Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (11)

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

Exercise 11.1. A perturbed vortex sheet nominally located at y = 0 separates flows of differing density in the presence of gravity with downward acceleration g. The upper stream is semi-infinite and has density ρ1 and horizontal velocity U1. The lower stream has thickness h density ρ2, and horizontal velocity U2. A smooth flat impenetrable surface located at y = –h lies below the second layer. The interfacial tension between the two fluids is σ. Assume a disturbance occurs on the vortex sheet with wave number k = 2π/λ, and complex wave speed c, i.e.

€

y[ ]sheet = f (x, t) = foRe eik(x−ct ){ }. The four boundary conditions are:

1) u1, v1 → 0 as y → +∞ 2) v2 = 0 on y = –h. 3)

€

u1 ⋅n =

€

u2 ⋅n = normal velocity of the vortex sheet on both sides of the vortex sheet.

4)

€

p1 − p2 =σ∂ 2 f∂x 2

on the vortex sheet (σ = interfacial surface tension)

a) Following the development in Section 11.3, show that:

€

c =ρ1U1 + ρ2U2 coth(kh)ρ1 + ρ2 coth(kh)

±(g /k)(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)

−ρ1ρ2 U1 −U2( )2 coth(kh)

ρ1 + ρ2 coth(kh)( )2%

& ' '

(

) * *

1 2

.

b) Use the result of part a) to show that the vortex sheet is unstable when:

€

tanh(kh) +ρ2ρ1

#

$ %

&

' ( gk(ρ2 − ρ1)

ρ2+σkρ2

#

$ %

&

' ( < U1 −U2( )2

c) Will the sheet be stable or unstable to long wavelength disturbances (

€

k→ 0) when ρ2 > ρ1 for a fixed velocity difference? d) Will the sheet be stable or unstable to short wavelength disturbances (

€

k→∞) for a fixed velocity difference? e) Will the sheet ever be unstable when U1 = U2? f) Under what conditions will the thickness h matter? Solution 11.1. Set up the problem with U1, U2 mean horizontal velocities and u1, u2, v1, and v2, perturbation velocities due to the small disturbances on the vortex sheet. Above the vortex sheet:

€

∇2φ1 = 0 and below the vortex sheet:

€

∇2φ2 = 0. By inspection or separation of variables:

€

φ1 = Ae+ky + Be−ky( )eik(x−ct ) +U1x (above the vortex sheet)

€

φ2 = Ce+ky + De−ky( )eik(x−ct ) +U2x (below the vortex sheet) where the constants must be evaluated using the boundary conditions (BCs), and c may be complex. BC 1) requires A = 0. BC 2) requires

€

Cke−kh + Dke+kh = 0, or that

€

Ce+ky + De−ky = 2Ce−kh cosh k(y + h)[ ] Thus,

€

φ1 = Beik(x−ct )−ky +U1x , and

€

φ2 = 2Ce−kheik(x−ct ) cosh k(y + h)[ ] +U2x . BC 3) may be linearized since fo is small (this amounts to ignoring the perturbation velocities compared to U1 and U2 and requiring the matching to take place on y = 0 instead of y = f(x,t). The linearized version of boundary condition 3) can be evaluated using:

€

v1 =U1∂f∂x

+∂f∂t

, and

€

v2 =U2∂f∂x

+∂f∂t

on y = 0.

Plug in the appropriate derivatives for v1, and v2 and evaluate:


€

−Bk = U1(ik) − ikc[ ] fo , and

€

2kCe−kh sinh kh[ ] = U2(ik) − ikc[ ] fo. The common factor of

€

eik(x−ct ) has been divided out of all four terms. Solve for A and B:

€

B = −i U1 − c[ ] fo , and

€

C = i U2 − c[ ] fo 2e−kh sinh kh[ ] . BC 4) requires use of the Bernoulli equation above and below the vortex sheet:

€

ρ1∂φ1∂t

+ρ12u1

2+ ρ1gy + p1 = Po1 , and

€

ρ2∂φ2∂t

+ρ22u2

2+ ρ2gy + p2 = Po2

Linearize these using

€

u j2≅U j

2 + 2U ju j (j = 1,2 and no summation implied), evaluate the gravity terms on y = f(x,t) (the other terms are eventually evaluated on y = 0), and set

€

p1 = p2 +σ∂ 2 f∂x 2

= p2 − k2σf (x,t)

€

Po1 − ρ1∂φ1∂t

−ρ12U12 + 2U1u1[ ] − ρ1gf (x, t) = Po2 − ρ2

∂φ2∂t

−ρ22U22 + 2U2u2[ ] − ρ2gf (x,t) − k 2σf (x, t)

When there is no disturbance on the vortex sheet, the flow is steady and u2 = u1 = 0. Therefore:

€

Po1 −ρ12U12 = Po2 −

ρ22U22

Subtract this from the full linearized pressure boundary condition and multiply by –1 to get:

€

ρ1∂φ1∂t

+ ρ1U1u1 + ρ1gf (x, t) = ρ2∂φ2∂t

+ ρ2U2u2 + ρ2gf (x,t) + k 2σf (x, t)

Now plug in the known forms for f(x,t) and the potentials, evaluate on y = 0 as appropriate, and cancel common terms:

€

ρ1[B(−ikc + ikU1) + gfo] = ρ2[2Ce−kh cosh(kh)(−ikc + ikU2) + gfo]+ k

2σfo Insert B and C from above, and divide by fo:

€

ρ1[k(U1 − c)2 + g] = ρ2[−k coth(kh)(U2 − c)

2 + g]+ k 2σ . This is a quadratic for c. Divide by k and put the equation in a more standard form:

€

ρ1 + ρ2 coth(kh)( )c 2 − 2 ρ1U1 + ρ2U2 coth(kh)( )c −σk + ρ1U12 + ρ2U2

2 coth(kh) + g k( )(ρ1 − ρ2) = 0 . Solving for c using the quadratic formula eventually produces:

€

c = cr + ici =ρ1U1 + ρ2U2 coth(kh)ρ1 + ρ2 coth(kh)

±g k( )(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)


ρ1 + ρ2 coth(kh)( )2%

& ' '

(

) * *

1 2

b) Here we see that ci = 0 (i.e. the right side has no imaginary part) unless the terms inside the square root together are negative. This means that the vortex sheet is unstable when the terms inside the [,]-brackets together are negative:

€

g k( )(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)


ρ1 + ρ2 coth(kh)( )2 < 0, or

€

tanh(kh) +ρ2ρ1

#

$ %

&

' ( gk(ρ2 − ρ1)

ρ2+σkρ2

#

$ %

&

' ( < U1 −U2( )2

c) When ρ2 > ρ1 and

€

k→ 0, the factor with g in it will be positive and unbounded. Thus, the inequality will be not be satisfied and the sheet will be neutrally stable. d) When

€

k→∞, the factor with σ in it will be positive and unbounded. Thus, the inequality will be not be satisfied and the sheet will be neutrally stable.


e) When U1 = U2 = U,

€

c =U ±g k( )(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)

%

& '

(

) *

1 2

, which specifies neutrally stable flow when

€

g k( )(ρ2 − ρ1) +σk > 0, and unstable flow when

€

g k( )(ρ2 − ρ1) +σk < 0. Thus, even when the fluid densities are not stably arranged, non-zero surface tension prevents high wave number fluctuations from being unstable. In addition, note that when U1 = U2 = ρ1 = 0, we recover the dispersion relation for water waves on pool of finite depth:

€

ω = ck = gk +σk 3 ρ2( ) tanh(kh)[ ]1 2

. f) Given that

€

tanh(kh)→1 for kh >> 1, the thickness h will only matter when kh << 1 and

€

ρ2 ρ1 is small enough so that 1 +

€

ρ2 ρ1 is noticeably different from

€

ρ2 ρ1 . When considered in terms of the original answer given for part a), ci ~ [kh]1/2 for kh << 1 (since coth(kh) ~ 1/kh for kh << 1), decreasing h causes a sinusoidal perturbation of the vortex sheet to have a smaller growth rate, thereby making it less unstable. The motivation for this problem is the stability of a layer of air sandwiched between a large reservoir of water (below) and a hard surface (above); such as when a layer of air is injected underneath a flat-bottom ship to reduce skin friction and/or change the ship hull's mechanical coupling with the water.


Exercise 11.2. Consider a fluid layer of depth h and density ρ2 lying under a lighter infinitely-deep fluid of density ρ1 < ρ2. By setting U1 = U2 = 0, in the results of the Exercise 11.1, the following formula for the phase speed is found:

€

c = ±(g /k)(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)

%

& '

(

) *

1 2

Now invert the sign of gravity and consider why drops form when a liquid is splashed on the underside of a flat surface. Are long or short waves more unstable? Does a professional painter want interior ceiling paint with high or low surface tension? For a smooth finish should the painter apply thin or thick coats of paint? Assuming the liquid has the properties of water (surface tension ≈ 72 mN/m, density ≈ 103 kg/m) and that the lighter fluid is air, what is the longest neutrally stable wavelength on the underside of a horizontal surface? [This is the Rayleigh–Taylor instability and it occurs when density and pressure gradients point in opposite directions. It may be readily observed by accelerating rapidly downward an upward-open cup of water.] Solution 11.2. Changing the sign of gravity in the given formula produces:

€

c = ±−(g /k)(ρ2 − ρ1) +σkρ1 + ρ2 coth(kh)

%

& '

(

) *

1 2

,

where the symbol "g" is presumed to specify a positive acceleration magnitude. Here, the heavy liquid with density ρ2 is accelerated into the lighter liquid with density ρ1, so ρ2 – ρ1 > 0. Thus, the first term in the numerator of the fraction under the square root is negative and while the second is positive. For the wave speed to have a non-zero imaginary component and thereby indicate instability, the first term, which is negative, must dominate. When k is small (long wavelengths) the first term must dominate, when k is large the second term must dominate. The wave number kb (presumed positive) that lies at the boundary between instability and neutral stability is given by:

€

−(g /kb )(ρ2 − ρ1) +σkb = 0 , or

€

kb = g(ρ2 − ρ1) σ , where k < kb implies instability, and k ≥ kb implies neutral stability. Thus, long wavelengths (small k) are most unstable. When the surface tension σ is higher, then kb decreases, and the range of k leading to instability is smaller (higher σ increases the realm of stability). Thus, a professional painter wants interior ceiling paint with a high surface tension so that the paint is less likely to spontaneously produce ripples if it is applied in a smooth layer. Incidentally, a professional painter also wants a paintbrush with many fine bristles so that only high-wave-number residual disturbances are left by brush strokes as the paint is applied. For small kh, coth(kh) ~ 1/kh, while for large kh, coth(kh) ~ 1. Thus,

€

c = ± kh −(g /k)(ρ2 − ρ1) +σkρ2

%

& '

(

) *

1 2

for kh << 1, and

€

c = ±−(g /k)(ρ2 − ρ1) +σk

ρ1 + ρ2

%

& '

(

) *

1 2

for kh >> 1.

In a painting scenario, ρ2 corresponds to the paint and ρ1 corresponds to the air, so ρ2 >> ρ1. Thus, the small-kh formula for c is smaller than the large-kh formula for c, by a factor of [kh]1/2. This means that instability in a layer of paint on a ceiling develops more slowly when the paint layer is thin. Thus, a smooth finish is more likely to be obtained via multiple thin coats than one thick one. For the given parameters, the longest neutrally stable wavelength will be:


€

λ = 2π kb =2π

g(ρ2 − ρ1) σ= 2π σ

g(ρ2 − ρ1)'

( )

*

+ ,

1 2

= 2π 0.0729.81(103)'

( ) *

+ ,

1 2

≈1.7cm .

Of course, considerations of paint viscosity are also important for achieving a high-quality paint finish. However, it is potentially remarkable how many physical features of paint application can be deduced from this inviscid analysis.


Exercise 11.3. Inviscid horizontal flow in the half space y > 0 moves at speed U over a porous surface located at y = 0. Here the fluid density ρ is constant and gravity plays no roll. A weak vertical velocity fluctuation occurs at the porous surface:

€

v[ ]surface = voRe eik(x−ct ){ }, where vo <<

U. a) The velocity potential for the flow may be written

€

˜ φ =Ux + φ , where φ leads to [v]surface at y = 0 and φ vanishes as y → +∞. Determine the perturbation potential φ in terms of vo, U, ρ, k, c, and the independent variables (x,y,t). b) The porous surface responds to pressure fluctuations in the fluid via:

€

p − ps[ ]y= 0 = −γ v[ ]surface , where p is the pressure in the fluid, ps is the steady static pressure that is felt on the surface when the vertical velocity fluctuations are absent, and γ is a real material parameter that defines the porous surface’s flow resistance. Determine a formula for c in terms of U, γ, ρ, and k. c) What is the propagation velocity, Re{c}, of the surface velocity fluctuation? d) What sign should γ have for the flow to be stable? Interpret your answer.

Solution 11.3. a) To match the porous wall boundary condition, the potential must be of the following form:

€

˜ φ =Ux + φ =Ux + F(y)eik(x−ct ). Setting

€

∇2φ = 0 and canceling common factors produces:

€

−k 2F + # # F = 0 , so that

€

F = Ae+ky + Be−ky . Here, A must be zero for |φ| = F(y) to vanish as y → +∞. The second boundary condition implies:

€

∂φ ∂y( )y= 0 = v[ ]surface = voeik(x−ct ), or

€

−kA = vo so that

€

φ = − vo k( )eik(x−ct )−ky .

b) Without gravity, the Bernoulli equation for this flow is:

€

ρ∂φ∂t

+ρ2u 2 + p = Po . Linearize this

equation with the perturbation potential to get:

€

ρ∂φ∂t

+ρ2U 2 + 2U ∂φ

∂x%

& '

(

) * + p = Po. For flow without

the perturbation this linearized equation is:

€

ρ2U 2 + ps = Po . Subtract these two Bernoulli

equations to find:

€

p − ps = −ρ∂φ∂t

+U ∂φ∂x

&

' (

)

* + , thus the linearized dynamic boundary condition is:

€

−ρ∂φ∂t

+U ∂φ∂x

&

' (

)

* + y= 0

= −γ v[ ]surface = −γvoeik(x−ct ).

Evaluate the left side, multiply by –1, and cancel the common exponential factors:

€

ρ −ikc + ikU( ) −ν o k( ) = γvo, solve for:

€

c = cr + ici =U − i γρ

.

c) The phase velocity is U. d) For the flow to be stable, ci should be negative so γ must be positive. This makes sense too; with γ positive, the exterior flow is allowed to penetrate (or expand) into the porous surface when


the local surface pressure is high and this cushioning effect is stabilizing. If γ was negative, then a local surface pressure maxima would induce an outward vertical flow from the porous surface.


Exercise 11.4. Repeat exercise 11.3 for a compliant surface nominally lying at y = 0 that is perturbed from equilibrium by a small surface wave: y[ ]surface =ζ (x, t) =ζo Re eik (x−ct ){ } .

a) Determine the perturbation potential φ in terms of U, ρ, k, and c by assuming that φ vanishes as y → +∞, and that there is no flow through the compliant surface. Ignore gravity. b) The compliant surface responds to pressure fluctuations in the fluid via: p− ps[ ]y=0 = −γζ (x, t) , where p is the pressure in the fluid, ps is the steady pressure that is felt on the surface when the surface wave is absent, and γ is a real material parameter that defines the surface’s compliance. Determine a formula for c in terms of U, γ, ρ, and k. c) What is the propagation velocity, Re{c}, of the surface waves? d) If γ is positive, is the flow stable? Interpret your answer.

Solution 11.4. a) Start with

€

˜ φ =Ux + φ . The x- and t- dependence in the problem will have to match that of the surface wave, therefore we must have

€

˜ φ =Ux + F(y)eik(x−ct ), where F(y) must be determined. Plugging this form for φ into

€

∇2φ = 0, produces

€

−k 2F + # # F = 0 after common factors have been cancelled. This equation has solutions of growing and decaying exponentials as y increases. The growing exponential is discarded because it is not meaningful as y → +∞. Thus:

€

φ = Ae−kyeik(x−ct ) where A is a constant that can be determined from the linearized kinematic boundary condition on the surface:

€

U ∂ζ∂x

+∂ζ∂t

=∂φ∂y%

& '

(

) * y= 0

→

€

ikUζ o + −ikc( )ζ o = −kA , or

€

A = i c −U( )ζ o

Thus:

€

φ = i c −U( )ζ oe−kyeik(x−ct ) b) Without gravity, the Bernoulli equation for this flow is:

€

ρ∂φ∂t

+ρ2u 2 + p = Po .

Linearize this equation with the perturbation potential to get:

€

ρ∂φ∂t

+ρ2U 2 + 2U ∂φ

∂x%

& '

(

) * + p = Po.

For flow without the surface perturbation this perturbation Bernoulli equation is:

€

ρ2U 2 + ps = Po .

Subtract these two Bernoulli equations to find:

€

p − ps = −ρ∂φ∂t

+U ∂φ∂x

&

' (

)

* + .

Thus, using the surface's constitutive relationship, the linearized dynamic boundary condition is:

€

−ρ∂φ∂t

+U ∂φ∂x

&

' (

)

* + y= 0

= −γζ (x, t).

Substitute in the result of part a) for φ:


€

−ρi c −U( ) −ikc + ikU( )ζ o = −γζ o . where the common factor of

€

eik(x−ct ) has been divided out. Reduce this equation to a quadratic form and solve for c.

€

c −U( )2 =γkρ

or

€

c =U ±γkρ

.

c) When γ is positive, the disturbance phase velocity can have two values:

€

cr =U ±γkρ

When γ is negative, the phase velocity = U. d) At this point all that can be said is that the flow is neutrally stable when γ is positive. When γ is positive, a suction pressure on the surface lifts it up, while an elevated pressure pushes it down. This behavior matches that of gravity. In fact, replacing γ in the surface's constitutive equation (specified in the statement of part b) by ρg produces:

€

p − ps = −ρgζ , which describes a hydrostatic balance, and

€

cr =U ±gk

,

which are the possible phase speeds of surface waves in deep water that is moving at speed U.


Exercise 11.5. As a simplified version of flag waving, consider the stability of a simple membrane in a uniform flow. Here, the undisturbed membrane lies in the x-z plane at y = 0, the flow is parallel to the x-axis at speed U, and the fluid has density ρ. The membrane has mass per unit area = ρm and uniform tension per unit length = T. The membrane satisfies a dynamic equation based on pressure forces and internal tension combined with its local surface curvature:

€

ρm∂ 2ζ∂t 2

= p2 − p1 + T ∂ 2ζ∂x 2

+∂ 2ζ∂z2

&

' (

)

* + .

Here, the vertical membrane displacement is given by y =

€

ζ (x,z,t), and p1 and p2 are the pressures acting on the membrane from above and below, respectively. The velocity potentials for the undisturbed flow above (1) and below (2) the membrane are

€

φ1 = φ2 =Ux . For the following items, assume a small amplitude wave is present on the membrane

€

ζ (x, t) = ζ oRe eik(x−ct ){ } with k a real parameter, and assume that all deflections and other

fluctuations are uniform in the z-direction and small enough for the usual linear simplifications. In addition, assume the static pressures above and below the membrane, in the absence of membrane motion, are matched. a) Using the membrane equation, determine the propagation speed of the membrane waves, Re{c}, in the absence of fluid loading (i.e. when p2 – p1 = ρ = 0). b) Assuming inviscid flow above and below the membrane, determine a formula for c in terms of T, ρm, ρ, U, and k. c) Is the membrane more or less unstable if U, T, ρ, and ρm are individually increased with the others held constant? d) What is the propagation speed of the membrane waves when U = 0? Compare this to your answer for part a) and explain any differences.

Solution 11.5. a) Plug

€

ζ (x, t) = ζ oRe eik(x−ct{ } into

€

ρm∂ 2ζ∂t 2

= p2 − p1 + T ∂ 2ζ∂x 2

+∂ 2ζ∂z2

&

' (

)

* + with p2 – p1

= ρ = 0) to find:

€

−(kc)2ρmζ = T −k 2ζ + 0( ), or

€

c = ± T ρm . b) For this geometry, a separation-of-variables solution will produce the following velocity potentials above and below the membrane:

€

φ1(x,y,t) = Aeik(x−ct )−ky +Ux , and

€

φ1(x,y,t) = Beik(x−ct )+ky +Ux where the boundary conditions at y → ±∞ have already been applied to eliminate the growing exponential terms in each direction. The linearized kinematic condition on the membrane

requires:

€

U ∂ζ∂x

+∂ζ∂t

= v(x,y = 0,t). Apply this requirement both above and below the

membrane,


€

ikUζ o − ikcζ o = −kA , and

€

ikUζ o − ikcζ o = +kB, to find:

€

A = −B = i c −U( )ζ o. The pressure fluctuation on the membrane is found by linearizing the Bernoulli equation and subtracting out the part that persists when there is no disturbance on the membrane. This leaves:

€

" p 1 = p1 − ps = −ρ∂φ1∂t

+ U " u 1'

( )

*

+ , , and

€

" p 2 = p2 − ps = −ρ∂φ2∂t

+ U " u 2'

( )

*

+ , ,

where ps is the steady pressure on both sides of the membrane when there is no disturbance and

€

" u 1,

€

" u 2 are the horizontal disturbance velocities above and below the membrane. Thus:

€

" p 2 − " p 1 = p2 − p1 = ρ∂φ1∂t

−∂φ2∂t

+ U( " u 1 − " u 2)'

( )

*

+ ,

Plug all this into the membrane equation and start working out the algebra to reach a quadratic equation:

€

ρm∂ 2ζ∂t 2

= ρ∂φ1∂t

−∂φ2∂t

+ U( ' u 1 − ' u 2)(

) *

+

, -

y= 0

+ T ∂ 2ζ∂x 2(

) *

+

, -

€

−(kc)2ρmζ o = ρ 2i(−ikc) c −U( )ζ o + 2ikUi c −U( )ζ o( ) − k 2Tζ o

€

−(kc)2ρm + k 2T = 2ρk c −U( )2

€

c 2 +2ρkρm

c −U( )2 − Tρm

= 0

€

c 2 1+2ρkρm

#

$ %

&

' ( −4Uρkρm

c − Tρm

+2ρU 2

kρm= 0

Now use the quadratic formula and keep simplifying.

€

c =

4Uρkρm

±4Uρkρm

#

$ %

&

' (

2

− 4 1+2ρkρm

#

$ %

&

' ( 2ρU 2

kρm−Tρm

#

$ %

&

' (

2 1+2ρkρm

#

$ %

&

' (

€

c =

2Uρ ± kρm T ρm( ) +2ρkρm

T ρm( ) −U 2( )2ρ + kρm

Now set

€

co = T ρm = phase speed of the membrane waves in a vacuum.

€

c =2ρU ± kρm co

2 + 2ρ kρm( ) co2 −U 2( )2ρ + kρm

c) The membrane is more unstable as the flow speed U increases. The membrane is less unstable as the membrane tension (or

€

co = T ρm ) increases. The membrane is more unstable as the fluid density increases when U > co. The membrane is less unstable as the fluid density increases when U < co. The influence of the membrane mass is more easily deciphered with an algebraic revision

of the result of part b):

€

c =2Uρ ± Tk 2ρm + 2kρ T − ρmU

2( )2ρ + kρm

. Thus, we see that the membrane is


more unstable with increasing membrane mass when

€

U > kρm 2ρ( ) −1[ ] T ρm( ) , but that the

membrane is less unstable with increasing membrane mass when

€

U < kρm 2ρ( ) −1[ ] T ρm( ) .

d) When U = 0:

€

c = ±co

1+ 2ρ kρm. The waves move slower with a quiescent fluid present than

in a vacuum because of the added mass (inertial loading) provided by the liquid.


Exercise 11.6. Prove that σr > 0 for the thermal instability discussed in Section 11.4 via the following steps that include integration by parts and use of the boundary conditions (11.38). a) Multiply (11.36) by

€

ˆ T * and integrate the result from z = –1/2 to z = +1/2, where z is the dimensionless vertical coordinate) to find:

€

σI1 + I2 = ˆ T *Wdz∫ where

€

I1 ≡ ˆ T 2dz∫ ,

€

I2 ≡ d ˆ T /dz2

+ K 2 ˆ T 2#

$ % & ' ( dz∫ , and the limits of the integrations have been suppressed for clarity.

b) Multiply (11.37) by W* and integrate from z = –1/2 to z = +1/2 to find:

€

σPr

J1 + J2 = RaK 2 W * ˆ T dz∫ where

€

J1 ≡ dW /dz 2 + K 2W 2[ ]dz∫ ,

€

J2 ≡ d2W /dz22

+ 2K 2 dW /dz 2 + K 4W 2[ ]dz∫ , and again the limits of the integrations have been

suppressed. c) Combine the results of a) and b) to eliminate the mixed integral of W &

€

ˆ T , and use the result of this combination to show that σi = 0 for Ra > 0. [Note: the integrals I1, I2, J1, and J2 are all positive definite].

Solution 11.6. a) Equation (11.36) is

€

σ + K 2 −d2

dz2

$

% &

'

( ) ̂ T = W . As directed, multiply by

€

ˆ T * and

integrate through the thickness of the layer. The term-by-term results on the left side are:

€

σ + K 2( ) ˆ T * ˆ T ∫ dz = σ + K 2( ) ˆ T 2

∫ dz, and

€

− ˆ T * d2 ˆ T dz2∫ dz = − ˆ T * d ˆ T

dz

$

% &

'

( ) −1 2

1 2

+d ˆ T *

dzd ˆ T dz∫ dz =

d ˆ T dz

2

∫ dz ,

where the boundary terms in the integration by parts are zero because of the boundary conditions on

€

ˆ T , and all the integrals are from –1/2 to +1/2. Thus, the remnant of (11.36) is:

€

σ + K 2( ) ˆ T 2

∫ dz +d ˆ T dz

2

∫ dz = ˆ T *Wdz∫ ,

which is readily rearranged to

€

σI1 + I2 = ˆ T *Wdz∫ ,

where

€

I1 ≡ ˆ T 2dz∫ , and

€

I2 ≡ d ˆ T /dz2

+ K 2 ˆ T 2#

$ % & ' ( dz∫ .

b) Equation (11.37) is

€

σPr

+ K 2 −d2

dz2

$

% &

'

( )

d2

dz2 −K 2$

% &

'

( ) W = −RaK 2 ˆ T , which expands to:

€

−σPr

+ K 2$

% &

'

( ) K 2 +

σPr

+ 2K 2$

% &

'

( )

d2

dz2 −d4

dz4

*

+ ,

-

. / W = −RaK 2 ˆ T .

As directed, multiply by W* and integrate through the thickness of the layer. The term-by-term results on the left side are:

€

−σPr

+ K 2$

% &

'

( ) K 2 W *W∫ dz = −

σPr

+ K 2$

% &

'

( ) K 2 W 2∫ dz ,


€

σPr

+ 2K 2#

$ %

&

' ( W * d2W

dz2∫ dz =σPr

+ 2K 2#

$ %

&

' ( W * dW

dz*

+ , -

. / −1 2

1 2

−dW *

dzdWdz∫ dz

1 2 3

4 5 6

= −σPr

+ 2K 2#

$ %

&

' (

dWdz

2

∫ dz ,

and

€

− W * d4Wdz4∫ dz = −W * d3W

dz3

$

% &

'

( ) −1 2

1 2

+dW *

dzd3Wdz3∫ dz

= +dW *

dzd2Wdz2

$

% &

'

( ) −1 2

1 2

−d2W *

dz2d2Wdz2∫ dz = −

d2Wdz2

2

∫ dz.

where all the boundary terms in the various integrations by parts are zero because of the boundary conditions on W, and all the integrals are from –1/2 to +1/2. Thus, the remnant of (11.37) multiplied by –1 is:

€

σPr

+ K 2#

$ %

&

' ( K 2 W 2∫ dz +

σPr

+ 2K 2#

$ %

&

' (

dWdz

2

∫ dz +d2Wdz2

2

∫ dz = RaK 2 W *∫ ˆ T dz ,

which is readily rearranged to

€

σPr

J1 + J2 = RaK 2 W * ˆ T dz∫

where

€

J1 ≡ dW /dz 2 + K 2W 2[ ]dz∫ , and

€

J2 ≡ d2W /dz22

+ 2K 2 dW /dz 2 + K 4W 2[ ]dz∫ .

c) The complex conjugate of the part a) result is:

€

σ *I1 + I2 = ˆ T W *dz∫ , Use this to eliminate the mixed integral in the result of part b) to find:

€

σPrJ1 + J2 = RaK 2 σ *I1 + I2( ) .

The imaginary part of this equation is:

€

σ iJ1Pr#

$ %

&

' ( = −RaK 2I1σ i, or

€

σ iJ1Pr

+ RaK 2I1#

$ %

&

' ( = 0 .

When Ra > 0 (the flow is top heavy), the contents of the final parentheses is positive definite so the final equation requires σi = 0.


Exercise 11.7. Consider the thermal instability of a fluid confined between two rigid plates, as discussed in Section 11.4. It was stated there without proof that the minimum critical Rayleigh number of Racr = 1708 is obtained for the gravest even mode. To verify this, consider the gravest odd mode for which

€

W = Asin(q0z) + Bsinh(qz) + C sinh(q*z) . (1) (Compare this with the gravest even mode structure: W = Acosq0z + Bcoshqz + Ccoshq*z.) Following Chandrasekhar (1961, p. 39), show that the minimum Rayleigh number is now 17,610, reached at the wave number Kcr = 5.365. Solution 11.7. From Section 12.4 the six roots for the vertical wave number q are:

€

±iqo = ±iK(s−1)1 2 ,

€

±q = ±K 1+ s 1+ i 3( ) 2[ ]1 2

, and

€

±q* = ±K 1+ s 1− i 3( ) 2[ ]1 2

, (2)

with

€

s ≡ Ra K 4( )1 3. The boundary conditions on W are:

€

W =dWdz

=d2

dz2−K 2

#

$ %

&

' (

2

W = 0 at z = ± 1/2. (3)

From the mode shape specification, equation (1), the first, second, and fourth derivatives of the mode shape are:

€

dWdz

= Aq0 cos(q0z) + Bqcosh(qz) + Cq* cosh(q*z) , (4)

€

d2Wdz2

= −Aq02 sin(q0z) + Bq2 sinh(qz) + Cq*2 sinh(q*z) , and (5)

€

d4Wdz4

= Aq04 sin(q0z) + Bq4 sinh(qz) + Cq*4 sinh(q*z) . (6)

From (5) and (6),

€

d2

dz2 −K2

#

$ %

&

' (

2

W =d4

dz4 − 2K 2 d2

dz2 + K 4#

$ %

&

' ( W

= A q02 + K 2( )

2sin(q0z) + B q2 −K 2( )

2sinh(qz) + C q*2 −K 2( )

2sinh(q*z).

Application of all three boundary conditions (3) at either z = +1/2 or z = –1/2 leads to:

€

sin(q0 /2) sinh(q /z) sinh(q* /2)q0 cos(q0 /2) qcosh(q /2) q* cosh(q* /2)

q02 + K 2( )2 sin(q0 /2) q2 −K 2( )2 sinh(q /2) q*2 −K 2( )2 sinh(q* /2)

#

$

% % %

&

'

( ( (

ABC

#

$

% % %

&

'

( ( (

= 0 .

For nontrivial solutions, the determinant of the 3x3 matrix has to be zero. Dividing by the first row of the matrix produces:

€

1 1 1q0 cot(q0 /2) qcoth(q /2) q* coth(q* /2)q02 + K 2( )2 q2 −K 2( )2 q*2 −K 2( )2

#

$

% % %

&

'

( ( (

ABC

#

$

% % %

&

'

( ( (

= 0. (7)

From (2) and definition of s given above:

€

q02 = K 2(s−1),

€

q02 + K 2( )

2= K 2(s−1) −K 2( )

2= K 4s2


€

q2 −K 2( )2

= K 2 1+ 12 s(1+ i 3)[ ] −K 2( )

2= 1

4 K4 s(1+ i 3)[ ]

2, and

€

q*2 −K 2( )2

= 14 K

4 s(1− i 3)[ ]2

So, (6) becomes

€

1 1 1q0 cot(q0 /2) qcoth(q /2) q* coth(q* /2)

1 14 1+ i 3[ ]

2 14 1− i 3[ ]

2

#

$

% % % %

&

'

( ( ( (

ABC

#

$

% % %

&

'

( ( (

= 0. (8)

after cancelling the common factor of K4s2 from the bottom row. Expanding the determinant, and using (2) and the definition of s given above produces a relationship between Ra and K which is sketched below.

This relationship has a minimum of Racr = 17,610 at K = 5.36. The critical Ra for the gravest even mode is 1708, signifying that it is more unstable.

Ra

105

K2 4 6


Exercise 11.8. Consider the centrifugal instability problem of Section 11.6. Making the narrow-gap approximation, work out the algebra of going from (11.50) to (11.51). Solution 11.8. The perturbation equations (11.50) are:

€

∂uR∂t

−2UϕuϕR

= −1ρ∂p∂R

+ ν ∇2uR −uRR2

(

) *

+

, - ,

€

∂uϕ∂t

+dUϕ

dR+Uϕ

R$

% &

'

( ) uR = ν ∇2uϕ −

uϕR2

$

% &

'

( ) ,

€

∂uz∂t

= −1ρ∂p∂z

+ ν∇2uz , and

€

∂∂R

RuR( ) +∂uz∂z

= 0 . (1)

Substituting in

€

uR = u(R)eσt coskz ,

€

uϕ = v(R)eσt coskz ,

€

uz = w(R)eσt sinkz , and

€

p ρ = ˆ p (R)eσt coskz , the equation set (1) becomes

€

νd

dRd

dR+

1R

#

$ %

&

' ( − k 2 −

σν

#

$ %

&

' ( u + 2

Uϕ

Rv =

dˆ p dR

,

€

νddR

ddR

+1R

#

$ %

&

' ( − k 2 −

σν

#

$ %

&

' ( v −

dUϕ

dR+Uϕ

R#

$ %

&

' ( u = 0 ,

€

νd

dR+

1R

#

$ %

&

' (

ddR

− k 2 −σν

#

$ %

&

' ( w = −kˆ p , and

€

ddR

+1R

"

# $

%

& ' u = −kw . (2)

Eliminating w between the third and fourth equations of set (2) produces:

€

νk 2

ddR

+1R

#

$ %

&

' (

ddR

− k 2 −σν

#

$ %

&

' (

ddR

+1R

#

$ %

&

' ( u = ˆ p .

Inserting this equation for

€

ˆ p into the first equation of set (2), and working on the algebra eventually leads to:

€

νk 2

ddR

ddR

+1R

#

$ %

&

' ( − k 2 −

σν

#

$ %

&

' ( ddR

ddR

+1R

#

$ %

&

' ( − k 2

#

$ %

&

' ( u = 2

Uϕ

Rv . (3)

The second equation of set (2) and equation (3) are a pair of equations relating u and v. Using the radius of the outer cylinder R2, define new dimensionless variables and parameters:

r = R/R2,

€

k 2 = K 2 R22 , and

€

ω =σR22 ν ,

so that the relevant equation pair becomes:

€

ddr

ddr

+1r

"

# $

%

& ' −K 2 −ω

"

# $

%

& ' ddr

ddr

+1r

"

# $

%

& ' −K 2"

# $

%

& ' u = 2 BK

2

ν1r2

+AR2

2

B"

# $

%

& ' v , and

€

ddr

ddr

+1r

"

# $

%

& ' −K 2 −ω

"

# $

%

& ' v = 2 A

νR22u ,

where Uϕ = Ar + B/r. It is convenient to make the transformation

€

2 AR22

νu→ u ,

so that the equations take the more convenient forms:

€

ddr

ddr

+1r

"

# $

%

& ' −K 2 −ω

"

# $

%

& ' ddr

ddr

+1r

"

# $

%

& ' −K 2"

# $

%

& ' u = −TaK 2 1

r2−κ

"

# $

%

& ' v , and


€

ddr

ddr

+1r

"

# $

%

& ' −K 2 −ω

"

# $

%

& ' v = u,

where

€

Ta = −4 ABν 2

R22 = 4Ω1

2R14

ν 2(1−µ) 1−µ η2( )

1−η2( )2,

€

κ = −ABR22 =1−µ η2

1−µ,

€

µ =Ω2

Ω1

,

€

η =R1R2

,

€

A = −Ω1η2 1−µ η2

1−η2, and

€

B =Ω1R12 1−µ1−η2

.

The no-slip and boundary conditions at the walls require: u = v = 0 and (d/dr)u = 0 at r = η and 1,

where the last of the three conditions is equivalent to w = 0 (see the final equation of set (1)). Now consider the narrow gap approximation that is valid when

€

R2 − R1 << 12 (R2 + R1) .

When this is true, d/dr >> 1/r so

€

ddr

+1r≅ddr

, and

€

A +Br2≅ Ω1 1− (1−µ) r − R1

R2 − R1

%

& '

(

) * .

Now convert the independent radial coordinate (R) to one (x) that starts on the inner cylinder using the gap dimension,

€

d = R2 − R1, as the length scale, and let k = K/d, and w = σd2/ν to find:

€

d2

dx 2−K 2 −ω

$

% &

'

( ) d2

dx 2−K 2

$

% &

'

( ) u =

2Ω1d4

νK 2 1− (1−µ)x( )v , and

€

d2

dx 2−K 2 −ω

$

% &

'

( ) v =

2Ad4

νu ,

as the relevant equation set. By the further transformation

€

u→ 2Ω1d2K 4

νu , these equations

become:

€

d2

dx 2−K 2 −ω

$

% &

'

( ) d2

dx 2−K 2

$

% &

'

( ) u = 1+αx( )v , and

€

d2

dx 2−K 2 −ω

$

% &

'

( ) v = −TaK 2u,

where

€

Ta = −4AΩ1

ν 2d4 , and α = –(1 – µ).

These are the same equations as (11. 51).


Exercise 11.9. Consider the centrifugal instability problem of Section 11.6. From (11.51) and (11.53), the eigenvalue problem for determining the marginal state (σ = 0) is

€

d2 dR2 − k 2( )2 ˆ u R = (1+αx) ˆ u ϕ ,

€

d2 dR2 − k 2( )2 ˆ u ϕ = −Tak 2 ˆ u R , (11.92,93)

with

€

ˆ u R = d ˆ u R dR = ˆ u ϕ = 0 at x = 0 and 1. Conditions on

€

ˆ u ϕ are satisfied by assuming solutions of the form

€

ˆ u ϕ = Cm sin(mπx)m=1

∞

∑ . (11.94)

Inserting this into (11.92), obtain an equation for

€

ˆ u R , and arrange so that the solution satisfies the four remaining conditions on

€

ˆ u R . With

€

ˆ u R determined in this manner and

€

ˆ u ϕ given by (11.94), (11.93) leads to an eigenvalue problem for Ta(k). Following Chandrasekhar (1961, p. 300), show that the minimum Taylor number is given by (11.54) and is reached at kcr = 3.12. Solution 11.9. Continuing the effort from Exercise 11.8, insert (11.94) into (11.92) to find:

€

d2 dx 2 −K 2( )2u = (1+αx)v = (1+αx) Cm sin(mπx)

m=1

∞

∑ , (5)

where a switch has been made to the dimensionless variables of Exercise 11.8. Now arrange that the solution satisfies the four remaining boundary conditions on u. With u determined in this fashion and v given by (11.94), (11.93) will lead to an equation for Ta. The solution of (5) is straightforward. The general solution can be written in the form:

€

u =Cm

(m2π 2 + K 2)2

A1(m ) coshKx + B1

(m ) sinhKx + A2(m )x coshKx + B2

(m )x sinhKx

+ (1+αx)sin(mπx) +4αmπ

m2π 2 + K 2 cos(mπx)

$

% &

' &

(

) &

* & m=1

∞

∑ , (6)

where the constants are determined by the boundary conditions u = (d/dx)u = 0 at x = 0 and 1. These conditions lead to the four equations:

€

A1(m ) = −

4αmπm2π 2 + K 2 ,

€

KB1(m ) + A2

(m ) = −mπ ,

€

A1(m ) coshK + B1

(m ) sinhK + A2(m ) coshK + B2

(m ) sinhK = (−1)m+1 4αmπm2π 2 + K 2 , (7)

€

A1(m )K sinhK + B1

(m )K coshK + A2(m ) coshK + K sinhK( ) + B2

(m ) sinhK + K coshK( ) = (−1)m+1(1+α)mπThe solution of these equations is:

€

A1(m ) = −

4αmπm2π 2 + K 2

€

B1(m ) =

mπΔ

K + βm sinhK + K coshK( ) − γm sinhK{ }

€

A2(m ) = −

mπΔ

sinh2K + βmK sinhK + K coshK( ) − γmK sinhK{ }

€

B2(m ) =

mπΔ

sinhK coshK −K( ) + βmK2 sinhK − γm K coshK − sinhK( ){ } (8)

where:

€

Δ = sinh2K −K 2,

€

βm =4α

m2π 2 + K 2 (−1)m+1 + coshK{ }, and

€

γm = (−1)m+1(1−α) +4α

m2π 2 + K 2 K sinhK .


Substituting u from (6) and v from (11.94) into (11.93) produces:

€

Cn n2π 2 + K 2( )

n=1

∞

∑ sinnπx

= TaK 2 Cm

(m2π 2 + K 2)2

A1(m ) coshKx + B1

(m ) sinhKx + A2(m )x coshKx + B2

(m )x sinhKx

+ (1+αx)sin(mπx) +4αmπ

m2π 2 + K 2 cos(mπx)

&

' (

) (

*

+ (

, ( m=1

∞

∑ . (9)

Multiply this equation by sin(nπx) and integrate from x = 0 to x = 1, to obtain a system of linear homogeneous equations for the constants Cm. The requirement that these constants are not all zero leads to the equation:

€

nπn2π 2 + K 2

1+ (−1)n+1 coshK[ ]A1(m ) + (−1)n+1 sinhK[ ]B1

(m ) + (−1)n+1 coshK −2K

n2π 2 + K 2 sinhK$

% & '

( ) A2

(m )

+ (−1)n+1 sinhK −2K

n2π 2 + K 2 1+ (−1)n+1 coshK{ }$

% & '

( ) B2

(m )

*

+ , ,

- , ,

.

/ , ,

0 , ,

+αxnm +12δnm −

12n2π 2 + K 2( )3 δnm

K 2Ta= 0 (10)

where

€

xnm =

0 if m + n is even and m ≠ n1/4 if m = n

4nmn2 −m2

2m2π 2 + K 2 −

1π 2(n2 −m2)

% & '

( ) *

if m + n is odd

%

&

+ +

'

+ +

(

)

+ +

*

+ +

.

On using the first two equations of (7), equation (10) simplifies to:

€

nπn2π 2 + K 2

4mπαm2π 2 + K 2 (−1)m+n −1[ ] − 2K

n2π 2 + K 2 (−1)n+1A2(m ) sinhK + (−1)n+1B2

(m ) coshK + B2(m )[ ]%

& '

( ) *

+αxnm +12δnm −

12n2π 2 + K 2( )

3 δnmK 2Ta

= 0 (11)

After substituting for the constants

€

A2(m ) and

€

B2(m ) given by (8), (11) becomes:

€

4mnπ 2αn2π 2 + K 2( ) m2π 2 + K 2( )

(−1)m+n −1[ ]

−2Kmnπ 2

n2π 2 + K 2( ) sinh2K −K 2( )

(sinhK coshK −K) 1+ (1+α)(−1)m+n[ ]+(sinhK −K coshK) (−1)n+1 + (1+α)(−1)m+1[ ]−

4kα sinhKm2π 2 + K 2 sinhK + K(−1)m+1[ ] (−1)m+n −1[ ]

%

&

' '

(

' '

)

*

' '

+

' '

+αxnm +12δnm −

12n2π 2 + K 2( )

3 δnmK 2Ta

= 0 (12)

A first approximation to the solution of (12) is obtained by setting the (1,1)-element of the determinant zero. This implies:


€

12π 2 + K 2( )

3 δnmK 2Ta

=14α +

12−

2Kπ 2(2 +α)π 2 + K 2( ) sinh2K −K 2( )

(sinhK coshK −K) + (sinhK −K coshK)[ ]

and this simplifies to:

€

Ta =2

2 +α

π 2 + K 2( )3

K 2 1−16Kπ 2 cosh2(K /2) π 2 + K 2( )2(sinhK + K)[ ]{ }

A plot of Ta(2 + α) as a function of K from this solution shows that the minimum value of the Taylor number is:

€

Tacr = 3430 (2 +α), and is reached at Kcr = 3.12.


Exercise 11.10. For a Kelvin–Helmholtz instability in a continuously stratified ocean, obtain a globally integrated energy equation in the form

€

12ddt

u2 + w2 + g2ρ2 ρ02N 2( )dV = − uw ∂U

∂z∫∫ dV .

(As in Figure 11.25, the integration in x takes place over an integer number of wavelengths.) Discuss the physical meaning of each term and the mechanism of instability. Solution 11.10. From (11.57), multiplying the perturbation equations for the Kelvin-Helmholtz instability by u, w, and

€

g2ρ ρ02N 2 and add them together

€

u ∂u∂t

+U ∂u∂x

+ w ∂U∂z

+1ρ0

∂p∂x

$

% &

'

( ) + w

∂w∂t

+U ∂w∂x

+ρρ0g +

1ρ0

∂p∂z

$

% &

'

( ) +

g2ρρ02N 2

∂ρ∂t

+U ∂ρ∂x

− ρ0wN 2

g$

% &

'

( ) = 0 .

This produces:

€

∂∂t

+U ∂∂x

#

$ %

&

' ( u2

2+w2

2+

g2ρ2

2ρ02N 2

#

$ %

&

' ( + uw

∂U∂z

+1ρ0

∂p∂x

+∂p∂z

#

$ %

&

' ( = 0

The final term can be rewritten:

€

∂∂t

+U ∂∂x

#

$ %

&

' ( u2

2+w2

2+

g2ρ2

2ρ02N 2

#

$ %

&

' ( + uw

∂U∂z

+1ρ0

∂(pu)∂x

+∂(pw)∂z

#

$ %

&

' ( − p

∂u∂x

+∂w∂z

#

$ %

&

' ( = 0.

After global integration over an integer number of x-direction wavelengths, the terms involving U(∂/∂x) will be zero, and the third term, which is equal to

€

ρ0−1 ∇ ⋅ (pu)dV∫ , transforms into a

surface integral,

€

ρ0−1 pu ⋅ndA∫ , which vanishes because no flow crosses the channel walls and an

there is an equal in-flux and out-flux across the open boundaries on the CV (see Figure 11.25). The last term vanishes because of the continuity equation. Thus, the resulting CV equation is:

€

12ddt

u2 + w2 + g2ρ2 ρ02N 2( )dV = − uw ∂U

∂z∫∫ dV .

This equation shows that the rate of change of the sum of the kinetic and potential energies is equal to the energy extracted from the correlation of the stream-wise (u) and vertical (w) velocity fluctuations with the average shear (∂U/∂z). However, in order for the perturbations to extract energy from the mean shear, they must be anti-correlated so that uw is negative on average when ∂U/∂z is positive. This is typical of shear instabilities. And, when u and w represent turbulent fluctuations the average correlation of u and w is called a Reynolds shear stress (see Ch. 12).


Exercise 11.11. In two-dimensional (x,y)-Cartesian coordinates, consider the inviscid stability of

horizontal parallel shear flow defined by two linear velocity gradients:

€

U(y) =S+y for y > 0S−y for y ≤ 0

$ % &

' ( )

,

where S+ and S– are real constants. Assume an infinitesimal velocity perturbation with vertical component v = f (y)exp ik(x − ct){ } , where k is positive real but ω may be complex.

a) Use the Rayleigh equation !!f − k2 f − f !!UU − c

= 0 with

€

f (y)→ 0 as

€

y →∞ to find f(y).

b) Require the pressure perturbation associated with v to be continuous across y = 0, and determine a single equation for the disturbance phase speed c in terms of the other parameters. c) For what values of S+, S–, and k, is this flow stable, unstable, or neutrally stable? d) What is special about the case S+ = S–?

Solution 11.11. a) Here d2U/dy2 = 0, so the Rayleigh equation simplifies to d2f/dy2 – k2f = 0 which has solutions A±exp(±ky). To satisfy the given boundary conditions for k positive & real, the decaying exponential must be chosen for y > 0 and y < 0, so f(y) = Aexp(–k|y|).

b) The continuity equation is

€

∂∂x

U(y) + # u ( ) +∂ # v ∂x

= 0 . For v = Aexp ik(x − ct)− k y{ } , this

implies: u = − ∂v ∂y( )dx∫ = −1 ik( ) df dy( )exp ik(x − ct){ }= −A ik( ) ∓k( )exp ik(x − ct)− k y{ } where the upper & lower signs go with y > 0 & y < 0, respectively. The linearized horizontal

momentum equation for the disturbance is: ∂u∂t+U(y)∂u

∂x+ v∂U

∂y= −

1ρ∂ "p∂x

. Evaluate the

derivatives on the left side and integrate in x to find: −ikc−ik

∓k( )+ S±y ik−ik"

#$

%

&' ∓k( )+ S±

(

)*

+

,- Aexp ik(x − ct)− k y{ }dx∫ = −

/pρ

, or

c ∓k( )− S±y ∓k( )+ S±"# $% Aexp ik(x − ct)− k y{ }dx∫ = −'pρ

.

Evaluate this equation above and below y = 0 and set the two results equal to find: −ck + S+ = +ck + S− or c = (S+ – S–)/2k.

c) Here c is always real; this flow is neutrally stable in all cases. d) When S+ = S–, then c = 0, and this implies zero phase speed for the disturbance.

x!

y! U(y)!


Exercise 11.12. Consider the inviscid stability of a constant vorticity layer of thickness h between uniform streams with flow speeds U1 and U3. Region 1 lies above the layer, y > h/2 with U(y) = U1. Region 2 lies within the layer, |y| ≤ h/2,

€

U(y) = 12 (U1 +U3) + (U1 −U3) y h( ) .

Region 3 lies below the layer, y < –h/2 with U(y) = U3.

a) Solve the Rayleigh equation,

€

" " f − k 2 f − f " " U U − c

= 0 , in each region, then use appropriate

boundary and matching conditions to obtain:

€

f1(y) = Acosh kh 2( ) + Bsinh kh 2( )( )e−k y−h 2( ) for y > +h/2,

€

f2(y) = Acosh ky( ) + Bsinh ky( ) for |y| ≤ h/2,

€

f3(y) = Acosh kh 2( ) − Bsinh kh 2( )( )e+k y+h 2( ) for y < –h/2. where f defines the spatial extent of the disturbance:

€

" v = f (y)eik(x−ct ) and

€

" u = − " f ik( )eik(x−ct ), and A and B are undetermined constants.

b) The linearized horizontal momentum equation is:

€

∂ # u ∂t

+ U ∂ # u ∂x

+ # v ∂U∂y

= −1ρ∂ # p ∂x

.

Integrate this equation with respect to x, require the pressure to be continuous at y = ± h/2, and simplify your results to find two additional constraint equations:

€

c −U1( ) # f 1(+h /2) = c −U1( ) # f 2(+h /2) +U1 −U3

hf2(+h /2) , and

€

c −U3( ) # f 3(−h /2) = c −U3( ) # f 2(−h /2) +U1 −U3

hf2(−h /2)

c) Define

€

co = c − 12 (U1 +U3) (this is the phase speed of the disturbance waves in a frame of

reference moving at the average speed), and use the results of parts a) and b) to determine a single equation for co:

€

co2 =

U1 −U3

2kh#

$ %

&

' ( 2

kh −1( )2 − e−2kh{ }

[This part of this problem requires patience and algebraic skill.] d) From the result of part c), co will be real for kh >> 1 (short wave disturbances), so the flow is stable or neutrally stable. However, for kh << 1 (long wave disturbances), use the result of part c) to show that:

€

co ≅ ±i U1 −U3

2$

% &

'

( ) 1−

43kh + ...

e) Determine the largest value of kh at which the flow is unstable. Solution 11.12. Here

€

" " U = 0 in all three regions, so the Rayleigh equation,

€

" " f − k 2 f − f " " U U −ω k

= 0 , reduces to:

€

" " f − k 2 f = 0 in all three regions. The solutions of this

equation are growing and decaying exponentials, or hyperbolic functions. Thus,

€

f1(y) = Ce−ky + De+ky for y > +h/2,


€


€

f3(y) = Ee−ky + Fe+ky for y < –h/2. Where A, B, C, D, E, and F are all constants. The disturbance must disappear as y → ±∞. This means that D = E = 0. To ensure continuity of f at y = ± h/2, the following must be true:

€

Ce−kh 2 = Acosh kh 2( ) + Bsinh kh 2( ) and

€

Fe−kh 2 = Acosh kh 2( ) − Bsinh kh 2( ) . Using these two equations to eliminate C and F produces:

€

f1(y) = Acosh kh 2( ) + Bsinh kh 2( )( )e−k y−h 2( ) for y > +h/2,

€


€

f3(y) = Acosh kh 2( ) − Bsinh kh 2( )( )e+k y+h 2( ) for y < –h/2. b) The only x-variation in the whole problem enters through the assumed form of the disturbance:

€

" v = f (y)eik(x−ct ) and

€

" u = − " f ik( )eik(x−ct ), so and x-integration is the same as dividing

by +ik. Therefore:

€

−1ρ

∂ % p ∂x

dx∫ = −% p ρ

=∂ % u ∂t

+ U ∂ % u ∂x

+ % v dUdy

'

( )

*

+ , ∫ dx =

1ik

∂ % u ∂t

+ U ∂ % u ∂x

+ % v dUdy

'

( )

*

+ , , and

there is no integration constant since the pressure fluctuation must disappear when

€

" u and

€

" v disappear. Now plug in the relationships for

€

" u and

€

" v and require continuity of pressure at y = ± h/2:

€

c " f 1 −U1 " f 1 + f1 " U 1( )y= +h 2 = c " f 2 −U2 " f 2 + f2 " U 2( )y= +h 2 , and

€

c " f 2 −U2 " f 2 + f2 " U 2( )y=−h 2 = c " f 3 −U3 " f 3 + f3 " U 3( )y=−h 2 Using these two equations and U1(y) = U1 for y > h/2,

€

U2(y) = 12 (U1 +U3) + (U1 −U3) y h( ) for |y|

≤ h/2, and U3(y) = U3 for y < –h/2 produces:

€

c −U1( ) # f 1(+h /2) = c −U1( ) # f 2(+h /2) +U1 −U3

hf2(+h /2) , and

€

c −U3( ) # f 3(−h /2) = c −U3( ) # f 2(−h /2) +U1 −U3

hf2(−h /2)

c) First evaluate rewrite the equations from part b) in terms of

€

co = c − (U1 +U3)2

:

€

co −ΔU2

$

% &

'

( ) * f 1(+h /2) = co −

ΔU2

$

% &

'

( ) * f 2(+h /2) +

ΔUh

f2(+h /2) , and

€

co +ΔU2

#

$ %

&

' ( ) f 3(−h /2) = co −

ΔU2

#

$ %

&

' ( ) f 2(−h /2) +

ΔUh

f2(−h /2)

where ΔU = U1 – U3. These equations can be made a little more tidy:

€

co −ΔU2

$

% &

'

( ) * f 1(+h /2) − * f 2(+h /2)( ) =

ΔUh

f2(+h /2) , and

€

co +ΔU2

#

$ %

&

' ( ) f 3(−h /2) − ) f 2(−h /2)( ) =

ΔUh

f2(−h /2) .

Now evaluate the functions and derivatives in the various regions:

€

f2(+h 2) = Acosh kh 2( ) + Bsinh kh 2( ),

€

f2(−h 2) = Acosh kh 2( ) − Bsinh kh 2( ),

€

f1" +h 2( ) = −kAcosh kh 2( ) − kBsinh kh 2( ),

€

f2"(+h 2) = kAsinh kh 2( ) + kBcosh kh 2( ),

€

f2"(−h 2) = −kAsinh kh 2( ) + kBcosh kh 2( ) , and

€

f3" −h 2( ) = kAcosh kh 2( ) − kBsinh kh 2( ) Stuff these evaluations into the modified part b) results to get:


€

co −ΔU2

$

% &

'

( ) −kAcosh kh 2( ) − kBsinh kh 2( ) − kAsinh kh 2( ) − kBcosh kh 2( )( )

=ΔUh

Acosh kh 2( ) + Bsinh kh 2( )( ),

and

€

co +ΔU2

#

$ %

&

' ( kAcosh kh 2( ) − kBsinh kh 2( ) + kAsinh kh 2( ) − kBcosh kh 2( )( )

=ΔUh

Acosh kh 2( ) − Bsinh kh 2( )( ).

Start simplifying these by collecting terms while noting that

€

e±z = cosh(z) ± sinh(z)

€

−kh co −ΔU2

$

% &

'

( ) A + B( )e+kh 2 = ΔU Acosh kh 2( ) + Bsinh kh 2( )( )

€

+kh co +ΔU2

#

$ %

&

' ( A − B( )e+kh 2 = ΔU Acosh kh 2( ) − Bsinh kh 2( )( )

Isolate A and B in each equation.

€

kh co −ΔU2

$

% &

'

( ) e+kh 2 + ΔU cosh kh

2$

% &

'

( )

*

+ ,

-

. / A + kh co −

ΔU2

$

% &

'

( ) e+kh 2 + ΔU sinh kh

2$

% &

'

( )

*

+ ,

-

. / B = 0

€

kh co +ΔU2

#

$ %

&

' ( e+kh 2 −ΔU cosh kh

2#

$ %

&

' (

*

+ ,

-

. / A + −kh co +

ΔU2

#

$ %

&

' ( e+kh 2 + ΔU sinh kh

2#

$ %

&

' (

*

+ ,

-

. / B = 0

These are two homogeneous linear algebraic equations. They will have a non-trivial solution when their determinant is zero.

€

kh co −ΔU2

$

% &

'

( ) e+kh 2 + ΔU cosh kh

2$

% &

'

( )

*

+ ,

-

. / × −kh co +

ΔU2

$

% &

'


2$

% &

'

( )

*

+ ,

-

. / −

kh co −ΔU2

$

% &

'


2$

% &

'

( )

*

+ ,

-

. / × kh co +

ΔU2

$

% &

'

( ) e+kh 2 −ΔU cosh kh

2$

% &

'

( )

*

+ ,

-

. / = 0

The remaining job is nothing less than a straight out algebraic effort. First carefully multiply the factors in square brackets together.

€

0 = −2k 2h2 co2 −

ΔU( )2

4

$

% & &

'

( ) ) e

kh + kh co −ΔU2

$

% &

'

( ) e+kh 2ΔU sinh kh

2$

% &

'

( )

−kh co +ΔU2

$

% &

'

( ) e+kh 2ΔU cosh kh

2$

% &

'

( ) + 2 ΔU( )2 sinh kh

2$

% &

'

( ) cosh

kh2

$

% &

'

( )

+kh co −ΔU2

$

% &

'

( ) e+kh 2ΔU cosh kh

2$

% &

'

( ) − kh co +

ΔU2

$

% &

'

( ) e+kh 2ΔU sinh kh

2$

% &

'

( )

Collect and simplify all the terms with hyperbolic functions.

€

0 = −2k 2h2 co2 −

ΔU( )2

4

$

% & &

'

( ) ) e

kh + kh co −ΔU2

$

% &

'

( ) e+khΔU − kh co +

ΔU2

$

% &

'

( ) e+khΔU + 2 ΔU( )2 e

+kh − e−kh

4$

% &

'

( )

Merge the two middle terms.

€

0 = −2k 2h2 co2 −

ΔU( )2

4

$

% & &

'

( ) ) e

kh − kh ΔU( )2e+kh + 2 ΔU( )2 e+kh − e−kh

4$

% &

'

( )


Solve for

€

co2 and continue simplification.

€

co2 =

ΔU( )2

4−ΔU( )2

2kh+ΔU( )2

k 2h21− e−2kh

4$

% &

'

( ) =

ΔU2kh$

% &

'

( ) 2

k 2h2 − 2kh +1− e−2kh( )

Make the last few adjustments to reach the final form:

€

co2 =

ΔU2kh#

$ %

&

' ( 2

kh −1( )2 − e−2kh( ).

d) Here we need to expand e–2kh out to the third order term:

€

e−2kh =1− 2kh + 12 (2kh)

2 − 16 (2kh)

3 + ....

€

co2 =

ΔU2kh#

$ %

&

' ( 2

k 2h2 − 2kh +1−1+ 2kh − 12(2kh)2 +

16(2kh)3 + ...

#

$ %

&

' ( =

ΔU2kh#

$ %

&

' ( 2

−k 2h2 +43k 3h3 + ...

#

$ %

&

' (

Thus:

€

co = ±i ΔU2

#

$ %

&

' ( 1−

43kh + ...

#

$ %

&

' ( 1 2

e) The limiting value of kh for instability will occur when the imaginary part of

€

co2 just reaches

zero, i.e. when

€

kh −1( )2 − e−2kh = 0 or

€

kh −1= e−kh which can be solved numerically via iteration on a hand calculator: kh ≈ 1.2785. For kh larger than this value the flow will be neutrally stable.


Exercise 11.13. Consider the inviscid instability of parallel flows given by the Rayleigh equation

€

(U − c) d2 ˆ v dy 2 − k 2 ˆ v #

$ %

&

' ( −

d2Udy 2

ˆ v = 0 , (11.95)

where the y-component of the perturbation velocity is

€

v = ˆ v (y)exp ik(x − ct){ }. (i) Note that this equation is identical to the Rayleigh equation (11.81) for the stream

function amplitude φ, as it must because

€

ˆ v (y) = −ikφ . For a flow bounded by walls at y1 and y2, note that the boundary conditions are identical in terms of φ and

€

ˆ v . (ii) Show that if c is an eigenvalue of (11.95), then so is its conjugate c* = cr – ici. What

aspect of (11.95) allows this result to be valid? (iii) Let U(y) be antisymmetric, so that U(y) = –U(–y). Demonstrate that if c(k) is an

eigenvalue, then –c(k) is also an eigenvalue. Explain the result physically in terms of the possible directions of propagation of perturbations in such an antisymmetric flow.

(iv) Let U (y) be symmetric so that U(y) = U(–y). Show that in this case

€

ˆ v is either symmetric or antisymmetric about y = 0.

[Hint: Letting y → – y, show that the solution

€

ˆ v (−y) satisfies (11.95) with the same eigenvalue c. Form a symmetric solution

€

S(y) = ˆ v (y) + ˆ v (−y) = S(−y) , and an antisymmetric solution

€

A(y) = ˆ v (y) − ˆ v (−y) = −A(−y) . Then write A[S-eqn] – S[A-eqn] = 0 where S-eqn indicates the differential equation (11.95) in terms of S. Canceling terms this reduces to (SAʹ′ – ASʹ′)ʹ′ = 0, where the prime (ʹ′) indicates a y-derivative. Integration gives SAʹ′ – ASʹ′ = 0, where the constant of integration is zero because of the boundary conditions. Another integration gives S = bA, where b is a constant of integration. Because the symmetric and antisymmetric functions cannot be proportional, it follows that one of them must be zero.] Comments: If v is symmetric, then the cross-stream velocity has the same sign across the entire flow, although the sign alternates every half wavelength along the flow. This mode is consequently called sinuous. On the other hand, if v is antisymmetric, then the shape of the jet expands and contracts along the length. This mode is now generally called the sausage instability because it resembles a line of linked sausages. Solution 11.13. (i) Since the stream function is ψ = φexp[ik(x – ct)], the cross stream velocity is:

€

v = −∂ψ∂x

= −ikφ exp ik(x − ct)[ ] = ˆ v exp ik(x − ct)[ ] so that

€

ˆ v = −ikφ .

Thus,

€

ˆ v and φ will satisfy identical equations and boundary conditions. (ii) The complex conjugate of (11.95) is:

€

(U − c*) d2 ˆ v *

dy 2 − k 2 ˆ v *#

$ %

&

' ( −

d2Udy 2

ˆ v * = 0,

and this equation is identical to (11.95) except that c* replaces c and

€

ˆ v * replaces

€

ˆ v . The boundary conditions on

€

ˆ v * and

€

ˆ v are also identical, namely

€

ˆ v * =

€

ˆ v = 0 at y1 or y2. Thus, if

€

ˆ v is an eigenfunction with eigenvalue c for some k, then

€

ˆ v * is and eigenfunction with eigenvalue c* = cr – ici for the same k. This property is only possible since (11.95) does not directly involve the imaginary root i. (iii) When the basic flow is antisymmetric, U(y) = –U(–y), so that all derivatives of U change sign when y is oppositely directed. That is, U´(y) = –U´(–y) and U´´(y) = –U´´(–y), where a prime denotes a derivative. Changing y into –y, (11.95) becomes:


€

(−U − c) d2 ˆ v dy 2 − k 2 ˆ v #

$ %

&

' ( +

d2Udy 2

ˆ v = 0 , or

€

(U + c) d2 ˆ v dy 2 − k 2 ˆ v #

$ %

&

' ( −

d2Udy 2

ˆ v = 0.

Considering the second of these two equations, all terms in are the same as in (11.95), except that c has been replaced by –c. Thus, if c is an eigenvalue, then –c is also an eigenvalue for the same k. This means that for each propagating mode that moves in the positive x-direction there is a corresponding mode that moves in the negative x-direction. This is obvious from the following sketch of the antisymmetric basic flow, and from the observation that the two directions of propagation are "similar" with respect to the basic velocity.

(iv) When the basic flow is symmetric, U(y) = U(–y), then U´´(y) = U´´(–y), so changing y into –y in (11.95) produces:

€

(U − c) d2 ˆ v dy 2 − k 2 ˆ v #

$ %

&

' ( −

d2Udy 2

ˆ v = 0 ,

so that

€

ˆ v (−y) satisfies (11.95) with the same eigenvalue c. By adding and subtracting the two solutions, we get symmetric and anti-symmetric solutions:

€

S(y) = ˆ v (y) + ˆ v (−y) = S(−y) and

€

A(y) = ˆ v (y) − ˆ v (−y) = −A(−y) . Obviously S and A satisfy the differential equation, therefore:

€

A (U − c) d2Sdy 2

− k 2S#

$ %

&

' ( −

d2Udy 2

S)

* +

,

- . − S (U − c) d

2Ady 2

− k 2A#

$ %

&

' ( −

d2Udy 2

A)

* +

,

- . = 0 ,

since the contents of both sets of {,]-brackets are zero. Canceling terms and dividing by U – c leaves:

€

Ad2Sdy 2

− S d2Ady 2

= 0 or

€

ddy

AdSdy

− S dAdy

#

$ %

&

' ( = 0 which integrates to

€

AdSdy

− S dAdy= const.

Evaluating the constant term at the wall, it turns out to be zero. So, divide the final equation by the product SA to find:

€

1SdSdy

=1AdAdy

which integrates to:

€

ln(S) = ln(A) + const., or

€

S = (const.)A .

However, symmetric and anti-symmetric functions cannot be proportional, so S or A must be zero. If both are zero this is the trivial case. And, the final relationship does not require both S and A to be zero if one is zero, because the division step with the first-derivative equation is not valid when either A or S is zero. The unstable waves on a symmetric jet are the either sinuous or sausage-like, but not a combination of both.

c

c

y


Exercise 11.14. Derive (11.88) starting from the incompressible Navier-Stokes momentum equation for the disturbed flow:

€

∂∂t

Ui + ui( ) + U j + u j( ) ∂∂x j

Ui + ui( ) = −1ρ∂∂xi

P + p( ) + ν∂ 2

∂x j∂x j

Ui + ui( ), (11.96)

where Ui and ui represent the basic flow and the disturbance, respectively. Subtract the equation of motion for the basic state from (11.96), multiply by ui and integrate the result within a stationary volume having stream-wise control surfaces chosen to coincide with the walls where no-slip conditions are satisfied or where ui

€

→ 0, and having a length (in the stream-wise direction) that is an integer number of disturbance wavelengths.

Solution 11.14. The basic flow momentum equation is:

€

∂Ui

∂t+U j

∂Ui

∂x j

= −1ρ∂P∂xi

+ ν∂ 2Ui

∂x j∂x j

.

Subtracting this from (11.96) produces:

€

∂ui∂t

+ u j∂Ui

∂x j

+U j∂ui∂x j

+ u j∂ui∂x j

= −1ρ∂p∂xi

+ ν∂ 2ui∂x j∂x j

.

Multiply by ui and integrate each term over the CV.

€

ui∂ui∂t

+ uiU j∂ui∂x j

+ uiu j∂ui∂x j

+ uiu j∂Ui

∂x j

= −uiρ∂p∂xi

+ uiν∂ 2ui∂x j∂x j

& ' (

) * + CV

∫ dV . (%)

Consider each term individually.

€

ui∂ui∂tCV

∫ dV =∂∂t

12ui2$

% &

'

( )

CV∫ dV =

ddt

12ui2

CV∫ dV .

Here the final equality follows because the CV is stationary and the volume integration leaves no spatial dependence.

€

uiU j∂ui∂x jCV

∫ dV =12U j

∂ui2

∂x j

+ ui2 ∂U j

∂x j

$

% & &

'

( ) )

CV∫ dV =

12∂(ui

2U j )∂x j

$

% & &

'

( ) )

CV∫ dV =

12ui2U jn j

CS∫ dA = 0

€

uiu j∂ui∂x jCV

∫ dV =12u j∂ui

2

∂x j

+ ui2 ∂u j

∂x j

$

% & &

'

( ) )

CV∫ dV =

12∂(ui

2u j )∂x j

$

% & &

'

( ) )

CV∫ dV =

12ui2u jn j

CS∫ dA = 0

Here CS is the control surface, and nj are the components of the outward normal vector from the control surface. And, ∂uj/∂xj = 0 = ∂Uj/∂xj has been used along with Gauss' theorem, and the fact that velocity fluctuations are zero on the stream wise control surfaces (no slip), or equal on the stream-normal control surfaces where nj changes sign between the inlet and the outlet (periodic) leading to cancellation.

€

ui∂p∂xiCV

∫ dV =∂(pui)∂xi

+ p∂ui∂xi

$

% &

'

( )

CV∫ dV = puini

CS∫ dA = 0 .

€

ui∂ui∂x j

2CV∫ dV =

∂∂x j

ui∂ui∂x j

$

% & &

'

( ) )

CV∫ dV −

∂ui∂x j

∂ui∂x jCV

∫ dV = ui∂ui∂x j

n jCS∫ dA − ∂ui

∂x j

∂ui∂x jCV

∫ dV = −∂ui∂x j

∂ui∂x jCV

∫ dV .

Here again incompressibility, Gauss' theorem, and the no slip and periodic boundary conditions lead to the simplifications. Thus, (%) becomes:

€

ddt

12ui2

CV∫ dV = −ρ uiu j

∂Ui

∂x jCV∫ dV −ν

∂ui∂x j

∂ui∂x jCV

∫ dV ,

which is (11.88).


Exercise 11.15. The process of transition from laminar to turbulent flow may be driven both by exterior flow fluctuations and nonlinearity. Both of these effects can be simulated with the simple nonlinear logistic map

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xn+1 = Axn (1− xn ) and a computer spread-sheet program. Here, xn can be considered to be the flow speed at the point of interest with A playing the role of the nonlinearity parameter (Reynolds number), x0 (the initial condition) playing the role of an external disturbance, and iteration of the equation playing the role of increasing time. The essential feature illustrated by this problem is that increasing the nonlinearity parameter or changing the initial condition in the presence of nonlinearity may fully alter the character of the resulting sequence of xn values. Plotting xn vs. n should aid understanding in parts b) through e). a) Determine the background solution of the logistic map that occurs when

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xn+1 = xn in terms of A. Now, get into a spread sheet program, and set up a column that computes

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xn+1 for n = 1 to 100 for user selectable values of x0 and A for 0 < x0 < 1, and 0 < A < 4. b) For A = 1.0, 1.5, 2.0, 2.9, choose a few different x0’s and numerically determine if the background solution is reached by n = 100. Is the flow stable for these values of A, i.e. does it converge toward the background solution? c) For the slightly larger value, A = 3.2, choose x0 = 0.6875, 0.6874, and 0.6876. Is the flow stable or oscillatory in these three cases? If it is oscillatory, how many iterations are needed for it to repeat? d) For A = 3.5, is the flow stable or oscillatory? If it is oscillatory, how many iterations are needed for it to repeat? Does any value of x0 lead to a stable solution? e) For A = 3.9, is the flow stable, oscillatory, or chaotic? Does any value of x0 lead to a stable solution? Solution 11.15. a) Setting xn = xn+1 = x in

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xn+1 = Axn (1− xn ) , produces:

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x = Ax(1− x) . Thus for x ≠ 0, the equilibrium or steady-flow solution for x is

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x =1− 1 A( ) . b) For A = 1.0, the solution is not converged but it is heading monotonically toward x = 0. This behavior is very much like a viscosity-dominated flow. For values of A above 1.0 up to about 2.0, the steady flow value of x is reached in a monotonic fashion. This “flow” is also stable and laminar. The following plot is for A = 1.1, xo = 0.5.

For values of A between 2.0 and 2.9, the iteration process still converges but there are oscillations that become more persistent as A increases. The signature of an instability is beginning to show. The plot below is for A = 2.9, xo = 0.9

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c) For A = 3.2, choose x0 = 0.6875, the equilibrium solution is maintained as shown in the following plot.

However, for A = 3.2 with x0 = 0.6874, or 0.6876 the iteration process becomes oscillatory with a period of 2 iterations as shown on the two following plots. This value of A models a transitional flow situation where both laminar and oscillatory flow states are possible.

d) For A = 3.5 the flow is always oscillatory no matter what value of x0 is chosen, and the period is 4 iterations. This “flow” is on its way to becoming turbulent because there are no stable

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equilibrium solutions and the period of oscillation is longer than for A = 3.2. The plot below is for A = 3.2 with x0 = 0.25.

e) For A = 3.9, fluctuations occur no matter what value of x0 is chosen, but these fluctuations are chaotic and do not repeat; the oscillation period is too large to be defined. This “flow” is effectively turbulent. The plot below is for A = 3.9 with x0 = 0.25

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Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (11)

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Transcript of Fluid Mechanics Kundu Cohen 6th edition solutions Sm ch (11)