Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (4)

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

    Exercise 4.1. Let a one-dimensional velocity field be u = u(x, t), with v = 0 and w = 0. The density varies as = 0(2 cos t). Find an expression for u(x, t) if u(0, t) = U. Solution 4.1. Here u = u(x,t)ex, and the density field is given, so a solution for u(x,t) might be found from the continuity equation:

    t

    + u( ) = 0, or specifically for this problem:

    t

    + ux

    + ux

    = 0.

    The given density field only depends on time so /x = 0, and this leads to:

    ux

    = 1t

    = 0 sin(t)

    0 2 cos(t)( ) u = sin(t)

    2 cos(t)'

    ( )

    *

    + , x + C(y,z,t) .

    where C is function of integration that does not depend on x. The initial condition requires: u(0, t) = U = C(y,z,t),

    so the final answer for u(x, t) is

    u =U sin(t)2 cos(t)

    $

    % &

    '

    ( ) x .

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

    Exercise 4.2. Consider the one-dimensional Cartesian velocity field: u = x t, 0, 0( ) where is a constant. a) Find a spatially uniform, time-dependent density field, = (t), that renders this flow field mass conserving when = o at t = to. b) What are the unsteady (u/t), advective ([u ]u ), and particle (Du/Dt) accelerations in this flow field? What does = 1 imply? Solution 4.2. a) Use the continuity equation and the given velocity field with = (t):

    t

    + u( ) = 0 implies

    ddt

    + t

    = 0.

    Separate variables and integrate:

    d

    = t dt >

    ln = ln t + C .

    Exponentiate and evaluate the integration constant at t = to to find:

    = o t to( ) .

    b) For the given flow field: ut

    = xt2ex , (u )u = u

    uxex =

    xt

    $

    %&

    '

    ()tex =

    2xt2ex , and

    DuDt

    =ut+ (u )u = ( 1)x

    t2ex .

    When = 1 the unsteady and advective acclerations are non-zero, but they are equal and opposite so that Du/Dt is zero.

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

    Exercise 4.3. Find a non-zero density field (x,y,z,t) that renders the following Cartesian velocity fields mass conserving. Comment on the physical significance and uniqueness of your solutions. a)

    u = U sin(t kx),0,0( ) where U, , k are positive constants. [Hint: exchange the independent variables x,t for a single independent variable = tkx] b)

    u = (y,+x,0) with = constant. [Hint: switch to cylindrical coordinates] c)

    u = A x ,B y ,C z( ) where A, B, C are constants. Solution 4.3. a)

    u = U sin(t kx),0,0( ) where U, , k are positive constants. Use the expanded

    form of the continuity equation for a unidirectional velocity,

    u = u,0,0( ) :

    t

    + ux

    + ux

    = 0 ,

    and plug in the given velocity field to find:

    t kU cos(t kx) +U sin(t kx)

    x= 0 .

    Now follow the hint and change from independent variables (x, t) to , where = tkx:

    t

    =t

    dd

    =dd

    , and

    x

    =x

    dd

    = k dd

    ,

    to find a first-order differential equation:

    dd

    kU sin dd

    = kU cos that can be

    separated and integrated:

    d

    = kU cos kU sin d >

    ln = ln kU sin( ) + C(y,z) .

    Here, the constant C must be used to make the solution dimensionally sound. Noting that /k has units of velocity, define M = kU/. Since the original equation did not contain any y or z dependence the constant of integration might depend on these variables. So, the final solution can be obtained by exponentiating the last equation:

    (x,y,z,t) = o(y,z)1M sin

    =o(y,z)

    1 kU ( )sin(t kx),

    where o(y,z) is an undetermined function. Thus, this solution is not fully determined; it is not unique. This is the density field that corresponds to a traveling-wave disturbance in a stationary medium. In the limit as M 0 with o = constant, this wave becomes an acoustic plane wave with /k = the speed of sound and M = the Mach number of the fluid particle motions. b)

    u = (y,+x,0) with = constant. The change to cylindrical coordinates is straight forward using:

    x = Rcos &

    y = Rsin , and

    uR = ucos + v sin &

    u = usin + v cos .

    uR = y cos +x sin = Rsin cos +Rcos sin = 0, and

    u = +y sin +x cos =Rsin2 +Rcos2 =R.

    Thus,

    u =Re , so the continuity equation is:

    t

    +1R

    R( ) = 0 or

    t

    +

    = 0 .

    The solution of this equation can be obtained from the method of characteristics. The idea is to determine special or characteristic directions or paths in the r--z-t space along which the solution for is easy to find. Start by postulating the existence of such a path

    = R(t),(t),z(t),t( ) so that the total derivative of with respect to time is:

    ddt

    =R

    dRdt

    +

    ddt

    +z

    dzdt

    +t

    .

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

    Comparing this equation with the one above shows that the characteristic paths are defined by

    dR dt = 0 ,

    d dt =, and

    dz dt = 0 . Thus, after a single time integration of each equation, the characteristic paths are determined to be:

    R = Ro,

    =t +o , and

    z = zo. where the quantities with subscript zero are constants. Along these paths the original equation for becomes d/dt = 0, which implies the density is constant along these paths. If the density at Ro, o, zo and t = 0 is

    o(Ro,o,zo), then the density at all later times can be obtained by substituting for ro, o, zo from the three equations that specify the characteristic path. Hence, the solution for the density is:

    = o(R, t,z) . Here again o is an undetermined function so this solution is not fully determined; it is not unique. In this case, the velocity field corresponds to solid-body rotation about the z-axis at angular rate , so conservation of mass implies that the density variations must revolve around the z-axis at the angular rate as well. c)

    u = A x ,B y ,C z( ) where A, B, C are constants.

    Use the expanded form of the continuity equation:

    t

    + u+ u = 0 , and plug in the given

    velocity field to find: t

    Ax2+By2+Cz2

    "

    #$

    %

    &'+

    Axx

    +Byy

    +Czz

    = 0 . This equation is linear in

    , so try a separation of variables solution: i.e.

    = X(x)Y (y)Z(z)T(t) . Putting in the trial solution, dividing the whole equation by the trial solution, and grouping terms yields:

    " T T

    +Ax

    " X X

    Ax 2

    $

    % & '

    ( ) +

    By

    " Y Y

    By 2

    $

    % &

    '

    ( ) +

    Cz

    " Z Z

    Cz2

    $

    % & '

    ( ) = 0

    where ( ) denotes derivative of ( ) with respect to is argument. Because each group of terms depends on only one of the independent coordinates, this equation can only be satisfied if each term is equal to a constant, and the 4 constants sum to zero. This means setting:

    " T T

    = d ,

    Ax

    " X X

    Ax 2

    = a,

    By

    " Y Y

    By 2

    = b, and

    Cz

    " Z Z

    Cz2

    = c where a + b + c + d = 0

    The solution of the first equation is:

    T(t) = Toedt where To is a constant, while that of the second

    can be found from:

    " X X

    =axA

    +1x lnX = ax

    2

    2A+ ln x + const., or

    X(x) = Xox expax 2

    2A" # $

    % & '

    ,

    where Xo is a constant. The solutions of the third and fourth equations are similar to that of the second. Combining the solutions of these equations and condensing the leading product of constants to

    C1 = ToXoYoZo produces:

    (x,y,z,t) = C1xyzexpax 2

    2A+by 2

    2B+cz2

    2C (a + b + c)t

    $ % &

    ' ( )

    where C1, a, b, c are undetermined constants, and the above restriction on a, b, c, and d has been used to eliminate d. Any particular version of this solution is acceptable as long as C1 0. Here, the density field is zero everywhere that

    u. The constants C1, a, b, and c are not determined so again this solution is not fully determined; it is not unique.

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

    Overall, conservation of mass is typically inadequate to fully specify a fluid density or velocity field; conservation of momentum and initial & boundary conditions are needed for complete flow field solutions.

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

    Exercise 4.4. A proposed conservation law for , a new fluid property, takes the following form: ddt

    dVV (t ) + Q ndS

    A(t ) = 0 , where V(t) is a material volume that moves with the fluid velocity

    u, A(t) is the surface of V(t), is the fluid density, andQ = . a) What partial differential equation is implied by the above conservation statement?

    b) Use the part-a) result and the continuity equation to show: t+u = 1

    ( ) .

    Solution 4.4. a) Start with the given material CV equation, and apply Reynolds Transport Thm. to the first term on the left side and Gausss Divergence Thm. to the other term:

    ddt

    dVV (t ) + Q ndS

    S(t ) = 0

    t( )dV

    V (t ) + u n( )dV

    A(t ) + QdV

    V (t ) = 0 .

    Now apply Gausss divergence theorem to the remaining surface integral, subsitute in the specified relationship Q = , and combine all the terms into one volume integral to find:

    t

    ( )+ u( ) ( )$%&

    '()dV

    V (t ) = 0 .

    Here V(t) is an arbitrary material volume, so the integrand must be zero. Thus, the partial differential equation implied by the given CV equation is:

    t

    ( )+ u( ) = ( ) .

    b) Expand the left side of the part a) result and group the terms that have as a coefficient:

    t+

    t+ u( )+ u =

    t+ u +

    t+ u( )

    #

    $%

    &

    '( .

    The contents of the parentheses is zero because of the continuity equation, so dividing by yields:

    t+u = 1

    ( ) ,

    which can be recognized as the advection-diffusion equation for a conserved passive scalar.

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

    Exercise 4.5. The components of a mass flow vector u are u = 4x2y, v = xyz, w = yz2. a) Compute the net mass outflow through the closed surface formed by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. b) Compute

    (u) and integrate over the volume bounded by the surface defined in part a) c) Explain why the results for parts a) and b) should be equal or unequal. Solution 4.5. a) The specified volume is a cube, and the mass outflow will have six contributions (one for each side):

    u ndA =cube surface 4x 2y[ ]x=1dydz

    z= 0

    1

    y= 0

    1

    4x 2y[ ]x= 0dydzz= 0

    1

    y= 0

    1

    + xyz[ ]y=1dxdz z= 0

    1

    x= 0

    1

    xyz[ ]y= 0dxdzz= 0

    1

    x= 0

    1

    + yz2[ ]z=1dxdy y= 0

    1

    x= 0

    1

    yz2[ ]z= 0dxdyy= 0

    1

    x= 0

    1

    The integral evaluations are straightforward and unremarkable, and lead to:

    u ndA =cube surface 2 + 0 + 14

    + 0 + 12

    + 0 = 114

    .

    b) First compute the divergence of the mass flow:

    (u) = x

    u( ) + y

    v( ) + z

    w( ) = 8xy + xz + 2yz ,

    then integrate it in the cubical volume:

    (u)cube dV = 8xy + xz + 2yz( )dxdydz

    z= 0

    1

    y= 0

    1

    x= 0

    1

    = 8 12&

    ' (

    )

    * + 12

    &

    ' (

    )

    * + +

    12

    &

    ' (

    )

    * + 12

    &

    ' (

    )

    * + + 2

    12

    &

    ' (

    )

    * + 12

    &

    ' (

    )

    * + =114

    .

    c) The two answers should be the same because of Gauss' Divergence Theorem.

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

    Exercise 4.6. Consider a simple fluid mechanical model for the atmosphere of an ideal spherical star that has a surface gas density of o and a radius ro. The escape velocity from the surface of the star is ve. Assume that a tenuous gas leaves the stars surface radially at speed vo uniformly over the stars surface. Use the steady continuity equation for the gas density and fluid velocity u = (ur,u ,u ) in spherical coordinates

    1r2

    r

    r2ur( )+ 1rsin

    u sin( )+1

    rsin

    u( ) = 0

    for the following items

    a) Determine when vo ve so that u = (ur,u ,u ) = vo 1 ve2 vo

    2( ) 1 ro r( )( ), 0, 0( ) . b) Simplify the result from part a) when vo >> ve so that: u = (ur,u ,u ) = (vo, 0, 0) . c) Simplify the result from part a) when vo = ve . d) Use words, sketches, or equations to describe what happens when vo < ve . State any assumptions that you make. Solution 4.6. a) ur is the only non-zero velocity component so only the continuity equation simplifies to

    1r2

    r

    r2ur( ) = 0, multiply by r2 and integrate to find:

    r2ur = C(,) , where C is the function of integration that cannot depend on r. Thus,

    =C(,)r2ur

    =oro

    2

    r21 ve

    2 vo2( ) 1 ro r( )( )[ ]

    1 2,

    where

    C(,) = oro2vo = const. is determined from the boundary conditions: = o & v = vo at r

    = ro. b) When

    vo >> ve , the square root factor in the result of part a) becomes unity, so

    = o ro2 r2 .

    c) When

    vo = ve , the square root factor simplifies to

    ro r( )1 2, so

    = o ro r[ ]3 2

    d) Here, the gas leaving the stars surface does not escape the stars gravitational field, so it must fall back to stars surface. For a tenuous gas, it might be acceptable to ignore molecular collisions so that the velocity field of the outward moving gas,

    ur+ = vo 1 ve

    2 vo2( ) 1 ro r( )( ) for

    r <

    rmax = ro 1 vo2 ve

    2[ ] , is equal and opposite to the velocity field of the inward falling gas so that

    ur = ur

    + . Both outward and inward mass fluxes must balance because the net mass flux must be zero. Thus, the resulting density field will have equal parts contributed by inward and outward flowing stellar gas. Under these conditions, the density field will be double that from part a) with a radial limitation:

    = 2oro

    2

    r21 ve

    2 vo2( ) 1 ro r( )( )[ ]

    1 2 for r < rmax

    0 for r rmax

    % & '

    ( '

    ) * '

    + ' where

    rmax = ro 1 vo2 ve

    2[ ] .

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

    Exercise 4.7. Consider the three-dimensional flow field ui = xi or equivalently u = rer, where is a constant with units of inverse time, xi is the position vector from the origin, r is the distance from the origin, and er is the radial unit vector. Find a density field that conserves mass when: a) (t) depends only on time t and = o at t = 0, and b) (r) depends only on the distance r and = 1 at r = 1 m. c) Does the sum (t) + (r) also conserve mass in this flow field? Explain your answer. Solution 4.7. The full continuity equation is: t + (u ) + u = 0 . a) When depends only on t, (u ) = 0 so the cont. equation reduces to: d dt + u = 0 . For the given velocity field,u = ui xi = xi xi = x1 x1 +x2 x2 +x3 x3( ) = 3 , so d dt = 3 . This equation has the exponential solution: (t) = oe

    3t when = o at t = 0. b) When only depends on distance r, /t = 0 so the continuity equation reduces to: (rer ) +3 = 0 , or r d dr( ) = 3 . This equation has the power law solution

    (r) = 1 1m r( )3when = 1 at r = 1 m.

    c) The proposed sum solution, (t) + (r), does conserve mass, and this is readily shown by substitution of (t) + (r) into the full continuity equation. This sum solution conserves mass because the continuity equation is linear in when u is specified independently of . When when u are both dependent field variables, the continuity equation is non-linear.

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

    Exercise 4.8. The definition of the stream function for two-dimensional constant-density flow in the x-y plane is:

    u = ez , where ez is the unit vector perpendicular to the x-y plane that determines a right-handed coordinate system. a) Verify that this vector definition is equivalent to

    u = y , and

    v = x in Cartesian coordinates. b) Determine the velocity components in r- polar coordinates in terms of r- derivatives of . c) Determine an equation for the z-component of the vorticity in terms of . Solution 4.8. a) Start from the definition of the cross product:

    u = ez = ex ey ez0 0 1

    x y 0= ex

    y

    '

    ( )

    *

    + , ey

    x

    '

    ( )

    *

    + , = ex

    y

    '

    ( )

    *

    + , + ey

    x

    '

    ( )

    *

    + , .

    Setting components equal from the extreme ends of this extended equality produces:

    u = y , and

    v = x . b) In r- polar coordinates u = urer + ue,

    ez er = e , and

    ez e = er . Plus,

    = err

    + e1r

    . Therefore, using the same definition in r- polar coordinates produces:

    u = urer + ue = ez = ez err

    + e1r

    (

    ) *

    +

    , - = e

    r

    + er1r

    .

    Matching components implies:

    ur =1r

    and

    u = r

    .

    c) The z-component of the vorticity is:

    z =vx

    uy

    =x

    x

    &

    ' (

    )

    * +

    y

    y

    &

    ' (

    )

    * + =

    2x 2

    + 2y 2

    &

    ' (

    )

    * + =

    2 .

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

    Exercise 4.9. A curve of

    (x,y) = C1 (= a constant) specifies a streamline in steady two-dimensional constant-density flow. If a neighboring streamline is specified by

    (x,y) = C2, show that the volume flux per unit depth into the page between the streamlines equals C2 C1 when C2 > C1. Solution 4.9. Start with a picture of the two streamlines and let Q = volume flow rate (per unit depth) between them.

    By definition,

    Q = |u | d1

    2

    where

    d is an increment of length that lies perpendicular to the flow

    in the stream tube. Again by definition,

    lies in the direction perpendicular to the lines of constant . Thus:

    d = | |

    dx = x( )dx + y( )dy x( )2 + y( )2

    =vdx + udyv 2 + u2

    =d|u |

    , so

    |u | d = d .

    This means that:

    Q = |u | d1

    2

    = d = C2 C11

    2

    , which completes the proof.

    = C1

    = C2

    d

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

    Exercise 4.10. Consider steady two-dimensional incompressible flow in r- polar coordinates where u = (ur,u ) , ur = + r

    2( )cos , and is positive constant. Ignore gravity. a) Determine the simplest possible u . b) Show that the simplest stream function for this flow is = r( )sin . c) Sketch the streamline pattern. Include arrowheads to show stream direction(s). d) If the flow is frictionless and the pressure far from the origin is p, evaluate the pressure p(r, ) on = 0 for r > 0 when the fluid density is . Does the pressure increase or decrease as r increases?

    Solution 4.10. a) Place the given ur = + r2( )cos into u = 1r

    r

    rur( )+1ru

    = 0 , to find:

    1rr

    rur( ) =1rr

    +rcos

    #

    $%

    &

    '(=

    r3cos = 1

    ru

    , or u

    = +r2cos u = +

    r2sin ,

    where the final constant of integration has been dropped to produce the simplest possible u . b) By definition: ur = 1 r( ) ( ) , and u = r . Thus,

    1r

    = +r2cos so

    = +

    rcos =

    rsin + f (r) , and

    r

    = +r2sin so =

    rsin + g( ) .

    The simplest stream function is recovered when f = g = 0: = r( )sin .

    c) Start from the result of b) and switch to Cartesian coordinates:

    =

    x2 + y2y

    x2 + y2 , which implies x2 + y2 = ( ) y , or

    x2 + y 2( )2 = 2( )2 . Thus, the streamlines are circles centered on the y-axis that are tangent to the x-axis at the origin of coordinates. The stream direction is determined from the radial velocity ur, which must be positive when is near zero. d) Use the ordinary Bernoulli-equation to find:

    p = p(r, )=0 +12 vr

    2 + v2( )

    =0= p(r, 0)+ 1

    2

    2

    r4cos2(0)+

    2

    r4sin2(0)

    #

    $%

    &

    '( , or

    p(r, 0) = p 122

    r4.

    The pressure increases as r increases.

    x!

    y!

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

    Exercise 4.11. The well-known undergraduate fluid mechanics textbook by Fox et al. (2009) provides the following statement of conservation of momentum for a constant-shape (non-rotating) control volume moving at a non-constant velocity U = U(t).

    ddt

    ureldVV *(t ) + urel urel n( )

    A*( t ) dA = gdV

    V *(t ) + fdA

    A*(t ) dUdt

    dVV *(t ) .

    Here

    urel = uU(t) is the fluid velocity observed in a frame of reference moving with the control volume while u and U are observed in a non-moving frame. Meanwhile, (4.17) states this law as

    ddt

    udVV *(t ) + u uU( ) n

    A*(t ) dA = gdV

    V *( t ) + fdA

    A*( t )

    where the replacement b = U has been made for the velocity of the accelerating control surface A*(t). Given that the two equations above are not identical, determine if these two statements of conservation of fluid momentum are contradictory or consistent. Solution 4.11. The goal here is to derive one result from the other. Start with the equation from the undergraduate textbook, substitute using

    urel = uU , and put the forces on the left and move the extra acceleration term to the other side of the equation.

    gdVV *(t ) + fdA

    A*(t ) = dUdt

    dVV *(t ) + ddt

    uU( )dVV *(t ) + uU( ) uU( ) n

    A*( t ) dA

    Now expand the terms on the terms on the right side.

    gdVV *(t ) + fdA

    A*(t ) =

    dUdt

    dVV *(t ) + ddt

    udVV *(t ) dUdt

    dVV *(t ) U ddt

    dVV *(t ) + u uU( ) n

    A*(t ) dA U uU( ) n

    A*(t ) dA

    Here, U and dU/dt depend on time only that is, they are uniform over V*(t) and A*(t) so they can move inside or outside of the volume and surface integrals over V*(t) and A*(t), but not through time differentiations. Thus, the first and third terms on the right side are equal and opposite. Rearrange the other terms to find:

    gdVV *(t ) + fdA

    A*(t ) = ddt

    udVV *(t ) + u uU( ) n

    A*( t ) dA U ddt

    dVV *(t ) + uU( ) n

    A*(t ) dA

    &

    ' (

    )

    * + .

    Conservation of mass requires the contents of the [,]-brackets to equal zero. After removing these terms, the remaining simplified equation,

    ddt

    udVV *(t ) + u uU( ) n

    A*(t ) dA = gdV

    V *( t ) + fdA

    A*( t ) ,

    is the same as (4.17). The two statements of conservation of momentum are consistent. The essential difference between the two CV formulations of conservation of momentum amounts to a transformation between an inertial and an accelerating frame of reference when mass is also conserved.

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

    Exercise 4.12. A jet of water with a diameter of 8 cm and a speed of 25 m/s impinges normally on a large stationary flat plate. Find the force required to hold the plate stationary. Compare the average pressure on the plate with the stagnation pressure if the plate is 20 times the area of the jet. Solution 4.12. Choose a rectangular CV with one side that covers and the flat stationary plate. Denote the plate area by A, and the jet velocity & area by Ujet and Ajet, respectively. In this problem only the x-component of the integral momentum equation is required, and x-direction fluid momentum only enters the CV of its left side. Assume that atmospheric pressure Po acts on the three sides of the CV that are not in contact with the plate.

    U jet2 A jet + 0 = PoA PdA

    A = Fx , or

    Fx = U jet2 A jet = (10

    3kg /m3)(25m /s)2 4(0.08m)2 = 3.142kN

    The average pressure is:

    Pave =Fx

    20A jet=U jet

    2

    20=(103kg /m3)(25m /s)2

    20= 31.25kPa .

    The stagnation pressure is:

    Ps =12U jet

    2 =(103kg /m3)(25m /s)2

    2= 312.5kPa , so

    PavePs

    =110

    Here Ps and Pave are reported as gauge pressures.

    Ujet

    Ajet

    x

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

    Exercise 4.13. Show that the thrust developed by a stationary rocket motor is F = AU2 + A(p patm), where patm is the atmospheric pressure, and p, , A, and U are, respectively, the pressure, density, area, and velocity of the fluid at the nozzle exit. Solution 4.13. Use the control volume shown, where F is the force that holds the rocket motor in place.

    Assuming the flow is steady and that control volume is not moving (b = 0), the horizontal component of (4.17) becomes:

    U 2A = F + patmA pA , or

    F = U 2A + A(p patm ) .

    F

    patm

    Up

    !A

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

    Exercise 4.14. Consider the propeller of an airplane moving with a velocity U1. Take a reference frame in which the air is moving and the propeller [disk] is stationary. Then the effect of the propeller is to accelerate the fluid from the upstream value U1 to the downstream value U2 > U1. Assuming incompressibility, show that the thrust developed by the propeller is given by F = A(U2

    2 U12 ) 2, where A is the projected area of the propeller and is the density (assumed

    constant). Show also that the velocity of the fluid at the plane of the propeller is the average value U = (U1 + U2)/2. [Hint: The flow can be idealized by a pressure jump of magnitude p = F/A right at the location of the propeller. Also apply Bernoullis equation between a section far upstream and a section immediately upstream of the propeller. Also apply the Bernoulli equation between a section immediately downstream of the propeller and a section far downstream. This will show that p = (U2

    2 U12 ) / 2 .]

    Solution 4.14. The CV and nominal propeller-edge streamlines are:

    where the mass flux within the curved propeller-edge streamlines is constant and equal to

    m = U1A1 = U2A2. Apply the steady constant density Bernoulli equation from "1" to "a", and from "b" to "2":

    p1 +12 U1

    2 = pa +12 Ua

    2 , and

    pb +12 Ub

    2 = p2 +12 U2

    2 . Since p1 = p2, and Ua = Ub, these equations become:

    pb pa = +12 U2

    2 U12( ).

    Now apply the momentum principle across the propeller plane:

    Ua2 + Ub

    2 = 0 = paA pbA + F , or

    F = A(pb pa ) =12 A U2

    2 U12( ) .

    where Ua = Ub, and F is the force applied to the propeller (positive to the right) by the device that holds it in place. To find the velocity at the propeller plane, use the stationary rectangular control volume shown above, and start with (4.5),

    U1A1 + U2A2 + U1(A1 A2) + m sides = 0, where the four terms correspond to the inlet, fast flow at the outlet, ordinary speed flow at the outlet, and stream-wise sides of the CV. Here the inlet flow is uniform at velocity U1. Using the definition of

    m , the conservation of mass statement becomes:

    U1(A1 A2) + m sides = 0 . where A2 is the fast-flow area on the outlet side of the CV. Now evaluate the horizontal component of (4.17)

    m U1 + m U2 + U12(A1 A2) + U1 m sides = (p1 p2)A1 + F .

    1 2a b

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

    Substitute in the conservation of mass result and note that p1 = p2 to find:

    m U1 + m U2 = F , or

    F = UA(U2 U1) . where

    m = UA , while U (= Ua = Ub) and A are flow velocity and area inside the curved lines at the propeller plane. Equate the relationship for F found from the Bernoulli equations with the one found from conserving mass and momentum to develop an equation for U:

    F = 12 A U22 U1

    2( ) = UA(U2 U1) , Use the second equality, cancel common factors, and solve for U to find:

    U = 12 (U2 +U1) .

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

    Exercise 4.15. Generalize the control volume analysis of Example 4.1 by considering the control volume geometry shown for steady two-dimensional flow past an arbitrary body in the absence of body forces. Show that the force the fluid exerts on the body is given by the Euler momentum integral,

    Fj = uiu j ij( )A1

    nidA , and that

    0 = uiA1

    nidA .

    Solution 4.15. For steady flow through a stationary control volume (b = 0) and no body force, (4.5) and (4.17) simplify to:

    A*(t ) u n( )dA = 0 uini

    A*( t ) dA = 0, and

    uA*(t ) u n( )dA = f(n,x,t)

    A*(t ) dA u juini

    A*(t ) dA = ni ijdA

    A*(t ) ,

    where the second equality in each case is the same as first after conversion to index notation, and (2.15) or (4.20b) has been used to write the surface force f in terms of the stress tensor ij. Here the control surface A* is composed of three pieces,

    A* = A1 + A2 + A3, where A1 is the outer surface, A2 conforms to the body's surface, and A3 is the connection between A1 and A2. Here n (with components ni) is the outward normal on the combined surface. Thus, n points outward on A1 (shown as n1 in the figure) and points into the body on A2 (shown as n2 in the figure). Thus, the conservation laws require:

    uiniA1 +A2 +A3

    dA = 0 and

    u jui ij( )niA1 +A2 +A3

    dA = 0.

    Now consider the contribution of each surface individually. In general, there are no simplifications to be made on A1 and all the integrand terms must be retained. The surface A2 coincides with the surface of the solid body, thus

    u n = uini = 0 on A2. This fact eliminated the contribution from A2 in the conservation of mass equation, and causes the first integrand term to dropout of the conservation of momentum equation. On A3, as the width of the opening goes to zero, the magnitude of the integrands on the upper and lower surfaces become equal but the normal on these two surfaces is opposite; therefore, A3 provides no net contribution to either conservation law. Thus, the integral laws become:

    uiniA1

    dA = 0 and

    ijniA2

    dA + u jui ij( )niA1

    dA = 0.

    The first of these is sought as part of the solution to this exercise. The second becomes the Euler momentum integral after considering the lone integral over A2 and orientation of n on A2. If n pointed outward from the body, this integral would represent the force, F, the fluid applies to the body. However on A2, n (= n2) points inward so n points outward from the body, thus

    F = ijniA2

    dA .

    Therefore the final result (the Euler momentum integral) is:

    F = u jui ij( )niA1

    dA .

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

    Exercise 4.16. The pressure rise

    p = p2 p1 that occurs for flow through a sudden pipe-cross-sectional-area expansion can depend on the average upstream flow speed Uave, the upstream pipe diameter d1, the downstream pipe diameter d2, and the fluid density and viscosity . Here p2 is the pressure downstream of the expansion where the flow is first fully adjusted to the larger pipe diameter. a) Find a dimensionless scaling law for p in terms of Uave, d1, d2, and . b) Simplify the result of part a) for high-Reynolds-number turbulent flow where does not matter. c) Use a control volume analysis to determine p in terms of Uave, d1, d2, and for the high Reynolds number limit. [Hints: i) a streamline drawing might help in determining or estimating the pressure on the vertical surfaces of the area transition, and ii) assume uniform flow profiles wherever possible.] d) Compute the ideal flow value for p and compare this to the result from part d) for a diameter ratio of d1/d2 = . What fraction of the maximum possible pressure rise does the sudden expansion achieve?

    Solution 4.16. a) This is a dimensional analysis task. Construct the units matrix. p Uave d1 d2 M 1 0 0 0 1 1 L -1 1 1 1 -3 -1 T -2 -1 0 0 0 -1 The matrix has rank three so there are 6 3 = 3 dimensionless groups. These can be found by inspection; thus:

    p Uave2( ) = f d1 d2 ,Uaved1 ( ) where f is an undetermined function.

    b) At high Reynolds number the viscosity will not be a parameter, so the Reynolds number can be dropped from the part a) result:

    p Uave2( ) = f d1 d2( ) . The remainder of this exercise

    involves finding f. c) Place a cylindrical CV in the duct that abuts the area change on the upstream side and extends far enough downstream to fully enclose the flows reattachment zone. For high-Re turbulence, its OK to assume uniform inflow and outflow profiles. Denote the outflow velocity by U2.

    Cons. of mass implies:

    4d12Uave +

    4d22U2 = 0 ; (a)

    Cons. of horizontal momentum:

    4d12Uave

    2 +4d22U2

    2 = p14d22 p2

    4d22. (b)

    Uave!

    d1 !d2!

    p2!p1!

    Separation Reattachment

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

    Here, the cross-stream momentum equation suggests that p1 acts on the flat surfaces of the area transition because the flow separates at the area junction and is parallel to the x-direction and this implies that

    p r = 0 . Use the above equations to determine

    p = p2 p1. Equation (b) implies:

    p2 p1 = d12Uave

    2 d22 U2

    2 , and equation (a) reduces to:

    U2 = d12 d2

    2( )Uave . Eliminate U2 to find:

    p2 p1 = d12Uave

    2 d22 d1

    2 d22( )2Uave2 , or

    p2 p1 = Uave2 d1

    2 d22( ) 1 d12 d22( )( ) .

    d) For an ideal flow:

    p2 p1 =12 Uave

    2 12 U22 = 12 Uave

    2 1 d14 d2

    4( ) = 15 32( )Uave2 From part d):

    p2 p1 = Uave2 d1

    2 d22( ) 1 d12 d22( )( ) = 3 16( )Uave2 . Thus, the pressure rise in a

    sudden expansion to four times the area with separated flow is only 40% of that possible in an ideal flow. Abrupt area changes are exceptionally common in automotive piston-engine exhaust systems where they are used for low-frequency noise control. However, they lead to pumping losses that reduce the engine's horsepower, especially under wide-open-throttle conditions, because of the principles illustrated in this exercise.

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

    Exercise 4.17. Consider how pressure gradients and skin friction develop in an empty wind tunnel or water tunnel test section when the flow is incompressible. Here the fluid has viscosity and density , and flows into a horizontal cylindrical pipe of length L with radius R at a uniform horizontal velocity Uo. The inlet of the pipe lies at x = 0. Boundary layer growth on the pipes walls induces the horizontal velocity on the pipes centerline to be UL at x = L; however, the pipe-wall boundary layer thickness remains much smaller than R. Here, L/R is of order 10, and UoR/ >> 1. The radial coordinate from the pipe centerline is r. a) Determine the displacement thickness,

    L* , of the boundary layer at x = L in terms of Uo, UL ,

    and R. Assume that the boundary layer displacement thickness is zero at x = 0. [The boundary layer displacement thickness, *, is the thickness of the zero-flow-speed layer that displaces the outer flow by the same amount as the actual boundary layer. For a boundary layer velocity profile u(y) with y = wall-normal coordinate and U = outer flow velocity, * is defined by:

    *= 1 u U( )( )0 dy .]

    b) Determine the pressure difference, P = PL Po, between the ends of the pipe in terms of , Uo, and UL. c) Assume the horizontal velocity profile at the outlet of the pipe can be approximated by:

    u(r,x = L) =UL 1 r R( )n( ) and estimate average skin friction,

    w, on the inside of the pipe

    between x = 0 and x = L in terms of , Uo, UL, R, L, and n. d) Calculate the skin friction coefficient,

    c f = w12 Uo

    2 , when Uo = 20.0 m/s, UL = 20.5 m/s, R = 1.5 m, L = 12 m, n = 80, and the fluid is water, i.e. = 103 kg/m3.

    Solution 4.17. a) Apply the principle of conservation of mass. The effective flow area will be

    R2 at the inlet (x = 0), but will be

    R L*( )2 at the outlet (x = L):

    UoR2 =UL R L

    *( )2.

    Solve for

    L* = R 1 Uo UL( ).

    b) Start from the steady Bernoulli equation (4.19) for horizontal flow (no body force):

    12Uo

    2 + Po =12UL

    2 + PL , and rearrange it to find:

    PL Po =12Uo

    2 12UL

    2 =12 Uo

    2 UL2( ).

    c) Choose a CV that encloses all the fluid in the tube between x = 0 and x = L, and apply the CV form of conservation of momentum (4.17). Here the flow is steady and there is no net body force, thus:

    2 Uo2rdr +

    r= 0

    R

    2 UL2 1 r R( )n( )2rdr = Po PL( )

    r= 0

    R

    R2 2RL w Cancel common factors, evaluate the integrals, and substitute in the part b) result for the pressure difference:

    Uo2R2 + 2 UL

    2 1 2 r R( )n + r R( )2n( )rdr = Po PL( )r= 0

    R

    R2 2RL w

    x = 0! x = L!2R!

    Uo! UL!r!

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

    Uo2R2 + 2UL

    2 R2

    22Rn

    Rn+2

    n + 21R2n

    R2n+2

    2n + 2$

    % &

    '

    ( ) = Po PL( )R2 2RL w

    Uo2R2 + UL

    2R2 1 4n + 2

    +1

    n +1$

    % &

    '

    ( ) =12 UL

    2 Uo2( )R2 2RL w

    Divide by R2 and collect terms:

    Uo2 +UL

    2 1 4n + 2

    +1

    n +1#

    $ %

    &

    ' ( =12UL2 Uo

    2( ) 2 L wR

    12

    4n + 2

    +1

    n +1#

    $ %

    &

    ' ( UL

    2 12Uo2 = 2 L w

    R

    Solve for the average shear stress:

    w =R4L

    Uo2 1 8

    n + 2+

    2n +1

    %

    & '

    (

    ) * UL

    2+

    , -

    .

    / 0

    d) Use the result of part c) to find:

    c f =2 wUo

    2 =R2L

    1 1 8n + 2

    +2

    n +1%

    & '

    (

    ) * UL2

    Uo2

    +

    , -

    .

    / 0 . Evaluate:

    c f =1.5m2(12m)

    1 1 882

    +281

    #

    $ %

    &

    ' ( 20.520.0

    #

    $ %

    &

    ' ( 2)

    * +

    ,

    - . = 0.00162

    The numbers provided here are approximately applicable to the William B. Morgan Large Cavitation Channel in Memphis, Tennessee, the world's largest low-turbulence water tunnel.

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

    Exercise 4.18. An acid solution with density flows horizontally into a mixing chamber at speed V1 at x = 0 where it meets a buffer solution with the same density moving at speed V2. The inlet flow layer thicknesses are h1 and h2 as shown, the mixer chamber height is constant at h1 + h2, and the chamber width into the page is b. Assume steady uniform flow across the two inlets and the outlet. Ignore fluid friction on the interior surfaces of the mixing chamber for parts a) and b). a) By conserving mass and momentum in a suitable control volume, determine the pressure difference, p = p(L) p(0), between the outlet (x = L) and inlet (x = 0) of the mixing chamber in terms of V1, V2, h1, h2, and . Do not use the Bernoulli equation. b) Is the pressure at the outlet higher or lower than that at the inlet when V1 V2? c) Explain how your answer to a) would be modified by friction on the interior surfaces of the mixing chamber.

    Solution 4.18. a) Choose a stationary control volume (b = 0) that captures the fluid between x = 0 and x = L. Here the inflows and outflows are presumed steady. Thus, conservation of mass from (4.5) reduces to:

    u ndA = 0A*(t ) .

    At x = 0 (the inlet CV surface), n = ex but the velocities of both streams are positive. At x = L (the outlet CV surface), n = +ex and the velocity of the mixed stream is positive. On both surfaces dA = bdy. Thus, the reduced form of (4.5) can be written:

    u ndA = ubdyinlet + ubdy

    outlet = V1bh1 V2bh2 + V3b(h1 + h2 ) = 0

    A*(t ) ,

    where V3 is the outlet flow speed. Here, the density and flow width b are constants, so this result can be simplified to:

    V1h1 +V2h2 =V3(h1 + h2 ) . () The same control volume should be used to conserve of momentum. Again the flow is presumed steady, so the horiztonal component of (4.17) reduces to:

    uA*(t ) (u n)dA = fx (n,x, t)

    A*(t ) dA ,

    where fx is the horizontal surface force on the control volume. The body force term is absent here because gravity acts vertially. As for conservation of mass with this control volume, only the inflow and outflow surfaces contribute. Evaluating the reduced form of (4.17) produces:

    uA*(t ) (u n)dA = u2bdy

    inlet + u2bdy

    outlet = V12bh1 V22bh2 + V32b(h1 + h2 )

    = fx (n,x, t)A*(t ) dA = + p(0)bdy

    inlet p(L)bdy

    outlet = p(0) p(L)( )b(h1 + h2 ),

    where the friction terms on the upper and lower CV boundaries have been neglected. Divide out the common factor of b, and rearrange this to isolate the pressure difference on the left:

    h1!h2!

    V1!

    V2!x = 0! x = L!

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

    p(0) p(L) = V32

    h1 + h2

    V12h1 +V2

    2h2( ) . Use () to eliminate V3,

    p(0) p(L) = V1h1 +V2h2h1 + h2

    "

    #$

    %

    &'

    2

    h1 + h2V12h1 +V2

    2h2( ) , and after some alegbra this can be simplified to:

    p(L) p(0) = h1h2(h1 + h2 )

    2 V1 V2( )2 .

    b) Interestingly, the pressure always rises when the two streams have different speeds: p(L) p(0), and it does not matter which velocity is larger. c) If friction were included in the calculation the pressure rise would be less.

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

    Exercise 4.19. Consider the situation depicted below. Wind strikes the side of a simple residential structure and is deflected up over the top of the structure. Assume the following: two-dimensional steady inviscid constant-density flow, uniform upstream velocity profile, linear gradient in the downstream velocity profile (velocity U at the upper boundary and zero velocity at the lower boundary as shown), no flow through the upper boundary of the control volume, and constant pressure on the upper boundary of the control volume. Using the control volume shown: a) Determine h2 in terms of U and h1, and b) Determine the direction and magnitude of the horizontal force on the house per unit depth into the page in terms of the fluid density , the upstream velocity U, and the height of the house h1. c) Evaluate the magnitude of the force for a house that is 10 m tall and 20 m long in wind of 22 m/sec (approximately 50 miles per hour).

    Solution 4.19. 3. a) Use the control volume formulation of the continuity equation to find that:

    Uh1 = u(y)dy0

    h2

    = Uyh2dy

    0

    h2

    = Uh2h22

    2=Uh22

    h2 = 2h1.

    b) The control volume form of the x-momentum equation is

    U 2dy0

    h1

    + u20

    h2

    dy = pdy0

    h1

    + pdy0

    h2

    + + p(n ex )dy + xtop .

    where x is the force on the fluid per unit depth into the page. It is delivered through the foundation of the house. Here the flow is assumed inviscid so there's no friction on the lower CV surface. Evaluating the integrals yields:

    U 2h1 +13U 2h2 = p +h1 h2 + (h2 h1)( ) +x

    x = 13U 2h1.

    Therefore, the force on the house in the +x direction per unit depth into page = +

    13U 2h1.

    c) (Force on the house) = (1.2 kg/m3)(22 m/sec)2(10 m)(20 m)/3 = 38.7 kN 8700 pounds. The force attempts to push the house in the positive x-direction, that is: downwind.

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

    Exercise 4.20. A large wind turbine with diameter D extracts a fraction of the kinetic energy from the airstream (density = = constant) that impinges on it with velocity U. a) What is the diameter of the wake zone, E, downstream of the windmill? b) Determine the magnitude and direction of the force on the windmill in terms of , U, D, and . c) Does your answer approach reasonable limits as 0 & 1?

    Solution 4.20. Place a control volume around the stream tube that hits the windmill with vertical inflow-outflow surfaces well upstream and downstream of the wind turbine. a) For = const, the volume flux in the stream tube must be constant. Therefore: V = A1U/A2. The efficiency of the wind turbine, , implies: (1 )U2/2 = V2/2, or V = U(1 )1/2. Eliminate V with the continuity result to find: A1U/A2 = U(1 )1/2 , or E = D/(1 )1/4 . b) For the chosen CV, cons. of horizontal momentum implies:

    U2A1 + V2A2 = +x, where x is force the wind turbine applies to the fluid. Here the flow speed outside the CV is assumed to be U everywhere. Therefore, the pressure on the CV surface is P everywhere because of the steady constant-density horizontal flow Bernoulli equation. Thus, the pressure integration drops out of the momentum equation because P acts on the entire control surface. Substituting in the results from part a) produces:

    Force on the windmill = x = U2 D2

    4 U2(1) D

    2

    4(1)1 2= U2 D

    2

    41 (1)1 2#$ %& .

    The force on the windmill points in the downstream direction, and is dimensionally sound (U2D2 is has units of force). c) This answer approaches reasonable limits. When as 0 the force on the wind turbine also goes to zero. So, when there is no change in airspeed, there should be no force on the wind turbine. When 1 the force on the windmill approaches U2D2/4, which precisely balances the momentum flux in the incoming stream tube.

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

    Exercise 4.21. An incompressible fluid of density flows through a horizontal rectangular duct of height h and width b. A uniform flat plate of length L and width b attached to the top of the duct at point A is deflected to an angle as shown. a) Estimate the pressure difference between the upper and lower sides of the plate in terms of x, , Uo, h, L and when the flow separates cleanly from the tip of the plate. b) If the plate has mass M and can rotate freely about the hinge at A, determine a formula for the angle in terms of the other parameters. You may leave your answer in terms of an integral.

    Solution 4.21. a) The question says to estimate the pressure difference. Thus, reasonable simplifying assumptions should be acceptable. The first of these is to ignore the two-dimensional character of the flow field and assume uniform horizontal flow at each x location. The second is to assume steady flow. So, for the conditions stated (incompressible), the pressure difference between the upper and lower side the plate can be estimated from the Bernoulli equation applied between a mid-duct point comfortably upstream of the plate, where the flow speed is Uo and the pressure is po, and a location x that is connected to the first point by a streamline, where the flow speed and pressure below the plate are U(x) and p(x).

    po +12Uo

    2 = p(x) + 12U 2(x)

    Conservation of mass in the duct requires:

    Uoh =U(x)h(x) =U(x) h x tan( ) . Combining these two equations to eliminate U(x) produces:

    p(x) po =12 Uo

    2 U 2(x)( ) = 12 Uo2

    Uo2h2

    (h x tan)2%

    & '

    (

    ) * =12Uo

    2 1 11 (x /h)2 tan2

    %

    & '

    (

    ) * .

    When the flow separates from the plate as shown, the pressure above the dotted separating streamline will be the same as that below it. This pressure is

    pupper = p(x = Lcos) po =12Uo

    2 1 11 (L /h)2 sin2

    %

    & '

    (

    ) * ,

    so that

    p(x) pupper =12Uo

    2 1 11 (x /h)2 tan2

    1+ 11 (L /h)2 sin2

    %

    & ' (

    ) *

    =12Uo

    2 11 (L /h)2 sin2

    1

    1 (x /h)2 tan2%

    & ' (

    ) * .

    b) The angle can be determined requiring the net moment on the plate to be zero. For a small element of the plate of length dl = dx/cos located at horizontal location x, the moment around A due to gravity is: x(M/L)g(dx/cos). For this same element, the moment due to fluid mechanical

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

    pressure is: +(x/cos)(p(x) pupper)b(dx/cos). Thus, the angle that solves the problem is specified by:

    Net moment on the plate

    = 0 = g MLcos

    + p(x) pupper( )b

    cos2$

    % & '

    ( ) x= 0

    x=L cos

    xdx .

    Using the above relationship for p(x) pupper, and pressing on with the integration and a little algebra yields a final transcendental equation:

    0 = g MLcos2

    +14Uo

    2bL2 1L h( )2 sin2

    ln 1 L h( )2 sin2{ } + 11 L h( )2 sin2

    %

    & ' '

    (

    ) * * ,

    which could be solved for specific values of the various parameters. Although complicated, this formula does reach the correct limits. When the fluid dynamic force is negligible, the second term can be ignored, the plate hangs straight down, and the answer = /2 is provided by the first term. When the plate's weight is negligible, the first term can be ignored so the plate's angle of deflection should be very small and this answer is recovered from the formula above when the terms inside the big parentheses go to zero, which occurs when 0.

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

    Exercise 4.22 A pipe of length L and cross sectional area A is to be used as a fluid distribution manifold that expels a steady uniform volume flux per unit length of an incompressible liquid from x = 0 to x = L. The liquid has density , and is to be expelled from the pipe through a slot of varying width, w(x). The goal of this problem is to determine w(x) in terms of the other parameters of the problem. The pipe-inlet pressure and liquid velocity at x = 0 are Po and Uo, respectively, and the pressure outside the pipe is Pe. If P(x) denotes the pressure on the inside of the pipe, then the liquid velocity through the slot Ue is determined from:

    P(x) Pe =12 Ue

    2 . For this problem assume that the expelled liquid exits the pipe perpendicular to the pipes axis, and note that wUe = const. = UoA/L, even though w and Ue both depend on x. a) Formulate a dimensionless scaling law for w in terms of x, L, A, , Uo, Po, and Pe. b) Ignore the effects of viscosity, assume all profiles through the cross section of the pipe are uniform, and use a suitable differential-control-volume analysis to show that:

    A dUdx

    + wUe = 0 , and

    ddxU 2 = dP

    dx .

    c) Use these equations and the relationships stated above to determine w(x) in terms of x, L, A, , Uo, Po, and Pe. Is the slot wider at x = 0 or at x = L?

    Solution 4.22. a) Create the parameter matrix. w x L A Uo Po Pe M 0 0 0 0 1 0 1 1 L 1 1 1 2 -3 1 -1 -1 T 0 0 0 0 0 1 -2 -2 There will be 8 3 = 5 dimensionless groups

    By inspection the groups are:

    wL

    ,

    xL

    ,

    AL2

    ,

    Uo2

    Po, and

    PoPe

    , thus:

    wL

    = fn xL, AL2, Uo

    2

    Po,PoPe

    #

    $ %

    &

    ' (

    b) For a differential slice of the pipe located between x and x+dx, cons. of mass implies:

    U(x)A + Ue (x)w(x)dx + U(x + dx)A = 0. Group the U-terms together and divide by dx:

    A U(x + dx) U(x)( ) dx + Ue (x)w(x) = 0. Divide by and take the limit as dx goes to zero to find:

    AdU dx +Uew = 0. Using the same CV conserv horizontal momentum

    U 2(x)A + U 2(x + dx)A = P(x)A P(x + dx)A Divide by A & dx, and take the limit as dx goes to zero to find:

    dU 2 dx = dP dx . c) Integrate the two equations found for part b):

    U = Uew A( )x + C1, and

    U 2 = P + C2

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

    The boundary conditions for U are: U(0) = Uo, and U(L) = 0, and from the problem statement: wUe = const. = UoA/L, so

    U(x) =Uo 1 x L( ) . Noting that P(0) = Po and U(0) = Uo, the integrated momentum equation becomes:

    P(x) + U 2(x) = Po + Uo2.

    Put this together with the COMA result and the exit velocity condition,

    P(x) Pe =12 Ue

    2 = 12 UoA wL( )2 ,

    to find:

    Pe +12 UoA wL( )

    2+ Uo

    2 1 x L( )2 = Po + Uo2, which can be solved for w(x) to get

    w(x) = A L( ) (Po Pe )12 Uo

    2( ) + 2 1 1 x L( )2( )[ ]1 2

    .

    The slot is wider at x = 0.

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

    Exercise 4.23. The take-off mass of a Boeing 747-400 may be as high as 400,000 kg. An Airbus A380 may be even heavier. Using a control volume (CV) that comfortably encloses the aircraft, explain why such large aircraft do not crush houses or people when they fly low overhead. Of course, the aircrafts wings generate lift but they are entirely contained within the CV and do not coincide with any of the CVs surfaces; thus merely stating the lift balances weight is not a satisfactory explanation. Given that the CVs vertical body-force term,

    g dVCV , will exceed

    4x106 N when the airplane and air in the CVs interior are included, your answer should instead specify which of the CVs surface forces or surface fluxes carries the signature of a large aircrafts impressive weight. Solution 4.23. Assume that the airplane is in steady level flight so that the flow is steady in the frame of reference of the airplane. Therefore, choose a frame of reference attached to the airplane and select a control volume (CV) that comfortably encloses the aircraft. In this frame of reference the CV is not moving but is instantaneously aligned above the house. The overall problem is really three dimensional, but will be simplified here. Therefore, take the CV to be a cube of volume V = L3 with z = 0 at mean roof level. Retaining the roof-peak in the solution is not essential since there is no flow through the bottom of the control volume. Denote the speed and mass of the aircraft by Uac and Mac, let the Cartesian-coordinate fluid velocity with respect to the control volume be

    u = (u,v,w) , and choose the front of the control volume to coincide with the y-z plane.

    For steady flow of a perfect fluid with density (viscous forces are irrelevant here because of the enormous Reynolds number), the integral momentum equation in the z-direction is:

    wudSfront + wudS + w2dS

    top

    back wvdS

    close side + wvdS

    far side

    = g(airV + Mac ) PdStop + PdS

    bottom

    where g is the acceleration of gravity, P is pressure, and the volume of the aircraft has been ignored compared V. There is no flux term on the bottom of the CV because the rooftop is assumed to be solid and impenetrable. On the front and back of the control volume u is approximately equal to Uac. The contribution from sides and top of the control volume will be small since |w|,|v| Uac at any reasonable distance from the aircraft. For low-Mach-number subsonic flight, such as during take-off and landing, the fluid density can be treated as constant. If the length of one side of the cubical control volume is L, then momentum conservation simplifies to :

    Uac wback w front( )dydz0

    L

    0

    L

    = g(airV + Mac ) + Pbottom Ptop( )dxdy0

    L

    0

    L

    . (a) In the absence of a difference between the front and back vertical velocity on the CV surface, the weight of the aircraft and air in the CV would be borne by the ground through the pressure

    x

    y

    z

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

    difference Pbottom Ptop. Here we can set Pbottom = Patm + Pextra, where Patm is the atmospheric pressure that supports the air in the CV and counter acts the pressure on the top of the CV, and Pextra is the extra pressure on the ground caused by the aircraft. In particular, evaluate (a) when the aircraft is absent to find a hydrostatic balance,

    0 = gairV + Patm Ptop( )dxdy0

    L

    0

    L

    ,

    and subtract this from (a) to find:

    gMac = Uac w front wback( )dydz0

    L

    0

    L

    + Pextradxdy0

    L

    0

    L

    . (b)

    Thus, the weight of the aircraft is borne by moving air (the first term in b) and an extra pressure on the ground (the second term in b). In reality, the pressure distribution on the aircraft's wings causes wfront wback to be both nonzero and positive (i.e. wfront is an updraft and wback is a downdraft), and this term primarily balances the impressive weight of the aircraft. For a landing speed of 100 m/s, L 100 m (~50% larger than the aircrafts wingspan), gMac = 4 mega-Newton, we can estimate the vertical velocity difference necessary to support the aircraft:

    w front wback 4 106N

    (1.2kg /m3)(100m /s)(104m2)~ 3.3m /s.

    Such wind speeds are not much of a hazard to most people, buildings, or other structures. Further consideration of the flow-field produced by a flying aircraft shows that heavy aircraft do not crush houses and people because the plane's weight ends up being spread over an extremely large area. A flying aircraft's pressure distribution may exist for a kilometer or more in all directions, so the supporting area for a plane with a weight of 4 mega-Newtons might easily exceed 1 km2 or 106 square meters. Therefore the extra pressure on the ground will be at most a few Pa (less than 0.01% of an atmosphere).

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

    Exercise 4.24. An inviscid incompressible liquid with density flows in a wide conduit of height H and width B into the page. The inlet stream travels at a uniform speed U and fills the conduit. The depth of the outlet stream is less than H. Air with negligible density fills the gap above the outlet stream. Gravity acts downward with acceleration g. Assume the flow is steady for the following items. a) Find a dimensionless scaling law for U in terms of , H, and g. b) Denote the outlet stream depth and speed by h and u, respectively, and write down a set of equations that will allow U, u, and h to be determined in terms of , H, and g. c) Solve for U, u, and h in terms of , H, and g. [Hint: solve for h first.] Solution 4.24. a) Density is the only parameter with units of mass. Thus, it cannot be part of a dimensionless law for U. The remaining parameters (U, g, H) can only be assembled into one dimensionless group,

    1 =U2 gH ,

    thus:

    U = const gH . b) Place the CV around the change in liquid level as shown above, ignore friction, set atmospheric pressure to zero, and assume uniform inflow and outflow. Cons. of mass:

    UH = uh (1) Cons. of horizontal momentum

    U 2H + u2h = P0H +12 gH

    2 12 gh2 (2)

    Bernoulli from 0 to 1:

    P0 +12 U

    2 + gH = 0 + 0 + gH (3) Bernoulli from 1 to 2:

    0 + 0 + gH = 0 + 12 u2 + gh (4)

    The zeros arise in the Bernoulli equations because point 1 is a stagnation point and the pressure there is atmospheric pressure. There are four unknowns (U, u, h, P0) so these four equations should be sufficient for a complete solution. c) Use (4) to find

    u = 2g(H h) . Use this to eliminate u from (1) and (2), and use (3) to eliminate P0 from (2). The remaining equations are:

    UH = h 2g(H h) , and

    U 2H + 2g(H h)h = 12 U2H + 12 g(H

    2 h2) . Divide each equation by . Square the first and divide by H, and simplify the second to get:

    U 2H = 2g(H h)h2 H , and

    12U2H + 2g(H h)h = + 12 g(H

    2 h2) . Use the first of these to eliminate

    U 2H in the second, and then divide by (Hh) to find:

    gh2 H + 2gh = + 12 g(H + h) , or

    h2 32Hh +12H

    2 = 0 = (h H)(h 12H). Thus,

    h = H 2 so (4) implies

    u = gH , and these together with (1) produce:

    U = gH 2.

    !

    !" #" !

    "#$!$"

    %! &! '!

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

    Exercise 4.25. A hydraulic jump is the shallow-water-wave equivalent of a gas-dynamic shock wave. A steady radial hydraulic jump can be observed safely in ones kitchen, bathroom, or backyard where a falling stream of water impacts a shallow pool of water on a flat surface. The radial location R of the jump will depend on gravity g, the depth of the water behind the jump H, the volume flow rate of the falling stream Q, and streams speed, U, where it impacts the plate. In your work, assume

    2gh

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

    Here the conservation of momentum equation specifies a balance between momentum fluxes and hydrostatic pressure terms. First eliminate V from the second equation using the first.

    U 2ho + UhoH

    $

    % &

    '

    ( ) 2

    H = g ho2

    2 g H

    2

    2

    Divide by , and manipulate this equation to eliminate a common factor of H ho.

    U 2 hoH(H ho) =

    g2H ho( ) H + ho( ) >

    U 2 hoH

    =g2H + ho( ) >

    ho 2U 2

    gH1

    #

    $ %

    &

    ' ( = H

    Now use the relationship

    ho = h(R) =Q (2RU) to eliminate ho and solve for R:

    Q2RU

    2U2

    gH1

    $

    % &

    '

    ( ) = H >

    R = Q2UH 2

    2U2

    gH

    $

    % &

    '

    ( )

    c) Putting this result into dimensionless form just takes a little algebra. The final result is:

    RH

    =12

    QgH 5

    2 UgH

    gHU

    $

    % &

    '

    ( )

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

    Exercise 4.26. A fine uniform mist of an inviscid incompressible fluid with density settles steadily at a total volume flow rate (per unit depth into the page) of 2q onto a flat horizontal surface of width 2s to form a liquid layer of thickness h(x) as shown. The geometry is two-dimensional. a) Formulate a dimensionless scaling law for h in terms of x, s, q, , and g. b) Use a suitable control volume analysis, assuming u(x) does not depend on y, to find a single cubic equation for h(x) in terms of h(0), s, q, x, and g. c) Determine h(0).

    Solution 4.26. a) There appear to be six parameters, but the density is the only parameter that includes the units of mass so it can be set aside. Thus, there will only be two independent dimensions, length (L) and time (T). Therefore, the units matrix is: h x s q g L 1 1 1 2 1 T 0 0 0 -1 -2 There will be 5 2 = 3 dimensionless groups. By inspection:

    1 = h s,

    2 = x s , and

    3 = q2 gs3 , so

    h s = fn x s,q2 gs3( ) . b) Conserve mass using the differential CV:

    u(x)h(x) + u(x + dx)h(x + dx) q s( )dx = 0, or

    d(uh) dx = q s. Conserve horizontal momentum in same differential CV with hydrostatic pressure forces:

    u2(x)h(x) + u2(x + dx)h(x + dx) = 12 gh2(x) 12 gh

    2(x + dx), or

    d(u2h) dx = g 2( )dh2 dx . Integrate these equations and evaluate the constants with u(0) = 0 to get

    uh = qx s, and

    u2h = g 2( ) h2 h2(0)( ). Eliminate u(x) between these two equations to find:

    q2x 2 hs2 = g 2( ) h2 h2(0)( ) , which is a cubic equation so h(x) cannot be put into a simple form. c) Global conservation of mass implies:

    q = u(s)h(s) . If Pa = atmospheric pressure, a Bernoulli streamline from x = 0 to x = s on the flat surface (the liquid surface is not a streamline) produces:

    Pa + gh(0) = Pa + gh(s) +12 u

    2(s), or

    u2(s) = 2g h(0) h(s)( ) = q2 h2(s) . This final equality and the result of part b) evaluated at x = s,

    q2 h(s) = g 2( ) h2(s) h2(0)( ) , represent two equations in two unknowns, h(0) & h(s), that can be solved simultaneously to determine h(0) in terms of q and g:

    h(0) = 3 q2 4g( )1 3

    = 3h(s).

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

    Exercise 4.27. A thin-walled pipe of mass mo, length L, and cross sectional area A is free to swing in the x-y plane from a frictionless pivot at point O. Water with density fills the pipe, flows into it at O perpendicular to the x-y plane, and is expelled at a right angle from the pipes end as shown. The pipes opening also has area A and gravity g acts downward. For a steady mass flow rate of

    m , the pipe hangs at angle with respect to the vertical as shown. Ignore fluid viscosity. a) Develop a dimensionless scaling law for in terms of mo, L, A, ,

    m , and g. b) Use a control volume analysis to determine the force components, Fx & Fy, applied to the pipe at the pivot point in terms of , mo, L, A, ,

    m , and g. c) Determine in terms of mo, L, A, ,

    m , and g. d) Above what value of

    m will the pipe rotate without stopping? Solution 4.27. a) The units matrix is: mo L A

    m g M 0 1 0 0 1 1 0 L 0 0 1 2 -3 0 1 T 0 0 0 -1 0 -1 -2 This matrix has rank three and there are seven parameters so four dimensionless groups should be constructed. By inspection the dimensionless groups are: ,

    m 2L gmo2 ,

    L2 A , and

    AL mo , so the dimensionless scaling law is:

    = f m 2L gmo2 ,L2 A,AL mo( ), where f is an undetermined

    function. b) Note that this part of the problem involves steady flow. Place the CV around the outside of the pipe except at the pivot point O where the CV slices through the pipe parallel to the x-y plane. Assume that the flow is in the z-direction across this control surface slice at O. Here, conservation of mass merely requires

    m = const. The control surface pressure is atmospheric everywhere outside the pipe, and pressure forces can only act in the z-direction on the control surface slice at O. Thus, there is no pressure contribution to the reaction forces, Fx and Fy, which arise from the portion of the control surface that passes through the pipe structure. The momentum flux of water at O is perpendicular to the x-y plane so only the outflow momentum flux is relevant for determining Fx and Fy. For the chosen stationary control volume, this outflow flux term can be readily evaluated because the waters discharge velocity and the CVs outward normal are parallel. Assume the outflow velocity is uniform too;

    u = m A( )n, so conservation of momentum implies:

    u(u n)dAoutlet =

    m An m

    A%

    & '

    (

    ) * A =

    m 2

    Aex cos + ey sin( ) = mo + AL( )gey + Fxex + Fyey ,

    where

    n = ex cos ey sin . Equating x & y components produces the final answers:

    Fx = m 2 A( )cos , and

    Fy = (mo + AL)g m 2 A( )sin .

    c) This part of the problem again involves STEADY flow. A small CV placed inside the pipe at its end shows that the force resulting from the flux of fluid discharged from the pipe primarily acts at end of the pipe. Based on this, equate the gravity- and flow-induced moments on the pipe

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

    about O so that Fx and Fy are not needed:

    (mo + AL)g L 2( )sin = m 2 A( )L . Solve for the angle:

    sin = 2 m 2 A( ) (mo + AL)g . d) When

    2 m 2 A( ) (mo + AL)g >1 there is no answer for since sin 1. Thus, when

    m > A(mo + AL)g 2 , no steady solution is possible, and the pipe will spin without stopping.

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

    Exercise 4.28. Construct a house of cards, or light a candle. Get the cardboard tube from the center of a roll of paper towels and back away from the cards or candle several feet so that by blowing you cannot knock down the cards or blow out the candle unaided. Now use the tube in two slightly different configurations. First, place the tube snugly against your face encircling your mouth, and try to blow down the house of cards or blow out the candle. Repeat the experiment while moving closer until the cards are knocked down or the candle is blown out (you may need to get closer to your target than might be expected; do not hyperventilate; do not start the cardboard tube on fire). Note the distance between the end of the tube and the card house or candle at this point. Rebuild the card house or relight the candle and repeat the experiment, except this time hold the tube a few centimeters away from your face and mouth, and blow through it. Again, determine the greatest distance from which you can knock down the cards or blow out the candle. a) Which configuration is more effective at knocking the cards down or blowing the candle out? b) Explain your findings with a suitable control-volume analysis. c) List some practical applications where this effect might be useful. Solution 4.28. a) The most effective configuration is when the tube is held a few centimeters in front of the face and mouth. b) The difference between the two cases is the greater volume flux through the tube when it is held a few centimeters in front of the mouth. Since the exit area of the tube does not change, the volume flux change results entirely from an increased exit velocity. Therefore, the control volume solution should determine the exit velocity from the tube in terms of other parameters. In the following, the subscript o will refer to conditions at the inlet end of the tube, the subscript e will refer to conditions at the exit or downstream end of the tube, and the subscript a will refer to ambient (motionless) conditions some distance from the tube. Consider a control volume that encloses the air inside the tube, ignore the frictional losses between the tube and the flowing air, and align the tube with the x-axis (see picture). For the air speeds and temperature differences under consideration the flow is essentially incompressible and isothermal which means, const.

    Let Af be the cross sectional area of the fast moving air blown out of the mouth, and Uf be the velocity of this air stream where it enters the tube. When = const, conservation of mass reduces to conservation of volume implying:

    AfU f + (Ao Af )Uo = AeUe (1)

    Tube Walls

    Control Volume Boundary

    UeUf

    Uo

    AfAo

    pope

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

    The horizontal-component of the momentum equation is then

    AfU f2 (Ao Af )Uo

    2 + AeUe2 = PodA

    Ao

    PedAAe

    (2)

    In both cases, Ao = Ae = A. Using these relationships and assuming that the inlet and exit pressure integrals can be approximated by the PoAo and PeAe, produces:

    AfU f + (A Af )Uo = AUe (3)

    (Po Pe )A + AfU f2 + (A Af )Uo

    2 = AUe2 (4)

    These are two equations in three unknowns: (PoPe), Uo, and Ue. The third equation is determined by the tube configuration. CASE I. When the tube is held against the face, the third relationship is Uo = 0, so (3) implies:

    Ue = Af A( )Uf = sU f , (5) where s = Af/A. CASE II. When the tube is held away from the mouth. The remaining relationship comes from a pressure balance outside the tube. The pressure of the air drawn into the tube from the ambient condition can be estimated from the steady Bernoulli equation:

    Pa +12 Ua

    2 = Pa = Po +12 Uo

    2 (6) where, of course, Ua 0. In addition, if the exit velocity is parallel to the tube axis, then the component of the momentum equation perpendicular to the tube axis suggests that Pa Pe so that

    Pe Po =12 Uo

    2 . (7) Put (7) into (4), then the equations to be solved for Ue are (3) and

    12Uo2A + AfU f

    2 + (A Af )Uo2 = AUe

    2 . (8) where the common factor of has been divided out. Rearrange (3) and (8) with s = Af/A to find:

    Uo =Ue sU f1 s

    , and

    ( 12 s)Uo2 =Ue

    2 sU f2 . (9a,b)

    Eliminate Uo and solve for Ue in terms of Uf and s (this involves some algebra).

    UeU f

    =s(2s1) + 2s(1 s)3

    (1 s)2 + s2 (10)

    Although it is not obvious by looking at it, this result, (10), is always larger than the result for Case I, (5), especially when s is small. For example, when s = 1/10: Case I > Ue = 0.100Uf, while Case II > Ue = 0.368Uf, almost 4 times larger! Or, when s = 1/4: Case I > Ue = 0.250Uf, while Case II > Ue = 0.535Uf, more than twice as large! c) For this problem, the fast-moving air from the mouth and the cardboard tube form a crude ejector pump. Applications for ejector pumps occur wherever rapid or steady pumping without moving parts is required for inflation or evacuation of a container or chamber. Examples: air bags, emergency exit ramps on commercial aircraft, emergency life support (ventilation of the lungs), and a variety of industrial processes involving corrosive materials. In addition, if the fast-moving stream carries a fuel and the entrained fluid carries the oxidizer (for example), the shape (i.e. the area ratio) of the ejector pump can be used to mix reactants to a set the mixture ratio for combustion.

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

    Exercise 4.29. Attach a drinking straw to a 15-cm-diameter cardboard disk with a hole at the center using tape or glue. Loosely fold the corners of a standard piece of paper upward so that the paper mildly cups the cardboard disk (see drawing). Place the cardboard disk in the central section of the folded paper. Attempt to lift the loosely folded paper off a flat surface by blowing or sucking air through the straw. a) Experimentally determine if blowing or suction is more effective in lifting the folded paper. b) Explain your findings with a control volume analysis.

    Solution 4.29. a) It would seem that the only way to lift the paper would be to suck on the straw. However, blowing is found to be just as successful! b) This can be explained with a control volume analysis. Choose a control volume that encloses a portion of the fluid between the paper and the cardboard disk. For simplicity assume that the flow is axisymmetric, constant density, and inviscid. Also assume that the paper and the cardboard are perfectly planar surfaces. Denote the distance between the paper and the cardboard disk by h, the diameter of the cardboard disk by R, and the radius & cross sectional area of the straw by rs & As.

    Conservation of mass implies: 2rhur = UiAs for r >> rs, where r is the radius of the control volume, ur is the local radial component of the flow at the circular edge of the control volume, and Ui is the inlet velocity from the straw. Therefore:

    ur =UiAs 2hr( ) for r >> rs. What we need to know is the pressure distribution above the piece of paper in terms of the parameters of the problem and p, the pressure below the piece of paper. The Bernoulli equation in this case is:

    p +12 uR

    2 = p(r) + 12 ur2 = pi +

    12 Ui

    2 where the 'i' subscript refers to conditions at the lower tip of the straw. For the paper to remain suspended we need the following to be true:

    2rsh

    ur

    r

    control volumeCardboard

    Paper

    R

    p

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

    Wp 2 p(r) + p[ ]rdrr= o

    r=R

    () where Wp is the weight of the paper. This pressure integral can be estimated from the vertical momentum equation for a control volume like that shown above with radius R:

    UiAs

    Uiez ez( )dA = Ui2As = 2 p(r)rdrr= rs

    r=R

    piAs + 2 p(r)rdrr= 0

    r=R

    .

    The first and second integrals on the right side come from the cardboard and paper surfaces, respectively. Rearrange this, use the Bernoulli equation, and then approximate the flow under the cardboard as purely radial for r > rs.

    2 p(r)rdrr= 0

    r=R

    = Ui2As + 2 p(r)rdrr= rs

    r=R

    + piAs

    2 p(r)rdrr= 0

    r=R

    = Ui2As + 2 p + 12 uR2 12 ur2( )rdrr= rs

    r=R

    + piAs.

    Now start reconstructing the integral of the pressure difference ():

    2 p p(r)[ ]rdrr= 0

    r=R

    = Ui2As 2 12 uR2 12 ur2( )rdrr= rs

    r=R

    piAs + pAs

    Inserting

    ur = uR R r( ) yields:

    2 p p(r)[ ]rdrr= 0

    r=R

    = Ui2As 12 uR2 R2 rs2( ) + uR2R2 lnR ln rs( ) piAs + pAs . Collect terms and introduce Wp:

    Wp p pi Ui2( )As + 12 uR2 rs2 + 2R2 ln R rs( ) R2( ).

    Simplifying further using

    As = rs2, the Bernoulli relationships, and

    uR =UiAs 2hR( ) =Uirs2 2hR( ) produces:

    Wp 12 Ui

    2As +12 uR

    2R2 2ln R rs( ) 1( ) , or,

    Wp 12Ui

    2As 1+rs2

    4h22ln R rs( ) 1( )

    %

    & '

    (

    ) * .

    For a straw diameter rs = 3 mm, h = 1 mm, and R = 80 mm, the factor inside the large parentheses is about +11.5. With these dimensions and Ui 20 m/sec, Wp can be as large as a 0.08 N. This strange lifting effect is used in manufacturing processes to pick up computer chips without touching them. Note also that Ui enters the final answer as

    Ui2 so it doesnt matter which

    ways the flow goes; suction or blowing through the straw should yield identical results.

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

    Exercise 4.30. A compression wave in a long gas-filled constant-area duct propagates to the left at speed U. To the left of the wave, the gas is quiescent with uniform density 1 and uniform pressure p1. To the right of the wave, the gas has uniform density 2 (> 1) and uniform pressure is p2 (> p1). Ignore the effects of viscosity in this problem. a) Formulate a dimensionless scaling law for U in terms of the pressures and densities. b) Determine U in terms of 1, 2, p1, and p2 using a control volume. c) Put your answer to part b) in dimensionless form and thereby determine the unknown function from part a). d) When the density and pressure changes are small, they are proportional:

    p2 p1 = c2 2 1( ) for 2 1( ) 2

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

    Exercise 4.31. A rectangular tank is placed on wheels and is given a constant horizontal acceleration a. Show that, at steady state, the angle made by the free surface with the horizontal is given by tan = a/g. Solution 4.31. Make a drawing and add some dimensions to the tank. Place the control volume around the water inside the rolling tank. Here the continuity equation merely ensures that the mass of water in the tank,

    M = V * dV , doesnt change as the tank

    accelerates. The horizontal component of the momentum equation is:

    ddt

    udVV *(t ) + u u at( )ex n

    S*(t ) dA = fxdA

    S*(t )

    where the CV velocity, b, has been set to b = atex, so that db/dt = aex is the specified steady horizontal acceleration. The body force term does not appear in the above equation since gravity acts in the vertical direction. At steady state, the fluid velocity will equal the velocity of the moving tank so the flux terms will be zero, and the pressure forces will be hydrostatic:

    p = g , where p is the pressure and g is the body force acceleration. In this situation, g has both horizontal and vertical components. However, the vertical component of the hydrostatic equation,

    p y = g, is not changed by the horizontal acceleration. This equation integrates to:

    p = A(x) gy where A(x) is the function of integration; it cannot depend on y. Here we know that atmospheric pressure occurs on the surface of the water so the pressure on the left and right sides of the tank are

    p1(y) = po + g(h1 y) and

    p2(y) = po + g(h2 y), respectively. Thus, the horizontal momentum conservation equation becomes:

    Ma = g(h1 y)Bdy0h1 g(h2 y)Bdy =

    gB20

    h2 h12 h22( ) , where B is the dimension of the tank into the page, and the CV-surface integration of po leads (as usual) to no net force. Now, use the two ends of this extended equality and divide by M noting that

    M = BL (h1 + h2) 2 where L is the horizontal length of the tank:

    a = gLh1 h2( ) = gtan , so

    tan = ag

    ! g!x!

    y!

    h2!h1! L!

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

    Exercise 4.32. Starting from rest at t = 0, an airliner of mass M accelerates at a constant rate

    a = aex into a headwind,

    u = uiex . For the following items, assume that: i) the x-component of the fluid velocity is ui on the front, sides, and back upper half of the control volume (CV), ii) the x-component of the fluid velocity is uo on the back lower half of the CV, iii) changes in M can be neglected, iv) changes of air momentum inside the CV can be neglected, and v) frictionless wheels. In addition, assume constant air density and uniform flow conditions exist on the various control surfaces. In your work, denote the CVs front and back area by A. (This approximate model is appropriate for real commercial airliners that have the engines hung under the wings). a) Find a dimensionless scaling law for uo at t = 0 in terms of ui, , a, M, and A. b) Using a CV that encloses the airliner (as shown) determine a formula for uo(t), the time-dependent air velocity on the lower half of the CVs back surface. c) Evaluate uo at t = 0, when M = 4 x 105 kg, a = 2.0 m/s2, ui = 5 m/s, = 1.2 kg/m3, and A = 1200 m2. Would you be able to walk comfortably behind the airliner?

    Solution 4.32. a) The units matrix is: uo ui a M A M 0 0 1 0 1 0 L 1 1 -3 1 0 2 T -1 -1 0 -2 0 0 This matrix has rank three and there are six parameters so three dimensionless groups should be constructed. These are found easily by inspection. The final dimensionless scaling law can be

    stated:

    uoui

    = fn aA1 2

    ui2 ,

    MA3 2

    #

    $ %

    &

    ' ( .

    b) For constant acceleration from rest, the CVs velocity is b = atex. Conserve mass:

    ddt

    M + Mair( ) + upper back +

    lower back +

    front +

    sides

    $ % &

    ' ( ) (ub) ndA = 0

    Drop the time derivative terms and evaluate the first three surface integrations:

    (ui at)ex (ex )A2

    + (uo at)ex (ex )A2

    + (ui at)ex (ex )A + (ub) ndSsides = 0

    Simplify to find:

    (ub) ndSsides = (ui uo)

    A2

    . (*)

    Note that ui must equal uo if the integral over the sides of the CV is not included. Now conserve horizontal momentum noting that there are no external forces on the CV.

    ddt

    Mat + Mairuair( ) + upper back +

    lower back +

    front +

    sides

    $ % &

    ' ( ) (u ex )(ub) ndS = 0

    Drop the air term from the time derivative, set dM/dt = 0, and evaluate the first three surface integra