fluid mechanics kundu cohen 6th edition solutions sm ch (3)

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

Exercise 3.1. The gradient operator in Cartesian coordinates (x, y, z) is:

= ex x( ) + ey y( ) + ez z( ) where

ex ,

ey , and

ez are the unit vectors. In cylindrical polar coordinates (R, , z) having the same origin, (see Figure 3.3b), coordinates and unit vectors are related by:

R = x 2 + y 2 ,

= tan1 y x( ) , and z = z; and

eR = ex cos + ey sin ,

e = ex sin + ey cos , and

ez = ez . Determine the following in the cylindrical polar coordinate system. a)

eR and

e b) the gradient operator c) the divergence of the velocity field u d) the Laplacian operator

2 e) the advective acceleration term (u)u Solution 3.1. The Cartesian unit vectors do not depend on the coordinates so the unit vectors from the cylindrical coordinate system can be differentiated when they are written in terms of ex, ey, and ez. a) First work with eR, use the given unit vector definition, and proceed with straightforward differentiation. The variables R and z do not appear in the formula for eR, so

eR R = eR z = 0 . However eR does depend on the angle . Thus,

eR

=

ex cos + ey sin( ) = ex sin + ey cos = e . Proceed to determine the derivatives of e. Again note that the variables R and z do not appear in the formula for e, so

e R = e z = 0. However, like eR, e does depend on the angle . Thus,

e

=

ex sin + ey cos( ) = ex cos ey sin = eR . The third unit vector, ez, is the same as the Cartesian unit vector and does not depend on the coordinates. b) Start by constructing the expressions for ex, ey, and ez in terms of eR, e, and ez. This can be

done my inverting the linear system

cos sin 0sin cos 00 0 1

$

% &

' &

(

) &

* &

exeyez

$

% &

' &

(

) &

* &

=

eReez

$

% &

' &

(

) &

* &

to find

ex = eR cos e sin ,

ey = eR sin + e cos , and

ez = ez (1,2,3) The next step is to use the coordinate definitions:

R = x 2 + y 2 ,

= tan1 y x( ) , and z = z (4,5,6) to transform the Cartesian partial derivatives.

x

=Rx

R

+x

+zx

z

=x

x 2 + y 2R

1

1+ y x( )2yx 2

%

& '

(

) *

+ (0) z

= cos R

sinR


y

=Ry

R

+y

+zy

z

=y

x 2 + y 2R

+1

1+ y x( )21x

$

% &

'

( )

+ (0) z

= sin R

+cosR

z

=Rz

R

+z

+zz

z

= (0) R

+ (0)

+ (1) z

=z

Now reassemble the gradient operator

= exx

+ eyy

+ ezz

using the cylindrical coordinate

unit vectors and differentiation definitions:

= eR cos e sin( ) cosR

sinR

&

' (

)

* + + eR sin + e cos( ) sin

R

+cosR

&

' (

)

* + + ez

z

.

Collect all of the terms with like unit vectors and differential operators together:

= eR cos2 + sin2( )

R+ eR cos sin + sin cos( )

1R

e sin cos + cos sin( )R

+ e sin2 + cos2( ) 1R

+ ezz

The terms in (,)-parentheses are either +1 or 0. When evaluated they produce:

= eRR

+ e1R

+ ezz

.

c) In cylindrical coordinates, the divergence of the velocity is:

u = eRR

+ e1R

+ ezz

&

' (

)

* + eRuR + eu + ezuz( ).

Further simplification requires that both the unit vectors and the u's be differentiated. Completing this task term by term produces:

eRR

eRuR( ) = eR eRuRr

+ uReR eRR

=uRR

,

eRR

eu( ) = eR euR

+ ueR eR

= 0 ,

eRR

ezuz( ) = eR ezuzR

+ uzeR ezr

= 0 ,

eR

eRuR( ) =e eRR

uR

+uRRe

eR

= 0 + uRRe e =

uRR

,

eR

eu( ) =e eR

u

+uRe

e

=1Ru

uRe eR =

1Ru

,

eR

ezuz( ) =e ezR

u

+uzRe

ez

= 0

ezz

eRuR( ) = ez eRuRz

+ uRez eRz

= 0,

ezz

eu( ) = ez euz

+ uez ez

= 0 , and

ezz

ezuz( ) = ez ezuzz

+ uzez ezz

=uzz

.


Reassembling the equation produces:

u = uRR

+uRR

+1Ru

+uzz

.

The 1st & 2nd terms on the right side are commonly combined to yield:

u = 1RR

RuR( ) +1Ru

+uzz

. (10)

d) The Laplacian operator is

2 , and its form in cylindrical coordinates can be found by evaluating the dot product. Fortunately, the results of part c) can be used via the following replacements for the second gradient operator of the dot product:

uR R

,

u 1R

, and

uzz

. (7,8,9)

Inserting the replacements (7,8,9) into (10) produces:

2 =1R

R

R R

$

% &

'

( ) +

1R2

2

2+

2

z2.

e) Start with the answer to part b) and compute the first dot product to find:

u = uRR

+ u1R

+ uzz

This is the scalar operator applied to u = uReR +ue +uzez to find the advective acceleration:

u ( )u = uRR

+ u1R

+ uzz

&

' (

)

* + uReR + ue + uzez( ) .

Here the components of u and the unit vectors eR and e depend on the angular coordinate.

u ( )u = eR uRuRR

+ u1RuR

+ uzuRz

&

' (

)

* + + uRu

1ReR

+

e uRuR

+u1Ru

+uzuz

!

"#

$

%&+u

2 1Re

+ ez uRuzR

+u1Ruz

+uzuzz

!

"#

$

%&

Use the results of part a) to evaluate the unit vector derivatives.

u ( )u = eR uRuRR

+ u1RuR

+ uzuRz

&

' (

)

* + + uRu

eR

+

e uRuR

+ u1Ru

+ uzuz

$

% &

'

( ) u

2 eRR

+ ez uRuzR

+ u1Ruz

+ uzuzz

$

% &

'

( )

Collect components

u ( )u = eR uRuRR

+ u1RuR

+ uzuRz

u2

R

'

( )

*

+ , +

e uRuR

+ u1Ru

+ uzuz

+uRuR

$

% &

'

( ) + ez uR

uzR

+ u1Ruz

+ uzuzz

$

% &

'

( )


Exercise 3.2. Consider Cartesian coordinates (as given in Exercise no. 1) and spherical polar coordinates (r, , ) having the same origin (see Figure 3.3c). Here coordinates and unit vectors are related by:

r = x 2 + y 2 + z2 ,

= tan1 x 2 + y 2 z( ) , and

= tan1 y x( ) ; and

er = ex cos sin + ey sin sin + ez cos ,

e = ex cos cos + ey sin cos ez sin , and

e = ex sin + ey cos . In the spherical polar coordinate system, determine the following items. a)

er ,

er ,

e ,

e , and

e b) the gradient operator c) the divergence of the velocity field u d) the Laplacian

2 e) the advective acceleration term (u)u Solution 3.2. The Cartesian unit vectors do not depend on the coordinates so the unit vectors from the spherical coordinate system can be differentiated when they are written in terms of ex, ey, and ez. a) First work with er, use the given unit vector definition, and proceed with straightforward differentiation. The variable r doesnt even appear in the formula for er, so

er r = 0. However er does depend on both angles. Thus,

er

=

ex cos sin + ey sin sin + ez cos( ) = ex cos cos + ey sin cos ez sin = e and,

er

=

ex cos sin + ey sin sin + ez cos( ) = ex sin sin + ey cos sin = e sin . Proceed to determine the derivatives of e. Again note that the variable r doesnt appear in its formula, so

e r = 0 . However, like er, e does depend on both angles. Thus,

e

=

ex cos cos + ey sin cos ez sin( ) = ex cos sin ey sin sin ez cos = er and,

e

=

ex cos cos + ey sin cos ez sin( ) = ex sin cos + ey cos cos = e cos . Now consider e and note that the variables r and dont appear in its formula, so

e r = e = 0 . However, e does depend on . Thus,

e

=

ex sin + ey cos( ) = ex cos ey sin . ($) The question now is how to relate the right side of this equation back to er, and e [note: because

e e ( ) = 0 ,

e can only be a linear combination of er and e]. Assuming

e = aer + be , then ($) and the unit vector definitions require:

acos sin + bcos cos = cos ,

asin sin + bsin cos = sin , and

acos bsin = 0 . After dividing out common factors, the first two equations are the same:

asin + bcos = 1. When this simplified equation is combined with the third equation, we obtain:

a = sin and

b = cos . Thus,

e = er sin e cos .


b) Start by constructing the expressions for ex, ey, and ez in terms of

er,

e , and

e . This can be

done my inverting the linear system

cos sin sin sin coscos cos sin cos sinsin cos 0

%

& '

( '

)

* '

+ '

exeyez

%

& '

( '

)

* '

+ '

=

eree

%

& '

( '

)

* '

+ '

to find

ex = er cos sin + e cos cos e sin

ey = er sin sin + e sin cos + e cos (1,2,3)

ez = er cos e sin . The next step is to use the coordinate definitions:

r = x 2 + y 2 + z2 ,

= tan1 x 2 + y 2 z( ) , and

= tan1 y x( ) (4,5,6) to transform the Cartesian partial derivatives.

x

=rx

r

+x

+x

=xrr

+1

1+ x 2 + y 2 z( )2

2x2z x 2 + y 2

y

x 2 + y 2

= cos sin r

+cos cos

r

sinrsin

y

=ry

r

+y

+y

=yrr

+1

1+ x 2 + y 2 z( )2

2y2z x 2 + y 2

+x

x 2 + y 2

= sin sin r

+sin cos

r

+cosrsin

z

=rz

r

+z

+z

=zrr

+1

1+ x 2 + y 2 z( )2 x 2 + y 2

z2

= cos rsinr

Now reassemble the gradient operator

= exx

+ eyy

+ ezz

using the spherical coordinate unit

vectors and differentiation definitions:

= er cos sin + e cos cos e sin( ) cos sinr

+cos cos

r

sinrsin

'

( )

*

+ ,

+ er sin sin + e sin cos + e cos( ) sin sinr

+sin cos

r

+cosrsin

%

& '

(

) *

+ er cos e sin( ) cosr

sinr

%

& '

(

) * .

Collect all of the terms with like unit vectors and differential operators together:

= er cos2 sin2 + sin2 sin2 + cos2( )

r

+er cos2 cos sin + sin2 cos sin cos sin( )1r

+er cos sin sin + sin cos sin( )1

rsin

+e cos2 cos sin + sin2 cos sin sin cos( )

r


+e cos2 cos2 + sin2 cos2 + sin2( )1r

+e cos sin cos + sin cos cos( )1

rsin

+e sin cos sin + cos sin sin( )r

+e sin cos cos + cos sin cos( )1r

+e sin2 + cos2( ) 1rsin

The terms in (,)-parentheses are either +1 or 0. When evaluated they produce:

= err

+ e1r

+ e1

rsin

.

c) In spherical coordinates, the divergence of the velocity is:

u = err

+ e1r

+ e1

rsin

'

( )

*

+ , erur + e u + eu( ).

Further simplification requires that the unit vectors and the u's be differentiated. Completing this task term by term produces:

err

erur( ) = er erurr

+ urer err

=urr

,

err

e u( ) = er eur

+ uer er

= 0 ,

err

eu( ) = er eur

+ uer er

= 0 ,

er

erur( ) =e err

ur

+urre

er

= 0 + urre e =

urr

,

er

e u( ) =e er

u

+ure

e

=1ru

ure er =

1ru

,

er

eu( ) =e er

u

+ure

e

= 0

ersin

erur( ) =e errsin

ur

+ur

rsine

er

=ur

rsine e sin( ) =

urr

,

ersin

e u( ) =e ersin

u

+u

rsine

e

=u

rsine e cos =

ur tan

,

ersin

eu( ) =1

rsine e

u

+ ue e

&

' (

)

* +

=1

rsinu

ue er sin + e cos( )'

( )

*

+ , =

1rsin

u

Reassembling the equation produces:

u = urr

+2urr

+1ru

+u

r tan+

1rsin

u

The 1st & 2nd terms, and the 3trd and 4th terms on the right side are commonly combined to yield:


u = 1r2

r

r2ur( ) + 1rsin

sin u( ) +1

rsinu

. (10)

d) The Laplacian operator is

2 , and its form in spherical polar coordinates can be found by evaluating the dot product. Fortunately, the results of part c) can be used via the following replacements for the second gradient operator of the dot product:

urr

,

u 1r

, and

u 1

rsin

. (7,8,9)

Inserting (7,8,9) into (10), the Laplacian then becomes:

2 =1r2

r

r2 r

$

% &

'

( ) +

1r2 sin

sin

$

% &

'

( ) +

1r2 sin2

2

2.

e) Start with the answer to part b) and compute the first dot product to find:

u = urr

+ u1r

+ u1

rsin

This is the scalar operator applied to

u = urer + ue + ue to find the advective acceleration:

u ( )u = urr

+ u1r

+ u1

rsin

'

( )

*

+ , urer + ue + ue( ) .

Here the components of u and the unit vectors depend on the angular coordinates.

u ( )u = er ururr

+ u1rur

+ u1

rsinur

'

( )

*

+ , + uru

1rer

+ uru1

rsiner

+

e urur

+ u1ru

+ u1

rsinu

%

& '

(

) * + u

2 1re

+ u u1

rsine

+

e urur

+ u1ru

+ u1

rsinu

%

& '

(

) * + uu

1re

+ u2 1rsin

e

Use the results of part a) to evaluate the unit vector derivatives.

u ( )u = er ururr

+ u1rur

+ u1

rsinur

'

( )

*

+ , + uru

er

+ urue sinrsin

+

e urur

+ u1ru

+ u1

rsinu

%

& '

(

) * u

2 err

+ u ue cosrsin

e urur

+ u1ru

+ u1

rsinu

%

& '

(

) * + 0 + u

2 1rsin

er sin e cos( )

Collect components

u ( )u = er ururr

+ u1rur

+ u1

rsinur

u2 + u

2

r

(

) *

+

, - +

e urur

+ u1ru

+ u1

rsinu

+urur

u2

rcot

&

' (

)

* +

e urur

+ u1ru

+ u1

rsinu

+urur

+u urcot

%

& '

(

) *


Exercise 3.3. In a steady two-dimensional flow, Cartesian-component particle trajectories are given by: x(t) = ro cos (t to )+o( ) and y(t) = ro sin (t to )+o( ) where ro = xo2 + yo2 and o = tan

1 yo xo( ) . a) From these trajectories determine the Lagrangian particle velocity components u(t) = dx/dt and v(t) = dy/dt, and convert these to Eulerian velocity components u(x,y) and v(x,y). b) Compute Cartesian particle acceleration components, ax = d2x/dt2 and ay = d2y/dt2, and show that they are equal to D/Dt of the Eulerian velocity components u(x,y) and v(x,y). Solution 3.3. a) Differentiate as suggested to find:

u(t) = dx(t)dt

= ro sin (t to )+o( ) and v(t) =dy(t)dt

= ro cos (t to )+o( ) .

Now use the original trajectory equations to eliminate the trig-functions: u = y and v = x .

b) Again differentiate as suggested to find:

ax (t) =d 2x(t)dt2

= 2ro cos (t to )+o( ) = 2x and

ay (t) =d 2y(t)dt2

= 2ro sin (t to )+o( ) = 2y .

Compute Du/Dt and Dv/Dt from the final two answers of part a): DuDt

=ut+uu

x+ v u

y= 0 y(0)+ x( ) = 2x and

DvDt

=vt+u v

x+ v v

y= 0 y( )+ x(0) = 2y .

The final equalities match as appropriate: ax = Du/Dt, and ay = Dv/Dt.


Exercise 3.4. In a steady two-dimensional flow, polar coordinate particle trajectories are given by: r(t) = ro and (t) = (t to )+o . a) From these trajectories determine the Lagrangian particle velocity components ur(t) = dr/dt and, u (t) = rd/dt, and convert these to Eulerian velocity components ur(r,) and u (r, ) .

b) Compute polar-coordinate particle acceleration components, ar = d2r dt2 r d dt( )

2 and a = rd

2 dt2 + 2 dr dt( ) d dt( ) , and show that they are equal to D/Dt of the Eulerian velocity with components ur(r,) and u (r, ) . Solution 3.4. a) Differentiate as suggested to find:

ur (t) =dr(t)dt

= 0 and u (t) = rd(t)dt

= r .

These equations are readily interpreted as Eulerian velocity components: ur = 0 and u = r .

b) Start with the particle accelerations: ar = d

2r dt2 r d dt( )2= 0 r( )2 = 2r and a = rd

2 dt2 + 2 dr dt( ) d dt( ) = 0+ 2(0) = 0 . Compute Dur/Dt and Du/Dt from the forms given in Appendix B and the two answers of part a):

DurDt

=urt

+ururr

+urur

u2

r= 0+ 0 (r)

2

r= 2r and

DuDt

=ut

+urur

+uru

+urur

= 0+ 0( )+ (0)+ 0 = 0 .

The final equalities match as appropriate: ar = Dur/Dt, and a = Du/Dt.


Exercise 3.5. if ds = (dx, dy, dz) is an element of arc length along a streamline (Figure 3.5) and u = (u, v, w) is the local fluid velocity vector, show that if ds is everywhere tangent to u then

dx u = dy v = dz w . Solution 3.5. If ds = (dx, dy, dz) and u are parallel, then they must have the same unit tangent vector t:

t = dsds

=(dx,dy,dz)

(dx)2 + (dy)2 + (dz)2=

(u,v,w)u2 + v 2 + w2

=uu

.

The three components of this equation imply:

dxds

=uu

,

dyds

=vu

, and

dzds

=wu

.

But these can be rearranged to find:

dsu

=dxu

=dyv

=dzw

.


Exercise 3.6. For the two-dimensional steady flow having velocity components u = Sy and v = Sx, determine the following when S is a positive real constant having units of 1/time. a) equations for the streamlines with a sketch of the flow pattern b) the components of the strain rate tensor c) the components of the rotation tensor d) the coordinate rotation that diagonalizes the strain rate tensor, and the principal strain rates. e) How is this flow field related to that in Example 3.5. Solution 3.6. a) For steady streamlines in two dimensions:

dxu

=dyv

or

dydx

=vu

=SxSy

=xy

, which implies:

ydy = xdx y 2 2 = x 2 2 + const. Solving for y(x) produces:

y = x 2 + const . These are hyperbolae that asymptote to the lines y = x. b) Compute the strain rate tensor from its definition:

Sij =12

uix j

+u jxi

#

$ % %

&

' ( (

=u x 12 u y + v x( )

12 v x + u y( ) v y

# $ %

& ' (

=0 12 S + S( )

12 S + S( ) 0

" # $

% & '

=0 SS 0

" # $

% & '

c) Compute the rotation tensor from its definition:

Rij =uix j

u jxi

=0 u y v x

v x u y 0$ % &

' ( )

=0 S S

S S 0$ % &

' ( )

=0 00 0$ % &

' ( )

d) From Example 2.4, a = 45 coordinate rotation diagonalizes the strain rate tensor. The

direction cosine matrix is:

Cij =cos sinsin cos

$ % &

' ( )

=121 11 1$ % &

' ( )

, and the rotated strain rate matrix

S is:

" S = CT S C = 12

1 11 1

% & '

( ) *

0 SS 0

% & '

( ) *

121 11 1% & '

( ) *

=12S SS S

% & '

( ) *

1 11 1% & '

( ) *

=S 00 S

% & '

( ) *

.

e) When this flow field is rotated 45 in the clockwise direction, it is the same as the flow field in Example 3.5.

x

y


Exercise 3.7. At the instant shown in Figure 3.2b, the (u,v)-velocity field in Cartesian coordinates is

u = A(y 2 x 2) (x 2 + y 2)2 , and

v = 2A xy (x 2 + y 2)2 where A is a positive constant. Determine the equations for the streamlines by rearranging the first equality in (3.7) to read

udy vdx = 0 = y( )dy + x( )dx and then looking for a solution in the form (x,y) = const. Solution 3.7. Rearrange the two-dimensional streamline condition, dx/u = dy/v, to obtain udy vdx = 0 as the description of a streamline. Assume this differential equation is solved by the function (x,y) = const, so that (/x)dx + (/y)dy = 0. Comparing the two equations requires:

u = /y , and v = /x. Now use the given velocity field to find:

y = A(y 2 x 2) (x 2 + y 2)2 , and

x = +2A xy (x 2 + y 2)2 . (a,b) Integrate (b) treating y as a constant:

x

= Ay 2x(x 2 + y 2)2

o = Ay2xdx

(x 2 + y 2)2= Ay 1

x 2 + y 2'

( )

*

+ , ,

where o may depend on y. Differentiate this result with respect to y to determine o:

y

o( ) =y

Ayx 2 + y 2

%

& '

(

) * =

Ax 2 + y 2

(Ay)

(x 2 + y 2)2(2y) = A(y

2 x 2)(x 2 + y 2)2

= u.

This result and equation (a) implies o/y = 0, so it is at most a constant. Thus, the streamlines are given by:

(x,y) = const.= Ayx 2 + y 2

.


Exercise 3.8. Determine the equivalent of the first equality in (3.7) for two dimensional (r,)-polar coordinates, and then find the equation for the streamline that passes through (ro, o) when u = (ur, u) = (A/r, B/r) where A and B are constants. Solution 3.8. The two-dimensional streamline condition in Cartesian coordinates is dx/u = dy/v, and is obtained from considering the streamline-tangent vector t:

t = dsds

=exdx + eydy

(dx)2 + (dy)2=exu + eyv

u2 + v 2=uu

.

In two-dimensional polar coordinates this becomes:

t = dsds

=erdr + e rd(dr)2 + (rd)2

=erur + e uur2 + u

2=uu

.

Equating components produces two equations:

drds

=uru

and

rdds

=uu

, or

dsu

=drur

=rdu

.

Thus, using the last equality and the given velocity field:

1rdrd

=uru

=A rB r

=AB

ln(r) = AB + const.

The initial condition allows the constant to be evaluated:

ln(ro) =ABo + const., which leads to

ln rro

"

# $

%

& ' =

AB

o( ) or

r = ro expAB

o( )$

% &

'

( ) .


Exercise 3.9. Determine the streamline, path line, and streak line that pass through the origin of coordinates at t = t when u = Uo + ocos(t) and v = osin(t) in two-dimensional Cartesian coordinates where Uo is a constant horizontal velocity. Compare your results those in Example. 3.3 for

Uo 0. Solution 3.9. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find:

dydx

=vu

=o sin(t)

Uo +o cos(t)= m(t) ,

where m is the streamline slope. Since m does not depend on the spatial coordinate, this equation is readily integrated to find straight time-dependent streamlines: y = m(t)x + const. Thus, the streamline that passes through (0,0) at t = t is:

y = o sin( $ t )Uo +o cos( $ t )

x .

(ii) For the path line, use both components of (3.8):

dxdt

=Uo +o cos(t) and

dydt

=o sin(t) ,

and integrate in time to find:

x xo =Uot + o sin(t) and

y yo = o cos(t) . Determine xo and yo by requiring the path line to pass through the origin at at t = t:

0 xo = Uo # t + o sin( # t ) and

0 yo = o cos( % t ) . The final component equations are:

x = Uo(t # t ) + o sin(t) sin( # t )( ) and

y = o cos(t) cos( % t )( ) . These two parametric equations for x(t) and y(t) can be combined to eliminate some of the t-dependence:

x Uo(t # t ) + o sin( # t )( )2

+ y o cos( # t )( )2

= o2 ,

which describes a moving circle with center located at

Uo(t # t ) o sin( # t ),o cos( # t )( ) . (iii) For the streak line, use the path line results but this time evaluate the constants at t = to instead of at t = t to find:

x =Uo(t to) + o sin(t) sin(to)( ) and

y = o cos(t) cos(to)( ). Now evaluate these equations at t = t to produce two parametric equations for the streak line coordinates x(to) and y(to):

x = Uo( " t to) + o sin( " t ) sin(to)( ) and

y = o cos( $ t ) cos(to)( ) . Some of the to dependence can be eliminated by combining the equations:

x Uo( # t to) o sin( # t )( )2

+ y + o cos( # t )( )2

= o2,

which describes a circle with a to-dependent center located at

Uo( " t to) + o sin( " t ),o cos( " t )( ). These results differ from those in Example 3.3 by the uniform translation velocity Uo so they can be put into correspondence with a Galilean transformation x = x Uo(t t).


Exercise 3.10. Compute and compare the streamline, path line, and streak line that pass through (1,1,0) at t = 0 for the following Cartesian velocity field u = (x, yt, 0). Solution 3.10. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find:

dydx

=vu

= ytx

dyy

= t dxx

ln y = t ln x + const., or y = const.xt.

Evaluating at x = y = 1 and t = 0 requires the constant to be unity, so the streamline is: y = 1. (ii) For the path line, use both components of (3.8):

dxdt

= x and

dydt

= yt ,

and integrate these in time to find:

x = C1et and

y = C2 exp t2 2{ },

where C1 and C2 are constants. Evaluating at x = y = 1 and t = 0 requires C1 = C2 = 1. Eliminate t from the y-equation using t = ln(x) to find the path line as:

y = exp (ln x)2 2{ } . (iii) For the streak line, use the path line results but this time evaluate the constants at t = to instead of at t = 0 to find:

x = exp t to{ } and

y = exp (to2 t 2) 2{ } .

Evaluate at t = 0, and eliminate to from the resulting equations to find:

y = exp +(ln x)2 2{ }.


Exercise 3.11. Consider a time-dependent flow field in two-dimensional Cartesian coordinates where

u = t 2 ,

v = xy , and

and are constant length and length time scales, respectively. a) Use dimensional analysis to determine the functional form of the streamline through x at time t. b) Find the equation for the streamline through x at time t and put your answer in dimensionless form. c) Repeat b) for the path line through x at time t. d) Repeat b) for the streak line through x at time t. Solution 3.11. a) The streamline y(x) will depend on x, t, t, x= (x,y),

, and . There are eight

parameters and two dimensions, thus there are six dimensionless groups:

y

= x, # x , # y , t, # t

%

& '

(

) * .

Here there are too many variables and parameters for dimensional analysis to be really useful. However, this effort provides a reminder to check units throughout the remainder of the solution. i) For the streamline, time is a constant. Use the first equality of (3.7) to find:

dydx

=vu

=xy t 2

dyy

=t 2

2xdx2

ln y = t2

2x 2

22+ const.

The initial condition requires, x = x and y = y at t = t, and this allows the constant to be determined, yielding:

ln y" y

#

$ %

&

' ( =(t 2x 2 " t 2 " x 2)

22 2.


dxdt

=t 2

and

dydt

=xy

,

and integrate the first of these in time and use the initial condition to find:

x = t + const. or

x # x = t1 # t 1( ) . Use this result for x(t) in the second equation for y(t):

dydt

=y

1t1$ t

%

& '

(

) * + $ x

%

& '

(

) * or

dyy

=1" t 1t

+1

" x %

& '

(

) * dt .

The last expression can be integrated to find:

ln y" y

#

$ %

&

' ( =

t" t 1+ " x

ln t

" t #

$ %

&

' ( =

" x

+1" t

#

$ %

&

' ( (t " t ) ln

t" t

#

$ %

&

' ( or

y" y =

t" t

#

$ %

&

' ( 1

exp " x " t

+1#

$ %

&

' (

t" t 1

#

$ %

&

' (

+

, -

.

/ 0 .

Now use the final equations for x(t) and y(t) to eliminate t. The equation for x(t) can be rearranged to find:

" t t

=1 (x " x ) " t

,

so the equation for y becomes:

y" y = 1 (x " x ) " t

%

& '

(

) * exp

" x " t

+1%

& '

(

) * 1

(x " x ) " t

%

& '

(

) * 1

1%

& ' '

(

) * *

+

, - -

.

/ 0 0 .

(iii) For the streak line, use the path line results but this time evaluate the integration constants at t = to instead of at t = t to find:


x # x = t1 to1( ) , and

y" y =

tto

#

$ %

&

' (

1

exp " x to

+1#

$ %

&

' (

tto

1#

$ %

&

' (

+

, -

.

/ 0 .

Now eliminate to find:

y" y = 1+ (x " x )t

%

& '

(

) * 1

exp " x

1t

+(x " x )

%

& '

(

) * 1

+1%

& ' '

(

) * * (x " x )t

%

& '

(

) *

+

, - -

.

/ 0 0 .

And evaluate at t = t to reach:

y" y

= 1+ (x " x ) " t

%

& '

(

) * 1

exp " x

1" t +(x " x )

%

& '

(

) * 1

+1%

& ' '

(

) * * (x " x ) " t

%

& '

(

) *

+

, - -

.

/ 0 0 .


Exercise 3.12. The velocity components in an unsteady plane flow are given by

u = x (1+ t) and

v = 2y (2 + t). Determine equations for the streamlines and path lines subject to x = x0 at t = 0. Solution 3.12. i) For the streamline, time is a constant. Use the first equality of (3.7) to find:

dydx

=vu

=2y (2 + t)x (1+ t)

dyy

=2(1+ t)(2 + t)

dxx ln y = 2(1+ t)

(2 + t)ln x + const.

Use of the initial condition produces:

ln y0 =2(1+ 0)(2 + 0)

ln x0 + const.,

so the final answer is:

ln yy0

"

# $

%

& ' =2(1+ t)(2 + t)

ln xx0

"

# $

%

& ' or

yy0

=xx0

"

# $

%

& '

2(1+ t )(2+ t )

.


dxdt

=x1+ t

and

dydt

=2y2 + t

,

and integrate the these in time to find:

ln x = ln(1+ t) + const. and

ln y = 2ln(2 + t) + const. Use the initial condition to determine the two constants, and exponentiate both equations:

x = x0(1+ t) and

y = y0 1+ t 2( )2 .

To determine the path line, eliminate t to find:

y = y0 1+ (x x0) 2x0( )2 .


Exercise 3.13. Using the geometry and notation of Fig. 3.8, prove (3.9). Solution 3.13. Before starting this problem, it is worthwhile to note that the acceleration of a fluid particle is invariant under the specified Galilean transformation so the components of U cannot be part of the final answer. Thus, transformation errors can be readily detected if terms are missing in the final results or extra ones have appeared. Figure 3.8 supports the following vector addition formula:

x = Ut + " x o + " x . Thus, the Cartesian-coordinate transformations in this case are given by:

" x = x ex U( )t " x o ,

" y = y ey U( )t " y o ,

" z = z ez U( )t " z o , and

" t = t . The transformation of the spatial derivatives between the stationary frame of reference, Oxyz, and the steadily moving frame, Oxyz is straightforward mathematics:

x

= # x x

# x

+ # y x

# y

+ # z x

# z

+ # t x

# t

= # x

,

y

= # x y

# x

+ # y y

# y

+ # z y

# z

+ # t y

# t

= # y

, and

z

= # x z

# x

+ # y z

# y

+ # z z

# z

+ # t z

# t

= # z

,

where the final equality on each line follows from differentiating the definitions of the moving coordinate variables given above. The time derivative requires more effort

t

= # x t

# x

+ # y t

# y

+ # z t

# z

+ # t t

# t

= ex U # x

ey U # y

ez U # z

+ # t

.

The first three equations imply:

= # and the fourth implies:

U $ + $ t

. The velocities will

be related by: u = U + u. Now use these results to assemble the fluid particle acceleration starting in the stationary coordinate system, and converting each velocity and differential operation to the moving coordinate system.

ut

+ u ( )u = U & + & t

'

( )

*

+ , U + & u [ ] + U + & u [ ] & ( ) U + & u [ ]

= U $ U + U $ t

U $ $ u + $ u $ t

+ U $ ( )U + $ u $ ( )U + U $ ( ) $ u + $ u $ ( ) $ u

= U $ $ u + $ u $ t + U $ ( ) $ u + $ u $ ( ) $ u =

$ u $ t

+ $ u $ ( ) $ u

Here most of the simplifications occur because all derivatives of U are zero; it is a constant. Thus, as expected, the form of the fluid particle acceleration is frame invariant for coordinate systems related by Galilean transformations.


Exercise 3.14. Determine the unsteady, u/t, and advective, (u)u, fluid acceleration terms for the following flow fields specified in Cartesian coordinates. a)

u = u(y,z,t),0,0( ) b)

u = x where

= 0,0,z (t)( ) c)

u = A(t) x,y,0( ) d) u = (Uo + uosin(kx t), 0, 0) where Uo, uo, k and are positive constants Solution 3.14. a) Here there is only one component of the fluid velocity; thus

u t = t( ) u(y,z,t),0,0( ) = u t ,0,0( ) , and

u [ ]u = u(y,z,t),0,0( ) x , y , z( )[ ] u(y,z,t),0,0( ) = u( x)[ ] u(y,z,t),0,0( ) = 0 . b) Here the fluid velocity has two components:

u = x = (zy, +zx, 0), so

u t = t( ) z y,z x,0( ) = ydzdt,x dz

dt,0

%

& '

(

) * =

dzdt

y,x,0( ) , and

u [ ]u = z y,z x,0( ) x , y , z( )[ ] z y,z x,0( )

= z y( x) +z x( y)[ ] z y,z x,0( ) = z2x,z2y,0( ) = z2 x,y,0( ) . c) Again the fluid velocity has two components: (Ax, Ay, 0), so

u t = t( ) Ax,Ay,0( ) = x dAdt,y dA

dt,0

$

% &

'

( ) =

dAdt

x,y,0( ) , and

u [ ]u = Ax,Ay,0( ) x , y , z( )[ ] Ax,Ay,0( )

= Ax( x) Ay( y)[ ] Ax,Ay,0( ) = A2x,A2y,0( ) = A2 x,y,0( ). d) Here again there is only one component of the fluid velocity; thus

u t = t( ) Uo + uo sin(kx t),0,0( ) = uo cos(kx t),0,0( ) , and

u [ ]u = Uo + uo sin(kx t),0,0( ) x , y , z( )[ ] Uo + uo sin(kx t),0,0( ).

= Uo + uo sin(kx t)( )( x)[ ] Uo + uo sin(kx t),0,0( )

= Uo + uo sin(kx t)( )kuo cos(kx t),0,0( )

= kUouo cos(kx t) +12 kuo

2 sin 2(kx t)[ ],0,0( ).


Exercise 3.15. Consider the following Cartesian velocity field

u = A(t) f (x),g(y),h(z)( ) where A, f, g, and h are non-constant functions of only one independent variable. a) Determine u/t, and (u)u in terms of A, f, g, and h, and their derivatives. b) Determine A, f, g, and h when Du/Dt = 0, u = 0 at x = 0, and u is finite for t > 0. c) For the conditions in b), determine the equation for the path line that passes through xo at time to, and show directly that the acceleration a of the fluid particle that follows this path is zero. Solution 3.15. a) Here there are three components of the fluid velocity; thus

ut

=t

Af ,Ag,Ah( ) = f dAdt,g dAdt,h dA

dt#

$ %

&

' ( =

dAdt

f ,g,h( ) , and

u [ ]u = A f ,g,h( ) x, y, z

%

& '

(

) *

+

, -

.

/ 0 Af ,Ag,Ah( )

= A2 f x

+ g y

+ h z

#

$ %

&

' ( f ,g,h( ) = A2 f

dfdx,g dgdy,h dhdz

)

* +

,

- . ,

where the partial derivatives have become total derivatives because f, g, and h are only functions of one variable. b) For the given velocity field, Du/Dt = 0 implies:

f dAdt

+ A2 f dfdx

= 0 ,

g dAdt

+ A2g dgdy

= 0 , and

h dAdt

+ A2h dhdz

= 0.

Start with the first equation, assume A2f is not zero, and divide by it to find:

1A2

dAdt

+dfdx

= 0 .

The first term in this equation only depends on t while the second one only depends on x, thus, each must be equal and opposite, and constant (= C). So,

1A2

dAdt

= C dAA2

= Cdt 1A

= C(t to) A =1

C(t to), and

dfdx

= C f = C(x xo) ,

where to and xo are constants of integration. Similarly, for the other two component directions:

g = C(y yo) and

h = C(z zo). Here, we presume to 0 so that A is finite for t > 0. c) The x-component of the path line is defined by:

dxdt

= u = Af = C(x xo)C(t to)

dx

x xo=

dtt to

ln(x xo) = ln(t to) + const. x xo =U(t to),

where U is a constant. Similarly for the other Cartesian directions:

y yo =V (t to), and

z zo =W (t to) , where V and W are constants. So defining the constant vector U = (U, V, W), the path line of interest is:

x xo = U(t to). Thus, x(t) is linear function of t, so dx(t)/dt = U, and a = d2x(t)/dt2 = 0. This exercise illustrates how the unsteady and advective acceleration terms may be equal and opposite.


Exercise 3.16 If a velocity field is given by u = ay and v = 0, compute the circulation around a circle of radius ro that is centered on at the origin. Check the result by using Stokes theorem. Solution 3.16. In plane polar coordinates, the vector path-length element, ds, on a circle of radius ro is ds =

tds = e rod . From Example 2.1, the radial and angular velocity component are: ur = ucos + vsin = arsincos + 0, and u = usin + vcos = arsin2 , where u = ay = arsin and v = 0 has been used. Thus, circulation is:

= u ds = u e rod = 0 = 2 = u rod = 0

= 2 = aro2 sin2d = 0

= 2 = aro2 .

Now use Stokes' theorem to reach the same result. Start by computing the vorticity:

u =ex ey ez x y zay 0 0

= aez .

Insert this into the Stokes' theorem area integral using n = ez, and dA = rdrd:

= ( u) ndAA = (aez )r= 0

r= ro = 0 = 2 ezrdrd = aro2 ,

and this matches the result of the prior circulation calculation.


Exercise 3.17. Consider a plane Couette flow of a viscous fluid confined between two flat plates a distance b apart. At steady state the velocity distribution is u = Uy/b and v = w = 0, where the upper plate at y = b is moving parallel to itself at speed U, and the lower plate is held stationary. Find the rates of linear strain, the rate of shear strain, and vorticity in this flow. Solution 3.17. Here there is only one velocity component: u = Uy/b. The strain rate tensor is:

Sij =12

uix j

+u jxi

#

$ % %

&

' ( ( Sij =

u x 12 u y + v x( )12 v x + u y( ) v y

* + ,

- . /

=0 U 2b

U 2b 0* + ,

- . /

.

Thus, the linear strain rates are both zero, and the shear strain rate is U/2b. The vorticity vector has one non-zero component:

z =vx

uy

= Ub

.


Exercise 3.18. The steady two-dimensional flow field inside a sloping passage is given in (x,y)-Cartesian coordinates by u = (u,v) = 3q 4h( ) 1 y h( )2( ) 1, y h( ) dh dx( )( ) where q is the volume flow rate per unit length into the page, and h is the passage's half thickness. Determine the streamlines, vorticity, and strain rate tensor in this flow away from x = 0 when h = x where is a positive constant. Sample profiles of u(x,y) vs. y are shown at two x-locations in the figure. What are the equations of the streamlines along which the x- and y-axes are aligned with the principal axes of the flow? What is the fluid particle rotation rate along these streamlines?

Solution 3.18. For planar flow in Cartesian coordinates, the streamlines are determined from:

dydx

=vu=y h( ) dh dx( )

1= y x( ) = y x .

Separate the differentials and integrate to find: ln(y) = ln(x) + const. Exponentiate to find: y = Cx, where C is a constant. Thus, the streamlines are straight lines through the origin of coordinates. The vorticity z is determined from:

z =vx

uy

=x

3q4h

#

$%

&

'( 1

y2

h2#

$%

&

'(yhdhdx

)

*+

,

-.

y

3q4h

#

$%

&

'( 1

y2

h2#

$%

&

'(

)

*+

,

-.

= 3q4

x

1x

1 y2

2x2#

$%

&

'(yx

)

*+

,

-.

3q4

y

1x

1 y2

2x2#

$%

&

'(

)

*+

,

-.

= 3q4

2yx3

+4y3

3x5)

*+

,

-.

3q4

2y 3x3

)

*+,

-.=

3qy2x3

1+ 2y2

2x2+

1 2

)

*+

,

-.

The strain-rate tensor Sij is computed from the following velocity derivatives:

S11 =ux

=x

3q4h

"

#$

%

&' 1

y2

h2"

#$

%

&'

)

*+

,

-.=3q4

x

1x

1 y2

2x2"

#$

%

&'

)

*+

,

-.=3q4

1x2

+3y2

3x4)

*+

,

-.=

3q4x2

1+ 3y2

2x2)

*+

,

-.

S22 =vy

=y

3q4h

"

#$

%

&' 1

y2

h2"

#$

%

&'yhdhdx

)

*+

,

-.=3q4

y

1x

1 y2

2x2"

#$

%

&'yx

)

*+

,

-.=

3q4x2

1 3y2

2x2"

#$

%

&' , and

S12 = S21 =12

uy+vx

"#$

%&'=12

y

3q4h

(

)*

+

,- 1

y2

h2(

)*

+

,-

/

01

2

34+12

x

3q4h

(

)*

+

,- 1

y2

h2(

)*

+

,-yhdhdx

/

01

2

34=

3qy4x3

1 2

1+ 2y2

2x2/

01

2

34 ,

where the differentiation details for S12 are available above in the calculation of z. The principle axes occur then the off-diagonal strain rate components are zero. This occurs when

x!

y!

h(x)!


S12 = S21 =3qy4x3

1 2

1+ 2y2

2x2"

#$

%

&'= 0 .

This equality is satisfied along the x-axis where y = 0, and when the contents of the [,]-brackets are zero:

1 2

+1= 2y2

2x2 or y = x (1+ 2 ) 2

These streamlines only occur inside flow wedge when > 1. The fluid particle rotation rate is z/2, so from the results above for z:

z2

!

"#

$

%&y=0

= 0 , and z2

!

"#

$

%&y=x (1+2 ) 2

= 3q (1+ 2 ) 2

2 3x2.


Exercise 3.19. For the flow field

u =U+ x , where U and are constant linear- and angular-velocity vectors, use Cartesian coordinates to a) show that Sij is zero, and b) determine Rij. Solution 3.19. Since no simplifications are given, all the components of U = (U1, U2, U3) and = (1, 2, 3) should be treated as being non-zero. In Cartesian coordinates, the velocity field is

u =U+ x =U1e1 +U2e2 +U3e3 +e1 e2 e31 2 3x1 x2 x3

= U1 +2x3 3x2( )e1 + U2 +3x1 1x3( )e2 + U3 +1x2 2x1( )e3 a) Use this velocity field result to compute the velocity gradient tensor, and its transpose (indicated with a superscript "T" below) to sort out which derivatives are zero and which ones are not.

uix j

=

u1 x1 u1 x2 u1 x3u2 x1 u2 x2 u2 x3u3 x1 u3 x2 u3 x3

#

$ %

& %

'

( %

) %

=

0 3 +2+3 0 12 +1 0

#

$ %

& %

'

( %

) %

u jxi

=uix j

#

$ % %

&

' ( (

T

=

u1 x1 u2 x1 u3 x1u1 x2 u2 x2 u3 x2u1 x3 u2 x3 u3 x3

)

* +

, +

-

. +

/ +

=

0 +3 23 0 +1+2 1 0

)

* +

, +

-

. +

/ +

This result can be used to construct the strain rate tensor Sij:

Sij =12

uix j

+u jxi

#

$ % %

&

' ( ( =12

uix j

+uix j

#

$ % %

&

' ( (

T#

$

% %

&

'

( (

=

0 12 3 + 3( )12 +2 2( )

12 +3 3( ) 0

12 1 + 1( )

12 2 + 2( )

12 +1 1( ) 0

+

, -

. -

/

0 -

1 -

=

0 0 00 0 00 0 0

"

# $

% $

&

' $

( $

.

b) Similarly for the rotation tensor:

Rij =uix j

u jxi

=uix j

uix j

$

% & &

'

( ) )

T

=

0 3 3 +2 + 2+3 + 3 0 12 2 +1 + 1 0

+

, -

. -

/

0 -

1 -

=

0 23 +22+23 0 2122 +21 0

$

% &

' &

(

) &

* &

.


Exercise 3.20. Starting with a small rectangular volume element V = x1x2x3, prove (3.14). Solution 3.20. The volumetric strain rate for a fluid element is:

1V

DDt

V( ) = 1x1x2x3

DDt

x1x2x3( ) =1x1

DDt

x1( ) +1x2

DDt

x2( ) +1x3

DDt

x3( ).

From Section 3.4 in the text, the linear strain rate corresponding to elongation or contraction of a fluid element in the first direction is:

1x1

DDt

x1( ) = S11 =u1x1

,

and this can be immediately extended to the other two directions,

1x2

DDt

x2( ) = S22 =u2x2

and

1x3

DDt

x3( ) = S33 =u3x3

,

because its geometric derivation (see Figure 3.10) did not rely on any special properties of the first direction. Substitution of these relationships into the final version of the volumetric strain rate given above produces:

1V

DDt

V( ) = S11 + S22 + S33 =u1x1

+u2x2

+u3x3

=uixi

= Sii.


Exercise 3.21. Let Oxyz be a stationary frame of reference, and let the z-axis be parallel with the fluid vorticity vector in the vicinity of O so that

= u =zez in this frame of reference. Now consider a second rotating frame of reference

O " x " y " z having the same origin that rotates about the z-axis at angular rate ez. Starting from the kinematic relationship,

u = (ez) x + $ u , show that in the vicinity of O the vorticity

" = " " u in the rotating frame of reference can only be zero when 2 = z, where

" is the gradient operator in the primed coordinates. The following unit vector transformation rules may be of use:

" e x = ex cos(t) + ey sin(t) ,

" e y = ex sin(t) + ey cos(t) , and

" e z = ez . Solution 3.21. The approach here is to compute

" = " " u in the stationary frame of reference and then determine the parameter choice(s) necessary for to be zero. The vector x must have a consistent representation in either frame, so

x = xex + yey = " x " e x + " y " e y = " x ex cos(t) + ey sin(t)( ) + " y ex sin(t) + ey cos(t)( ). Equating components in the stationary frame of reference produces:

x = " x cos(t) " y sin(t) , and

y = " x sin(t) + " y cos(t) . The remaining independent variables are the same in either frame:

t = " t , and

z = " z . Thus, spatial derivatives are related by:

# x

=x # x

x

+y # x

y

+z # x

z

+t # x

t

= cos(t) x

+ sin(t) y

,

# y

=x # y

x

+y # y

y

+z # y

z

+t # y

t

= sin(t) x

+ cos(t) y

, and

# z

=x # z

x

+y # z

y

+z # z

z

+t # z

t

=z

.

Using the above information, the gradient operator in the rotating coordinates can be rewritten in terms of the stationary frame coordinates and unit vectors:

" = " e x " x

+ " e y " y

+ " e z " z

= ex cos(t) + ey sin(t)( ) cos(t)x

+ sin(t) y

$

% &

'

( )

+ ex sin(t) + ey cos(t)( ) sin(t)x

+ cos(t) y

%

& '

(

) * + ez

z

" = ex cos2(t)

x+ ex sin(t)cos(t)

y

+ ey sin(t)cos(t)x

+ ey sin2(t)

y&

' (

)

* +

+ ex sin2(t)

x ex sin(t)cos(t)

y

ey sin(t)cos(t)x

+ ey cos2(t)

y%

& '

(

) * + ez

z

" = ex cos2(t)

x+ ex sin(t)cos(t)

y

+ ey sin(t)cos(t)x

+ ey sin2(t)

y&

' (

)

* +

+ ex sin2(t)

x ex sin(t)cos(t)

y

ey sin(t)cos(t)x

+ ey cos2(t)

y%

& '

(

) * + ez

z

" = exx

+ eyy

+ ezz

.


This result seems too simple, but should not be a surprising to a routine user of vector calculus. Now resolve rotating-frame velocity components in the stationary frame of reference:

" u = u (ez) x = uex + vey + wez +yex xey . Using the stationary frame components and unit vectors, the vorticity in the rotating frame is:

" = " " u =ex ey ez x y zu +y v x w

=

ex ey ez x y zu v w

+

ex ey ez x y zy x 0

= u+ex ey ez x y zy x 0

=zez + ex (0) + ey (0) + ez ( ) = ez z 2( )

Thus, will be zero when z = 2.


Exercise 3.22. Consider a plane-polar area element having dimensions dr and rd. For two-dimensional flow in this plane, evaluate the right-hand side of Stokes theorem

ndA = u ds and thereby show that the expression for vorticity in plane-polar coordinates

is:

z =1rr

ru( ) 1rur

.

Solution 3.22. Using the element shown with angular width d,

application of Stokes' theorem provides:

ndA = u ds zrddr = urdr + u +ur

dr(

) *

+

, - r + dr( )d ur +

ur

d(

) *

+

, - dr u rd , or

zrddr = urdr + ru d + rur

drd + u drd urdr ur

ddr u rd + ...

= r ur

drd + u drd ur

ddr + ...

where + ... indicates the presence of higher order terms in dr and d. Division by rddr and passing to the limit where dr and d go to zero produces:

z =ur

+1ru

1rur

=1r

ru( ) ur

&

' (

)

* + .

uru!

u! + (!u!/!")dr

ur + (!ur/!!)d!

dr

!


Exercise 3.23. The velocity field of a certain flow is given by

u = 2xy 2 + 2xz2 ,

v = x 2y , and

w = x 2z. Consider the fluid region inside a spherical volume x2 + y2 + z2 = a2. Verify the validity of Gauss theorem

udV = u ndAA

V by integrating over the sphere.

Solution 3.23. First compute the divergence of the velocity field.

u = ux

+vy

+wz

=x

2xy 2 + 2xz2( ) + y

x 2y( ) + z

x 2z( ) = 2y 2 + 2z2 + 2x 2 = 2r2 . The volume integral of

u is:

udV = 2r2( )4r2dr =r= 0

r= a

V 85

a5 .

Now work on the surface integration using spherical coordinates (see Figure 3.3d, and Appendix B). Here,

n = er = ex cos sin + ey sin sin + ex cos , so

u n = ucos sin + v sin sin + wcos

= 2xy 2 + 2xz2( )cos sin + x 2y( )sin sin + x 2z( )cos Unfortunately, this result is in mixed variables so convert everything to spherical polar coordinates using

x = rcos sin ,

y = rsin sin ,

z = rcos . This conversion produces:

u n = 2r3 cos sin2 sin3 + cos sin cos2( )cos sin

+r3 cos2 sin sin3( )sin sin + r3 cos2 sin2 cos( )cos

= 2r3 sin2 sin2 + cos2( )cos2 sin2 + r3 sin2 sin2( )cos2 sin2 + r3 cos2( )cos2 sin2

= r3 3sin2 sin2 + 3cos2( )cos2 sin2 = r3 34 sin2(2) 1 cos2( )2

+ 3cos2 cos2 1 cos2( )( ) So, the surface integral produces:

u ndAA = a3 34 sin

2(2) 1 cos2( )2

+ 3cos2 cos2 1 cos2( )( ) = 0

=

= 0

= 2

a2 sindd

= a5 34 sin2(2)d 1 cos2( )

2

= 0

=

= 0

= 2

sind + a5 3cos2d cos2 1 cos2( ) = 0

=

= 0

= 2

sind

= a5 34

#

$ %

&

' ( 1 2( )

2

1

+1

d + a5 3( ) 2 1 2( )1

+1

d

= a5 34

#

$ %

&

' ( 1615

+ a5 3( ) 415

=1215

+1215

#

$ %

&

' ( a5 =

85a5,

which is the same as the result of the volume integration.


Exercise 3.24. A flow field on the xy-plane has the velocity components u = 3x + y and v = 2x 3y. Show that the circulation around the circle (x 1)2 + (y 6)2 = 4 is 4. Solution 3.24. The circle is centered at (1,6) and its radius is 2. So, if (x,y) is a point on the circle then x = 1 + 2cos, and y = 6 + 2sin, where is the angle from the horizontal. The velocity component tangent to the circle will be

u = e u = sin,cos( ) (u,v) = usin + v cos . Evaluate this velocity component.

u = (3x + y)sin + (2x 3y)cos = (3sin + 2cos)x (sin + 3cos)y

= (3sin + 2cos)(1+ 2cos) (sin + 3cos)(6 + 2sin)= 3sin 6sin cos + 2cos + 4cos2 6sin 2sin2 18cos 6cos sin= 9sin 12sin cos 16cos + 4cos2 2sin2

Now compute the circulation:

= u = 0

2

rd = 9sin 12sin cos 16cos + 4cos2 2sin2( ) = 0

2

2d

= 0 + 0 + 0 + 4 2( )2 = 4


Exercise 3.25. Consider solid-body rotation about the origin in two dimensions: ur = 0 and u = 0r. Use a polar-coordinate element of dimension rd and dr, and verify that the circulation is vorticity times area. (In Section 5 this was verified for a circular element surrounding the origin.) Solution 3.25. Using the element shown with angular width d,

the circulation is around the element is the sum or four terms:

= u ds = urdr + u +ur

dr'

( )

*

+ , r + dr( )d ur +

ur

d'

( )

*

+ , dr u rd .

Now substitute in the velocity field: ur = 0 and u = 0r.

= 0r +0dr( ) r + dr( )d 0r2d = 20rdrd +0(dr)2d = (20)rdrd , where the final equality holds in the limit as the differential elements become small and (dr)2d is negligible compared to rdrd. Thus, since the voriticity is 20 and rdrd is the area of the element, the final relationship states: circulation = (vorticity) x (area).

uru!

u! + (!u!/!")dr

ur + (!ur/!!)d!

dr

!


Exercise 3.26. Consider the steady Cartesian velocity field

u = Ayx 2 + y 2( )

, +Axx 2 + y 2( )

,0$

%

& &

'

(

) ) .

a) Determine the streamline that passes through

x = (xo,yo,0) b) Compute Rij for this velocity field. c) For A > 0, explain the sense of rotation (i.e. clockwise or counter clockwise) for fluid elements for < 1, = 1, and > 1.

Solution 3.26. a) Use the definition of a streamline:

dydx

=vu

=Ax x 2 + y 2( )

Ay x 2 + y 2( )=

xy

. Use the

two ends of this extended equality to find:

ydy = xdx , and integrate the resulting differential relationship to get: y2 2 = x2 2+ const . Evaluate the constant using the required condition:

x 2 + y 2 = xo2 + yo

2 . This is a circle with radius

xo2 + yo

2 . Therefore the streamlines are circles.

b) The rotation tensor is: Rij =uix j

ujxi

=

0 u yv x 0v x u y 0 0

0 0 0

"

#$

%$$

&

'$

($$

, where the

second equality comes from putting the specified velocity field into the definition of Rij with u = (u,v,w) = (u1,u2,u3) = ui as usual. Evaluating the derivatives produces:

Rij =A

x2 + y2( )+1

0 x2 y2 + 2y2 x2 + y2 2x2( ) 0x2 + y2 2x2 x2 y2 + 2y2( ) 0 0

0 0 0

"

#

$$

%

$$

&

'

$$

(

$$

= 2A(1)x2 + y2( )

0 1 01 0 00 0 0

"

#$

%$

&

'$

($

c) The following answers are based on A > 0. For < 1, fluid particles rotate counter clockwise (positive z). For = 1, fluid particles do not rotate. For > 1, fluid particles rotate clockwise (negative z). Interestingly, the streamlines are the same (circular!) for all three possibilities.


Exercise 3.27. Using indicial notation (and no vector identities) show that the acceleration a of a fluid particle is given by:

a = u t + 12 u2( ) + u where is the vorticity.

Solution 3.27. The acceleration of a fluid particle is a = Du Dt u t + u ( )u . Thus, the task is to prove

u ( )u = 12 u2( ) + u. Start from the advective acceleration written in index

notation and force the rotation tensor to appear:

u juix j

= u juix j

u jxi

$

% & &

'

( ) ) u j

u jxi

= u jRij +12

xi

u j2( ) .

From (3.15),

Rij = ijkk , so this becomes

u juix j

= ijku jk +12

xi

u j2( ) = ikjku j + 12

xi

u j2( ) = u+ 12 u

2(

) *

+

, - ,

where the final equality follows from the index notation definitions of the cross product (2.21), gradient (2.22), and vector magnitude (

u j2 = u 2 ).


Exercise 3.28. Starting from (3.29), show that the maximum u in a Gaussian vortex occurs when

1+ 2(r2 2) = exp(r2 2) . Verify that this implies r 1.12091. Solution 3.28. Differentiate the u equation from (3.29) with respect to r and set this derivative equal to zero.

ddr

u (r)( ) =2

ddr

1 exp r2 2( )r

'

( ) )

*

+ , , =

2

1 exp r2 2( )

r2exp r2 2( )

r2r 2

'

( )

*

+ ,

'

( ) )

*

+ , , = 0.

Eliminate common factors assuming r 0.

0 = 1+ exp r2 2( ) + 2r2 2( )exp r2 2( ) = 1+ 1+ 2r2 2( )exp r2 2( ). This can be rearranged to:

exp r2 2( ) =1+ 2r2 2 , which is the desired result. When r/ 1.12091, then

exp r2 2( ) = 3.51289 and

1+ 2r2 2 = 3.51288 , which is suitable numerical agreement.


Exercise 3.29. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the parallelogram shown is hl(d/dt)cos when depends on time while h and l are constants.

Solution 3.29. With F = 1, the volume integral of F/t on the right side of (3.35) is zero, and the integrands are simplified, so (3.35) simplifies to:

ddt

dVV*(t ) = + b ndAA*(t ) .

(Here the volume integration is two dimensional, and produces the parallelogram's area, hlsin, which can be time differentiated, d(hlsin)/dt, to reach the desired result. However, this is not the intended solution path for this problem.) When is time-dependent, the parallelogram has three moving sides (a left side of length h, a right side of length h, and a top side of length l). Thus, the simplified version of (3.35) reduces to:

ddt

dVV*(t ) = +left side +right side top side{ }b ndA .

Here we note that the left and right sides move identically, so b will be the same. However, n points in opposite directions on these two sides, so the contributions from these two sides cancel. The x-coordinates of points on the parallelogram's top side are hcos x(t) l + hcos. The y-coordinate of the parallelogram's top side is y(t) = hsin. Time differentiate the location of any point on the parallelogram's top side to find b:

b = (dx/dt, dy/dt) = (hsin, hcos)(d/dt). And, on the parallelogram's top side, n is ey, so

b n = hcos( ) d dt( ) , and in two dimensions dA is just dx. Thus, the simplified version of (3.35) is:

ddt

dVV*(t ) = b ntop side dA = hcos

hcos

l+hcos

ddt dx = hl cosddt

,

and this is the desired result.

x!

y!l!

h!(t)!


Exercise 3.30. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the triangle shown is 12 b dh dt( ) when h depends on time and b is constant.


ddt

dVV*(t ) = + b ndAA*(t ) .

(Here the volume integration is two dimensional, and produces the triangle's area, hb/2, which can be time differentiated, (d/dt)(hb/2) to reach the desired result. However, this is not the intended solution path for this problem.) When h is time-dependent, the triangle has two moving sides (a top side of length b, and a hypotenuse of length [h2 + b2]1/2). Thus, the simplified version of (3.35) reduces to:

ddt

dVV*(t ) = +top side hypotenuse{ }b ndA .

The x-coordinates of the triangle's top side are 0 x b. The y-coordinate of the triangle's top side is y(t) = h(t). Time differentiate the location of any point on the triangles's top side to find b:

b = (dx/dt, dy/dt) = (0, dh/dt). And, on the triangle's top side, n is ey, so

b n = dh dt , The x-y coordinates of the triangle's hypotenuse fall on the line y = (h/b)x for 0 x b. The motion of points on the hypotenuse can be described by a purely vertical velocity. So, for any constant x-location, differentiate the equation of this line to find b:

b = (dx/dt, dy/dt) = (0, (x/b)dh/dt). Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, 1), respectively. Thus, the outward unit normal on the hypotenuse of the triangle is:

n =h b,1( )(h / b)2 +1

=(h,b)h2 + b2

, so b n = x dh dt( )h2 + b2

.

Thus, the simplified version of (3.35) can be written: ddt

dVV*(t ) = b n

top side dA+ b n

hypotenuse dA = dhdt0

b

dx x dh dt( )h2 + b20

b

1+ (h / b)2dx$%&' ,

where, in two dimensions, dA is just dx on the top side and dA is a path length element along the hypotenuse, ds = [1 + (dy/dx)2]1/2dx. The factor in [,]-brackets is this path length element in terms of dx. Perform the integrations to find:

ddt

dVV*(t ) = b

dhdt

1bdhdt

x2

2#

$%

&

'(0

b

=b2dhdt

,


x!

y!b!

h(t)!


Exercise 3.31. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the ellipse shown is b da dt( ) when a depends on time and b is constant.


ddt

dVV*(t ) = + b ndAA*(t ) .

(Here the volume integration is two dimensional, and produces the ellipse's area, ab, which can be time differentiated, (d/dt)( ab) to reach the desired result. However, this is not the intended solution path for this problem.) When a is time-dependent, the contour that defines the ellipse, (y/b)2 + (x/a)2 = 1, is also time dependent. However, the symmetry of the ellipse allows the analysis to completed in the first quadrant alone and then multiplied by 4. Thus, the simplified version of (3.35) reduces to:

ddt

dVV*(t ) = 4 b n

first quadrant dA .

The equation for the ellipse in the first quadrant is:

x = +a 1 y b( )2 or y = +b 1 x a( )2 . The motion of points on this curve can be described by a purely horizontal velocity when a varies but b is constant. So, for any constant y-location, differentiate the first equation to find b:

b = dxdt, 0

!

"#

$

%&=

dadt

1 y b( )2 , 0!

"#

$

%&=

xadadt, 0

!

"#

$

%& ,

where the final equality comes from changing the independent variable from y to x. Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, 1), respectively. Thus, the outward unit normal on the ellipse in the first quadrant is:

n =dy / dx,1( )(dy / dx)2 +1

, so b n = x(dy / dx)a (dy / dx)2 +1

dadt

.

Thus, the simplified version of (3.35) can be written: ddt

dVV*(t ) = 4 b n

first quadrant dA = 4 x(dy / dx)

a (dy / dx)2 +1dadt0

a

1+ (dy / dx)2dx$%&' ,

where, in two dimensions, dA is a path length element, ds = [1 + (dy/dx)2]1/2dx, along the first-quadrant portion of the ellipse. This path length element is the factor in [,]-brackets above.

x!

y!

b!

a(t)!


Simplify the integrand, insert dy/dx = (bx/a2)/[1 (x/a)2]1/2, and perform the integration using the substitution x = asin to find:

ddt

dVV*(t ) = 4

x(dy / dx)a

dadt0

a

dx = 4adadt

xb(x / a2 )1 (x / a)20

a

dx

= 4b dadt

sin21 sin20

2

cosd = 4b dadt sin2

0

2

d = 4b dadt4= b da

dt,



Exercise 3.32. For the following time-dependent volumes V*(t) and smooth single-valued integrand functions F, choose an appropriate coordinate system and show that

d dt( ) FdVV *( t ) obtained from (3.30) is equal to that obtained from (3.35). a) V*(t) = L1(t)L2L3 is a rectangular solid defined by 0 xi Li, where L1 depends on time while L2 and L3 are constants, and the integrand function F(x1,t) depends only on the first coordinate and time. b) V*(t) = (/4)d2(t)L is a cylinder defined by 0 R d(t)/2 and 0 z L, where the cylinders diameter d depends on time while its length L is constant, and the integrand function F(R,t) depends only on the distance from the cylinders axis and time. c) V*(t) = (/6)D3(t) is a sphere defined by 0 r D(t)/2 where the spheres diameter D depends on time, and the integrand function F(r,t) depends only on the radial distance from the center of the sphere and time.

Solution 3.32. The two equations are (3.30)

ddt

F(x, t)dxx=a( t )

x=b( t )

= Ftdx

a

b

+ dbdtF b,t( ) da

dtF a,t( ),

and (3.35)

ddt

F(x,t)dVV *(t ) = F(x,t)

tdV

V *(t ) + F(x,t)b ndA

A*(t ) .

a) Use Cartesian coordinates with the origin at xi = 0. The cross sectional area of the rectangular solid, L2L3, is constant, so dV = L2L3dx1. The volume integral proceeds from x1 = 0 (= a) to x1 = L1(t) (= b) so (3.30) implies:

ddt

F(x,t)dVV *(t ) = ddt

F(x1,t)L2L3dx1x1 =0

x=L1 ( t )

= Ft

L2L3dx10

L1

+ dL1dtF L1,t( )L2L3 . (a1)

Now start from (3.35) using a control volume that encloses the rectangular solid. In this case the only control surface that moves lies at x1 = L1, has outward normal n = e1, and moves with velocity b = (dL1/dt)ex. First evaluate the left side term from (3.35).

ddt


F(x1,t)x3= 0

L3

x2= 0

L2

dx1dx2dx3x1= 0

L1 ( t )

= L2L3ddt

F(x1,t)dx1x1= 0

L1 ( t )

.

Now evaluate the right side terms from (3.35).

ddt

F(x,t)dVV *(t ) = F(x1,t)

tx3 = 0

L3

x2 = 0

L2

dx1dx2dx3x1 = 0

L1 ( t )

+ F(L1,t)x3 = 0

L3

x2 = 0

L2

dL1dte1

$

% &

'

( ) e1dx2dx3

= L2L3F(x1,t)

tdx1

x1 = 0

L1 (t )

+ L2L3dL1dt

F(L1,t)

Setting the left and right side terms equal produces

L2L3ddt

F(x1,t)dx1x1= 0

L1 ( t )

= L2L3F(x1,t)

tdx1

x1= 0

L1 ( t )

+ L2L3dL1dt

F(L1,t) , (a2)

which is identical to (a1). b) Use cylindrical coordinates with the origin at R = z = 0. The length area of the cylinder, L, is constant, so dV = L(2RdR). The volume integral proceeds from R = 0 (= a) to R = d(t)/2 (= b) so (3.30) implies:

ddt

F(x,t)dVV *(t ) = 2L ddt

F(R,t)RdRR =0

d ( t ) / 2

= 2L Ft

RdR0

d / 2

+ 2Ld d 2( )dt

F d /2,t( ) d2

. (b1)


Now start from (3.35) using a control volume that encloses the cylinder. In this case the only control surface that moves lies at R = d/2, has outward normal n = eR, and moves with velocity b = (d(d/2)/dt)eR. First evaluate the left side term from (3.35).

ddt


F(R,t)= 0

2

R= 0

d / 2

RddRdzz= 0

L

= 2L ddtF(R,t)RdR

0

d / 2

.


ddt

F(x,t)dVV *(t ) = F(R,t)

t= 0

2

R= 0

d / 2

RddRdzz= 0

L

+ F(d/2,t)= 0

2

z= 0

L

d(d /2)dteR

&

' (

)

* + eR

d2ddz

= 2L F(R,t)t

RdRR= 0

d / 2

+ 2L d(d /2)dtF(d/2,t) d

2


2L ddt

F(R,t)RdR0

d / 2

= 2L F(R,t)t

RdRR= 0

d / 2

+ 2L d(d /2)dtF(d/2,t) d

2 , (b2)

which is identical to (b1). c) Use spherical coordinates with the origin at r = 0. The sphere expands symmetrically so the volume element is dV = 4r2dr. The volume integral proceeds from r = 0 (= a) to r = D(t)/2 (= b) so (3.30) implies:

ddt

F(x,t)dVV *(t ) = 4 ddt

F(r,t)r2drr=0

D(t ) / 2

= 4 Ftr2dr

0

D / 2

+ 4d D 2( )dt

F D /2,t( ) D2

%

& '

(

) * 2

. (c1)

Now start from (3.35) using a control volume that encloses the sphere. In this case the moving control surface lies at r = D/2, has outward normal n = er, and moves with velocity b = (d(D/2)/dt)er. First evaluate the left side term from (3.35).

ddt


F(R,t)= 0

2

= 0

r2drsinddr= 0

D / 2

= 4 ddtF(r,t)r2dr

0

D / 2

.


ddt

F(x,t)dVV *(t ) = F(r,t)

t= 0

2

= 0

r2drsinddr= 0

D / 2

+ F(d/2,t)= 0

2

= 0

d(D /2)dteR

'

( )

*

+ , eR

D2

'

( )

*

+ ,

2

sindd

= 4 F(r,t)t

r2drr= 0

D / 2

+ 4 d(D /2)dtF(D/2,t) D

2'

( )

*

+ ,

2


4 ddt

F(r,t)r2dr0

D / 2

= 4 F(r,t)t

r2drr= 0

D / 2

+ 4 d(D /2)dtF(D/2,t) D

2%

& '

(

) * 2

, (c2)

which is identical to (c1).


Exercise 3.33. Starting from (3.35), set F = 1 and derive (3.14) when b = u and V*(t) = V

0. Solution 3.33. With F = 1, b = u, and V*(t) = V with surface A, (3.35) becomes:

ddt

dVV = 0 + u ndA

A .

The first integral is merely V. Use Gauss' divergence theorem on the second term to convert it to volume integral.

ddt

V( ) = udVV .

As V

0 the integral reduces to a product of V and the integrand evaluated at the center point of V. Divide both sides of the last equation by V and take the limit as V

0:

limV0

1V

ddt

V( ) = limV0

1V

udVV = lim

V0

1V

u( )V + ...[ ] = u = Sii, and this is (3.14).


Exercise 3.34. For a smooth single valued function F(x) that only depends on space and an arbitrarily-shaped control volume that moves with velocity b(t) that only depends on time, show that

d dt( ) F(x)dVV *( t ) = b F(x)dVV *( t )( ) . Solution 3.34. Start from Reynolds transport theorem (3.35):

ddt

F(x,t)dVV *(t ) = F(x,t)

tdV

V *(t ) + F(x,t)b ndA

A*(t ) .

When F does not depend on time, the first term on the right drops out.

ddt

F(x,t)dVV *(t ) = F(x,t)b ndA

A*(t ) .

When b does not depend on location, it can be taken outside the surface integral.

ddt

F(x,t)dVV *(t ) = b F(x,t)ndA

A*( t )

$

% &

'

( ) .

Apply Gauss' theorem to the integral in large parentheses, to reach the desired form:

ddt

F(x,t)dVV *(t ) = b F(x,t)dA

V *( t )

%

& '

(

) * .


3.35. Show that (3.35) reduces to (3.5) when V*(t) = V

0 and the control surface velocity b is equal to the fluid velocity u(x,t). Solution 3.35. When V*(t) = V with surface A, V is small, and b = u, V represents a fluid particle. Under these conditions (3.35) becomes:

ddt

F(x,t)dVV = F(x,t)

tdV

V + F(x,t)u ndA

A ,

and the time derivative is evaluated following V. Use Gauss' divergence theorem on the final term to convert it to a volume integral,

F(x,t)u ndAA = F(x,t)u( )

V dV ,

so that (3.35) becomes:

ddt

F(x,t)dVV = F(x,t)

t+ F(x,t)u( )

'

( ) *

+ , dV

V = F(x,t)

t+ F(x,t) u+ u ( )F(x,t)

'

( ) *

+ , dV

V ,

where the second equality follows from expanding the divergence of the product Fu. As V

0 the various integrals reduce to a product of V and the integrand evaluated at the center point of V. Divide both sides of the prior equation by V and take the limit as V

0to find:

limV0

1V

ddt

F(x,t)dVV = lim

V0

1V

F(x,t)t

+ F(x,t) u+ u ( )F(x,t)(

) * +

, - dV

V ,

limV0

1V

ddt

F(x,t)V + ...[ ] = limV0

1V

F(x,t)t

+ F(x,t) u+ u ( )F(x,t)'

( )

*

+ , V + ...

-

. /

0

1 2 , or

ddtF(x,t) + F(x,t) lim

V0

1V

ddt

V( ) = F(x,t)t

+ F(x,t) u+ u ( )F(x,t) ,

where the product rule for derivative has been used on product FV in [,]-braces on the left.

From (3.14) or Exercise 3.33:

limV0

1V

ddt

V( ) = u , so the second terms on both sides of the

last equation are equal and may be subtracted out leaving:

ddtF(x,t) = F(x,t)

t+ u ( )F(x,t) ,

and this is (3.5) when the identification

D Dt d dt is made.

fluid mechanics kundu cohen 6th edition solutions sm ch (3)

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