fluid mechanics kundu cohen 6th edition solutions sm ch (13)

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

Exercise 13.1. The Gulf Stream flows northward along the east coast of the United States with a surface current of average magnitude 2 m/s. If the flow is assumed to be in geostrophic balance, find the average slope of the sea surface across the current at a latitude of 45°N. [Answer: 2.1 cm per km] Solution 13.1. Given v = 2 m/s, and f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1. For geostrophic balance:

fv = 1ρ∂p∂x

= g∂η∂x

,

so the sea-surface slope is: ∂η∂x

=fvg=(1.03×10−4s−1)(2ms−1)

9.81ms−2= 2.1×10−5 ,

which is equivalent to an eastward surface rise of 2.1 cm per km.


Exercise 13.2. A plate with water (ν = 10−6 m2/s) above it rotates at a rate of 10 revolutions per minute. Find the depth of the Ekman layer, assuming that the flow is laminar. Solution 13.2. Given Ω = 10 rpm = 20π rad./min. = 1.047 s–1, the Coriolis frequency is f = 2Ω = 2.09 s–1. From (13.28), the Ekman layer thickness is:

δ = 2ν f = 2(10−6 m2s−1) (2.09s−1) = 0.978×10−3m ≅1 mm .


Exercise 13.3. Assume that the atmospheric Ekman layer over the earth’s surface at a latitude of 45ºN can be approximated by an eddy viscosity of ν V = 10 m2/s. If the geostrophic velocity above the Ekman layer is 10 m/s, what is the Ekman transport across isobars? [Answer: 2203 m2/s] Solution 13.3. Given ν V = 10 m2/s, f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, and U = 10 m/s, the Ekman layer height from (13.28) is

δ = 2ν f = 2(10m2s−1) (1.03×10−4s−1) = 440.7m . Using the equation at the top of page 720, the net transport perpendicular to the geostrophic stream is:

12Uδ = 0.5(10ms–1)(440.7m) = 2,203m2s–1 .


Exercise 13.4. a) From the set (13.45) – (13.47), develop the following equation for the water surface elevation η(x,y,t):

∂∂t

∂2

∂t2+ f 2 − gH ∂2

∂x2+∂2

∂y2#

$%

&

'(

)*+

,-.η(x, y, t) = 0

b) Using η(x, y, t) = η̂ exp i(kx + ly−ωt){ } show that that the dispersion relationship reduces to ω = 0 or (13.82). c) What type of flows have ω = 0? Solution 13.4. The starting-point equation set is:

∂η∂t

+H ∂u∂x

+∂v∂y

!

"#

$

%&= 0 , ∂u

∂t− fv = −g∂η

∂x, ∂v∂t+ fu = −g∂η

∂y. (13.45, 13.46, 13.47)

Apply ∂/∂t to (13.46), multiply (13.47) by f, and add the results to find: ∂2

∂t2+ f 2

"

#$

%

&'u = −g

∂2η∂x∂t

+ f ∂η∂y

"

#$

%

&' . (a)

Multiply (13.46) by –f, apply ∂/∂t to (13.47), and add the results to find: ∂2

∂t2+ f 2

"

#$

%

&'v = −g

∂2η∂y∂t

− f ∂η∂x

"

#$

%

&' . (b)

Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach: ∂2

∂t2+ f 2

"

#$

%

&'∂u∂x+∂v∂y

"

#$

%

&'=

∂2

∂t2+ f 2

"

#$

%

&' −

1H∂η∂t

"

#$

%

&'= −g

∂2

∂x2+∂2

∂y2"

#$

%

&'∂η∂t

, (c)

where the first equality follows from (13.45). Using the final equality, multiply by –H, and collect all the terms to left side of the equation to find:

∂2

∂t2+ f 2

"

#$

%

&'∂η∂t

"

#$

%

&'− gH

∂2

∂x2+∂2

∂y2"

#$

%

&'∂η∂t

= 0 . (d)

Factor out the common differentiation operation ∂/∂t to reach: ∂∂t

∂2

∂t2+ f 2 − gH ∂2

∂x2+∂2

∂y2#

$%

&

'(

)*+

,-.η(x, y, t) = 0 .

b) Substitute the trial solution η(x, y, t) = η̂ exp i(kx + ly−ωt){ } into the part a) differential equation to find:

−iω −ω 2 + f 2 − gH −k2 − l2( ){ }η̂ = 0 .

For nonzero η̂ , this equation has solutions ω = 0 and ω 2 = f 2 + gH k2 + l2( ) = f 2 + gHK 2 ,

where K2 = k2 + l2, and the final equation is (13.82). c) The specification ω = 0 implies steady flow, and the steady flow solutions of (13.45) – (13.47) are geostrophic flows.


Exercise 13.5. Find the axis ratio of a hodograph plot for a semidiurnal tide in the middle of the ocean at a latitude of 45°N. Assume that the mid-ocean tides are rotational surface gravity waves of long wavelength and are unaffected by the proximity of coastal boundaries. If the depth of the ocean is 4 km, find the wavelength, the phase velocity, and the group velocity. Note, however, that the wavelength is comparable to the width of the ocean, so that the neglect of coastal boundaries is not very realistic. Solution 13.5. Given f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, ω = 2 rev./day = 1.45 x10–4 s–1, and H = 4 km, the axis ratio (as given at the top of page 733) is

axis ratio = ω/f = 1.45/1.03 = 1.414. At the given depth the phase speed is:

c = [gH]1/2 = [(9.81 m2/s)(4,000m)]1/2 = 198 m/s, and the group speed is the same because it is a shallow water wave. The wavelength is:

λ = c(period) = (198 m/s)(12 hrs)(3600 s/hr) = 8554 km. This wavelength is long enough so that the neglect of coast boundaries is an unrealistic approximation.


Exercise 13.6. An internal Kelvin wave on the thermocline of the ocean propagates along the west coast of Australia. The thermocline has a depth of 50 m and has a nearly discontinuous density change of 2 kg/m3 across it. The layer below the thermocline is deep. At a latitude of 30°S, find the direction and magnitude of the propagation speed and the decay scale perpendicular to the coast. Solution 13.6. The given information is: f = –2Ωsin30° = –0.73x10–4 s–1, H = 50 m, and Δρ = 2 kg/m3. This is the case of a shallow layer of lighter water, overlying a deep sea. From equation (8.115), the internal gravity wave speed is:

c = !g H = g Δρρo

H = (9.81ms−1) 2kgm−3

103kgm−3 (50m) = 0.99ms−1 .

These waves propagate southward along the west coast of Australia. The decay scale perpendicular to the coast is the Rossby radius:

Λ =cf=

0.99ms−1

0.73×10−4s−1=1.36×104m =13.6km .


Exercise 13.7. Derive (13.96) for the vertical velocity w from (4.10), (13.48), (13.49), (13.51), (13.95) by eliminating all other dependent variables. Solution 13.7. The five starting-point equations are:

∂u∂x+∂v∂y+∂w∂z

= 0 , (4.10)

∂u∂t− fv = − 1

ρ0

∂ #p∂x

, ∂v∂t+ fu = − 1

ρ0

∂ #p∂y

, (13.48, 13.49)

∂ "ρ∂t

− ρ0N 2

gw = 0 , and (13.51)

∂w∂t

= −1ρ0

∂ #p∂z

−g #ρρ0

. (13.95)

Apply ∂/∂t to (13.48), multiply (13.49) by f, and add the results to find: ∂2

∂t2+ f 2

"

#$

%

&'u = −

1ρ0

∂2 )p∂x∂t

+ f ∂ )p∂y

"

#$

%

&' . (a)

Multiply (13.48) by –f, apply ∂/∂t to (13.49), and add the results to find: ∂2

∂t2+ f 2

"

#$

%

&'v = −

1ρ0

∂2 )p∂y∂t

− f ∂ )p∂x

"

#$

%

&' . (b)

Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach: ∂2

∂t2+ f 2

"

#$

%

&'∂u∂x+∂v∂y

"

#$

%

&'=

∂2

∂t2+ f 2

"

#$

%

&' −

∂w∂z

"

#$

%

&'= −

1ρ0

∂2

∂x2+∂2

∂y2"

#$

%

&'∂ )p∂t

, (c)

where the first equality follows from (4.10). Now apply ∂/∂t to (13.95), and use (13.51) to substitute for ∂ρ´/∂t:

∂2w∂t2

= −1ρ0

∂2 #p∂z∂t

−gρ0

∂ #ρ∂t

= −1ρ0

∂2 #p∂z∂t

−gρ0

ρ0N 2

gw

$

%&

'

()= −

1ρ0

∂2 #p∂z∂t

− N 2w . (d)

Use the definitions ∇H2 ≡ ∂2 ∂x2 +∂2 ∂y2 to rewrite the final equality of (c), and rewrite the ends

of the extended equality (d): ∂2

∂t2+ f 2

"

#$

%

&'∂w∂z

=1ρ0∇H2 ∂ )p∂t

, and ∂2

∂t2+ N 2"

#$

%

&'w =

1ρ0

∂2 (p∂z∂t

(e,f)

Apply ∂/∂z to (e) and ∇H2 to (f), and add the results to find:

∂2

∂t2+ f 2

"

#$

%

&'∂2w∂z2

+∇H2 ∂2

∂t2+ N 2"

#$

%

&'w =

1ρ0∇H2 ∂

2 )p∂z∂t

−1ρ0∇H2 ∂

2 )p∂z∂t

= 0 .

Rearrange the terms noting that ∇H2 +∂2 ∂z2 ≡ ∇2 :

∇2 ∂2w∂t2

+ f 2 ∂2w∂z2

+ N 2∇H2 w = 0 ,

which is (13.96).


Exercise 13.8. Using the dispersion relation m2 = k2(N2 − ω2)/(ω2 − f2) for internal waves, show

that the group velocity vector is given by cgx,cgz!" #$=(N 2 − f 2 )km

(m2 + k2 )3 2 (m2 f 2 + k2N 2 )1 2m,−k[ ] .

[Hint: Differentiate the dispersion relation partially with respect to k and m.] Show that cg and c are perpendicular and have oppositely directed vertical components. Verify that cg is parallel to u.] Solution 13.8. Start with m2 = k2(N2 − ω2)/(ω2 − f2), and algebraically rearrange it to find:

ω2 = f2 + k2(N2 − ω2)/m2. (1)

Here we note that c = (cx, cz) = ωk2 +m2 k,m( ) , so the components of the group velocity must be

found next. To find cgx, differentiate (1) with respect to the horizontal wave number k:

2ω ∂ω∂k

=2km2 N 2 −ω 2( )− k

2

m2 2ω∂ω∂k

#

$%

&

'( .

Set cgx = ∂ω/∂k, to find:

2ωcgx =2km2 N 2 −ω 2( )− k

2

m2 2ωcgx( ) ,

or ωcgx m2 + k2( ) = k N 2 −ω 2( ) ,

or cgx =k N 2 −ω 2( )ω m2 + k2( )

. (2)

From here ω can be eliminated from (2) using (1) to find:

cgx =km2 N 2 − f 2( )

m2 + k2( )3 2

m2 f 2 + k2N 2( )1 2 .

To find cgy, differentiate (1) with respect to the vertical wave number m:

2ω ∂ω∂m

=k2

m2 −2ω ∂ω∂m

#

$%

&

'(− 2k2 N 2 −ω 2( ) 1m3 .

Set cgz = ∂ω/∂m, and rearrange the last equation to find:

cgz = −k2 N 2 −ω 2( )mω m2 + k2( )

. (3)

Here again, ω can be eliminated from (3) using (1) to find:

cgz = −k2m N 2 − f 2( )

m2 + k2( )3 2

m2 f 2 + k2N 2( )1 2 .

Thus we can write: cg = cgx,cgz( ) = (N 2 − f 2 )km(m2 + k2 )3 2 (m2 f 2 + k2N 2 )1 2

m,−k( ) .


To show that cg and c are perpendicular, it is sufficient to show that their dot product is zero:

cg ⋅ c =(N 2 − f 2 )km

(m2 + k2 )3 2 (m2 f 2 + k2N 2 )1 2m,−k( ) ⋅ ω

k2 +m2 k,m( ) = 0 .

Here the sign of cz is opposite that of cgz , so it follows that the vertical components of c and cg are oppositely directed. Finally, cg and u are parallel because:

cgxcgz

= −mk=uw

,

where (13.111) has been used to reach the final equality.


Exercise 13.9. Suppose the atmosphere at a latitude of 45°N is idealized by a uniformly stratified layer of height 10 km, across which the potential temperature increases by 50°C.

a) What is the value of the buoyancy frequency N? b) Find the speed of a long gravity wave corresponding to the n = 1 baroclinic mode. c) For the n = 1 mode, find the westward speed of nondispersive (i.e., very large wavelength) Rossby waves. [Answer: N = 0.01279 s−1; c1 = 40.71 m/s; cx = −3.12 m/s]

Solution 13.9. The given info. is f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, H = 10 km, and ΔT = 50 °C. At absolute temperature T, the coefficient of thermal expansion for a perfect gas is 1/T. a) Therefore the buoyancy frequency is given by:

N 2 = −gρo

dρdz

= gα dTdz

=gTdTdz

=9.81ms−2

300K50°104m

=1.635×10−4s−2 , so N = 0.01279 s–1.

b) From equation (13.71), the internal gravity wave speed corresponding to the nth normal mode is cn = NH/nπ, so

c1 = NH/π = (0.01279 s–1)(104 m)/π = 40.71 m/s. c) From Section 13.13, the westward speed of a non-dispersive Rossby wave (corresponding to the n = 1 mode) is:

cx ≅ −βc12 f 2 = −(2×10−11s−1m−1)(40.71ms−1)2 (1.03×10−4 s−1)2 = −3.12ms−1 .


Exercise 13.10. Consider a steady flow rotating between plane parallel boundaries a distance L apart. The angular velocity is Ω and a small rectilinear velocity U is superposed. There is a protuberance of height h << L in the flow. The Ekman and Rossby numbers are both small: Ro << 1, E << 1. Obtain an integral of the relevant equations of motion that relates the modified pressure and the stream function for the motion, and show that the modified pressure is constant on streamlines. Solution 13.10. For Rossby and Ekman numbers both much less than unity (Ro << 1 and E << 1), the Taylor-Proudman theorem gives

2Ω×u = −∇pρ

for the momentum equation, where p includes the hydrostatic pressure. And, for incompressible flow, we must have: ∇⋅u = 0 . The boundary conditions w = 0 on z = 0 and L require w = 0 everywhere and ∂u/∂z = 0. Thus, we can set u = (u, v), and since all streamlines are in z = constant planes we can set u = −ez ×∇ψ where ψ is the 2D stream function. Here Ω = ezΩ, so the momentum equation becomes

2Ωez × −ez ×∇ψ( ) = −∇pρ

, or −2Ω ez ez ⋅∇ψ( )−∇ψ(ez ⋅ ez )%& '(= −∇pρ

.

In the second equation, the first term in (,)-parentheses is zero because ez is perpendicular to the plane of the flow. Thus, the above equation reduces to:

2Ω∇ψ = −∇pρ

.

Since all streamlines fall in z = constant planes, the above equation can be written: ∇(p+ 2ρΩψ) = 0 ,

which implies p + 2ρΩψ = constant in z = constant planes. And, since ψ is constant on streamlines, p is constant on streamlines, too.


Exercise 13.11. Consider an atmosphere of height H that initially contains quiescent air and N different cyclonic disks of height H and radius Ri inside which the air rotates at rate Ωi. After some time, the various cyclonic disks merge into one because of the reverse energy cascade of geostrophic turbulence. Show that the radius Rf and rotation rate Ωf of the single final disk is

<Begin Equation>

Rf2 = Ωi

2Ri4

i=1

N∑ Ωi

2Ri2

i=1

N∑ and, Ω f

2 = Ωi2Ri

2

i=1

N∑( )

2

Ωi2Ri

4

i=1

N∑ .

</End Equation> by conserving energy and enstrophy. How are these answers different if all of the energy but only a fraction ε (0 < ε < 1) of the enstrophy is retained after the merging process? Assume the relevant horizontal area is the same at the start and end of the disk-merging process.

Solution 13.11. This exercise may solved similarly to Example 13.14. For an atmosphere of height H, the kinetic energy KE of rotating disks of air with radii Ri undergoing solid body rotation with rates Ωi is:

KE = π4ρoH

i∑ Ωi

2Ri4 =

π4ρoHΩ f

2Rf44 .

where ρo is the altitude-averaged density and the second equality applies to the final fully-merged rotating disk of air. The vorticity inside each of the initial disks is 2Ωi, so conservation of enstrophy requires

Enstropy = 1A

πRi2

i∑ (2Ωi )

2 =πRf

2

A(2Ω f )

2 .

where A is the relevant horizontal area for averaging the square of the vorticity. Simplify the two equations by dividing out common factors:

Ωi2Ri

4

i∑ =Ω f

2Rf44 and Ri

2

i∑ Ωi

2 = Rf2Ω f

2 .

Simultaneous solution of these two equations leads to:

Rf2 = Ωi

2Ri4

i=1

N∑ Ωi

2Ri2

i=1

N∑ is , Ω f

2 = Ωi2Ri

2

i=1

N∑( )

2

Ωi2Ri

4

i=1

N∑ .

When only a fraction ε of the enstropy is retained, then two equations for simultaneous solution are:

Ωi2Ri

4

i∑ =Ω f

2Rf44 and ε Ri

2

i∑ Ωi

2 = Rf2Ω f

2 ,

and these lead to:

Rf2 = Ωi

2Ri4

i=1

N∑ ε Ωi

2Ri2

i=1

N∑( ) is , Ω f

2 = ε Ωi2Ri

2

i=1

N∑( )

2

Ωi2Ri

4

i=1

N∑ .

Thus, when enstropy is lost the final rotating disk is larger and rotates slower than when enstropy is conserved, as was found in Example 13.14 for two cyclonic disks.

fluid mechanics kundu cohen 6th edition solutions sm ch (13)

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