Chapter 1 Governing Equations

Chapter 1

Governing Equations

Fluid flow (turbulent and laminar) are governed by the Navier-Stokes equations.The incompressible Navier-Stokes equations assuming constant fluid properties canbe written as the continuity

∂ui

∂xi

= 0 (1.1)

the momentum

ρ∂ui

∂t+ ρuj

∂ui

∂xj

= − ∂p

∂xi

+∂tij∂xj

(1.2)

and the energy∂θ

∂t+ uj

∂θ

∂xj

= κ∇2θ (1.3)

equations. ui, p and θ are the instantaneous velocity, pressure and temperaturerespectively. ρ is the density of the fluid and tij is the viscous stress tensor definedby

tij = 2µsij (1.4)

where sij is the strain-rate tensor defined as

sij =1

2

(∂ui

∂xj

+∂uj

∂xi

)(1.5)

Exercise 1.1: Show that∂

∂xj

(uiuj) = uj∂ui

∂xj

and thus Eq. (1.2) can be re-written as

ρ∂ui

∂t+ ρ

∂

∂xj

(uiuj) = − ∂p

∂xi

+∂tij∂xj

1

2 CHAPTER 1. GOVERNING EQUATIONS

In the study of turbulence, it is common the decompose velocity, pressure andtemperature into its mean and fluctuating components,

ui = 〈Ui〉+ u′

i

p = 〈P 〉+ p′

(1.6)

θ = 〈Θ〉+ θ′

Note that you can assume that the mean and differentiation operators commute. Insome cases, Reynolds averaging is similar to time averaging.

〈φ(x, t)〉 = φ(x, t) =1

T

∫ t+T

t

φ(x, t)dt.

so

⟨∂φ

∂x

⟩=

1

T

∫ t+T

t

∂φ(x, t)

∂xdt

Since the integral is only over time, one can take the spatial (x) derivative out ofthe integral

⟨∂φ

∂x

⟩=

∂

∂x

1

T

∫ t+T

t

φ(x, t)dt

=∂

∂x〈φ(x, t)〉

3

Exercise 1.2: Substitute Eq. (1.6) into Eqs. (1.1) - (1.3) and taking the meanof the equations, we obtain the following

∂ 〈Ui〉∂xi

= 0 (1.7)

∂ 〈Ui〉∂t

+ 〈Uj〉∂ 〈Ui〉∂xj

= −1

ρ

∂ 〈P 〉∂xi

+∂

∂xj

(ν∂ 〈Ui〉∂xj

−⟨u

′

iu′

j

⟩)(1.8)

∂ 〈Θ〉∂t

+ 〈Uj〉∂ 〈Θ〉∂xj

= κ∇2 〈Θ〉 − ∂

∂xj

⟨u′jθ

′⟩ (1.9)

Note that ⟨∂(uiuj)

∂xj

⟩=

∂

∂xj

〈(uiuj)〉

=∂

∂xj

⟨(〈Ui〉+ u′i)(〈Uj〉+ u′j)

⟩=

∂

∂xj

⟨〈Ui〉〈Uj〉+ 〈Uj〉u′i + 〈Ui〉u′j + u′iu

′j

⟩=

∂

∂xj

⟨〈Ui〉〈Uj〉+ u′iu

′j

⟩=

∂

∂xj

(〈Ui〉〈Uj〉) +∂

∂xj

⟨u′iu

′j

⟩= 〈Uj〉

∂

∂xj

〈Ui〉+∂

∂xj

⟨u′iu

′j

⟩Equations (1.7)-(1.8) are called usually called the Reynolds Averaged Navier-

Stokes equations. Equation (1.9) is the Reynolds averaged equation for energy. Theterm

⟨u

′iu

′j

⟩is usually called the Reynolds stresses. This is because one can rewrite

Eq. (1.8) as

∂ 〈Ui〉∂t


= −1

ρ

∂

∂xj

(〈P 〉 δij +


−⟨u

′

iu′

j

⟩))(1.10)

The first term on the right hand side is the isotropic stress from the mean pressurefield, the second term is the molecular viscous stress and the last term is the ”ap-parent” stress due to the turbulent field. It is the transfer of momentum from theturbulent fluctuating field to the mean momentum field.

The Reynolds stress tensor is a symmetric (i.e.⟨u

′iu

′j

⟩=⟨u

′ju

′i

⟩) second order

tensor. The diagonal components,⟨u

′αu

′α

⟩=⟨u

′2α

⟩, are normal stresses while the

off-diagonal components,⟨u

′ju

′i

⟩for i 6= j, are shear stresses. The turbulent kinetic

energy is usually defined to be half the trace of the Reynolds stress tensor


k =1

2

⟨u

′

iu′

i

⟩=

1

2

(⟨u

′21

⟩+⟨u

′22

⟩+⟨u

′23

⟩)(1.11)

k can be used to to decompose the Reynolds stress tensor⟨u

′

iu′

j

⟩= aij +

2

3kδij (1.12)

into an isotropic part ((2/3)kδij) and an deviatoric part (aij). By rearranging, Eq.(1.12), one can arrive at the definition of aij,

aij =⟨u

′

iu′

i

⟩− 2

3kδij (1.13)

We can now write Eq. (1.8) as

∂ 〈Ui〉∂t


= −1

ρ

∂

∂xi

(〈P 〉+

2

3k

)+

∂

∂xj


− aij

)(1.14)

The above equation shows that only the deviatoric component, aij, is responsiblefor transporting momentum because the isotropic component of the Reynolds stress,(2/3)k, can be thought of as a modification to the mean pressure, 〈P 〉.

Sometimes, it is also useful to define the normalized anisotropy tensor bij

bij =aij

2k(1.15)

Exercise 1.3: The following exercise was taken from [3]. Each of the followingsecond order tensor is incorrect. Why?

1. ⟨u

′

iu′

j

⟩=

0.5 0.1 00.1 0.3 0.10 0.1 −0.1

2. ⟨

u′

iu′

j

⟩=

0.5 0.1 0−0.2 0.3 0.1

0 0.1 0.1

3.

aij =

1.8 0.2 00.2 −1.6 0.10 0.1 −0.3

5

Exercise 1.4: An irrotational fluid will satisfy the following equations

∂ui

∂xj

− ∂uj

∂xi

= 0 (1.16)

1. Show that for an irrotational fluid,⟨u

′

i

(∂u

′i

∂xj

−∂u

′j

∂xi

)⟩= 0 (1.17)

2. Thus show that

⟨u

′

i

(∂u

′i

∂xj

−∂u

′j

∂xi

)⟩=

∂

∂xj

(1

2

⟨u

′

iu′

i

⟩)− ∂

∂xi

(⟨u

′

iu′

j

⟩)= 0 (1.18)

From the above equation, it is easy to see that

∂

∂xi

⟨u

′

iu′

j

⟩=

∂k

∂xj

(1.19)

Thus the Reynolds stress has the same effect as the isotropic stress in anirrotational fluid. Note that k can be absorbed into the pressure term ofthe governing equations. So for an irrotational fluid, the governing equationis closed. We need not need to model aij in order to close the mean flowequations.

The main problem with turbulence modelling is to be able to relate the Reynoldsstress term

⟨u

′iu

′j

⟩to the mean flow. One can get an evolution equation for

⟨u

′iu

′j

⟩=

τij.

∂τij∂t

+ 〈Uk〉τij∂xk

= −τik∂ 〈Uj〉∂xk

− τjk∂ 〈Ui〉∂xk

+ εij

−Πij +∂

∂xk

[ν∂τij∂xk

+ Cijk

](1.20)

where

Πij =

⟨p

′

ρ

(∂u

′i

∂xj

+∂u

′j

∂xi

)⟩(1.21)

εij = 2ν

⟨∂u

′i

∂xk

∂u′j

∂xk

⟩(1.22)


ρCijk = ρ⟨u

′

iu′

ju′

k

⟩+⟨p

′u

′

i

⟩δjk +

⟨p

′u

′

j

⟩δik (1.23)

Details of the derivation can be found in [6]. The important thing to note about Eq.(1.20) is that it involves third order correlations which are unknown. Herein liesthe closure problem of turbulence. Due to the nonlinear nature of the Navier-Stokesequations, we generate more and more unknowns as we try to obtain equations forthe higher order correlations.

If we assume that turbulence is statistically stationary, then Reynolds averagingis similar to time averaging.

〈f(x, t)〉 = f(x, t) = limT→∞

∫ t+T

t

f(x, t)dt.

If we assume that turbulence is homogeneous, then Reynolds averaging is similar toperforming spatial averaging.

〈f(x, t)〉 = [f(x, t)] = limV→∞

∫V

f(x, t)dV.

Note that f(x, t) is independent of time and [f(x, t)] is independent of any spatialvariation.

In this course, we will assume that turbulence is statistically stationary so wewill mainly be using time averages. Thus the Reynolds averaged equations will beof the form

∂Ui

∂xi

= 0 (1.24)

Uj∂Ui

∂xj

= −1

ρ

∂P

∂xi

+∂

∂xj

(ν∂Ui

∂xj

− u′iu

′j

)(1.25)

∂Θ

∂t+ Uj

∂Θ

∂xj

= κ∇2Θ (1.26)

The above 5 equations (remember that Eq. (1.25) is actually 3 equations) contains

11 unknowns (U1, U2, U3, P , u′1u

′1, u

′1u

′2, u

′1u

′3, u

′2u

′2, u

′2u

′3, u

′3u

′3 and Θ). So it is not

possible to solve the equations. This is the classical closure problem in turbulence.The “science” of turbulence modelling is the methodology whereby one tries toexpress the Reynolds stress u

′iu

′j with expressions involving Ui.

1.1 Gradient diffussion (turbulent viscosity) hy-

pothesis

In both the mean scalar and mean momentum equations, there are terms that needto be modelled. In the scalar equation, in order to close the equation, one needs to

1.2. MODELLING OF REYNOLDS SHEAR STRESS 7

model the scalar flux⟨u

′jθ

′⟩and in the momentum equation, one needs a model for

the deviatoric Reynolds stress term aij.Let’s look at the scalar flux term first. The Gradient diffusion hypothesis says

that the scalar flux can be modelled as⟨u

′

jθ′⟩

= −κT∂ 〈Θ〉∂xj

(1.27)

κT is usually called the turbulent diffusivity. It is a positive quantity and it is usuallya function of space and time. Equation (1.27) says that the scalar flux vector

⟨u

′θ

′⟩is in the direction of the mean scalar gradient −∇〈Θ〉. This is not a very goodassumption because it has been shown by [5] that the angle between

⟨u

′θ

′⟩and

−∇〈Θ〉 can be as big as 65o. Note that this assumption is analogous to Fourier’slaw of heat conduction. Substituting Eq. (1.27) into Eq. (1.9) will give a closed setof equation for the mean scalar 〈Θ〉

∂ 〈Θ〉∂t

+ 〈Uj〉∂ 〈Θ〉∂xj

=∂

∂xj

(κ+ κT )∂ 〈Θ〉∂xj

(1.28)

Similarly, for the momentum equation, it is possible to model the deviatoricstress tensor, aij, as

aij = −νT

(∂ 〈Ui〉∂xj

+∂ 〈Uj〉∂xi

)= −2νT 〈Sij〉 (1.29)

Substituting Eq. 1.29 into Eq. (1.14), it is possible to obtain the followin equa-tion for the mean momentum field

∂ 〈Ui〉∂t


= −1

ρ

∂

∂xi

(〈P 〉+

2

3k

)+

∂

∂xj

(ν + νT )∂ 〈Ui〉∂xj

(1.30)

The above is a closed set of equation which can be solved to obtain the mean velocityfield.

1.2 Modelling of Reynolds Shear Stress

In relating the Reynolds stress the the mean velocity, it is common to draw theanalogy with the viscous stress in a Newtonian fluid

− (τij + Pδij) /ρ = −2ν

(∂ui

∂xj

+∂uj

∂xi

)(1.31)

Thus the Reynolds stress is assumed to be related to the mean velocity gradients as(see [3])


u′iu

′j −

2

3kδij = νT

(∂Ui

∂xj

+∂Uj

∂xi

)(1.32)

where k is the turbulent kinetic energy

k = u′iu

′i

and νT is called the eddy (turbulent) viscosity. This very simplified assumption (Eq.1.32) is usually referred to as the Boussinesq eddy-viscosity approximation ([6]). Inorder to close the RANS equations, all that remains is to obtain an appropriateexpression for the turbulent viscosity, νT (x, t). νT is usually written as a product ofa velocity and time scales ([3])

νT = l∗v∗ (1.33)

Note that if we assume that the flow is statistically stationary, then l∗ and v∗ are ingeneral a function of space and thus νT is only a function of space alone.

Turbulence models based on the Boussinesq eddy-viscosity approximation canbroadly be classified into

1. Algebraic models

2. Two-Equation models.

In algebraic models, l∗ is dependent on the flow that you are solving. In One-Equation and Two-Equation models, transport equations need to be solved in orderto provide an approximation for l∗ and v∗.

1.2.1 Algebraic models

One of the more popular algebraic models is the mixing length model. In the mixinglength model, we express

l∗ = lmix v∗ = lmix

(2SijSij

)1/2

where

Sij =1

2

(∂U i

∂xj

+∂U j

∂xi

).

Substituting into Eq. (1.33) gives the following expression for the eddy viscosity forthe mixing length model

νT = l2mix

(2SijSij

)1/2(1.34)

1.3. SPECIFIC FLOWS 9

There is no general expression for the mixing length, lmix, that is valid for allflows. For different flows, you have to specify a different expression for lmix in orderto obtain good agreement with experimental observations (see later). Therefore,algebraic models are not useful for practical applications.

1.2.2 Two-Equation models

Arguably, the most popular of the two-equation models is the k-ε model. In the k-εmodel, we express

l∗ = Cµk3/2

εv∗ = k1/2

Substituting into Eq. (1.33) gives the following expression for the eddy viscosity forthe k-ε model

νT = Cµk2

ε(1.35)

where Cµ = 0.09 is a constant. In general, the turbulent kinetic energy, k, andits dissipation rate, ε are functions of space and time. We would have to solve thefollowing partial differential equations to obtain the values of k and ε

ρ∂k

∂t+ ρU j

∂k

∂xj

= ρτij∂U i

∂xj

− ρε+∂

∂xj

[(µ+ µT/σk)

∂k

∂xj

](1.36)

ρ∂ε

∂t+ ρU j

∂ε

∂xj

= ρCε1ε

kτij∂U i

∂xj

− ρCε2ε2

k+

∂

∂xj

[(µ+ µT/σε)

∂ε

∂xj

](1.37)

where τij = u′iu′j is given by Eq. (1.32). The model constants are Cε1 = 1.44,

Cε2 = 1.92, σk = 1.0 and σε = 1.3. Thus the k-ε model is more expensive becauseyou need to solve two more equations in order to get the values of k and ε in yourflow field. However, it is believed that the constants in Eqs. (1.37) and (1.37) areuniversal across all possible flow configurations. Hence it can be used for all flowsand the k-ε model is the most widely used turbulence model in commercial CFDpackages.

1.3 Specific Flows

There are a few basic flows which are of interest in engineering. They are

• Free shear flows

– Mixing layer


– Wakes

– Jets

• Boundary Layer

• Channel

Free shear flows are flows which do not have a boundary. The mean flows areusually two-dimensional and usually the gradients in the streamwise (x) direction issmall compared to the gradient in the cross-stream (y) direction.

Exercise 1.5:

(a) Starting from Eq. (1.25), convince yourself that the two-dimensional RANSequations can be written as

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x+ ν

(∂2U

∂x2+∂2U

∂y2

)− ∂

∂xu′u′ − ∂

∂yu′v′ (1.38)

U∂V

∂x+ V

∂V

∂y= −1

ρ

∂P

∂y+ ν

(∂2V

∂x2+∂2V

∂y2

)− ∂

∂xu′v′ − ∂

∂yv′v′ (1.39)

(b) By assuming that V and all streamwise gradients are small and integratingfrom a crosstream y location to ∞, show that Eq. (1.39) can be simplified to

P (x, y)

ρ=P0(x)

ρ− v′v′ (1.40)

where P0(x) is the free stream pressure at y = ∞ . Thus

1

ρ

∂P (x, y)

∂x=

1

ρ

dP0(x)

dx− ∂v′v′

∂x(1.41)

(c) Substitute Eq. (1.41) into Eq. (1.38) and show that it can be simplified to be

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P0

∂x+ ν

(∂2U

∂y2

)− ∂

∂yu′v′ (1.42)

1.3.1 Free Shear Flows

The mean velocity profiles for a mixing layer, wakes and jets are shown in Fig. 1.1.Let us now look at how we can use the RANS equations in free shear flows. For these


Flow

Flow

Meanvelocityprofileof a jet

Meanvelocityprofileof a wake

FlowmagnitudeU1

FlowmagnitudeU2

Meanvelocityprofileof a mixinglayer

Figure 1.1: Schematic of the mean velocity profiles of jets, wakes and mixing layers.


flows, the free stream is uniform, hence there is no pressure gradient at infinity, i.e.dP0/dx = 0. Thus for free shear flows, the equations of motions can be simplified to

∂U

∂x+∂V

∂y= 0 (1.43)

U∂U

∂x+ V

∂U

∂y=∂τxy

∂y(1.44)

where

τxy = ν∂U

∂y− u′v′

We need to approximate −u′v′. This term is usually called the turbulent shearstress.

1.3.2 Boundary layer flows

Boundary layer flows are semi-infinite, i.e. the 0 ≤ y ≤ ∞. These flows are usuallydriven by some kind of pressure gradient dP0(x)/dx at infinity. Thus the governingequation for boundary layer flows given by

∂U

∂x+∂V

∂y= 0 (1.45)

U∂U

∂x+ V

∂U

∂y= −dP0

dx+ ν

∂2U

∂y2− ∂u′v′

∂y(1.46)

1.3.3 Fully Developed Channel Flow

Channel flow is the flow between two infinite plates. The cross-stream domain is0 < y < H where H is the channel height. In fully developed channel flow, themean velocities is assumed to be not changing in the x direction. This means thatthe continuity equation simplifies to

dV

dy= 0 (1.47)

Since V = 0 at the walls of the channel, Eq. (1.47) implies that V (y) = 0.The lateral momentum equation Eq. (1.39) becomes

−1

ρ

∂P

∂y+− ∂

∂yv′v′ = 0

Integrating with respect to y gives


∫ y

0

1

ρ

∂P

∂y′dy′ +

∫ y

0

∂

∂yv′v′dy′ = 0

1

ρ

(P (x, y)− P (x, 0)

)+ v′v′(y) = 0 (1.48)

Remember that the flow is assumed to be fully developed so that all velocitycomponents (and their fluctuations) are assumed to be functions of y only. Ingeneral pressure can be both functions of x and y. P (x, 0) is the pressure at thechannel wall which we can define as

P (x, 0) = Pw(x)

If we substitute the above into Eq. (1.48) and differentiate with respect to x gives

∂

∂x

(1

ρ

(P (x, y)− P (x, 0)

)+ v′v′(y)

)= 0

1

ρ

(∂P (x, y)

∂x− ∂Pw(x)

∂x

)+∂v′v′(y)

∂x= 0

∂P (x, y)

∂x=

dPw(x)

dx(1.49)

Equation (1.49) implies that the pressure gradient is uniform across the whole flow([3]). Substitute Eq. (1.49) into Eq. (1.38) and ignoring the x derivatives (becausewe are assuming fully developed flow) and V = 0 gives

1

ρ

dPw

dx=

∂

∂y

(ν∂U

∂y− u′v′

)dPw

dx=

dτ

dy(1.50)

where τ is the total (laminar and turbulent) shear shear stress defined to be

τ = ρν∂U

∂y− ρu′v′

From Eq. (1.50) it should be clear that since dPw/dx is only a function of x and τis only a function of y, that dPw/dx and dτ/dx must be a constant.

In order to solve Eq. (1.50), note that

τw = τ(y = 0)

= ρν∂U

∂y(y = 0)− ρu′v′(y = 0)

= ρν∂U

∂y(y = 0) (1.51)


Since the problem is antisymmetric about the mid plane of the channel of height H,

τ(y = H) = ρν∂U

∂y(y = H)− ρu′v′(y = H)

= −ρν ∂U∂y

(y = 0)

= −τw (1.52)

Integrating Eq. (1.50) from y = 0 to y = H and making use of Eq. (1.52) gives

∫ y=H

y=0

dPw

dxdy =

∫ y=H

y=0

dτ

dydy

HdPw

dx= −2τw

dPw

dx= − 2

Hτw = constant (1.53)

Substituting Eq. (1.53) into Eq. (1.50) and integrating to y gives

∫ y′=y

y′=0

− 2

Hτwdy

′ =

∫ y′=y

y′=0

dτ(y′)

dy′dy′

− 2

Hτwy = τ(y)− τw

τ(y) = τw

(1− 2y

H

). (1.54)

Chapter 2

Analysis

In this chapter, analysis will be carried out on the governing equations for basicturbulent flows. In many of the problems, a similarity solution , in terms of thesimilarity variable, η, will be sought. We will need to transform the equations from(x, y) space to (x, η) space where

η = y/δ(x)

With this change of variables, we will need to see how the partial derivatives withrespect to x and y transform.(

∂

∂x

)y

=

(∂x

∂x

)y

(∂

∂x

)η

+

(∂η

∂x

)y

(∂

∂η

)x

(2.1)

(2.2)

Exercise 2.1: Show that (∂η

∂x

)y

= −δ′(x)

δ(x)η (2.3)

Substituting Eq. (2.3) back into Eq. (2.1) gives(∂

∂x

)x

=

(∂

∂x

)η

− δ′(x)

δ(x)η

(∂

∂η

)η

(2.4)

Similarly, one can show that(∂

∂y

)y

=

(∂x

∂y

)x

(∂

∂x

)η

+

(∂η

∂y

)x

(∂

∂η

)x

=

(∂η

∂y

)x

(∂

∂η

)x

=1

δ(x)

(∂

∂η

)x

(2.5)

15

16 CHAPTER 2. ANALYSIS

2.1 The far wake

The mean velocity of a wake is symmetrical about the x axis. So we only need tosolve for the domain 0 ≤ y ≤ ∞. The boundary conditions for the problem is

U(x, y) = U∞

and∂U(x, y)

∂y= 0

when y = ∞ and y = 0 respectively. Following [4] we find the solution to Eq. (1.43)and (1.44) using first linearising the equations by assuming a solution of the form

U = U∞ − u(x, y)

V = 0− v(x, y)

Assuming that U∞ >> u and U∞ >> v solution that needs to be solved for thewake is given by

U∞∂u

∂x= −∂τxy

∂y

Using the algebraic mixing length model, the governing equation becomes

U∞∂u

∂x=

∂

∂y

((ν + νT )

∂u

∂y

).

where νT = l2mix|∂u/∂y|. For turbulent flows, νT >> ν, thus the above equation canbe written as

U∞∂u

∂x=

∂

∂y

(l2mix

∣∣∣∣∂u∂y∣∣∣∣ ∂u∂y

). (2.6)

The mixing length, lmix, is usually written as

lmix = αδ(x) (2.7)

where δ(x) is the width of the wake. Substitute Eq. (2.7) into Eq. (2.6) will give

U∞∂u

∂x=

∂

∂y

(α2δ(x)2

∣∣∣∣∂u∂y∣∣∣∣ ∂u∂y

). (2.8)

The boundary conditions for a plane wake are given by

u→ 0 as y →∞ (2.9)

and

2.1. THE FAR WAKE 17

∂u

∂x= 0 at y = 0 (2.10)

Thus, the unknowns in this problems are α, δ(x) and u. In order to find thesequantities, one would need to solve Eq. (2.8) subject to the boundary conditionsspecified by Eqs. (2.9) and (2.10). The following solution shown below was extractedfrom (Wilcox [6]). To obtain a solution, assume a solution of the form

u(x, y) = u0(x)F (η). (2.11)

η is the similarity variable

η =y

δ(x)(2.12)

and u0(x) is a velocity scale function which is yet to be determined. With theintroduction of all these variables, we need to find the function u0(x) and F (η) inorder to determine u(x, y). Before we can substitute Eqs. (2.11) and (2.12) into Eq.(2.8) observe that

(∂

∂x

)y

=

(∂x

∂x

)y

(∂

∂x

)η

+

(∂η

∂x

)y

(∂

∂η

)x

=

(∂

∂x

)η

− y

δ(x)2

dδ(x)

dx

(∂

∂η

)x

=

(∂

∂x

)η

− ηδ′(x)

δ(x)

(∂

∂η

)x

(2.13)

(∂

∂y

)x

=

(∂x

∂y

)x

(∂

∂x

)η

+

(∂η

∂y

)x

(∂

∂η

)x

=1

δ(x)

(∂

∂η

)x

(2.14)

Using Eqs (2.13) and (2.14), observe that

(∂u

∂x

)y

=

(∂u

∂x

)η

− ηδ′(x)

δ(x)

(∂u

∂η

)x

= F (η)du0(x)

dx− η

δ′(x)

δ(x)u0(x)

dF (η)

dη

= F (η)u′0(x)− ηδ′(x)

δ(x)u0(x)F

′(η) (2.15)


(∂u

∂y

)x

=1

δ(x)

(∂u

∂η

)x

=u0(x)

δ(x)

dF

dη

=u0(x)

δ(x)F ′(η) (2.16)

Substituting Eqs. (2.15) and (2.16) into Eq. (2.8) gives

U∞

(F (η)u′0(x)− η

δ′(x)

δ(x)u0(x)F

′(η)

)=

1

δ(x)

∂

∂η

(α2δ(x)2

∣∣∣∣u0(x)

δ(x)F ′(η)

∣∣∣∣ u0(x)

δ(x)F ′(η)

).

Remembering that u0(x) and δ(x) are both positive quantities, the above equationcan be simplified to

U∞

(F (η)u′0(x)− η

δ′(x)

δ(x)u0(x)F

′(η)

)=

u20(x)α

2

δ(x)

d

dη(|F ′(η)|F ′(η))(

U∞u′0(x)δ(x)

u20(x)

)F (η)−

(U∞δ

′(x)

u0(x)

)ηF ′(η) = α2 d

dη(|F ′(η)|F ′(η)) . (2.17)

In order for Eq. (2.17) to have a similarity solution, all the terms in brackets on theleft hand side must be constants, i.e.

U∞u′0(x)δ(x)

u20(x)

= a1 = constant (2.18)

U∞δ′(x)

u0(x)= a2 = constant (2.19)

We would like to use Eqs. (2.18) and (2.19) to obtain δ(x) and u0(x). However, itturns out that this cannot be done by only using Eqs. (2.18) and (2.19). In orderto solve for δ(x) and u0(x), one needs to use the following integral constraint

D

2=

∫ ∞

0

ρU(U∞ − U)dy

=

∫ ∞

0

ρ(U∞ − u)udy

By neglecting u2, the above equation can be written as


D

2=

∫ ∞

0

ρU∞udy

=

∫ ∞

0

ρU∞u0(x)F (η)δ(x)dη

D

2ρU∞u0(x)δ(x)=

∫ ∞

0

F (η)dη

Hence, we can say that

D

2ρU∞u0(x)δ(x)= 1 (2.20)

Note that we could have introduced another constant here but it is possible to justsimply absorb the constant into the definition of δ(x). From Eqs. (2.18), (2.19) and(2.20), it is possible to solve for u0(x) and δ(x). Rearranging Eq. (2.20) will give

u0(x) =D

2ρU∞δ(x)(2.21)

Substitute into Eq. (2.19) gives

U∞δ′(x)2ρU∞δ(x)

D= a2

dδ(x)

dx=

a2D

δ(x)2ρU2∞

δ(x)dδ(x) =a2D

2ρU2∞dx

Integrating both sides will give

δ(x)2

2=

a2D

2ρU2∞x

δ(x) =

√a2D

ρU2∞x (2.22)

From experimental data (see [4]) suggest that

δ(x) ≈ 0.805

√a2D

ρU2∞x (2.23)

Comparing Eqs. (2.22) and (2.23) will give you


a2 = 0.8052

= 0.648 (2.24)

Substituting back into Eq. (2.21) will give you

u0(x) =D

2ρU∞

√ρU2

∞x

a2D

=1

2

√D

a2ρx(2.25)

In order to obtain a relationship between a1 and a2, divide Eq. (2.18) by Eq. (2.19)and use Eq. (2.21). This gives

a1

a2

=δ(x)

δ′(x)

u′0(x)

u0(x)

=δ(x)

δ′(x)

(D

2ρU∞

)−δ′(x)δ2(x)

(2ρU∞δ(x)

D

)= −1 (2.26)

At this point, we have u0(x) (Eq. 2.25) and δ(x) (Eq. 2.22). The only quantityleft to find is F (η). We can find F (η) by substituting Eqs. (2.26), (2.18) and (2.19)back into Eq. (2.17) to give

−a2F (η)− a2ηF′(η) = α2 d

dη(|F ′(η)|F ′(η)) (2.27)

Note that F (η) is a monotonically decreasing function. It has a maximum value ofη = 0 and a minimum value of zero when η = ∞. Thus F ′(η) < 0 and |F ′(η)| =−F ′(η). So Eq. (2.27) can be written as

−a2F (η)− a2ηF′(η) = α2 d

dη(−F ′(η)F ′(η))

a2d

dη(ηF (η)) = α2 d

dη(F ′(η)F ′(η))

Integrating both sides will give

a2ηF (η) = α2F ′(η)F ′(η) (2.28)

Note that the constant of integration is zero because the symmetry of the problemrequires that F ′(η) = 0 when η = 0. Continuing from Eq. (2.28)


F ′(η) = −√a2

α

√ηF (η)

The equation above is separable. So

F−1/2dF = −√a2

αη1/2dη

F 1/2 = −1

3

√a2

αη3/2 + C

F =

(C − 1

3

√a2

αη3/2

)2

(2.29)

The function F (η) has a maximum value when η = 0 and slowly decreases to negativeinfinity when η = ∞. But the boundary condition requires that F (∞) = 0. Hence,we will use the expression for F (η) only for 0 ≤ η ≤ 1. For η > 1, we will defineF (η) = 0. This will satisfy the boundary condition F (∞) = 0. In order to ensurethat the function is continuous, we must set F (η) = 0 when η = 1. Substitutingthis into Eq. (2.29) gives

0 =

(C − 1

3

√a2

α1

)2

C =

√a2

3α

Substituting back into Eq. (2.29) gives

F (η) =

{(a2/(9α

2))(1− η3/2

)2if 0 ≤ η ≤ 1

0 otherwise(2.30)

A relationship between α and a2 can be found by implemeting the integral con-straint


1 =

∫ ∞

0

F (η)dη

=

∫ 1

0

F (η)dη

=a2

9α2

∫ 1

0

(1− η3/2)2dη

=a2

9α2

∫ 1

0

(1− 2η3/2 + η3)dη

=a2

9α2

[η − 4

5η5/2 +

1

4η4

]1

0

=a2

9α2

(1− 4

5+

1

4

)α =

√a2

20

Using Eq. (2.24) will give a numerical value of α,

α =

√0.648

20= 0.18 (2.31)

Using Eqs. (2.11), (2.25), (2.30) will give (for 0 ≤ η ≤ 1)

u(x, y) = u0F (η)

=1

2

√D

a2ρxF (η)

=1

2

√D

a2ρx

a2

9α2

(1− η3/2

)2=

1

2

√a2

9α2

√D

ρx

(1− η3/2

)2= 1.38

√D

ρx

(1− η3/2

)2Thus

U(x, y) = U∞ − u(x, y)

= U∞ − 1.38

√D

ρx

(1− η3/2

)2(2.32)

2.2. MIXING LAYER 23

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

10.

805

η

0.45 F(η)

Figure 2.1: Comparison of prediction of mixing layer model with experimental datafor a plane wake. ◦ experimental data, theoretical prediction.

A plot of F (η) versus η is shown in Fig. 2.1. This plot compares with avail-able experimental data and it shows that the model prediction compares well withexperimental data.

2.2 Mixing layer

In the mixing layer, the only mean velocity gradient that is important is ∂U/∂y.We will now analyse the mixing layer with the mixing length model

Exercise 2.2: Show that Eq. (1.34) reduces to

νT = l2mix

∣∣∣∣∂U∂y∣∣∣∣ (2.33)

Using Eq. (2.33) and assuming that ν << νT , the governing equation for theturbulent mixing layer is

∂U

∂x+∂V

∂y= 0 (2.34)


U∂U

∂x+ V

∂U

∂y=

∂

∂y

(l2mix

∣∣∣∣∂U∂y∣∣∣∣ ∂U∂y

)(2.35)

the boundary conditions are

U(x, y) = U1 as y → +∞ (2.36)

U(x, y) = U2 as y → −∞ (2.37)

Since the flow is incompressible, we can express the velocity components in termsof streamfunction

U =∂ψ

∂y(2.38)

and

V = −∂ψ∂x

(2.39)

Substituting Eqs. (2.38) and (2.39) into Eqs. (2.35) - (2.37) (remembering that onecan remove the modulus sign because ∂U/∂x ≥ 0 for mixing layer if you assumethat U1 > U2) gives

∂ψ

∂y

∂2ψ

∂x∂y− ∂ψ

∂x

∂2ψ

∂y2=

∂

∂y

(l2mix

(∂2ψ

∂y2

)2)

(2.40)

∂ψ

∂y= U1 as y → +∞ (2.41)

∂ψ

∂y= U2 as y → −∞ (2.42)

We will seek a similarity solution to Eq. (2.40) of the form

ψ(x, y) = ψ0(x)F (η) (2.43)

where

η =y

δ(x)

With this change of variables, some of the derivatives can be expressed as

∂ψ

∂y=

1

δ(x)

∂ψ

∂η

=ψo(x)

δ(x)F ′(η)


∂ψ

∂x=

∂ψ

∂x− δ′(x)

δ(x)η∂ψ

∂η

= F (η)ψ′o(x)−δ′(x)

δ(x)ηF ′(η)ψo(x)

∂2ψ

∂y2=

∂

∂y

∂ψ

∂y

=∂

∂y

1

δ(x)

∂ψ

∂η

=1

δ(x)

∂

∂η

(1

δ(x)

∂ψ

∂η

)=

1

δ2(x)

∂2ψ

∂η2

=1

δ2(x)ψ0(x)F

′′(η)

∂2ψ

∂x∂y=

∂

∂x

∂ψ

∂y

=

(∂

∂x

)y

(1

δ(x)

∂ψ

∂η

)=

(∂

∂x

)η

(1

δ(x)

∂ψ

∂η

)− δ′(x)

δ(x)η∂

∂η

(1

δ(x)

∂ψ

∂η

)=

(∂

∂x

)η

(1

δ(x)ψ0(x)F

′(η)

)− δ′(x)

δ(x)η∂

∂η

(1

δ(x)ψ0(x)F

′(η)

)=

(− δ

′(x)

δ2(x)ψ0(x) +

ψ′0(x)

δ(x)

)F ′(η)− δ′(x)

δ(x)η

(1

δ(x)ψ0(x)F

′′(η)

)=

(− δ

′(x)

δ2(x)ψ0(x) +

ψ′0(x)

δ(x)

)F ′(η)− δ′(x)ηψ0(x)

δ2(x)F

′′(η)

As with all free shear layer flows, we will assume that lmix = αδ(x)

∂

∂y

(l2mix

(∂2ψ

∂y2

)2)

=1

δ(x)

∂

∂η

(l2mix

1

δ4(x)ψ2

0(x)(F

′′)2)

=1

δ(x)

∂

∂η

(α2δ2(x)

1

δ4(x)ψ2

0(x)(F

′′)2)

=α2ψ2

0(x)

δ3(x)

∂

∂η

(F

′′)2


Exercise 2.3: Substitute Eq. (2.43) into all the terms in Eq. (2.40) and showthat(

−δ′(x) +

ψ′0(x)

ψ0(x)δ(x)

)(F

′(η))2

−(ψ

′0(x)

ψ0(x)δ(x)

)F (η)F ′′(η) = α2 ∂

∂η

(F

′′(η))2

(2.44)

In general, the coefficients of all terms on the left hand side of Eq. (2.44) arefunctions of x. The coefficient on the right hand side of Eq. (2.44) is a constant. Thecondition for existence of the similarity solution is that all coefficients be independentof x ([6]).

Exercise 2.4:

(a) Applying the boundary condition in Eq. (2.41) and setting

limη→∞

d

dηF (η) = 1, (2.45)

show that

ψ0(x)

δ(x)= U1 (2.46)

(b) Using the results above, show that the boundary condition at −∞ in the (x, η)space can be written as

limη→−∞

d

dηF (η) =

U2

U1

(2.47)

(b) Show that similarity solution exist for Eq. (2.44) provided that

ψ0(x) = AU1x (2.48)

δ(x) = Ax (2.49)

where A is a constant.

(c) Substitute Eq. (2.48) and (2.49) into Eq. (2.44) and show that you will getthe following linear ordinary differential equation

2α2d3F (η)

dη3+ AF (η) = 0 (2.50)

Equation (2.50) is a third order linear ordinary differential equation. Since it is thirdorder ODE, we would need three boundary conditions. Thus far the only boundary


Figure 2.2: Comparison of prediction of mixing layer model with experimental data.Figure taken from [6]

conditions available are Eqs. (2.45) and (2.47). Remember that the absolute valueof the streamfunction is not important. Thus we can just set ψ(x, y = 0) = 0. Thiswould lead us to (see Eq. (2.43))

F (0) = 0 (2.51)

We can use elementary numerical methods to obtain a solution to Eq. (2.50)with boundary conditions given by Eqs. (2.45) (2.47) and (2.51). The numericalsolution has been obtain by [6]. In his analysis, and he has found that the best fitwith experimental data of [2] occur when

A = 0.247 and α = 0.071

This is shown in Fig. 2.2


2.2.1 Numerical method

Instead of the partly analytical method outline in the previous section, one can alsoattempt to solve Eqs. (2.34) and (2.35) using numerical method. First transform theindependent variables from (x, y) to (x, ψ(x, y)) where ψ(x, y) is the streamfunctiondefined by

∂ψ

∂x= −V (2.52)

∂ψ

∂y= U (2.53)

This transformation is called the von Mises transformation. The derivativeoperators

∂

∂x=

∂

∂x+∂ψ

∂x

∂

∂ψ

=∂

∂x− V

∂

∂ψ(2.54)

∂

∂y=

∂ψ

∂y

∂

∂ψ

= U∂

∂ψ(2.55)

when we use Eqs. (2.52) and (2.53). If we substitute Eqs. (2.54) and (2.55) intoEq. (2.35) one will obtain

U

(∂U

∂x− V

∂U

∂ψ

)+ V

(U∂U

∂ψ

)= U

∂

∂ψ

(νTU

∂U

∂ψ

)U∂U

∂x= U

∂

∂ψ

(νTU

∂U

∂ψ

)∂U

∂x=

∂

∂ψ

(νTU

∂U

∂ψ

)(2.56)

Equation (2.56) is a simple diffusion (parabolic) equation that one can use standardnumerical method to solve.

2.3 Channel Flow

Predicting the near wall velocity profile is one of the most challenging task facingturbulence modellers today. The velocity profile close to a smooth wall is shown in

2.3. CHANNEL FLOW 29

100 101 102 1030

5

10

15

20

25

30

U+

y+

U+=y+

U+=(1/κ) log(y+)+C

Viscous region

Logregion

Wakeregion

Figure 2.3: Velocity profile of experimentally measured data ( ) Experimentaldata taken at Reτ ≈ 4706. It was found that the log law has κ = 0.41 and C = 5.0.


Fig. 2.3. In general, there are three separate regions, the viscous sublayer, the loglayer and the defect layer. In the viscous region, viscous forces are dominant andthe normalized velocity, U+ varies linearly with y+ where

U+ =U

uτ

and

y+ =yuτ

ν.

uτ =

√ν∂U

∂y(y = 0) (2.57)

is the friction velocity. In the log layer, the viscous stress is negligible comparedto the Reynolds stress. The velocity profile in the log layer can be approximatedby a logarithmic function (although this is currently under much debate in thescientific community). At the end of the log layer, the velocity profile deviates froma logarithmic function. This region is called the defect layer.

From the analysis in the previous section, it has been shown that the mixinglength model can be used as a model for turbulent free shear flows. In all theprevious analysis, it was assumed that the mixing length is

• constant across the layer (i.e. lmix is only a function of x) and

• proportional to the thickness of the shear layer (i.e. lmix = αδ(x) )

These assumptions hold if the flow have no boundaries. For flows near a solidboundary, the behaviour turbulence is different and has to be treated accordingly.

Let’s now analyse the governing equation for the channel flow. The momentumequation for a channel is given by

∂

∂y

(ν∂U

∂y− u′v′

)− 1

ρ

dPw

dx= 0 (2.58)

Using the Boussinesq assumption,

−u′v′ = νT∂U

∂y

Equation (2.58) can be written as

∂

∂y

([ν + νT ]

∂U

∂y

)− 1

ρ

dPw

dx= 0 (2.59)

Since the problem is symmetrical, about the centreline of the channel, we only reallyneed to look at the domain 0 ≤ y ≤ H/2, where H is the total height of the channel.In order to solve Eq. (2.59), integrate with respect to y and using Eq. (1.53) gives,


∫ y

0

∂

∂y

([ν + νT ]

∂U

∂y

)dy =

1

ρ

∫ y

0

dPw

dxdy

= −∫ y

0

2

Hτwdy[

(ν + νT )∂U

∂y

]y

0

= − 2

Hτwy

ν∂U

∂y(y) + νT

∂U

∂y(y)− ν

∂U

∂y(0) = − 2

Hτwy (2.60)

If we assume that y is small and ν << νT , then the first term on the left hand sideand the term of right hand side of the above equation can be neglected. Thus,

νT∂U

∂y(y) = ν

∂U

∂y(0)

=τwρ

= u2τ (2.61)

Applying the mixing length model

νT = l2mix

∣∣∣∣∂U∂y∣∣∣∣

into Eq. (2.61) gives

l2mix

(∂U

∂y(y)

)2

= u2τ (2.62)

Recall that lmix is a function of y. In order to see the form of lmix we need to lookat experimental data. From experimental measurements, it has been observed thatthere is a region close to the wall of the channel where the velocity profile can bewritten in terms of a logarithmic function

U

uτ

=1

κlog(yuτ

ν

)+ C (2.63)

where κ = 0.41 and C = 5.0 are constants obtain from experimental data. Substi-tuting Eq. (2.63) into Eq. (2.62) gives

l2mix

(uτ

κy

)2

= u2τ

lmix = κy (2.64)

Thus in order to for the modelled mean velocity profile to look like Eq. (2.63) inthe log layer, one has to let the mixing length, lmix to be directly proportional to y,


the distance from the nearest wall. In the viscous sublayer, viscous forces are moreimportant than turbulent forces. From Eq. (2.60)

ν∂U

∂y(y) = ν

∂U

∂y(0)

ν∂U

∂y(y) = u2

τ

ν

∫ y

0

∂U

∂y(y)dy =

∫ y

0

u2τdy

ν[U]y0

= u2τy

νU(y) = u2τy

U

uτ

=yuτ

ν

U+ = y+ (2.65)

Thus in the viscous sublayer, lmix must approach zero. [1] came up with thefollowing expression for lmix which is valid for both the viscous sublayer and the loglayer

lmix = κy(1− e−y+/A+

0

)(2.66)

where A+0 = 26. The comparison of the model with experimental data at Reτ = 4706

is shown in Fig. 2.4


100

101

102

103

0

5

10

15

20

25

U+

y+

U+=y+

U+=(1/κ) log(y+)+C

Figure 2.4: Comparison of the mixing layer model prediction ( ) with experi-mental data (◦ ). Experimental data taken at Reτ = 4706. κ = 0.41 and C = 5.0.


Bibliography

[1] E. V. Driest. On turbulent flow near a wall. Journal of the Aeronautical Sciences,23:1007, 1956.

[2] H. Liepmann and J. Laufer. Investigation of free turbulent mixing. NACA TN1247, 1947.

[3] S. Pope. Turbulent Flows. Cambridge University Press, 2000.

[4] H. Schlicting. Boundary Layer Theory. McGraw-Hill, 1979.

[5] S. Tavoularis and S. Corrsin. Experiments in nearly homogeneous turbulentshear flow with uniform mean temperature gradient. part i. J. Fluid Mech.,104:311–347, 1981.

[6] D. Wilcox. Turbulence Modelling for CFD. DCW Industries, Inc., 1998.

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Chapter 1 Governing Equations

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