# 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,

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