Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (7)

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

Exercise 7.1. a) Show that (7.7) solves (7.5) and leads to u = (U,V). b) Integrate (7.6) within circular area centered on (x´, y´) of radius

€

" r = (x − " x )2 + (y − " y )2 to show that (7.8) is a solution of (7.6). Solution 7.1. a) Start with

€

ψ = −Vx +Uy . Insert this into the Laplace equation:

€

∇2ψ =∂ 2

∂x 2+∂ 2

∂y 2%

& '

(

) * −Vx +Uy( ) = 0 .

The final equality follows because ψ is just a linear function so both of the indicated second derivatives are zero. The velocity field is found via differentiation.

€

u =∂ψ∂y,−∂ψ∂x

%

& '

(

) * =

∂(−Vx +Uy)∂y

,−∂(−Vx +Uy)∂x

%

& '

(

) * = U,V( ) .

b) First apply a simple shift transformation that places x´ = (x´, y´) at the origin of coordinates. Define these new coordinates by:

€

X = x − # x ,

€

Y = y − # y and set

€

" r = x − " x = X 2 + Y 2 . The gradient operator

€

∇XY in the shifted coordinates X = (X, Y) is the same as

€

∇ in the unshifted coordinates (x, y), so the (7.6) becomes:

€

∇XY2 ψ = −Γδ(x − ' x )δ(y − ' y ) = −Γδ(X)δ(Y ) .

Integrate this equation inside a circle of radius r´:

€

∇XY2 ψ

circular area∫∫ dA = ∇XYψ ⋅n

circle∫ d = −Γ δ(X)δ(Y )

x=− ) r 2−y 2

x= + ) r 2−y 2

∫y=− ) r

y= + ) r

∫ dXdY ,

where the first equality follows from Gauss' divergence theorem in two dimensions, and the double integral on the right side includes the location X = 0 (aka x = x´) so a contribution is collected from both delta functions. Thus, the double integral on the right side of this equation is unity. The dot product in the middle portion of the above equation simplifies to ∂ψ/∂r´ because n = er on the circle and

€

er ⋅ ∇XY = ∂ ∂ % r . Plus, in an unbounded uniform environment, there are no preferred directions so ψ = ψ(r´) alone (no angular dependence). Thus, the integrated field equation simplifies to

€

∂ψ∂ $ r %

& '

(

) *

θ = 0

2π

∫ $ r dθ = 2π $ r ∂ψ∂ $ r

= −Γ ,

which implies

€

∂ψ∂ $ r

= −Γ2π $ r

or

€

ψ = −Γ2πln( & r ) = −

Γ2πln (x − & x )2 + (y − & y )2 ,

and this is (7.8); thus, (7.8) is a solution of (7.6). Here, the constant from the final integration has been suppressed (set equal to zero) because it has no impact on the velocity field, which depends on derivatives of ψ.


Exercise 7.2. For two-dimensional ideal flow, show separately that: a)

€

∇ψ ⋅∇φ = 0 , b)

€

−∇ψ ×∇φ = u 2ez, c)

€

∇ψ2

= ∇φ2 , and d)

€

∇φ = −ez ×∇ψ . Solution 7.2. In terms of components, the velocity u = (u, v) = (∂ψ/∂y, –∂ψ/∂x) = (∂φ/∂x, ∂φ/∂y), and these equations can be used to complete the various parts of this problem.

a)

€

∇ψ ⋅∇φ =∂ψ∂x,∂ψ∂y

'

( )

*

+ , ⋅

∂φ∂x,∂φ∂y

'

( )

*

+ , = −v,u( ) ⋅ u,v( ) = −vu + uv = 0 .

b)

€

−∇ψ ×∇φ = −

ex ey ez∂ψ ∂x ∂ψ ∂y 0∂φ ∂x ∂φ ∂y 0

= −

ex ey ez−v u 0u v 0

= −ez −v2 − u2( ) = ez u

2

c)

€

∇ψ2

=∂ψ∂x%

& '

(

) * 2

+∂ψ∂y%

& '

(

) *

2

= (−v)2 + u2 = u2 + v 2 =∂φ∂x%

& '

(

) * 2

+∂φ∂y%

& '

(

) *

2

= ∇φ2

d)

€

−ez ×∇ψ = −

ex ey ez0 0 1

∂ψ ∂x ∂ψ ∂y 0

€

= −ex 0 − ∂ψ∂y

%

& '

(

) * − ey

∂ψ∂x

− 0%

& '

(

) * − ez 0 − 0( )

= ex∂ψ∂y%

& '

(

) * − ey

∂ψ∂x%

& '

(

) * = exu + eyv = ex

∂φ∂x%

& '

(

) * + ey

∂φ∂y%

& '

(

) * =∇φ


Exercise 7.3. a) Show that (7.14) solves (7.12) and leads to u = (U,V). b) Integrate (7.13) within circular area centered on (x´, y´) of radius !r = (x − !x )2 + (y− !y )2 to show that (7.15) is a solution of (7.13). c) For the flow described by (7.15), show that the volume flux (per unit depth into the page) = u ⋅n

C∫ ds computed from a closed contour C that encircles the point (x´, y´) is qs. Here n is

the outward normal on C and ds is a differential element of C. Solution 7.3. a) Start with

€

φ =Ux +Vy . Insert this into the Laplace equation:

€

∇2φ =∂ 2

∂x 2+∂ 2

∂y 2%

& '

(

) * Ux +Vy( ) = 0

The final equality follows because φ is a linear function so both of the indicated second derivatives are zero. The velocity field is found via differentiation.

€

u =∂φ∂x,∂φ∂y

$

% &

'

( ) =

∂(Ux +Vy)∂x

,∂(Ux +Vy)∂y

$

% &

'

( ) = U,V( ) .

b) First apply a simple shift transformation that places x´ = (x´, y´) at the origin of coordinates. Define these new coordinates by:

€

X = x − # x ,

€

Y = y − # y and set

€

" r = x − " x = X 2 + Y 2 . The gradient operator

€

∇XY in the shifted coordinates X = (X, Y) is the same as

€

∇ in the unshifted coordinates (x, y), so the (7.13) becomes:

∇XY2 φ = qsδ(x − #x )δ(y− #y ) = qsδ(X)δ(Y ) .

Integrate this equation inside a circle of radius r´:

€

∇XY2 φ

circular area∫∫ dA = ∇XYφ ⋅n

circle∫ d = m δ(X)δ(Y )

x=− ( r 2−y 2

x= + ( r 2−y 2

∫y=− ( r

y= + ( r

∫ dXdY ,

where the first equality follows from Gauss' divergence theorem in two dimensions, and the double integral on the right side includes the location X = 0 (aka x = x´) so a contribution is collected from both delta functions. Thus, the right-side double integration is unity. The dot product in the middle portion of the above equation simplifies to ∂φ/∂r´ because n = er´ on the circle and

€

e " r ⋅ ∇XY = ∂ ∂ " r . Plus, in a unbounded uniform environment, there are no preferred directions so φ = φ(r´) alone (no angular dependence). Thus, the integrated field equation simplifies to

∂φ∂ !r"

#$

%

&'

θ=0

2π

∫ !r dθ = 2π !r ∂φ∂ !r

= qs , which implies

∂φ∂ !r

=qs2π !r

or φ = qs2πln( !r ) = qs

2πln (x − !x )2 + (y− !y )2 ,

and this is (7.15). Thus, (7.15) is a solution of (7.13). Here, the constant from the final integration has been suppressed (set equal to zero) because it has no impact on the velocity field, which depends on derivatives of φ. c) For the flux integration, evaluate the dot product of u and n = er´:

€

u ⋅n =∇φ ⋅ e % r = ∂φ ∂ % r = u % r , where (7.21) has been used to reach the last equality. Thus,


u ⋅nC∫ ds = ur

0

2π

∫ rdθ = ∂φ∂ #r

#r dθ0

2π

∫ = 2π qs2π #r$

%&

'

() #r = qs ,

where ∂φ/∂r´ comes from the end of part b).


Exercise 7.4. Show that (7.1) reduces to

€

∂φ∂t

+12∇φ

2+pρ

= const. when the flow is described by

the velocity potential φ.

Solution 7.4. Start from (7.1),

€

∂u∂t

+ u ⋅ ∇( )u+ 1 ρ( )∇p = 0 , and use the vector identity (B3.9),

€

u ⋅ ∇( )u =∇ 12 u

2( ) −u× ∇ ×u( ) to replace the advective acceleration with a gradient and cross product term:

€

∂u∂t

+∇12u 2

$

% &

'

( ) −u× ∇ ×u( ) +

1ρ∇p = 0.

Use the definition of the potential,

€

∇φ = u , to eliminate u from this equation:

€

∂∂t∇φ +∇

12∇φ

2%

& '

(

) * −∇φ × ∇ ×∇φ( ) +

1ρ∇p = 0 .

The third term is zero because

€

∇ ×∇φ = 0 for any scalar function φ. The remaining terms can all be grouped under one gradient operation:

€

∇∂φ∂t

+12∇φ

2+pρ

&

' (

)

* + = 0 .

Thus, the grouping of terms in the parentheses can at most be a function of time. Let this function be B(t),

€

∂φ∂t

+12∇φ

2+pρ

= B(t) ,

Now define a mildly revised potential that includes B(t):

€

φ = # φ + B(τ ) − const.( )dτ0

t∫ , so that

€

∂φ∂t

=∂ $ φ ∂t

+ B(t) − const.

This does not change the velocity field because

€

∇ # φ =∇φ = u . Therefore, the remnant of (7.1) becomes:

€

∂ # φ ∂t

+ B(t) − const.+ 12∇ # φ

2+pρ

= B(t), or

€

∂ # φ ∂t

+12∇ # φ

2+pρ

= const.,

which is the requisite equation when the primes are dropped from φ.


Exercise 7.5. Consider the following two-dimensional Cartesian flow fields: (i) solid body rotation (SBR) at angular rate Ω about the origin: (u, v) = (–Ωy, Ωx); and (ii) uniform expansion (UE) at linear expansion rate Θ: (u, v) = (Θx, Θy). Here Ω, Θ, and the fluid density ρ are positive real constants and there is no body force. a) What is the stream function ψSBR(x,y) for solid body rotation? b) Is there a potential function φSBR(x,y) for solid body rotation? Specify it if it exists. c) What is the pressure, pSBR(x,y), in the solid body rotation flow when pSBR(0,0) = po? d) What is the potential function φUE(x,y) for uniform expansion? e) Is there a stream function ψUE(x,y) for uniform expansion? Specify it if it exists. f) Determine the pressure, pUE(x,y), in the uniform expansion flow when pUE(0,0) = po. Solution 7.5. a) Start with the stream-function-based velocity definitions,

u = ∂ψ ∂y = −Ωy and v = −∂ψ ∂x =Ωx , and integrate each second equality once to find: ψ = −(Ω 2)y2 + f1(x) and ψ = −(Ω 2)x2 + f2 (y) . The two equations are consistent when: ψ = −(Ω 2)(x2 + y2 )+ const. , and this result makes sense; the streamlines are circles. b) There is no potential function for SBR flow because it is everywhere rotational. c) Start with (u, v) = (–Ωy, Ωx) and use (7.1) simplified for steady flow where ∂/∂t = 0:

€

ui∂u j

∂xi= −

1ρ∂p∂x j

.

Evaluate the left side of this equation using the given velocity field:

ρ −Ωyex +Ωxey( ) ⋅ ex∂∂x

+ ey∂∂y

$

%&

'

() −Ωyex +Ωxey( ) = −ex

∂ p∂x

− ey∂ p∂y

.

This reduces to:

−ρΩ2xex − ρΩ2yey = −ex

∂ p∂x

− ey∂ p∂y

.

Considering each component separately,

ρΩ2x = ∂ p∂x

and ρΩ2y = ∂ p∂y

,

and integrate to find: pSBR (x, y) = ρΩ

2 x2 2( )+ f (y) and pSBR (x, y) = ρΩ2 y2 2( )+ g(x) , where

f and g are functions of integration. Combining these equations into one by appropriately choosing f and g, and then applying the boundary condition on x = y = 0 to determine the final constant, yields:

pSBR (x, y) = po +ρΩ2

2x2 + y2( ) , or in cylindrical coordinates

€

p(R) = po +ρΩ3

2

2R2.

This answer makes sense because the necessary centripetal acceleration of the rotating fluid must be provided by pressure forces. So, increasing pressure with increasing R is physically meaningful, even when viewed from a rotating dynamics point of view. This result can be checked with the swirling flow of a liquid in a cylindrical container (see Section 5.1). The height of the free surface is proportional to the pressure in the fluid, and the free surface of a liquid in solid body rotation is parabolic. This effect was utilized by astronomers to cast telescope mirror blanks on rotating tables in the early part of the 1900’s. The radial pressure gradient found


above is also used to separate constituents in fluid mixtures in a centrifuge, a technique widely employed in the production of weapons-grade uranium through the separation of isotopes of UF6 (a gas), and in the biological sciences to obtain nucleic material from cells. d) Start with the potential-function-based velocity definitions,

u = ∂φ ∂x =Θx and v = ∂φ ∂y =Θy , and integrate each equation once to find: φ = (Θ 2)x2 + f3(y) and φ = (Θ 2)y2 + f4 (x) . The two equations are consistent when: φ = (Θ 2)(x2 + y2 )+ const. e) There is no stream function for UE flow because it is everywhere not mass conserving in constant density flow. f) Use the momentum equation as in part c) or the Bernoulli equation between the origin and the point (x,y):

pUE (x, y)+12 ρ u2 + v2( ) = po + 0 , or pUE (x, y) = po −

12 ρΘ

2 x2 + y2( ) . This answer is, of course, unusual since the flow is everywhere non-mass-conserving.


Exercise 7.6. Determine u and v, and sketch streamlines for a) ψ = A(x2 – y2), and b) φ = A(x2 – y2). Solution 7.6. a) From the definition of the stream function:

€

u = ∂ψ ∂y = −2Ay ,

€

v = −∂ψ ∂x = −2Ax , and the streamlines are given by x2 – y2 = const. For large x and y, these streamlines asymptote to the lines y = ±x. Assuming positive A, the signs for u and v indicate that the flow is toward the origin along y = x in the first quadrant. Therefore, the streamlines look like the upper drawing to the right. b) From the definition of the

potential function:

€

u =∂φ∂x

= 2Ax ,

and

€

v =∂φ∂y

= −2Ay . Here the

streamlines are given by

€

dydx

=vu

=−2Ay2Ax

= −yx

which can be

integrated to find: ln(y) = –ln(x) + const., or

€

xy = const. For large x and y, these streamlines asymptote to the lines y = 0 or x = 0 (the coordinate axes). Assuming positive A, the signs of u and v indicate the flow is away from the origin along y = 0 (the x-axis) and toward the origin along x = 0 (the y-axis). Therefore, the streamlines look like the lower drawing to the right.

x

y

x

y


Exercise 7.7. For the following two-dimensional stream and potential functions, find the fluid velocity u = (u,v), and determine why these are or are not ideal flows.

a) ψ = A(x2 + y2) b) φ = A(x2 + y2)

Solution 7.7. a) From the definition of the stream function: u =∂ψ ∂y = 2Ay , v = −∂ψ ∂x = −2Ax , so ∂u/∂x + ∂v/∂y = 0 and the flow conserves mass. Here, the streamlines are given by x2 + y2 = const, which are circles. However, ∇2ψ = 4A = const. , so (7.4) implies that ωz = –4A = const. so the given stream function corresponds to solid body rotation at rate –2A about the z-axis. Thus, this flow is not an ideal flow because fluid elements rotate. b) From the definition of the potential function: u =∂φ ∂x = 2Ax , and v =∂φ ∂y = 2Ay , so ∂u/∂x + ∂v/∂y = 4A and this flow does not conserve mass. Here the streamlines are given by dydx

=vu=2Ax2Ay

=yx

, and this can be integrated to find: ln(y) = ln(x) + const., or y = (const.)x .

These are straight lines through the origin of coordinates. This potential function corresponds to uniform expansion in the x-y plane. This is not an ideal flow because it does not conserve mass.


Exercise 7.8. Assume ψ = ax3 + bx2y + cxy2 + dy3 where a, b, c, and d are constants; and determine two independent solutions to the Laplace equation. Sketch the streamlines for both flow fields. Solution 7.8. Determine the implications of

€

∇2ψ = 0 by evaluating derivatives:

€

∂ 2ψ∂x 2

= 6ax + 2by , and

€

∂ 2ψ∂y 2

= 2cx + 6dy , so

€

∂ 2ψ∂x 2

+∂ 2ψ∂y 2

= 0 = 6ax + 2by + 2cx + 6dy .

The first non-trivial solution ψ1 occurs when 6a + 2c = 0 or c = –3a, and b = d = 0:

€

ψ1 = a x 3 − 3xy 2( ) = ax x + 3y( ) x – 3y( ) . Thus, the streamlines corresponding to ψ1 = 0 are three lines specified by x = 0 and

€

y = ± x 3 . The velocity components determined from ψ1 are:

€

u = ∂ψ1 ∂y = −6axy , and

€

v = −∂ψ1 ∂x = −a(3x 2 − 3y 2) = −3a(x + y)(x − y). Assuming positive a, the streamlines for this flow field appear in the upper figure to the right. The second non-trivial solution ψ2 occurs when a = c = 0, and 2b + 6d = 0 or d = –3d:

€

ψ2 = d −3x 2y + y 3( ) = dy y + 3x( ) y – 3x( ) . Thus, the streamlines corresponding to ψ2 = 0 are three lines specified by y = 0 and

€

y = ±x 3 . The velocity components determined from ψ2 are:

€

u = ∂ψ2 ∂y = −3dx 2 + 3dy 2 = −3d(x + y)(x − y) , and

€

v = −∂ψ2 ∂x = −d(−6xy) = 6dxy . Assuming positive d, the streamlines for this flow field appear in the lower figure to the right. The two flow fields are identical when a = d and the streamlines are rotated by 30°.

x

y

x

y


Exercise 7.9. Repeat exercise 7.8. for ψ = ax4 + bx3y + cx2y2 + dxy3 + ey4 where a, b, c, d, and e are constants. Solution 7.9. Determine the implications of

€

∇2ψ = 0 by evaluating derivatives:

€

∂ 2ψ∂x 2

=12ax 2 + 6bxy + 2cy 2, and

€

∂ 2ψ∂y 2

= 2cx 2 + 6dxy +12ey 2, so

€

∂ 2ψ∂x 2

+∂ 2ψ∂y 2

= 0 =12ax 2 + 6bxy + 2c(y 2 + x 2) + 6dxy +12ey 2 .

Therefore, non-trivial solutions occur when 12a + 2c = 0 or c = –6a, 6b + 6d = 0 or b = –d, and 2c + 12e = 0 or c = –6e. Choosing a and then b as the free parameter, leads to:

€

ψ1 = a(x 4 − 6x 2y 2 + y 4 ) = a (x 2 − y 2)2 − 4x 2y 2( ) = a x 2 − y 2 + 2xy( ) x 2 − y 2 − 2xy( ) = a (x + y)2 − 2y 2( ) (x − y)2 − 2y 2( ) = a x + (1+ 2)y( ) x + (1− 2)y( ) x − (1− 2)y( ) x − (1+ 2)y( ),

and

€

ψ2 = b(x 3y − xy 3) = bxy(x + y)(x − y) . Consider each solution in turn. The streamlines corresponding to ψ1 = 0 are four lines specified all possible sign combinations of

€

y = ± x 1± 2( ). The velocity components determined from ψ1 are:

€

u = ∂ψ1 ∂y = −12ax 2y + 3y 3 , and

€

v = −∂ψ1 ∂x = −a(4x 3 −12xy 2). Assuming positive a, the streamlines for this flow field appear in the upper figure to the right.

The streamlines corresponding to ψ2 = 0 are four lines specified by x = 0, y = 0, y = x, and y = –x. The velocity components determined from ψ2 are:

€

u = ∂ψ2 ∂y = bx 3 − 3xy 2 , and

€

v = −∂ψ2 ∂x = −3ax 2y + y 3 . Assuming positive a, the streamlines for this flow field appear in the lower figure to the left.

x

y

x

y


Exercise 7.10. Without using complex variables, determine: a) The potential φ for an ideal vortex of strength Γ starting from (7.8) b) The stream function for an ideal point source of strength qs starting from (7.15) c) Is there any ambiguity in your answers to parts a) and b)? If so, does this ambiguity influence the fluid velocity? Solution 7.10. a) The starting point is the stream function for a two-dimensional point vortex of strength Γ located at x = x´ = (x´, y´):

€

ψ = −Γ2πln (x − & x )2 + (y − & y )2 .

The velocity components are:

€

u =∂ψ∂y

= −Γ2π

y − ' y (x − ' x )2 + (y − ' y )2

=∂φ∂x

, and

€

v = −∂ψ∂x

=Γ2π

x − ' x (x − ' x )2 + (y − ' y )2

=∂φ∂y

,

where the final equality in each case comes from the definition of the potential function. Integrate to find φ(x, y) using these final equalities to find

€

φ(x,y) = −Γ2π

y − & y (x − & x )2 + (y − & y )2

dx + f (y)∫ , and

€

φ(x,y) =Γ2π

x − & x (x − & x )2 + (y − & y )2

dy + g(x)∫ ,

where f(y) and g(x) are single-variable functions that appear because of the integrations. Rearrange the first integral, and create the integration variable tanθ = (y – y´)/(x – x´), noting that sec2θdθ = –[(y – y´)/(x – x´)2]dx.

€

φ(x,y) = −Γ2π

(y − & y ) (x − & x )2

1+ (y − & y ) (x − & x )[ ]2dx + f (y)∫ =

Γ2π

sec2θdθ1+ tan2θ∫ + f (y) =

Γ2π

θ + f (y) .

Use the same integration variable in the second integral to find:

€

φ(x,y) =Γ2π

sec2θdθ1+ tan2θ∫ + g(x) =

Γ2π

θ + g(x) .

The only way for the two results to be consistent is for f(y) = g(x) = const. Thus,

€

φ(x,y) =Γ2πtan−1 y − & y

x − & x '

( )

*

+ , + const.

b) For this part, the starting point is the potential for a two-dimensional point source of strength qs located at x = x´ = (x´, y´):

φ =qs2πln (x − "x )2 + (y− "y )2 .


u = ∂φ∂x

=qs2π

x − "x(x − "x )2 + (y− "y )2

=∂ψ∂y

, and v = ∂φ∂y

=qs2π

y− "y(x − "x )2 + (y− "y )2

= −∂ψ∂x

,

where the final equality in each case comes from the definition of the stream function. Integrate to find ψ(x, y) using these final equalities to find

ψ(x, y) = qs2π

x − "x(x − "x )2 + (y− "y )2

dy+ f (x)∫ , and ψ(x, y) = − qs2π

y− "y(x − "x )2 + (y− "y )2

dx + g(y)∫ ,

where f(x) and g(y) are single-variable functions that appear because of the integrations. These integrations are the same as in part a); thus using the same integration variable produces:

ψ(x, y) = qs2π

θ + f (x) = qs2π

θ + g(y) .


The only way for the two results to be consistent is for f(x) = g(y) = const, so

ψ(x, y) = qs2πtan−1 y− "y

x − "x#

$%

&

'(+ const.

c) In both cases, there is an undetermined constant in the answer. However, the fluid velocity field depends on derivatives of φ and ψ, so this constant does not influence the velocity field.


Exercise 7.11. Determine the stream function of a doublet starting from (7.29) and show that the streamlines are circles having centers on the y-axis that are tangent to x-axis at the origin. Solution 7.11. The potential for a doublet is:

€

φ =d2πcosθr

=d2π

xx 2 + y 2

,

so the velocity field is:

€

u =∂φ∂x

=d2π

∂∂x

xx 2 + y 2%

& '

(

) * =

d2π

1x 2 + y 2

−x

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

%

& '

(

) * =

d2π

y 2 − x 2

(x 2 + y 2)2%

& '

(

) * =

∂ψ∂y

, and

€

v =∂φ∂y

=d2π

∂∂y

xx 2 + y 2%

& '

(

) * =

d2π

−x

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

%

& '

(

) * =

d2π

−2xy(x 2 + y 2)2%

& '

(

) * = −

∂ψ∂x

,

where the final equalities for each component follow from the definition of the stream function. Integrate the v-equation using x2 + y2 = β and 2xdx = dβ to find:

€

ψ(x,y) =d2π

2xy(x 2 + y 2)2

dx∫ + f (y) =d2π

y dββ 2∫ + f (y) = −

d2π

yβ

+ f (y) = −d2π

yx 2 + y 2

+ f (y).

When this result is partial-differentiated with respect y, the result is:

€

∂∂yψ(x, y) = −

d2π

∂∂y

yx 2 + y 2 + f (y)&

' (

)

* +

= −d

2π1

x 2 + y 2 −y

(x 2 + y 2)2 (2y) + , f (y)&

' (

)

* + = −

d2π

x 2 − y 2

(x 2 + y 2)2 + , f (y)&

' (

)

* +

which matches the u-equation result above when f´(y) = 0. Thus, f = const. and the stream function is:

€

ψ(x,y) = −d2π

yx 2 + y 2

+ const.

Curves of constant ψ are streamlines, so rename const. as ψo, and rearrange the above equation to determine the streamline curves in a standard functional form.

€

ψ −ψo = −d2π

yx 2 + y 2

, or

€

x 2 + y 2 +d

2π ψ −ψo( )y +

d 2

16π 2 ψ −ψo( )2=

d 2

16π 2 ψ −ψo( )2,

which simplifies to:

€

x 2 + y +d

4π ψ −ψo( )

%

& '

(

) *

2

=d 2

16π 2 ψ −ψo( )2.

These are circles of radius a =

€

d 4π ψ −ψo( ) that are centered at y = –a, thus they are tangent to the origin, as shown in the drawing.

x

y


Exercise 7.12. Consider steady horizontal flow at speed U past a stationary source of strength qs located at the origin of coordinates in two dimensions, (7.30) or (7.31). To hold it in place, an external force per unit depth into the page, F, is applied to the source. a) Develop a dimensionless scaling law for F = |F|. b) Use a cylindrical control volume centered on the source with radius R and having depth B into the page, the steady ideal-flow momentum conservation equation for a control volume,

<Begin Equation> ρu u ⋅n( )

A*∫ dA = − pndA

A*∫ +F ,

</End Equation> and an appropriate Bernoulli equation to determine the magnitude and direction of F without using Blasius Theorem. c) Is the direction of F unusual in anyway? Explain it physically. Solution 7.12. a) There are only 4 parameters: F, qs, U, and ρ, and these span all three dimensions. Thus, there is only one dimensionless parameter, F/ρUqs, so it must be constant. This implies F = (const.)ρUqs, so rest of this problem merely involves determining the constant and the direction of F. b) It is convenient to use x-y and r-θ velocities expressed in polar coordinates. Using the potential

φ =Ux + qs2πln x2 + y2 =Urcosθ + qs

2πln r ,

and performing the requisite differentiations produces:

u = (u,v) = U +qs2πr

cosθ, qs2πr

sinθ!

"#

$

%& , and u = (ur,uθ ) = U cosθ + qs

2πr,−U sinθ

"

#$

%

&'

Again, choose the control volume to be a cylinder of radius R and span B centered on the point source. An evaluation of the conservation of mass equation merely validates that a source of strength m is present in the flow field. To find the force from the given CV momentum equation, first identify unit vectors and evaluate dot products. Here,

€

n = er so that

€

u ⋅n = ur . Therefore:

€

F = ρ(uex + vey )urBRdθθ = 0

θ = 2π

∫ + p(ex cosθ + ey sinθ)BRdθθ = 0

θ = 2π

∫

The pressure at r can be evaluated via a Bernoulli equation:

p∞ +12 ρU

2 = p+ 12 ρ u2 + v2( ) = p+ 1

2 ρ U 2 +Uqs cosθ

πr+

qs2

4π 2r2"

#$

%

&' , or p = p∞ −

ρUqs cosθ2πr

−ρqs

2

8π 2r2.

With this replacement, the x-component of the momentum equation is FxB= ρ U +

qs2πR

cosθ!

"#

$

%& U cosθ +

qs2πR

!

"#

$

%&Rdθ

θ=0

θ=2π

∫ + p∞ −ρUqs cosθ2πR

−ρqs

2

8π 2R2!

"#

$

%&cosθRdθ

θ=0

θ=2π

∫

The only integrand pieces that contribute involve even powers of like trigonometric functions. FxB= ρ

qsU2π

1+ cos2θ( )dθθ=0

θ=2π

∫ − ρqsU2π

cos2θ dθ =θ=0

θ=2π

∫ ρqsU2π

dθθ=0

θ=2π

∫ = +ρqsU


The sign of this result is a paradox. It implies that a downstream-pointing force (per unit depth) is applied to the fluid in the control volume. Thus, if this force were not present, the point source would move upstream! A point source in a uniform stream is an elementary rocket. Manipulations similar for the y-component of the momentum equation are similar to those above with the final finding that: Fy = 0. Interestingly, these results do not depend on the size of the control volume. NOTE: The force components must be resolved in a Cartesian coordinate system, since any net radial force on the CV is accounted for by fluid pressure, and any net angular "force" on the CV is a torque, not a force. c) The force points in the positive x-direction, and therefore prevents the source from moving upstream! Thus, an untethered point source is an elementary a rocket. This rocket-type character can be visually ascertained by considering a finite-length control volume that follows the dividing streamline and encloses the source as is shown in the drawing below.

x

y


Exercise 7.13. Repeat all three parts of Exercise 7.12 for steady ideal flow past a stationary irrotational vortex located at the origin when the control volume is centered on the vortex. The stream function for this flow is: ψ =Ursinθ − Γ 2π( ) ln(r) Solution 7.13. a) There are only 4 parameters: F, Γ, U, and ρ, and these span all three dimensions. Thus, there is only one dimensionless parameter, F/ρUΓ, so it must be constant. This implies F = (const.)ρUΓ, so rest of this problem merely involves determining the constant and the direction of F. b) For this problem, it is convenient to use a mixture of x-y and r-θ velocities expressed in polar coordinates. The potential for this flow field is

€

φ =Urcosθ + Γθ 2π . Performing the requisite differentiations produces:

€

u = (u,v) = U −Γ2πr

sinθ, Γ2πr

cosθ&

' (

)

* + , and

€

u = (ur,uθ ) = U cosθ,−U sinθ +Γ2πr

&

' (

)

* +

Choose the control volume to be a cylinder of radius R centered on the point vortex and having depth B into the page. The equation for conservation of mass is identically satisfied for all potential flows without sources. To find the force from the given CV momentum equation, first identify unit vectors and evaluate dot products. Here,

€

n = er so that

€

u ⋅n = ur . Therefore:

€

F = ρ(uex + vey )urBRdθθ = 0

θ = 2π

∫ + p(ex cosθ + ey sinθ)BRdθθ = 0

θ = 2π

∫

The pressure at r can be evaluated via a Bernoulli equation:

€

p∞ + 12 ρU

2 = p + 12 ρ u2 + v 2( ) = p + 1

2 ρ U2 −2UΓsinθ2πr

+Γ2

4π 2r2(

) *

+

, - , so

€

p = p∞ +ρUΓsinθ2πr

−ρΓ2

8π 2r2.

Evaluate this at r = R, and place it into the x-component of the momentum equation.

€

FxB

= ρ U −Γ2πR

sinθ'

( )

*

+ , U cosθRdθ

θ = 0

θ = 2π

∫ + p∞ +ρUΓsinθ2πR

−ρΓ2

8π 2R2'

( )

*

+ , cosθRdθ

θ = 0

θ = 2π

∫ = 0

In each case the various trigonometric functions produce a net zero for each term. The y-component of the equation is:

€

FyB

= ρΓ2πR

cosθ&

' (

)

* + U cosθRdθ

θ = 0

θ = 2π

∫ + p∞ +ρUΓsinθ2πR

−ρΓ2

8π 2R2&

' (

)

* + sinθRdθ

θ = 0

θ = 2π

∫ .

Evaluate the integrals. The non-trivial ones become:

€

FyB

= ρUΓ2π

cos2θdθθ = 0

θ = 2π

∫ + ρUΓ2π

sin2θdθθ = 0

θ = 2π

∫ = ρUΓ .

Hence, the force on control volume is directed along the positive y-axis (perpendicular to the uniform stream) for positive Γ. When R → 0, this force remains the same. Therefore, to keep the vortex stationary, an equal and opposite force must be produced by the hydrodynamic interaction of the point vortex with the incoming flow. This hydrodynamic force is called the lift (= L = –Fy), because it can support the weight of an object. The final result, which is true for all two-dimensional lifting bodies, is:

€

L B = −ρUΓ. The minus sign appears here because of the choice of the positive direction of circulation.


NOTE: The force components must be resolved in a Cartesian coordinate system, since any net radial force on the CV is accounted for by fluid pressure, and any net angular "force" on the CV leads to a torque, not a force. c) In this case, the flow upstream of the vortex is drawn downward and the flow downstream of the vortex is deflected upward. This asymmetry in the flow leads to the lift force.


Exercise 7.14. Use the principle of conservation of mass (4.5) and an appropriate control volume to show that maximum half thickness of the half body described by (7.30) or (7.31) is hmax = qs/2U. Solution 7.14. Place a control surface on the streamline that divides the fluid from the source from the fluid that originates upstream. The velocity u is:

u = (u,v) = U +qs2πr

cosθ, qs2πr

sinθ!

"#

$

%& ,

but this simplifies to u = (U, 0) far from the source (

€

r→∞). Therefore, far downstream of the source, the volume flux (per unit depth into the page) inside the CV must still be qs and its flow speed will be U. Therefore, conservation of mass implies:

ρqs = 2ρUhmax, or hmax = qs/2U.


Exercise 7.15. By integrating the surface pressure, show that the drag on a plane half-body (Figure 7.7) is zero. Solution 7.15. For the half-body produced by a free stream of speed U and source of strength m, the potential and velocity field are:

φ =Ux + qs2πln x2 + y2 =Urcosθ + qs

2πln r , and u = (ur,uθ ) = U cosθ + qs

2πr,−U sinθ

"

#$

%

&' .

The stream function is given by:

ψ =Ursinθ + qs2π

θ ,

so the streamline r(θ) that divides the fluid that originates upstream from the fluid that comes from the source is given by:

qs2=Ursinθ + qs

2πθ , or r(θ ) = qs (π −θ )

2πU sinθ.

The coefficient of pressure on the dividing streamline will be:

Cp(θ ) = 1− ur2 +uθ

2

U 2

"

#$

%

&'r(θ )

= 1−U 2 cos2θ + qsU πr( )cosθ + qs 2πr( )2

+U 2 sin2θ

U 2

"

#$$

%

&''r(θ )

= −qsπUr

cosθ − qs2

4π 2U 2r2

"

#$

%

&'r(θ )

= −qs

2πU2 cosθ

r+

qs2πU

1r2

"

#$

%

&'r(θ )

= − qs2πU

2 cosθqs (π −θ )

2πU sinθ + qs2πU

4π 2U 2 sin2θqs

2 (π −θ )2

"

#$

%

&'= − 2 cosθ sinθ

(π −θ )+

sin2θ(π −θ )2

"

#$

%

&'

The drag force on the body will be:

€

Drag = ex ⋅ −(p − p∞)ndAsurface∫ = −ex ⋅

12 ρU

2Cp (θ)nBdξθ = 0

θ = 2π∫ = − 1

2 ρU2B Cp (θ)ex ⋅ndξ0

2π∫ ,

where B is the spanwise dimension,

€

n = ∇ψ ∇ψ[ ]r(θ ) is the outward normal from the dividing streamline, and dξ is a contour increment along the dividing streamline. Fortunately, some of the intricacies of the geometry cancel out. First consider the dot product:

€

ex ⋅n = ex ⋅∇ψ∇ψ

=∂ψ ∂xur2 + uθ

2=

−v

uθ 1+ ur2 uθ

2( )= −

vuθ

1

1+ 1 r( )2 dr dθ( ).

Here v is the vertical velocity component, and the final equality follows from the streamline condition,

€

dr ur = rdθ uθ , in r-θ polar coordinates (see Exercise 3.8). Now develop dξ in r-θ polar coordinates:

€

dξ = (dr)2 + (rdθ)2 = rdθ 1+ 1 r( )2 dr dθ( ) . Therefore:

ex ⋅ndξ = −νuθ

1

1+ 1 r( )2 dr dθ( )rdθ 1+ 1 r( )2 dr dθ( ) = − ν

uθrdθ = − qs

2πUdθ .

where the final equality follows because v = qs 2πr( )sinθ , and uθ = –Usinθ. Thus, the drag integral becomes:


DragB

= 12 ρU

2 qs2πU

Cp(θ )dθ0

2π

∫ = −qsρU4π

sin(2θ )(π −θ )

+sin2θ(π −θ )2

#

$%

&

'(dθ

0

2π

∫ .

Change the integration variable to β = π – θ, and note that the integrand is not singular because of the behavior of the sine function for small argument:

DragB

= −qsρU4π

−sin(2β)

β+sin2 ββ 2

"

#$

%

&'dβ

−π

+π

∫ = −qsρU4π

−sin(2β)

β+12β 2

−cos(2β)2β 2

"

#$

%

&'dβ

−π

+π

∫ .

In the last expression, leave the first term alone, integrate the second term to get –1/2β, and integrate the third term by parts to reach:

DragB

= −qsρU4π

−12β

+12βcos(2β)

"

#$

%

&'−π

+π

+ −sin(2β)

β−12−2β

(

)*

+

,-sin(2β)

(

)*

+

,-dβ

−π

+π

∫/01

21

341

51= 0 ,

and this is the desired result.


Exercise 7.16. Ideal flow past a cylinder (7.33) is perturbed by adding a small vertical velocity without changing the orientation of the doublet:

€

ψ = −Uγx +Uy − Ua2yx 2 + y 2

= −Uγrcosθ +U r − a2

r&

' (

)

* + sinθ .

a) Show that the stagnation point locations are rs = a and θs = γ/2, π + γ/2 when γ << 1. b) Does this flow include a closed body? Solution 7.16. Use the polar-coordinate version of the stream function to find the radial (ur), and angular (uθ) velocity components.

€

−∂ψ∂r

= uθ =Uγ cosθ −U 1+a2

r2'

( )

*

+ , sinθ , and

€

1r∂ψ∂θ

= ur =Uγ sinθ +U 1− a2

r2'

( )

*

+ , cosθ .

At a stagnation point both velocity components will be zero, so

€

0 =Uγ cosθs −U sinθs −Ua2

rs2 sinθs , and

€

0 =Uγ sinθs +U cosθs −Ua2

rs2 cosθs,

where the subscript s denotes stagnation point variables. These are two non-linear equations in two unknowns. First eliminate θs by creating tanθs in each equation:

€

0 = γ − 1+a2

rs2

$

% &

'

( ) tanθs, and

€

0 = γ tanθs + 1− a2

rs2

%

& '

(

) * .

Now eliminate tanθs between these two equations, and solve for rs to find:

€

rsa

= 1+ γ 2[ ]−1 4 , and

€

tanθs =γ

1+ 1+ γ 2.

When γ << 1, these two results can be expanded:

€

rsa≅1− 1

4γ 2 + ..., and

€

tanθs ≅γ

1+1+ 12 γ

2 + ...=γ21− 14γ 2 + ...

&

' (

)

* + .

Thus, to lowest order, the two stagnation point locations are r = a, and θ = γ/2, π + γ/2. b) No, the flow no longer includes a closed body. The sketch below shows what happens.

x

y


Exercise 7.17. For the following flow fields (b, U, Q, and Γ are positive real constants), sketch streamlines. a)

€

ψ = b r cos θ 2( ) for |θ| < 180° b)

€

ψ =Uy + Γ 2π( ) ln x 2 + (y − b)2( ) − ln x 2 + (y + b)2( )[ ]

c)

€

φ = Q 2π( ) ln x 2 + (y − 2na)2( )n=−∞

n= +∞∑ for |y| < a.

Solution 7.17. There are several ways to attack this problem. The most direct is to take a few derivatives to determine the velocity components and then plot the results. However, detailed plots are not required here (the problem merely says sketch the streamlines); thus, flow directions can be ascertained by considering limiting values of the independent variables. In particular, for parts b) and c), very near a point vortex or a point source the streamlines will be circular or radial, respectively, because the 1/r–factor will cause the point vortex or point source to dominate the local flow field. Similarly, very far from a point vortex or a point source the streamlines will be little changed from those that would exist in the absence of the point source or point vortex. So, in parts b) and c), start with a few appropriate streamlines near the vortex or point source, and then join them up with a few horizontal parallel segments for |x| large that represent a parallel stream. In joining-up the streamlines from various parts of the sketches, the requirement that volume flux (per unit depth) remain constant between streamlines can be useful in determining where streamlines come closer together (increasing flow speed) and where the bend away from each other (decreasing flow speed). Here, the Bernoulli equation can also lend insight; streamlines will converge where pressure decreases and diverge where the pressure increases. There is an alternative drawing for part b) where the two stagnation points occur on the y-axis. a) for b > 0, b)

c)

y

x

y

x

y

x

y = a

y = -a


Exercise 7.18. Take a standard sheet of paper and cut it in half. Make a simple airfoil with one half and a cylinder with the other half that are approximately the same size as shown. a) If the cylinder and the airfoil are dropped from the same height at the same time with the airfoil pointed toward the ground in its most streamlined configuration, predict which one reaches the ground first. b) Stand on a chair and perform this experiment. What happens? Are your results repeatable? c) Can you explain what you observe?

Solution 7.18. a) If the airfoil remains aligned in the vertical direction, its drag is less than the cylinder. Therefore, it should reach the ground first. b) The airfoil that was tested for this solution manual, reached the ground first about 1 out of 10 ten times. In most cases, the airfoil refused to travel straight down and fluttered off sideways so that it reached the ground after the cylinder. c) Consider a frame of reference in which the flow about the airfoil is initially steady. A symmetric airfoil that is misaligned with respect to a uniform stream feels a lift force and a pitching moment. These are the result of an uneven pressure distribution on the airfoil surfaces caused by the fluid being forced to speed up unevenly as it travels around the pitched airfoil. This uneven velocity leads to an uneven pressure distribution through the Bernoulli equation. The lift force is proportional to the miss-alignment angle (commonly called the angle of attack). So, the airfoil is dynamically unstable, that is, any small perturbation in its angle of attack leads to lift forces that tend to increase the angle of attack. This phenomenon is one of the reasons that a rear stabilizer – with angle of attack opposite of that of the main wing – is part of the tail structure of most aircraft.

Tape


Exercise 7.19. Consider the following two-dimensional stream function composed of a uniform horizontal stream of speed U and two vortices of equal and opposite strength in (x,y)-Cartesian coordinates.

€

ψ(x,y) =Uy + Γ 2π( ) ln x 2 + (y − b)2 − Γ 2π( ) ln x 2 + (y + b)2 a) Simplify this stream function for the combined limit of

€

b→ 0 and

€

Γ→∞ when 2bΓ = C = constant to find:

€

ψ(x,y) =Uy 1− C 2πU( ) x 2 + y 2( )−1( )

b) Switch to (r,θ)-polar coordinates and find both components of the velocity using the simplified stream function. c) For the simplified stream function, determine where ur = 0. d) Sketch the streamlines for the simplified stream function, and describe this flow. Solution 7.19. a) The square roots may be simplified for the limit

€

b→ 0:

€

x 2 + (y ± b)2 = r 1+ (±2yb+ b2) r2 ≈ r 1± by r2( )

where

€

r2 = x 2 + y 2. Thus:

€

ln x 2 + (y ± b)2 ≈ ln(r) + ln 1± by r2( ) ≈ ln(r) ± by r2 , so the stream function becomes:

€

ψ(x,y) =Uy +Γ2π

ln(r) − byr2

&

' (

)

* + −

Γ2π

ln(r) +byr2

&

' (

)

* + =Uy −

Γbyπr2

=Uy 1− CUπr2

&

' (

)

* + .

b) In polar coordinates y = rsinθ so:

€

ψ(r,θ) =Ursinθ 1−C Uπr2( ) . Thus:

€

ur = 1 r( ) ∂ψ ∂θ( ) =U cosθ 1−C Uπr2( ) , and

€

uθ = − ∂ψ ∂r( ) = −U sinθ 1+ C Uπr2( )

c) ur = 0 when

€

θ = ±π 2 , or

€

r = C Uπ . d) This flow as two stagnation points at θ = 0 and π, and

€

r = C Uπ . In terms of

€

a = C Uπ ,

€

ur =U cosθ 1− a2 r2( ) and

€

uθ = −U sinθ 1+ a2 r2( ) , which is ideal 2D flow past a round cylinder. Thus the streamlines appear as in Figure 7.9.


Exercise 7.20. Graphically generate the streamline pattern for a plane half-body in the following manner. Take a source of strength qs = 200 m2/s and a uniform stream U = 10 m/s. Draw radial streamlines from the source at equal intervals of Δθ = π/10, with the corresponding stream function interval : Δψsource = (qs/2π)Δθ = 10 m2/s. Now draw streamlines of the uniform flow with the same interval, that is, Δψstream = UΔψ = 10 m2/s. This requires Δy = 1 m, which can be plotted assuming a linear scale of 1 cm = 1m. Connect points of equal ψ = ψ source + ψstream to display the flow pattern. Solution 7.20. Let subscripts "1" and "2" represent the source and the free stream, respectively.

ψ1 =qs2π

θ → Δψ1 =qs2π

Δθ =qs20

=10m2 / s , and

€

ψ2 =Uy → Δψ2 =UΔy =10m2 /s.

With the above intervals for the stream functions, draw the streamlines for the source and the uniform stream by connecting values of constant total ψ = ψ source + ψstream. The drawing below shows this construction where the heavier lines are the streamlines.

The radial lines from the point-source location are 18° apart. The point-source stream-function values are listed around the edge of the drawing, while the free-stream function values are listed at or near the downstream edge of the drawing.

!1 = 100

!1 = 90

!1 = 80

!1 = 70 !

1 = 60

!1 = 50

!1 = 20

!1 = 40 !

1 = 30

!1 = 10

!1 = 0!

2 = 0

!2 = 10

!2 = 20

30

40

60

70

80

90

100


Exercise 7.21. Consider the two-dimensional steady flow formed by combining a uniform stream of speed U in the positive x-direction, a source of strength qs > 0 at (x,y) = (–a, 0), and a sink of strength qs at (x,y) = (+a, 0) where a > 0. The pressure far upstream of the origin is p∞. a) Write down the velocity potential and the stream function for this flow field. b) What are the coordinates of the stagnation points, marked by s in the figure? c) Determine the pressure in this flow field along the y-axis. d) There is a closed streamline in this flow that defines a Rankine body. Obtain a transcendental algebraic equation for this streamline, and show that the half-width, h, of the body in the y-direction is given by: h/a = cot(πUh/qs). (The introduction of angles may be useful here.) Solution 7.21. a) For both ideal flow functions, there will three terms: the free stream, source at x = –a, and the sink at x = +a.

Velocity potential: φ(x, y) =Ux + qs2πln (x + a)2 + y2 − qs

2πln (x − a)2 + y2

Stream function: ψ =Uy+ qs2π

θ1 −qs2π

θ2 =?Uy+ qs

2πtan−1 y

x + a"

#$

%

&'−

qs2πtan−1 y

x − a"

#$

%

&'

where θ1 and θ2 are angles determined from the x-axis having vertices that lie at x = –a and +a, respectively. Later on the correct branches of the inverse tangent function will have to be chosen. b) First compute the velocities and look for places where both components are zero. Cartesian velocity components are:

u = ∂φ∂x

=U +qs2π

x + a(x + a)2 + y2!

"#

$

%&−

qs2π

x − a(x − a)2 + y2!

"#

$

%&

v = ∂φ∂y

=qs2π

y(x + a)2 + y2!

"#

$

%&−

qs2π

y(x − a)2 + y2!

"#

$

%&

By examination, v is zero on x = 0 (the y-axis) and on y = 0 (the x-axis). However, u is non-zero on x = 0 (the y-axis). Thus, the stagnation points must occur on y = 0 (the x-axis), when

u(x, 0) = 0 =U +qs2π

x + a(x + a)2 + 02!

"#

$

%&−

qs2π

x − a(x − a)2 + 02!

"#

$

%&=U +

qs2π

1x + a!

"#

$

%&−

qs2π

1x − a!

"#

$

%&

Solve for x using the final equality.

0 =U +qs2π

x − ax2 − a2"

#$

%

&'−

qs2π

x + ax2 − a2"

#$

%

&'=U −

qsaπ

1x2 − a2"

#$

%

&' , or x = ± a2 + qsa

πU.

Thus, the two stagnation points are located at: (x, y) = ± a2 + qsaπU

, 0!

"#

$

%& .

x!

y!

y = h!

a! a!

U!

s! s! +qs! –qs!


c) Using the information above, (u,v)x=0 = U +qsπ

aa2 + y2!

"#

$

%&, 0

!

"#

$

%& on the y-axis. Thus, use the

steady ideal flow Bernoulli equation,

€

p(x,y) + 12 ρ u2 + v 2( ) = p∞ + 1

2 ρU2 , with the above

replacements for u and v.

p(x, y)+ 12 ρ U +

qsπ

aa2 + y2!

"#

$

%&

!

"#

$

%&

2

= p∞ +12 ρU

2

Simplify this expression to find:

p(0, y)− p∞ = +12 ρU

2 − 12 ρ U +

qsπ

aa2 + y2#

$%

&

'(

#

$%

&

'(

2

= −ρqsUπ

aa2 + y2#

$%

&

'(−

ρqs2

2π 2a

a2 + y2#

$%

&

'(

2

.

d) The streamline that lies on the x-axis divides to become the closed streamline. Upstream of the source and sink, all the angles are π. Thus the stream-function constant, can be evaluated

ψ =Ursinθ + qs2π

θ2 −qs2π

θ1 =Ursinπ +qs2π

π −qs2π

π = 0

So the equations for the shape of the Rankine body in polar and Cartesian coordinates are:

0 =Ursinθ + qs2π

θ1 −θ2( ) , or 0 =Uy+ qs2πtan−1 y

x + a"

#$

%

&'−

qs2πtan−1 y

x − a"

#$

%

&' ,

where the branches of the inverse tangent function must be chosen correctly. To find an equation for the half width, h, of the body, a combination of the above formulae is needed. From the symmetry of the flow, the widest part of the Rankine body will occur at x = 0, so the body contour will cross the y-axis at (0,±h). Consider the point (0,h). The angles associated with this point are:

€

θ = π 2 ,

€

θ1 = tan−1 h a( ) , and

€

θ2 = π − tan−1 h a( ) , where the usual range, –π/2 to + π/2, of the inverse tangent has been chosen. Thus, the polar-coordinate body-shape equation produces:

0 =Uh+ qs2π

2 tan−1 ha"

#$%

&'−π

"

#$

%

&' , or πUh

qs=π2− tan−1 h

a"

#$%

&' , or

Now take the cotangent of both sides of the final equation and use the trigonometric identity:

€

cot π2− z

$

% &

'

( ) = tan(z) to find: cot πUh

qs

!

"#

$

%&= cot

π2− tan−1 h

a!

"#$

%&

!

"#

$

%&= tan tan−1

ha!

"#$

%&

!

"#

$

%&=

ha

, and the two

ends of this extended equality are the desired result.


Exercise 7.22. A stationary ideal two-dimensional vortex with clockwise circulation Γ is located at point (0, a), above a flat plate. The plate coincides with the x-axis. A uniform stream U directed along the x-axis flows past the vortex. a) Sketch the flow pattern and show that it represents the flow over an oval-shaped body when Γ/πa > U. [Hint: Introduce the image vortex and locate the two stagnation points on the x-axis.] b) If the pressure far from the origin just above the x-axis is p∞ show that the pressure p at any

location on the plate is: p∞ − p =ρΓ2a2

2π 2 (x2 + a2 )2−

ρUΓaπ (x2 + a2 )

.

c) Using the result of part b), show that the total upward force F on the plate per unit depth into the page is F = –ρUΓ + ρΓ2/4πa when the pressure everywhere below the plate is p∞. Solution 7.22. a) The stream function for this flow field is:

€

ψ =Uy + Γ 2π( ) ln x 2 + (y − a)2( ) − ln x 2 + (y + a)2( )[ ] ,

where the image vortex is represented by the second natural log function. For y = 0 and Γ/πa > U the flow field looks like:

b) The horizontal velocity on y = 0 is:

€

u(x,0) =∂ψ∂y$

% &

'

( ) y= 0

= U +Γ2π

y − ax 2 + (y − a)2

−Γ2π

y + ax 2 + (y + a)2

$

% &

'

( ) y= 0

=U −Γπ

ax 2 + a2

.

The Bernoulli equation evaluated on y = 0 determines the pressure p on the plate:

€

p∞ +12ρU 2 = p +

12ρ U −

Γπ

ax 2 + a2

'

( )

*

+ , 2

→ p − p∞ =12ρ 2U Γ

πa

x 2 + a2−Γ2

π 2a2

x 2 + a2( )2'

(

) )

*

+

, , .

This pressure difference can be mildly simplified to:

€

p − p∞ =ρUΓa

π (x 2 + a2)−

ρΓ2a2

2π 2(x 2 + a2)2.

c) The net vertical force F (positive upward) on the plate (per unit depth into the page) is:

€

F = (p∞ − p)dx−∞

+∞

∫ = −ρUΓa

π (x 2 + a2)+

ρΓ2a2

2π 2(x 2 + a2)2(

) *

+

, - dx

−∞

+∞

∫ .

The integrals can be evaluated with the change of variables x = atanβ, with dx = –asec2β dβ.

€

F = −ρUΓa2

πa2sec2 βdβ1+ tan2 β−π 2

+π 2

∫ +ρΓ2a3

2π 2a4sec2 βdβ

(1+ tan2 β)2−π 2

+π 2

∫ = −ρUΓπ

dβ−π 2

+π 2

∫ +ρΓ2

2π 2acos2 βdβ

−π 2

+π 2

∫

=ρUΓπ

(π ) +ρΓ2

2π 2aπ2(

) *

+

, - =

ρΓ2

4πa− ρUΓ.

y

x


Exercise 7.23. Consider plane flow around a circular cylinder. Use the complex potential and Blasius theorem (7.60) to show that the drag is zero and the lift is L = ρUΓ. (In Section 7.3, these results were obtained by integrating the surface pressure distribution.) Solution 7.23. The complex potential and its derivative for ideal flow past a circular cylinder with circulation –Γ are:

€

w =U z +a2

z"

# $

%

& ' +

iΓ2πlog z , and

€

dwdz

=U 1− a2

z2#

$ %

&

' ( +

iΓ2πz

.

Use these in Blasius' theorem (7.60):

€

D− iL =12iρ dw

dz$

% &

'

( )

2

dz∫ =12iρ U 1− a

2

z2

$

% &

'

( ) +

iΓ2πz

$

% &

'

( )

2

dz∫

=12iρ U 2 1− a

2

z2

$

% &

'

( )

2

+iΓπzU 1− a

2

z2

$

% &

'

( ) −

Γ2

4π 2z2

$

% & &

'

( ) )

2

dz∫ =12iρ iΓ

πzUdz∫

=12iρ iΓπU(2πi) = −iρUΓ.

where the last two equalities follow from the residue theorem. Thus, equating real and imaginary parts produces:

D = 0, and L = ρUΓ.


Exercise 7.24. For the doublet flow described by (7.29) and sketched in Figure 7.6, show u < 0 for y < x and u > 0 for y > x. Also, show that v < 0 in the first quadrant and v > 0 in the second quadrant. Solution 7.24. Start from (7.30)

€

φ =d2πcosθr

=d2π

xx 2 + y 2

.

The velocity field is obtained by direct differentiation.

€

u =∂φ∂x

=d2π

∂∂x

xx 2 + y 2%

& '

(

) * =

d2π

1x 2 + y 2

−x

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

%

& '

(

) * =

d2π

y 2 − x 2

(x 2 + y 2)2%

& '

(

) * , and

€

v =∂φ∂y

=d2π

∂∂y

xx 2 + y 2%

& '

(

) * =

d2π

−x

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

%

& '

(

) * =

d2π

−2xy(x 2 + y 2)2%

& '

(

) * ,

Given the numerator of the final expression for u, clearly u < 0 for y < x and u > 0 for y > x. Similarly, given the numerator of the final expression for v, it will be negative when x and y have the same sign (first and third quadrants) and it will be positive when x and y have opposite signs (second and fourth quadrants).


Exercise 7.25. Hurricane winds blow over a Quonset hut, that is, a long half-circular cylindrical cross-section building, 6 m in diameter. If the velocity far upstream is U∞ = 40 m/s and p∞ = 1.003 × 105 N/m, ρ∞ = 1.23 kg/m3, find the force per unit depth on the building, assuming the pressure inside the hut is a) p∞, and b) stagnation pressure, p∞ +

12 ρ∞U∞

2 . Solution 7.25. If the flow over the hut is ideal, it can be modeled using the potential for flow past a cylinder without circulation, and compute the velocity components:

€

φ =U r +a2

r#

$ %

&

' ( cosθ , and

€

ur =∂φ∂r

=U 1− a2

r2%

& '

(

) * cosθ &

€

uθ =1r∂φ∂θ

= −U 1+a2

r2&

' (

)

* + sinθ .

The ideal flow pressure ps on the surface of the hut (r = a) can be determined from the steady Bernoulli equation:

€

p∞ +12ρU 2 = ps +

12ρ ur

2 + uθ2( )r= a → ps − p∞ =

12ρ U 2 − uθ

2[ ]r= a( ) =12ρU 2 1− 4sin2θ( ),

where

€

uθ[ ]r= a = −2U sinθ . a) For ideal flow there is no drag so Fx = 0. When the pressure inside the hut is p∞, the vertical force Fy (per unit depth) on the hut is

€

Fy = (p∞ − ps)ey ⋅n0

π

∫ adθ = −12ρU 2a 1− 4sin2θ( )sinθ

0

π

∫ dθ = −12ρU 2a −3 + 4 cos2θ( )sinθ

0

π

∫ dθ

= −12ρU 2a 3cosθ − 4

3cos3θ

)

* + ,

- . 0

π

= −12ρU 2a −6 +

83

)

* + ,

- . =

53ρU 2a.

For ρ = 1.23 kg/m3, U = 40 m/s, and a = 3m, then Fy = 9.84 kN/m. b) Again, for ideal flow there is no drag so Fx = 0. When the pressure inside the hut is

€

p∞ + 12 ρ∞U∞

2 , the vertical force Fy (per unit depth) on the hut is

€

Fy = (p∞ + 12 ρU

2 − ps)ey ⋅n0

π

∫ adθ =12ρU 2a 4 sin2θ( )sinθ

0

π

∫ dθ = 2ρU 2a 1− cos2θ( )sinθ0

π

∫ dθ

= 2ρU 2a −cosθ +13

cos3θ)

* + ,

- . 0

π

= 2ρU 2a 2 − 23

)

* + ,

- . =

83ρU 2a.

For ρ = 1.23 kg/m3, U = 40 m/s, and a = 3m, then Fy = 15.7 kN/m.


Exercise 7.26. In a two-dimensional ideal flow, a source of strength qs is located a meters above an infinite plane. Find the fluid velocity on the plane, the pressure on the plane, and the reaction force on the plane assuming constant pressure p∞ below the plane. Solution 7.26. The potential function for this flow field is:

φ = qs 2π( ) ln x2 + (y− a)2( )+ ln x2 + (y+ a)2( )"#$

%&',


u(x, y) = ∂φ∂x

=qs2π

xx2 + (y− a)2

+x

x2 + (y+ a)2"

#$

%

&' , and

v(x, y) = ∂φ∂y

=qs2π

y− ax2 + (y− a)2

+y+ a

x2 + (y+ a)2"

#$

%

&' .

On y = 0 these become:

u(x, 0) = qsπ

xx2 + a2!

"#$

%& , and

€

v(x,0) = 0 .

The pressure on the plane pp can be determined from the steady flow Bernoulli equation:

p∞ = pp +12ρqsπ

xx2 + a2

"

#$

%

&'2

→ p∞ − pp =12ρqs2

π 2x2

(x2 + a2 )2.

The net vertical force F (positive upward) on the plate (per unit depth into the page) will be:

F = (p∞ − pp )dx−∞

+∞

∫ =12ρqs2

π 2x2

(x2 + a2 )2dx

−∞

+∞

∫ =12ρqs2

π 21

x2 + a2−

a2

(x2 + a2 )2$

%&

'

()dx

−∞

+∞

∫ ,

where the final equality follows from a partial fractions decomposition of the integrand. The integrals can be evaluated with the change of variables x = atanβ, with dx = –asec2β dβ.

F = 12ρqs2

π 2asec2 β1+ tan2 β

−sec2 β

(1+ tan2 β)2"

#$

%

&'dβ

−π 2

+π 2

∫ =12ρqs2

π 2aπ −

π2

"

#$

%

&'=14ρqs2

πa.

Interestingly, the force on the plane is positive, independent of the sign of qs.


Exercise 7.27. Consider a two-dimensional ideal flow over a circular cylinder of radius r = a with axis coincident with a right angle corner, as shown in the figure. Assuming that ψ = Axy (with A = constant) when the cylinder is absent, solve for the stream function and velocity components. Solution 7.27. When the cylinder is present, it will only effect the flow near the origin, so the stream function far from the origin will be ψo = Axy. And, the field equation for the stream function,

€

∇2ψ = 0 , is linear, so a solution in the form of a simple superposition can be sought:

€

ψ(x,y) =ψo +ψ1 = Axy +ψ1 = Ar2 sinθ cosθ +ψ1(r,θ) = A 2( )r2 sin(2θ) + R(r)Θ(θ), where ψ1 is the modification of the stream function necessary near the origin, x = rcosθ, y = rsinθ, and the further assumption that ψ1 can be found via separation of variables in polar coordinates has been made. Placing this two term trial solution into the field equation produces:

€

0 =1r∂∂r

r ∂ψ1∂r

$

% &

'

( ) +

1r2∂ 2ψ1∂θ 2

, (&)

and the boundary condition on ψ1 is determined from:

€

1r∂ψ∂θ

%

& ' (

) * r= a=1r∂ψo

∂θ+1r∂ψ1∂θ

%

& ' (

) * r= a= ur(a,θ) = 0 .

For

€

ψo = A 2( )r2 sin(2θ) , this implies:

€

Aacos(2θ) +1a∂ψ1∂θ

%

& ' (

) * r= a= 0 , or

€

∂ψ1∂θ

%

& ' (

) * r= a= −Aa2 cos(2θ) → ψ1(a,θ) = −

Aa2

2sin(2θ).

Thus, the form Θ(θ) = sin(2θ) has been determined so that

€

ψ1 = R(r)sin(2θ), and the boundary condition implies

€

R(a) = −Aa2 2. Placing the trial solution for ψ1 into (&) produces:

€

0 =1rddr

r dRdr

"

# $

%

& ' −

4r2R

)

* +

,

- . sin(2θ) ,

where the partial derivatives have been changed total derivatives because R(r) only depends on the variable r. This equation is equi-dimensional and therefore has power law solutions of the form R = Brm. Substituting in this solution form leads to the algebraic equation:

€

m2rm−2 − 4rm−2= 0 , which implies m = ±2. The positive root reproduces ψo, so the negative root should be chosen here. The boundary condition allows B to be evaluated:

€

R(a) = B r2[ ]r= a = B a2 = −Aa2 2 → B = −Aa4 2 . Therefore:

€

ψ(x,y) =ψo +ψ1 =Aa2

2r2

a2−a2

r2$

% &

'

( ) sin(2θ), and

€

ur =1r∂ψ∂θ

= Aa2 ra2−a2

r3&

' (

)

* + cos(2θ) and

€

uθ = −∂ψ∂r

= Aa2 ra2

+a2

r3&

' (

)

* + sin(2θ).

To recover the flow directions shown in the drawing for this exercise, the constant A must be negative.


Exercise 7.28. Consider the following two-dimensional velocity potential consisting of two sources and one sink, all of strength qs:

<Begin Equation> φ(x, y) = qs 2π( ) ln (x − b)2 + y2 + ln (x − a2 b)2 + y2 − ln x2 + y2( ) .

</End Equation> Here a and b are positive constants and b > a. a) Determine the locations of the two stagnation points in this flow field. b) Sketch the streamlines in this flow field. c) Show that the closed streamline in this flow is given by x2 + y2 = a2. Solution 7.28. a) Here φ only depends on y2 when x = 0. Thus,

€

v = ∂φ ∂y = 0 on y = 0, the x-axis, so finding the stagnation points means, finding the x-axis locations where

€

u = ∂φ ∂x = 0 ; ∂φ∂x!

"#$

%&y=0=qs2π

x − b(x − b)2 + y2

+x − a2 b

(x − a2 b)2 + y2−

xx2 + y2

!

"#

$

%&y=0

=qs2π

1x − b

+1

x − a2 b−1x

(

)*

+

,-= 0

Working with the final equality produces:

€

x(x − a2 b) + x(x − b) − (x − b)(x − a2 b) = 0, which simplifies to:

€

x 2 − a2 = 0 . Thus, the two stagnation points are at (±a,0). b) The two sources and the sink all lie on the x-axis with the sink at the origin. The stagnation points lie between the two sources at x = +a and to left of the origin at x = –a. A closed body is obtained, and it is shown in part c) that it is circular. To allow labeling, the streamlines are as shown for y ≥ 0 only. They are symmetric about y = 0. c) First convert everything to the usual polar coordinates, and define:

€

r1 ≡ (x − b)2 + y 2 = (rcosθ − b)2 + r2 sin2θ = r2 − 2brcosθ + b2 , and

€

r2 ≡ (x − a2 b)2 + y 2 = (rcosθ − a2 b)2 + r2 sin2θ = r2 − 2(a2 b)rcosθ + (a2 b)2 . Thus, the potential can be written:

φ = qs 2π( ) ln r1 + ln r2 − ln r( ) , so the radial velocity is:

ur =∂φ∂r

=qs2π

r − 2brcosθr12 +

r − 2(a2 b)rcosθr22 −

1r

"

#$

%

&' ,

where the chair rule

€

∂∂rln(ri) =

∂ ln(ri)∂ri

∂ri∂r

to reach the form shown. To find the location where ur

= 0, the above equation implies:

€

rr22 r − bcosθ( ) + rr1

2 r − (a2 b)cosθ( ) − r12r22 = 0 , or in expanded form:

x

y

–a +a +b +a2/b


€

r2 − 2(a2 b)rcosθ + (a2 b)2( ) r2 − brcosθ( ) + r2 − 2brcosθ + b2( ) r2 − (a2 b)rcosθ( ) − r2 − 2brcosθ + b2( ) r2 − 2(a2 b)rcosθ + (a2 b)2( ) = 0.

Patience with the multiplications produces at total of 21 terms:

€

r4 − br3 cosθ − 2(a2 b)r3 cosθ + 2a2r2 cos2θ + (a4 b2)r2 − (a4 b)rcosθr4 − (a2 b)r3 cosθ − 2br3 cosθ + 2a2r2 cos2θ + b2r2 − a2brcosθ−r4 + 2br3 cosθ − b2r2 + 2(a2 b)r3 cosθ − 4a2r2 cos2θ + 2a2brcosθ−(a4 b2)r2 + 2(a4 b)rcosθ − a4 = 0.

Group like terms according to the power of the cosine starting with cos0θ:

€

r4 + (a4 b2)r2 + r4 + b2r2 − r4 − b2r2 − (a4 b2)r2 − a4[ ] +

−br3 − 2(a2 b)r3 − (a4 b)r − (a2 b)r3 − 2br3 − a2br + 2br3 + 2(a2 b)r3 + 2a2br + 2(a4 b)r[ ]cosθ +

2a2r2 + 2a2r2 − 4a2r2[ ]cos2θ = 0

Cancel the equal and opposite terms:

€

r4 − a4[ ] + −br3 − (a2 b)r3 + a2br + (a4 b)r[ ]cosθ = 0 . Rearrange and factor:

€

r4 − a4 + − b + (a2 b)( )r3 + a2 b + (a2 b)( )r[ ]cosθ = 0.

€

r4 − a4 + b + (a2 b)( ) −r2 + a2[ ]rcosθ = 0 .

€

r2 − a2( ) r2 + a2( ) − b + (a2 b)( ) r2 − a2[ ]rcosθ = 0.

€

r2 − a2( ) r2 + a2 − b + (a2 b)( )rcosθ[ ] = 0 . The final equality shows that ur = 0 at r = a.


Exercise 7.29. Without using complex variables, derive the results of the Kutta–Zhukhovsky lift theorem (7.62) for steady two-dimensional irrotational constant-density flow past an arbitrary-cross-section object by considering the clam-shell control volume (shown as a dashed line) in the limit as

€

r→∞. Here A1 is a large circular contour, A2 follows the object’s cross section contour, and A3 connects A1 and A2. Let p∞ and Uex be the pressure and flow velocity far from the origin of coordinates, and denote the flow extent perpendicular to the x-y plane by B. Solution 7.29. In ideal flow, the only surface forces are pressure forces. Therefore, the drag (D) and lift (L) forces on the body are defined by:

€

Dex + Ley = − p − p∞( )n2dAA2

∫ = p − p∞( )ndAA2

∫ , (a)

where the free stream velocity is U = Uex, and n is the outward normal on the total control volume; thus it points into the body on surface A2 so the usual minus sign is missing from the final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure integration because

€

p∞ndAclosed surface∫ = 0 .

The starting point for the derivation is the integral form of the steady ideal flow momentum equation applied to the entire clam-shell control volume:

€

ρu(u ⋅n)dAA1 +A2 +A3

∫ = − p − p∞( )ndAA1 +A2 +A3

∫

with the implied limit that the separation between the surfaces of A3 goes to zero. When this limit is taken, the net contribution is zero because the integrands of the two surfaces of A3 will be identical while the normal vectors will point in opposite directions; thus the integrations on the upper and lower surfaces of A3 cancel out. This leaves:

€

ρu(u ⋅n)dAA1

∫ + ρu(u ⋅n)dAA2

∫ = ρu(u ⋅n)dAA1

∫ + 0 = − p − p∞( )ndAA1 +A2

∫ ,

where the first equality follows because

€

u ⋅n = 0 on the solid surface A2. Now substitute in (a) for the pressure integration over A2, and rearrange:

€

ρu(u ⋅n)dAA1

∫ = − p − p∞( )ndAA1

∫ − Dex + Ley( ) or

€

Dex + Ley = − p − p∞( )ndAA1

∫ − ρu(u ⋅n)dAA1

∫ .

Thus, the drag and lift on the body can be obtained from integrals over a circular surface (A1) that is distant from the body. For this circular surface, the area element will be dA = Brdθ, and n = er = excosθ + eysinθ, so the last equation can be written in component form:

€

DB

= − pcosθ + ρuur( )0

2π

∫ rdθ and

€

LB

= − psinθ + ρvur( )0

2π

∫ rdθ .

where u = (u, v) = (ur, uθ) in Cartesian and polar coordinates, respectively. When r is large the potential for the flow field can be expanded in inverse powers of r:

φ =Urcosθ + qs2πln(r)− Γ

2πθ +

d cosθ2πr

+...


where qs is the net source strength, Γ is the total clockwise circulation, d is the total dipole strength, etc. for the flow around the body. For a closed body, qs must be zero. Thus, the velocities on A1 at large r are

€

ur =∂φ∂r

=U cosθ − dcosθ2πr2

+ ... ,

€

u =∂φ∂x

=U +Γ2πsinθr

+ ... , and

€

v =∂φ∂y

= −Γ2πcosθr

+ ... .

Using these results, the Bernoulli equation allows the pressure p to be determined at large r:

€

p +12ρ u2 + v 2( ) = p∞ +

12ρU 2 → p − p∞ =

12ρ U 2 − u2 − v 2( ) , or

€

p − p∞ =12ρ U 2 −U 2 −

ΓUπ

sinθr

−Γ2

4π 2sin2θr2 + ...− Γ2

4π 2cos2θr2 + ...

(

) *

+

, -

=12ρ −

ΓUπ

sinθr

−Γ2

4π 2r2 + ...(

) *

+

, - .

Thus, the drag integral becomes:

€

DB

= −limr→∞

12ρ −

ΓUπsinθr

−Γ2

4π 2r2+ ...

)

* +

,

- . cosθ + ρ U +

Γ2πsinθr

+ ...)

* +

,

- . U cosθ −

dcosθ2πr2

+ ...)

* +

,

- .

)

* +

,

- .

0

2π

∫ rdθ .

where the limit

€

r→∞ is now explicitly written. Collect terms under the integrand according to their power of r:

€

DB

= −limr→∞

ρU 2rcosθ − ρΓU2π

sinθ cosθ +ρΓU2π

sinθ cosθ − ρΓ2 cosθ8π 2r

−ρUdcosθ2πr

+ ...)

* +

,

- .

0

2π

∫ dθ .

The first integrand term makes no contribution because the integral of the cosine over one period is zero. The second and third integrand terms cancel. The fourth and fifth integrand terms are proportional to 1/r and disappear when the limit is taken, and all remaining integrand terms are even smaller; therefore D = 0. Now consider the lift integral with the large-r forms for the pressure and velocity substituted in:

€

LB

= −limr→∞

12ρ −

ΓUπsinθr

−Γ2

4π 2r2+ ...

)

* +

,

- . sinθ + ρ −

Γ2πcosθr

)

* +

,

- . U cosθ −

dcosθ2πr2

+ ...)

* +

,

- .

)

* +

,

- .

0

2π

∫ rdθ .

where the limit

€

r→∞ is now explicitly written. As before, collect terms under the integrand according to their power of r:

€

LB

= −limr→∞

−ρΓU2π

sin2θ − ρΓU2π

cos2θ − Γ2

8π 2rsinθ + ...

)

* +

,

- .

0

2π

∫ dθ = limr→∞

ρΓU2π

+Γ2

8π 2rsinθ + ...

)

* +

,

- .

0

2π

∫ dθ .

The first integrand term contributes ρΓU. The second term and the higher order terms are proportional to 1/r or are even smaller, and disappear when the limit is taken. Thus, the final results are:

D = 0 and L = ρUΓ, which proves Blasius' theorem.


Exercise 7.30. Pressure fluctuations in wall-bounded turbulent flows are a common source of flow noise. Such fluctuations are caused by turbulent eddies as they move over the bounding surface. A simple ideal-flow model that captures some of the important phenomena involves a two-dimensional vortex that moves above a flat surface in a fluid of density ρ. Thus, for the following items, use the potential:

€

φ(x,y,t) = −Γ2πtan−1 y − h

x −Ut&

' (

)

* + +

Γ2πtan−1 y + h

x −Ut&

' (

)

* +

where h is the distance of the vortex above the flat surface, Γ is the vortex strength, and U is the convection speed of the vortex. a) Compute the horizontal u and vertical v velocity components and verify that v = 0 on y = 0. b) Determine the pressure at x = y = 0 in terms of ρ, t, Γ, h, and U. c) Based on your results from part b), is it possible for a fast-moving high-strength vortex far from the surface to have the same pressure signature as a slow-moving low-strength vortex closer to the surface?

Solution 7.30. a) Start with the given potential,

€

φ(x,y,t) = −Γ2πtan−1 y − h

x −Ut&

' (

)

* + +

Γ2πtan−1 y + h

x −Ut&

' (

)

* + ,

and differentiate to find:

€

u =∂φ∂x

= −Γ2π

1+(y − h)2

(x −Ut)2'

( )

*

+ ,

−1

−y − h

(x −Ut)2'

( )

*

+ , +

Γ2π

1+(y + h)2

(x −Ut)2'

( )

*

+ ,

−1

−y + h

(x −Ut)2'

( )

*

+ ,

€

=Γ2π

y − h(x −Ut)2 + (y − h)2

−y + h

(x −Ut)2 + (y + h)2% & '

( ) *

€

v =∂φ∂y

= −Γ2π

1+(y − h)2

(x −Ut)2'

( )

*

+ ,

−11

x −Ut'

( )

*

+ , +

Γ2π

1+(y + h)2

(x −Ut)2'

( )

*

+ ,

−11

x −Ut'

( )

*

+ ,

€

= −Γ2π

x −Ut(x −Ut)2 + (y − h)2

−x −Ut

(x −Ut)2 + (y + h)2% & '

( ) *

At y = 0, the two factors inside the {,}-braces are equal and opposite so v(x, y=0) = 0. b) The pressure at x = y = 0 can be determined from the Bernoulli equation:

€

ρ∂φ∂t

+12ρ(u2 + v 2) + p

%

& '

(

) * x= y= 0

= p∞ or

€

p(0,0,t) − p∞ = − ρ∂φ∂t

+12ρ(u2 + v 2)

'

( )

*

+ , x= y= 0

First determine

€

∂φ ∂t ,

€

∂φ∂t

= −Γ2π

1+(y − h)2

(x −Ut)2'

( )

*

+ ,

−1

−y − h

(x −Ut)2'

( )

*

+ , −U( ) +

Γ2π

1+(y + h)2

(x −Ut)2'

( )

*

+ ,

−1

−y + h

(x −Ut)2'

( )

*

+ , −U( )

€

=−UΓ2π

y − h(x −Ut)2 + (y − h)2

−y + h

(x −Ut)2 + (y + h)2% & '

( ) *

,

and then evaluate it at x = y = 0:

€

∂φ∂t$

% &

'

( ) x= y= 0

=UhΓ π(Ut)2 + h2

. Now evaluate u and v at x = y = 0:

€

u =−Γh πU 2t 2 + h2

& v = 0, and put all of this into the Bernoulli relationship:


€

p(0,0,t) − p∞ρ

= −UhΓ πU 2t 2 + h2

−Γ2h2 2π 2

U 2t 2 + h2( )2. (1)

c) The question here is whether or not U can be replaced with C1U, h can be replaced with C2h and Γ replaced with C3Γ with the three constants (C1, C2, C3) chosen so that p(0,0,t) remains unchanged. In other words can (C1, C2, C3) be found so that (2) is identical to (1).

€

p(0,0,t) − p∞ρ

=?−C1C2C3UhΓ πC22U 2t 2 + C1

2h2−C22C3

2Γ2h2 2π 2

C22U 2t 2 + C1

2h2( )2 (2)

For there to be any hope of making (2) a genuine equality, the denominator factors imply that C1 = C2. With this simplification (2) becomes:

€

p(0,0,t) − p∞ρ

=?−C3UhΓ πU 2t 2 + h2

−C32Γ2h2 2π 2

C12 U 2t 2 + h2( )2

(3)

So, to recover (1), (3) implies: C3 = 1, and C1 = C3; therefore the surface pressure signature of an ideal moving line vortex is unique.


Exercise 7.31. A pair of equal strength ideal line vortices having axes perpendicular to the x-y plane are located at

€

xa (t) = xa (t),ya (t)( ) , and

€

xb (t) = xb (t),yb (t)( ) , and move in their mutually-induced velocity fields. The stream function for this flow is given by:

€

ψ(x,y,t) = −Γ2π

ln x − xa (t) + lnx − xb (t)( ). Explicitly determine xa (t) and xb(t) given that

€

xa (0) = (−ro,0) and

€

xb (0) = (ro,0) . Switching to polar coordinates at some point in your solution may be useful. Solution 7.31. This problem can quickly get very tedious without clear organization. For notational simplicity, let

€

(a) = x − xa (t)( )2 + y − ya (t)( )2 , and

€

(b) = x − xb (t)( )2 + y − yb (t)( )2. Differentiate the stream function to find:

€

u(x,y,t) =∂ψ∂y

= −Γ2π

y − ya(a)

+y − yb(b)

' ( )

* + ,

, and

€

v(x,y,t) = −∂ψ∂x

= +Γ2π

x − xa(a)

+x − xb(b)

' ( )

* + ,

.

This velocity field leads to 4 nonlinear first-order differential equations for point vortex coordinate positions.

€

dxadt

= −dxbdt

=Γ2π

yb − ya(xb − xa )

2 + (yb − ya )2

% & '

( ) *

, and

€

dyadt

= −dybdt

=Γ2π

xa − xb(xb − xa )

2 + (yb − ya )2

% & '

( ) *

.

The first two equalities yield:

€

xa (t) + xb (t) = constant, and

€

ya (t) + yb (t) = constant. With the initial conditions, the constants can be evaluated:

€

xa (t) + xb (t) = 0, and

€

ya (t) + yb (t) = 0. Use these two equations to eliminate xb and yb. The remaining two equations for xa and ya:

€

dxadt

=Γ2π

−2ya(−2xa )

2 + (−2ya )2

% & '

( ) *

, and

€

dyadt

=Γ2π

2xa(−2xa )

2 + (−2ya )2

% & '

( ) *

.

Taking a ratio of these implies:

€

dyadxa

= −xaya

, integrate →

€

xa2 + ya

2 = const., and initial conditions →

€

xa2 + ya

2 = ro2.

Hence the vortices move in a circle with radius ro. Based on this, let xa(t) = rocosθ(t) and evaluate the differential equation for xa(t) in terms of θ(t): dxadt

= −ro sinθdθdt

= −Γ4πro

2 ro2 − ro

2 cos2θ(t) = −Γsinθ(t)4πro

→

€

dθdt

=Γ4πro

2 or

€

θ(t) =Γt4πro

2 + θo .

The initial conditions set θo = π. With this relationship, reconstruction of the whole solution is possible:

€

xa (t) = −ro cos Γt 4πro2( ),

€

ya (t) = −ro sin Γt 4πro2( ) , and

€

xb (t) = ro cos Γt 4πro2( ) ,

€

yb (t) = ro sin Γt 4πro2( ) .

Hence, the pair of vortices circles the origin with a constant angular velocity. This type of free vortex motion is seen any time vortices of the same sign interact (for example in the trailing vortices left by the multiple wing-tips on biplanes, or the multiple tip feathers of birds).


Exercise 7.32. Two unequal strength ideal line vortices having axes perpendicular to the x-y plane are located at x1(t) = x1(t), y1(t)( ) with circulation Γ1, and x2 (t) = x2 (t), y2 (t)( ) with circulation Γ2, and move in their mutually induced velocity fields. The stream function for this flow is given by: ψ(x, y, t) = − Γ1 2π( ) ln x− x1(t) − Γ2 2π( ) ln x− x2 (t) . Explicitly determine x1(t) and x2 (t) in terms of Γ1, Γ2, h1, h2, and h, given x1(0) = (h1, 0) , x2 (0) = (h2, 0) , and h2 – h1 = h > 0 [Hint: choose a convenient origin of coordinates, and switch to polar coordinates after finding the shape of the trajectories.] Solution 7.32. This problem can quickly get very tedious without clear organization. For notational simplicity, let (1) = x − x1(t)( )2 + y− y1(t)( )2 , and (2) = x − x2 (t)( )2 + y− y2 (t)( )2 . Differentiate the stream function to find:

u = ∂ψ∂y

= −12π

Γ1(y− y1)(1)

+Γ2(y− y2 )(2)

#$%

&'(

, and v = −∂ψ∂x

= +12π

Γ1(x − x1)(1)

+Γ2(x − x2 )(2)

#$%

&'(

.

This velocity field leads to 4 nonlinear first-order differential equations for point vortex coordinate positions. dx1dt

= u(x1, y1, t) = −Γ22π

y1 − y2(x2 − x1)

2 + (y2 − y1)2

#$%

&'(

, dx2dt

= u(x2, y2, t) = −Γ12π

y2 − y1(x2 − x1)

2 + (y2 − y1)2

#$%

&'(

,

dy1dt

= v(x1, y1, t) =Γ22π

x1 − x2(x2 − x1)

2 + (y2 − y1)2

#$%

&'(

, and dy2dt

= u(x2, y2, t) =Γ12π

x2 − x1(x2 − x1)

2 + (y2 − y1)2

#$%

&'(

.

From these, it is clear that: 2πΓ2

dx1dt

= −2πΓ1

dx2dt

=y2 − y1

(x2 − x1)2 + (y2 − y1)

2 , and − 2πΓ2

dy1dt

=2πΓ1

dy2dt

=x2 − x1

(x2 − x1)2 + (y2 − y1)

2 ,

so: ddt

x1Γ2

+x2Γ1

"

#$

%

&'= 0 , and d

dty1Γ2

+y2Γ1

"

#$

%

&'= 0 .

Thus, the contents of the parentheses are constants. The initial conditions then imply: x1Γ2

+x2Γ1

=h1Γ2

+h2Γ1

, and y1Γ2

+y2Γ1

= 0 . (&)

Now choose the origin of coordinates so that h1/Γ2 + h2/Γ1 = 0, and use h2 – h1 = h to find:

h1 =−hΓ2Γ1 +Γ2

and h2 =hΓ1

Γ1 +Γ2. (%)

With this repositioning, equations (&) become: x1Γ2

+x2Γ1

= 0 , and y1Γ2

+y2Γ1

= 0 . ($)

Now return to the differential equations for x1 and y1, and compute the ratio:

−1Γ2

dy1dt

1Γ2

dx1dt

= −dy1dx1

=x2 − x1y2 − y1

=−Γ1Γ2

x1 − x1

−Γ1Γ2

y1 − y1=x1y1

.


where the third equality follows from ($). Integrating the second and final members of this extended equality and using the initial conditions produces: x1

2 + y12 = h1

2 . Similar steps for the second vortex lead to: x2

2 + y22 = h2

2 . Hence the vortices move in circles with radii h1 and h2. In addition, the distance between them remains constant:

(x2 − x1)2 + (y2 − y1)

2 = x22 1+ Γ2

Γ1

#

$%

&

'(

2

+ y22 1+ Γ2

Γ1

#

$%

&

'(

2

= h22 1+ Γ2

Γ1

#

$%

&

'(

2

= h2 ,

where the first equality comes from substituting for x1 and y1 from ($), and the final equality comes from the second equation of (%). Based on this, let x1(t) = h1cosθ(t), y1(t) = h1sinθ(t), x2(t) = h2cosθ(t), and y2(t) = h2sinθ(t). With these substitutions, the distance between the vortices remains constant (= h):

(x2 − x1)2 + (y2 − y1)

2 = (h2 − h1)2 (cos2θ + sin2θ ) = h2 ,

so the only remaining task is to find θ(t). Evaluate the differential equation for x1(t) in terms of θ(t):

dx1dt

= −h1 sinθdθdt

= −Γ22π

y1 − y2(x2 − x1)

2 + (y2 − y1)2

#$%

&'(=Γ22π

hsinθh2

.

Use the second and final terms to reach: dθdt

= −Γ22πhh1

=Γ2 +Γ12πh2

, or θ(t) = Γ2 +Γ12πh2

!

"#

$

%&t ,

where the initial conditions have been used to find that the constant of integration is zero. With this relationship, reconstruction of the whole solution is possible:

x1(t) = −hΓ2

Γ1 +Γ2cos Γ2 +Γ1

2πh2#

$%

&

'(t

)

*+

,

-. , y1(t) = −

hΓ2Γ1 +Γ2

sin Γ2 +Γ12πh2

#

$%

&

'(t

)

*+

,

-. ,

x2 (t) =hΓ1

Γ1 +Γ2cos Γ2 +Γ1

2πh2"

#$

%

&'t

(

)*

+

,- , and y2 (t) =

hΓ1Γ1 +Γ2

sin Γ2 +Γ12πh2

"

#$

%

&'t

(

)*

+

,- .

Hence, the pair of vortices circles the origin with a constant angular velocity. When the two vortices are of equal strength, this solution reverts to that for Exercise 7.31. In the limit as Γ1→−Γ2 (i.e. the two vortices are of equal and opposite sign), x2 − x1→ const. = h , and the two

y-coordinates become equal: y1, y2 →−Γ2t2πh

, and this matches the discussion of interacting

vortices in Section 5.6.


Exercise 7.33. Consider the unsteady potential flow of two ideal sinks located at

€

xa (t) = xa (t),0( ) and

€

xb (t) = xb (t),0( ) that are free to move along the x-axis in an ideal fluid that is stationary far from the origin. Assume that each sink will move in the velocity field induced by the other.

φ(x, y, t) = − qs2π

ln x − xa (t)( )2 + y2 + ln x − xb(t)( )2 + y2"#$

%&'

, with qs > 0.

a) Determine

€

xa (t) and

€

xb (t) when

€

xa (0) = −L,0( ) and

€

xb (0) = +L,0( ) b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x = y = 0 as function of p∞, ρ, qs, and xa(t). Solution 7.33. a) All the action takes place in the x-direction so ignore the vertical (v) component of velocity.

u(x, y, t) = ∂φ∂x!

"#

$

%&= −

qs2π

x − xa (t)x − xa (t)( )2 + y2

+x − xb(t)

x − xb(t)( )2 + y2!

"##

$

%&&

For

€

x = xa (t) and y = 0: u(xa, 0, t) = −qs2π

xa − xbxa − xb( )2 + y2

"

#$$

%

&''y=0

= −qs2π

1xa − xb

"

#$

%

&'=

dxadt

, and

for

€

x = xb (t) and y = 0: u(xb, 0, t) = −qs2π

xb − xaxb − xa( )2 + y2

"

#$$

%

&''y=0

= −qs2π

1xb − xa

"

#$

%

&'=

dxbdt

= −u(xa, 0, t) .

Thus,

€

dxa dt + dxb dt = 0, or

€

xa + xb = const.= 0 where final equality comes from the initial conditions. So, dxa dt = −qs 4π xa( ) or xa = ± const.− qst 2π . The xa-initial condition

requires the minus sign and const. = L2; thus, xa (t) = −xb(t) = − L2 − qst 2π .

b) Use the unsteady Bernoulli equation:

€

∂φ ∂t + 12 u

2+ p ρ = p∞ ρ . From the symmetry of the

flow, x = y = 0 is a stagnation point (u = 0), so

€

p = p∞ − ρ∂φ ∂t .

∂φ∂t!

"#

$

%&x=y=0

= −qs2π

− x − xa (t)( ) dxa dt( )x − xa (t)( )2 + y2

−x − xb(t)( ) dxb dt( )x − xb(t)( )2 + y2

(

)**

+

,--x=y=0

= −qs2π

dxa dtxa

+dxb dtxb

!

"#

$

%&

∂φ∂t!

"#

$

%&x=y=0

= −qsπ

dxa dtxa

!

"#

$

%&=

qs2

4π 21xa2

!

"#

$

%& , so p = p∞ −

ρqs2

4π 2xa2

Thus, the pressure at the origin becomes very negative as the two sinks approach each other.


Exercise 7.34. Consider the unsteady potential flow of an ideal source and sink located at

€

x1(t) = x1(t),0( ) and

€

x2(t) = x2(t),0( ) that are free to move along the x-axis in an ideal fluid that is stationary far from the origin. Assume that the source and sink will move in the velocity field induced by the other.

φ(x, y, t) = qs2π

ln x − x1(t)( )2 + y2 − ln x − x2 (t)( )2 + y2"#$

%&'

, with qs > 0.

a) Determine

€

x1(t) and

€

x2(t) when x1(0) = −, 0( ) and x2 (0) = +, 0( ) . b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x = y = 0 as function of p∞, ρ, qs, and x1(t). Solution 7.34. a) All the action takes place in the x-direction so ignore the vertical (v) component of velocity.

u(x, y, t) = ∂φ∂x!

"#

$

%&=

qs2π

x − x1(t)x − x1(t)( )2 + y2

−x − x2 (t)

x − x2 (t)( )2 + y2!

"##

$

%&&

For

€

x = x1(t) and y = 0: u(x1, 0, t) =qs2π

−x1 − x2

x1 − x2( )2 + y2"

#$$

%

&''y=0

= −qs2π

1x1 − x2

"

#$

%

&'=

dx1dt

, and

for

€

x = x2(t) and y = 0: u(x2, 0, t) =qs2π

x2 − x1x2 − x1( )2 + y2

"

#$$

%

&''y=0

=qs2π

1x2 − x1

"

#$

%

&'=

dx2dt

= u(x1, 0, t) .

Thus,

€

dx1 dt = dx2 dt , or x1 − x2 = const. = −2 where final equality comes from the initial conditions. So, dx1 dt = qs 4π( ) or x1 = qs 4π( ) t + const. The x1-initial condition requires const. = – ; thus, x1(t) = qs 4π( ) t − , and x2 (t) = qs 4π( ) t + .

b) Use the unsteady Bernoulli equation:

€

∂φ ∂t + 12 u

2+ p ρ = p∞ ρ . From the symmetry of the

flow, the vertical velocity component will be zero at the origin, so

€

p = p∞ − ρ∂φ ∂t − 12 ρu

2 .

∂φ∂t!

"#

$

%&x=y=0

=qs2π

− x − x1(t)( ) dx1 dt( )x − x1(t)( )2

+ y2−− x − x2 (t)( ) dx2 dt( )

x − x2 (t)( )2+ y2

(

)**

+

,--x=y=0

=qs2π

dx1 dtx1

−dx2 dtx2

!

"#

$

%&

= qs2π

qs 4πqs 4π( ) t −

−qs 4π

qs 4π( ) t + !

"##

$

%&&=

qs2π

2qs4π

!

"#

$

%&

1qs 4π( )2 t2 − 2

= qs2

4π 21

qs 4π( )2 t2 − 2


∂φ∂x!

"#

$

%&x=y=0

= u(0, 0) = qs2π

x − x1(t)x − x1(t)( )2

+ y2−

x − x2 (t)x − x2 (t)( )2

+ y2

!

"##

$

%&&x=y=0

=qs2π

−1x1(t)

+1

x2 (t)!

"#

$

%&

= qs2π

−1qs 4π( ) t −

+1

qs 4π( ) t + !

"##

$

%&&=

qs2π

−2( ) 1qs 4π( )2 t2 − 2

= − qsπ

1qs 4π ( )2 t2 − 2

so p = p∞ − ρqs2

4π 21

qs 4π( )2 t2 − 2− ρ

qs22

2π1

qs 4π( )2 t2 − 2#$

%&2 .


Exercise 7.35. Consider a free ideal line vortex oriented parallel to the z-axis in a 90° corner defined by the solid walls θ = 0 and θ = 90°. If the vortex passes the through the plane of the flow at (x, y), show that the vortex path is given by: x–2 + y–2 = constant. [Hint: Three image vortices are needed at points (−x, −y), (−x, y) and (x, −y). Carefully choose the directions of rotation of these image vortices, show that dy/dx = v/u = −y3/x3, and integrate to produce the desired result.] Solution 7.35. The induced velocity at point (x, y) will include contributions from the three image vortices. If the first vortex at (x, y) has strength +Γ, then the second vortex at (−x, y) and the third vortex at (x, −y) will have strength –Γ. The fourth vortex at (−x, −y) will have strength +Γ. Therefore:

u = Γ2π

−12xey +

12yex +

12 x2 + y2

−yx2 + y2

ex +x

x2 + y2ey

#

$%%

&

'((

#

$%%

&

'((

which is the contribution of the second, third, and fourth vortices, respectively. This velocity can be rewritten:

€

u =Γ4π

1y−

yx 2 + y 2

%

& '

(

) * ex + −

1x

+x

x 2 + y 2%

& '

(

) * ey

+

, -

.

/ 0 =

Γ

4π x 2 + y 2( )x 2

yex −

y 2

xey

+

, -

.

/ 0 = uex + vey

The path lines for the location x = (x, y) of the first vortex are:

€

dxdt

= u =Γ

4π x 2 + y 2( )x 2

y, and

€

dydt

= v = −Γ

4π x 2 + y 2( )y 2

x.

Divide the second equation by the first to find:

€

dydx

=dydt

dxdt

= −Γ

4π x 2 + y 2( )y 2

xΓ

4π x 2 + y 2( )x 2

y= −

y 3

x 3.

Using the two ends of this equality, separate and integrate the equation:

€

−dyy 3

=dxx 3

→1y 2

+1x 2

= const.

This is the desired result.

12

34


Exercise 7.36. In ideal flow, streamlines are defined by dψ = 0, and potential lines are defined by dφ = 0. Starting from these relationships, show that streamlines and potential lines are perpendicular. a) in plane flow where x and y are the independent spatial coordinates, and b) in axisymmetric flow where R and z are the independent spatial coordinates. [Hint: For any two independent coordinates x1 and x2, the unit tangent to the curve x2 = f(x1) is

€

t = e1 + (df dx1)e2( ) 1+ (df dx1)2 ; thus, for a) and b) it is sufficient to show

€

t( )ψ= const ⋅ t( )φ= const = 0]

Solution 7.36. a) In two-dimensional ideal flow,

€

dψ =∂ψ∂x

dx +∂ψ∂y

dy = −vdx + udy. Thus a curve

defined by dψ = 0, has a slope

€

dy dx = v u . Using the hint, the tangent vector to this curve is:

€

t( )ψ= const = ex + (v u)ey( ) 1+ (v u)2 = uex + vey( ) u2 + v 2 .

Similarly,

€

dφ =∂φ∂x

dx +∂φ∂y

dy = udx + vdy , so a curve defined by dφ = 0, has a slope

€

dy dx = −u v . Using the hint, the tangent vector to this curve is:

€

t( )φ= const = ex + (−u v)ey( ) 1+ (u v)2 = vex − uey( ) v 2 + u2 . Forming the dot product of the two unit vectors produces:

€

t( )ψ= const ⋅ t( )φ= const =uex + vey( )u2 + v 2

⋅vex − uey( )v 2 + u2

=uv − vuv 2 + u2

= 0 .

b) In three-dimensional axisymmetric flow,

€

dψ =∂ψ∂R

dR +∂ψ∂z

dz = RuzdR − RuRdz. Thus a curve

defined by dψ = 0, has a slope

€

dR dz = uR uz . Using the hint, the tangent vector to this curve is:

€

t( )ψ= const = ez + (uR uz)eR( ) 1+ (uR uz )2 = uzez + uReR( ) uz

2 + uR2 .

Similarly,

€

dφ =∂φ∂R

dR +∂φ∂zdz = uRdR + uzdz , so a curve defined by dφ = 0, has a slope

€

dR dz = −uz uR . Using the hint, the tangent vector to this curve is:

€

t( )φ= const = ez + (−uz uR )ey( ) 1+ (uz uR )2 = uRez − uzeR( ) uR

2 + uz2 .

Forming the dot product of the two unit vectors produces:

€

t( )ψ= const ⋅ t( )φ= const =uzez + uReR( )uz2 + uR

2⋅uRez − uzeR( )uR2 + uz

2=uzuR − uRuzuR2 + uz

2 = 0 .


Exercise 7.37. Consider a three-dimensional point source of strength Q (m3/s). Use a spherical control volume and the principle of conservation of mass to argue that the velocity components in spherical coordinates are uθ = 0 and ur = Q/4πr2 and that the velocity potential and stream function must be of the form φ = φ(r) and ψ = ψ(θ). Integrate the velocity, to show that φ = −Q/4πr and ψ = −Qcosθ /4π . Solution 7.37. For a point source of strength Q (m3/s), the tangential velocity is zero because of symmetry. The radial velocity times the area 4πr2 equals Q so ur = Q/ 4πr2. Therefore,

€

ur =∂φ∂r

=1

r2 sinθ∂ψ∂θ

=Q4πr2

, and

€

uθ =1r∂φ∂θ

= −1

rsinθ∂ψ∂r

= 0.

The uθ equations, requires φ = φ(r) and ψ = ψ(θ). Using these results and integrating the ur equations produces:

€

φ(r) = −Q4πr

and

€

ψ(θ) = −Qcosθ4π

.


Exercise 7.38. Solve the Poisson equation

€

∇2φ =Qδ(x − & x ) in a uniform unbounded three-dimensional domain to obtain the velocity potential φ = –Q/4π|x – x´| for an ideal point source located at x´. Solution 7.38. This exercise is similar to Exercise 5.9. First apply a simple shift transformation that places x´ at the origin of coordinates. Define these new coordinates by:

€

X = x − # x ,

€

Y = y − # y ,

€

Z = z − # z , and set

€

r = x − # x = X 2 +Y 2 + Z 2 . The gradient operator

€

∇XYZ in the shifted coordinates X = (X, Y, Z) is the same as

€

∇ in the unshifted coordinates (x, y, z), so the field equation for φ becomes:

€

∇XYZ2 φ =Qδ(X) =Qδ(x − & x ) .

Integrate this equation inside a sphere of radius r:

€

∇XYZ2 φdV

sphere∫∫∫ = ∇XYZφ ⋅ndA

spherical surface∫∫ =Q δ(X)

x=− r 2−y 2−z 2

x=−+ r 2−y 2−z 2

∫y=− r 2−z 2

y= + r 2−z 2

∫z=−r

z= +r

∫ dXdYdZ ,

where the first equality follows from Gauss' divergence theorem, and the triple integral on the right side includes the location X = 0 (aka x = x´) so a contribution is collected from the three-dimensional delta function. Thus, the right side of this equation is Q (times unity). The dot product in the middle portion of the above equation simplifies to ∂φ/∂r because n = er on the spherical surface and

€

er ⋅ ∇XYZ = ∂ ∂r . Plus, in an unbounded uniform environment, there are no preferred directions so φ = φ(r) alone (no angular dependence). Thus, the integrated field equation simplifies to

€

∂φ∂r$

% &

'

( ) r2 sinθdθd

ϕ= 0

2π

∫θ = 0

π

∫ ϕ = r2 ∂φ∂r

sinθdθdϕ= 0

2π

∫θ = 0

π

∫ ϕ = 4πr2 ∂φ∂r

=Q ,

which implies

€

∂φ∂r

=Q4πr2

or

€

φ = −Q4πr

= −Q

4π x − % x .


Exercise 7.39. Using

€

(R,ϕ,z)-cylindrical coordinates, consider steady three-dimensional potential flow for a point source of strength Q at the origin in a free stream flowing along the z-axis at speed U:

€

φ(R,ϕ,z) =Uz − Q4π R2 + z2

.

a) Sketch the streamlines for this flow in any R-z half-plane. b) Find the coordinates of the stagnation point that occurs in this flow. c) Determine the pressure gradient, ∇p , at the stagnation point found in part b). d) If R = a(z) defines the stream surface that encloses the fluid that emerges from the source, determine a(z) for

€

z→ +∞. e) Use Stokes’ stream function to determine an equation for a(z) that is valid for any value of z. f) Use the control-volume momentum equation,

€

ρu u ⋅n( )dS =S∫ − pndS + F

S∫ where n is the

outward normal from the control volume, to determine the force F applied to the point source to hold it stationary. g) If the fluid expelled from the source is replaced by a solid body having the same shape, what is the drag on the front of this body? Solution 7.39. a) A simple sketch appears to the right. b) From symmetry, the stagnation point will lie on R = 0, thus only the axial velocity needs to be considered to find its axial location zs.

€

∂φ∂z$

% &

'

( ) r= 0

= uz(zs,0) =U +Qzs

4π zs2( )3 2

. Solve

for zs using

€

zs2( )3 2

= zs3 to find

€

zs = − Q 4πU . c) Take the gradient of the Bernoulli equation or consider the steady ideal-flow momentum

equation to find:

€

u ⋅ ∇( )u = −1ρ∇p . The fluid velocity is zero at a stagnation point, so

€

∇p = 0 at

a stagnation point. d) Far downstream of the source, all of the fluid will be moving horizontally at speed U. Thus, the fluid that comes from the source will emerge from a stream tube of radius a∞ determined from:

€

πa∞2U =Q , which implies:

€

a∞ = Q πU . e) Set the velocity relationships for the Stokes stream function ψS equal to those from the potential:

€

uR = −1R∂ψS

∂z=

QR4π R2 + z2( )3 2

=∂φ∂R

, and

€

uz =1R∂ψS

∂R=U +

Qz4π R2 + z2( )3 2

=∂φ∂z

.

Use variable substitutions

€

tanβ = R z in the first equation and

€

γ = R2 + z2 in the second equation to integrate these relationships to find:

€

ψS = − Qz 4π( ) R2 + z2( )−1 2

+ f (R) , and

€

ψS =UR2 2 − Qz 4π( ) R2 + z2( )−1 2

+ g(z)

z

R


So,

€

ψS =UR2 2 − Q 4π( ) 1+ z R2 + z2( ) where the two functions f(r) and g(z) have been chosen

so that ψs = 0 on R = 0 when z < 0. The equation for a(z) is found by setting R = a & ψs = 0:

€

0 =Ua2 2 − Q 4π( ) 1+ z a2 + z2( ). This implicit equation for a can be solved to find:

€

a(z) = 12a∞2 − z2 + z z2 + 2a∞

2( )1 2

where

€

a∞ = Q πU . f) From the symmetry of the flow, only a z-direction force is anticipated and the simplest possible control volume is a sphere of radius a. Thus, using spherical coordinates where

€

r = R2 + z2 denotes the distance from the source and θ is the polar angle, the vector momentum equation can be simplified:

€

Fz = ρuz u ⋅ er( )dAS∫ + per ⋅ ezdA

S∫ ,

where the replacement

€

n = er has been made. For the given potential,

€

φ =Urcosθ −Q 4πr , the radial and angular velocities are

€

ur =∂φ∂r

=U cosθ +Q4πr2

and

€

uθ =1r∂φ∂θ

= −U sinθ .

The pressure can be obtained from the Bernoulli equation:

€

p(r,θ) +12ρ ur

2 + uθ2( ) = p(r,θ) +

12ρ U 2 cos2θ +

QU cosθ2πR2

+Q2

16π 2R4+U 2 sin2θ

%

& '

(

) * = p∞ +

12ρU 2 .

Thus,

€

p(r,θ) = p∞ +12ρ −

QU cosθ2πr2

−Q2

16π 2r4'

( )

*

+ , , so the pressure integral becomes:

€

per ⋅ ezdSS∫ = p∞ +

12ρ −

QU cosθ2πr2

−Q2

16π 2r4)

* +

,

- .

/

0 1

2

3 4

θ = 0

π

∫ (cosθ)2πr2 sinθdθ = −12ρQU cos2θ

θ = 0

π

∫ sinθdθ

€

= −13ρQU

The terms with odd-powers of the cosine do not contribute to the angular integration on the interval θ = 0 to π. To calculate the flux integral, first find uz and change it into the spherical coordinate system

€

uz =∂φ∂z

=U +Qz

4π R2 + z2[ ]3 2=U +

Qcosθ4πr2

.

Then proceed with:

€

ρuz u ⋅ er( )dSS∫ = ρ U +

Qcosθ4πr2

'

( )

*

+ ,

θ = 0

π

∫ U cosθ +Q4πr2

'

( )

*

+ , 2πr2 sinθdθ

€

= ρQU2

1+ cos2θ( )θ = 0

π

∫ sinθdθ = −ρQU2

β +β 3

3(

) *

+

, - 1

−1

=43ρQU

Thus,

€

F = ρQU 43−13

$

% &

'

( ) ez = ρQUez ; again, the odd-powers of the cosine do not contribute. This

force points downstream, so if it were not applied the source would move upstream! This result matches the finding for a source in a free stream in two dimensions. g) Here, consider a spherical control volume that is very large so that the velocity is nearly

€

Uez on the surface of the control volume. In this case, the pressure integral will be unchanged as


long as the pressure within the body is p∞. However, the flux integral will not include the contribution from the fluid that emerges from the source. Thus, the force in this case will be equal to that obtained for part d) minus the momentum flux that is no longer present:

€

F = F( )d ) − ρUQ( )ez = 0 Therefore, there will be no drag on the front of the body when the body is long compared to its diameter.


Exercise 7.40. In (R, ϕ, z) cylindrical coordinates, the three-dimensional potential for a point source at (0,0,s) is given by:

€

φ = − Q 4π( ) R2 + (z − s)2[ ]−1 2 .

a) By combining a source of strength +Q at (0,0,–b), a sink of strength –Q at (0,0,+b), and a uniform stream with velocity Uez, derive the potential (7.85) for flow around a sphere of radius a by taking the limit as Q → ∞, and b → 0, such that

€

d = −2bQez = –2πa3Uez = constant. Put your final answer in spherical coordinates in terms of U, r, θ, and a. b) Repeat part a) for the Stokes stream function starting from

€

ψ = − Q 4π( )(z − s) R2 + (z − s)2[ ]−1 2 .

Solution 7.40. In cylindrical coordinates, the three-dimensional potential for a uniform stream of

€

U = (0,0,U), a point source of strength +Q at

€

x = (0,0,−b), and a point-sink of strength –Q at

€

x = (0,0,+b) is:

€

φ =Uz − Q4π R2 + (z + b)2

+Q

4π R2 + (z − b).

Expand the square roots for b → 0, letting

€

r = R2 + z2

€

1R2 + (z ± b)2

≈1

r 1± (2zb r2)≈

1r 1± (zb r2)( )

≈1r1 (zb r2)( ).

Make replacements in the equation for φ:

€

φ =Uz − Q4πr

1− zbr2−1− zb

r2%

& ' (

) * =Uz +

Qzb2πr3

=U ⋅ x − d ⋅ x4π | x |3

.

where the last equality was obtained by using the given definition

€

d = −2bQez . Using z = rcosθ (where θ = polar angle) and

€

|d |= 2πUa3, the potential becomes:

€

φ = Ur +|d |4πr2

$

% &

'

( ) cosθ , or

€

φ =Ur 1+a3

2r3#

$ %

&

' ( cosθ .

b) In cylindrical coordinates, the Stokes stream function for a uniform stream of

€

U = (0,0,U), a point source of strength +Q at

€

x = (0,0,−b) , and a point-sink of strength –Q at

€

x = (0,0,+b) is:

€

ψ =12UR2 − Q

4π(z + b)

R2 + (z + b)2+Q4π

(z − b)R2 + (z − b)2

.

As in part a), expand the square roots for b → 0, letting

€

r = R2 + z2

€

1R2 + (z ± b)2

≈1

r 1± (2zb r2)≈

1r 1± (zb r2)( )

≈1r1 (zb r2)( ).

Make replacements in the equation for ψ:

€

ψ =12UR2 −

Q4π

z + br

1− zbr2

%

& '

(

) *

%

& '

(

) * +

Q4π

z − br

1+zbr2

%

& '

(

) *

%

& '

(

) *

=12UR2 +

Q4π

−zr−br

+z2br3 +

zb2

r3 +zr−br

+z2br3 +

zb2

r3

%

& '

(

) * =

12UR2 +

Q4π

−2br

+2z2br3

%

& '

(

) * .

Using R = rsinθ and z = rcosθ (where θ = polar angle), and

€

|d |= 2πUa3, as given in the problem statement, the stream function becomes:


€

ψ =12Ur2 sin2θ +

2Qb4π

−1r

+r2 cos2θr3

&

' (

)

* + =12Ur2 1− a

3

r3&

' (

)

* + sin2θ .


Exercise 7.41. a) Determine the locus of points in uniform ideal flow past a circular cylinder of radius a without circulation where the velocity perturbation produced by the presence of the cylinder is 1% of the free stream value. b) Repeat for uniform ideal flow past a sphere. c) Explain the physical reason(s) for the differences between the answers for a) and b). Solution 7.41. a) From (7.34), the velocity components for ideal flow about a cylinder with radius a are:

€

ur =U 1− a2

r2#

$ %

&

' ( cosθ , and

€

uθ = −U 1+a2

r2$

% &

'

( ) sinθ .

Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the perturbation is:

€

u−Uex = U 2 a2 r2( )2cos2θ +U 2 a2 r2( )

2sin2θ =U a2 r2( ) .

Therefore, a 1% perturbation implies:

€

0.01=u−UexU

=a2

r2→ r =10a.

b) From (7.86), the velocity components for ideal flow about a sphere with radius a is:

€

ur =U 1− a3

r3#

$ %

&

' ( cosθ , and

€

uθ = −U 1+a3

2r3$

% &

'

( ) sinθ .

Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the perturbation is:

€

u−Uex = U 2 a3 r3( )2cos2θ +U 2 a3 2r3( )

2sin2θ =U a3 r3( ) cos2θ + 1

4 sin2θ .

Therefore, a 1% perturbation implies:

0.01=u−UexU

= a3 r3( ) cos2θ + 14 sin

2θ → r = 4.64a 34 cos

2θ + 14( )1 6

.

c) For the sphere, the 1%-perturbation distance is smaller because the sphere's projected area in the direction of the flow (πa2) is smaller than that of the cylinder (2a x span); the sphere's blockage is smaller. Thus, the flow has an easier time getting around the sphere. This effect also decreases the 1%-perturbation distance at θ = π/2 in the case of the sphere.


Exercise 7.42. Using the figure for Exercise 7.29 with

€

A3 → 0 and

€

r→∞ , expand the three-dimensional potential for a stationary arbitrary-shape closed body in inverse powers of the distance r and prove that ideal flow theory predicts zero drag on the body. Solution 7.42. The figure is reproduced here and a portion of this solution follows that given for Exercise 7.29. However, this time the geometry is three-dimensional and the figure must be interpreted as a slice through three dimensional control volume so that A1 is a spherical surface. In ideal flow, the only surface forces are pressure forces. Therefore, the hydrodynamic force F on the body is defined by:

€

F = − p − p∞( )n2dAA2

∫ = p − p∞( )ndAA2

∫ , (a)

where the free stream velocity is U = Uex, and n is the outward normal on the total control volume; thus it points into the body on surface A2 so the usual minus sign is missing from the final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure integration because

€

p∞ndAclosed surface∫ = 0 .

The starting point for the derivation is the integral form of the steady ideal flow momentum equation applied to the entire clam-shell control volume:

€

ρu(u ⋅n)dAA1 +A2 +A3

∫ = − p − p∞( )ndAA1 +A2 +A3

∫

with the implied limit that the tube denoted by A3 goes to zero. When this limit is taken, the net contribution of A3 is zero because the surface area of A3 goes to zero. This leaves:

€

ρu(u ⋅n)dAA1

∫ + ρu(u ⋅n)dAA2

∫ = ρu(u ⋅n)dAA1

∫ + 0 = − p − p∞( )ndAA1 +A2

∫ ,

where the first equality follows because

€

u ⋅n = 0 on the solid surface A2. Now substitute in (a) for the pressure integration over A2, and rearrange:

€

ρu(u ⋅n)dAA1

∫ = − p − p∞( )ndAA1

∫ −F or

€

F = − p − p∞( )ndAA1

∫ − ρu(u ⋅n)dAA1

∫ .

Thus, the force on the body can be obtained from integrals over a spherical surface (A1) that is distant from the body. The pressure can be written in term of the velocity using the Bernoulli equation:

€

p∞ +12ρU 2 = p +

12ρu 2 = p +

12ρ u ⋅u( ) ,

and this allows the pressure integral to be written:

€

− p − p∞( )ndAA1

∫ = −12ρ U 2 −u ⋅u( )ndAA1

∫ =12ρ u ⋅u( )ndAA1

∫ ,

where the final equality follows because U2 is a constant so

€

U 2ndAclosed surface∫ = 0 . So, the force on the

body can be obtained entirely from considerations of the velocity field:

€

F = ρ 12 u ⋅u( )n−u(u ⋅n)( )dA

A1

∫ .


Far from the body, the fluid velocity will primarily be U = Uex = constant, so it can be written u = U + u´, where u´ is the velocity perturbation that comes from the presence of the body. Inserting this decomposition into the last equation produces:

€

F = ρ 12U

2n + U ⋅ $ u ( )n + 12 $ u ⋅ $ u ( )n−U(U ⋅n) −U( $ u ⋅n) − $ u (U ⋅n) − $ u ( $ u ⋅n)( )dA

A1

∫

= ρ U ⋅ $ u ( )n + 12 $ u ⋅ $ u ( )n−U( $ u ⋅n) − $ u (U ⋅n) − $ u ( $ u ⋅n)( )dA

A1

∫ .

As before the second equality follows by dropping the terms that involve the quadratic factors of the constant vector U. The five remaining integrand terms all involve the perturbation velocity. An arbitrary shape closed body may be represented in potential flow by a collection of sources and sinks. Thus, for the limit as

€

r→∞ , the potential may be expanded in inverse powers of r:

€

φ =U ⋅ x + $ φ =U ⋅ x +φ1r

+φ2r2

+ ...

where the coefficients φm may be functions of the angular variables. The first term in this expansion corresponds to the free stream. The remaining term produce u´. However, the body is closed so the φ1 term must be zero because it represents the net source strength of the potential. The fluid velocity is the gradient of the potential, therefore the largest term representing the velocity perturbation will involve another factor of 1/r.

€

" u ~ φ2r3

+ ...

Using this scaling of the velocity perturbation with radial distance, the magnitude of the five integrand terms above can be assessed using dA = r2sinθdθdϕ,

€

F ~ limr→∞

ρUφ2

r3 +φ2

2

2r6 +Uφ2

r3 +Uφ2

r3 + +φ2

2

2r6

&

' (

)

* + r2

all angles∫ sinθdθdφ ∝ lim

r→∞

constr

= 0 .

Thus, pure potential flow cannot predict drag. This another statement of D'Alembert's paradox. Interestingly, if a "φ0" term is added to the expanded φ above, non-zero lift and drag are possible. However, this involves the introduction of vorticity in three-dimensions and elementary treatment of this topic for aircraft wings is postponed to Ch. 14.


Exercise 7.43. Consider steady ideal flow over a hemisphere of constant radius a lying on the y-z plane. For the spherical coordinate system shown, the potential for this flow is:

€

φ(r,θ,ϕ) =Ur(1+ a3 2r3)cosθ where U is the flow velocity far from the hemisphere. Assume gravity acts downward along the x-axis. Ignore fluid viscosity in this problem. a) Determine all three components of the fluid velocity on the surface of the hemisphere, r = a, in

spherical polar coordinates:

€

(ur,uθ ,uϕ ) =∇φ =∂φ∂r,1r∂φ∂θ, 1rsinθ

∂φ∂ϕ

'

( )

*

+ , .

b) Determine the pressure, p, on r = a. c) Determine the hydrodynamic force, Rx, on the hemisphere assuming stagnation pressure is felt everywhere underneath the hemisphere. [Hints:

€

er ⋅ ex = sinθ cosϕ ,

€

sin2θdθ = π 20

π∫ ,

and

€

sin4 θdθ = 3π 80

π∫ ].

d) For the conditions of part c) what density ρh must the hemisphere have to remain on the surface. Solution 7.43. a) Differentiate

€

φ(r,θ,ϕ) =Ur 1+ a3 2r3( )cosθ and evaluate on r = a.

€

ur =∂φ∂r

=U 1− a3

r3%

& '

(

) * cosθ –>

€

ur(r = a) =U 1− a3

a3#

$ %

&

' ( cosθ = 0

€

uθ =1r∂φ∂θ

= −U 1+a3

2r3&

' (

)

* + sinθ –>

€

uθ (r = a) = −3U2sinθ

€

uϕ =1

rsinθ∂φ∂ϕ

= 0 –>

€

uϕ (r = a) = 0

b) The steady Bernoulli equation applies here:

€

p∞ + 12 ρU 2 = p(r = a) + 1

2 ρ u (r = a) 2 . Use the

results of part a) to find:

€

p(r = a) − p∞ρ

= 12U

2 − 12 ρ u (r = a) 2 = 1

2U2 − 9

8U2 sin2θ , or

€

p(r = a) − p∞ = 12 ρU

2 1− 94 sin

2θ( ) = 12 ρU

2 94 cos

2θ − 54( )

c) The hydrodynamic force will be determined by the pressure-force difference between the top and bottom of the hemisphere. Here the stagnation pressure is

€

p∞ + 12 ρU

2 , thus: Rx = (pressure force pushing up on the flat side) – (pressure force pushing down on the curved side)

€

= p∞ + 12 ρU

2( )πa2 − p(r = a) er ⋅ ex( )dShemisphere∫

€

= p∞ + 12 ρU

2( )πa2 − p∞ + 12 ρU

2 1− 94 sin

2θ( )[ ]ϕ=−π 2

+π 2

∫θ = 0

π

∫ cosϕ sinθ( )a2 sinθdϕdθ

€

= p∞ + 12 ρU

2( )πa2 − a2 p∞ + 12 ρU

2 1− 94 sin

2θ( )[ ]sin2θdθ cosϕϕ=−π 2

+π 2

∫θ = 0

π

∫ dϕ

θ!ϕ!

r!

z!

x!

y!


€

= p∞ + 12 ρU

2( )πa2 − 2a2 p∞ + 12 ρU

2 1− 94 sin

2θ( )[ ]sin2θdθθ = 0

π

∫

€

= p∞ + 12 ρU

2( )πa2 − 2a2 p∞ + 12 ρU

2[ ] sin2θdθθ = 0

π

∫ − 2a2 12 ρU

2( ) − 94( ) sin4 θdθθ = 0

π

∫

€

= p∞ + 12 ρU

2( )πa2 − 2a2 p∞ + 12 ρU

2[ ] π2&

' (

)

* + + a2ρU 2 9

43π8

&

' (

)

* + =27π32

ρU 2a2

where the second-to-last equality has required use of the two integral hints. d) The hemisphere will remain on the surface when its weight is greater than or equal to the pressure force and the buoyant force:

€

ρh23πa3

$

% &

'

( ) g ≥

27π32

ρU 2a2 + ρ23πa3

$

% &

'

( ) g , and this implies:

€

ρh ≥ 1+8164

U 2

ga$

% &

'

( ) ρ .


Exercise 7.44. The flow-field produced by suction flow into a round vacuum cleaner nozzle held above a large flat surface can be easily investigated with a simple experiment, and analyzed via potential flow in (R, ϕ, z)-cylindrical coordinates with the method of images. a) Do the experiment first. Obtain a vacuum cleaner that has a hose for attachments. Remove any cleaning attachments (brush, wand, etc) or unplug the hose from the cleaning head, and attach an extension hose or something with a round opening (~4 cm diameter is recommended). Find a smooth dry flat horizontal surface that is a ~0.5 meter or more in diameter. Sprinkle the central 1/3 of the surface with a light granular material that is easy to see (granulated sugar, dry coffee grounds, salt, flour, talcum powder, etc. should work well). The grains should be 1/2 to 1 mm apart on average. Turn on the vacuum cleaner and lower the vacuum hose opening from ~0.25 meter above the surface toward the surface with the vacuum opening facing toward the surface. When the hose gets to within about one opening diameter of the surface or so, the granular material should start to move. Once the granular material starts moving, hold the hose opening at the same height or lift the hose slightly so that grains are not sucked into it. If many grains are vacuumed up, distribute new ones in the bare spot(s) and start over. Once the correct hose-opening-to-surface distance is achieved, hold the hose steady and let the suction airflow of the vacuum cleaner scour a pattern into the distributed granular material. Describe the shape of the final pattern, and measure any relevant dimensions. Now see if ideal flow theory can explain the pattern observed in part a). As a first approximation, the flow field near the hose inlet can be modeled as a sink (a source with strength –Q) above an infinite flat boundary since the vacuum cleaner outlet (a source with strength +Q) is likely to be far enough away to be ignored. Denote the fluid density by ρ, the pressure far away by p∞, and the pressure on the flat surface by p(R). The potential for this flow field will be the sum of two terms:

€

φ(R,z) =+Q

4π R2 + (z − h)2+ K(R,z)

b) Sketch the streamlines in the y-z plane for z > 0. c) Determine K(R,z). d) Use dimensional analysis to determine how p(R) – p∞ must depend on ρ, Q, R, and h. e) Compute p(R) – p∞ from the steady Bernoulli equation. Is this pressure distribution consistent with the results of part a)? Where is the lowest pressure? (This is also the location of the highest speed surface flow). Is a grain at the origin of coordinates the one most likely to be picked up by the vacuum cleaner? Solution 7.44. a) The pattern scoured in the granular material is axisymmetric. The scoured region is a ring that is a little larger than the vacuum cleaner opening. A pile of the granular material is left directly below the center of the vacuum cleaner opening.


c) Based on the method of images, K must represent a sink at z = –h, so:

€

K(R,z) =+Q

4π R2 + (z + h)2

d) 5 parameters - 3 dimensions = 2 groups. Clearly Π1 = R h . A little more effort yields:

€

Π2 = P(R) − P∞( )h4 ρQ2 . Therefore:

€

P(R) − P∞ = ρQ2 h4( ) f R h( ) e) Use the potential and the Bernoulli equ. There is no velocity very far from the origin and there will be no vertical velocity on z = 0, so:

€

P∞ = P(R) +12ρ uR

2[ ]z= 0. Here:

€

uR =∂φ∂R

=−QR

4π R2 + (z − h)2[ ]3 2+

−QR4π R2 + (z + h)2[ ]3 2

, so

€

uR[ ]z= 0 =−QR

2π R2 + h2[ ]3 2,

and

€

P(R) − P∞ =−ρQ2R2

8π 2 R2 + h2[ ]3= −

18π 2

ρQ2

h4R h( )2

1+ R h( )2[ ]3 .

This pressure distribution is consistent with results of part a). The radial location of the minimum pressure (maximum surface flow speed) is found from:

€

ddR

P(R) − P∞( ) = 0 = −18π 2

ρQ2

h41

1+ R h( )2[ ]3 − 3

R h( )2

1+ R h( )2[ ]4

&

' (

) (

*

+ (

, (

2Rh2

.

Dividing out all the non-zero factors and parameters yields:

€

R 1− 2 R h( )2( ) = 0, which implies: R

= 0, or

€

R = h 2 . The second answer matches (within experimental accuracy) the radius of the ring on the surface where the scouring was most complete in the experiment. The first answer, R = 0, is a stagnation point (a pressure maximum) so a grain located there is not likely to be picked up. In reality, when the vacuum cleaner hose opening is held close enough to the surface, even the r = 0 grains are picked up because of the finite size of the grains and because of unsteady flow processes. Overall, when the hose opening is comfortably above the surface, the primary difference between the experiment and the analysis is the size of the sink. In the experiment, air suction takes place across a finite area whereas the potential flow sink is a point in space. Thus, some minor differences between theoretical and experimental results are expected, but such differences might not be easily detected given the simplicity and qualitative nature of the experiments conducted for part a).

y

z1. a)b)


Exercise 7.45. There is a point source of strength Q (m3/s) at the origin, and a uniform line sink of strength k = Q/a extending from z = 0 to z = a. The two are combined with a uniform stream U parallel to the z-axis. Show that the combination represents the flow past a closed surface of revolution of airship shape, whose total length is the difference of the roots of:

€

z2

a2za

±1"

# $

%

& ' =

Q4πUa2

Solution 7.45. From the symmetry of the flow, the stagnation points that define the length of the body must lie on the z-axis. Consider first the stagnation point, P, to the right of the source and line sink. The fluid velocity, uz, on the z-axis will be that due to the free stream, the source, and the line sink.

€

uz(0,z) =U +Q4πz2

−k

4π (z −ζ )20

a

∫ dζ .

Substitute in for the value of k, perform the integration and set the result equal to zero to get:

€

0 =U +Q4πz2

1z2−

1z(z − a)

$

% & '

( ) , which implies:

€

z2

a2za−1

#

$ %

&

' ( =

Q4πUa2

.

When P is to the left of the source and line sink, the on-axis fluid velocity is:

€

uz(0,z) =U −Q4πz2

+k

4π (z −ζ )20

a

∫ dζ .

Repeating the prior steps produces:

€

z2

a2za

+1"

# $

%

& ' =

Q4πUa2

.

z

d!!

U

aQ

P


Exercsie 7.46. Using a computer, determine the surface contour of an axisymmetric half-body formed by a line source of strength k (m2/s) distributed uniformly along the z-axis from z = 0 to z = a and a uniform stream. The nose of this body is more pointed than that formed by the combination of a point source and a uniform stream. From a mass balance, show that far downstream the radius of the half-body is

€

r = ak πU . Solution 7.46. The Stokes stream function for this flow is given by:

€

ψ =12UR2 − k

4π(z −ζ )dζ(z −ζ )2 + R20

a

∫ =12UR2 +

k4π

(z − a)2 + R2 − z2 + R2( ),

where the final equality follows from evaluating the integral. The body contour will follow the dividing streamline that starts from the stagnation point located at R = 0 on the negative part of the z-axis. On the negative z-axis, the difference of square roots above is +a, so the body contour will follow the R-z trajectory implicitly specified by:

€

ka4π

=12UR2 +

k4π

(z − a)2 + R2 − z2 + R2( ) .

For computer evaluation dimensionless variable are best. Therefore, define z* = z/a and R*= R/a, and find:

€

1=ΩR*2 + (z* −1)2 + R*2 − z*2 + R*2 , where Ω = 2πaU/k is the dimensionless flow speed. This implicit relationship is plotted below for Ω = 2.0. The flow is from left to right, the R* axis is vertical, and the z* axis is horizontal.

The flow becomes uniform with speed U far downstream of the body. If r is the asymptotic radius of the half body, then a mass balance gives:

Uπr2 = flow out of the line source = ka, which implies:

€

r = ka πU

!"

!#$"

%"

&%" &!#$" !" !#$" %" %#$" '"


Exercise 7.47. Consider the radial flow induced by the collapse of a spherical cavitation bubble of radius R(t) in a large quiescent bath of incompressible inviscid fluid of density ρ. The pressure far from the bubble is p∞. Ignore gravity. a) Determine the velocity potential φ(r,t) for the radial flow outside the bubble. b) Determine the pressure p(R(t), t) on the surface of the bubble. c) Suppose that at t = 0 the pressure on the surface of the bubble is p∞, the bubble radius is Ro, and its initial velocity is

€

− ˙ R o (i.e. the bubble is shrinking), how long will it take for the bubble to completely collapse if its surface pressure remains constant?

Solution 7.47. a) All of the flow will be in the radial direction. Thus:

€

∇2φ =1r2

∂∂r

r2 ∂φ∂r

%

& '

(

) * = 0.

Away from r = 0, this equation can be integrated directly to find:

€

φ = −A /r , where A may be a function of time. To evaluate A, match the radial fluid velocity at the surface of the sphere to the

surface velocity of the sphere:

€

∂φ∂r$

% &

'

( )

r= R( t )

= ˙ R (t) =A(t)R2 . Solve for A to get:

€

φ = −R2 ˙ R

r.

b) Use the Bernoulli equation:

€

ρ∂φ∂t

+12ρu 2 + p = p∞ , to get:

€

p − p∞ρ

= +2R ˙ R 2

r+

R2 ˙ ̇ R r

−R4 ˙ R 2

2r4 .

Evaluate this on the surface of the bubble r = R(t) to get:

€

p(R,t) − p∞ρ

=32

˙ R 2 + R˙ ̇ R

c) Set p(R,t) = p∞ to get:

€

0 =32

˙ R 2 + R˙ ̇ R = R−1/ 2 ddt

(R3 / 2 ˙ R ) , and integrate:

€

R3 / 2 ˙ R = C . Integrate

again to find:

€

2R5 / 2 5 = Ct + D . Now evaluate the constants from the given information:

€

−Ro3 / 2 ˙ R o = C , and

€

2Ro5 / 2 5 = D . Thus,

€

R(t) = Ro5 / 2 − 5 2( )Ro

3 / 2 ˙ R ot[ ]2 5

, so the collapse time is:

€

t = 2Ro (5 ˙ R o).


Exercise 7.48. Derive the apparent mass per unit depth into the page of a cylinder of radius a that travels at speed

€

Uc (t) = dxc dt along the x-axis in a large reservoir of an ideal quiescent fluid with density ρ. Use an appropriate Bernoulli equation and the following time-dependent two-dimensional

potential:

€

φ(x,y,t) = −a2Uc (x − xc )(x − xc )

2 + y 2, where xc(t) is location of the center of the cylinder, and the

Cartesian coordinates are x and y. [Hint: steady cylinder motion does not contribute to the cylinder’s apparent mass; keep only the term (or terms) from the Bernoulli equation necessary to determine apparent mass]. Solution 7.48. In an ideal incompressible fluid, the hydrodynamic loads on the moving cylinder will occur through pressure forces and the pressure will be set by the unsteady Bernoulli equation (UBE).

€

ρ∂φ∂t

+12ρ∇φ

2+ ρgy + p = p∞ or

€

p∞ − p = ρ∂φ∂t

+12ρ∇φ

2+ ρgy

The force that the fluid applies to the cylinder, Fc, is

€

Fc = (p∞ − p)ndS =surface∫ (p∞ − p)erBadθθ = 0

θ = 2π∫

where B is the length of the cylinder along its axis,

€

er = ex cosθ + ey sinθ , and θ is an angle measured from the positive x-axis with its apex at the center of the cylinder. In addition, define the distance from the center of the cylinder as

€

R(t) = x − xc (t)( )2 + y 2 so that

€

cosθ = x − xc (t)( ) R(t). Now the potential can be written:

€

φ(R,θ) = − a2Uc R( )cosθ where R, θ, and Uc are all functions of time for a fixed (x,y) location. Only the unsteady term in the UBE will contribute a term that involves

€

dUc dt ≡ ˙ U c , thus:

€

p∞ − p = −ρ a2 ˙ U c R( )cosθ + terms leading to no net force( ). Evaluate this one unsteady term on R = a,

€

p∞ − p( )R= a = −ρa ˙ U c cosθ + ..., and this leads to:

€

Fc B = −ρa2 ˙ U c cosθ(ex cosθ + ey sinθ)dθθ = 0

θ = 2π∫ = −ρa2 ˙ U c (πex + 0ey ) = −πa2ρ ˙ U cex

So, the apparent mass of the cylinder is equal to the mass of the fluid displaced by the cylinder. The streamlines for this flow when Uc < 0 are shown in Figure 3.2b.


Exercise 7.49. A stationary sphere of radius a and mass m resides in inviscid fluid with constant density ρ. a) Determine the buoyancy force on the sphere when gravity g acts downward. b) At t = 0, the sphere is released from rest. What is its initial acceleration? c) What is the sphere’s initial acceleration if it is a bubble in a heavy fluid (i.e. when m → 0)? Solution 7.49. a) The buoyancy force arises from the static pressure acting on the surface of the sphere. Choose the spherical coordinate system so that ez points opposite gravity and the hydrostatic pressure is

€

p = po − ρgz . Thus the buoyancy force on sphere will be:

€

FB = − (p − po)era2 sinθdθdϕ

0

π

∫0

2π

∫ = (po − p) ex cosϕ sinθ + ey sinϕ sinθ + ez cosθ( )a2 sinθdθdϕ0

π

∫0

2π

∫

Performing the ϕ integration eliminates two of the coordinate directions, so

€

FB = 2π (+ρgz)eza2 cosθ sinθdθ

0

π

∫ = 2πρga3ez cos2θ sinθdθ0

π

∫ ,

where

€

p − po = −ρgz, and z = acosθ have been used. The θ-integration is completed via a straightforward change of variable:

€

FB = +2πρga3ez cos2θ sinθdθ0

π

∫ = −2πρga3ez β 2dβ1

−1

∫ = +43πρga3ez .

b) Newton’s second law for the sphere will be:

€

FB −mgez = (m + madded )dusdt

, where madded is 1/2

the mass of the displaced fluid. Thus,

€

dusdt

=FB −mgezm + madded

=(4 /3)ρa3g −mgm + 2 /3( )ρa3

ez =(4 /3)ρa3 −m2 /3( )ρa3 + m

gez

c) If m → 0, then

€

dusdt

= 2gez .


Exercise 7.50. A sphere of mass m and volume V is attached to the end of a light thin flexible cable of length L. In vacuum, with gravity g acting, the natural frequencies for small longitudinal (bouncing) and transverse (pendulum) oscillations of the sphere are ωb and ωp. Ignore the effects of viscosity, and estimate these natural frequencies when the same sphere and cable are submerged in water with density ρw. What is ωp when

€

m << ρwV ? Solution 7.50. For the longitudinal (bouncing) vibrations, the natural frequency will be

determined from the simple linear differential equation:

€

M d2zdt 2

+σz = 0 where z is the coordinate

of the sphere’s center of mass in longitudinal direction, and σ is the extensional stiffness of the cable. In a vacuum, there will be no fluid dynamic effects so the longitudinal natural frequency is

€

ωb = σ M . When the sphere is placed in water, it will have added mass equal to one half the mass of the displaced fluid, so that small longitudinal oscillations will follow:

€

M + 12 ρwV[ ] d

2zdt 2

+ γdzdt

+σz = 0 ,

where γ is the viscous damping coefficient that occurs because of the viscosity of water. The buoyant force does enter here because it is steady and it does not contribute to the restoring force. However, the buoyant force does make a difference in the equilibrium length of the cable. When the effects of viscosity are ignored, the in-water natural frequency is seen to be

€

ωb,w = σ M + 12 ρwV( ) =ωb 1+ ρwV 2M ,

which is lower than the in-vacuum natural frequency for any ρw, V, and M. Small transverse (pendulum) oscillations in vacuum will be determined by

€

M d2θdt 2

+ M gLθ = 0, where θ is the angle of departure from the vertical, so that

€

ω p = g L .

When the sphere is submerged in water, both added mass and buoyancy influence its pendulum frequency because both contribute to the restoring force when θ ≠ 0. First consider the case when the sphere sinks in the water (

€

M > ρwV ). Here, the differential equation governing small

changes in θ is

€

M + 12 ρwV[ ] d

2θdt 2

+ γdθdt

+ M − ρwV( ) gLθ = 0 ,

where again γ is the viscous damping coefficient. Therefore, the inviscid natural frequency

estimate is

€

ω p,w =gL#

$ % &

' ( M − ρwVM + 1

2 ρwV=ω p

1− ρwV M1+ ρwV 2M

.

If

€

M < ρwV , pendulum oscillations will still occur but the sphere is now a float and the cable restrains it from rising to the water surface, thus, the governing differential equation is:

€

M + 12 ρwV[ ] d

2θdt 2

+ γdθdt

+ −M + ρwV( ) gLθ = 0

so the natural frequency is

€

ω p,w =gL#

$ % &

' ( −M + ρwVM + 1

2 ρwV=ω p

−1+ ρwV M1+ ρwV 2M

.

Combining the cases produces:

€

ω p,w =ω p1− ρwV M1+ ρwV 2M

, so

€

ω p,w =ω p 2 when

€

M << ρwV .


Exercise 7.51. Determine the ideal-flow force on a stationary sphere for the following unsteady flow conditions. a) The free stream of velocity Uez is constant but the sphere’s radius a(t) varies. b) The free stream velocity magnitude changes, U(t)ez, but the sphere’s radius a is constant. c) The free stream velocity changes direction U(excosΩt+ eysinΩt), but its magnitude U and the sphere’s radius a are constant. Solution 7.51. a) The potential for a stationary sphere with a constant radius in a uniform flow parallel to the z-axis is given by (7.85):

€

φ =Ur 1+a3

2r3#

$ %

&

' ( cosθ .

When the sphere's radius varies, a time dependent source term proportional to 1/r must be added to this potential to produce the requisite radial velocity at r = a. The strength of this source term is Q = 4πa2(da/dt), thus the appropriate time dependent potential for this situation is:

€

φ =Ur 1+a3

2r3#

$ %

&

' ( cosθ −

a2

rdadt

, so

€

∂φ∂t

=32U a2

r2$

% &

'

( ) dadtcosθ − 2 a

rdadt

$

% &

'

( ) 2

−a2

rd2adt 2

,

€

∂φ∂z

= uz =U 1+a3

2r3$

% &

'

( ) −32U a3z2

r5+a2zr3

dadt

=U 1+a3

2r31− 3cos2θ( )

$

% &

'

( ) +

a2

r2dadtcosθ ,

€

∂φ∂r

= ur =U 1− a3

r3%

& '

(

) * cosθ +

a2

r2dadt

and

€

1r∂φ∂θ

= uθ = −U 1+a3

2r3&

' (

)

* + sinθ .

Now consider a CV that encloses the sphere and expands and contracts with it. In this case, the control surface velocity will be b = (da/dt)er. The integral momentum equation (4.17) becomes:

€

ddt

ρudVCV∫ + ρu(u−b) ⋅ndA

CS∫ =

ddt

ρudVCV∫ + ρu ur −

dadt

&

' (

)

* + dA

CS∫ = − p

CS∫ erdA + Fs.

Here the control surface is spherical so n = er, the flux integral is zero because ur = da/dt on the control surface (r = a), and Fs is an externally applied force that holds the sphere stationary. The ideal flow force on the sphere, FIF, will be equal and opposite to Fs so

€

FIF = −Fs = − p[ ]r=aera2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ −ddt

ρur2 sinθdϕdθϕ= 0

2π

∫θ = 0

π

∫r= 0

a

∫ dr .

From the symmetry of the flow, the only non-zero force component is in the z-direction:

€

FIF( )z = ez ⋅FIF = − p[ ]r=a cosθa2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ −ddt

ρuzr2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫r= 0

a

∫ dr .

The pressure is obtained from the unsteady Bernoulli equation:

€

ρ∂φ∂t

+12ρ∇φ

2+ p = p∞ +

12ρU 2 , or

€

p = p∞ +12ρ U 2 − ∇φ

2( ) − ρ ∂φ∂t .

On r = a this simplifies to:

€

p[ ]r= a = p∞ +12ρ U 2 −

dadt

%

& '

(

) * 2

−94U 2 sin2θ

%

& ' '

(

) * * − ρ

32U dadtcosθ − 2 da

dt%

& '

(

) * 2

− a d2adt 2

%

& ' '

(

) * * .


This can be rearranged into terms that do not lead to a net force, those in [,]-brackets below, and the one term that does, the final term:

€

p[ ]r= a = p∞ +32ρdadt

$

% &

'

( ) 2

+12ρU 2 1− 9

4sin2θ

$

% &

'

( ) + ρa

d2adt 2

,

- .

/

0 1 − ρ

32U dadtcosθ .

Therefore, the pressure contribution to the ideal-flow force is:

€

− p[ ]r=aera2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ = 3πρUa2 dadt

cos2θ sinθdθθ = 0

π

∫ = 2πρUa2 dadt

,

which is a drag force when da/dt > 0. Similarly, the unsteady term's contribution to the ideal-flow force is:

−ddt

ρuzr2 sinθ dϕ dθ

ϕ=0

2π

∫θ=0

π

∫r=0

a

∫ dr

= −2πρ ddt

U 1+ a3

2r3 1−3cos2θ( )#

$%

&

'(+

a2

r2dadt

cosθ#

$%

&

'(sinθ dθ

θ=0

π

∫r=0

a

∫ r2dr

= −2πρ ddt

U 1+ a3

2r3 1−3β 2( )#

$%

&

'(+

a2

r2dadtβ

#

$%

&

'(dβ

β=−1

+1

∫r=0

a

∫ r2dr

= −2πρ ddt

U β +a3

2r3 β −β 3( )#

$%

&

'(+

a2

r2dadtβ 2

2

)

*+

,

-.−1

+1

r=0

a

∫ r2dr = −2πρU ddt

2r=0

a

∫ r2dr

= − 4π3ρU d

dta3( ) = −4πρUa2 da

dt.

This term represents momentum change that occurs within the sphere. Combining the pressure and internal-momentum-change forces leads to:

€

FIF( )z = −2πρUa2 dadt

(sphere internal flow included).

which is a thrust force when da/dt > 0; the sphere is pushed upstream when its radius increases. In this case, momentum change in the sphere's interior provides the decisive contribution to the ideal flow force. However, if the sphere's interior is considered to be massless or motionless, or to have zero or constant momentum, then the unsteady term's contribution is zero, and the ideal-flow force is just the drag that results from the pressure distribution on the sphere's surface:

€

FIF( )z = +2πρUa2 dadt

(sphere-internal flow excluded).

b) In this case the sphere's geometry doesn't change, thus the potential is:

€

φ =U(t)r 1+a3

2r3#

$ %

&

' ( cosθ ,

and this leads to a ∂φ/∂t term in the pressure integral that is not present in the steady flow case. The requisite time derivative and the velocity components are:

€

∂φ∂t

=dUdt

r 1+a3

2r3$

% &

'

( ) cosθ ,

€

∂φ∂r

= ur =U 1− a3

r3%

& '

(

) * cosθ and

€

1r∂φ∂θ

= uθ = −U 1+a3

2r3&

' (

)

* + sinθ .

This time consider a fixed-size CV that encloses the sphere. Use the integral momentum equation (4.17) to find:


€

FIF( )z = − p[ ]r=a cosθa2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ −ddt

ρuzr2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫r= 0

a

∫ dr .

The flux term considered in part a) is zero here because the sphere doesn't change size; its surface is motionless (ur = 0 at r = a). The unsteady term's contribution to the ideal-flow force is

€

−ddt

ρuzr2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫r= 0

a

∫ dr = −4π3ρa3 dU

dt,

and can be obtained by repeating the part a) analysis of this term with a = const. and U = U(t). To evaluate the pressure term, use the Bernoulli equation

€

ρ∂φ∂t

+12ρ∇φ

2+ p = const.

from the stagnation point at r = a and θ = π to any other point on the surface of the sphere to find the pressure p(a,θ) on the surface of the sphere:

€

ρ32a dUdtcosθ

$

% &

'

( ) +12ρ U 2 −

94U 2 sin2θ

$

% &

'

( ) + p(a,θ) = −ρ

32a dUdt

+12ρU 2 + p(a,π ), or

€

p[ ]r= a ≡ p(a,θ) = p(a,π ) − ρ 32a dUdt(1+ cosθ) +

12ρU 2 9

4U 2 sin2θ

'

( )

*

+ , .

Only the term involving the cosine leads to a net force:

€

− p[ ]r=a cosθa2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ = −2π ρ −32a dUdtcosθ

(

) *

+

, - cosθa2 sinθdθ

θ = 0

π

∫ = 2πρa3 dUdt

,

which is a drag force; the flow pushes the sphere downstream when dU/dt > 0. Thus, as in part a), there are two possible answers:

€

FIF( )z = +23πρa3 dU

dt (sphere internal flow included), and

€

FIF( )z = +2πρa3 dUdt

(sphere-internal flow excluded).

When sphere's internal flow is neglected, the resulting drag arises from two sources. First, the pressure gradient in the accelerating uniform stream (∂p/∂z = –ρdU/dt) leads to a force similar to buoyancy with dU/dt replacing the acceleration of gravity, (4/3)πρa3dU/dt, that pushes the sphere toward lower pressure (downstream). Second, the accelerating free stream must go around the sphere and this leads to an apparent mass effect, (2/3)πρa3dU/dt, that again pushes the sphere toward lower pressure. The sum of these two effects correctly recovers the result above, 2πρa3dU/dt. c) Although it is tempting, the switch to a rotating coordinate frame with same origin and z-axis, such a switch is not necessarily helpful because of the kinematic relationship

€

u[ ]inertial frame =Ω× x + u[ ]rotating frame , where Ω is the angular rotation rate of the rotating frame. With z-axis vertical, Ω = Ωez, x = (x, y, z), and an inertial-frame velocity of u = U(excosΩt+ eysinΩt), this kinematic relationship becomes:

€

u[ ]inertial frame = U(ex cosΩt + ey sinΩt) =Ω× x + u[ ]rotating frame =Ω(− % y % e x + % x % e y ) + % u , where the primes denote coordinates, velocities, and unit vectors in the rotating frame; and the cross product has been computed in the rotating frame. Here, the x´-direction can be chosen to simplify the inertial-frame velocity:

€

" e x = ex cosΩt + ey sinΩt , and this leaves:


€

" u = u[ ]inertial frame −Ω(− " y " e x + " x " e y ) = U " e x −Ω(− " y " e x + " x " e y ) , which is only a clear simplification of

€

u[ ]inertial frame =U(ex cosΩt + ey sinΩt) along the z-axis in the rotating frame where (x´, y´) = (0, 0). Therefore, use an inertial frame of reference with the z-axis vertical, so the flow far from sphere lies in the x-y plane. In this case, the potential can be determined starting from (7.88):

€

φ = U(t) − d(t)4π x 3

%

& ' '

(

) * * ⋅ x .

where the free stream velocity changes direction,

€

U =U(ex cosΩt + ey sinΩt), and the dipole strength opposes the free stream and sets the sphere's diameter,

€

d = −2πUa3(ex cosΩt + ey sinΩt). Using spherical coordinates,

€

x = exrcosϕ sinθ + eyrsinϕ sinθ + ezrcosθ , performing the dot products, and using the two-angle sum formula for the cosine function allows the potential to be written entirely in terms of scalars:

€

φ =U r +a3

2r2#

$ %

&

' ( cos ) ϕ sinθ ,

where ϕ´ = ϕ – Ωt. The radial and angular velocities, and ∂φ/∂t are:

€

∂φ∂r

= ur =U 1− a3

r3%

& '

(

) * cos + ϕ sinθ ,

€

1r∂φ∂θ

= uθ =U 1+a3

2r3%

& '

(

) * cos + ϕ cosθ ,

€

1rsinθ

∂φ∂ϕ

= uϕ = −U 1+a3

2r3'

( )

*

+ , sin - ϕ , and

€

∂φ∂t

=ΩU r +a3

2r2%

& '

(

) * sin + ϕ sinθ .

As in parts a) and b), the ideal flow force on the sphere, FIF, can be obtained from a surface integral of the pressure, and a volume integral over the sphere's interior.

€

FIF = − p[ ]r=aera2 sinθdϕdθ

ϕ= 0

2π

∫θ = 0

π

∫ −ddt


2π

∫θ = 0

π

∫r= 0

a

∫ dr .

To evaluate the pressure term, use the Bernoulli equation,

€

ρ∂φ∂t

+12ρ∇φ

2+ p = const.,

from the stagnation point at r = a, θ = π/2, and ϕ´ = π to any other point on the surface of the sphere to find the pressure

€

p(a,θ, # ϕ ) ≡ p[ ]r= a on the surface of the sphere:

€

32ρΩUasin $ ϕ sinθ +

12ρU 2 9

4cos2 $ ϕ cos2θ + sin2 $ ϕ ( ) + p(a,θ, $ ϕ ) = p(a,π /2,π ).

Rearrange this and denote p(a, π/2, π) by ps.

€

p(a,θ, # ϕ ) = ps −98ρU 2 1− cos2 # ϕ sin2θ( ) − 32 ρΩUasin # ϕ sinθ .

Only the final term on the right leads to a net pressure force:

€


ϕ= 0

2π

∫θ = 0

π

∫

=32ρΩUa2 sin ) ϕ sin2θ ex cosϕ sinθ + ey sinϕ sinθ + ez cosθ( )dϕdθ.

ϕ= 0

2π

∫θ = 0

π

∫


The θ-integration renders the z-component zero. For the other two terms, the θ-integration is

€

sin3θdθ0

π∫ = 4 3, so the integrated pressure force simplifies to:

€


ϕ= 0

2π

∫θ = 0

π

∫ = 2ρΩUa2 sin ) ϕ ex cosϕ + ey sinϕ( )dϕ.ϕ= 0

2π

∫

Expand

€

sin " ϕ = sinϕ cos(Ωt) − cosϕ sin(Ωt), and integrate to find:

€


ϕ= 0

2π

∫θ = 0

π

∫ = 2πρΩUa3 −ex sin(Ωt) + ey cos(Ωt)( ).

This force pushes the sphere perpendicular to the instantaneous free stream direction. To evaluate the sphere's internal-flow contribution to the force, use the volume integral results from part a) but adjust them for the direction of the free-stream in this problem:

€


2π

∫θ = 0

π

∫r= 0

a

∫ dr =4π3ρa3U ex cos(Ωt) + ey sin(Ωt)( ), so that:

€

−ddt


2π

∫θ = 0

π

∫r= 0

a

∫ dr = −4π3ρa3ΩU −ex sin(Ωt) + ey cos(Ωt)( ) .

This is the force that results from sphere's internal flow contribution. So, as in parts a) and b), there are two possible answers:

€

FIF =23πρΩUa3 −ex sin(Ωt) + ey cos(Ωt)( ) (sphere internal flow included), and

€

FIF = 2πρΩUa3 −ex sin(Ωt) + ey cos(Ωt)( ) (sphere-internal flow excluded). The direction of the force can be partially explained by the pressure gradient in the accelerating

uniform stream,

€

∇p = −ρdUdt

= −ρΩU −ex sin(Ωt) + ey cos(Ωt)( ), which leads to a force similar to

buoyancy with

€

ΩU −ex sin(Ωt) + ey cos(Ωt)( ) replacing the acceleration of gravity. This buoyancy-mimicking force pushes the sphere toward lower pressure.


Exercise 7.52. Consider the flow field produced by a sphere of radius a that moves in the x-direction at constant speed U along the x-axis in an unbounded environment of a quiescent ideal fluid with density ρ. The pressure far from the sphere is p∞ and there is no body force. The velocity potential for this flow field is:

φ(x, y, z, t) = − a3U2

x −Ut

(x −Ut)2 + y2 + z2"# $%3 2 .

a) If u = (u,v,w) , what is u(x, 0, 0, t) , the velocity along the x-axis as a function of time? [Hint: consider the symmetry of the situation before differentiating in all directions.] b) What is p(x,0,0,t), the pressure along the x-axis as a function of time? c) What is the pressure on the x-axis at x = Ut ± a? d) If the plane defined by z = h is an impenetrable flat surface and the sphere executes the same motion, what additional term should be added to the given potential? e) Compared to the sphere's apparent mass in an unbounded environment, is the sphere's apparent mass larger, the same, or smaller when the impenetrable flat surface is present?

Solution 7.52. a) To save writing let, r(t) = (x −Ut)2 + y2 + z2"# $%1 2

, so φ = − a3U2(x −Ut)r3(t)

.

Compute velocity components from ∇φ = u . Start with: u = ∂φ∂x

= −a3U2

r2 (t)−3(x −Ut)2

r5(t)#

$%

&

'( ,

evaluate this on the x-axis (y = z = 0) where r(t) = x −Ut to find: u(x, 0, 0, t) = a3Ux −Ut 3

. Here φ

depends on y2 so: v = ∂φ∂y

=∂φ∂y2

∂y2

∂y= 2y ∂φ

∂y2; thus, v(x,0,0,t) = 0. Similarly, φ depends on z2 so

w(x,0,0,t) = 0. Thus, the fluid velocity on the x-axis has only an x-component: u(x, 0, 0, t) = a3U x −Ut 3( )ex .

b) The appropriate Bernoulli equation is: ρ ∂φ∂t+12ρu 2+ p = p∞ , so p− p∞ = −ρ

∂φ∂t−12ρu 2 .

Time differentiate the potential but utilize the chain-rule and the part a) results: ∂φ∂t

=∂φ

∂(x −Ut)∂(x −Ut)

∂t=∂φ∂x

−U( ) = −Uu .

Evaluate on the x-axis: ρ ∂φ∂t

"

#$%

&'(x,0,0,t )= −

ρa3U 2

x −Ut 3, so:

p(x, 0, 0, t)− p∞ = +ρa3U 2

x −Ut 3−12ρ

a6U 2

x −Ut 6=12ρU 2 2a3

x −Ut 3−

a6

x −Ut 6#

$%%

&

'(( .

x!

y !

z !

h !

x = Ut !

U !

2a !


c) At x = Ut ± a the factor in large (,)-parentheses in the part b) result is unity, so p = p∞ + (1/2)ρU2 at these locations. This answer can also be reached by noting that the specified locations are stagnation points on the sphere when it is stationary and the flow moves past it. d) The method of images applies to this situation, so an image sphere must move along the line defined by z = 2h and y = 0. Thus, the additional term for the potential is:

−a3U2

x −Ut

(x −Ut)2 + y2 + (z− 2h)2"# $%3 2 .

e) The impenetrable flat surface requires the fluid to move faster above the real sphere to get out of its way as it travels so the sphere's apparent mass is larger when the impenetrable flat surface is present. Here it should be noted that apparent mass is a property of the sphere's geometry, the geometry of its environment, and the fluid density. It exists independently of the sphere's motion. When the sphere accelerates both its actual and apparent mass play a role. If the sphere does not accelerate then neither its actual nor its apparent mass plays a role. However, the sphere does not loose its actual mass when moving at a constant velocity, so it does not lose its apparent mass either.


Exercise 7.53. In three dimensions, consider a solid object moving with velocity U near the origin of coordinates in an unbounded quiescent bath of inviscid incompressible fluid with density ρ. The kinetic energy of the moving fluid in this situation is:

€

KE =12ρ ∇φ

2

V∫ dV

where φ is the velocity potential and V is a control volume that contains all of the moving fluid but excludes the object. (Such a control volume is shown in the figure for Exercise 7.29 when

€

A3 → 0 and U = 0.)

a) Show that

€

KE = −12ρ φ ∇φ ⋅n( )

A∫ dA where A encloses the body and is coincident with its

surface, and n is the outward normal on A. b) The apparent mass, M, of the moving body may be defined by

€

KE = 12 MU

2 . Using this definition, the result of a), and (7.97) with xs = 0, show that M = 2πa3ρ/3 for a sphere. Solution 7.53. Start from the given equation, use the recommended CV, add

€

φ∇2φ = 0 to the integrand, and apply Green's divergence theorem.

€

KE =ρ2

limA3 →0

∇φ2

V∫ dV =

ρ2

limA3 →0

∇φ ⋅∇φ( )V∫ dV

=ρ2

limA3 →0

∇φ ⋅∇φ + φ∇2φ( )V∫ dV =

ρ2

limA3 →0

∇ ⋅ φ∇φ( )V∫ dV

=ρ2

limA3 →0

φ∇φ ⋅niAi

∫ dAi=1

3

∑

where n1 is the normal on A1, n2 is the normal on A2, and n3 is the normal on A3. When the limit is taken, the net contribution of A3 is zero because the surface area of A3 goes to zero. This leaves:

€

KE =ρ2

φ∇φ ⋅n1A1

∫ dA +ρ2

φ∇φ ⋅n2A2

∫ dA ,

where n1 points away from the origin on A1 and n2 points inward on A2. By definition, the volume V enclosed by A1 contains all the moving fluid, thus

€

∇φ ⋅n1 = u ⋅n1 = 0 on A1, so the first integral in the last equation is equal to zero. Now define the outward normal on A2 as n = – n2, and drop the subscript from the body-conforming surface A2 to reach:

€

KE = −ρ2

φ∇φ ⋅nA∫ dA .

b) The potential for a sphere of radius a moving at velocity U that is instantaneously centered at the origin of a spherical coordinate system is:

€

φ = −a3

2r3U ⋅ x = −

a3

2r2U cosθ ,

where the second equality follows when the direction of the z-axis is chosen to coincide with the sphere's velocity at the moment of interest. For a sphere, n = er and surface area element is dA = a2sinθdθdϕ, so the kinetic energy integral becomes:


€

KE = −ρ2

φ∂φ∂r

&

' ( )

* + r= aϕ= 0

2π

∫θ = 0

π

∫ a2 sinθdθdϕ = −ρ2

−a2U 2 cos2θ

&

' ( )

* + r= aϕ= 0

2π

∫θ = 0

π

∫ a2 sinθdθdϕ

= πρa3

2U 2 cos2θ

θ = 0

π

∫ sinθdθ = −πρa3

2U 2 β 2dβ =

1

−1

∫ πρa3

2U 2 2

31

2 3 4

5 6

=12ρ

2πa3

31

2 3

4

5 6 U 2 .

The parentheses in the final equality contain the apparent mass M.

Engineering

Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (7)