fluid mechanics kundu cohen 6th edition solutions sm ch (8)

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

Exercise 8.1. Starting from (8.5) and working in (x,y,z) Cartesian coordinates, determine an equation that specifies the locus of points that defines a wave crest. Verify that the travel speed of the crests in the direction of K = (k, l, m) is c = ω/|K|. Can anything be determined about the wave crest travel speed in other directions? Solution 8.1. The phase of the waveform in (8.5) is

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K ⋅ x −ωt . Thus wave crests are specified by:

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K ⋅ x crest −ωt = 2nπ , since cos(2nπ) = 1. In (x,y,z) Cartesian coordinates with K = (k, l, m), this is the equation for a series of planes:

€

kxcrest + lycrest + mzcrest = 2nπ +ωt , (*) so (8.5) describes plane waves. The unit vector perpendicular to these planes,

€

e⊥ , is determined from the gradient of the phase:

€

e⊥ =∇ K ⋅ x −ωt( )∇ K ⋅ x −ωt( )

=KK

= eK .

To determine the travel speed of the crests, time differentiate equation (*) to find:

€

k ddtxcrest + l d

dtycrest + m d

dtzcrest =ω ,

since k, l, m, and 2nπ are constants. In terms of a dot product, this last equation is:

€

K ⋅ddtx crest =ω , or

€

KK⋅ddtx crest = eK ⋅

ddtx crest =

ωK

= c .

Thus, the travel speed of wave crests, dxcrest/dt, in the direction of K is c. And, the phase relationship (8.5) does not provide any information about the travel speed in other directions. However, it is not possible to determine the wave speed other directions. Consider the following hypothesis:

€

ddtx crest = ce⊥ + de||,

where d the crest-parallel wave speed, and e|| is a unit vector that is parallel to the wave crests with

€

e⊥ ⋅ e|| = 0 = eK ⋅ e||. However,

€

e|| ⋅ ∇ K ⋅ x −ωt( ) = Ke|| ⋅ eK = 0 ; thus, there are no changes in the phase of the wave (as represented by

€

∇ K ⋅ x −ωt( )) in the crest-parallel direction(s). Hence it is impossible to determine anything about the crest-parallel wave speed d by observing the wave.


Exercise 8.2. For ka << 1, use the potential for linear deep water waves,

€

φ(z,x,t) = a ω k( )ekz sin(kx −ωt) and the waveform

€

η(x, t) = acos kx −ωt( ) +αka2 cos 2 kx −ωt( )[ ] to show that: a) with an appropriate choice of the constant α, the kinematic boundary condition (8.16) can be satisfied for terms proportional to (ka)0 and (ka)1 once the common factor of aω has been divided out, and b) with an appropriate choice of the constant γ, the dynamic boundary condition (8.19) can be satisfied for terms proportional to (ka)0, (ka)1, and (ka)2 when ω2 = gk(1 + γk2a2) once the common factor of ag has been divided out. Solution 8.2. a) First determine the horizontal (u) and vertical (w) velocities from the given velocity potential:

€

u =∂φ∂x

=aωkekz k cos(kx −ωt)[ ] = aωekz cos(kx −ωt),

€

w =∂φ∂z

=aωk

kekz[ ]sin(kx −ωt) = aωekz sin(kx −ωt),

and the derivatives of the surface waveform:

€

∂η∂t

=ωasin kx −ωt( ) + 2ωαka2 sin 2 kx −ωt( )[ ] ,

€

∂η∂x

= −kasin kx −ωt( ) − 2αk 2a2 sin 2 kx −ωt( )[ ] .

Use these results and the surface waveform in (8.16),

€

∂φ∂z$

% &

'

( ) z=η

=∂η∂t

+∂η∂x

∂φ∂x$

% &

'

( ) z=η

:

€

aωekη sin(kx −ωt) =ωasin kx −ωt( ) + 2ωαka2 sin 2 kx −ωt( )[ ] + −kasin kx −ωt( ) − 2αk 2a2 sin 2 kx −ωt( )[ ]( )aωekη cos(kx −ωt).

To simplify, let β = kx – ωt, and divide out common factors to find:

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ekη sinβ = sinβ + 2αkasin(2β) − kasinβ + 2αk 2a2 sin(2β)( )ekη cosβ . Since matching up through terms proportional to ka is necessary, the exponential function should be expanded through its first-order term,

€

ekη ≅1+ kacosβ for ka << 1, so

€

1+ kacosβ( )sinβ = sinβ + 2αkasin(2β) − kasinβ + 2αk 2a2 sin(2β)( ) 1+ kacosβ( )cosβ . Perform the various multiplications, and group terms that are proportional to (ka)0 and (ka)1.

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sinβ + ka cosβ sinβ( ) = sinβ + ka 2α sin(2β) − sinβ cosβ( ) + k 2a2 ...( ). The (ka)0 terms match. The (ka)1 terms will match if:

€

cosβ sinβ = 2α sin(2β) − sinβ cosβ → 2cosβ sinβ = 2α sin(2β), and this equation is satisfied when α = 1/2. b) First determine the time derivative of the velocity potential and use the notation of part a):

€

∂φ∂t

=aωkekz −ω cos(kx −ωt)[ ] = −

aω 2

kekz cosβ .

After using the Bernoulli equation, the dynamic boundary condition is:

€

∂φ∂t

+12u 2 + gz

$

% &

'

( ) z=η

= 0→−aω 2

kekη cosβ +

ρ2a2ω 2e2kη cos2 β + a2ω 2e2kη sin2 β( ) + gη = 0 .


Substitute in the linear approximation for

€

ekη ≅1+ kacosβ , the part a) result for η = acosβ + αka2cos(2β), and ω2 = gk(1 + γk2a2) to find:

€

−agk(1+ γk 2a2)

k(1+ kacosβ)cosβ +

a2

2gk(1+ γk 2a2)(1+ 2kacosβ)

+ g acosβ +αka2 cos(2β)( ) = 0.

Cancel common factors and simplify:

€

−(1+ γk 2a2)(1+ kacosβ)cosβ +ka2(1+ γk 2a2)(1+ 2kacosβ) + cosβ +αkacos(2β)( ) = 0 .

Perform the various multiplications, and group terms that are proportional to (ka)0, (ka)1 and (ka)2.

€

−cosβ + cosβ + ka −cos2 β +12

+α cos(2β)%

& '

(

) * + k 2a2 −γ cosβ + cosβ( ) + k 3a3(...) = 0 .

The three terms inside the larger parentheses sum to zero when α = 1/2 based on trigonometric identities, and the k2a2 term is zero when γ = 1.


Exercise 8.3. The field equation for surface waves on a deep fluid layer in two dimensions (x,z) is: ∂ 2φ ∂x2 +∂ 2φ ∂z2 = 0 , where φ is the velocity potential, ∇φ = (u,w) . The linearized free-surface boundary conditions and the bottom boundary condition are:

∂φ ∂z( )z=0 ≅∂η ∂t , ∂φ ∂t( )z=0 + gη ≅ 0 , and ∂φ ∂z( )z→−∞= 0 ,

where z = η(x,t) defines the free surface, gravity g points downward along the z-axis, the undisturbed free surface lies at z = 0. The goal of this problem is to develop the general solution for these equations without assuming a sinusoidal form for the free surface as was done in Section 8.1and 8.2. a) Assume φ(x, z, t) = Λ(x, t)Z(z) , and use the field equation and bottom boundary condition to show that φ(x, z, t) = Λ(x, t)exp(+kz) , where k is a positive real constant. b) Use the results of part a) and the remaining boundary conditions to show:

∂ 2Λ∂t2

+ gkΛ = 0 and ∂2Λ∂x2

+ k2Λ = 0 .

c) For a fixed value of k, find Λ(x,t) in terms of four unknown amplitudes A, B, C, and D. d) For the initial conditions: η = h(x) and ∂η/∂t =

€

˙ h (x) at t = 0, determine the general form of φ(x, z, t).

Solution 8.3. a) Insert

€

φ(x.z,t) = Λ(x, t)Z(z) into the field equation

€

∂2φ ∂x2 + ∂2φ ∂z2 = 0 and

divide by φ to find:

€

1Λ∂2Λ∂x2

+1Zd2Zdz2

= 0 . The first term depends only on x and t while the second

depends only on z; thus each must be constant to ensure the equation is satisfied for all possible (x, z, t). Therefore set the first term equal to –k2 and the second term equal to +k2. Then, the equation involving Z(z) is

€

d2Z dz2 − k2Z = 0 which has the solution: Z = A+e+kz + A–e–kz, where A± are constants. The bottom boundary condition requires

€

dZ dz→ 0 as

€

z→−∞ , so the A– must be zero. Therefore,

€

φ(x.z,t) = Λ(x, t)e+kz , and the constant A+ has been incorporated into Λ. b) Substituting the result of part a) into the first surface boundary condition produces:

€

∂φ ∂z( )z=0 = Λkekz( )z=0 = Λk = ∂η ∂t . Substituting the result of part a) into the second surface boundary condition produces:

€

∂φ ∂t( )z=0 + gη = ekz (∂Λ /∂t)( )z=0 + gη = ∂Λ ∂t + gη = 0 Take the partial derivative of the last equality with respect to time and use the result of the first surface boundary condition to eliminate ∂η/∂t to find:

€

∂2Λ ∂t2 + gkΛ = 0 . From part a) the equation involving Λ, is

€

∂2Λ ∂x2 + k2Λ = 0. c) Assume a separation-of-variables form for Λ = X(x)T(t). The two equations derived for part b) then imply:

€

d2T dt2 + gkT = 0 , and

€

d2X dx2 + k2X = 0 . The first equation has solutions:

€

sin t gk( ) and

€

cos t gk( ), while the second has solutions sin(kx) and cos(kx). Allowing for all four possibilities for Λ then implies:


€

Λ(x, t) = # A sin(kx)sin t gk( ) + # B sin(kx)cos t gk( ) + # C cos(kx)sin t gk( ) + # D cos(kx)cos t gk( ) . Unfortunately, this form hides the wave-propagation character of the solution, but with the use of trigonometric angle-sum and angle-difference formulae for the sine and cosine functions, the above form for Λ can be written:

€

Λ(x, t) = Asin kx − t gk( ) + Bcos kx − t gk( ) + C sin kx + t gk( ) + Dcos kx + t gk( ) , where 2A = B´ – C´, 2B = A´ + D´, 2C = B´ + C´, and 2D = –A´ + D´. d) The above form for Λ is only valid when k ≥ 0, so k may take on any non-negative value. Thus, the final solution must be of the form:

€

φ(x,z,t) =A(k)sin kx − t gk( ) + B(k)cos kx − t gk( ) +

C(k)sin kx + t gk( ) + D(k)cos kx + t gk( )

$

% &

' &

(

) &

* & ekzdk

0

+∞

∫ .

Now combine the initial and boundary conditions. The first boundary condition evaluated at t = 0 implies:

€

∂φ ∂z( )(z,t)=0 = ∂η ∂t( )t=0 = ˙ h (x) , or

€

˙ h (x) = A(k) + C(k)[ ]sin kx( ) + B(k) + D(k)[ ]cos kx( ){ }kdk0

+∞

∫ .

The second boundary condition evaluated at t = 0 implies:

€

∂φ ∂t( )(z,t)=0 = −gη( )t=0 = −gh(x), or

€

−gh(x) = gk −A(k) + C(k)[ ]cos kx( ) + B(k) −D(k)[ ]sin kx( ){ }dk0

+∞

∫

The final formulae for A, B, C, and D are found from considerations of symmetry about x = 0:

€

˙ h (x) − ˙ h (−x) = 2k A(k) + C(k)[ ]sin kx( )dk0

+∞

∫ ,

€

˙ h (x) + ˙ h (−x) = 2k B(k) + D(k)[ ]cos kx( )dk0

+∞

∫ ,

€

−g h(x) + h(−x)( ) = gk −A(k) + C(k)[ ]cos kx( )dk0

+∞

∫ , and

€

−g h(x) − h(−x)( ) = gk B(k) −D(k)[ ]sin kx( )dk0

+∞

∫

and can be evaluated in formal terms via inverse sine and cosine transforms.


Exercise 8.4. Derive (8.37) from (8.27). Solution 8.4. Use the velocity components specified in (8.27), and the definition of the stream function in two-dimensions (x, z):

€

u =∂ψ∂z

= aωcosh k(z + H)( )sinh kH( )

cos(kx −ωt), and

€

w = −∂ψ∂x

= aωsinh k(z + H)( )sinh kH( )

sin(kx −ωt).

Integrate each equation to find:

€

ψ =aωksinh k(z + H)( )sinh kH( )

cos(kx −ωt) + f (x), and

€


cos(kx −ωt) + g(z) .

where f and g are single-variable functions of integration. These are consistent when f(x) = g(y) = constant, and this constant can be set to zero without loss of generality. The final result is (8.37):

€


cos(kx −ωt) .


Exercise 8.5. Consider stationary surface gravity waves in a rectangular container of length L and breadth b, containing water of undisturbed depth H. Show that the velocity potential φ = Acos(mπx/L)cos(nπy/b)cosh[k(z + H)]e–iωt, satisfies the Laplace equation and the wall boundary conditions, if (mπ/L)2 + (nπ/b)2 = k2. Here m and n are integers. To satisfy the linearized free-surface boundary condition, show that the allowable frequencies must be ω2 = gk tanh(kH). [Hint: combine the two boundary conditions (8.18) and (8.21) into a single equation ∂2φ/∂t2 = −g ∂φ/∂z at z = 0.] Solution 8.5. Start from

€

φ = Acos mπx L( )cos nπy b( )cosh k(z + H)[ ]e−iω t and compute velocities.

€

u =∂φ∂x

= −A mπ L( )sin mπx L( )cos nπy b( )cosh k(z + H)[ ]e− iω t which implies u = 0 at x = 0 and x = L, so the x-direction BCs are satisfied. Similarly,

v = ∂φ∂y

= −A nπ b( )cos mπ x L( )sin nπ y b( )cosh k(z+H )[ ]e−iω t which implies v = 0 at y = 0 and y = b, so the y-direction BCs are satisfied. The Laplacian is:

€

∂ 2φ∂x 2

+∂ 2φ∂y 2

+∂ 2φ∂z2

= −m2π 2

L2−n2π 2

b2+ k 2

&

' (

)

* + Acos mπx L( )cos nπy b( )cosh k(z + H)[ ]e− iω t ,

and it equals zero when:

€

k 2 =m2π 2

L2+n2π 2

b2.

The linearized free surface boundary conditions are:

€

∂η ∂t = ∂φ ∂z and

€

∂φ ∂t = −gη at z = 0, which can be combined to get:

€

∂ 2φ ∂t 2 = −g∂η ∂t = −g∂φ ∂z at z = 0. Using the given potential, this requires

€

A(−iω)2 cos mπx L( )cos nπy b( )cosh k(z + H)[ ]e−iω t[ ]z= 0

= −g Ak cos mπx L( )cos nπy b( )sinh k(z + H)[ ]e−iω t[ ]z= 0

,

or

€

−ω 2 cosh kH[ ] = −gk sinh kH[ ] which can be written:

€

ω 2 = gk tanh kH[ ] .


Exercise 8.6. A lake has the following dimensions L = 30 km, b = 2 km, and H = 100 m. If the wind sets up the mode m = 1 and n = 0, show that the period of the oscillation is 32 min. Solution 8.6. Use the results of Exercise 8.5:

€

k 2 =m2π 2

L2+n2π 2

b2 and

€

ω 2 = gk tanh kH[ ] ,

and eliminate k to find:

€

ω 2 = g m2π 2

L2+n2π 2

b2$

% &

'

( )

1 2

tanh m2π 2

L2+n2π 2

b2$

% &

'

( )

1 2

H*

+ , ,

-

. / / .

Thus, the frequency ω10 of the m = 1, n = 0 mode is given by:

€

ω10 = g πLtanh π

LH

$

% &

'

( ) = (9.81) π

30 ×103tanh π (100)

30 ×103$

% &

'

( ) = 3.27 ×10−3 rad /s,

so its period T10 is:

€

T10 =2πω10

= 32 minutes.


Exercise 8.7. Fill a square or rectangular cake pan half-way with water. Do the same for a round frying pan of about the same size. Agitate the water by carrying the two pans (outside) while walking briskly at a consistent pace on a horizontal surface. a) Which shape lends itself better to spilling? b) At what portion of the perimeter of the rectangular pan does spilling occur most readily? c) Explain your observations in terms of standing wave modes. Solution 8.7. a) For the pans tested by the third author, the water usually spilled out of the square or rectangular pan first. b) The water almost always left the square or rectangular pan from a corner. c) All of the resonant modes of the rectangular pan have maximum vertical surface displacement at the pan's corner (this is not true of the circular pan since it has no corners). When side-to-side and front-to-back waves combine in the rectangular pan they can constructively interfere in the corners of the pan and cause spilling. This exercise provides is a simple demonstration of how tides may affect nearly-straight and jagged coastlines. When an inlet or bay with interior corners (analogous to the square pan) is excited at its resonant frequency by tidal action (analogous to your random shaking), the tidal fluctuations can be enormous. The extreme tides (low-to-high change of 14 m or more) of the Bay of Fundy on southern coast of Newfoundland are an extreme example. Smooth coastlines do not experience such large tidal swings.


Exercise 8.8. Use (8.27), (8.28), and (8.38), to prove (8.39). Solution 8.8. Equation (8.38) specifies a double integration,

Ek =ρ2λ

u2 +w2( )−H

0

∫0

λ

∫ dzdx , (8.38)

involving the two velocity components given by (8.27):

u = aωcosh k(z+H )( )sinh kH( )

cos(kx −ωt) , and w = aωsinh k(z+H )( )sinh kH( )

sin(kx −ωt) . (8.27)

Substituting (8.27) into (8.38) produces:

Ek =ρa2ω 2

2λ sinh2 (kH )cos2 (kx −ωt)dx cosh2 k(z+H )( )

−H

0

∫0

λ

∫ dz

+ ρa2ω 2

2λ sinh2 (kH )sin2 (kx −ωt)dx sinh2 k(z+H )( )

−H

0

∫0

λ

∫ dz

The two x-integrations produce λ/2 since the average values of sin2() and cos2() are both 1/2. This leaves the integrations of the hyperbolic functions.

cosh2 k(z+H )( )−H

0

∫ dz = 14

e2k (z+H ) + 2+ e−2k (z+H )( )−H

0

∫ dz

= e2kH

4e2kz

−H

0

∫ dz+ 12

dz−H

0

∫ +e−2kH

4e−2kz

−H

0

∫ dz

= e2kH

4e2kz

2k!

"#

$

%&−H

0

+H2+e−2kH

4e−2kz

−2k!

"#

$

%&−H

0

= e2kH

41

2k−e−2kH

2k!

"#

$

%&+

H2−e−2kH

41

2k−e+2kH

2k!

"#

$

%&

= 18k

e2kH −1+ 4kH − e−2kH +1( ) = 18k

e2kH + 4kH − e−2kH( )

sinh2 k(z+H )( )−H

0

∫ dz = 14

e2k (z+H ) − 2+ e−2k (z+H )( )−H

0

∫ dz

= e2kH

4e2kz

−H

0

∫ dz− 12

dz−H

0

∫ +e−2kH

4e−2kz

−H

0

∫ dz

= e2kH

4e2kz

2k#

$%

&

'(−H

0

−H2+e−2kH

4e−2kz

−2k#

$%

&

'(−H

0

= e2kH

41

2k−e−2kH

2k#

$%

&

'(−

H2−e−2kH

41

2k−e+2kH

2k#

$%

&

'(

= 18k

e2kH −1− 4kH − e−2kH +1( ) = 18k

e2kH − 4kH − e−2kH( )

Thus, wave kinetic energy becomes:


Ek =ρa2ω 2

2λ sinh2 (kH )λ2!

"#

$

%&

18k

e2kH + 4kH − e−2kH + e2kH − 4kH − e−2kH( )

= ρa2ω 2

4sinh2 (kH )18k!

"#

$

%& 2e2kH − 2e−2kH( ) = ρa2ω 2

4sinh2 (kH )1k!

"#$

%&ekH + e−kH

2!

"#

$

%&ekH − e−kH

2!

"#

$

%&

= ρa2ω 2

4sinh2 (kH )1k!

"#$

%&cosh(kH )sinh(kH ) = ρa2ω 2

4k tanh(kH )

Now use the dispersion relationship (8.28) ω = gk tanh(kH ) (8.28)

and recognize that η2 = a2/2:

Ek =ρa2 gk tan(kH )( )4k tanh(kH )

=12ρgη2 ,

and this is (8.39).


Exercise 8.9. Show that the group velocity of pure capillary waves in deep water, for which the gravitational effects are negligible, is cg = (3/2)c. Solution 8.9. For surface waves in the presence of gravity and surface tension, the dispersion relationship is given by (8.56):

€

ω = k g +σk 2

ρ

%

& '

(

) * tanh kH( ) .

When gravity is negligible, this simplifies to:

€

ω =σk 3

ρtanh kH( ) .

The group velocity in this case is:

€

cg =∂ω∂k

=∂∂k

σk 3 ρ( ) tanh kH( ) =1

2 σk 3 ρ( ) tanh kH( )3σk 2

ρtanh kH( ) +

σk 3

ρsech2 kH( )

&

' (

)

* + .

For deep water, kH is large enough so that tanh(kH) ≈ 1, and sech(kH) << 1.

With these simplifications the group velocity becomes:

€

cg ≅1

2 σk 3 ρ( )3σk 2

ρ

%

& '

(

) * =32

σkρ

.

For pure capillary waves in deep water, the phase speed c is:

€

c =ωk≅1k

σk 3 ρ =σkρ

,

so the group speed reduces to: cg = (3/2)c.


Exercise 8.10. Assuming deep water, plot the group velocity of surface gravity waves, including surface tension σ, as a function of λ for water at 20 °C (ρ = 1000 kg/m3 and σ = 0.074 N/m).

a) Show that the group velocity is cg =12

gk1+3σ k2 ρg1+σ k2 ρg

.

b) Show that this becomes minimum at a wave number given by σ k2

ρg=23−1.

c) Verify that cg min = 17.8 cm/s.

Solution 8.10. a) Start from (8.56):

€

ω = k g +σk 2

ρ

%

& '

(

) * tanh kH( ) . In deep water tanh(kH) ≈ 1, so

this simplifies to:

€

ω ≅ gk +σk 3 ρ . The group velocity in this case is:

€

cg =∂ω∂k

=∂∂k

gk +σk 3 ρ =1

2 gk +σk 3 ρg +

3σk 2

ρ

&

' (

)

* + =12

gk1+ 3σk 2 ρg1+σk 2 ρg

.

This relationship is plotted here; group speed is on the vertical axis in m/s and wavelength is on the horizontal axis in cm.

b) The minimum group speed is found by differentiating cg with respect to k, setting this derivative equal to zero, solving for kmin, and then evaluating cg at kmin.

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€

0 =∂cg∂k

=∂∂k

g21+ 3σk 2 ρgk +σk 3 ρg

%

& ' '

(

) * * =

g2

6σk ρg( ) k +σk 3 ρg( ) − 12 1+ 3σk 2 ρg( )

2

k +σk 3 ρg( )3 2%

&

' '

(

)

* * .

Simplify and set

€

x =σk 2 ρg to find:

€

0 = 6σk ρg( ) k +σk 3 ρg( ) − 12 1+ 3σk 2 ρg( )

2= 6x 1+ x( ) − 1

2 1+ 3x( )2 or 3x2 + 6x – 1 = 0.

The final quadratic implies:

€

xmin = −1+ 2 3 = 0.1547. The corresponding wave number is:

€

kmin = ρgxmin σ = 103 ⋅ 9.81⋅ 0.1547 0.074 =143.2 /m . c) Thus:

€

cg( )min =12

gkmin

1+ 3xmin1+ xmin

=12

9.81143.2

1+ 3(0.1547)1+ 0.1547

= 0.178m /s .


Exercise 8.11. The energy propagation characteristics of sinusoidal deep-water capillary-gravity waves follow those of pure gravity waves when corrections are made for the influence of surface tension. Assume a waveform shape of η(x,t) = acos(kx – ωt) for the following items. a) Show that the sum of the wave's kinetic and potential energies (per unit surface area) can be written: E = Ek +Ep =

12 ρa

2 g where g = g+ k2σ ρ . b) Determine the wave energy flux EF (per unit length) from

<Begin Equation>

EF = ω2π

!p u−∞

0

∫ dz−σ t ⋅ u[ ]z=0&

'(

)

*+dt

0

2π ω

∫ ,

</End Equation> where t is the surface tangent vector, and the extra term involving σ represents energy transfer via surface tension. c) Use the results of parts a) and b) to show that EF = Ecg, where cg is the group velocity determined in Exercise 8.10 part a). Solution 8.11. a) The kinetic and potential energies of sinusoidal deep-water capillary gravity waves can be written:

Ek =ρ2λ

u2 +w2( )−H

0

∫0

λ

∫ dzdx , and Ep =ρgλ

z0

η

∫ dzdx0

λ

∫ +σλ

1+ ∂η∂x#

$%

&

'(2

−1#

$

%%

&

'

((dx

0

λ

∫ ,

where the second term in the potential energy is the surface tension contribution. Using (8.47) for u and w in kinetic energy integration leads to:

Ek =ρ

2λa2ω 2 e2kz cos2 (kx −ωt)+ sin2 (kx −ωt)( )

−∞

0

∫0

λ

∫ dzdx

= ρ2a2ω 2 e2kz

−∞

0

∫ =ρ4ka2ω 2 =

ρ4ka2 kg+σ k

3

ρ

$

%&

'

()

= ρga2

4+σ k2a2

4=ρ4g 1+σ k

2

ρg$

%&

'

()a2,

where the dispersion relationship has been used to eliminate ω2. The potential energy integration can also be performed for small slopes (ka << 1):

Ep =ρgλ

z0

η

∫ dzdx0

λ

∫ +σλ

1+ ∂η∂x#

$%

&

'(

2

−1#

$

%%

&

'

((dx

0

λ

∫ ≅ρgλ

η2 dx0

λ

∫ +σλ

12∂η∂x#

$%

&

'(

2#

$%%

&

'((dx

0

λ

∫

= ρg2λ

a2 cos2 (kx −ωt)dx0

λ

∫ +σ2λ

a2k2 sin2 (kx −ωt)dx0

λ

∫

= ρga2

4+σ4a2k2 =

ρ4g 1+σ k

2

ρg#

$%

&

'(a2

Thus, the total wave energy per unit area is:

E = Ek +Ep =ρ4g 1+σ k

2

ρg!

"#

$

%&a2 +

ρ4g 1+σ k

2

ρg!

"#

$

%&a2 =

12ρa2 g where g = g 1+ k2σ ρg( ) .


b) From the problem statement, the energy flux crossing any fixed x-location is:

EF = ω2π

!p u−∞

0

∫ dz−σ t ⋅ u[ ]z=0&

'(

)

*+dt

0

2π ω

∫ .

The pressure fluctuation p´ is given by (8.30) so the linearized Bernoulli equation implies:

!p = −ρ ∂φ∂t

= −ρ∂∂t

aωke+kz sin(kx −ωt)

$

%&

'

()= ρa

ω 2

ke+kz cos(kx −ωt) .

where the form for φ is obtained from (8.26) in the limit kH→∞ . The horizontal velocity in this case is:

u = ∂φ∂x

=∂∂x

aωke+kz sin(kx −ωt)

#

$%

&

'(= aωe+kz cos(kx −ωt) ,

and is also given by (8.47). Thus, the vertical integral is:

!p u−∞

0

∫ dz = ρaω2

ke+kz cos(kx −ωt)

%

&'

(

)* aωe+kz cos(kx −ωt)( )

−∞

0

∫ dz

= ρa2 ω3

kcos2 (kx −ωt) e+2kz

−∞

0

∫ dz = ρa2 ω3

2k2 cos2 (kx −ωt)

Before moving to the time average, consider the surface tension term. Here, the unit vector tangent to the surface is:

t = 1

1+ ∂η ∂x( )2ex + ∂η ∂x( )ez( ) where ∂η/∂x = –aksin(kx – ωt),

and the fluid velocity at the surface can be obtained from (8.47) evaluated at z = 0: u[ ]z=0 = aωex cos(kx −ωt)+ aωez sin(kx −ωt) .

Thus,

−σ t ⋅ u[ ]z=0=

−σ

1+ a2k2 sin2 (kx −ωt)aω cos(kx −ωt)− a2kω sin2 (kx −ωt)( )

≅ −σ aω cos(kx −ωt)− a2kω sin2 (kx −ωt)( )

and the approximation is valid when ka << 1, which is the case here. At any location, the time average of cos(kx – ωt) is zero, while the time average of cos2(kx – ωt) and sin2(kx – ωt) are both 1/2. Thus,

EF = ρa2 ω3

4k2+σa2kω2

.

c) To show that EF = Ecg, where cg is the group velocity determined in Exercise 8.10 part a), show that ratio EF/E leads to the group velocity.

EFE

=ρa2 ω

3

4k2+σa2kω2

ρ2g 1+σ k

2

ρg!

"#

$

%&a2

=

ωω 2

2k2+σ kρ

!

"#

$

%&

g+σ k2

ρ

!

"#

$

%&

=

k g+σ k2

ρ

!

"#

$

%&

12k2

k g+σ k2

ρ

!

"#

$

%&+

σ kρ

!

"#

$

%&

g+σ k2

ρ

!

"#

$

%&

,

where the third equality comes from using the dispersion relationship (8.56) in the limit of kH→∞ . Continue simplifying


EFE

=

k g2k

+σ k2ρ

+σ kρ

!

"#

$

%&

g+σ k2

ρ

=

kg 12k

+3σ k2ρg

!

"#

$

%&

1+σ k2

ρg

=12

gk

1+3σ k2

ρg!

"#

$

%&

1+σ k2

ρg

.

The final form is the group velocity specified by Exercise 8.10 part a).


Exercise 8.12. The effect of viscosity on the energy of linear deep-water surface waves can be determined from the wave motion’s velocity components and the viscous dissipation (4.58). a) For incompressible flow, the viscous dissipation of energy per unit mass of fluid is

€

ε = 2 µ ρ( )Sij2 where Sij is the strain rate tensor and µ is the fluid's viscosity. Determine ε using (8.47). b) The total wave energy per unit surface area, E, for a linear sinusoidal water wave with amplitude a is given by (8.42). Assume that a is function time, set dE/dt = –ε, and show that a(t) = a0exp[–2(µ/ρ)k2t] where a0 is the wave amplitude at t = 0. c) Using a nominal value of µ/ρ = 10–6 m2/s for water, determine the time necessary for an amplitude reduction of 50% for water-surface waves having λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m, and 100 m. d) Convert the times calculated in c) to travel distances by multiplication with an appropriate group speed. Remember to include surface tension. Can a typhoon located near New Zealand produce increased surf on the coast of North America? [The circumference of the earth is approximately 40,000 km]. Solution 8.12. a) First calculate the stress tensor using the velocity components specified by (8.47):

€

u = aωekz cos(kx −ωt) and

€

w = aωekz sin(kx −ωt) .

€

Sij =12∂ui∂x j

+∂u j

∂xi

#

$ % %

&

' ( ( =

∂u ∂x 12 ∂u ∂z + ∂w ∂x( )

12 ∂w ∂x + ∂u ∂z( ) ∂w ∂z

)

* +

,

- .

= aωkekz−sin(kx −ωt) cos(kx −ωt)cos(kx −ωt) sin(kx −ωt))

* +

,

- . .

Now compute the viscous kinetic-energy dissipation per unit mass:

€

ε = 2 µ ρ( )Sij2 = 2 µ ρ( ) S112 + S12

2 + S212 + S22

2( ) = 2 µ ρ( )a2ω 2k 2e2kz 2cos2(kx −ωt) + 2sin2(kx −ωt)( ) = 4 µ ρ( )a2ω 2k 2e2kz .

b) From (8.42) E = (1/2)ρga2. The energy dissipation per unit volume is ρε, so the dissipation per unit surface area is obtained by integrating ρε through the depth. Therefore:

€

dEdt

= − ρεdz−∞

0

∫ = −4µa2ω 2k 2 e2kzdz−∞

0

∫ = −4µa2ω 2k 2 e2kz

2k(

) *

+

, - −∞

0

= −2µa2ω 2k = −4µk 2 Eρ

,

where the deep-water dispersion relationship ω2 = gk has been used to reach the final form. Integrating the differential equation represented by the extreme ends of this extended equality produces:

€

E = E0 exp −4 µ ρ( )k 2t{ } , or in terms of wave amplitude:

€

a = a0 exp −2 µ ρ( )k 2t{ }, where E0 and a0 are the wave energy and amplitude, respectively, at t = 0. c) For a factor of two amplitude decrease, the time t1/2 will be:

€

2 µ ρ( )k 2t1 2 = ln(2)→ t1 2 = ln(2) 2 µ ρ( )k 2 = ln(2)λ2 8π 2 µ ρ( ) .

For the given viscosity/density ratio,

€

ln(2)λ2 8π 2 µ ρ( ) = 8.78 ×103 λ m( )2 λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m, 100 m. t1/2 = 8.78 ms 0.878 s 87.8 s

€

8.78 ×103 s

€

8.78 ×105 s

€

8.78 ×107 s d) The group speeds can be calculated from the results of exercise 8.10. Let d1/2 be the distance traveled in deep water for a factor of two amplitude decrease.


λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m, 100 m. cg = 1.02 m/s 0.311 m/s 0.212 m/s 0.625 m/s 1.98 m/s 6.25 m/s d1/2 = 8.96 mm 0.273 m 18.6 m 5.49 km 1.74 Mm 549 Mm (where Mm = mega-meters). The circumference of the earth is only 40 Mm, so a typhoon off the coast of New Zealand that generates waves with wavelengths of approximately 30 m or longer is able to produce increased surf on the coast of North America.


Exercise 8.13. Consider a deep-water wave train with a Gaussian envelope that resides near x = 0 at t = 0 and travels in the positive-x direction. The surface shape at any time is a Fourier superposition of waves with all possible wave numbers:

€

η(x, t) = ˜ η (k)exp i kx − g k( )1/ 2t( )$

% & ' ( ) dk

−∞

+∞

∫ , (†)

where

€

˜ η (k) is the amplitude of the wave component with wave number k, and the dispersion relation is ω = (gk)1/2. For the following items assume the surface shape at t = 0 is:

€

η(x,0) =12πα

exp −x 2

2α 2 + ikd x& ' (

) * +

.

Here, α sets the initial horizontal extent of the wave train, with larger α producing a longer wave train. a) Plot Re{η(x,0)} for |x| ≤ 40 m when α = 10 m and kd = 2π/λd = 2π/10 m–1. b) Use the inverse Fourier transform at t = 0,

€

˜ η (k) = 1 2π( ) η(x,0)exp −ikx[ ]dx−∞

+∞∫ , to find the

wave amplitude distribution:

€

˜ η (k) = 1 2π( )exp − 12 (k − kd )2α 2{ } , and plot this function for 0 < k <

2kd using the numerical values from part a). Does the dominant contribution to the wave activity come from wave numbers near kd for the part a) values? c) For large x and t, the integrand of (†) will be highly oscillatory unless the phase

€

Φ≡ kx − g k( )1/ 2 t happens to be constant. Thus, for any x and t, the primary contribution to η will come from the region where the phase in (†) does not depend on k. Thus, set dΦ/dk = 0, and solve for ks (= the wave number where the phase is independent of k) in terms of x, t, and g. d) Based on the result of part b), set ks = kd to find the x-location where the dominant portion of the wave activity occurs at time t. At this location, the ratio x/t is the propagation speed of the dominant portion of the wave activity. Is this propagation speed the phase speed, the group speed, or another speed altogether? Solution 8.13. a) Below is a plot of Re{η(x,0)} vs. x for |x| ≤ 40 m when α = 10 m and kd = 2π/λd = 2π/10 m–1.

!"#"$%

!"#"&%

!"#"'%

!"#"(%

"%

"#"(%

"#"'%

"#"&%

"#"$%

"#")%

!$"% !&"% !'"% !("% "% ("% '"% &"% $"%


b) Start from the given formula for

€

˜ η (k) , and insert the given initial wave form η(x,0):

€

˜ η (k) =1

2πη(x,0)exp −ikx[ ]

−∞

+∞

∫ dx =1

2π1

2παexp −

x 2

2α 2 + ikd x( ) *

+ , -

exp −ikx[ ]−∞

+∞

∫ dx .

Complete the square in the exponent, and use the integration variable

€

β =12α

x −α 2i(kd − k)[ ]

to find:

€

˜ η (k) =1

2π 2παexp −

12α 2 x 2 − 2α 2i(kd − k)x +α 4 (kd − k)2 −α 4 (kd − k)2[ ]&

' (

) * + −∞

+∞

∫ dx

=1

2π 2παexp −

12α 2 x −α 2i(kd − k)[ ]

2−α 2(kd − k)2

2& ' (

) * + −∞

+∞

∫ dx

=1

2π 2παexp −

α 2(kd − k)2

2& ' (

) * +

2α exp −β 2{ }−∞

+∞

∫ dβ =1

2πexp −

α 2(kd − k)2

2& ' (

) * + ,

where the final definite integral over b is

€

π . Below is a plot of

€

˜ η (k)vs. k for when α = 10 m and kd = 2π/λd = 2π/10 m–1 ≈ 0.628 m–1. Clearly the wave-packet carries most of its energy in wave numbers near kd, and

€

˜ η (k) is negligible outside 0.2 < k < 1.0.

c) From the plot above, only positive wave numbers will matter because

€

˜ η (k) is small away from kd, so the absolute value can be dropped from k in the formula for the phase. Thus, the stationary phase wave number ks is determined from:

€

ddk

Φ( )k= ks=ddk

kx − gk( )1/ 2 t( )k= ks

= x − 12

gkst = 0.

Solving for ks produces:

€

ks = gt 2 4x 2 . d) Thus, at time t, the dominant portion of the wave packet plotted in part a) will reside at a location x determined from:

€

ks = kd = gt 2 4x 2 , and this location is:

€

x = 12 g kd t . Therefore, the

propagation speed, x/t, of the dominant portion of the wave packet is the group speed,

€

cg = 12 g kd .

!"

!#!$"

!#!%"

!#!&"

!#!'"

!#("

!#($"

!#(%"

!#(&"

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


Exercise 8.14. Show that the vertical component of the Stokes’ drift is zero. Solution 8.14. Start with the vertical-direction particle velocity equation (8.84b) expanded to include first order-variations in both fluid velocity components.

€

dzp (t)dt

= w(xp ,zp,t) = w(x0,z0,t) + ξ∂w∂x$

% &

'

( ) x0 ,z0

+ ζ∂w∂z$

% &

'

( ) x0 ,z0

+ ...

where ξ = xp – x0, and ζ = zp – z0, and (x0, z0) is the fluid element location in the absence of wave motion. The vertical component of the Stokes' drift will be the time average of this equation over one wave period (indicated here by angle brackets).

€

dzp (t)dt

= w(x0,z0,t) + ξ∂w∂x$

% &

'

( ) x0 ,z0

+ ζ∂w∂z$

% &

'

( ) x0 ,z0

+ ...= 0 + ξ∂w∂x$

% &

'

( ) x0 ,z0

+ ζ∂w∂z

$

% &

'

( ) x0 ,z0

+ ...

From (8.47),

€

w = aωekz sin(kx −ωt) so

€

aωekz sin(kx −ωt) = 0 and this is reflected in the second equality above. The two time averages involving displacements and velocity gradients can be calculated using the deep-water ka << 1 results for ξ, ζ, u, and w given by (8.46) and (8.47):

€

ξ ≅ −aekz0 sin(kx0 −ωt),

€

ζ ≅ aekz0 cos(kx0 −ωt), and

€

u = aωekz cos(kx −ωt) , plus the result for w given above. Substituting in these relationships produces:

€

dzp (t)dt

= 0 −aekz0 sin(kx0 −ωt)∂∂xaωekz sin(kx −ωt)

%

& '

(

) * x0 ,z0

+ aekz0 cos(kx0 −ωt)∂∂zaωekz sin(kx −ωt)

%

& '

(

) * x0 ,z0

+ ...

Simplify and consolidate terms.

€

dzp (t)dt

= −a2ωke2k0 sin(kx0 −ωt)cos(kx0 −ωt) + a2ωke2kz0 cos(kx0 −ωt)sin(kx −ωt) + ...

=12a2ωke2kz0 − sin(2kx0 − 2ωt) + sin(2kx0 − 2ωt)[ ] = 0.

Thus, the vertical component of Stokes' drift is zero.


Exercise 8.15. Extend the deep water Stokes’ drift result (8.85) to arbitrary depth to derive (8.86). Solution 8.15. Start with the horizontal-direction particle velocity equation (8.84a) expanded to include first order-variations in both fluid velocity components.

€

dxp (t)dt

= u(xp ,zp,t) = u(x0,z0,t) + ξ∂u∂x$

% &

'

( ) x0 ,z0

+ ζ∂u∂z$

% &

'

( ) x0 ,z0

+ ... ,

where ξ = xp – x0, and ζ = zp – z0, and (x0, z0) is the fluid element location in the absence of wave motion. The Stokes' drift will be the time average of this equation over one wave period (indicated here by angle brackets).

€

dxp (t)dt

= u(x0,z0,t) + ξ∂u∂x$

% &

'

( ) x0 ,z0

+ ζ∂u∂z$

% &

'

( ) x0 ,z0

+ ...= 0 + ξ∂u∂x$

% &

'

( ) x0 ,z0

+ ζ∂u∂z$

% &

'

( ) x0 ,z0

+ ...

From (8.28),

€

u = aωcosh k(z + H)( )sinh kH( )

cos(kx −ωt) so

€

u = 0 because

€

cos(kx −ωt) = 0 , and this

is reflected in the second equality above. The two time averages involving displacements and velocity gradients can be calculated using the deep-water ka << 1 results for ξ, ζ, u, and w given by (8.27) and (8.35):

€

ξ ≅ −acosh k(z0 + H)( )sinh kH( )

sin(kx0 −ωt) ,

€

ζ ≅ asinh k(z0 + H)( )sinh kH( )

cos(kx0 −ωt) , and

€

w = aωsinh k(z + H)( )sinh kH( )

sin(kx −ωt) .

plus the result for u given above. Substituting in these relationships produces:

€

dxp (t)dt

= 0 −acosh k(z0 + H)( )

sinh kH( )sin(kx0 −ωt)

∂∂xaω

cosh k(z + H)( )sinh kH( )

cos(kx −ωt)%

& '

(

) * x0 ,z0

+ asinh k(z0 + H)( )

sinh kH( )cos(kx0 −ωt)

∂∂zaω

cosh k(z + H)( )sinh kH( )

cos(kx −ωt)%

& '

(

) * x0 ,z0

+ ...

Simplify and consolidate terms.

€

dzp (t)dt

= a2ωkcosh2 k(z0 + H)( )

sinh2 kH( )cos2(kx0 −ωt) + a2ωk

sinh2 k(z0 + H)( )sinh2 kH( )

cos2(kx0 −ωt) + ...

= a2ωkcosh2 k(z0 + H)( )

sinh2 kH( )12

+sinh2 k(z0 + H)( )

sinh2 kH( )12

$

% &

'

( ) + ...

=a2ωk

2sinh2 kH( )cosh2 k(z0 + H)( ) + sinh2 k(z0 + H)( )[ ] + ...

=a2ωk

2sinh2 kH( )cosh 2k(z0 + H)( ) + ...

The final result is (8.86).


Exercise 8.16. Explicitly show through substitution and differentiation that (8.88) is a solution of (8.87). Solution 8.16. Equation (8.87) involves time and space differentiations of (8.88). Let

€

γ = 3a 4H 3( )1 2

to save a little writing. The necessary derivatives are:

€

∂η∂t

=∂∂t

a cosh2 3a 4H 3( )1 2(x − ct){ }%

& ' ( ) *

= −2asinh γ(x − ct){ }(−γc)cosh3 γ(x − ct){ }

= 2aγcsinh γ(x − ct){ }cosh3 γ(x − ct){ }

,

€

∂η∂x

=∂∂x

acosh2 γ(x − ct){ }

&

' (

)

* + = −

2asinh γ(x − ct){ }(γ)cosh3 γ(x − ct){ }

= −2aγsinh γ(x − ct){ }cosh3 γ(x − ct){ }

,

€

∂ 2η∂x 2 =

∂∂x

−2aγsinh γ(x − ct){ }

cosh3 γ(x − ct){ }

&

' (

)

* + = −2aγ 2 cosh γ(x − ct){ }

cosh3 γ(x − ct){ }− 3

sinh2 γ(x − ct){ }cosh4 γ(x − ct){ }

&

' (

)

* +

= −2aγ 2 cosh2 γ(x − ct){ }− 3sinh2 γ(x − ct){ }cosh4 γ(x − ct){ }

&

' (

)

* + = −2aγ 2 1− 2sinh2 γ(x − ct){ }


&

' (

)

* + ,

and

€

∂ 3η∂x 3 = −2aγ 2 ∂

∂x1− 2sinh2 γ(x − ct){ }


&

' (

)

* +

= −2aγ 3 −4sinh γ(x − ct){ }cosh γ(x − ct){ }cosh4 γ(x − ct){ }

− 41− 2sinh2 γ(x − ct){ }( )sinh γ(x − ct){ }


&

' ( (

)

* + +

= 8aγ 3 sinh γ(x − ct){ }cosh2 γ(x − ct){ }cosh5 γ(x − ct){ }

+1− 2sinh2 γ(x − ct){ }( )sinh γ(x − ct){ }


&

' ( (

)

* + +

= 8aγ 3 sinh γ(x − ct){ } 3− cosh2 γ(x − ct){ }( )cosh5 γ(x − ct){ }

&

' ( (

)

* + +

It is now clear that the arguments of all the hyperbolic functions are the same, so, to save writing, continue with the implicit notation that {} = {γ(x – ct)}. The next step is to start reconstructing (8.87), the Korteweg-DeVries equation. Consider the first two terms together:

€

∂η∂t

+ c0∂η∂x

= 2aγ(c − c0)sinh{ }cosh3{ }

= a2γ c0Hsinh{ }cosh3{ }

,

where the final equality follows from the propagation speed relationship

€

c = c0 1+ a 2H( ) . The third and fourth terms are:

€

32c0ηH∂η∂x

+c0H

2

6∂ 3η∂x 3 = −

3c0γa2

Hsinh{ }

cosh5{ }+

4c0H2

3aγ 3 sinh{ } 3− cosh2{ }( )

cosh5{ }

&

' ( (

)

* + +

= −3c0γa

2

H+

4c0H2

33a

4H 3

,

- .

/

0 1 3aγ

&

' (

)

* +

sinh{ }cosh5{ }

−4c0H

2

33a

4H 3

,

- .

/

0 1 aγ

sinh{ }cosh3{ }

&

' (

)

* + ,


€

= −3c0γa

2

H+c0

H3a2γ

$

% &

'

( )

sinh{ }cosh5{ }

− a2γc0

Hsinh{ }

cosh3{ }

$

% &

'

( )

= 0 − a2γc0

Hsinh{ }

cosh3{ }

$

% &

'

( )

Thus, adding together all four terms produces:

€

∂η∂t

+ c0∂η∂x

+32c0ηH∂η∂x

+c0H

2

6∂ 3η∂x 3

= a2γ c0Hsinh{ }cosh3{ }

− a2γ c0H

sinh{ }cosh3{ }

&

' (

)

* + = 0 ,

which shows that (8.88) is a solution of (8.87).


Exercise 8.17. A sinusoidal long-wavelength shallow-water wave with amplitude A and wave number k1 = 2π/λ1 travels to the right in water of depth H1 until it encounters a mild depth transition at x = 0 to a slightly shallower depth H2. A portion of the incident wave continues to the right with amplitude B and a portion is reflected and propagates to the left. a) By requiring the water surface deflection and the horizontal volume flux in the water column to be continuous at x = 0, show that B A = 2 1+ H2 H1( ) .

b) If a tsunami wave starts with A = 2.0 m in water 5 km deep, estimate its amplitude when it reaches water 10 m deep if the ocean depth change can be modeled as a large number of discrete depth changes. [Recommendation: consider multiple steps of H2/H1 = 0.90, since (0.90)59 ≈ 0.002 = (10 m)/(5 km)]. c) Redo part b) when the wave's energy flux remains constant at both depths when λ1 = 100 km (as was done in Example 8.6). d) Compare the results of parts b) and c). Which amplitude in shallow water is lower and why is it lower?

Solution 8.17. The surface elevation on the left will involve two waves:

η(x, t) = Aexp ik1 x − t gH1( ){ }+C exp ik1 −x − t gH1( ){ } for x < 0,

where C is the amplitude of the reflected wave. The surface elevation on the right will only involve the transmitted wave with amplitude B:

η(x, t) = Bexp ik2 x − t gH2( ){ } for x > 0.

To determine the relationships between wave amplitudes, match the surface elevations and the volume flows at x = 0. Start with the surface elevation:

η(0−, t) =η(0+, t) or A+C( )exp −ik1t gH1{ }= Bexp −ik2t gH2{ } .

Here, the wave frequencies must be the same. This implies: ω = k1 gH1 = k2 gH2 and A+C = B . (1,2)

From (8.51), the horizontal velocity of the incident wave is u = Aωk1H1

exp ik1 x − t gH1( ){ }

so the horizontal volume flux (per unit depth perpendicular to the page) is

q = uH1 =Aωk1exp ik1 x − t gH1( ){ } . Thus, continuity of the volume flux at x = 0 requires:

q(0−, t) = q(0+, t) or Aωk1

−Cωk1

=Bωk2

, (3)

where (1) has been used to eliminate the complex exponentials and the reflected wave carries a minus sign because it propagates in the negative-x direction. Now, use (2) and (3) to eliminate C and reach:

x!

z!

H1! H2!

A! B!


2A = k1k2+1

!

"#

$

%&B or B

A= 2 1+ k1

k2

!

"#

$

%&

−1

= 2 1+ H2

H1

!

"##

$

%&&

−1

.

where the final equality follows from (1). b) For one step with H2/H1 = 0.90, B/A = 1.026334. For 59 such steps, (B/A)59 = (1.026334)59 = 4.635. Thus the wave amplitude will be approximately 9.27 m in 10 m of water. This result must be approximate since the wave amplitude is now comparable to the water depth and this does not satisfy the requirements of the linear theory. c) Follow the approach in Example 8.6. First, determine the wave speed and ka value when H = 5 km: c = gH = (9.81ms−2 )(5, 000m) = 221.5 ms–1, ka = 2πa/λ = 2π(2.0m)/(100km) = 1.257×10−4 .

Here a/H = 4.0×10−4 and ka are both small enough to use linear wave results. So, from (8.44), the energy flux, EF, for these shallow water waves is:

EF = ρga2

2c = (10

3kgm−3)(9.81ms−2 )(2.0m)2

2(221.5ms−1) = 4.346 MW/m.

The wave period here is λ/c = (100km)/(221.5 ms–1) = 451.5s, and this will be the same (or nearly so) when the wave train reaches shallower water where its speed will be

c = gH = (9.81ms−2 )(10m) = 9.90 ms–1. The wavelength in shallow water will be (9.90 ms–1)(451.5s) = 4.47 km. For constant energy flux, the wave amplitude at the shallower depth will be:

a = 2EFρgc

!

"#

$

%&

1 2

=2 ⋅ 4.346×106Wm−1

(103kgm−3)(9.81ms−2 )(9.90ms−1)!

"#

$

%&

1 2

= 9.46 m,

Here again, the amplitude-depth ratio is too large for the linear theory; a nonlinear theory is needed to better determine the wave shape and its amplitude. d) The answer to part b) is slightly lower because some wave energy is reflected at each depth transition, while no such energy reflection is accounted for in part c).


Exercise 8.18. A thermocline is a thin layer in the upper ocean across which water temperature and, consequently, water density change rapidly. Suppose the thermocline in a very deep ocean is at a depth of 100 m from the ocean surface, and that the temperature drops across it from 30 to 20 °C. Show that the reduced gravity is gʹ′ = 0.025 m/s2. Neglecting Coriolis effects, show that the speed of propagation of long gravity waves on such a thermocline is 1.58 m/s. Solution 8.18. This is the case of an internal wave at the interface between a shallow layer overlying an infinitely deep fluid (Section 8.7). Here the layer depth is 100 m, the temperature difference is ΔT = 10°C, and the average temperature is 25°C so the thermal expansion coefficient α is approximately

€

2.5 ×10−4 /°C (from Appendix A3). The relative density change will be Δρ/ρ = αΔT =

€

2.5 ×10−3. Thus, the reduced gravity and wave speed can be determined from (8.115) and (8.116):

€

" g = gαΔT = 9.81(0.0025) = 0.0245m /s2 , and

€

c = 0.0245(100) =1.58m /s.


Exercise 8.19. Consider internal waves in a continuously stratified fluid of buoyancy frequency N = 0.02 s–1 and average density 800 kg/m3. What is the direction of ray paths if the frequency of oscillation is ω = 0.01 s–1? Find the energy flux per unit area if the amplitude of the vertical velocity is

€

ˆ w = 1 cm/s and the horizontal wavelength is π meters. Solution 8.19. Here, N = 0.02 s–1, ω = 0.01 s–1, k = 2π/λ = 2 m–1,

€

ˆ w = 1 cm/s, and ρo = 800 kg/m3. From (8.138), ω = Ncosθ, solve for the angle:

€

θ = cos−1 ω N( ) = cos−1 0.01 0.02( ) = 60°. To determine the energy flux, calculate the vertical wave number:

m = ktanθ = 2tan(60°) = 3.464 m–1, and use (8.158)

€

F =ρ0ωm ˆ w 2

2k 2 exmk− ez

%

& '

(

) * =

800(0.01)(3.464)(0.01)2

2(2)2 ex3.464

2− ez

%

& '

(

) *

= 3.464 ×10−4 1.732ex − ez( )W /m2.

The magnitude of the energy flux is:

€

F = 3.464 ×10−4 1.7322 +1( )1 2

= 6.92 ×10−4W /m2.


Exercise 8.20. Consider internal waves at a density interface between two infinitely deep fluids, and show that the average kinetic energy per unit horizontal area is Ek = (ρ2 − ρ1)ga2/4. Solution 8.20. From Section 8.7, the velocity potentials above and below the interface are:

€

φ1 =iωake−kz exp i(kx −ωt){ } , and

€

φ2 = −iωake+kz exp i(kx −ωt){ }.

respectively. The (real) velocity components are:

€

u1 = Re ∂φ1∂x

$

% &

'

( ) = Re

iωak(ik)e−kz exp i(kx −ωt){ }

$

% &

'

( ) = −ωae−kz cos(kx −ωt),

€

u2 = Re ∂φ2∂x

$

% &

'

( ) = Re −

iωak(ik)e+kz exp i(kx −ωt){ }

$

% &

'

( ) =ωae+kz cos(kx −ωt) ,

€

w1 = Re ∂φ1∂z

$

% &

'

( ) = Re

iωak(−k)e−kz exp i(kx −ωt){ }

$

% &

'

( ) =ωae−kz sin(kx −ωt), and

€

w2 = Re ∂φ2∂z

$

% &

'

( ) = Re −

iωak(k)e+kz exp i(kx −ωt){ }

$

% &

'

( ) =ωae+kz sin(kx −ωt).

The kinetic energy per unit surface are then involves integrating through the depth:

€

KE =12ρ1 u1

2 + w12( )

0

∞

∫ dz +12ρ2 u2

2 + w22( )

−∞

0

∫ dz

=12ρ1 ω 2a2 cos2() +ω 2a2 sin2()( )

0

∞

∫ e−2kzdz +12ρ2 ω 2a2 cos2() +ω 2a2 sin2()( )e+2kz

−∞

0

∫ dz

=ω 2a2

2ρ1 e−2kz

0

∞

∫ dz +ω 2a2

2ρ2 e+2kz

−∞

0

∫ dz =ω 2a2

4kρ1 + ρ2( ).

where () = kx – ωt. The dispersion relationship for these waves is (8.95)

€

ω 2 = gk ρ2 − ρ1ρ2 + ρ1

%

& '

(

) * .

Use this to eliminate ω from the final form for the kinetic energy per unit surface area to find:

€

KE =ω 2a2

4kρ1 + ρ2( ) = gk ρ2 − ρ1

ρ2 + ρ1

%

& '

(

) * a2

4kρ1 + ρ2( ) =

(ρ2 − ρ1)ga2

4.


Exercise 8.21. Consider waves in a finite layer overlying an infinitely deep fluid. Using the constants given in (8.105) through (8.108), prove the dispersion relation (8.109). Solution 8.21. Equations (8.105)-(8.108) are:

€

A = −ia2

ωk

+gω

$

% &

'

( ) ,

€

B =ia2

ωk−gω

$

% &

'

( ) ,

€

C = −ia2

ωk

+gω

$

% &

'

( ) −

ia2

ωk−gω

$

% &

'

( ) e2kH , and

€

b = i kωCe−kH ,

respectively. Applying BC (8.100)

€

ρ1∂φ1∂t

+ ρ1gζ = ρ2∂φ2∂t

+ ρ2gζ at z = –H.

leads to:

€

ρ1 −iω( ) Ae−kH + Be+kH( ) + ρ1gb = ρ2 −iω( ) Ce−kH( ) + ρ2gb. Substituting for b in terms of C produces:

€

−ωρ1 Ae−kH + Be+kH( ) = Ce−kH −ωρ2 +

gkω(ρ2 − ρ1)

%

& ' (

) * .

Now eliminate A, B, and C using (8.105)-(8.107) and let x = ω2/gk to ease the algebra. These steps lead to:

ρ1 −(x +1)e−kH + (x −1)e+kH( ) = e−kH (x +1)e−kH + (x −1)e+kH"# $%

ρ2 − ρ1x

− ρ2"

#&$

%'.

This can be rearranged and stated in terms of hyperbolic functions:

€

(x −1) x ρ1 sinh(kH) + ρ2 cosh(kH)[ ] − (ρ2 − ρ1)sinh(kH){ } = 0, which is (8.109).


Exercise 8.22. A simple model of oceanic internal waves involves two ideal incompressible fluids (ρ2 > ρ1) trapped between two horizontal surfaces at z = h1 and z = –h2, and having an average interface location of z = 0. For traveling waves on the interface, assume that the interface deflection from z = 0 is

€

ξ = ξoRe exp i(ωt − kx)( ){ }. The phase speed of the waves is c = ω/k.

a) Show that dispersion relationship is:

€

ω 2 =gk(ρ2 − ρ1)

ρ2 coth(kh2) + ρ1 coth(kh1) where g is the

acceleration of gravity. b) Determine the limiting form of c for short (i.e. unconfined) waves, kh1 and kh2 → ∞. c) Determine the limiting form of c for long (i.e. confined) waves, kh1 and kh2 → 0. d) At fixed wavelength λ (or fixed k = 2π/λ), do confined waves go faster or slower than unconfined waves? e) At a fixed frequency, what happens to the wavelength and phase speed as ρ2 – ρ1 → 0? f) What happens if ρ2 < ρ1?

Solution 8.22. a) For this problem there at two velocity potentials that must be matched at the moving interface. Based on the form of the interface wave,

€

ξ = ξoRe exp i(ωt − kx)( ){ } and the development given in the chapter, the form of the two potentials can be set:

€

φ1(x,z,t) = Z1(z)ei(ωt−kx ) , and

€

φ2(x,z,t) = Z2(z)ei(ωt−kx )

Here the boundary conditions and the field equation are:

€

∇2φ = 0:

€

Z1(z) = A1+e+kz + A1−e

−kz , and

€

Z2(z) = A2+e+kz + A2−e

−kz

€

∂φ∂z

=∂ξ∂t

on z = 0:

€

kA1+ − kA1– = iωξo = kA2+ − kA2– (1,2)

€

ρ1∂φ1∂t

+ ρ1gξ = ρ2∂φ2∂t

+ ρ2gξ on z = 0:

€

iωρ1 A1+ + A1−( ) + gρ1ξo = iωρ2 A2+ + A2−( ) + gρ2ξo (3)

€

∂φ1∂z

$

% &

'

( ) z= h1

=∂φ2∂z

$

% &

'

( ) z=−h2

= 0

€

kA1+e+kh1 − kA1–e

−kh1 = 0 , and

€

kA1+e−kh2 − kA1–e

+kh2 = 0 (4,5)

The three boundary conditions yield five equations, enough to determine all four A’s and the dispersion relationship. First use the (1) & (4), and (2) & (5) to find:

€

A1,± = −iωξo2k

e kh1

sinh kh1( ) , and

€

A2,± = +iωξo2k

e±kh2

sinh kh2( )

Plug these into (3), to find:

€

ωρ1ωξokcosh(kh1)sinh kh1( )

%

& '

(

) * + gρ1ξo = −ωρ2

ωξokcosh(kh2)sinh kh2( )

%

& '

(

) * + gρ2ξo


Cancel the common factor of ξo, simplify and rearrange:

€

ω 2 =gk(ρ2 − ρ1)

ρ1 coth(kh1) + ρ2 coth(kh2).

b) Using the results of part a), the phase speed is

€

c =ωk

= ±g(ρ2 − ρ1) k

ρ1 coth(kh1) + ρ2 coth(kh2).

For short waves, kh1 and kh2 → ∞, and the hyperbolic co-tangent functions both approach unity so:

€

c =ωk

= ±g(ρ2 − ρ1)k(ρ1 + ρ2)

c) For long waves, kh1 and kh2 → 0, and the hyperbolic co-tangent functions both approach the

inverse of their arguments:

€

c =ωk

= ±g(ρ2 − ρ1) k

ρ1(1/kh1) + ρ2(1/kh2)= ±

g(ρ2 − ρ1)h1h2ρ1h2 + ρ2h1

Note that when ρ1 → 0, the answers to parts b) and c) both recover the ordinary linear water wave results. d) For simplicity, take h1 = h2 = h, then for short (unconfined) waves:

c = ± g(ρ2 − ρ1)(ρ1 + ρ2 )

1k

,

and for long (confined) waves:

€

c = ±g(ρ2 − ρ1)ρ1 + ρ2

h = ±g(ρ2 − ρ1)ρ1 + ρ2

khk

.

For k = const., as

€

kh→ 0, the confined waves travel slower than the unconfined waves with the same wavelength (same wave number k). e) As ρ2 – ρ1 → 0, the phase speed and wavelength both go to zero. f) If ρ2 < ρ1, then the two fluid layers are not stably stratified. The dispersion relationship then

requires ω to be imaginary, i.e.

€

ω = ±i gk(ρ1 − ρ2)ρ2 coth(kh2) + ρ1 coth(kh1)

. This means that there is an

interface wave solution that is growing exponentially with increasing time. The situation is unstable, but not much more can be determined from the linearized theory. In reality, the two fluids will switch places but the linearized theory considered here is not valid throughout this process.


Exercise 8.23. Consider the long-wavelength limit of surface and interface waves with amplitudes a and b, respectively, that occur when two ideal fluids with densities ρ1 and ρ2 (> ρ1) are layered as shown in Figure 8.28. Here, the velocity potentials – accurate through second order in kH1 and KH2 – in the two fluids are:

φ1 ≅ A1 1+12 k

2 (z− z1)2( )ei(kx−ωt ) and φ2 ≅ A2 1+

12 k

2 (z− z2 )2( )ei(kx−ωt ) for kH1, KH2 << 1,

where b, A1, z1, A2, and z2 are constants to be found in terms of a, g, k, ω, H1 and H2. From analysis similar to that in Section 8.7, find the dispersion relationship ω = ω(k). When H1 = H2 = H/2, show that the surface and interface waves are in phase with a > b for the barotropic mode, and out-of-phase with b > a for the baroclinic mode. For this case, what are the phase speeds of the two modes? Which mode travels faster? What happens to the baroclinic mode's phase speed and amplitude as ρ2 − ρ1→ 0 ?

Figure 8.28.

Solution 8.23. The form of the two potentials are given in the problem statement, and the surface and interface elevations from equilibrium can be taken from (8.101) and (8.102):

η = aei(kx−ωt ) and ζ = bei(kx−ωt ) . The boundary conditions are:

∂φ2/∂z = 0 on z = –(H1 + H2) (i) ∂φ1/∂z = ∂η/∂t on z = 0 (ii) ∂φ1/∂t + gη = 0 on z = 0 (iii)

∂φ1/∂z = ∂φ2/∂z = ∂ζ/∂t on z = –H1 (iv) ρ1∂φ1/∂t + ρ1gζ = ρ2∂φ2/∂t + ρ2gζ on z = –H1 (v)

Apply these in turn produces the following algebraic equations: (i): A2k2(z – z2) = 0 on z = –(H1 + H2), or A2k2(–H1 – H2 – z2) = 0 (ii): A1k2(z – z1) = –iωa on z = 0, or A1k2z1 = iωa (iii) −iωA1 1+

12 k

2 (z− z1)2( )+ ga = 0 on z = 0, or iωA1 ≅ ga

(iv) A1k2(z – z1) = A2k2(z – z2) = –iωb on z = –H1, or A1k2(H1 + z1) = A2k2(H1 + z2) = iωb (v) −iωρ1A1 1+

12 k

2 (z− z1)2( )+ ρ1gb = −iωρ2A2 1+ 1

2 k2 (z− z2 )

2( )+ ρ2gb on z = –H1, or −iωρ1A1 + ρ1gb ≅ −iωρ2A2 + ρ2gb where the second order terms involving [k.(vertical distance)]2 appearing in (iii) and (v) have been dropped compared to unity in the equations on the right. Noting that (iv) provides two equations, the first five equations on the right form a non-linear algebraic system that can be solved to find:

z2 = –(H1 + H2), z1 = –ω2/k2g, A1 = –iga/ω, A2 = −iaωk

1kH2

−gkH1

ω 2H2

"

#$

%

&' , and b = a 1− gk

2H1

ω 2

"

#$

%

&' .

z!

b!

Barotropic!

x!

a!

H1!

H2!

ρ1!

ρ2!

z! a!

Baroclinic!

x!

b!


Substitute these into the final approximate equation provided by (v):

−iωρ1−igaω

"

#$

%

&'+ ρ1ga 1−

gk2H1

ω 2

"

#$

%

&' ≅ −iωρ2

−iaωk

"

#$

%

&'

1kH2

−gkH1

ω 2H2

"

#$

%

&'+ ρ2ga 1−

gk2H1

ω 2

"

#$

%

&' .

Divide out the common factor of a, and simplify:

−ρ1g2k2H1

ω 2 ≅−ρ2ω

2

k1kH2

−gkH1

ω 2H2

#

$%

&

'(+ ρ2g− ρ2

g2k2H1

ω 2 , or

0 ≅ − ρ2ω2

k2H2

+ ρ2g 1+H1

H2

#

$%

&

'(− ρ2 − ρ1( ) g

2k2H1

ω 2

Multiply –ω2k2H2/ρ2 and rearrange to find:

0 ≅ω 4 − gk kH2 + kH1( )ω 2 +(ρ2 − ρ1)

ρ2k2g2kH1kH2 .

This is a quadratic equation for ω2. The two solutions are:

ω 2 =k2g(H1 +H2 )

21± 1− ρ2 − ρ1

ρ2

"

#$

%

&'4H1H2

(H1 +H2 )2

"

#

$$

%

&

'' .

Set H1 = H2 = H/2 to reach:

ω 2 =k2gH2

1± ρ1ρ2

!

"##

$

%&& .

Consider the "+" sign for the barotropic mode. For H1 = H2 = H/2, the relationship between the surface deflection amplitude a and the interface deflection amplitude b is:

b = a 1− gk2H1

ω 2

"

#$

%

&'= a 1−

gk2Hk2gH 1+ ρ1 ρ2( )

"

#

$$

%

&

''= a

ρ1 ρ21+ ρ1 ρ2

"

#$$

%

&'' .

The factor in parentheses is positive with a value near 1/2 when the densities are nearly equal. Thus, a and b are in phase. Now consider the "–" sign for the baroclinic mode.

b = a 1− gk2H1

ω 2

"

#$

%

&'= a 1−

gk2Hk2gH 1− ρ1 ρ2( )

"

#

$$

%

&

''= a

− ρ1 ρ21− ρ1 ρ2

"

#$$

%

&'' .

The factor in parentheses is negative with a magnitude much larger than unity when the densities are nearly equal. Thus, a and b are out of phase. The phase speeds of the two modes when H1 = H2 = H/2 are obtained from the dispersion relationships above:

cp =ωk=gH21± ρ1

ρ2

!

"##

$

%&&

'

())

*

+,,

1 2

,

with barotropic mode (corresponding to the "+" sign) traveling faster than the baroclinic mode (corresponding to the "–" sign). As ρ2 − ρ1→ 0 , the baroclinic mode's phase speed approaches zero, and its amplitude becomes unbounded. This mode amplitude result is inconsistent with the long-wavelength & shallow-water approximations, so a more refined theory is needed to truly explain the baroclinic mode amplitude in this limit.

fluid mechanics kundu cohen 6th edition solutions sm ch (8)

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