fluid mechanics kundu cohen 6th edition solutions sm ch (5)

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

Exercise 5.1. A closed cylindrical tank 4 m high and 2 m in diameter contains water to a depth of 3 m. When the cylinder is rotated at a constant angular velocity of 40 rad/s, show that nearly 0.71 m2 of the bottom surface of the tank is uncovered. [Hint: The free surface is in the form of a paraboloid of revolution. For a point on the free surface, let h be the height above the (imaginary) vertex of the paraboloid and r be the local radius of the paraboloid. From Section 5.1, h =ω0

2r2 2g , where ω0 is the angular velocity of the tank. Apply this equation to the two points where the paraboloid cuts the top and bottom surfaces of the tank.] Solution 5.1. Denote the local radius of the paraboloid of revolution as r(h), and let h = 0 at vertex of this parabaloid. The goal of this problem is to determine πr2(h1), where h1 coincides with the bottom of the tank. Start with the geometric constraints. The volume V of water in the tank is: V = π(1.0m)2(3m) = 3π m3, where the tank radius is R = 1 m. This volume is conserved. Thus, when the tank is rotating, the following must be true:

€

V = π R2 − r2(h)( )h= h1

h= h2

∫ dh ,

where h2 – h1 = 4 m. From the hint we can write:

€

h =ω02r2 2g , or

€

r2 = 2gh ω02 ,

so that the volume relationship becomes:

€

V = π R2 −2gω0

2 h%

& '

(

) *

h= h1

h= h2

∫ dh = π R2h − 2gω0

2h2

2%

& '

(

) * h1

h2

= π R2 h2 − h1( ) − gω0

2 h22 − h1

2( )%

& '

(

) *

= π h2 − h1( ) R2 −gω0

2 h2 + h1( )%

& '

(

) * .

Substitute in V = 3π, R = 1, ω0 = 40, g = 9.81, and h2 – h1 = 4 m to find:

€

3π = 4π 1− 9.811600

2h1 + 4( )$

% &

'

( ) or

€

h1 =1216009.81

1− 34

#

$ %

&

' ( − 4

)

* +

,

- . =18.39m ,

so the uncovered area at the bottom of the tank is:

€

πr2(h1) = 2πgh ω02 = 2π (9.81)(18.39) 1600 = 0.0708m2.


Exercise 5.2. A tornado can be idealized as a Rankine vortex with a core of diameter 30 m. The gauge pressure at a radius of 15 m is −2000 N/m2 (i.e.,, the absolute pressure is 2000 N/m2 below atmospheric). (a) Show that the circulation around any circuit surrounding the core is 5485 m2/s. [Hint: Apply the Bernoulli equation between infinity and the edge of the core.] (b) Such a tornado is moving at a linear speed of 25 m/s relative to the ground. Find the time required for the gauge pressure to drop from −500 to −2000 N/m2. Neglect compressibility effects and assume an air temperature of 25°C. (Note that the tornado causes a sudden decrease of the local atmospheric pressure. The damage to structures is often caused by the resulting excess pressure on the inside of the walls, which can cause a house to explode.) Solution 5.2. At 25°C and one atmosphere, air density is ρair = (101.3 kPa)/(287m2s–2K–1)(298K) = 1.18 kgm–3. a) The flow is irrotational outside the Rankine vortex core, so the steady constant density Bernoulli equation implies at any distance r from the center of vortex core:

€

12ρairU∞

2 + p∞ =12ρairU

2(r) + p(r), or

€

U(r) = 2(p∞ − p(r)) ρair[ ]1 2 ,

where the "∞" subscript refers to conditions very far from the vortex (U∞ ≈ 0). From equation (3.28), U(r) = Γ/2πr, so where r = rc = the core radius:

€

Γ = 2πrc 2(p∞ − pc ) ρair[ ]1 2 = 2π (15) 2(2000) 1.18[ ]1 2 = 5485m2s−1. b) The radial distance at which the gauge pressure is –500 Pa can be determined from the results developed for part a) for U(r):

U(r) = Γ/2πr =

€

= 2(p∞ − p(r)) ρair[ ]1 2 , or

€

r = Γ 2π( ) 2(p∞ − p(r)) ρair[ ]−1 2 . Evaluate to find r, and divide by (r – rc) by 25 m/s to determine the time for the gauge pressure to drop from −500Pa to −2000Pa.

€

r = Γ 2π( ) 2(p∞ − p(r)) ρair[ ]−1 2 = 5485m2s−1 2π( ) 2(500Pa) 1.18kgm−3[ ]−1 2

= 30.0m , so time = (30m – 15m)/25ms–1 = 0.60 s.


Exercise 5.3. The velocity field of a flow in cylindrical coordinates (R, ϕ, z) is u = (uR, uϕ, uz) = (0, aRz, 0) where a is a constant. (a) Show that the vorticity components are ω = (ωR, ωϕ, ωz) = (–aR, 0, 2az). (b) Verify that

€

∇ ⋅ω = 0. (c) Sketch the streamlines and vortex lines in an (R,z)-plane. Show that the vortex lines are given by zR2 = constant. Solution 5.3. a) The vorticity components in cylindrical coordinates are provided in Appendix B:

€

ωR =1R∂uz∂ϕ

−∂uϕ∂z

= 0 – aR = −aR

€

ωϕ =∂uR∂z

−∂uz∂R

= 0 – 0 = 0 , and

€

ωz =1R∂(Ruϕ )∂R

−1R∂uR∂ϕ

= 2az – 0 = 2az .

b)

€

∇ ⋅ω =1R∂(RωR )∂R

+1R∂ωϕ

∂ϕ+∂ωz

∂z=1R∂(−aR2)∂R

+1R∂(0)∂ϕ

+∂(2az)∂z

= −2a + 2a = 0

c) The only non-zero velocity component is uϕ, so the streamlines are circles centered on the z axis. Vortex lines are given by:

€

dRωR

=dzωz

→dR−aR

=dz2az

→ − ln(R) =12ln(z) + const.

Exponentiate the last expression to find:

€

1 R = const. z , or R2z = const. Thus the vortex lines asymptote to the plane z = 0 and to the z-axis.

R

z


Exercise 5.4. Starting from the flow field of an ideal vortex (5.2), compute the viscous stresses τrr, τrθ, and τθθ, and show that the net viscous force on a fluid element, (∂τij/∂xi), is zero. Solution 5.4. In two dimensions, the ideal vortex flow field is u =(ur, uθ) = (0, Γ/2πr). This flow field is incompressible since

€

∇ ⋅u = 1 r( ) ∂uθ ∂θ( ) = 0 . Therefore, the viscous stress listings for plane polar coordinates in Appendix B apply:

τ rr = 2µSrr = 2µ∂ur∂r

= 0 , τθθ = 2µSθθ = 2µ1r∂uθ∂θ

+urr

!

"#

$

%&= 0 , and

τ rθ = 2µSrθ = µ r ∂∂r

uθr

!

"#

$

%&+1r∂ur∂θ

!

"#

$

%&= µr

Γ2πr3!

"#

$

%&(−2) = −µ

Γπr2

.

For incompressible flow (∂ui/∂xi = 0) with constant viscosity: ∂τ ij∂xi

= 2µ ∂∂xi

Sij = µ∂∂xi

∂ui∂x j

+∂uj∂xi

!

"##

$

%&&= µ

∂∂x j

∂ui∂xi

+µ∂ 2uj∂xi

2 = µ∂ 2uj∂xi

2 =∇2u .

Using the velocity field given above, and remembering to differentiate unit vectors, this becomes:

€

∇2u =1r∂∂r

r ∂∂r

uθeθ( )%

& '

(

) * +

1r2

∂ 2

∂θ 2 uθeθ( ) =1r∂∂r

reθ∂uθ∂r

%

& '

(

) * +

1r2

∂∂θ

−uθer( )

= eθ1r∂uθ∂r

+ eθ∂ 2uθ∂r2 −

uθr2 eθ = −eθ

µΓπ

−1r3 +

2r3 −

1r3

%

& '

(

) * = 0.


Exercise 5.5. Consider the situation depicted in Figure 5.6. Use a Cartesian coordinate system with a horizontal x-axis that puts the barrier at x = 0, a vertical y-axis that puts the bottom of the container at y = 0 and the top of the container at y = H, and a z-axis that points out of the page. Show that, at the instant the barrier is removed, the rate of baroclinic vorticity production at the interface between the two fluids is:

€

Dωz

Dt=2(ρ2 − ρ1)g(ρ2 + ρ1)δ

,

where the thickness of the density transition layer just after barrier removal is δ << H, and the density in this thin interface layer is assumed to be (ρ1 + ρ2)/2. If necessary, also assume that fluid pressures match at y = H/2 just after barrier removal, and that the width of the container into the page is b. State any additional assumptions that you make. Solution 5.5. At the instant the barrier is removed hydrostatic pressure will persist in each liquid. Let this pressure be p2 for the fluid on the left and p1 for the fluid on the right. If p1 and p2 match at the center of the square container, then:

€

p2 = po − ρ2g y −H 2( ) , and

€

p1 = po − ρ1g y −H 2( ) . The final answer does not depend on this matching point. Now consider a direct application of Newton's second law in to a small element, δ wide and dy tall, located at height y above the bottom of the container and straddling the interface. At the instant of interest the fluid is not yet moving so the forces come from the hydrostatic pressures. The horizontal equation of motion is:

€

ρ2 + ρ12

#

$ %

&

' ( bδdy( ) du

dt= p2 − p1( )bdy ,

where u is the element's horizontal velocity, and b is the dimension of the container into the page. Because the pressure forces are hydrostatic, there is no vertical pressure force imbalance so the vertical component of Newton's second law for the same fluid element is:

€

ρ2 + ρ12

#

$ %

&

' ( bδdy( ) dv

dt= 0 .

Thus, since v is initially zero along the interface, it remains that way until the fluid starts moving. Returning to the horizontal equation, canceling common factors, and inserting the relationships for p1 and p2 produces:

€

ρ2 + ρ12

#

$ %

&

' ( δ( ) du

dt= po − ρ2g y −H 2( ) − po + ρ1g y −H 2( ) = ρ2 − ρ1( )g −y + H 2( ) .

Simplify and solve for du/dt to find:

€

dudt

=2 ρ2 − ρ1( )gρ2 + ρ1( )δ

−y + H 2( ).

In this circumstance the z-component of vorticity is ωz = (∂v/∂x) – (∂u/∂y). At the instant of interest, v = dv/dt = 0, so

€

ddtωz =

ddt

−∂u∂y

%

& '

(

) * = −

∂∂y

dudt

%

& '

(

) * =2 ρ2 − ρ1( )gρ2 + ρ1( )δ

.

Thus, when the heavier fluid is on the left, positive vorticity directed along the z-axis is produced when the barrier is removed.

y = H

y = 0

dy

!

"2 "1


Exercise 5.6. At t = 0 a constant-strength z-directed vortex sheet is created in the x-z plane (y = 0) in an infinite pool of a fluid with kinematic viscosity ν, that is ω(y,0) = ezγδ(y). The symmetry of the initial condition suggests that ω = ωzez and that ωz will only depend on y and t. Determine ω(y,t) for t > 0 via the following steps. a) Determine a dimensionless scaling law for ωz in terms of γ, ν, y, and t. b) Simplify the general vorticity equation (5.13) to a linear field equation for ωz for this situation. c) Based on the fact that the field equation is linear, simplify the result of part a) by requiring ωz to be proportional to γ, plug the simplified dimensionless scaling law into the equation determined for part b), and solve this equation to find the undetermined function to reach:

€

ωz(y, t) =γ

2 πνtexp −

y 2

4νt' ( )

* + ,

Solution 5.6. a) This part of the exercise is dimensional analysis. The parameters are: ωz, γ, ν, y, and t. Use these to create the parameter matrix: ωz γ ν y t –––––––––––––––––––––––––––– Mass: 0 0 0 0 0 Length: 0 1 2 1 0 Time: -1 -1 -1 0 -1 This rank of this matrix is two so 5 parameters – 2 dimensions = 3 groups. These are readily constructed by inspection: ∏1 = ωzt, ∏2 = γ[t/ν]1/2, ∏3 = y/[νt]1/2. Thus, a dimensionless scaling law is: ωzt = Φ(γy/ν, y/[νt]1/2), where Φ is an unknown function. b) The vorticity equation in an inertial frame of reference is:

€

∂ω∂t

+ u ⋅ ∇( )ω = ω ⋅∇( )u+ ν∇2ω .

Here ω = (0, 0, ωz), so this equation simplifies to:

€

∂ωz

∂t+ u∂ωz

∂x+ v ∂ωz

∂y+ w ∂ωz

∂z=ωz

∂w∂z

+ ν∂ 2ωz

∂x 2+∂ 2ωz

∂y 2+∂ 2ωz

∂z2%

& '

(

) * .

where u = (u, v, w). However, the exercise statement suggests that ωz only depends on y and t. In this case the vorticity equation simplifies further.

€

∂ωz

∂t+ v ∂ωz

∂y=ωz

∂w∂z

+ ν∂ 2ωz

∂y 2%

& '

(

) * .

The boundary conditions on ωz are independent of x and z. Therefore, all the dependent field variables should all be independent of these directions. This implies ∂u/∂x = ∂w/∂z = 0, so the incompressible flow continuity equation becomes ∂v/∂y = 0 which implies v = constant. A constant velocity can be removed from the field equations via a Galilean transformation to a moving coordinate system. Thus, v can be chosen equal to zero without loss of generality, so the

middle terms in the vorticity equation drop out completely leaving:

€

∂ωz

∂t= ν

∂ 2ωz

∂y 2%

& '

(

) * .

c) For ωz to be the solution of a linear equation it must be proportional to γ the initial field strength. Therefore set: ωz = (γ/[νt]1/2)F(y/[νt]1/2), where F is merely another undetermined


function. Compute the derivatives that compose the field equation using η = y/[νt]1/2 as the argument of F.

€

∂ωz

∂t= −

12

γ

νt 3F(η) +

γνt

( F (η) − y2 νt 3

)

* +

,

- . , and

€

ν∂ 2ωz

∂y 2=νγνt

& & F (η) 1νt

,

where the prime denotes differentiation of F with respect to its argument. Recreating the field equation produces:

€

−12

γ

νt 3F(η) +

γνt

& F (η) − y2 νt 3

'

( )

*

+ , =

νγνt

& & F (η) 1νt

, or

€

" " F (η) +12η " F (η) +

12

F(η) = 0 .

This last equation can be rewritten, and integrated once to find:

€

ddη

# F (η) +12ηF(η)

$

% &

'

( ) = 0

€

→

€

" F (η) +12ηF(η) = C ,

where C is a constant. The resulting first-order differential equation has an integrating factor of exp{+η2/4} that leads to:

€

ddη

e+η 2 4F(η)( ) = Ce+η 2 4 , so

€

F(η) = De−η2 4 + e−η

2 4C e+ξ 2 4

0

η

∫ dξ .

where D is another constant. The initial condition, ωz = 0 for y > 0 at t = 0, requires C = 0. This leaves:

€

ωz(y, t) =Dνtexp −

y 2

4νt% & '

( ) *

.

To match the initial spatial profile, ωz(y,0) = γδ(y), integrate in y and take the limit as

€

t→ 0.

€

γδ(y)dy−∞

+∞

∫ = γ = limt→0

ωz (y,t)dy−∞

+∞

∫)

* +

,

- . = lim

t→0

Dνtexp −

y 2

4νt0 1 2

3 4 5 dy

−∞

+∞

∫)

* +

,

- . = 2Dlim

t→0exp −ζ 2{ }dζ

−∞

+∞

∫)

* +

,

- . = 2D π .

Thus,

€

D = γ 2 π , so the final solution is:

€

ωz(y, t) =γ

2 πνtexp −

y 2

4νt' ( )

* + ,

.


Exercise 5.7. a) Starting from the continuity and Euler equations for an inviscid compressible fluid,

€

∂ρ ∂t +∇ ⋅ ρu( ) = 0 and

€

ρ Du Dt( ) = −∇p + ρg , derive the Vazsonyi equation:

€

DDt

ωρ

$

% &

'

( ) =

ωρ

$

% &

'

( ) ⋅ ∇u+

1ρ3∇ρ ×∇p ,

when the body force is conservative:

€

g = −∇Φ. This equation shows that ω /ρ in a compressible flow plays the nearly same dynamic role as ω in an incompressible flow [see (5.26) with Ω = 0 and ν = 0]. b) Show that the final term in the Vazsonyi equation may also be written:

€

1 ρ( )∇T ×∇s . c) Simplify the Vazsonyi equation for barotropic flow. Solution 5.7. a) Start by taking the curl of the Euler equation:

€

∇ ×∂u∂t

+ (u ⋅ ∇)u = −1ρ∇p+ g

(

) *

+

, - →

∂ω∂t

+∇ × (u ⋅ ∇)u = −∇ ×1ρ∇p

0

1 2

3

4 5 +∇ × g .

The second term can modified by using the vector identity:

€

(u ⋅ ∇)u =ω ×u+ 12∇u

2 ,

€

∇ × (u ⋅ ∇)u =∇ × ω ×u( ) +∇ × 12∇u

2=∇ × ω ×u( ) ,

since

€

∇ ×∇(any scalar) = 0. This can be further manipulated using the vector identity (B3.10):

€

∇ × ω ×u( ) = u ⋅ ∇( )ω – u ∇ ⋅ω( ) +ω ∇ ⋅u( ) − ω ⋅∇( )u. Here the second term on the right is zero because

€

∇ ⋅ω =∇ ⋅ ∇ ×u( ) = 0, and the third term on the right can be rewritten using the continuity equation (4.8):

€

∇ × ω ×u( ) = u ⋅ ∇( )ω −ωρDρDt

− ω ⋅∇( )u.

The term involving the pressure in the curl of the Euler equation is:

€

−∇ ×1ρ∇p

&

' (

)

* + = − ∇

1ρ

&

' (

)

* + ×∇p−

1ρ∇ ×∇p = − ∇

1ρ

&

' (

)

* + ×∇p .

The body force term in the curl of the Euler equation is:

€

∇ × g = −∇ × ∇Φ( ) = 0. Thus, the curl of the Euler equation becomes:

€

1ρ∂ω∂t

+ u ⋅ ∇( )ω +ωρDρDt

− ω ⋅∇( )u(

) *

+

, - = −

1ρ∇1ρ

(

) *

+

, - ×∇p

when each side is divided by ρ. Use of the definition of D/Dt simplifies the left side:

€

1ρDωDt

+ωD 1 ρ( )Dt

−ωρ⋅∇

'

( )

*

+ , u = −

1ρ∇1ρ

'

( )

*

+ , ×∇p .

Combine the first to terms, and perform two simple rearrangements to reach the requisite form:

€

DDt

ωρ

$

% &

'

( ) = +

ωρ⋅∇

$

% &

'

( ) u+

1ρ3∇ρ ×∇p .

b) Start from the thermodynamic property relationship: de = Tds – pd(1/ρ). Therefore:

€

∇e = T∇s− p∇ 1 ρ( ), and

€

∇ ×∇e = 0 =∇ × T∇s( ) −∇ × p∇ 1 ρ( )( ) =∇T ×∇s−∇p ×∇ 1 ρ( ), since

€

∇ ×∇s =∇ ×∇ 1 ρ( ) = 0 . Thus,

€

1 ρ3( )∇p ×∇ρ = 1 ρ( )∇T ×∇s. c) In barotropic flow,

€

∇p and

€

∇ρ will be parallel, so the Vazsonyi equation becomes

€

D Dt( ) ω ρ( ) = ω ρ( ) ⋅ ∇u


Exercise 5.8. Starting from the unsteady momentum equation for a compressible fluid with constant viscosities:

€

ρDuDt

+∇p = ρg + µ∇2u+ µυ + 13µ( )∇ ∇ ⋅u( ) ,

show that

€

∂u∂t

+ω ×u = T∇s−∇ h + 12 u

2+Φ( ) − µ

ρ∇ ×ω +

1ρ

µυ + 43 µ( )∇ ∇ ⋅u( )

where T = temperature, h = enthalpy per unit mass, s = entropy per unit mass, and the body force is conservative:

€

g = −∇Φ. This is the viscous Crocco-Vazsonyi equation. Simplify this equation for steady inviscid non-heat-conducting flow to find the Bernoulli equation (4.78)

€

h + 12 u

2+Φ = constant along a streamline,

which is valid when the flow is rotational and nonisothermal. Solution 5.8. Divide the given momentum equation by ρ and use a vector identity (B3.9 with u = F) for the advective acceleration term,

€

(u ⋅ ∇)u =ω ×u+ 12∇u

2 ,

€

DuDt

+1ρ∇p =

∂u∂t

+ω ×u+ 12∇u

2+1ρ∇p = g +

µρ∇2u+

1ρ

µυ + 13µ( )∇ ∇ ⋅u( ) .

Using the second equality,

€

g = −∇Φ , and the vector identity (B3.13) for the Laplacian of a vector,

€

∇2u =∇(∇ ⋅u) −∇ × (∇ ×u) =∇(∇ ⋅u) −∇ ×ω , allows the last equation to be rewritten:

€

∂u∂t

+ω ×u+ 12∇u

2+1ρ∇p = −∇Φ+

µρ∇(∇ ⋅u) −∇ ×ω( ) +

1ρ

µυ + 13µ( )∇ ∇ ⋅u( ) .

Collect and combine terms to see how close this is to the target equation.

€

∂u∂t

+ω ×u = −1ρ∇p−∇ 1

2 u2

+Φ( ) − µρ∇ ×ω +

1ρ

µυ + 43 µ( )∇ ∇ ⋅u( ).

Comparing this equation with the target equation shows that the term involving the thermodynamic variable must be modified. Use the property relationship, dh = Tds + (1/ρ)dp, to deduce:

€

∇h = T∇s+ 1 ρ( )∇p → − 1 ρ( )∇p = T∇s−∇h . Substituting this into the last equation produces:

€

∂u∂t

+ω ×u = T∇s−∇ h + 12 u

2+Φ( ) − µ

ρ∇ ×ω +

1ρ

µυ + 43 µ( )∇ ∇ ⋅u( ),

which is the requisite result. Steady flow implies (∂/∂t)(anything) = 0. Inviscid implies µ = µυ = 0. Inviscid (frictionless) non-heat conducting flows are isentropic, so

€

∇s = 0 . Thus, the Crocco-Vazsonyi equation simplifies to:

€

ω ×u = −∇ h + 12 u

2+Φ( ) .

Take the dot-product of this equation with the streamline direction,

€

es ≡ u u , noting that

€

ω ×u is perpendicular to es.

€

es ⋅ ω ×u( ) = 0 = −es ⋅ ∇ h + 12 u

2+Φ( ) →

∂∂s

h + 12 u

2+Φ( ) = 0,

where s is the path length along a streamline. Thus,

€

h + 12 u

2+Φ = constant along a streamline.


Exercise 5.9. a) Solve

€

∇2G(x, # x ) = δ(x − # x ) for G(x,x´) in a uniform unbounded three-dimensional domain, where δ(x – x´) = δ(x – x´)δ(y – y´)δ(z – z´) is the three dimensional Dirac delta function.

b) Use the result of part a) to show that:

€

φ(x) = −1

4πq( % x )x − % x all % x

∫ d3 % x is the solution of the Poisson

equation

€

∇2φ(x) = q(x) in a uniform unbounded three-dimensional domain. Solution 5.9. a) 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 G becomes:

€

∇XYZ2 G = δ(X) = δ(x − % x ) .

Integrate this equation inside a sphere of radius r:

∇XYZ2 GdV

sphere∫∫∫ = ∇XYZG ⋅ndA

spherical surface∫∫ = δ(X)

X=− r2−Y 2−Z 2

X=−+ r2−Y 2−Z 2

∫Y=− r2−Z 2

Y=+ r2−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 unity. The dot product in the middle portion of the above equation simplifies to ∂G/∂r because n = er on the spherical surface and

€

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

€

∂G∂r

#

$ %

&

' ( r2 sinθdθd

ϕ= 0

2π

∫θ = 0

π

∫ ϕ = r2 ∂G∂r

sinθdθdϕ= 0

2π

∫θ = 0

π

∫ ϕ = 4πr2 ∂G∂r

=1,

which implies

€

∂G∂r

=14πr2

or

€

G = −14πr

= −1

4π x − $ x .

b) Work in the unshifted coordinate system, and start with the given information about the Green’s function,

€

∇2G(x, # x ) = δ(x − # x ) , and multiply both sides by

€

q( " x ) to get:

€

∇2 q( # x )G(x, # x )[ ] = q( # x )δ(x − # x ). (1) The q-function can slide inside the differential operator,

€

∇2, because

€

q( " x ) depends on x´ while the operator acts on x. Now integrate (1) over all possible values of x´:

€

∇2 q( # x )G(x, # x )[ ]all # x ∫ d3 # x =∇2 q( # x )G(x, # x )

all # x ∫ d3 # x = q( # x )δ(x − # x )

all # x ∫ d3 # x = q(x) . (2)

Here, the first equality follows because the

€

∇2-operation and the integration can be exchanged since they act on different variables. The final equality follows from the properties of the three-dimensional Dirac delta-function. The formal solution for

€

φ(x) can now be found by comparing the original equation,

€

∇2φ(x) = q(x) , (3) to the second and fourth terms of (2):

€

∇2 q( # x )G(x, # x )all # x ∫ d3 # x

% & '

( ) *

= q(x). (4)


Here, both statements are true, so we must conclude that

€

φ(x) = q( # x )G(x, # x )all # x ∫ d3 # x = −

14π

q( # x )x − # x all # x

∫ d3 # x , (5)

where the final equality in (5) comes from the result of part (a). This conclusion is possible because solutions of Poisson’s equation with suitably defined boundary conditions are unique. However, existence and uniqueness proofs are beyond the scope of this text. Interestingly, the results of this primarily-mathematical exercise are useful for incompressible ideal fluid flows.


Exercise 5.10. Start with the equations of motion in the rotating coordinates, and prove (5.29) Kelvin’s circulation theorem for the absolute vorticity. Assume that the flow is inviscid and barotropic and that the body forces are conservative. Explain the result physically. Solution 5.10. The general idea is to redo the ordinary derivation of Kelvin's theorem in a steadily rotating coordinate system where the inviscid momentum equation is:

€

ρDuDt

= −∇p + ρ ge − 2Ω×u[ ] , ($)

Here all velocities are observed in the rotating frame so the primes used in Section 4.7 have been dropped, and

€

ge = g −Ω × Ω × x( ), as also discussed in Section 4.7. Start by time differentiating the definition of the circulation (3.18):

€

DΓDt

=DDt

u ⋅ dxC∫ =

DuDt

⋅ dxC∫ + u ⋅ D

Dtdx( )

C∫ , (#)

dx is an arc length element of C, a closed contour. The first term of the right-most equality of (#) simplifies as follows:

€

DuDt

⋅ dxC∫ = −

1ρ∇p + ge − 2Ω×u

)

* +

,

- .

C∫ ⋅ dx = − 2Ω×u( )

C∫ ⋅ dx ,

because the pressure-integral and body force terms are single valued, and will integrate to zero on a closed contour. The second term in the right-most equality of (#) is zero as shown in Section 5.2. Thus, (#) implies:

€

DΓDt

= − 2Ω×u( )C∫ ⋅ dx . (@)

Consider the product of vectors in the integrand

€

−2 Ω×u( ) ⋅ dx = −2εijkΩ jukdxi = 2Ω jε jikdxiuk = −2Ω⋅ u× dx( ). Now define n to be the unit vector perpendicular to the plane defined by u and dx, and let

€

u⊥ be the component of u perpendicular to dx so that:

€

−2 Ω×u( ) ⋅ dx = −2Ω⋅nu⊥dx . In an interval of time dt,

€

u⊥ will locally advect the contour element dx through an area

€

δA = u⊥dxdt . Since n is normal to δA, the directed area element is

€

nδA = nu⊥dxdt , and this implies

€

D Dt( ) nδA( ) = nu⊥dx . Thus

€

−2 Ω×u( ) ⋅ dx = −2Ω⋅ D Dt( ) nδA( ) , and when integrated around the whole contour this produces:

€

− 2Ω×u( )C∫ ⋅ dx = − 2Ω

C∫ ⋅

DDtndA( ) = −

DDt

2ΩC∫ ⋅ndA .

Thus, (@) can be written:

€

DΓaDt

=DDt

ω + 2Ω( )C∫ ⋅ndA = 0 .

This means that the circulation developed from vorticity observed in the flow and from the rotation of the coordinate frame together satisfy Kelvin's theorem in a steadily rotating frame of reference.

C

!

n dx u

u


Exercise 5.11. In (R,ϕ,z) cylindrical coordinates, consider the radial velocity uR = −R–1(∂ψ/∂z), and the axial velocity uz = R–1(∂ψ/∂R) determined from the axisymmetric stream function

€

ψ(R,z) =Aa4

10R2

a2#

$ %

&

' ( 1−

R2

a2−z2

a2#

$ %

&

' ( where A is a constant. This flow is known as Hill’s spherical

vortex. a) For R2 + z2 ≤ a2, sketch the streamlines of this flow in a plane that contains the z-axis. What does a represent? b) Determine u = uR(R,z)eR + uz(R,z)ez c) Given

€

ωϕ = ∂uR ∂z( ) − ∂uz ∂R( ), show that ω = AReϕ in this flow and that this vorticity field is a solution of the vorticity equation (5.13). d) Does this flow include stretching of vortex lines? Solution 5.11. a) The zero streamline is given by R = 0 and by R2 + z2 = a2. For R2 + z2 ≤ a2, the others appear as shown to the right. b) The velocity components are:

€

uR = −1R∂ψ∂z

= −1RAa4

10R2

a2%

& '

(

) * −2za2

%

& '

(

) * =

ARz5

, and

€

uz =1R∂ψ∂R

=1RAa4

102Ra2

−4R3

a4−z2

a22Ra2

%

& '

(

) * =

Aa2

51− 2 R

2

a2−z2

a2%

& '

(

) * .

c) Compute the components of the vorticity. Here, uϕ = 0 and neither uR or uz depend on ϕ, so

€

ωR =1R∂uz∂ϕ

−∂uϕ∂z

= 0,

€

ωϕ =∂uR∂z

−∂uz∂R

=AR5−Aa2

5−4 R

a2&

' (

)

* + = AR , and

€

ωz =1R∂(Ruϕ )∂R

−1R∂uR∂ϕ

= 0 .

Therefore, ω = AReϕ in this flow. The vorticity equation is:

€

∂ω ∂t + u ⋅ ∇( )ω = ω ⋅∇( )u+ ν∇2ω . The flow is steady so the ∂/∂t term is zero. For ω = ωϕeϕ = AReϕ, the remaining terms are:

advection term:

€

u ⋅ ∇( ) ωϕeϕ( ) = eϕuR∂ωϕ

∂R= AuReϕ ,

vortex-stretching term:

€

ωϕeϕ ⋅ ∇( ) u( ) =ωϕ

R∂∂ϕ

uReR + uzez( ) =ARRuR∂eR∂ϕ

= AuReϕ ,

and viscous term:

€

ν∇2 ωϕeϕ( ) =∂ 2

∂R2+1R∂∂R

+1R2

∂ 2

∂ϕ 2

'

( )

*

+ , ωϕeϕ( )

€

= eϕ∂ 2ωϕ

∂R2+ eϕ

1R∂ωϕ

∂R+ωϕ

1R2

∂ 2eϕ∂ϕ 2

= eϕ∂ 2ωϕ

∂R2+1R∂ωϕ

∂R−ωϕ

R2&

' (

)

* + = eϕ 0 +

1RA − AR

R2&

' (

)

* + = 0

So, the vorticity equation satisfied by a balance of the advection and vortex stretching terms. d) Yes, this flow includes stretching of vortex lines.

R

z


Exercise 5.12. In (R,ϕ,z) cylindrical coordinates, consider the flow field uR = –αR/2, uϕ = 0, and uz = αz. a) Compute the strain rate components SRR, Szz, and SRz. What sign of α causes fluid elements to elongate in the z-direction? Is this flow incompressible? b) Show that it is possible for a steady vortex (a Burgers’ vortex) to exist in this flow field by adding uϕ = (Γ/2πR)[1 – exp(–αR2/4ν)] to uR and uz from part a) and then determining a pressure field p(R,z) that together with u = (uR, uϕ, uz) solves the Navier-Stokes momentum equation for a fluid with constant density ρ and kinematic viscosity ν. c) Determine the vorticity in the Burgers’ vortex flow of part b). d) Explain how the vorticity distribution can be steady when α ≠ 0 and fluid elements are stretched or compressed. e) Interpret what is happening in this flow when α > 0 and when α < 0. Solution 5.12. a) Using the strain rates from Appendix B,

€

SRR =∂uR∂R

= −α2

,

€

Szz =∂uz∂z

=α , and

€

SRz =12∂uR∂z

+∂uz∂R

#

$ %

&

' ( = 0.

Therefore, when α > 0, fluid elements elongate in the z-direction. The divergence of the velocity field is:

€

1 R( ) ∂(RuR ) ∂R( ) + ∂uz ∂z = 1 R( ) −2αR 2( ) +α = 0 , so the flow is incompressible. b) Here: uR = –αR/2, uϕ = (Γ/2πR)[1 – exp(–αR2/4ν)], and uz = αz. None of these velocities depends on the angle ϕ. Consider the three components of the Navier-Stokes momentum equation: (i) Steady R-direction equation with ∂/∂ϕ = 0, from Appendix B:

€

uR∂uR∂R

+ uz∂uR∂z

−uϕ2

R= −

1ρ∂p∂R

+ ν1R

∂∂R

R ∂uR∂R

'

( )

*

+ , +

∂2uR∂z2

−uRR2

'

( )

*

+ , .

The various terms involving the velocity are:

€

uR∂uR∂R

= −αR2

−α2

%

& '

(

) * =

α 2R4

,

€

uz∂uR∂z

= 0 ,

€

1R∂∂R

R ∂uR∂R

#

$ %

&

' ( = −

α2R

,

€

∂2uR∂z2

= 0 , and

€

−uRR2

=α2R

.

So, the R-direction equation becomes:

€

α 2R4

−Γ2

4π 2R31− exp −αR

2

4ν'

( )

*

+ ,

-

. /

0

1 2

2

= −1ρ∂p∂R

,

which can formally be integrated

€

p(R,z) = −ρα 2 % R 4

−Γ2

4π 2 % R 31− exp −α % R 2

4ν)

* +

,

- .

/

0 1

2

3 4

25 6 7

8 7

9 : 7

; 7 0

R

∫ d % R + f (z) .

where f(z) is an unknown function. (ii) Steady ϕ-direction equation with ∂/∂ϕ = 0, from Appendix B, is

€

uR∂uϕ∂R

+ uz∂uϕ∂z

+uRuϕR

= ν1R

∂∂R

R∂uϕ∂R

%

& '

(

) * +

∂2uϕ∂z2

−uϕR2

%

& '

(

) * = ν

∂∂R

1R∂(Ruϕ )∂R

%

& '

(

) * +

∂2uϕ∂z2

%

& '

(

) * .

The various terms are:


€

uR∂uϕ∂R

= −αR2

Γ2π

−1R21− exp −αR

2

4ν)

* +

,

- .

/

0 1

2

3 4 +

α2νexp −αR

2

4ν)

* +

,

- .

)

* + +

,

- . . ,

€

uz∂uϕ∂z

= 0 ,

€

uRuϕR

= −αΓ4πR

1− exp −αR2

4ν(

) *

+

, -

.

/ 0

1

2 3 ,

€

ν∂∂R

1R∂(Ruϕ )∂R

%

& '

(

) * =

νΓ2π

∂∂R

1R

∂∂R

1− exp −αR2

4ν%

& '

(

) *

/

0 1

2

3 4

%

& ' '

(

) * * = −

Γ2π

α 2R4ν

exp −αR2

4ν%

& '

(

) * , and

€

∂2uϕ∂z2

= 0 .

So, the ϕ-direction equation becomes:

€

αΓ4π

1R1− exp −αR

2

4ν'

( )

*

+ ,

-

. /

0

1 2 −

αR2νexp −αR

2

4ν'

( )

*

+ ,

'

( ) )

*

+ , , −

αΓ4πR

1− exp −αR2

4ν'

( )

*

+ ,

-

. /

0

1 2 = −

Γ2π

α 2R4ν

exp −αR2

4ν'

( )

*

+ , .

Multiply by 4π/Γα, and collect like terms:

€

1R1−1( ) − 1

R1−1( )exp −αR

2

4ν%

& '

(

) * −

αR2νexp −αR

2

4ν%

& '

(

) * = −

αR2νexp −αR

2

4ν%

& '

(

) * ;

so the ϕ-direction momentum equation is satisfied. (iii) Steady z-direction equation with ∂/∂ϕ = 0, from Appendix B:

€

uR∂uz∂R

+ uz∂uz∂z

= −1ρ∂p∂z

+ ν1R

∂∂R

R ∂uz∂R

&

' (

)

* + +

∂2uz∂z2

&

' (

)

* + .

The various terms are:

€

uR∂uz∂R

= 0 ,

€

uz∂uz∂z

=α 2z,

€

1R∂∂R

R ∂uz∂R

#

$ %

&

' ( = 0, and

€

∂2uz∂z2

= 0.

So, the z-direction equation becomes:

€

α 2z = −1ρ∂p∂z

, or

€

p(z,R) = −ρα 2z2 2 + g(R) ,

and this determines the unknown function,

€

f (z) = −ρα 2z2 2, in the final equation of (i), so the pressure is:

€

p(R,z) − p(0,0) = −ρα 2

2z2 +

R2

4%

& '

(

) * + ρ

Γ2

4π 2 - R 31− exp −α - R 2

4ν%

& '

(

) *

/

0 1

2

3 4

25 6 7

8 7

9 : 7

; 7 0

R

∫ d - R .

c) There is only one non-zero component of the vorticity:

€

ωz =1R∂∂R

Ruϕ( ) − 1R∂uR∂ϕ

= +Γα4πν

exp −αR2

4ν*

+ ,

-

. / .

d) When α > 0, the vorticity distribution can be steady because the strain field concentrates the vorticity at precisely the correct rate to balance the vorticity's outward diffusion. e) When α > 0, then flow comes inward toward the z-axis along z = 0 and turns vertically upward and downward along the +z and –z axis, respectively. Here the vorticity is concentrated near the z-axis and diffuses outward against the incoming flow; it decays to zero with increasing R. When α > 0, the flow comes vertically upward and downward along the –z and +z axis, respectively, and then turns outward away from the z-axis as it approaches z = 0. Here the vorticity is very high at large R and it diffuses inward against the outflow.


Exercise 5.13. A vortex ring of radius a and strength Γ lies in the x-y plane as shown in the figure. a) Use the Biot-Savart law (5.13) to reach the following formula for the induced velocity along the x-axis:

u(x) = −Γez4π

(xcosϕ − a)adϕx2 − 2axcosϕ + a2#$ %&

3 2ϕ=0

2π

∫ .

b) What is u(0), the induced velocity at the origin of coordinates? c) What is u(x) to leading order in a/x when x >> a?

Solution 5.13. a) Start with:

u(x) = Γ4π

eϕ0

2π

∫ ×(x− %x )x− %x 3 adϕ ,

where eϕ = – exsinϕ + eycosϕ, x = xex, x´ = aeR = a(excosϕ + eysinϕ), eϕ × ex = −e z cosϕ ,

eϕ × eR = −e z , and x− "x = (x − acosϕ )2 + a2 sin2ϕ#$ %&1 2

. Insert these relationships, to find:

u(x) = Γ4π

(−xez cosϕ + aez )(x − acosϕ )2 + a2 sin2ϕ#$ %&

3 20

2π

∫ adϕ ,

Rearrange the numerator and simplify the denominator of the integrand to find:

u(x) = −Γez4π

(xcosϕ − a)adϕx2 − 2axcosϕ + a2#$ %&

3 20

2π

∫ ,

b) At the origin, x = 0, so the part a) result simplifies to:

u(0) = −Γez4π

−a2dϕa2#$ %&

3 20

2π

∫ =Γez2a

,

c) For x >> a, the denominator of the part a) result can be expanded in a two-term power series:

x2 − 2axcosϕ + a2"# $%−3 2

= x−3 1− 2axcosϕ + a

2

x2"

#&

$

%'

−3 2

≈ x−3 1+ 3axcosϕ

"

#&$

%' for a x→ 0 .

So, the induced velocity becomes:

z!

ϕ"

x!

Γ! y!a!


u(x, t) ≈ − Γez4π x3 2

(xcosϕ − a)0

2π

∫ 1+ 3axcosϕ

%

&'(

)*adϕ for a x→ 0 .

Multiplying the integrand factors together and evaluating the integrals leads to:

u(x, t) ≈ − Γez4π x3

−2πa+3πa( )a = −Γa2

4x3ez for a x→ 0 .


Exercise 5.14. An ideal line vortex parallel to the z-axis of strength Γ intersects the x-y plane at x = 0 and y = h. Two solid walls are located at y = 0 and y = H > 0. Use the method of images for the following. a) Based on symmetry arguments, determine the horizontal velocity u of the vortex when h = H/2. b) Show that for 0 < h < H the horizontal velocity of the vortex is:

€

u(0,h) =Γ4πh

1− 2 1(nH /h)2 −1n=1

∞

∑'

( )

*

+ , ,

and evaluate the sum when h = H/2 to verify your answer to part a). Solution 5.14. There are two flat solid walls at y = 0 and y = H. The combined effect of each requires many image vortices. a) When the vortex is located at x = (0, H/2), the induced velocities from the two closest image vortices cancel, and the induced velocities from the next closest image vortices cancel as well. In fact, the total induced velocity from the whole array of image vortices is zero because vortex contributions cancel in pairs. Therefore,

u(0, H/2) = 0. b) Construct the induced velocity by adding the contributions from each image vortex. When 0 < h < H, the first image vortex below the lower solid wall must be located at y = –h and it must have a strength of –Γ to mimic the effect of the wall at y = 0. Based on the drawing, this induced velocity will be in the positive x-direction, so

€

u(0,h) =Γ4πh

+ ...= Γ2π

12h

+ ...$

% & '

( )

Now consider the first image vortex above the upper wall. It must be located at y = 2H – h and it must have a strength of –Γ to mimic the effect of the wall at y = H. Based on the drawing, this induced velocity will be in the negative x-direction, so

€

u(0,h) =Γ2π

12h

−1

2H − 2h+ ...

%

& ' (

) *

However, the first upper image vortex spoils the lower wall boundary condition, so another image vortex must be added at y = –2H + h and it must have a strength of +Γ to mimic the effect of the wall at y = 0. Based on the drawing, this induced velocity will be in the negative x-direction, so

€

u(0,h) =Γ2π

12h

−1

2H − 2h−12H

+ ...%

& ' (

) *

But, now another vortex with strength +Γ needs to added at y = 2H + h to preserve the upper wall boundary condition, so

€

u(0,h) =Γ2π

12h

−1

2H − 2h−12H

+12H

+ ...%

& ' (

) * .

Continuing this construction leads to:

€

u(0,h) =Γ2π

12h

−1

2H − 2h−12H

+12H

+1

2H + 2h−

14H − 2h

−14H

+14H

+1

4H + 2h...

%

& ' (

) * ,

y = 0

y = H

y = 2H

y = 3H

y = –H

y = –2H

y

x


which can be simplified by removing terms that are equal and opposite:

€

u(0,h) =Γ2π

12h

−1

2H − 2h+

12H + 2h

−1

4H − 2h+

14H + 2h

−1

6H − 2h+

16H + 2h

...%

& ' (

) * .

The first term is unique while all the others follow in pairs, so

€

u(0,h) =Γ2π

12h

+ −1

2nH − 2h+

12nH + 2h

%

& '

(

) *

n=1

∞

∑-

. /

0

1 2 =

Γ4πh

1− 2 1n2 H h( )2 −1n=1

∞

∑-

. / /

0

1 2 2 .

where the second equality follows from algebraic manipulations within the big parentheses. When h = H/2, this formula becomes:

€

u(0,H /2) =Γ4πh

1− 2 14n2 −1n=1

∞

∑'

( )

*

+ , .

The sum is 1/2, and this can be found by looking in an appropriate mathematical reference (see Gradshteyn, I. S., and Ryzhik, I. M., Tables of Integrals, Series, and Products [Academic Press, New York, 1980], p. 8), or by considering the terms in the sum:

€

14n2 −1n=1

∞

∑ =12

12n −1

−1

2n +1%

& '

(

) *

n=1

∞

∑ =121− 13

%

& '

(

) * +

13−15

%

& '

(

) * +

15−17

%

& '

(

) * + ...

+

, -

.

/ 0

=121+ −

13

+13

%

& '

(

) * + −

15

+15

%

& '

(

) * + −

17

+17

%

& '

(

) * + ...

+

, -

.

/ 0 =12.

The pair-cancellation of terms is the mathematical signature of the symmetry discussed in part a), and series of this type are sometimes known as telescoping series. With either evaluation approach, the net induced velocity is

€

u(0,H /2) =Γ4πh

1− 2 12%

& ' (

) *

+

, -

.

/ 0 = 0,

which matches the part a) result.


Exercise 5.15. The axis of an infinite solid circular cylinder with radius a coincides with the z-axis. The cylinder is stationary and immersed in an incompressible inviscid fluid, and the net circulation around it is zero. An ideal line vortex parallel to the cylinder with circulation Γ passes through the x-y plane at x = L > a and y = 0. Here two image vortices are needed to satisfy the boundary condition on the cylinder’s surface. If one of these is located at x = y = 0 and has strength Γ. Determine the strength and location of the second image vortex. Solution 5.15. If the net circulation around the cylinder is zero, then the second image vortex must have strength –Γ. Therefore, the fluid velocity u at any point will be a sum of velocities induced by the vortex at (0, a), the first image vortex at (0, 0), and second image vortex at (x´, y´). This velocity must be tangent to the surface of the cylinder. Therefore, determine the fluid velocity at the cylinder-surface point (xs, ys). Using Cartesian unit vectors and the diagram to the right, with the circulation directions of the vortices shown, leads to:

€

u(xs,ys) =Γ

2πa−

ys

aex +

xs

aey

%

& '

(

) * +

Γ

2π (L − xs)2 + ys

2−

ys

(L − xs)2 + ys

2ex −

(L − xs)(L − xs)

2 + ys2ey

%

& ' '

(

) * *

+Γ

2π (xs − + x )2 + (ys − + y )2

ys − + y (xs − + x )2 + (ys − + y )2

ex −xs − + x

(xs − + x )2 + (ys − + y )2ey

%

& ' '

(

) * *

This can be simplified to:

€

u(xs,ys) =Γ

2π−

ys

a2 −ys

(L − xs)2 + ys

2 +ys − % y

(xs − % x )2 + (ys − % y )2

&

' (

)

* + ex

+Γ

2πxs

a2 −(L − xs)

(L − xs)2 + ys

2 −xs − % x

(xs − % x )2 + (ys − % y )2

&

' (

)

* + ey .

Here the outward normal to the surface of the cylinder is

€

er = xs a( )ex + ys a( )ey . Therefore to find (x´, y´), set the dot product of u and er to zero,

€

u(xs,ys) ⋅ er =Γ

2πa−

xsys

a2 −xsys

(L − xs)2 + ys

2 +xs(ys − & y )

(xs − & x )2 + (ys − & y )2

'

( )

*

+ ,

+Γ

2πaxsys

a2 −(L − xs)ys

(L − xs)2 + ys

2 −(xs − & x )ys

(xs − & x )2 + (ys − & y )2

'

( )

*

+ , = 0.

After dividing out common factors and cancelling equal and opposite terms this reduces to:

€

−Lys

(L − xs)2 + ys

2 +# x ys − xs # y

(xs − # x )2 + (ys − # y )2= 0 ,

which must be true for all values of (xs, ys). So, at xs = a and ys = 0, this equation becomes:

€

−a # y (a − # x )2 + # y 2

= 0 ,

which can only be true when y´ = 0, and this leaves a single equation for x´:

!

!!(0,L)(0,0)

(xs,ys)

(x´,y´)

x

y


€

−Lys

(L − xs)2 + ys

2 +# x ys

(xs − # x )2 + ys2 = 0.

Expand the denominators, divide by ys, and use

€

xs2 + ys

2 = a2, to find:

€

LL2 − 2Lxs + a2

=# x

a2 − 2xs # x + # x 2.

Cross multiply and cancel common terms to reach:

€

L " x 2 − (L2 + a2) " x + La2 = L " x − L( ) " x − a2 L( ) = 0. The root that leads to a vortex location inside the cylinder is: x´ = a2/L. So, the final answers are: • the second image vortex has strength –Γ, and • its location is (a2/L, 0).


Exercise 5.16. Consider the interaction of two vortex rings of equal strength and similar sense of rotation. Argue that they go through each other, as described near the end of section 5.6. Solution 5.16. Consider two vortex rings with a similar sense of rotation as shown in the figure to the right. The radii and speeds of the two vortices are equal. The motion at A is the resultant of VB, VC, and VD, while the motion at C is the resultant of VA, VB, and VD. Comparing the velocity components on A and C, it is clear that the net resultant is an enlargement of the vortex on the right, and a contraction of the vortex on the left, as indicated by the vertical arrows in the lower figure (a). Parts (b) and (c) of the lower figure show that left-side vortex accelerates to the right and passes through the right-side vortex. Part c) show that the new right side vortex grows and slows while the new left-side vortex contracts and accelerates. This results in part d), which is identical to the starting condition shown in the first figure except that the ring vortices have exchanged places. Thus, the entire process starts over.

AC

D B

VB

VD

VA

VD

VC

VB

(a) (b) (c) (d)


Exercise 5.17. A constant density irrotational flow in a rectangular torus has a circulation Γ and volumetric flow rate Q. The inner radius is r1, the outer radius is r2, and the height is h. Compute the total kinetic energy of this flow in terms of only ρ, Γ, and Q.

Solution 5.17. For an irrotational vortical flow with circulation Γ. The velocity in the angular direction will be:

€

uθ =Γ2πr

.

The volumetric flow rate will be:

€

Q = uθ dzdr0

h

∫r1

r2

∫ =Γ2πr

dzdr0

h

∫r1

r2

∫ =hΓ2πln r2

r1

&

' (

)

* + .

The total kinetic energy, KE, of the fluid in the in the torus

will be:

€

KE =12ρuθ

2

0

2π

∫ dzrdrdθ0

h

∫r1

r2

∫ =ρΓ2

4π1rdzdr

0

h

∫r1

r2

∫ =ρΓ2h4π

ln r2r1

'

( )

*

+ , =12ρΓQ .

hr1

r2


Exercise 5.18. Consider a cylindrical tank of radius R filled with a viscous fluid spinning steadily about its axis with constant angular velocity Ω . Assume that the flow is in a steady state. (a) Find

€

ω ⋅ndAA∫ where A is a horizontal plane surface through the fluid normal to the axis of

rotation and bounded by the wall of the tank, and n is the normal on A. (b) The tank then stops spinning. Find again the value of

€

ω ⋅ndAA∫ .

Solution 5.18. a) At steady state the flow inside the tank will be in solid body rotation, and the vorticity will be twice the rotation rate: ω = 2Ω = constant. Thus,

€

ω ⋅ndAA∫ = 2ΩπR2.

b) Let C be the bounding curve around A, and let ds be an element of C. Then,

€

ω ⋅ dAA∫ = u ⋅ ds

C∫ = uθ[ ]r=R Rdθ = 00

2π∫ , since uθ = 0 at r = R because of the no-slip boundary

condition.


Exercise 5.19. Using Figure 5.13, prove (5.32) assuming that: (i) the two vortices travel in circles, (ii) each vortex's speed along it's circular trajectory is constant, and (iii) the period of the motion is the same for both vortices.

Solution 5.19. When the two vortices travel in circles, with radii h1 and h2, at constant speeds with the same period T, the circumference of each circular trajectory divided by the induced speed of the vortex on that trajectory must match:

2πh1Γ2 2πh

= T = 2πh2Γ1 2πh

,

where the term on the left applies to the vortex on the left (1), and the term on the right applies to the vortex on the right (2). This relationship can be simplified to:

h1 Γ2 = h2 Γ1 . (#) Now use h = h1 + h2 to substitute for h2 using h2 = h – h1, and solve for h1 to find:

h1Γ2

=h− h1Γ1

, or h11Γ2

+1Γ1

"

#$

%

&'=

hΓ1

, which implies h1 =hΓ1

1Γ2

+1Γ1

"

#$

%

&'

−1

=Γ2h

Γ2 +Γ1 .

The relationship (#) then implies:

€

h2 =Γ1h

Γ1 + Γ2 for the other vortex.

Thus, G is located at the place where the diagonal line connecting the tips of the induced velocity vectors crosses the line segment that connects the vortices. So, G is halfway between the two vortices when they are of equal strength, but it lies closer to the stronger vortex when the vortices are of unequal strength. Here G is also the "center of vorticity". For example, when G is located at the origin of coordinates:

center of vorticity ≡ rnΓnn=1

2

∑

= (−h1, 0)Γ1 + (h2, 0)Γ2 = −h1Γ1 + h2Γ2, 0( ) = −Γ2hΓ1 +Γ1hΓ2

Γ1 +Γ2

, 0%

&'

(

)*= (0, 0).


Exercise 5.20. Consider two-dimensional steady flow in the x-y plane outside of a long circular cylinder of radius a that is centered on and rotating about the z-axis at a constant angular rate of Ωz. Show that the fluid velocity on the x-axis is u(x,0) = (Ωza2/x)ey for x > a when the cylinder is replaced by a) a circular vortex sheet of radius a with strength γ = Ωza, and b) a circular region of uniform vorticity ω = 2Ωzez with radius a. c) Describe the flow for x2 + y2 < a2 for parts a) and b). Solution 5.20. a) For a circular vortex sheet, an element of the circle of length adθ located at angle θ will have a circulation of dΓ = γadθ = Ωza2dθ. The distance between the vortex element and the location (x, 0) is

€

(x − acosθ)2 + a2 sin2θ . Thus, the magnitude and direction of the induced velocity increment du at (x, 0) are

€

dΓ 2π (x − acosθ)2 + a2 sin2θ( ) and

€

(asinθ)ex + (x − acosθ)ey(x − acosθ)2 + a2 sin2θ

.

The total induced velocity at (x, 0) is determined from integrating over all vortex elements.

€

u(x,0) =Ωza

2

2πθ =−π

+π

∫(asinθ)ex + (x − acosθ)ey(x − acosθ)2 + a2 sin2θ

'

( )

*

+ , dθ =

Ωza2

2π(asinθ)ex + (x − acosθ)ey

x 2 + a2 − 2ax cosθ'

( )

*

+ ,

−π

+π

∫ dθ .

Separate the components and consider them individually.

€

u(x,0) =Ωza

2ex2π

asinθdθx 2 + a2 − 2ax cosθ−π

π

∫ +Ωza

2ey2π

(x − acosθ)dθx 2 + a2 − 2ax cosθ−π

π

∫ .

The x-component of u is zero because it is specified as an integration of an odd function on an even interval. The remaining y-component integration can be rewritten to reach a tabulated integral form, and then evaluated:

€

u(x,0) =Ωza

2ey2π

12x

1+x 2 − a2

x 2 + a2 − 2ax cosθ&

' (

)

* + dθ

−π

π

∫

=Ωza

2ey2π

12x

θ +2(x 2 − a2)

x 2 + a2( )2− 4a2x 2

tan−1x 2 + a2( )

2− 4a2x 2

x 2 + a2 − 2axtanθ

2

&

'

( ( (

)

*

+ + +

&

'

( ( (

)

*

+ + + −π

+π

=Ωza

2ey2π

12x

θ + 2tan−1 x 2 − a2

(x − a)2 tanθ2

&

' (

)

* +

&

' (

)

* + −π

+π

=Ωza

2ey2π

12x

2π + 2 π2− 2 −

π2

&

' (

)

* +

&

' (

)

* + =

Ωza2eyx

.

[Thus,

€

u(x,0) =(x − acosθ)dθ

x 2 + a2 − 2ax cosθ−π

+π

∫ =2πx

and is independent of a!]

b) For a uniform distribution of vorticity within the circle specified by x2 + y2 < a2, an element of area dA = rdθdr will have a circulation dΓ = ωzdA = 2Ωzrdθdr. The distance between the vortex element and the location (x, 0) is

€

(x − rcosθ)2 + r2 sin2θ . Thus, the magnitude and direction of the induced velocity at (x, 0) are

!(acos!,asin!)

x

y

!

du


€

dΓ 2π (x − rcosθ)2 + r2 sin2θ( ) and

€

(rsinθ)ex + (x − rcosθ)ey(x − rcosθ)2 + r2 sin2θ

.

The total induced velocity at (x, 0) is determined from integrating over all vortex sheet elements.

€

u(x,0) =2Ωz

2π−π

+π

∫(rsinθ)ex + (x − rcosθ)ey(x − rcosθ)2 + r2 sin2θ

'

( )

*

+ ,

0

a

∫ rdθdr =Ωz

π

(rsinθ)ex + (x − rcosθ)eyx 2 + r2 − 2rx cosθ

'

( )

*

+ ,

−π

+π

∫0

a

∫ dθrdr.

The angular integration is the same as that completed in part a) with a replaced by r, so

€

u(x,0) =Ωzeyπ

2πx0

a

∫ rdr =Ωzeyπ

2πxa2

2=Ωza

2eyx

.

c) In part a) the fluid is stationary for x2 + y2 < a2. In part b), the fluid is in solid body rotation for x2 + y2 < a2.


Exercise 5.21. An ideal line vortex in a half space filled with an inviscid constant-density fluid has circulation Γ, lies parallel to the z-axis, and passes through the x-y plane at x = 0 and y = h. The plane defined by y = 0 is a solid surface. a) Use the method of images to find u(x,y) for y > 0 and show that the fluid velocity on y = 0 is u(x,0) =

€

Γhex π (x 2 + h2)[ ] . b) Show that u(0,y) is unchanged for y > 0 if the image vortex is replaced by a vortex sheet of strength

€

γ(x) = −u(x,0) on y = 0. c) (If you have the patience) Repeat part b) for u(x,y) when y > 0. Solution 5.21. a) If the actual vortex is at (0, h) and has circulation Γ, then an image vortex of with circulation –Γ at (0, –h) produces a no-through-flow boundary condition on y = 0. In Cartesian coordinates, the velocity u at any location (x, y) is the sum of the two induced velocities:

€

u(x,y) =Γ

2π x 2 + (y − h)2−(y − h)ex + xeyx 2 + (y − h)2

%

& ' '

(

) * * +

Γ

2π x 2 + (y + h)2(y + h)ex − xeyx 2 + (y + h)2

%

& ' '

(

) * * ,

and this form can be simplified to:

€

u(x,y) =Γ2π

−(y − h)ex + xeyx 2 + (y − h)2

+(y + h)ex − xeyx 2 + (y + h)2

%

& '

(

) * . (†)

When evaluated on y = 0, the velocity is:

€

u(x,0) =Γ2π

hex + xeyx 2 + h2

+hex − xeyx 2 + h2

%

& '

(

) * =

Γhexπ (x 2 + h2)

.

b) First of all, determine u(0,y) from the results of part a)

€

u(0,y) =Γ2π

−exy − h

+exy + h

%

& '

(

) * . (%)

For a flat vortex sheet lying on y = 0, an element of the length dx´ located at x´ will have a circulation of dΓ = γ(x´)dx´. The distance between this vortex element and the location (x, y) is

€

(x − # x )2 + y 2 . The magnitude and direction of the induced velocity increment du at (x, y) are

€

dΓ 2π (x − % x )2 + y 2( ) and

€

−yex + (x − # x )ey

(x − # x )2 + y 2.

When the image vortex is replaced by a vortex sheet, u(x,y) for y > 0 will be the sum of the induced velocity from the actual vortex and that from the vortex sheet, which is an integral.

€

u(x,y) =Γ2π

−(y − h)ex + xey

x 2 + (y − h)2+

−yex + (x − % x )ey

(x − % x )2 + y 2&

' (

)

* +

−∞

+∞

∫ γ( % x )d % x Γ

/ 0 1

2 3 4

. (£)

For this part of this exercise, finding u(0,y) from this formula is sufficient, so put x = 0 to find:

€

u(0,y) =Γ2π

−ex

y − h+

−yex − % x ey

% x 2 + y 2&

' (

)

* +

−∞

+∞

∫ γ( % x )d % x Γ

/ 0 1

2 3 4

.

Comparing this equation with (%), sets the remaining task as showing the equality of

€

−yex − # x ey

# x 2 + y 2$

% &

'

( )

−∞

+∞

∫ γ( # x )d # x Γ

=ex

y + h.

To do this set, γ(x´) = – u(x´, 0) from part a), and separate the left side into components:


€

−yex − # x ey

# x 2 + y 2$

% &

'

( )

−∞

+∞

∫ −hπ ( # x 2 + h2)

d # x =hyex

πd # x

( # x 2 + y 2)( # x 2 + h2)−∞

+∞

∫ +hey

π# x d # x

( # x 2 + y 2)( # x 2 + h2)−∞

+∞

∫ .

The y-component integral is zero because it involves an odd integrand and an even interval. This leaves the x-component:

€

hyex

πd # x

( # x 2 + y 2)( # x 2 + h2)−∞

+∞

∫ =hyex

π1

h2 − y 21

# x 2 + y 2 −1

# x 2 + h2

'

( )

*

+ , d # x

−∞

+∞

∫

=hyex

ππ

h2 − y 21y−

1h

-

. /

0

1 2 =

hyex

h2 − y 2h − y

hy-

. /

0

1 2 =

ex

y + h,

where first equality follows from a partial fractions decomposition of the integrand on the left, the second equality follows from:

€

I(a) = x 2 + a2( )−1dx = π a

−∞

+∞∫ , and the final equality achieves

the desired form. Therefore, the vortex sheet on y = 0 appropriately mimics an image vortex. c) The effort here is nearly a repeat of part b). This time start from (£) and compare it to (†). The task here is to show that the following equality is holds:

€

−yex + (x − # x )ey

(x − # x )2 + y 2$

% &

'

( )

−∞

+∞

∫ γ( # x )d # x Γ

=(y + h)ex − xey

x 2 + (y + h)2. (&)

First set, γ(x´) = – u(x´, 0) from part a), and begin working on L, the left side of (&):

€

L = Lxex + Lyey =−yex + (x − # x )ey

(x − # x )2 + y 2$

% &

'

( )

−∞

+∞

∫ −hd # x π ( # x 2 + h2)

=hπ

yex − (x − # x )ey( )d # x (x − # x )2 + y 2( )( # x 2 + h2)−∞

+∞

∫ .

Consider the x-component integration first:

€

Lx =hyπ

d # x (x − # x )2 + y 2( )( # x 2 + h2)−∞

+∞

∫ =hyπ

A( # x − x) + B( # x − x)2 + y 2

+C # x + D# x 2 + h2

' ( )

* + ,

d # x −∞

+∞

∫ ,

where A, B, C, and D are constants. Some algebra is required to find:

€

A = −2x E ,

€

B = (h2 + x 2 − y 2) E ,

€

C = +2x E , and

€

D = (x 2 + y 2 − h2) E , where

€

E = (x 2 + y 2 − h2)2 + 4x 2h2 . The integrations that include A and C are zero because they involve odd integrands and even intervals. The integrals involving B and D can be evaluated using the I(a) formula above, to determine:

€

Lx =hyπ

Bπy

+Dπh

#

$ %

&

' ( = Bh + yD =

h3 + hx 2 − hy 2 + yx 2 + y 3 − yh2

(x 2 + y 2 − h2)2 + 4x 2h2.

The numerator and denominator in can be rearranged:

€

Lx =h(h2 − y 2) + x 2(h + y) + y(y 2 − h2)

(x 2 + y 2 + h2)2 − 4y 2h2=

(h + y) h(h − y) + x 2 + y(y − h)[ ]x 2 + y 2 + h2 + 2yh( ) x 2 + y 2 + h2 − 2yh[ ]

.

Further rearrangement shows that the factors in square brackets are equal, so

€

Lx =(h + y) x 2 + (y − h)2[ ]

x 2 + (y + h)2( ) x 2 + y 2 + h2 − 2yh[ ]=

h + yx 2 + (y + h)2

,

which matches the x-component on the right side of (&). Now consider the y-component integration:

€

Ly =hπ

( # x − x)d # x (x − # x )2 + y 2( )( # x 2 + h2)−∞

+∞

∫ =hπ

F( # x − x) + G( # x − x)2 + y 2

+H # x + J# x 2 + h2

' ( )

* + ,

d # x −∞

+∞

∫ ,

where F, G, H, and J are constants. Some algebra is required to find:


€

F = (h2 + x 2 − y 2) E ,

€

G = 2xy 2 E ,

€

H = (−h2 − x 2 + y 2) E , and

€

J = −x(x 2 + y 2 + h2) E , The integrations that include F and H are zero because they involve odd integrands and even intervals. The integrals involving G and J can be evaluated using the I(a) formula above, to determine:

€

Ly =hπ

Gπy

+Jπh

#

$ %

&

' ( =G

hy

+ J =2xyh − x 3 − xy 2 − xh2

(x 2 + y 2 − h2)2 + 4x 2h2 = −x x 2 + y 2 − 2yh + h2[ ]

(x 2 + y 2 + h2)2 − 4y 2h2

= −x x 2 + y 2 − 2yh + h2[ ]

x 2 + y 2 + h2 + 2yh( ) x 2 + y 2 + h2 − 2yh[ ]= −

xx 2 + (y + h)2 ,

and the final form matches the y-component on the right side of (&). Thus, the equivalence of a variable strength vortex sheet and an image vortex is established.

fluid mechanics kundu cohen 6th edition solutions sm ch (5)

Engineering