A and B are constants. The boundary condition on the liquid surface (y = 0) is zero shear stress (u/y = 0), and this allows A to be evaluated:

    0 = uy#

    $ %

    &

    ' ( y= 0

    = gsin

    y + A#

    $ % &

    ' ( y= 0= A .

    The boundary condition, on the solid surface (y = h) is zero velocity (u = 0), and this allows B to be evaluated:

    0 = u(h) = gsin2

    h2 + B, or

    B = gsin2

    h2 .

    so the velocity profile is:

    u(y) = gsin2

    y 2 h2( ). The volume flow rate per unit width is:

    Q = u(y)dy0

    h = gsin2

    y 2 h2( )dy0h = gsin2

    h3

    3 h3

    &

    ' (

    )

    * + =

    gsin3

    h3 .

    The magnitude of the shear stress is:

    = uy

    = gsin

    &

    ' (

    )

    * + y = gy sin ,

    and the maximum shear stress, w = ghsin, occurs at the solid surface.

    u(y)

    yx

    h

    !

    stress

    distribution

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

    Exercise 9.8. In two-dimensional (x,y)-coordinates, the Navier-Stokes equations for the fluid velocity, u = (u,v) , in a constant-viscosity constant-density flow are: u x +v y = 0 ,

    ut+uu

    x+ v u

    y=

    1px+

    2ux2

    +2uy2

    #

    $%

    &

    '( , and

    vt+u v

    x+ v v

    y=

    1py+

    2vx2

    +2vy2

    #

    $%

    &

    '( .

    a) Cross differentiate and sum the two momentum equations to reach the following equation for z = v x u y , the vorticity normal to the x-y plane:

    zt

    +uzx

    + v zy

    =2zx2

    +2zy2

    "

    #$

    %

    &' .

    b) The simplest nontrivial solution of this equation is uniform shear or solid body rotation (z = constant). The next simplest solution is a linear function of the independent coordinates: z = ax + by, where a and b are constants. Starting from this vorticity field, derive the following velocity field:

    u = b2(ax + by)2

    a2 + b2+ c

    "

    #$

    %

    &' and v =

    a2(ax + by)2

    a2 + b2+ c

    !

    "#

    $

    %& .

    where c is an undetermined constant. c) For the part b) flow, sketch the streamlines. State any assumptions you make about a, b, and c. d) For the part b) flow when a = 0, b > 0, and u = (Uo, 0) at the origin of coordinates with Uo > 0, sketch the velocity profile along a line x = constant, and determine p . Solution 9.8. a) Apply /y to the x-direction momentum equation, and /x to the y-direction momentum equation to reach:

    t

    uy

    #

    $%

    &

    '(

    uy

    ux

    u x

    uy

    #

    $%

    &

    '(

    vy

    uy

    v y

    uy

    #

    $%

    &

    '(=1

    2pyx

    2

    x2+

    2

    y2#

    $%

    &

    '(

    uy

    #

    $%

    &

    '( , and

    t

    vx

    "

    #$

    %

    &'+

    ux

    vx+u

    xvx

    "

    #$

    %

    &'+

    vx

    vy+ v

    xvy

    "

    #$

    %

    &'=

    1

    2pxy

    +2

    x2+

    2

    y2"

    #$

    %

    &'

    vx

    "

    #$

    %

    &' .

    Add these two equations together noting that pressure terms cancel, and that the second and fourth terms in each equation sum to zero because of the continuity equation, u x +v y = 0 .

    t

    vx

    uy

    #

    $%

    &

    '(+u

    x

    vx

    uy

    #

    $%

    &

    '(+ v

    y

    vx

    uy

    #

    $%

    &

    '(=

    2

    x2+

    2

    y2#

    $%

    &

    '(

    vx

    uy

    #

    $%

    &

    '( .

    Substituting in the definition z = v x u y leads to the final form: zt

    +uzx

    + v zy

    =2zx2

    +2zy2

    "

    #$

    %

    &' .

    b) To find the velocity field start with: z = ax + by = v/x u/y. (1) The part a) equation implies: ua + vb = 0 or v = (a/b)u. (2) The continuity equation is also needed: u/x + v/y = 0. (3) Use (2) to eliminate v, from (1) and (3) to find:

    ax + by = abux

    uy

    , and ux

    abuy

    = 0 .

    Combine these twice, first to eliminate u/y, then to eliminate u/x to reach:

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

    ax + by = a2

    b2+1

    "

    #$

    %

    &'uy

    , and ax + by = ab+ba

    "

    #$

    %

    &'ux

    .

    Integrate to find: axy+ b y2

    2=

    a2

    b2+1

    "

    #$

    %

    &'u+ f (x) , and a

    x2

    2+ bxy = a

    b+ba

    "

    #$

    %

    &'u+ g(y) , where f(x)

    and g(y) are unknown functions. Solve both equations for u and absorb multiplicative constants into f(x) and g(y).

    u = b2

    f (x)+ 2abxy+ b2y2

    a2 + b2"

    #$

    %

    &' , and u =

    b2a2x2 + ab2xy+ g(y)

    a2 + b2"

    #$

    %

    &' .

    For consistency, the unknown functions must be: f(x) = a2x2 + const. and g(y) = b2y2 + const. so

    u = b2(ax + by)2

    a2 + b2+ c

    "

    #$

    %

    &' , and then from (2)

    v = abu = a

    2(ax + by)2

    a2 + b2+ c

    "

    #$

    %

    &' .

    where c is another undetermined constant. c) dysl/dxsl = v/u = a/b. Thus, if a and b are both positive constants, the streamlines are negatively-slopped straight lines. In this case, u < 0 and v > 0, so the arrows point upward and to the left.

    d) Evaluate the constants to find: u =Uo b 2( ) y2 and v = 0, so the horizontal velocity profile is parabolic. To get the pressure gradient, evaluate the 2D Navier-Stokes equations using the part b) velocity field:

    0 = 1px+ (b) , and 0 = 1

    py+ 0 , so

    p = b, 0( ) .

    x!

    y !

    b!a!

    Uo!y !

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

    Exercise 9.9. Consider circular Couette flow as described by (9.10) in the limit of a thin gap between the cylinders. Use the definitions: R = (R1 + R2 ) 2 , h = R2 R1 , = (1 +2 ) 2 , =2 1 , and R = R + y to complete the following items. a) Show that u (y) R +R y h( ) when y

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

    u (y) = R+ y( )+R 1 yR +y2

    R2 +...

    $

    %&

    '

    ()yh+12Rh

    3h2

    4+ y2

    $

    %&

    '

    ()+

    hy4R

    2 +12R

    3hy2

    4h3

    16$

    %&

    '

    ()

    *

    +,

    -

    ./ .

    Keeping only linear terms in y R and h R leads to the final series:

    u (y) =R+Ryh

    #

    $%

    &

    '(+R

    yR

    #

    $%

    &

    '(+... R+R

    yh

    #

    $%

    &

    '( ,

    where the final approximate equality follows because h

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

    Exercise 9.10. Room temperature water drains through a round vertical tube with diameter d. The length of the tube is L. The pressure at the tube's inlet and outlet is atmospheric, the flow is steady, and L >> d. a) Using dimensional analysis write a physical law for the mass flow rate

    m through the tube. b) Assume that the velocity profile in the tube is independent of the vertical coordinate, determine a formula for

    m , and put it in dimensionless form. c) What is the change in

    m if the temperature is raised and the water's viscosity drops by a factor of two? Solution 9.10. a) This part is just dimensional analysis. The solution parameter is

    m , and boundary condition & material parameters are: = density of the fluid, = viscosity of the fluid, g = acceleration of gravity, d = tube diameter, and L = tube length. The parameter matrix is:

    m g d L Mass: 1 1 1 0 0 0 Length: 0 -3 -1 1 1 1 Time: -1 0 -1 -2 0 0 and it has rank 3. The number of dimensionless groups is: 6 parameters - 3 dimensions = 3.

    These groups are: 1 =

    m d

    , 2 =

    g2d3

    2, and 3 =

    Ld

    . Therefore, the scaling law is:

    m d

    = F g2d3

    2, Ld

    #

    $ %

    &

    ' ( .

    b) When the velocity profile in the tube is steady and independent of the vertical coordinate, the vertical momentum equation is

    0 = g + R

    R

    RuzR

    %

    & '

    (

    ) *

    where the z-axis points upward (opposite gravity), R is the radial coordinate, and uz is the vertical velocity. This is the equation for Poiseuille flow with the pressure gradient replaced by g. Noting that downward velocity produces positive

    m , the results in Section 8.2 imply:

    m = Q = 8

    d2

    $

    % &

    '

    ( )

    4

    g = 128

    2gd4

    or

    m d

    =

    1282gd3

    21 =

    128

    2.

    c) The flow rate is inversely proportional to and = (T). Since the ratio

    room temphotter

    = 2 , the

    flow rate will approximately double (the density of water falls slightly as T increases toward the boiling point) but this will likely be a very small effect.

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

    Exercise 9.11. Consider steady laminar flow through the annular space formed by two coaxial tubes aligned with the z-axis. The flow is along the axis of the tubes and is maintained by a pressure gradient dp/dz. Show that the axial velocity at any radius R is

    uz(R) =14

    dpdz

    R2 a2 b2 a2

    ln b a( )ln Ra

    #

    $ %

    &

    ' ( ,

    where a is the radius of the inner tube and b is the radius of the outer tube. Find the radius at which the maximum velocity is reached, the volume flow rate, and the stress distribution. Solution 9.11. Use cylindrical coordinates, and assume uz = uz(R), where R is the radial coordinate. The axial (z-direction) momentum equation is:

    0 = pz

    + 1R

    R

    RuzR

    #

    $ %

    &

    ' ( .

    Integrate twice to find:

    RuzR

    =R2

    2pz

    + A , and

    uz(R) =R2

    4pz

    + A logR +B.

    The two boundary conditions are u = 0 at R = a, b and these allow the two constants A and B to be determined. Alternatively evaluating the above formula for uz(R) at R = a and b produces.

    uz(a) = 0 =a2

    4pz

    + A loga +B , and

    uz(b) =b2

    4pz

    + A logb +B.

    These two equations can be solved to find:

    A = 14

    pz

    $

    % &

    '

    ( ) b2 a2

    ln(b /a), and

    B = 14

    pz

    #

    $ %

    &

    ' ( b2 a2

    ln(b /a)log(a) a2

    *

    + ,

    -

    . / ,

    so that:

    u(R) = 14

    pz

    #

    $ %

    &

    ' ( R2 a2

    b2 a2

    ln(b /a)log R a( )

    *

    + ,

    -

    . / .

    The shear stress is:

    = uR

    =R2

    pz

    14

    pz

    "

    #$

    %

    &'b2 a2

    R ln(b / a).

    The maximum velocity occurs where = 0, and this condition implies:

    0 = Rmax2

    pz

    14

    pz

    $

    % &

    '

    ( )

    b2 a2

    Rmax ln(b /a) or

    Rmax =b2 a2

    2ln(b /a).

    The volume flow rate Q is:

    Q = 2 u(R)RdRa

    b = 8

    pz

    %

    & '

    (

    ) * b2 a2( )

    2

    ln(b /a)+ a4 b4

    ,

    -

    .

    .

    /

    0

    1 1 .

    a

    b

    Rmax

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

    Exercise 9.12. A long round wire with radius a is pulled at a steady speed U, along the axis of a long round tube of radius b that is filled with a viscous fluid. Assuming laminar fully-developed axial flow with

    p z = 0 in cylindrical coordinates

    (R,,z) with u = (0, 0, w(R)), determine w(R) assuming constant fluid density and viscosity with no body force. a) What force per unit length of the wire is needed to maintain the motion of the wire? b) Explain what happens to w(R) when

    b. Is this situation physically meaningful? What additional term(s) from the equations of motion need to be retained to correct this situation?

    Solution 9.12. The steady axisymmetric no-swirl equations are:

    1RR

    RuR( ) +wz

    = 0,

    uRuRR

    + w uRz

    = 1pR

    +1R

    R

    RuRR

    %

    & '

    (

    ) * +

    2uRz2

    uRR2

    %

    & '

    (

    ) * ,

    0 = 1R

    p

    , and

    uRwR

    + w wz

    = 1pz

    +1R

    R

    RwR

    %

    & '

    (

    ) * +

    2wz2

    %

    & '

    (

    ) * .

    Setting u = (0, 0, w(R)) automatically satisfies the cons. of mass equation. With

    p z = 0 , the other equations further simplify to:

    0 = 1pR

    = 1R

    p

    , and

    0 = 1R

    R

    RwR

    $

    % &

    '

    ( )

    $

    % &

    '

    ( )

    Thus, the pressure is constant everywhere because its gradient is zero. For nonzero viscosity and density, integrate the final equation once to find:

    C1 = R w R( ) where C1 is a constant. Divide by R and integrate again:

    C1 ln(R) + C2 = w(R) where C2 is a constant. The two constants can be determined from the boundary conditions on the wire and tube surfaces:

    w(R = a) =U = C1 ln(a) + C2, and

    w(R = b) = 0 = C1 ln(b) + C2 .

    Simultaneous solution of these two algebraic equations produces:

    w(R) = Uln b a( )

    ln b R( )

    a) The viscous shear stress on the wire, w, will act to retard its motion. The viscous force per unit length of wire will act in the z-direction on the wire and will be

    2aw = 2awR

    %

    & '

    (

    ) * R= a

    = 2a Uln b a( )

    1b R

    bR2

    %

    & '

    (

    ) *

    %

    & '

    (

    ) * R= a

    = 2Uln b a( )

    .

    Thus, the force necessary to maintain the motion will be:

    = +2Uez ln b a( ) b) When

    b, the solution indicates that the force necessary to keep the wire moving will approach zero. This is not physically meaningful for any finite wire length and time duration because the fluid is viscous. The problem(s) with this limiting situation can be corrected by including the unsteady terms,

    t , or terms involving uR and

    z in the conservation equations.

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

    Exercise 9.13. Consider steady unidirectional incompressible viscous flow in Cartesian coordinates, u = v = 0 with w = w(x,y) without body forces. a) Starting from the steady version of (8.1), derive a single equation for w assuming that

    p z is non-zero and constant. b) Guess w(x,y) for a tube with elliptical cross section

    x a( )2 + y b( )2 =1. c) Determine w(x,y) in for a tube of rectangular cross section: a/2 x +a/2, b/2 y +b/2. (Hint: find particular (a polynomial) and homogeneous (a Fourier series) solutions for w.) Solution 9.13. a) Start with the full z-direction momentum equation in Cartesian coordinates:

    wt

    + uwx

    + v wy

    + w wz

    = 1pz

    + 2wx 2

    + 2wy 2

    + 2wz2

    &

    ' (

    )

    * + .

    For steady flow with u = v = 0 and w = w(x,y), a lot terms that drop out of the above equation,

    and it simplifies to:

    pz

    = 2wx 2

    + 2wy 2

    #

    $ %

    &

    ' ( . Note that the density has dropped out because there is

    no fluid acceleration in this type of flow. b) The boundary condition for this flow is w = 0 on the wall of the tube. For an elliptical cross section given by

    x a( )2 + y b( )2 =1, a possible solution is

    w(x,y) =Wo 1 x a( )2 y b( )2[ ]

    where Wo is a constant that needs to be determined from the equation of motion. Plugging this

    guessed solution into the equation found for part a), produces:

    pz

    = Wo2a2

    +2b2

    $

    % &

    '

    ( ) . If

    p z is

    constant then:

    w(x,y) = 12

    pz

    1a2

    +1b2

    $

    % &

    '

    ( ) 1

    1 x2

    a2y 2

    b2*

    + ,

    -

    . / , and this reduces to the usual Poiseuille

    flow formula when a = b. c) The solution for a rectangular cross section is a bit more involved. The equation for w in part a) is inhomogeneous, so its solution will be made up of a particular solution and a homogeneous solution: w = wp + wh. Here, the particular solution must satisfy the governing equation, but it need not satisfy the boundary conditions. The constants in the homogeneous solution can be used to fix up the solution at the tube walls. Therefore, start by looking for a one-variable particular solution, i.e. wp = wp(x), and plug this into the field equation:

    pz

    = 2wpx 2

    + 2wpy 2

    #

    $ %

    &

    ' ( =

    d2wpdx 2

    .

    Integrate twice to find:

    wp (x) =12

    dpdzx 2 + C1x + C2 . The two constants can be chosen to zero-

    out the velocity at x = a/2;

    wp (x) = a2

    8dpdz1 x

    2

    (a /2)2#

    $ %

    &

    ' ( . Now, the homogeneous solution is

    needed. It solves

    2whx 2

    + 2why 2

    = 0. Assume a variables-separable form:

    wh (x,y) = X(x)Y (y),

    plug it in, and label the separation constant k2 to find two ordinary differential equations:

    " " X + k 2X = 0, and

    " " Y k 2Y = 0 . The symmetry of the situation implies that only the cos-part of X and the cosh-part of Y are needed:

    wh (x,y) = Acos(kx)cosh(ky) . To zero-out the velocity at x

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

    = a/2, the separation constant k must be restricted to a set of special values:

    k = kn = (2n 1) a . Thus, the overall solution takes the following form:

    w = wp + wh = a2

    8dpdz1 x

    2

    (a /2)2#

    $ %

    &

    ' ( + An cos

    (2n 1)xa

    #

    $ %

    &

    ' (

    n=1

    cosh (2n 1)ya#

    $ %

    &

    ' ( . (1)

    Here it is clear than w is zero on x = a/2. The Ans are determined by requiring w = 0 on y = b/2:

    a2

    8dpdz1 x

    2

    (a /2)2#

    $ %

    &

    ' ( = An cos

    (2n 1)xa

    #

    $ %

    &

    ' (

    n=1

    cosh (2n 1)b2a#

    $ %

    &

    ' ( .

    Multiply both sides by cos[(2m1)x/a], and integrate from x = a/2 to x = +a/2:

    a2

    8dpdz

    1 x2

    (a /2)2#

    $ %

    &

    ' ( cos

    (2m 1)xa

    #

    $ %

    &

    ' ( dx

    x=a / 2

    x= a / 2

    = Am cosh(2m 1)b

    2a#

    $ %

    &

    ' ( a2

    The integral on the left side is a little tedious but can be evaluated by parts:

    a2

    8dpdz

    (1)m+1 16a(2m 1)3 3

    $

    % &

    '

    ( ) = Am cosh

    (2m 1)b2a

    $

    % &

    '

    ( ) a2

    Thus:

    Am =4a2

    dpdz

    (1)m+1

    (2m 1)3 3$

    % &

    '

    ( ) cosh

    (2m 1)b2a

    $

    % &

    '

    ( ) . This result together with (1) completes

    the solution.

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

    Exercise 9.14. A long vertical cylinder of radius b rotates with angular velocity concentrically outside a smaller stationary cylinder of radius a. The annular space is filled with fluid of

    viscosity . Show that the steady velocity distribution is:

    u =R2 a2

    b2 a2b2R

    , and that the torque

    exerted on either cylinder, per unit length, equals 4a2b2/(b2 a2). Solution 9.14. Use cylindrical coordinates, and start from equation (8.10),

    u (R) =1

    R22 R1

    2 2R22 1R1

    2[ ]R 2 1[ ] R12R2

    2

    R% & '

    ( ) *

    ,

    and set R1 = a, R2 = b, 2 = , and 1 = 0 to find:

    u (R) =1

    b2 a2b2 0[ ]R 0[ ] a

    2b2

    R% & '

    ( ) *

    =b2

    b2 a2R2 a2

    R=R2 a2

    b2 a2b2R

    .

    From Appendix B the shear stress is

    R = RR

    uR

    %

    & '

    (

    ) * =

    Rb2b2 a2

    R

    1 a2

    R2%

    & '

    (

    ) * =2a2b2b2 a2

    1R2

    %

    & '

    (

    ) * .

    If the axial length of the cylinders is h, then

    torque on the inner cylinder = (R)R = a(area)(radius) =

    2a2b2b2 a2

    1a2

    $

    % &

    '

    ( ) 2ah( )a = 4a

    2b2hb2 a2

    ,

    torque on the outer cylinder = (R)R = b(area)(radius) =

    2a2b2b2 a2

    1b2

    $

    % &

    '

    ( ) 2bh( )b = 4a

    2b2hb2 a2

    .

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

    Exercise 9.15. Consider a solid cylinder of radius a, steadily rotating at angular speed in an infinite viscous fluid. The steady solution is irrotational: u = a2/R. Show that the work done by the external agent in maintaining the flow (namely, the value of 2Rur at R = a) equals the viscous dissipation rate of fluid kinetic energy in the flow field. Solution 9.15. Using the given velocity field, the shear stress is:

    R = RR

    uR

    %

    & '

    (

    ) * = a2R

    R

    1R2

    %

    & '

    (

    ) * = 2a2

    1R2

    .

    The work done per unit height =

    2aRu{ }R= a = 2a 2a = 4a22 .

    From (4.58) the viscous dissipation rate of kinetic energy per unit volume for an incompressible flow is

    = 2SijSij , where is the viscous dissipation of kinetic energy per unit mass. For the given flow field there is only one non-zero independent strain component:

    SR = SR =R2

    R

    uR

    $

    % &

    '

    ( ) =

    a2

    2R R

    1R2

    $

    % &

    '

    ( ) = a2

    1R2

    .

    Therefore:

    = 2SijSij = 2 SR2 + SR

    2( ) = 42 a4

    R4,

    so the kinetic energy dissipation rate per unit height is:

    a

    2RdR = 82a4 1R3

    dRa

    = 42a2 ,

    which equals the work done turning the cylinder.

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

    Exercise 9.16. Redo the lubrication-theory scaling provided in Section 9.3 for the situation when there is an imposed external time scale so that the appropriate dimensionless time is t* = t/, instead of that shown in (9.14). Show that this leads to the additional requirement h2/

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

    Exercise 9.17. For lubrication flow under the sloped bearing of Example 9.3, the assumed velocity profile was

    u(x,y) = 1 2( ) dP dx( )y h(x) y)( ) +Uy h(x) , the derived pressure was

    P(x) = Pe + 3U ho2L( )x(L x), and the load (per unit depth) carried by the bearing was

    W = UL2 2ho2 . Use these equations to determine the frictional force (per unit depth), Ff,

    applied to the lower (flat) stationary surface in terms of W, ho/L, and . What is the spatially-averaged coefficient of friction under the bearing? Solution 9.17. For lubrication flow under the sloped bearing, the velocity field was assumed to

    be:

    u(y) = 12

    px

    y(h y) + Uyh

    , so the wall shear stress is:

    w = uy

    $

    % &

    '

    ( ) y= 0

    = h2px

    + Uh

    .

    Integrate this from x = 0 to x = L to find the friction force Ff per unit depth. Here the pressure gradient will be:

    dp dx = 3U ho2L( )(L 2x) , so

    Ff = wdx = ho(1x L)

    23Uho2L

    (L 2x) + Uho(1x L)

    %

    & '

    (

    ) * 0

    L0

    L dx

    Both integrals are elementary, so Ff is found to be:

    Ff = U 2L4ho

    ULho

    ln(1) .

    However, because is small, only the 0 and 1 terms should be reported so,

    Ff ULho

    1+ 2

    $

    % &

    '

    ( ) ,

    or using

    W = UL2

    2ho2 ,

    Ff WhoL

    2 +( ) . The coefficient of friction is a ratio:

    FfW

    ho(2 +)L

    2hoL

    .

    Hence, we can see that it is not possible to obtain a small coefficient of friction since the maximum value of is ~ho/L. Thus, the purpose of this type of bearing would not necessarily be to produces a small coefficient of friction. Instead it prevents scuffing and wear since the presence of the oil film prevents the bearing pad and the surface from coming into contact.

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

    Exercise 9.18. A bearing pad of total length 2L moves to the right at constant speed U above a thin film of incompressible oil with viscosity and density . There is a step change in the gap thickness (from h1 to h2) below the bearing as shown. Assume that the oil flow under the bearing

    pad follows:

    u(y) = y(h j y)2

    dP(x)dx

    +Uyh j

    , where j = 1 or 2. The pad is instantaneously

    aligned above the coordinate system shown. The pressure in the oil ahead and behind the bearing is Pe. a) By conserving mass for the oil flow, find a relationship between , U, hj, dP/dx, and an unknown constant C. b) Use the result of part a) and continuity of the pressure at x = 0, to determine

    P(0) Pe =6UL(h1 h2)

    h23 + h1

    3

    c) Can this bearing support a externally applied downward load when h1 < h2? Solution 9.18. a) Use a CV attached to the bearing pad with vertical flux surfaces that span the gap. Conservation of mass implies:

    u(y) U( )dy0

    h j

    = C

    y(h j y)2

    dP(x)dx

    +Uyh j

    U#

    $ % %

    &

    ' ( ( dy

    0

    h j

    = C

    Performing the integration produces:

    12

    dP(x)dx

    h j3

    2h j3

    3

    #

    $ %

    &

    ' ( +U

    h j2

    Uh j = h j3

    12dP(x)dx

    Uh j2

    = C .

    b) Use part a), solve for dP/dx,

    dP(x)dx

    = 12h j3 C +

    Uh j2

    #

    $ %

    &

    ' ( , and integrate to find:

    P(x) P(0) = 12xh j3 C +

    Uh j2

    #

    $ %

    &

    ' ( .

    Evaluate P(x) at x = L and note that P(+L) = P(L) = Pe with P(x) continuous at x = 0:

    12Lh13 C +

    Uh12

    #

    $ %

    &

    ' ( = +

    12Lh23 C +

    Uh22

    #

    $ %

    &

    ' ( .

    Solve for

    C = Uh1h22

    h12 + h2

    2

    h13 + h2

    3

    # $ %

    & ' (

    , and then find:

    P(0) Pe =12Lh1

    3 C +Uh1

    2#

    $ %

    &

    ' ( =

    12Lh1

    3 Uh1h2

    2h1

    2 + h22

    h13 + h2

    3

    ) * +

    , - .

    +Uh1

    2

    #

    $ %

    &

    ' ( =

    6ULh1

    2 h2h1

    2 + h22

    h13 + h2

    3

    ) * +

    , - .

    +1#

    $ %

    &

    ' (

    = 6ULh1

    2h1

    2h2 h23 + h1

    3 + h23

    h13 + h2

    3

    #

    $ %

    &

    ' ( =

    6ULh1

    2h1

    2h2 + h13

    h13 + h2

    3

    #

    $ %

    &

    ' ( = 6UL

    h1 h2h1

    3 + h23

    #

    $ %

    &

    ' (

    c) When h1 < h2, the pressure at x = 0 is less than Pe. This means that the bearing pad is sucked toward the surface, so when h1 < h2 the bearing cannot support an external load.

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

    Exercise 9.19. A flat disk of radius a rotates above a solid boundary at a steady rotational speed of . The gap, h (

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

    Exercise 9.20. A circular block with radius a and weight W is released at t = 0 on a thin layer of an incompressible fluid with viscosity that is supported by a smooth horizontal motionless surface. The fluid layer's initial thickness is ho. Assume that flow in the gap between the block and the surface is quasi-steady with a parabolic velocity profile:

    uR (R,z,t) = dP(R) dR( )z h(t) z)( ) 2 where R is the distance from the center of the block, P(R) is the pressure at R, z is the vertical coordinate from the smooth surface, h(t) is the gap thickness, and t is time. a) By considering conservation of mass, show that:

    dh dt = (h3 6R)(dP dR) . b) If W is known, determine h(t) and note how long it takes for h(t) to reach zero. Solution 9.20. Consider a circular control volume under the block with radius of R and height h(t). The integral form of conservation of mass implies:

    R2 dhdt

    = 2R uR (R,z,t)dz0h( t ) = 2R 12

    dP(R)dR

    z h(t) z)( )dz0h(t ) = R

    dP(R)dR

    h3

    6

    which simplifies to:

    dhdt

    =h3

    6RdP(R)dR

    .

    b) Since h does not depend on R, this equation can be integrated in the radial direction from R to a:

    dhdt

    RdR = h3

    6dP

    R

    aR

    a which implies

    dhdt

    a2 R2

    2#

    $ %

    &

    ' ( =

    h3

    6Pe P(R)( ) ,

    where Pe is the exterior pressure on the circular block (perhaps equal to 1 atm). Assuming quasi-steady conditions, the weight of the block must be balanced by an integral of the excess-pressure underneath the block:

    W = 2 P(R) Pe( )0a RdR = dhdt

    6h3

    a2 R2( )0a RdR = dhdt

    6h3

    a2 a2

    2a4

    4%

    & '

    (

    ) * =

    dhdt3a4

    2h3.

    Now, treat this result as a first-order nonlinear differential equation for h(t) and integrate in time:

    h3dhho

    h( t ) = 12

    1h2

    1ho2

    $

    % &

    '

    ( ) =

    2Wt3a4

    which implies

    1h2(t)

    =1ho2 +

    4Wt3a4

    .

    Solving for h(t) yields:

    h(t) = ho 1+4Who

    2t3a4

    #

    $ %

    &

    ' (

    1 2

    . For this solution, h(t) 0 only when t .

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

    Exercise 9.21. Consider the inverse of the previous exercise. A block and smooth surface are separated by a thin layer of a viscous fluid with thickness ho. At t = 0, a force, F, is applied to separate them. If ho is arbitrarily small can the block and plate be separated easily? Perform some tests in your kitchen. Use maple syrup, peanut butter, liquid soap, pudding, etc. for the viscous liquid. The flat top-side of a metal jar lid or the flat bottom of a drinking glass make a good circular block. (Lids with raised edges and cups & glasses with ridges or sloped bottoms do not work well). A flat countertop or the flat portion of a dinner plate can be the motionless smooth surface. Can the item used for the block be more easily separated from the surface when tilted relative to the surface? Describe your experiments and try to explain your results. Solution 9.21. Using a Formica counter-top and the underside of a coffee-cup saucer, a compelling case can be made for the accuracy of the result of Exercise 9.20. If an upward force F is applied to the flat-bottomed block, the gap distance as a function of time is:

    h(t) = ho 14Fho

    2t3a4

    $

    % &

    '

    ( )

    1 2

    (this is the same as the answer for Exercise 9.20 with W replaced by F). This formula then predicts that the time for the plate to rise to z = +, t, is:

    t =3a4

    4Fho2 .

    So, t will increase with increasing a and , and decrease with increasing F and ho. Puddles of peanut butter and pancake syrup both resisted vertical motion of the saucer. The peanut butter (more viscous) was more effective in preventing the saucer from being lifted. The other parametric dependencies were qualitatively verified as well. In all cases, the adhesion between the saucer and the countertop failed when a small air-passage opened up under the saucer. This allowed air, which is much less viscous than the liquids tested, to rush in under the saucer allowing it to be lifted.

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

    Exercise 9.22. A rectangular slab of width 2L (and depth B into the page) moves vertically on a thin layer of oil that flows horizontally as shown. Assume

    u(y, t) = h2 2( ) dP dx( ) y h( ) 1 y h( ) where h(x,t) is the instantaneous gap between the slab and the surface, is the oils viscosity, and P(x,t) is the pressure in the oil below the slab. The slab is slightly misaligned with the surface so that

    h(x, t) = ho 1+ x L( ) + h ot where

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

    Integrate once:

    h3

    12Px

    = h ox + C , and then again with h(x,t) evaluated at t = 0,

    P =12 h ox + C

    ho3 1+ x L( )3

    dx + D 12ho3 h ox + C( ) 1 3 x L( ) dx + D

    P 12ho

    3 h o

    x 2

    2+ Cx h o

    3L

    x 3

    3C 3

    Lx 2

    2%

    & '

    (

    ) * + D =

    12ho

    3 h o

    x 2

    21 2

    Lx

    %

    & '

    (

    ) * + Cx 1

    32L

    x%

    & '

    (

    ) *

    %

    & '

    (

    ) * + D

    Here, C and D could depend on t, but the instant of interest is t = 0 so this dependence is suppressed. c) There are two boundary conditions on the pressure at either end of the slab:

    P(x = L) = P(x = +L) = Po. d) Using the results of part b) and c), leads to two equations for the constants C and D.

    P(L) = Po =12ho

    3 h o

    L2

    21+ 2( ) CL 1+ 3

    2$

    % &

    '

    ( )

    $

    % &

    '

    ( ) + D, and

    P(+L) = Po =12ho

    3 h o

    L2

    21 2( ) + CL 1 3

    2$

    % &

    '

    ( )

    $

    % &

    '

    ( ) + D.

    These can be solved to find:

    C = h oL and

    D = Po 6L2

    ho3

    h o ,

    where quadratic terms in have been dropped. Thus, the pressure distribution is:

    P(x) = 12ho

    3 h o

    x 2

    21 2

    Lx

    $

    % &

    '

    ( ) + h oLx 1

    32L

    x$

    % &

    '

    ( )

    $

    % &

    '

    ( ) + Po

    6L2

    ho3

    h o

    = Po +6L2

    ho3

    h ox 2

    L21 2

    Lx

    $

    % &

    '

    ( ) + 2

    xL

    1 32L

    x$

    % &

    '

    ( ) 1

    $

    % &

    '

    ( )

    = Po 6L2

    ho3

    h o 1x 2

    L2+

    2L3

    x 3 2L

    x + ...$

    % &

    '

    ( )

    = Po 6L2

    ho3

    h o 1x 2

    L2$

    % &

    '

    ( ) 1

    2L

    x$

    % &

    '

    ( ) + ...

    where again quadratic terms in have been dropped. e) When the slab is moving downward (so

    h o is negative), the highest pressure point lies in the region where the gap is the smallest and this pressure distribution causes a moment on the slab that tends to align the slab with the surface. This is beneficial in lubrication flows because the oil pressure acts to prevent contact between slab and the surface. When the slab is moving upward, the lowest pressure lies in the region where the gap is the smallest and this pressure distribution causes a moment on the slab that tends to misalign the slab with the surface. These phenomena are readily apparent when pressing or lifting a flat object from a wet or oily surface. In particular, it may be nearly impossible to lift the object straight up; however, once one corner or edge is tilted up, the object is much easier to lift completely.

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

    Exercise 9.23. Show that the lubrication approximation can be extended to viscous flow within narrow gaps h(x,y,t) that depend on two spatial coordinates. Start from (4.10) and (9.1), and use Cartesian coordinates oriented so that x-y plane is locally tangent to the center-plane of the gap. Scale the equations using a direct extension of (9.14):

    x* = x/L , y* = y/L , z* = z/h = y/L , t* = Ut/L , u* = u/U , v* = v/U , w* = w/U , and p* = p/Pa. where L is the characteristic distance for the gap thickness to change in either the x or y direction, and = h/L. Simplify these equation when 2ReL 0, but UL/Pah2 remains of order unity to find:

    0 1px

    + 2uz2

    ,

    0 1py

    + 2vz2

    , and

    0 1pz

    .

    Solution 9.23. In Cartesian coordinates the starting equations (4.10) and the three components of (8.1) are:

    ux

    +vy

    +wz

    = 0 ,

    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

    &

    ' (

    )

    * + .

    First insert the given scaling into the continuity equation:

    ux

    +vy

    +wz

    =ULu*

    x*+ULv*

    y*+Uh

    w*

    z*=UL

    u*

    x*+v*

    y*+w*

    z*$

    % &

    '

    ( ) = 0.

    Thus, there are no simplifications of the continuity equation as 2ReL 0, so it remains unchanged. Now insert the given scaling into the x-momentum equation, to find:

    UL U

    u*

    t*+U 2

    Lu* u

    *

    x*+U 2

    Lv* u

    *

    y*+U 2

    hw* u

    *

    z*=

    PaL

    p*

    x*+

    UL2

    2u*

    x*2+UL2

    2u*

    y*2+Uh2

    2u*

    z*2'

    ( )

    *

    + , ,

    Simplify using = h/L, and collect terms with like coefficients:

    U 2

    Lu*

    t*+ u* u

    *

    x*+ v* u

    *

    y*+ w* u

    *

    z*#

    $ %

    &

    ' ( =

    PaL

    p*

    x*+

    UL2

    2u*

    x*2+ 2u*

    y*2#

    $ %

    &

    ' ( +

    Uh2

    2u*

    z*2.

    Multiply through by h2/U, and use = h/L and ReL = UL/ to reach:

    2 ReLu*

    t*+ u* u

    *

    x*+ v* u

    *

    y*+ w* u

    *

    z*$

    % &

    '

    ( ) =

    Pah2

    ULp*

    x*+ 2

    2u*

    x*2+ 2u*

    y*2$

    % &

    '

    ( ) +

    2u*

    z*2.

    When 2ReL 0, but UL/Pah2 remains of order unity this equation simplifies to:

    0 Pah2

    ULp*

    x*+ 2u*

    z*2, or in dimensional form:

    0 1px

    + 2uz2

    .

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

    The procedure for the y-momentum equation is the same as that for the x-momentum equation with u replaced by v and p/x replaced by p/y. The final dimensionless and dimensional forms are:

    0 Pah2

    ULp*

    y*+ 2v*

    z*2, or in dimensional form:

    0 1py

    + 2vz2

    .

    The z-momentum equation is different that the other two because the scaling of w differs from that of u and v, and the pressure gradient involves z (not x or y). Inserting the given scaling leads to:

    UL U

    w*

    t*+U 2

    Lu* w

    *

    x*+U 2

    Lv* w

    *

    y*+2U 2

    hw* w

    *

    z*=

    PaL

    p*

    z*+

    UL2

    2w*

    x*2+UL2

    2w*

    y*2+Uh2

    2w*

    z*2'

    ( )

    *

    + ,

    As before, simplify using = h/L, and collect terms with like coefficients:

    U 2

    Lw*

    t*+ u* w

    *

    x*+ v* w

    *

    y*+ w* w

    *

    z*$

    % &

    '

    ( ) =

    PaL

    p*

    z*+

    UL2

    2w*

    x*2+ 2w*

    y*2$

    % &

    '

    ( ) +

    Uh2

    2w*

    z*2.

    Multiply through by h2/U, and use = h/L and ReL = UL/ to reach:

    4 ReLw*

    t*+ u* w

    *

    x*+ v* w

    *

    y*+ w* w

    *

    z*$

    % &

    '

    ( ) =

    Pah2

    ULp*

    z*+ 4

    2w*

    x*2+ 2w*

    y*2$

    % &

    '

    ( ) + 2

    2w*

    z*2.

    When 2ReL 0, but UL/Pah2 remains of order unity this equation simplifies to:

    0 Pah2

    ULp*

    z*, or in dimensional form:

    0 1pz

    .

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

    Exercise 9.24. A squeegee is pulled across a smooth flat stationary surface at a constant speed U. The gap between the squeegee and surface is h(x) = hoexp{+x/L} and this gap is filled with a fluid of constant density and viscosity . If the squeegee is wetted by the fluid for 0 x L, and the pressure in the surrounding air is pe, what is the pressure p(x) distribution under the squeegee and what force W perpendicular to the surface is needed to hold the squeegee in place? Ignore gravity in your work.

    Solution 9.24. Follow the solution approach in Example 9.3. Here the pressure gradient will be:

    dp(x)dx

    = 12h3(x)

    C1 6Uh2 (x)

    = 12ho3 C1e

    3 x L 6Uho2 e

    2 x L .

    Integrate to find the pressure distribution:

    p(x) = +12L3ho

    3C1e

    3 x L +6UL2ho

    2e2 x L +C2 =

    3ULho2

    43Uho

    C1e3 x L + e2 x L

    "

    #$

    %

    &'+C2 .

    The boundary conditions are p(0) = p(L) = pe, and these lead to two equations for the two constants C1 and C2.

    pe =3ULho2

    43Uho

    C1 +1!

    "#

    $

    %&+C2 , and pe =

    3ULho2

    43Uho

    C1e3 + e2

    "

    #$

    %

    &'+C2 . (1,2)

    Subtract (2) from (1) to find C1:

    0 = 3ULho2

    43Uho

    C1(1 e3 )+1 e2

    "

    #$

    %

    &' or C1 =

    3Uho41 e2

    1 e3 .

    Substitute this into (1) to find C2:

    C2 = pe 3ULho2

    e2 e3

    1 e3"

    #$

    %

    &' .

    Use these constants to construct p(x):

    p(x) pe =3ULho2

    1 e2

    1 e3e3 x L + e2 x L e

    2 e3

    1 e3"

    #$

    %

    &' .

    The force (or load) W is determined from the integral:

    W = p(x) pe( )0

    L

    dx = 3LUho21 e2

    1 e3e3 x L + e2 x L e

    2 e3

    1 e3#

    $%

    &

    '(dx

    0

    L

    .

    The integration of the exponentials may be done directly:

    W = 3LUho2

    1 e2

    1 e3e3 x L

    3 L"

    #$

    %

    &'0

    L

    +e2 x L

    2 L"

    #$

    %

    &'0

    L

    e2 e3

    1 e3x[ ]0

    L(

    )**

    +

    ,-- .

    Continue with the evaluation:

    U!

    x!x = L!x = 0!

    h(x)!pe!

    pe!W!

    p(x) = ?!

    ho!

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

    W = 3LUho

    2L

    31 e2

    1 e3e3 1( ) L2 e

    2 1( ) e2 e3

    1 e3L

    "

    #$

    %

    &'

    = 3LUho

    2L

    61 e2( ) e

    2 e3

    1 e3L

    "

    #$

    %

    &'

    And, when

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

    Exercise 9.25. A close-fitting solid cylinder with net weight W (= actual weight buoyancy), length L, and radius a is centered in and may slide along the axis of a long vertical tube with radius a + h, where h

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

    Q = 2 a+ 12 h( ) w(y)dy =0h

    2 a+ 12 h( ) h3

    12pa pbL

    #

    $%

    &

    '(

    #

    $%

    &

    '( .

    Eliminate the pressure difference from last two equations for Q and W to find:

    Q =W a+ 12 h( )h36L ah+ a2!" #$

    Wh3

    6aL,

    where the second more approximate form is valid as h a 0 .

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

    Exercise 9.26. A thin film of viscous fluid is bounded below by a flat stationary plate at z = 0. If the in-plane velocity at the upper film surface, z = h(x,y,t), is U = U(x,y,t)ex + V(x,y,t)ey, use the equations derived in Exercise 8.19 to produce the Reynolds equation for constant-density thin-flim lubrication:

    h3

    $

    % &

    '

    ( ) p

    *

    + ,

    -

    . / =12

    ht

    + 6 hU( )

    where

    = ex x( ) + ey y( ) merely involves the two in-plane dimensions. Solution 9.26. For the specified geometry, solutions for the x- and y-direction equations are given by appropriately evaluated versions of (8.19):

    u h2

    2px

    zh1 z

    h%

    & '

    (

    ) * +U

    zh

    , and

    v h2

    2py

    zh1 z

    h%

    & '

    (

    ) * +V

    zh

    .

    where h, U, and V are functions of x, y, and t. Now consider a stationary nearly-rectangular control volume of with elemental base area dxdy and height h(x,y,t). If the flat vertical sides of the CV are perpendicular to the coordinate directions, then an integral statement of conservation of mass will include one term involving the time-variation of h and four flux terms, one for each vertical side:

    htdxdy dy udz

    0

    h(x,y,t )

    dx vdz0

    h(x,y,t )

    + dy udz0

    h(x+dx,y,t )

    + dx vdz0

    h(x,y+dy,t )

    = 0.

    For constant density, this can be simplified and rearranged:

    ht

    +1dx

    udz0

    h(x+dx,y,t )

    udz0

    h(x,y,t )

    %

    & '

    (

    ) * +

    1dy

    vdz0

    h(x,y+dy,t )

    vdz0

    h(x,,t )

    %

    & '

    (

    ) * = 0 , or

    ht

    +x

    udz0

    h(x,y,t )

    $

    % &

    '

    ( ) +

    y

    vdz0

    h(x,,t )

    $

    % &

    '

    ( ) = 0 .

    The two integrations in [,]-brackets in the second equality can be completed using the equations above for u and v.

    u0

    h

    dz h2

    2px

    zh

    1 zh

    &

    ' (

    )

    * + dz

    0

    h

    +U zhdz

    0

    h

    = h3

    2px

    (1)d0

    1

    +Uh d0

    1

    = h3

    2px

    12

    13

    &

    ' (

    )

    * + +Uh

    12

    &

    ' (

    )

    * +

    = h3

    12px

    +Uh2

    ,

    , and

    v0

    h

    dz h2

    2py

    zh

    1 zh

    &

    ' (

    )

    * + dz

    0

    h

    +V zhdz

    0

    h

    = h3

    2py

    (1)d0

    1

    +Vh d0

    1

    = h3

    2py

    12

    13

    &

    ' (

    )

    * + +Vh

    12

    &

    ' (

    )

    * +

    = h3

    12py

    +Vh2

    .

    .

    The result of the continuity equation becomes:

    ht

    +x

    h3

    12px

    +Uh2

    $

    % &

    '

    ( ) +

    y

    h3

    12py

    +Vh2

    $

    % &

    '

    ( ) = 0.

    Rearrange this equation and use

    = ex x( ) + ey y( ) and U = Uex + Vey,

    12ht

    + 6 x

    Uh( ) + 6 y

    Vh( ) = x

    h3

    px

    #

    $ %

    &

    ' ( +

    y

    h3

    py

    #

    $ %

    &

    ' ( or

    12ht

    + 6 hU( ) = h3

    p

    %

    & '

    (

    ) * .

    Here it is worth noting that when U, V, and h are known, the solution of final equation specifies the pressure in the fluid film.

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

    Exercise 8.21. Fluid of density and viscosity flows inside a long tapered tube of length L and radius R(x) = (1 x/L)Ro where < 1, and Ro 0).

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

    b) The flow accelerates or decelerates inside a contracting or expanding pipe, respectively. The

    above solution will only be valid if fluid acceleration can be ignored compared to

    1dpdx

    or

    2u

    (recall that these two terms are set equal to obtain the solution for the velocity profile). Hence

    we need only consider one of them, and therefore must have:

    uux

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

    Exercise 9.28. A circular lubricated bearing of radius a holds a stationary round shaft. The bearing hub rotates at angular rate as shown. A load per unit depth on the shaft, W, causes the center of the shaft to be displaced from the center of the rotating hub by a distance ho, where ho is the average gap thickness and ho

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

    dPd

    =12aC1 + 6a

    2hh3

    =12aC1 + 6a

    2ho(1+ cos)ho3(1+ cos)3

    Even though integration of this equation can be completed exactly, it's tedious and unnecessary because

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

    Exercise 9.29. As a simple model of small-artery blood flow, consider slowly varying viscous flow through a round flexible tube with inlet at z = 0 and outlet at z = L. At z = 0, the volume flux entering the tube is

    Qo (t) . At z = L, the pressure equals the exterior pressure pe. The radius of the tube, a(z,t), expands and contracts in proportion to pressure variations within the tube so that (i)

    a ae = (p pe ) , where ae is the tube radius when the pressure, p(z,t), in the tube is equal to pe, and is a positive constant. Assume the local volume flux, Q(z,t), is related to

    p z by (ii)

    Q = a4 8( ) p z( ) . a) By conserving mass, find a partial differential equation that relates Q and a. b) Combine (i), (ii), and the result of part a) into one partial differential equation for a(z,t). c) Determine a(z) when Qo is a constant and the flow is perfectly steady.

    Solution 9.29. a) Conserve mass in a differential disk control volume perpendicular to the flow direction that has circular inlet and outlet surfaces located at z and z + z with radii of a(z,t) and a(z + z,t). There is no flow through the tube wall so:

    t

    dVCV u(r,t)

    0

    a(z,t )

    2rdr + u(r,t)0

    a(z+z,t )

    2rdr = 0 .

    Noting that

    Q(z,t) = u(r,t)2rdrr= 0

    r=R(z,t ) and that the CV volume is

    a2(z,t)z allows the equation above to be simplified to:

    t( ) a2( ) + Q(z + z,t) Q(z,t)( ) z = 0 , or taking the limit as

    z 0:

    t

    a2( ) + Qz

    = 0 .

    b) Differentiate (i) to find:

    a z = p z . Put this into (ii) to get:

    Q = a4 8( ) a z( ) .

    Insert this into the result of part a):

    t

    a2( ) = z

    a4

    8az

    $

    % &

    '

    ( ) .

    c) For steady flow, the part a) result is not needed. From part b), (i) and (ii) imply:

    Qo = a4 8( ) a z( ) . Integrate once to find:

    a5

    40= Qoz + C where C is a constant. At z =

    L, a = ae so

    C = ae5

    40+QoL . Thus the final answer is:

    a(z) = 40Qo

    (L z) + ae5%

    & ' (

    ) *

    1 5

    .

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

    Exercise 9.30. Consider a simple model of flow from a tube of toothpaste. A liquid with viscosity and density is squeezed out of a round horizontal tube having radius a(t). In your work, assume that a is decreasing and use cylindrical coordinates with the z-axis coincident with the centerline of the tube. The tube is closed at z = 0, but is open to the atmosphere at z = L. Ignore gravity. a) If w is the fluid velocity along the z-axis, show that:

    za dadt

    + w(z,R,t)RdR0

    a = 0 .

    b) Determine the pressure distribution, p(z) p(L), by assuming the flow in the tube can be

    treated within the lubrication approximation by setting

    w(z,R,t) = 14

    dpdz

    a2(t) R2( ) . c) Find the cross-section-average flow velocity

    wave (z,t) in terms of z, a, and

    da dt . d) If the pressure difference between z = 0 and z = L is P, what is the volume flux exiting the tube as a function of time. Does this answer partially explain why fully emptying a toothpaste tube by squeezing it is essentially impossible? Solution 9.30. a) Choose a CV that encloses the fluid inside the tube from the capped end to an axial distance z. Conservation of mass implies:

    ddt

    a2z( ) + w(z,R,t)2RdR = 00a or

    za dadt

    + w(z,R,t)RdR = 00

    a .

    b) Here p = p(z) alone; insert the given velocity profile into the result of part a) and integrate:

    za dadt

    =14

    dpdz

    a2(t) R2( )RdR0a = a

    4 (t)16

    dpdz

    .

    The ends of this equality form a differential equation for the pressure; integrate from z to L:

    p(z) p(L) = 8 a3( ) da dt( ) L2 z2( ). c) Multiply the result of part a) by 2a2, insert factors of , and recognize the definition of wave:

    2a2

    w(z,R,t)RdR0

    a = 1

    a2w(z,R,t)2RdR

    0

    a = wave (z,t) =

    2a2

    %

    & '

    (

    ) * za

    dadt

    = 2zadadt

    ,

    d) Use Q(t) = a2wave(L,t), and combine the results of parts b) and c) to eliminate da/dt. From

    part b):

    p(0) p(L) = P = 8a3

    dadt

    $

    % &

    '

    ( ) L2 02( ) , or

    dadt

    "

    # $

    %

    & ' =

    a3P8L2

    ; so

    Q(t) = a2wave (L,t) = a2 2Ladadt

    = 2La dadt

    = 2La a3P

    8L2%

    & '

    (

    ) * =

    4a4PL

    Yes, this answer explains a lot about flow from a toothpaste tube. Here, it is clear that for a finite P (the pressure one can exert by squeezing), the volume flux from the tube will be proportional to

    a4 . Thus, as the tube empties and its effective cross section decreases, the flow rate from the tube approaches zero. The only hope for better emptying the tube is to decrease L, and it is an easily observed fact that more toothpaste can be extracted from a nearly-empty tube by rolling it up from the sealed end.

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

    Exercise 9.31. A large flat plate below an infinite stationary incompressible viscous fluid is set in motion with a constant acceleration,

    u , at t = 0. A prediction for the subsequent fluid motion, u(y,t), is sought. a) Use dimensional analysis to write a physical law for u(y,t) in this flow. b) Starting from the x-component of (8.1) determine a linear partial differential equation for u(y,t). c) The linearity of the equation obtained for part c) suggests that u(y,t) must be directly proportional to

    u . Simplify your dimensional analysis to incorporate this requirement. d) Let = y/(t)1/2 be the independent variable, and derive a second-order ordinary linear differential equation for the unknown function f() left from the dimensional analysis. e) From an analogy between fluid acceleration in this problem and fluid velocity in Stokes first

    problem, deduce the solution

    u(y, t) = u 1 erf y 2 $ t ( )[ ]d $ t 0

    t

    and show that it solves the

    equation of part b). f) Determine f() and if your patience holds out show that it solves the equation found in part d). g) Sketch the expected velocity profile shapes for several different times. Note the direction of increasing time on your sketch.

    Solution 9.31. a) The problem parameters are:

    u , , y, t, and . The solution parameter is u. Create the parameter matrix: u

    u y t Mass: 0 0 1 0 0 1 Length: 1 1 -3 1 0 -1 Time: -1 -2 0 0 1 -1 This matrix has rank three. Determine the number of dimensionless groups: 6 parameters - 3 dimensions = 3 groups Construct the dimensionless groups: 1 = u/

    u t, 2 = y/(t)1/2, 3 =

    u ty/ Write a dimensionless law: u =

    u tF(y/(t)1/2,

    u ty/), where F is an unknown function. b) Assume that u = u(y,t) only!, then u/x = 2u/x2 = 0. In this case the continuity equation requires v/y = 0, so since v = 0 at the surface of the plate it is zero everywhere. If the plate is large, there will be no end effects so the pressure gradient must also be zero. This eliminates the nonlinear terms in the equation, the pressure gradient, and one of the viscous terms. The final

    result being:

    ut

    = 2uy 2

    .

    c) If the form of the solution must be linear with respect to

    u , then 3 can not be a parameter since 1 takes care of any linear relationship between u and

    u . This means: u =

    u tf(y/(t)1/2)

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

    d) First convert derivatives, then differentiate being sure to account for the pre-factor of

    u t.

    t

    =t

    = 12

    yt 3

    = 2t

    , and

    y

    =y

    =1t

    .

    Therefore:

    ut

    =t

    u tf ()[ ] = u f u t 2t

    dfd

    , and

    2uy 2

    = 2

    y 2 u tf ()[ ] = u t 1

    td2 fd2

    .

    Put these converted derivatives together and cancel common factors, to find:

    d2 fd2

    +2dfd

    f = 0.

    e) In Stokes' first problem the plate's velocity was impulsively raised from 0 to U and the velocity field equation and its solution are:

    ut

    = 2uy 2

    and

    u(y, t) =U 1 erf y 2 t( )[ ] . In the current problem, the plate's acceleration is impulsively raised from 0 to

    u , and the

    horizontal acceleration of fluid particles is

    a(y, t) = ut

    + uux

    =ut

    + 0 , because u/x = 0. Thus,

    by time differentiating the field equation for u, the field equation for a and its analogous solution is obtained:

    at

    = 2ay 2

    and

    a(y, t) = u 1 erf y 2 t( )[ ] . However, a = u/t implies

    u(y, t) = a(y, " t )d " t " t = 0

    " t = t , so the above analogy suggests the velocity

    field for the impulsively accelerated plate will be:

    u(y, t) = u 1 erf y 2 $ t ( )[ ]$ t = 0$ t = t d $ t . To show that this solution satisfies the field equation for u, determine the derivatives using the definition of the error function,

    erf( ) = 2 ( ) exp 2( )0 d .

    ut

    = u t

    1 erf y 2 % t ( )[ ]0

    t

    d % t = u 1 erf y 2 t( )[ ] ,

    uy

    = u y

    1 erf y 2 % t ( )[ ]0

    t

    d % t = u y

    t 2

    exp 2( )dd % t 0

    y 2 % t

    0

    t

    ) * +

    , - .

    = u 0 2

    exp y 2 4 % t ( ) 12 % t d % t 0t

    ) * +

    , - .

    = u 1

    exp y 2 4 % t ( ) % t

    d % t 0

    t

    ) * /

    + /

    , - /

    . / ,

    2uy 2

    = u

    y

    exp y 2 4 & t ( )

    & t d & t

    0

    t

    ( ) *

    + *

    , - *

    . * =

    u y2

    exp y 2 4 & t ( )& t 3 2

    d & t 0

    t

    .

    Manipulate the final relationship for the second y-derivative to introduce the integration variable

    = y 2 $ t using

    d = y 2 ( ) $ t 3 2 1 2( )d $ t = ydt 4 $ t 3 2 to find:

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

    2uy 2

    = 2 u

    exp y 2 4 & t ( ) yd & t 4 & t 3 2'

    ( )

    *

    + , =

    0

    t

    2 u

    exp 2( )d

    y 2 t

    = + 2 u

    exp 2( )dy 2 t

    = + 2 u

    exp 2( )d0

    2 u

    exp 2( )d0

    y 2 t

    = u 1 2

    exp 2( )d0

    y 2 t

    '

    ( ) )

    *

    + , , = u 1 erf y 2 t( )( ) = ut

    The final equality follows from the determination of u/t above; thus, the velocity solution obtained by analogy does solve the field equation. f) Start from the part e) solution and insert the definition of the error function to find:

    u(y, t) = u 1 2

    exp( 2)d0

    y 2 & t

    (

    ) * *

    +

    , - - 0

    t

    d & t .

    Set

    " = y " t so that

    d " = y 2 " t 3( )d " t = 2y 2( ) " 3d " t and

    d " t = 2 y 2 " 3( )d " , and use this in the equation for u(y, t):

    u(y, t) = u 1 2

    exp( 2)d0

    ' 2

    '

    ( )

    *

    + ,

    2y2

    / 30

    1 2

    3

    4 5 d / .

    Multiply by t outside the integral, divide by t inside the integral, and recognize the definition of to determine the form of the similarity function:

    u(y, t) u t

    = f () = 22 1 2

    exp( 2)d0

    ' 2

    '

    ( )

    *

    + ,

    d.

    . 3= 22 1 erf . 2( )[ ]

    d.

    . 3.

    The first and second derivatives of this function are needed to determine if it solves the part d) equation. These derivatives are:

    dfd

    = 4 1 erf $ 2

    %

    & '

    (

    ) *

    +

    , -

    .

    / 0

    d$

    $ 3 22 1 erf

    2%

    & '

    (

    ) *

    +

    , -

    .

    / 0 13

    =2

    f 1+ erf 2

    %

    & '

    (

    ) *

    %

    & '

    (

    ) * , and

    d2 fd2

    = 22

    f 1+ erf 2

    $

    % &

    '

    ( )

    $

    % &

    '

    ( ) +

    2

    dfd

    +2e

    2 4 12

    $

    % &

    '

    ( )

    = 22

    f 1+ erf 2

    $

    % &

    '

    ( )

    $

    % &

    '

    ( ) +

    2

    2

    f 1+ erf 2

    $

    % &

    '

    ( )

    $

    % &

    '

    ( ) +

    1e

    2 4$

    % &

    '

    ( )

    = 22

    f 22

    1 erf 2

    $

    % &

    '

    ( )

    ,

    - .

    /

    0 1 +

    2

    e2 4,

    where the second equality for the second derivative follows from substituting in the first derivative result. Use these derivative results to assemble the part d) equation:

    d2 fd2

    +2dfd

    f = 22

    f 22

    1 erf 2

    $

    % &

    '

    ( )

    *

    + ,

    -

    . / +

    2

    e2 4 + f 1+ erf

    2$

    % &

    '

    ( )

    $

    % &

    '

    ( ) f

    = 22

    f 22

    +1$

    % &

    '

    ( ) 1 erf

    2

    $

    % &

    '

    ( )

    *

    + ,

    -

    . / +

    2

    e2 4 .

    Consider the first term on the right and integrate by parts twice to find:

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

    22

    f = 4 1 erf $ 2

    %

    & '

    (

    ) *

    +

    , -

    .

    / 0

    d$

    $ 3=221 erf

    2%

    & '

    (

    ) *

    +

    , -

    .

    / 0

    2e

    2 4

    212

    e $ 2 4d $

    +

    , -

    .

    / 0 .

    The final term on the right can be converted into 1 erf(/2) by changing the integration variable to = /2 and breaking the range of integration, to , into two intervals, to 0, and 0 to . Thus, the reassembled part d) differential equation becomes:

    d2 fd2

    +2dfd

    f = 22

    1 erf 2

    $

    % &

    '

    ( )

    *

    + ,

    -

    . /

    2e

    2 4

    + 1 erf

    2$

    % &

    '

    ( )

    *

    + ,

    -

    . /

    22

    +1$

    % &

    '

    ( ) 1 erf

    2

    $

    % &

    '

    ( )

    *

    + ,

    -

    . / +

    2

    e2 4

    = 0.

    Here final equality occurs because there is term-by-term cancellation on the right side. Thus, the function f() determined above is a valid solution of the part d) equation. g) The velocity disturbance spreads linearly with time in flow direction, and in proportion to

    t in the vertical direction. The contours shown to the right are for quadratically spaced time increments.

    u(y,t)

    y

    Increasing Time

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

    Exercise 9.32. a) When z is complex, the small-argument expansion of the zeroth-order Bessel function Jo(z) =1

    14 z

    2 +... remains valid. Use this to show that (9.43) reduces to (9.6) as 0

    when dp/dz = p/L. The next term in the series is 164 z4 . At what value of a is the

    magnitude of this term equal to 5% of the second term. b) When z is complex, the large-argument expansion of the zeroth-order Bessel function Jo(z) 2 z( )

    1 2 cos z 14 #$ %& remains valid for |arg(z)| < . Use this to show that (9.43) reduces to the velocity profile of a viscous boundary layer on a plane wall beneath an oscillating flow as :

    uz (y, t) = pL

    sin(t) exp y 2

    #$%

    &'(sin t y

    2

    )

    *+

    ,

    -.

    /

    011

    2

    344,

    where y is the distance from the tube wall, R = a y, y

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

    cos z 4

    "

    #$

    %

    &'=

    12

    exp i 12+

    i2

    "

    #$

    %

    &'

    R

    i 4

    ()*

    +*

    ,-*

    .*+ exp i 1

    2+

    i2

    "

    #$

    %

    &'

    R

    + i 4

    ()*

    +*

    ,-*

    .*

    /

    011

    2

    344

    = 12

    exp i2

    12

    "

    #$

    %

    &'

    R

    i 4

    ()*

    +*

    ,-*

    .*+ exp i

    2+

    12

    "

    #$

    %

    &'

    R

    + i 4

    ()*

    +*

    ,-*

    .*

    /

    011

    2

    344

    When , the first term becomes exponentially small, so

    cos z 4

    "

    #$

    %

    &' 12exp i

    2+12

    "

    #$

    %

    &'

    R

    + i 4

    )*+

    ,+

    -.+

    /+ as .

    Now use R = a y in the above expression and collect like factors:

    cos z 4

    "

    #$

    %

    &' 12exp i a y

    2 +4

    "

    #$$

    %

    &''+

    a y2

    )*+

    ,+

    -.+

    /+ as .

    or:

    cos z 4

    "

    #$

    %

    &'

    ei 4

    2exp a(1+ i)

    2

    "

    #$$

    %

    &''exp

    (1+ i)2

    y"

    #$$

    %

    &'' as .

    So, in this limit:

    Joi3 2R

    !

    "##

    $

    %&&

    Joi3 2a

    !

    "##

    $

    %&&

    =

    2

    i3 2 (a y)

    ei 4

    2exp a(1+ i)

    2

    !

    "##

    $

    %&&exp

    (1+ i)2

    y!

    "##

    $

    %&&

    2

    i3 2a

    ei 4

    2exp a(1+ i)

    2

    !

    "##

    $

    %&&

    = aa y

    exp (1+ i)2

    y!

    "##

    $

    %&&

    exp (1+ i)2

    y!

    "##

    $

    %&&

    where the final approximate equality holds when y

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

    Exercise 9.33. A round tube bent into a U-shape having inner diameter d holds a column of liquid with overall length L. Initially the column of liquid is pushed upward on the right side of the U-tube and downward on the left side of the U-tube so that the two liquid surfaces are a vertical distance 2ho apart. If the liquid has density and viscosity , and the column is released from rest, find and solve an approximate ordinary differential equation that describes the subsequent damped oscillations of h(t), the liquid height above equilibrium in the right side of the U-tube, assuming that the flow profile at any time throughout the tube is parabolic. Under what condition(s) is this approximate solution valid? Will oscillations occur in this parameter regime?

    Solution 9.33. This is the fluid mechanical pendulum with viscous effects included. Use Newton's second law for the mass of fluid in the water column. The unbalanced gravitational force tends to decrease h(t) and is 2gh(d2/4) and the wall shear stress opposes the motion. Thus:

    4d 2L d

    2hdt2

    = dL w 2g4d 2h ,

    where w is presumed to be negative. Here dh/dt is the cross-section-average fluid velocity, so from the results for steady flow in a round tube:

    Vave =dhdt=

    d 2

    32dpdz

    = d 2

    324 wd

    "

    #$

    %

    &' ,

    where the final equality follows from (9.8). Use this results to eliminate w from the first equation to reach:

    4d 2L d

    2hdt2

    = dL 8ddhdt

    "

    #$

    %

    &' 2gh

    4d 2h .

    Divide by the coefficient of d2h/dt2, and rearrange the terms: d 2hdt2

    +32d 2

    dhdt+2gLh = 0 . ($)

    This equation can be solved by assuming an exponential solution: h(t) = hoemt, which leads to an algebraic equation for m:

    m2 32d 2

    "

    #$

    %

    &'m+

    2gL= 0 or m = 16

    d 21 1 gd

    4

    128 2L

    "

    #$$

    %

    &'' ,

    where = /. Thus, the general solution is:

    L!

    d!

    h(t)!

    g!

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

    h(t) = A+ exp16td 2

    1+ 1 gd4

    128 2L

    "

    #$$

    %

    &''

    ()*

    +*

    ,-*

    .*+ A exp

    16td 2

    1 1