Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (12)

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

    Exercise 12.1. Determine general relationships for the second, third, and four central moments (variance = 2, skewness = S, and kurtosis = K) of the random variable u in terms of its first four ordinary moments:

    u ,

    u2 ,

    u3 , and

    u4 . Solution 12.1. a) The variance = 2 is the second central moment, so:

    2 =1N

    u(n) u ( )2n=1

    N

    = 1Nu2(n) 2u(n)u + u 2( )

    n=1

    N

    = u2 2u u + u 2 = u2 u 2 .

    The skewness = S is the third central moment, so

    S = 1N

    u(n) u ( )3n=1

    N

    = 1Nu3(n) 3u2(n)u + 3u(n)u 2 u 3( )

    n=1

    N

    = u3 3u2u + 3u u 2 u 3 = u3 3u2u + 2u 3.

    The kurtosis = K is the fourth central moment, so

    K = 1N

    u(n) u ( )4n=1

    N

    = 1Nu4 (n) 4u3(n)u + 6u2(n)u 2 4u(n)u 3 + u 4( )

    n=1

    N

    = u4 4u3u + 6u2u 2 4u u 3 + u 4 = u4 4u3u + 6u2u 2 3u 4 .

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

    Exercise 12.2. Calculate the mean, mean square, variance, and rms value of the periodic time series

    u(t) = U + U0 cos t( ), where

    U , U0 and are positive real constants. Solution 12.2. The given time series is periodic so time averaging over one period will yield the desired results.

    Average:

    u = 12 ( )

    U + U0 cos(t)( )0

    2

    dt = 12 ( )U t + U0

    sin(t)

    %

    & ' (

    ) * 0

    2

    = U .

    Mean Square:

    u2 = 12 ( )

    U + U0 cos(t)( )2

    0

    2

    dt

    = 12 ( )

    U 2 + 2U0U cos(t) + U02 cos2(t)( )

    0

    2

    dt

    = 12 ( )

    U 2t + 2U0U

    sin(t) + U02

    2t + sin(2t)

    2%

    & '

    (

    ) *

    +

    , -

    .

    / 0

    0

    2

    = U 2 + U02

    2.

    Variance:

    u u ( )2 = 12 ( )

    U + U0 cos(t) U ( )2

    0

    2

    dt = U02 cos2(t)0

    2

    dt = U02

    2.

    rms value:

    u2 = U 2 + U02

    2.

    And for completeness,

    Standard deviation:

    u u ( )2 = U02

    The rms value and the standard deviation are not equal unless

    u = U = 0 .

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

    Exercise 12.3. Show that the autocorrelation function

    u(t)u(t + ) of a periodic series u = Ucos(t) is itself periodic. Solution 12.3. The given time series is periodic so time averaging over one period will yield the desired results.

    u(t)u(t + ) = U2

    2 ( )cos(t)cos (t + )( )

    0

    2

    dt

    = U2

    2 ( )cos(t) cos(t)cos() sin(t)sin()[ ]

    0

    2

    dt

    = U2

    2 ( )cos( ) cos2(t)

    0

    2

    dt U2

    2 ( )sin( ) cos(t)

    0

    2

    sin(t)dt

    = U2

    2 ( )cos( ) 1

    22

    '

    ( )

    *

    + , 0 =

    U 2

    2cos()

    And, since cos(t) is periodic, then

    u(t)u(t + ) is periodic.

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

    Exercise 12.4. Calculate the zero-lag cross-correlation

    u(t)v(t) between two periodic series u(t) = cos t and v(t) = cos(t + ) by performing at time average over one period = 2/. For values of = 0, /4, and /2, plot the scatter diagrams of u vs v at different times, as in Figure 12.8. Note that the plot is a straight line if = 0, an ellipse if = /4, and a circle if = /2; the straight line, as well as the axes of the ellipse, are inclined at 45 to the uv-axes. Argue that the straight line signifies a perfect correlation, the ellipse a partial correlation, and the circle a zero correlation. Solution 12.4. The given time series is periodic so time averaging over one period will yield the desired results.

    u(t)v(t) = U2

    2 ( )cos(t)cos t + ( )

    0

    2

    dt

    = U2

    2 ( )cos(t) cos(t)cos sin(t)sin[ ]

    0

    2

    dt

    = U2

    2 ( )cos cos2(t)

    0

    2

    dt U2

    2 ( )sin cos(t)

    0

    2

    sin(t)dt

    = U2

    2 ( )cos 1

    22

    '

    ( )

    *

    + , 0 =

    U 2

    2cos

    The scatter diagrams are obtained by placing sample points from different times in a two-dimensional (u,v)-coordinate plane. The locus of sample points is obtained by eliminating t from using the equations for u and v:

    v = cos(t + ) = cost cos sint sin = ucos 1 u2 sin . Use the two ends of this extended equality to find:

    v ucos = 1 u2 sin , or

    v 2 2uv cos + u2 cos2 = (1 u2)sin2 . Further simplify:

    v 2 2uv cos + u2 = sin2 . This quadratic relationship can be cast in a standard form by switching to sum, v + u, and difference, v u, coordinates:

    1 cos2

    $

    % &

    '

    ( ) (v + u)2 +

    1+ cos2

    $

    % &

    '

    ( ) (v u)2 = sin2 =1 cos2 ,

    which implies:

    (v + u)2

    2(1+ cos)+

    (v u)2

    2(1 cos)=1 when 0 or ,

    v = u when = 0, and v = u when = . The first possibility is the equation for an ellipse having major and minor axes rotated 45 from the u and v axes. The other two possibilities are just straight lines. When = 0 then v = u so the resulting distribution of sample points is a straight line with unity slope.

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

    In this case, each value of u is linked to a single and equal value of v. This is perfect correlation between u and v.

    When = /4, then the locus of possible points becomes

    (v + u)2

    2 + 2+(v u)2

    2 2=1, which is:

    In this case, each value of u is linked to a two values of v. Here positive v is more likely with positive u, and negative v is more likely with negative u. Thus, this situation corresponds to partial correlation between u and v. When = /2, then the locus of possible points becomes, which is a circle:

    In this case, each value of u is linked to a two values of v. Here positive v is equally likely with positive or negative u, and negative v is equally likely with positive or negative u. Thus, this situation corresponds to no correlation between u and v.

    u

    v

    u

    v

    u

    v

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

    Exercise 12.5. If u(t) is a stationary random signal, show that u(t) and

    du(t) dt are uncorrelated. Solution 12.5. a) For a stationary signal u(t), use the time-average definition of the correlation and evaluate the integral:

    limt

    1t

    u(t) dudtdt

    t 2

    +t 2

    = limT

    12t

    du2

    dtdt =

    t 2

    +t 2

    limt

    12t

    u2(t /2) u2(t /2)[ ] = 0 . The final equality occurs because u2 is stationary and remains finite at while the divisor of the [,]-brackets goes to infinity.

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

    Exercise 12.6. Let R() and S() be a Fourier transform pair. Show that S() is real and symmetric if R() is real and symmetric. Solution 12.6. Start with:

    S() = 12

    eiR( )d

    +

    ,

    and decompose into real and imaginary parts:

    S() = 12

    cos isin( )R( )d

    +

    .

    Since sin is odd and R() is even in , the integral over the imaginary part is zero. Thus,

    S() = 12

    cos()R()d =

    +

    12cos( )R( )

    +

    d = S(),

    which clearly shows that: (1) S() is real because it can be computed from the integral of the two real functions R() and cos(), and (2) S() is symmetric about = 0 because S() = S().

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

    Exercise 12.7. Compute the power spectrum, integral time scale, and Taylor time scale when

    R11() = u12 exp 2( )cos(o ) , assuming that and o are real positive constants.

    Solution 12.7.

    Se () =u12

    2e

    2

    cos(o)exp i{ }

    +

    d = u12

    4e

    2

    ei(o ) + ei(o )( )d

    +

    The exponents of the two terms in the integrand are:

    2 i( o) = 2 +

    i( o)

    +( o)

    2

    4 2

    ( o)2

    4 2&

    ' (

    )

    * +

    = + i( o)2

    &

    ' (

    )

    * +

    2

    ( o)

    2

    4,

    where the top sign belongs with the first term. Let

    = +i( o)2

    &

    ' (

    )

    * + be the new integration

    variable:

    Se () =u12

    4 exp ( o)

    2

    4& ' (

    ) * +

    e2

    d

    +

    + u12

    4 exp ( +o)

    2

    4& ' (

    ) * +

    e2

    d

    +

    .

    The integral in both terms is

    , so the energy spectrum is:

    Se () =u12

    4 exp ( o)

    2

    4& ' (

    ) * +

    + exp ( +o)2

    4& ' (

    ) * +

    ,

    - .

    /

    0 1 .