# 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 .