Wiley · Some solutions to exercises with apologies for any mistakes PART I Chapter 1 Exercise 1.1, Origin of the closure problem: The closure problem arises in any non-linear system

Some solutions to exerciseswith apologies for any mistakes

PART I

Chapter 1

Exercise 1.1, Origin of the closure problem:The closure problem arises in any non-linear system for which one attempts to derive anequation for the average value. Let ξ correspond to the result of coin tossing, as in the text,and let

u3 + 3u2 + 3u = (u + 1)3 − 1 = 7ξShow that if u3 = u3 and u2 = u2 were correct, then u = (9/2)1/3 − 1. Show that the correctvalue is u = 1/2. Explain why these differ, and how this illustrates the ‘closure problem’.

Solution to Ex. 1.1, If u3 = u3, averaging the equation would give

u3 + 3u2 + 3u = (u + 1)3 − 1 = 7ξ = 7/2

so u = (9/2)1/3 − 1. However, the equation has the exact solution u = (7ξ + 1)1/3 − 1. Sinceξ = 1 with probability 1/2 , and ξ = 0 with probability 1/2 ,

u = (81/3 − 1) × 1/2 + (0) × 1/2 = 1/2

Exercise 1.2, Eddies:Identify what you would consider to be large and small scale eddies in photographs 1.4 and1.7.

Solution to Ex. 1.2, Descriptive

Exercise 1.3, Turbulence in practice:Discuss practical situations where turbulent flows might be unwanted or even an advantage.Why do you think golf balls have dimples?

1

Chapter 2

Exercise 2.1, Dissipation range scaling:

In the legend of figure 2.1 let Rλ = R1/2T . Assuming that the energetic range begins where

the data leave the −5/3 line, do these data roughly confirm the RT scaling of η/L?Solution to Ex. 2.1,

η = (ν3/ε)1/4; η/L = (ν3/L4ε)1/4 = (ν3/k3/2L3)1/4 = R−3/4T

From the figure, at Rλ = 600, ηκ ≈ 7 × 10−4 and RT = 3.6 × 105. At Rλ = 1, 500,ηκ ≈ 2 × 10−4 and RT = 2.25 × 106. Assume κ ∝ 1/L; the data give

(η/L)600(η/L)1,500

= 7/2 = 3.5

while the dimensional analysis says this should be proportional to R−3/4T .(

(RT )1500(RT )600

)3/4= (22.5/3.6)3/4 = 3.95.

Good, order of magnitude, confirmation of the scaling.

Exercise 2.2, DNS:One application of dimensional analysis is to estimating the computer requirements for DirectNumerical Simulation of turbulence. The computational mesh must be fine enough to resolvethe smallest eddies, and contain enough points to resolve the largest. Explain why thisimplies that the number of grid points, N , scales as N ∝ (L/η)3 in 3-dimensions. Obtainthe exponent in N ∼ RnT . Estimate the number of grid points needed when RT = 104.

Solution to Ex. 2.2, L/η ∝ R3/4T −→ N ∼ R9/4T When Rt = 104, N ∼ 109.Exercise 2.3, Relative dispersion:The inertial range velocity (εr)1/3 can be described as the velocity at which two fluid elementsthat are separated by distance r move apart (provided their separation is in the inertial range,η � r � L). Deduce the power law for the time-dependence of the mean square separationr2(t). Use dimensional analysis. Also infer the result by integrating an ordinary differentialequation. The scaling v2 ∝ r2/3 is often called Richardson’s 2/3-law.

Solution to Ex. 2.3, Solution by dimensional analysis: r2 ∝ εt3.A solution by integrating an o.d.e.: From Kolmogoroff’s 2/3-law

dtr2 ∝ ε2/3r22/3

The solution to this equation is r21/3 ∝ ε2/3t + r21/30 . If ε2/3t >> r2

1/3

0 then r2 ∝ εt3.

2

Exercise 2.4, Averaging via p.d.f.’s:Let the p.d.f. of a random variable be given by the function

P (x) = Ax2e−x for the range ∞ > x ≥ 0P (x) = 0 for x < 0.

What is the value of the normalization constant A? Evaluate sin(ax) where a is a constant.

Solution to Ex. 2.4, A = 1/2.

sin(ax) = 1/2Im

∫ ∞0

x2e−x(1−ia)dx = Im[(1 + ia)3/(1 + a2)3] =3a − a3(1 + a2)3

Exercise 2.5, Taylor microscale:Let u be a statistically homogeneous function of x. A microscale, λ is defined by

limξ→0

d2ξ

[u(x)u(x + ξ)

]= −u

2

λ2,

if the correlation function is twice differentiable. Show that u2/λ2 = (dxu)2. Assume that(dxu)2 follows dissipation range scaling and obtain the RT dependence of λ/L.

Solution to Ex. 2.5,

−u2

λ2= u(x)d2ξu(x + ξ)|ξ=0 = ud2xu = dx(udxu) − dxudxu

= dx(udxu) − (dxu)2 = −(dxu)2

because homogeneity implies dx(any average) = 0. In dissipation range (dxu)2 ∼ ε/ν leadsto

λ/L ∼ R−1/2T

Exercise 2.6, Toor’s analogy, (2.2.13):To what does the term ‘analogy’ refer?

The p.d.f. P (m) = (1 + m)a−1(1 − m)b−1/B(a, b), for −1 < m < 1 and P (m) = 0 for|m| ≥ 1 is referred to as the ‘beta’ probability density. The normalization coefficient isB(a, b) = 2a+b−1Γ(a)Γ(b)/Γ(a+b) where Γ is the factorial function Γ(a) = (a−1)!, extendedto non-integer arguments.

The beta distribution is a popular model for the mixture fraction p.d.f. in reacting flows.Given that Γ(1/2) =

√π, evaluate γA for a = b = 1/2, a = b = 2 and a = b = 4. Plot

the β-distribution for these same three cases. Based on these cases, describe how the p.d.f.evolves as mixing and reaction proceed.

Solution to Ex. 2.6, The analogy is between non-reactive scalar mixing and reactantconsumption. The elegance is that one can predict properties of reacting flow withoutexplicitly analyzing the reaction process.

3

Mean reactant concentrations (normalized by initial value):

γA = Γ(2a)/Γ(a)Γ(a + 1)22a in general for a = b

γA = 1/π for a = b = 1/2

γA = 3/16 for a = b = 2

γA = 35/256 for a = b = 4

When a = b the variance is proportional to1/1+a+b, per p.22. As a increases, variancedecreases, in accord with time-evolution.

m

p(m

)

Beta distributions

-1.0 -0.5 0 0.5 1.00

0.5

1.0

1.5

2.0 a=b=2 a=b=4 a=b=1/2

Exercise 2.7, Langevin equation:Why does the expansion (2.2.20) begin with 1 and why is the next term negative?

In (2.2.16) and subsequent equations let s2 = (1 − r2)σ2. Derive the equation

1/2dtu2 = −u2

TL+

σ2

TL.

from (2.2.18). Solve for u2(t) with initial condition u2(0) = 0, σ and TL being constants.Is u2 a statistically stationary random variable? The exact solution for the variance that isderived here will be used to test a Monte-Carlo simulation in the next exercise.

Solution to Ex. 2.7, r(0) = 1 by (2.15), r is a correlation function so |r| ≤ 1. Solutionis u2 = σ2(1 − e−2t/TL); this is not statistically stationary because u2 is a function of time.Exercise 2.8, More on stochastic processes:Most computer libraries have a random number algorithm that generates values in the range{0, 1} with equal probability; i.e. P (ũ) = 1, 0 ≤ ũ ≤ 1.

Show that ũ = 1/2 and that u2 = 1/12 where u is the fluctuation ũ − ũ. Deduce thatξ ≡ √12 (ũ − 1/2) has ξ = 0 and ξ2 = 1, as is needed in the model (2.2.16). A Gaussianrandom variable can be approximated by summing N values ξi and normalizing by

√N . For

N = 16

ξg = 1/4

16∑1

ξi.

Program this and verify by averaging a large number (10,000 say) of values that ξg ≈ 0 andξ2g ≈ 1. Let s in (2.2.16) be as in exercise 2.7, normalize t by TL and u by σ and solve (2.2.16)numerically starting with u = 0 and integrating to t/TL = 10 by steps of 0.05. Computeu2(t) by averaging 100 such solutions, by averaging 1,000 such solutions and by averaging4,000 such solutions. Plot these estimates of u2 vs. t and compare to the exact result foundin exercise 2.7. Does it look like the average is converging like 1/

√N?

4

Solution to Ex. 2.8, ũ =∫ 1

0ũdũ = 1/2; u2 =

∫ 10(ũ − ũ)2dũ = ∫ 1

0(ũ − 1/2)2dũ = 1/12

t

u2(t

)

0 2.5 5.0 7.5 10.00

0.25

0.50

0.75

1.00

1.25

Analytical 4000 samples 1000 samples 100 samples

Langevin equation solutions.

Exercise 2.9, Isotropy:Verify that (2.3.35) is the most general fourth order, isotropic tensor.

Exercise 2.10, Solving equations via Cayley-Hamilton:Use the Cayley-Hamilton theorem to solve

bij = Sikbkj − 1/3δijSklblk + Sijin which S is a given, trace-free matrix (Skk = 0). Why is b also trace-free? This solution isonly valid if S is such that the solution is not infinite; what is this solvability criterion?

Solution to Ex. 2.10, In matrix form, solve

b = S · b − 1/3Trace(S · b)I + S

Let b = AS + B(S2 − 1/3IS2) where S2 = Trace(S2). Substituting this into the equationgives

AS + B(S2 − 1/3IS2) = AS2 + B(S3 − 1/3SS2) − 1/3(AS2 + BS3)I + S

Cayley-Hamilton is S3 = 1/3IS3 + 1/2S2S. Equating coefficients of S2 gives B = A.

Equating coefficients of S gives A = 1/(1 − 1/6S2). So

b =S + S2 − 1/3IS2

1 − 1/6S2

and the solvability condition is S2 �= 6.

5

Exercise 2.11, Solving equations via generalized Cayley-Hamilton:τ is the solution to

τ = τ · ΩΩ − ΩΩ · τ + Trace(S · τ )δ − Sin which S is symmetric and trace-free and ΩΩ is antisymmetric and trace-free. Show that τis symmetric. Use (2.3.46) to solve this equation in two dimensions.

6

Chapter 3

Exercise 3.1, Derivation of Reynolds stress transport equation:This exercise may seem laborious, but it is a good introduction to the use of Reynoldsaveraging, and to the Reynolds averaged Navier-Stokes (RANS) equations.

Derive (3.1.4) from (3.1.3) and (3.1.2) and then obtain (3.1.5). Symbolically the stepsare

RSij = uj

[NS(U + u)i − NS(U + u)i

]+ ui

[NS(U + u)j − NS(U + u)j

]where ‘NS’ represents the Navier-Stokes equations. How many equations and how manyunknowns are there? Why not also form the moment of the continuity equation?

Solution to Ex. 3.1, uj∂iui = 0 introduces a new unknown vector uj∂iui that is un-related to other unknowns in the RANS equation: so the continuity equation is simplyirrelevant to the single point, averaged equations. 6 equations, 28 unknowns (given U) – 22more unknowns than equations.

Exercise 3.2, Production of k:Using (3.1.3) show that the rate at which mean energy (per unit mass) 1/2UiUi is lost tothe turbulence is uiuk∂kUi = −P . (Conservation terms are not an ‘energy loss’.) Thisdemonstrates that the term ‘production’ is actually referring to the transfer of energy fromthe mean flow to the turbulence, and not to a net source of energy.

Solution to Ex. 3.2,

1/2DtU2 = ∂k(UkP ) − Ui∂kuiuk

= ∂k (UkP − Uiuiuk) + uiuk∂kUior, in vector notation

1/2DtU2 = ∇ · (U P − U · uu) − Pk

The first term on the right is in conservation form, and simply transports energy. Pk is theenergy lost to (or gained from) turbulence.

Exercise 3.3, The mixing length rationale:Consider the mean concentration C(xi) of a passive quantity that is convected by a turbulentvelocity vector uj(t). Derive an eddy diffusion formula analogous to (3.2.9). The eddydiffusivity should come out as a second order tensor, i.e., its a matrix of components. Isthis matrix symmetric in general? How about when the turbulence is statistically stationary(see the discussion of stationarity above (2.2.16))? Give your reasoning in answer to thesequestions.

Solution to Ex. 3.3,

uic ≈ uij∂jC =∫ t

0

ui(t)uj(t′) dt′∂jC

7

The eddy diffusion tensor is defined by uic = κTij∂jC, so

κTij =

∫ t0

ui(t)uj(t′) dt′

In general κTij �= κTji because ui is evaluated at the later time t and uj at the earlier time t′in κTij , while uj is at the later and ui the earlier time in κTji . However, if the turbulence is

statistically stationary ui(t)uj(t′) is a function only of |t − t′| and

κTij =

∫ t0

uiuj(|t − t′|) dt′ = κTji

so it doesn’t matter which component is earlier and which is later.

Exercise 3.4, Anisotropy equation:The Reynolds stress anisotropy tensor is defined as bij = uiuj/k − 2/3δij. Using (3.2.15)and (3.2.16) derive the evolution equation for bij in homogeneous turbulence. The equationshould involve bij , ∂jUi, φij, εij and k, with no explicit or implicit dependence on uiuj .

‘Isotropy’ means complete lack of any directional preference. Hence the identity matrix[δij ] is isotropic because all the diagonal components are equal; or more correctly, becauseif the coordinate system were rotated, the identity matrix would remain unchanged. Thetensor bij measures the departure of uiuj from isotropy.

Solution to Ex. 3.4,

ḃij = −φij + εijk

− 2/3(∂jUi + ∂iUj)−bik∂jUk − bjk∂iUk − (bij + 2/3δij)(P − ε)

where P = −bkl∂kUl.

8

Chapter 4

Exercise 4.1, Turbulent kinetic energy equation in the channel:Write the form of the turbulent kinetic energy equation (3.1.6) for this special case of parallelflow: U(y), homogeneity in x, z and stationarity in t. Show that∫ 2H

0

Pdy =∫ 2H

0

εdy.

What boundary condition on k is did you use, and why? Recall that P is the rate at whichturbulent energy is produced from the mean shear. This exercise suggests that in shearlayers, production and dissipation are similar in magnitude.

Solution to Ex. 4.1, The boundary condition ∂yk = 0 = k follows from the definitionk = 1/2u2 + v2 + w2 and the no-slip condition u = v = w = 0: k has a quadratic zero.

Exercise 4.2, Magnitude of Cf :Let RH = 10

4. Use formula (4.1.11) to estimate the value of Cf .

Solution to Ex. 4.2, A printout of√

(2/Cf) versus 2.43 log(104√

Cf/2)+5 gives Cf =5.7 × 10−3.Exercise 4.3, Pressure drop across a channel:Because of the favorable pressure gradient, the log-law can be considered valid practically tothe center of a fully developed, plane channel flow. The centerline velocity, UCL , and volumeflux, Q, are related by

Q ≈ 0.87UCLAwhere A is the cross-sectional area of the channel. Its height is 2H . The channel is longand narrow, so the sidewalls can be ignored. A certain volume flux, Q, of airflow is desired.Write a formula that can be used to estimate the necessary pressure drop across a channelof length L — do not try to solve it for ∆P ; that could be done numerically.

Solution to Ex. 4.3,From momentum conservation

ρu2∗ = 2H∆P/L

Drag formula

u∗ =UCLκ

log(u∗H/2) + Bκ=

Qκ

0.87A[log(u∗H/2ν) + Bκ]

Squaring, and using κ = 0.41 and B = 5.0

u2∗ =0.22Q2

A2[ 1/2 log(u2∗H2/4ν2) + 2.05]2

∆P

ρ=

0.44Q2L

HA2[log(∆PH3/2µνL) + 4.1]2

9

Exercise 4.4, Zones of the boundary layer:Make a sketch of U(y) as in figure 4.5. Indicate the law-of-the-wall and law-of-the-wakeregions in both log-linear and linear-linear coordinates. In zero pressure gradient, what isthe approximate magnitude of U/U∞ at the top of the log-layer? What is it for the adversepressure gradient data?

Solution to Ex. 4.4, 0.75 on ZPG. 0.5 APG

Exercise 4.5, The momentum integral:Derive the momentum integral equation (4.2.20) for the case where dxU∞ �= 0. Substituteδ∗ = Hθ. Will a favorable pressure gradient (dxU∞ > 0 and dxP < 0) increase or decreasethe growth rate of θ? Answer the same question for an adverse pressure gradient.

Exercise 4.6, Law of the wake:Infer a drag law by evaluating (4.2.16) at y = δ99. Use Coles’ wake function for w(y). SetU99 = U∞ and compare this to (4.2.15). Comment on the effect of pressure gradient on skinfriction.

Exercise 4.7, Drag law:The power law form U = Ud(y/d)

a, y < d is sometimes used to fit the mean flow profile. dis the boundary layer edge, so Ud = U∞ and U = U∞, y ≥ d. The exponent a is a smallnumber, about 1/7, although it varies with Reynolds number, as you will show. Calculateθ, δ∗ and H = δ/θ for this profile. For small a, (y/d)a = ea log(y/d) ≈ 1 + a log(y/d). If thisis matched to the log-law (4.1.7), how is a related to u∗/U∞? What is the correspondingskin friction law, written like (4.1.11) in terms of friction coefficient, Cf , and momentumthickness Reynolds number, Rθ = Udθ/ν?

Solution to Ex. 4.7,

Θ = ad/(1 + 3a + 2a2) → d = Θ(1 + 3a + 2a2)/a

Comparing U = U∞ + aU∞ log(y/d) to the log-law (4.1.7)

U =u∗κ

log(y+) + Bu∗

gives a = u∗/κU∞ =√

Cf/2/κ and U∞ + u∗/κ log(ν/u∗d) = Bu∗. The latter can berearranged to

U∞u∗

=

√2

Cf=

1

κlog(u∗d/ν) + B =

1

κlog

[κRθ(1 + 3a + 2a

2)]+ B

Alternatively, a depends on Reynolds number via

1/a = log[κRθ(1 + 3a + 2a

2)]+ Bκ

10

Exercise 4.8, Measurement of u∗:Measurements in a particular turbulent boundary layer give a thickness δ99 = 3.2cm. Thevelocity U is measured at three heights with the results, U = 40.1m/s at y = 2.05cm,U = 34.7m/s at y = 0.59cm and U = 30.7m/s at y = 0.22cm.

i Estimate the friction velocity u∗.ii Let the fluid be air with ν = 0.15cm2/s. Is this an adverse or favorable pressure

gradient boundary layer?

Solution to Ex. 4.8,The lower two points are in the log-layer; hence u∗ log(y2/y1)/κ = U1 − U2 =⇒ u∗ =

4. × 0.41/ log(.59/.22) = 1.66 m/s. The lowest point is at y+ = 243 which satisfies thecriterion 40 < y+ < 0.2δ99.

At y = 2.05cm, y+ = 2269, u∗(log y+/κ + 5.1) = 39.75m/s. Hence the law-of-the wakecorrection is (40.1 − 39.75)/1.66 = 0.21. The wake correction is 2Π/κ sin(π/2(2.05/3.2))2 =4.32Π. Hence Π(β) = 0.21/4.32 = 0.049. The empirical formula (4.2.17) in the text givesβ = −0.48. Since β < 0 the pressure gradient is favorable.Exercise 4.9, Governing equations for self-similar flow:Substitute the assumed forms

U = ∆U(x)f(y/δ(x)); V = ∆U(x)g(y/δ(x)); uv = ∆U2h(y/δ)

for the case of a 2-D jet into (4.3.25) using (4.3.31) but ignore (4.3.34). Let ζ = y/δ. Rewritethe equations that you derived as ordinary differential equations involving f(ζ), g(ζ) andh(ζ). In order for the independent variable to be ζ alone, the coefficients in this equationcannot be functions of x or y — because then the solution would be of the form f(ζ, x) orf(ζ, y), which are not self-similar. What condition must be imposed on ∆U and dxδ for thisto be the case? Do you think that all the results (4.3.34) can be derived by this approach?

Solution to Ex. 4.9, Substitution into momentum equation gives

δ

∆U

d∆U

dxf 2 − dδ

dxηff ′ + gf ′ = −h′

For similarity, the coefficients must have proportionate x dependence: in this case they mustbe constant:

dδ

dx= const. :

δ

∆U

d∆U

dx= const.

The former gives δ ∝ x; the latter would be satisfied for ∆U ∝ xn, for any n. n = −1/2 isobtained from the momentum integral.

All scaling can be derived by this more formal approach. In the case of the wake it showsthat similarity can only be approximate, under a small deficit assumption.

Exercise 4.10, The momentumless wake:For a zero momentum deficit, plane wake, the constant in (4.3.30) is zero. I.e., in the smalldeficit approximation, U∞

∫∞−∞(U − U∞)dy = 0.

11

Use the eddy viscosity approximation −uv = νT ∂yU ; assume νT to be constant; usethe small defect approximation for the convection velocity; and form the second moment of(4.3.28) with respect to y2, after making these approximations, to find the similarity scalingin this case. (Form the second moment means multiply by y2, then integrate between ±∞).

Solution to Ex. 4.10, Consider∫ ∞−∞

y2[∂xU(U − U∞) + ∂yV (U − U∞)dy =∫ ∞−∞

y2νT ∂2y(U − U∞)dy

In the present case of zero momentum flux the right side integrates to 0,∫ ∞−∞

y2νT ∂2y(U − U∞)dy = 2νT

∫ ∞−∞

(U − U∞)dy = 0.

Requiring U = U∞−udf(η) and ud

One integration gives

−a2f

∫fdη = cf ′

Let∫

fdη = h; note that h′ → 0 as η → ∞.

− 1/2ah′h = ch′′

Integrate:− 1/4a(h2 − h2∞) = ch′.

Integrate one last time, or intuit that the solution is of the form h = 1/k tanh(kη) (orf = sech2(kη))

−a4(tanh2 − 1) = ck2sech2.

The functions cancel, leaving k =√

a/4c.The solution is

U/UCL = sech2η√

a/4c

The 50% thickness is defined such that at η = 1, 0.5 = sech2(√

a/4c). Hence√a/4c = 0.8813 and a = 0.05 −→ c = .016

Exercise 4.12, Enthalpy thickness:Show that the enthalpy thickness evolves according to

d∆

dx= St

in analogy to the evolution equation (4.2.23) for momentum thickness. Compare the evo-lution of ∆(x) to that of Θ(x) for a zero pressure gradient boundary layer in air over therange 103 < Rθ < 10

4. The comparison can be in the form of computed curves using a datacorrelation for Cf .

Exercise 4.13, Reflected plume model:Because the equation of a passive scalar is linear, superposition is applicable. The meanconcentration of a sum of sources is the sum of their individual mean concentrations. TheGaussian cloud

C =Q√

(2π)3X2 Y 2exp

[− x

2

2X2− (y − ys)

2

2Y 2

]is the mean concentration of a line source located at x = 0, y = ys. If a no-flux wall exists aty = 0, a fictitous source can be added at y = −ys to satisfy the boundary condition ∂yC = 0on y = 0. Write the concentration distribution for this ‘reflected plume’ model. Show thatthe concentration integrated across the physical doman, which is y > 0, −∞ < x < ∞, is

13

constant in time. There is no mean flow, just turbulence, so dtX = u and dtY = v. Theheight of the plume centerline is defined as

YL(t) =

∫ ∞−∞

∫ ∞0

yC(y)dy dx/∫ ∞

−∞

∫ ∞0

C(y)dy dx.

Show that this increases with time, even though there is no mean velocity in the y-direction.For the exponential correlation analyzed in the text find the ‘Lagrangian mean velocity’,dtYL when ys = 0. Why is the Lagrangian mean velocity not zero, even though there is noEulerian mean velocity?

Solution to Ex. 4.13,

C =Q√

(2π)3X2 Y 2exp

−x22X2

[exp

−(y − ys)22Y 2

+ exp−(y + ys)2

2Y 2

]; y > 0

Integral is ∫ ∞0

exp−(y − ys)2

2Y 2+ exp

−(y + ys)22Y 2

dy

=

∫ ∞0

exp−(y − ys)2

2Y 2dy +

∫ 0−∞

exp−(y − ys)2

2Y 2dy =

√2πY 2

after changing y to −y in the second integral. x-integral is√

2πX2 so area integral isconstant.

In for non-zero ys, YL is expressed in incomplete Gamma-functions. However, for ys = 0

YL =

∫ ∞0

exp−y22Y 2

ydy/∫ ∞

0

exp−y22Y 2

dy =

√2Y 2/π.

and (4.4.52) can be substituted to find YL(t).

14

Chapter 5

Exercise 5.1, Useful mathematical relation:Prove that εilpεjnp = (δijδln − δinδlj), as used in the analysis of (9.2.31). The brute forceapproach is to write out all 81 components. However, all but 12 are zero. Use this identityto show that ν|ω|2 = ε in homogeneous, incompressible turbulence. The vorticity vector isωi = εijk∂juk.

Solution to Ex. 5.1,

ν|ω|2 = ν�pil�pjn∂iul∂jun = ν(δijδln − δinδjl)∂iul∂jun= ν∂iul∂iul − ∂iul∂lui = ε − ∂i(ul∂lui) = ε

Since ∂i(ul∂lui) = 0 due to homogeneity. So the dissipation and vorticity spectra are inti-mately connected. |ω|2 is often called ‘enstrophy’.Exercise 5.2, Vorticity and dissipation:Suppose that the scale of non-homogeneity is δ ∼ L. Show the magnitude of the first termon the right of (5.2.3) relative to the second is O(RT ).

Exercise 5.3, The balance between stretching and diffusion:Show by dimensional analysis of the stretching (ωj∂jui) and diffusion (ν∇2ωi) terms of thevorticity equation that a balance is reached when the vortex radius is O(η). I.e., if λ is theradius then λ ∼ η. Note that the rate of strain at small scales is of order √ε/ν.

Solution to Ex. 5.3, ωj∂jui ∼ ω√

ε/ν ≈ ν∇2ωi ∼ νω/λ2. So λ2 ∼√

ν3/ε = η2

Exercise 5.4, Introduction to RDT:Find a solution of the form

ω3 = A(t)ei(k1(t)x1+k2x2)

to the inviscid vorticity equation if the velocity is of the form u3 = αx3, u2 = v2 andu1 = v1 − αx1 and ∇ · v = 0. The initial condition is ω3(0) = ω0ei(κ1x1+κ2x2). Because v isthe velocity generated by the vorticity, it too is proportional to ei(k1(t)x1+k2x2).

Solution to Ex. 5.4, k̇1 = αk1; Ȧ = αA. v2 = k1ω3/|k|2; v1 = −k2ω3/|k|2Exercise 5.5, Burger’s vortex:Solve the two dimensional vorticity equation corresponding to (5.2.5) for ωx(y) when thestraining flow is U = αx and V = −αy. What is W (y)? Why is this an exact solution tothe Navier-Stokes equations?

Solution to Ex. 5.5,ωx = ωCLe

−αy2/2ν

W = W−∞ + ωCL

∫ y−∞

e−αy′2/2νdy′

15

Chapter 6

PART II

Exercise 6.1, Integral equation closure:Write a program to solve the model (6.1.1) through (6.1.3). Plot solutions for θ(x/a),H(x/a) and Cf(x/a) with U∞/U0 = (1 − x/a), where a is a characteristic length and U0 acharacteristic velocity. Start with Cf (0) = 3 × 10−3, H(0) = 1.35 and Ra = 105. (Whatis θ(0)/a?) Stop the computation when the boundary layer separates; at what x/a doesseparation occur? Also plot these variables with U∞/U0 = (1 + x/a) in the range 0 ≤ x/a ≤0.5. Discuss the effect of favorable and adverse pressure gradients on Cf and H .

Solution to Ex. 6.1, Separation at x = 0.227. APG causes Cf to drop; FPG increasesit. Separation criterion is H = 3.

x

APG

0 0.05 0.10 0.15 0.20 0.250

1

2

310ThH1000Cf

x

FPG

0 0.1 0.2 0.3 0.4 0.50

2

4

610ThH1000Cf

Exercise 6.2, Golf ball revisited:

16

In chapter 1, exercise 1.3 asked why golfballs have dimples. An extension to fig-ure 6.2 provides an answer: the dimplespromote transition. Provide a qualitativeexplanation of the figure at right.

Reynolds number

smooth sphere

golf ball

Dra

g co

effic

ient

Exercise 6.3, The mixing length model:Show that if m = κy then the mixing length formula (6.1.6) gives the log-law in the constantstress layer (−uv = u2∗). Johnson & King (1985) suggested using νT = u∗m, which gives thesame result in a constant stress layer. Compare the mean velocity obtained from these twoformulations in a linear stress layer, −uv = u2∗ + αy.

Solution to Ex. 6.3, In the constant stress layer (mdyU)2 = u2∗; hence dyU = u∗/κy,

which gives the log-law when integrated. For JK: u∗mdyU = u2∗, which gives the same result.In a linear stress layer, JK gives u∗κydyU = u2∗ + αy, or

U =u∗κ

log y+ +α

u∗κy + Bu∗

where the integration constant was chosen to recover the standard log-law as α → 0. Themixing length model gives

U = u∗

∫ √1 + αy/u2∗

κydy

=u∗κ

(2√

1 + αy/u2∗ + log

[√1 + αy/u2∗ − 1√1 + αy/u2∗ + 1

])+ u∗const.

To obtain the constant, let α → 0; the above becomes

U → u∗κ

(2 + log

[αy

4u2∗

])+ u∗const.

Setting this to u∗/κ log(yu∗/ν) + Bu∗ gives

const. = B − (2 + log(αν/4u3∗))/κ

Exercise 6.4, The Stratford boundary layer:Further to exercise 6.3: Stratford proposed a clever idea for reducing skin friction. Hesuggested that a carefully designed adverse pressure gradient could reduce u2∗ almost to zero(in the linear stress layer) along a substantial length. The limiting case u2∗ = 0 is called the

17

Stratford boundary layer. What mean flow profile does the mixing length model predict inthis case?

Solution to Ex. 6.4,

(mdU/dy)2 = αy −→ dU/dy = √αy/m =

√α/κ2y

using the fact that in the constant stress region m = κy. Since this owes to dimensionalanalysis, it is assumed valid in the linear stress layer as well. Thus we obtain the half powerlaw, U ∝ y1/2Exercise 6.5, Mixing length model with Van Driest damping:The log-law applies in the region 35

the k-ε model can predict such effects. There is an analogy between rotation and streamlinecurvature, so your conclusions apply to effects of curvature on the turbulence as well.

Solution to Ex. 6.6, The exact expression for production is P = −uiuj∂iUj. Themean velocity gradient is ∂iUj = Sij + Ωij is terms of the rate of strain and rate of rotationtensors. Note that Ωij is anti-symmetric: Ωij = −Ωji, uiujΩij = 0. (Proof:

uiujΩij = ujuiΩji = uiujΩji = −uiujΩijHence uiujΩij = 0.) So P = −uiujSij and with the eddy viscosity model P = 2νT SijSji −2/3k∂iUi. In incompressible flow ∂iUi = 0. Since Ω drops out of the equation rotation effectsare not represented.

Exercise 6.7, Realizability of k-ε:Consider the k-ε model (6.2.18, 6.2.19) for the case of homogeneous turbulence. Prove thatif k and ε are greater than zero initially they can not subsequently become negative.

Solution to Ex. 6.7, Let ξ = k/ε. Then

dtξ =dtk

ε− kdtε

ε2= 2CµS

2(1 − Cε1)ξ2 + (Cε2 − 1)

Since Cε2 > 1 were ξ to approach 0 this equation gives dtξ > 0. Hence ξ cannot becomenegative, or even reach 0. Then

dtk =

(2CµS

2ξ − 1ξ

)k

gives

k = k0 exp

[∫ t0

(2CµS2ξ − 1/ξ)dt′

]> 0

Exercise 6.8, Bounds on production:Show that

|P| ≤ 2k|λmax|where |λmax| is the eigenvalue of the rate of strain tensor S with maximum absolute value.This suggests that eddy viscosity models should be constrained by

2νT |S|2 ≤ 2k|λmax|.

Let the eigenvalues of S be λ1, λ2 and λ3. For incompressible flow λ1 + λ2 + λ3 = 0. Ifλ3 = 0, show that

|λmax|2 = |S|2/2and generally that

|S|2 ≥ 3|λmax|2/2

19

where λmax is the maximum eigenvalue of S. Hence conclude that the eddy viscosity oughtto satisfy

νT ≤ 2k3|λmax| .

For the k-ε model Cµ ≤ 2ε/(3k|λmax|) could be imposed.Solution to Ex. 6.8, Choose the coordinate system in which S is diagonal:

|P| = |u21λ1 + u22λ2 + u23λ3| < (u21 + u22 + u23)|λmax| = 2k|λmax|

In 2-D, S = diag(λ,−λ, 0) so S2 = 2λ2; or |λ|2 = |S|2/2.In 3-D, S = diag(λ1, λ2,−λ1 − λ2) so S2 = 2(λ21 + λ22 − λ1λ2). This last expression is

minimum when 2λ1 = λ2 if λ2 is the maximum eigenvalue. Hence S2 ≥ 3/2λ2max.

Exercise 6.9, Front solution:Show that the mean shear at the propagating front solution corresponding to (6.2.46) is∂yU ∼ (ct − y)(σε−1)/(2−σε), y < ct.

Solution to Ex. 6.9, From the solution in the text νT = Aµ(ct−y) and c = Aµ/(2−σε).Near the front

dtU = Aµ∂y[(ct − y)∂yU ]Thus U ∼ const + (ct − y)n with n = 1/(2 − σε) and dyU ∼ (ct − y)n−1.Exercise 6.10, Calibrate an eddy viscosity transport model:Eddy-viscosity transport models, like S-A, are an alternative to the k − ε model for full-blown CFD analysis. They solve a single equation for the dependent variable νT . Considerthe model equation

DtνT = C1|S|νT − (∂iνT )(∂iνT ) + ∂i[(νT

σν+ ν

)∂iνT

]or, in vector notation

DtνT = C1|S|νT − |∇νT |2 + ∇ ·[(νT

σν+ ν

)∇νT

].

Note that 2|S|2 = |∂yU |2 in parallel shear flow.C1 and σν are empirical constants. Use log-layer analysis to obtain a formula relating C1

and σν to the vonKarman constant, κ. What second experimental datum might be used todetermine values for C1 and σν?

Solution to Ex. 6.10, In the log-region νT = u∗κy, ∂yU = u∗/κy and the Reynoldsnumber is high, so ν can be ignored in comparison to νT . Substituting νT = u∗κy

0 =C1√

2− κ2 + κ

2

σν

20

This gives σν = 1/(1 − C1/κ2√

2). C1 could be obtained from data on homogeneous shearflow, for which dtνT = C1|S|νT .Exercise 6.11, Channel flow by k − ε:Use the velocity and length scales u∗ and H to set up the non-dimensional k-ε equationsthat must be solved in conjunction with (4.1.4) for fully developed channel flow. Non-dimensionalize the wall-function boundary conditions (6.2.35). [The U boundary conditionsimplifies to U(y/H = 40/Rτ) = log(40)/κ + B in this non-dimensionalization.] Solve theresulting problem numerically for steady channel flow with Rτ = 10, 000. Plot your solutionfor U+ and k+.

Solution to Ex. 6.11,

log-linear linear-linear

y+102 103 104

0

10

20

30

40U+10k+100ε+

y+0 2000 4000 6000 8000 10000

0

10

20

30

40U+10k+100ε+

Figure 1: Channel flow: k − ε with wall function.

See figure 1. The numerical centerline velocity is U+ ≈ 27. or Cf = 2.7 × 10−3.Exercise 6.12, Channel flow by 2-layer:Repeat previous exercise for the 2-layer k-ε model.

Exercise 6.13, Turbulent dispersion:Stochastic models provide a concrete physical representation of the ensemble averaged smooth-ing effect of random convection. The following is simpler than Taylor’s model: the positionof convected fluid element is given by

Y (t + dt) = Y (t) +√

α(t)dt ξ(t)

in which ξ(t) the random process defined below equation 2.2.16.1) Show that α(t) = 2KT where Kt is the eddy diffusivity.2) Use the k − ε formula KT = Cµk2/ε to show that in decaying grid turbulence

Y 2 ∝ (t + t0)(2Cε2−3)/(Cε2−1)

21

Exercise 6.14, Non-existence of 1/2-power law:In a linear stress layer the total shear stress varies as τ = αy, with α being a constant thathas dimensions /t2. By dimensional reasoning U ∝ (αy)1/2 and

k ∝ αy; ε ∝ α3/2y1/2; ω ∝ (αy)−1/2.

Show that the k − ε and k − ω models do not admit a power law solution of this form.

22

Chapter 7

Exercise 7.1, Three-dimensional boundary layers:Let αRS = u2u3/u1u2 and αU = S23/S12 denote the tangents of the direction of Reynoldsstress and mean rate of strain. Show that (7.1.1) implies the evolution equation

dtαRS =−2kCµC1S12

u1u2(αU − αRS)

If the initial state is an equilibrium shear flow, U(y), then at t = 0

−2kCµC1S12u1u2

=−CµC1k∂yU

uv

=−CµC1k

νT=

C1T

.

Suppose that initially αU = 0 = αRS but the flow subsequently veers so that it developsa component W (y) and αU > 0. Show that if αU increases monotonically with time thenαRS ≤ αU . Thus the angle of the Reynolds stress lags that of the mean rate of strain.Exercise 7.2, The anisotropy tensor:Why must the diagonal components, b11, b22, b33 of the anisotropy tensor (7.1.5) lie between− 2/3 and 4/3? In fact, generally the eigenvalues of bij have to lie between − 2/3 and 4/3 .

Solution to Ex. 7.2, u2 > 0 and k > 1/2u2. From the first b11 = u2/k − 2/3 > −2/3.From the second b11 < 4/3

Exercise 7.3, Return to isotropy:Use the Rotta model for Fij to solve (7.1.7) for the relaxation of bij in homogeneous turbu-lence with no mean flow, ∂jUi = 0, and initially bij = b

0ij . Obtain the turbulence time-scale,

T = k/ε, from the k-ε solution for decaying isotropic turbulence §6.2.1. Why is it OK touse the isotropic solution, even though bij �= 0? Rewrite your solution for bij as a solutionfor uiuj . Also explain why C1 > 1 is necessary.

Solution to Ex. 7.3,

dtbij = (1 − C1)bijT

For decaying grid turbulence k = t−n and ε = −dtk = nk/t; hence T = t/n. The aboveequation becomes dtbij = n(1 − C1)bij/t with solution bij = b0ijtn(1−C1). The decay law for kapplies to decaying non-isotropic turbulence as well. C1 > 1, or else the anisotropy wouldgrow unbounded with time (see the previous problem).

Exercise 7.4, Invariants of the anisotropy tensor:The second (II) and third (III) invariants of bij are defined in (2.3.41) as II = − 1/2bijbji =− 1/2b2ii and III = 1/3bijbjkbki = 1/3b3ii. (The first invariant bii is identically 0.) For the samecase of relaxation toward isotropy via Rotta’s model as in exercise 7.3, write the evolution

23

equations for II and III. The two dimensional plane with coordinates II and III is calledthe ‘invariant map’. Find the equation for III as a function of II for the Rotta model byforming the ratio

dIII

dII=

dtIII

dtII.

Then show that this model predicts III ∝ II3/2.Solution to Ex. 7.4, dtIII = −3(1−C1)III/T ; dtII = −2(1−C1)II/T so dIII/dII =

3III/2II.

Exercise 7.5, Solution to SSG for homogeneous shear:Derive (7.2.39) with (7.1.35) included. Substitute the SSG values for the coefficients of therapid model. Use either the empirical value PR = 1.6 or the k-ε value (6.2.24). Set theRotta constant to C1 = 1.7. Compare numerical values to the experimental results. (Thevalues cited in the text include a non-linear term in the slow model.)

Exercise 7.6, εijk:Show that

εijkεilm = δjlδkm − δjlδmk(if you have not done so in exercise 5.1) From this show that if Ωij = εijkΩk then ωi =εijkΩjk = 2Ωi.

Exercise 7.7, GLM:Show that (7.1.31) can be written in terms of the production tensor as

℘rapidij = [ 4/5 − 4/3(C2 + C3)]kSij−C2(Pij − 2/3δijP) − C3(Dij − 2/3δijP) (0.1)

where Dij ≡ −uiuk∂jUk − ujuk∂iUk. From this show that the IP model (C2 = 3/5 andC3 = 0) corresponds to canceling the terms involving Dij and Sij to leave

℘rapidij = − 3/5(Pij − 2/3δijP)This is a simple, and for that reason among others, popular model.

Solution to Ex. 7.7, This is essentially given in the text.

Exercise 7.8, Solution for axisymmetric, homogeneous rate of strain:Find the equilibrium solution to (7.2.36) for the incompressible, homogeneous straining flow,U = Sx, V = − 1/2Sy and W = − 1/2Sz.

Solution to Ex. 7.8, From

∂1U1 = S, ∂2U2 = ∂3U3 = − 1/2Sit can be seen that the only non-zero b’s are b11 and b22 = b33. Since bkk = 0, by definition,

b33 = b22 = − 1/2b11.

24

Substituting the two above equations into (7.2.36) with dtbij = 0 (equilibrium condition)gives

0 = (1 − C1)b11 − 2b11Sk/ε − (b11 + 2/3)PR − 8/15Sk/ε + (C2 + C3)(2b11Sk/ε + 2/3PR)PR (≡ P/ε) is known from (6.2.24b) in terms of the k-ε constants. Rearranging gives

b11 =8/15Sk/ε + 2/3(1 − C2 − C3)PR

(1 − C1) + 2(C2 + C3 − 1)Sk/ε − PRHowever, we also know that PR = −bijSijk/ε which is PR = − 3/2b11Sk/ε in the present

case. Hence, a consistent equilibrium requires that Sk/ε be the solution to the quadraticequation

PR =4/5(Sk/ε)2 + (1 − C2 − C3)PRSk/ε

(C1 − 1) + 2(1 − C2 − C3)Sk/ε + PRwhere PR = Cε2−1Cε1−1 is a known constant.Exercise 7.9, Equations in a rotating frame:If equations (7.2.36) are referred to a reference frame rotating about the x3-axis, the i, j-component of the mean velocity gradient becomes

∂jUi − εij3Ω3in the absolute frame. Here Ω3 is the rate of frame rotation and ∂jUi is the velocity gradi-ent in the rotating frame. Evaluate the components of the production tensor for rotating,homogeneous shear flow, Ui(xj) = δi1δj2Sx2 and compare to (7.2.41).

The time-derivative of bij in the stationary frame is replaced by

∂tbij + Ω3(bikεkj3 + bjkεki3)

relative to the rotating frame. Substitute this and the transformed velocity gradient tensorinto (7.2.36) to derive the evolution equation (8.3.25) for bij in the rotating frame. Thereare two choices for the last part of this exercise: either

(i) Solve this equation numerically for rotating, homogeneous shear flow. The initialcondition is isotropy, bij = 0. Non-dimensionalize time by S so that the equationscontain the ratio Ω/S. Investigate how the solution depends on this parameter—experiment with its value. The k-ε equations

dtk = −kSb12 − ε ; dtε = − (Cε1kSb12 + Cε2ε) εk

will be needed as well. Include a plot of b12 vs. time. Or(ii) Derive the closed-form equilibrium solution (8.3.33) for b11, b22 and b12 and Sk/ε as

functions of Ω/S in homogeneous shear flow. You can use (6.2.24) for the equilibriumvalue of P/ε.

In either case, discuss the stabilizing or destabilizing effects of rotation.

25

Exercise 7.10, Wall functions for SMC:State the wall function boundary conditions for k, ε, U and uiuj that can be applied some-where inside the log-layer. Write a program to solve plane channel flow with one of the SMCformulations and wall functions. Compare the solution to channel flow data at Rτ = 590presented in figure 4.3.

Exercise 7.11, Limiting budget:Consider the Reynolds stress budget (3.1.5). By examining the scaling of each term with yas y → 0 at a no-slip wall, derive the limiting behaviors (7.3.57)ff.Exercise 7.12, Near wall analysis:Verify that the first of equations 7.3.70 has the behavior

εuiujk

− ν ∂2uiuj∂y2

= ℘nn = kfij = O(y3)

as y → 0. Show that if fij(0) = O(1) and uiuj is not singular, the only possible behaviorsare uiuj = O(y

4) or uiuj = O(y2).

Solution to Ex. 7.12, Let uiuj ∼ Ayn and note k ∼ εy2/2ν. Then A[2 − n(n −1)]yn−2 ∼ y2. Hence n = 4 by equating the exponents. Otherwise the left side must vanishgiving n = 2. Negative n = −1 is singular.

26

Chapter 8

Exercise 8.1, Objective tensors:Show that the absolute vorticity tensor Ωij = 1/2(∂Vi/∂xj − ∂Vj/∂xi) + εjikΩFk transformsproperly under the arbitrary change of reference frame (8.1.5).

Exercise 8.2, Realizability by stochastic equations:If a model can be derived from a well defined random process, then by definition it isrealizable. Consider a random velocity vector that satisfies

ui(t + dt) =

(1 − a dt

T

)ui(t) +

√bkdt

Tξi(t).

Here ξ is a vector version of the random variable used in equation (2.2.16). It has theadditional property that

ξiξj = δij .

a and b are coefficients that will be determined and k is the turbulent kinetic energy, asusual.

i) Derive the evolution equation for the Reynolds stress tensor uiuj .ii) Assume the high Reynolds number formula T = k/ε and determine a and b such that

the Reynolds stress evolution equation is identical to the Rotta model for decayinganisotropic turbulence.

iii) What realizability constraint is implied by the square root in equation (8.5.71)?

Exercise 8.3, Bifurcation:Verify (8.3.35). Substitute constants and plot the bifurcation curve for the LRR model.

Exercise 8.4, Non-linear constitutive equations:Non-linear constitutive equations In constitutive modeling, the coefficients in equation (8.3.29)are regarded as unknown functions of |S| and |W | that can be formulated with completefreedom, provided they are dimensionally consistent. However, useful constraints can be ob-tained by invoking realizability. Consider a general two-dimensional mean flow field wherethe mean rate of strain and mean vorticity tensors can be written according to (8.3.24) andlet

b = 2CµS + B(Ω · S − S ·Ω) + C(S2 − 1/3 |S|2δ)Show that for the flow (8.3.24), the constraints

B2 ≤ [( 2/3 + 1/3cS2)2 − (2CµS)2] /(2ΩS)2and

C ≤ min [1/S2, 3 [ 4/3 − 2Cµ|S|] /S2]ensure that a (quadratically) non-linear model is realizable. S and Ω are understood to benon-dimensionalized by k/ε

27

Solution to Ex. 8.4, For this flow

uiujk

= 2/3

1 0 00 1 00 0 1

+2CµS 1 0 00 −1 0

0 0 0

−2ΩSb 0 1 01 0 0

0 0 0

+cS2 1/3 0 00 1/3 0

0 0 −2/3

The Schwartz inequality

uv2 ≤ u2 v2gives

(2ΩS)2b2 ≤ (2/3 + cS2/3)2 − (2CµS)2Hence

b2 ≤ [(2/3 + 1/3cS2)2 − (2CµS)2] /(2ΩS)2.The condition that u2 and v2 < 2k gives

2/3 + 1/3cS2 ± 2CµS < 2

orc ≤ 3( 4/3 − 2Cµ|S|)/S2

The condition w2 ≥ 0 gives2/3 − 2/3cS2 ≥ 0

orc ≤ 1/S2

A rigorous approach is the show that these conditions ensure that the eigenvalues of uiuj/kare non-negative and less than 2.

28

Chapter 9

PART III

Exercise 9.1, Familiarization with δ:Show that lim→0

∫∞−∞ e

−|x|eikxdx = 2πδ(k) by explicitly evaluating the integral, then apply-ing the definition of the δ-function.

Solution to Ex. 9.1, Just evaluate the integral∫ ∞−∞

e−|x|eikxdx =2�

�2 + k2

and reason: This → 0 as � → 0 if k �= 0; this → ∞ as � → 0 if k = 0; and ∫∞−∞ 2�/(k2+�2)dk =∫∞−∞ 2/(x

2 + 1)dx = 2π. So all the properties of 2πδ(k) are satisfied.

Exercise 9.2, Direction of independence:Let r = (rx, ry, rz) be the components of the separation between two hot-wire anemometers.

Suppose that Rii(r) = e−(r2x+r2y)/L2 , independently of rz. This means that the eddying

motion is independent of the z-axis. What is Φ(kx, ky, kz)? What is the kz-space analog toindependence of z in physical space?

Solution to Ex. 9.2,q2L2

4πδ(kz)e

−(k2x+k2y)L2/4

Exercise 9.3, Integral length scale:The integral length-scale provides a measure of the size of energetic eddies. This length-scalecan be defined as a vector with x1-component

L1 =

∫ ∞0

Rii(x1; x2 = 0, x3 = 0)dx1,

similarly for L2, L3. For instance Ly = 1/2∫∞−∞ Rii(0, y, 0)dy. Show that

q2Ly = π

∫ ∞∫−∞

Φ(kx, 0, kz)dkx dkz

The 1-D energy spectrum in the y-direction is defined by

Θ(ky) =

∫ ∞∫−∞

Φ(kx, ky, kz)dkx dkz

so Θ(0) = q2Ly/π. Often it is easier to measure, or compute, Θ(0) than to obtain L byintegrating the correlation function.

29

Solution to Ex. 9.3,∫ ∞∫−∞

Φ(kx, 0, kz)dkx dkz =

∫ ∞∫−∞

[q2

(2π)3

∫ ∞∫−∞

∫ei(kxx+kzz)Rii(r)dxdydz

]dkx dkz

=q2

2π

∫ ∞∫−∞

∫δ(x)δ(z)Rii(r)dxdydz =

q2

2π

∫ ∞−∞

Rii(0, y, 0)dy = q2Ly/π

Exercise 9.4, One dimensional spectrum:Derive a formula that relates Θij(k1) to the Fourier transform of a two-point correlationfunction.

Solution to Ex. 9.4, By definition

Φ(kx, ky, kz) =q2

(2π)3

∫ ∞∫−∞

∫ei(kxx+kzz)Rii(r)dxdydz

So it is just a matter of playing with the order of integration∫ ∞∫−∞

Φ(kx, 0, kz)dkx dkz =

∫ ∞∫−∞

{q2

(2π)3

∫ ∞∫−∞

∫ei(kxx+kzz)Rii(r)dxdydz

}dkx dkz

=q2

(2π)3

∫ ∞∫−∞

∫ {∫ ∞∫−∞

ei(kxx+kzz)dkx dkz

}Rii(r)dxdydz

=q2

2π

∫ ∞∫−∞

∫δ(x)δ(z)Rii(r)dxdydz

=q2

2π

∫ ∞−∞

Rii(0, y, 0)dy = q2Ly/π

Exercise 9.5, Practice with isotropy:Derive (9.2.35) in physical space by assuming that uiuj(r) is a function of rk and δkl: i) Writethe most general tensoral form and apply incompressibility; ii) Why is the incompressibilityconstraint now ∂riuiuj(r) = 0 = ∂rjuiuj(r)? In your derivation simply obtain the form(9.2.35) with f an unspecified function; you don’t have to relate it to the energy spectrum.

Exercise 9.6, Experimental test of isotropy:The 1-D spectra Θ11(k1) and Θ22(k1) can be measured with a fixed probe. The method is tomeasure frequency spectra and then use Taylor’s hypothesis to replace ω by k1Uc, where Ucis the mean convection velocity measured by the anemometer. Show that if the turbulenceis isotropic then

Θ22(k1) = 1/2 [Θ11(k1) − k1∂k1Θ11(k1)]

30

Where the 1-D spectrum is defined by

Θij(k1) =

∫ ∞∫−∞

Φij(k)dk2 dk3

This relation between spectral components has been used to test experimentally for isotropy.

Solution to Ex. 9.6, Let F (k) = E(k)/4πk4 then, integrating in cylindrical coordi-nates (r2 = k23 + k

22)

Φ22 =

∫ ∞∫−∞

k21+k22F (k)dk2 dk3 =

∫ 2π0

k21

∫ ∞0

F (√

r2 + k21)rdrdθ+

∫ 2π0

∫ ∞0

F (√

r2 + k21)r3drθ

= k21

∫ ∞0

F (√

r2 + k21)2πrdr + 1/2

∫ ∞0

F (√

r2 + k21)2πr3dr

Compare this to 1/2(Φ11(k1) − k1dk1Φ11(k1))

Φ11 =

∫ ∞∫−∞

k23 + k22F (k)dk2 dk3 =

∫ ∞0

F (√

r2 + k21)2πr3dr

One half of this is the second term in Φ22. Differentiating this

∂k1Φ11 = k1

∫ ∞0

drF (√

r2 + k21)2πr2dr = −2k1

∫ ∞0

F (√

r2 + k21)2πrdr

Since drF (√

r2 + k21) = rF′(√

r2 + k21) and dk1F (√

r2 + k21) = k1F′(√

r2 + k21). − 1/2k1times this is the first term in Φ22.

Exercise 9.7, The VonKarman spectrum:What is the proportionality between the length scale L that appears in the VonKarmanspectrum and the 1-D integral length-scale L111? This notation means ‘the integral scale inthe x1-direction of the 1-1 correlation’, L

111 =

∫∞0

u1(x)u1(x + rx)drx.

Solution to Ex. 9.7, Cs − CKε2/3Ls+5/3. ε ∝ q2/t ∝ q3/L → ε ∝ L2/t3.Exercise 9.8, One dimensional spectra:Evaluate the 1-D spectrum, Φ122(k1). Use this to show that the integral scale in the x1direction of the 22-correlation is one half the integral scale in the x1 direction of the 11-correlation. In general the transverse integral scale in isotropic turbulence is one-half thelongitudinal integral scale.

Exercise 9.9, Synthetic spectra:Another way to synthesize an isotropic, incompressible random field is

û =

[ξ − (k · ξ)k|k|2

]√E(k)

4πk2

31

Verify that (9.2.41) has an isotropic spectrum.

32

Chapter 10

Exercise 10.1, Triad interactions:The Navier-Stokes equations contain the term uj∂jui. Write the Fourier transform of thisterm as a convolution integral. Also write it as an integral over wave-number triangles.

Exercise 10.2, Energy decay:Compte-Bellot & Corrsin considered a crude model for the spectrum. Let

E(k) = Csks k < 1/L(t)

E(k) = CKε2/3k−5/3 1/η > k > 1/L(t)

E(k) = 0 k > 1/η

(0.2)

CK is supposed to be a universal constant. Consider the case η ≈ 0. How is Cs related toCK , ε and L(t)? How is q

2 related to ε and L(t)? Assume that ε and q2 follow power-lawdecays. Derive the formula relating the exponent in L ∝ tm to s.

Solution to Ex. 10.2, L ∝ t2/(3+s)Exercise 10.3, Two-point correlations:Consider the two point correlation q2Rii = ui(x)ui(x

′) in homogeneous turbulence. If r =x − x′ then

∂xjRii(r) = −∂x′jRii(r)and ui(x

′)∂xui(x) = ∂x[ui(x)ui(x′)] because x and x′ can be regarded as independent vari-ables. From such considerations, derive a formula relating ε = ν∂iuj∂iuj to the behavior ofthe two-point correlation function, Rii(r), as r → 0.

Solution to Ex. 10.3, ε = −νq2∇2rRii|0Exercise 10.4, Physical space:The spectral evolution equations have a physical space corollary. Correlations and spectraform Fourier transform pairs. Because of this, spectral closure models are called ‘two-point’closures. An equation for the two-point auto-correlation in homogeneous, isotropic turbu-lence, analogous to (10.2.9), is

∂tuiui(r) + 2ν∇2ruiui(r) = −2∂rjujuiu′i(r)

whereujuiu

′i(r) = uj(x)ui(x)ui(x

′)

Derive equation (10.2.16).

Exercise 10.5, A dissipation formula for isotropic turbulence:Streamwise derivatives of streamwise velocity (∂1u1) are the easiest to measure. Show thatthis derivative can be used to infer the rate of dissipation by deriving the formula (∂1u1)2 =

33

ε/15ν for isotropic turbulence. Your derivation must start from the energy spectrum tensor,equation (9.2.23).

This relation can also be derived by invoking the definition of isotropy in physical space,rather than via Fourier space. The general form for an isotropic fourth order tensor is (2.3.35)

∂iuj∂kul = Aδikδjl + Bδilδjk + Cδijδkl

The ‘velocity derivative skewness’ is defined by

Sd = (∂1u1)3/((∂1u1)2)3/2

High Reynolds number experimental values of Sd are about 0.4. In isotropic turbulence theexact equation for dtε contains a term proportional to ν(∂1u1)3 (the self-stretching term).

Show that this term non-dimensionalized by ε2/k is of order R1/2T , where RT = k

2/εν.

Solution to Ex. 10.5,For integral approach, in spherical coordinates

∂1u1∂1u1 =

∫ ∞∫−∞

∫k21

E(k)

4πk2

(1 − k

21

k2

)d3k

=

∫ 2π0

∫ π0

∫ ∞0

E

4π

(1 − cos2 φ) cos2 φ sin φdφdθk2dk

= 2/15

∫ ∞o

Ek2dk = ε/15ν

Symmetry demands that B = C; incompressibility demands that A+B+3C = 0. HenceA = −4C and

∂iuj∂kul = C[δilδjk + δijδkl − 4δikδjl]The coefficient C is related to the streamwise derivative of the streamwise velocity by the1111 component: −2C = (∂1u1)2. Upon contraction, this leads to

ε = ν∂iuj∂iuj = 15ν(∂1u1)2.

Exercise 10.6, Scale similarity:Strict scale similarity would require that E(k) be of the form q2(t)L(t)Ẽ(η) with η = kL(t)and where Ẽ is a non-dimensional, similarity solution. Assume that q2 ∝ t−n and find n byrequiring strict self-similarity in equation (10.2.9).

34

Piston

H(t)

U = 0

U = dH/dt

Figure 2: Schematic for compression of turbulence by a piston.

Chapter 11

Exercise 11.1, Turbulence in a piston; 1-D compression:RDT analysis can be be applied to a uni-directional compression (Hunt, 1977). This modelsthe effect of compression on turbulence in a piston, away from the walls because the analysisis of homogeneous turbulence. Consider the mean flow

U =xdH/dt

H(t)

corresponding to figure 11.9. Show that s1 = H/H0, i.e., that the total strain is just theexpansion ratio. This permits the solution to be written in terms of the physical dimensionH instead of time.

Show that the distorted wavevector is

κ1 =H0H

k1; κ2 = k2; κ3 = k3

and from Cachy’s formula, that

ω̂1 = ω̂01 ; ω̂2 =H0H

ω̂02 ; ω̂3 =H0H

ω̂03

By invoking (11.1.16) obtain a complete solution for the component û1 and obtain the inte-grand in

u21 =

∫ ∞∫−∞

∫û1û∗1d

3k

explicitly. Evaluate the integral for initially isotropic turbulence.

Solution to Ex. 11.1,

u21 =q20β

2

4

[1

β2 − 1 +(β2 − 2) tan−1√β2 − 1

(β2 − 1)3/2]

35

where β = H0/H and β ≥ 1 is assumed. As β → 1 this gives

u21 → 1/3q20β2[1 − 2/5(β2 − 1)].

As β → ∞ one findsu21 → q20βπ/8; u22 = u23 → q20βπ/16.]

Exercise 11.2, Homogeneous shear RDT:Use a two-term series solution to (11.2.31), starting with isotropic turbulence, to obtain theshort-time solutions (11.2.33) for v2 and uv.

36

Chapter 12

PART IV

Exercise 12.1, Energy conserving schemes:Consider the equation

∂tφ + u∂xφ + ∂x(uφ) = 0.

Show that φ2 is conserved. The velocity is stored at the half grid points: x = (j + 1/2)∆x,j = 1, 2, 3 . . . J and φ is stored at the grid points x = j. Consider the discretizations

uδxφ|j = 12

(u(j + 1/2)

φ(j + 1) − φ(j)∆x

+ u(j − 1/2)φ(j) − φ(j − 1)∆x

)

δx(uφ)|j = 1∆x

(u(j + 1/2)

φ(j + 1) + φ(j)

2− u(j − 1/2)φ(j) + φ(j − 1)

2

).

Show that φ2 is conserved for this discretization.

Solution to Ex. 12.1, Multiplying by φ(j) gives

1/2∂tφ2 = (u(j + 1/2)φ(j + 1)φ(j) − u(j − 1/2)φ(j)φ(j − 1))/∆x

= [F (j + 1) − F (j)]/∆x

Exercise 12.2, DNS via Burger’s equation:Solve Burger’s equation (12.2.16) numerically with ν = 0.01 and 0.001, imposing the initialcondition

1/2 +{

e−20x2

+ e−20(x+1)2

+ e−20(x−1)2}

(0.5 + 0.5ξ)

where ξ is a random number between 0 and 1. The flow domain is − 1/2 ≤ x ≤ 1/2 . Applyperiodic boundary conditions: u(1/2 + j∆x) = u(−1/2 + j∆x). Use about 160 grid points.Integrate to a non-dimensional time of 0.12, plotting the solution at t = 0, 0.04, 0.08, 0.12.Looking ahead to §13.1.1: compare the DNS velocity field to its filtered field at 4 equallyspaced times, t = 0, 0.4, 0.8 and 0.12. Try the running average filter (13.1.2) with N = 2and 3, and the Pad’e filter (13.1.4) with α = 0.45.

Note: DNS commonly employs the Runge-Kutta method for time integration. The discrete equations

∂tuj = −u2j+1 − u2j−1

4∆x+

1∆x

(νj+1/2

uj+1 − uj∆x

− νj−1/2 uj − uj−1∆x)

are a system of ordinary differential equations of the form

∂tuj = RHS(ui).

The second order Runge-Kutta method is

up(:) = RHS(u(:),t)u1(:) = u(:)+up(:)∆t/2up(:) = RHS(u1(:),t)u(:) = u(:)+up(:)∆t ; t = t+∆t.

37

At the end, u has been advanced one time step. This is a simple integration method, which facilitates testsof various convection schemes.

Exercise 12.3, Spectral methods:Use fast Fourier transforms (FFT) to evaluate ∂xu

2 and ∂2xu. In the pseudo-spectral methodu2 is evaluated in physical space to avoid convolution sums; derivatives are evaluated inFourier space, as in (12.3.25). Specify u by the initial condition (12.3.26) of the previousproblem. Optional: replace the finite difference discretization used in the previous problemby pseudo-spectral evaluation and solve Burger’s equation.

38

! Solve Burger’s equation by Runge-Kutta integration

! g95 -r8 burger_DNS.f90 -o burger_DNS

!*************************************************************

MODULE common1

REAL :: vnu=0.000,dx

LOGICAL :: lg

END MODULE common1

!*************************************************************

PROGRAM BURGER_EX

USE common1

PARAMETER (N=128)

REAL :: U(N),UF(N),S(N),US(N,30),x(N),t=0.,dt,DTprin,Tprint,CC

INTEGER :: i,itm,icur=1,NT

EXTERNAL RHS

pi = 4.d0*atan(1.)

! WRITE(*,"(a)",ADVANCE="NO")"READ c.f.l. " ; READ(*,*)cc

cc = 0.05

dx = 1.d0/dfloat(N-1)

dt = cc*dx

Tend = 0.2 ; DTprin = Tend/5. ; Tprint = DTprin

NT = Tend/dt + 1.01

!**************** Initial Condition ************************

DO i=1,N

x(i) = -0.5+(i-1)*dx

jj = 1000*rand(i/2)

U(i) = (exp(-20.*x(i)**2)+exp(-20.*(x(i)-1.)**2) &

+exp(-20.*(x(i)+1.)**2))*(0.5+0.5*rand(jj))+.5

ENDDO

U(N) = U(1)

US(:,1) = U(:)

!**************** Integrate in time ************************

time: DO itm = 1,NT

sum=0.

DO i=1,N-1

sum = sum+u(i)**2

ENDDO

IF((itm-1)/5*5.eq.itm-1)PRINT *,t,sum*dx

39

call rk4(N,RHS,t,U,t+dt)

IF(t.gt.Tprint)THEN

Tprint = Tprint+DTprin

icur = icur + 1

US(:,icur) = U(:)

ENDIF

ENDDO time

!********************* Output *****************************

OPEN (1,file="plotfil")

DO i=1,N

WRITE(1,’(12f9.4)’) x(i),(US(i,ic),ic=1,icur)

ENDDO

WRITE(1,’(/a)’) ’ ’

DO ic=1,icur

CALL FILTER(N,US(1,ic),UF)

US(:,ic)=UF(:)

ENDDO

DO i=1,N

WRITE(1,’(12f9.4)’) x(i),(US(i,ic),ic=1,icur)

ENDDO

WRITE(*,"(a,8i3)")"gplot plotfil ",(ic,ic=1,icur)

STOP

END PROGRAM BURGER_EX

!******************************************************

SUBROUTINE RHS(N,t,U,S)

!******************************************************

USE common1

REAL :: U(N),S(N),work(2*N),data(2,N)

INTEGER :: j,jm,jp,N

DO j=1,N

jp = mod(j-1,N-1)+2

jm = N-1-mod(N-j,N-1)

! S(j) = -(U(jp)**2-U(jm)**2)/(4.*dx) & ! central

S(j) = -(U(jp)**2-U(jm)**2)/(6.*dx) -U(j)*(U(jp)-U(jm))/(6.*dx) & ! energy conserv

! S(j) = -U(j)*(U(j)-U(jm))/(2.*dx) & ! upwind

+ (U(jp)-2.*U(j)+U(jm))*vnu/dx**2

ENDDO

RETURN

END SUBROUTINE RHS

40

!******************************************************

SUBROUTINE FILTER(N,U,UF)

!******************************************************

USE common1

REAL :: U(N),Uf(N)

INTEGER :: j,jm,jp,N,NF=3

DO j=1,N

Uf(j) = U(j)

DO i=1,NF-1

jp = mod(j+i-2,N-1)+2

jm = N-1-mod(N-j+i-1,N-1)

Uf(j) = Uf(j)+U(jp)+U(jm)

ENDDO

jp = mod(j+NF-2,N-1)+2

jm = N-1-mod(N-NF+i-1,N-1)

Uf(j) = (Uf(j)+.5*(U(jp)+U(jm)))/(2.*NF)

ENDDO

END SUBROUTINE FILTER

!*********************************************************

! 4th order Runge-Kutta routine *

!*********************************************************

SUBROUTINE rk4(N,RHS,t,y,tend)

REAL :: y(N),yp(N),y1(N),y1p(N),y2(N),y2p(N),y3(N),y3p(N) &

,h,t,tend,tmid

EXTERNAL RHS

h = tend-t

tmid = t+h/2.d0

CALL RHS(n,t,y,yp)

y1 = y+0.5d0*h*yp

CALL RHS(n,tmid,y1,y1p)

y2 = y+0.5d0*h*y1p

CALL RHS(n,tmid,y2,y2p)

y3 = y+h*y2p

CALL RHS(n,tend,y3,y3p)

y = y+h/6.0*(y3p+2.0d0*y2p+2.0d0*y1p+yp)

t = tend

RETURN

END

41

Chapter 13

Exercise 13.1, Filtering in physical space:The top-hat filter is

û =1

∆f

∫ ∆f /2−∆f/2

u(x + ξ) dξ

Let u = sin(kx). Evaluate û and ̂̂u. Letu =

{ |1 − x|, x < 10, |x| ≥ 1.

Evaluate û and compare it to the unfiltered function.

Solution to Ex. 13.1, For the sine function

sin(kx) =2sin(k∆f/2)

k∆fsin(kx); sin(kx) =

[2sin(k∆f/2)

k∆f

]2sin(kx)

For the triangular function, if 1 − ∆f/2 < x < 1 + ∆f/2 averaging will not alter the linearslope. If |x| > 1 + ∆f/2 filtering is over a region where u = 0. Hence

û =

{0 |x| > 1 + ∆f/2

1 − x −1 + ∆f/2 < x < 1 − ∆f/2

When x is within ∆f/2 of 1 or −1 the function is smoothed

û =

1

2∆f

[∆f2

+ 1 − x]2

1 − ∆f/2 < x < 1 + ∆f/21

2∆f

([∆f2

+ x]2

− 1)

−1 − ∆f/2 < x < −1 + ∆f/2

Exercise 13.2, Filter in Fourier space:The Gaussian filter is expressed by the running average

û =1√π∆f

∫ ∞−∞

e−ξ2/∆2f u(x + ξ) dξ

What is Fk of this filter in Fourier space?

Solution to Ex. 13.2,

Fk =1√π∆f

∫ ∞−∞

e−ξ2/∆2f e−ikξ dξ =

e−k2∆2f/4√π∆f

∫ ∞−∞

e−(ξ/∆f +ik∆f/2)2

dξ

42

=e−k

2∆2f/4√π

∫ ∞−∞

ex2

dx = e−k2∆2f/4

Exercise 13.3, Filtered spectrum:Convince yourself that the filtered and unfiltered one-dimensional energy spectra (see page242) are related by

Θ̂11 = FkF∗k Θ11.

On a single graph, plot the filtered and unfiltered spectrum (9.2.28) with p = 17/6 for variousratios of ∆f/L. Use either the top-hat or Gaussian filter.

Solution to Ex. 13.3, ûi(k) = Fku(k) → ûi(k)û∗j(k) = FkF ∗k ui(k)u∗j(k) see (9.2.13)and (9.2.18) to relate this to spectra.

Exercise 13.4, Filtering and vorticity:Let

u = A cos kx sin ky cos kz

v = −A sin kx cos ky cos kzw = 0.

This is called the Taylor-Green vortex. Evaluate the vorticity vector field. Average |u|2 and|ω2| over one period of the sine waves in each direction. Let

A2 =k3

(1 + (kη)2)2.

Filter the velocity and vorticity fields with a filter width equal to 5η. Average |û|2 and |ω̂2|over a period and compare to their unfiltered values.

Exercise 13.5, WALE subgrid model:Show that

SijSji = 1/6(S2S2 + Ω2Ω2) − 2/3Ω2S2 + 2S2ijΩ2ji

where Sij is defined in (13.1.19). Ω2ij = ΩikΩkj , Ω2 = Ω2ii and similarly for S. The Cayley-

Hamilton theorem (2.3.38) with the invariants defined in equation (2.3.40) is needed. Whyis Ω2 ≤ 0? Show that SijSji is zero in parallel shear flow, u(y).

Solution to Ex. 13.5,

|S 2| = |S4| + |Ω4| + 2S2 · Ω2 + 1/3(|S2| + |Ω2|)2 − 2/3(|S2| + |Ω2|)2

= |S4| − 1/3S2S2 + |Ω4| − 1/3Ω2Ω2 + 2S2 · Ω2 − 2/3 |S2||Ω2|where |(·)| denotes the trace of a matrix, not an absolute value. From Cayley-Hamilton

S4 = S · S3 = 1/2 |S3|S + 1/2 |S2|S2

43

Thus|S4| = 1/2S2S2 and |Ω4| = 1/2Ω2Ω2

which gives the desired result. In parallel shear flow Ω2 = −S2 so Ω = −S and |S 2| = 0Exercise 13.6, LES via Burger’s equation:Repeat 12.3, adding the Smagorinsky model (13.1.12) and using a 4 times coarser grid.Select a value of cburger to optimize agreement with the filtered DNS field. Compare the LESsolution to the filtered DNS data at several times.

44