Turbulence Modelling - Engineering - Fluid Mechanics, Lecture notes of Fluid Mechanics

Download Turbulence Modelling - Engineering - Fluid Mechanics and more Lecture notes Fluid Mechanics in PDF only on Docsity!

Turbulence Modelling

and Turbulent Flows

Predictions of turbulent flows

What flow parameters or quantities do we want to predict?

a) Integral parameters: θ , δ

, H, etc.

▬▬▬► Integral equations

b) Transition or flow separation points

▬▬▬► Integral equations or turbulence models,

depending on the required accuracy

c) Pressure distribution over lifting bodies (drag and lift)

▬▬▬► Potential (ideal) flow equations

d) Time averaged quantities: U , u',uvetc.

▬▬▬► Turbulence models

e) Instantaneous values: u, v, w, p, etc. and turbulence structure

▬▬▬► Direct numerical simulation (DNS) or

Large eddy simulation (LES)

f) Other models include pdf and hybrid models****.

There are other considerations in choosing the computational method,

including:

Cost of computation

Range of applicability

Accuracy of the result

and Turbulent Flows

Solving the Navier-Stokes equations

In order to solve the Navier-Stokes equations directly, one must resolve the

smallest eddies of turbulence η. As we discussed previously that the ratio of η to

the energy-containing eddy size l is given by

4 −^3 = Rl l

η

For example, for flows with the Reynolds number Rl = 10

6 , η /l ≈ 3.2 x 10

• . This

indicates that we need to have extremely fine computational meshes to resolve the

Kolmogorov microscale of turbulence.

A similar argument also holds for the time step of computation since we need to

resolve the time scale of velocity variation across the Kolmogorov microscale.

The time required to perform DNS simulation is estimated as.

3 Tg ∝R l

estimated time with giga-flop computers IBM Blue Gene

NEC Earth Simulator

and Turbulent Flows

The Reynolds equations

The Reynolds equations are part of the Navier-Stokes equations for turbulent

flows. These equations are useful in the predictions of time-averaged (mean)

quantities when the flow is turbulent.

A typical physical quantity in turbulent flow shows fluctuations with time.

For example, the velocity can be expressed as

forthelaminarflowsince 0

lim

U =U u =

U(t)dt T

U=U+u, U =

T

T (^0)

→∞

The Reynolds equations can be obtained by substituting

U =U+u V=V+v W=W+w P=P+ p

in the Navier-Stokes equations, and take time average over the entire expression:

and Turbulent Flows

Therefore,

+ U

x

1 P

+ U=-

x

1 P

RHS =-

z

w

y

v

x

u

• u z

uw

y

uv

x

u

z

U

+W

y

U

+V

x

U

+U

t

U

z

U

+W

y

U

+V

x

U

+U

t

U

LHS =

2 2

2

ν ρ

ν ρ

0

Applying similar techniques in the rest of the equations, we have the Reynolds

equations.

z

w

y

vw

x

uw

z

W

y

W

x

W

z

1 P

z

W

+W

y

W

+V

x

W

+U

t

W

z

vw

y

v

x

uv

z

V

y

V

x

V

y

1 P

z

V

+W

y

V

+V

x

V

+U

t

V

z

uw

y

uv

x

u

z

U

y

U

x

U

x

1 P

z

U

+W

y

U

+V

x

U

+U

t

U

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

ν ρ

ν ρ

ν ρ

Reynolds Streesses

and Turbulent Flows

Taking a time average over the instantaneous Continuity equation

z

W w

y

V v

x

U u

∂

the Continuity equation for mean velocities is obtained as

z

W

y

V

x

U

Also, the Continuity equation for fluctuating velocities is given by

z

w

y

v

x

u

∂

To explain the Reynolds-stress terms we take a simple example (see below).

In the laminar flows, the shear stress is given by y

u l = ∂

τ μ. In turbulent flows,

stresses are also produced by turbulent mixing. Since the mass flow rate of fluid

blob with component velocity fluctuations u and -v is - ρ v per unit area, the

momentum that this blob carries in the vertical direction is – ρ uv per unit area per

unit time. Therefore, the turbulent stress (averaged over a long time) is

τt=-^ ρ^ uv

and Turbulent Flows

This situation has been caused as we lost information on turbulence during the

process of time averaging. The original Navier-Stokes equations contain all the

information required, but they are not very useful in practice, because:

• The Navier-Stokes equations are very difficult and time-consuming to solve.
• They contain enormous amount of information, all of which may not be

required.

Turbulence Modelling

We now have a situation where the number of unknown variables exceeds the

number of equations available, which is called the closure problem. In order to

overcome this problem, we have to model^ the turbulence.

Modelling turbulence means that we should express the unknown variables in

terms of known variables. A common approach is to treat the turbulence as a

property of fluid (phenomenological analysis):

y

U

• uv t ∂

so that

y

U

= ( + t ) ∂

where μt is called the turbulent viscosity (Boussinesq 1988).

The zero, one and two-equation models are called the first-order closures, since

equations for first-order moments (mean values) are modelled.

and Turbulent Flows

Turbulence models

One-equation models … μ t expressed as a

function of variables, one of which will be

solved simultaneously.

Two-equation models … μ t expressed as a

function of variables, two of which will be

solved simultaneously

Zero-equation models … μ t specified as a

constant or algebraically

Second-order closures … Equations for the

Reynolds stresses (second-order moments) will

be solved simultaneously with the equations for

U , V,Wand P.

The first-order closures

The Reynolds stress is given by

y

U

t =^ - uv= t ∂

τ ρ^ μ

where, Prandtl (1925) hypothesised that the velocity fluctuations can be given by

y

U

u v lm ∂

The mixing length lm comes from the analogy with the mean-free path of

molecules.

and Turbulent Flows

Zero-equation models

In zero-equation turbulence models, the mixing length l m is set either a constant

value or given as an algebraic expression.

• plane mixing layers: lm = 0.07 δ , where δ is the layer width
• plane jets: lm = 0.09 δ , where δ is the half width
• round jets: lm = 0.075 δ , where δ is the half width
• pipe flows: (^) ⎟ ⎠

R

y

• 0.06 1 - R

y =0.14-0.08 1 - R

l

2 4 m

and Turbulent Flows

• boundary layers (near-wall region): van Driest's formula (1956)

=0.40 A= 26.

l u u = l

yu y =

A

y l = .y^1 - exp -

• m m

• w

m

κ ρ ν

τ

ν

κ

where, lm = 0 as y

→ 0 and lm = κ y as y

→ ∞

One-equation models

In zero-equation models, y

U

t ∂

μ ∝ therefore^ μ t = 0 when^ =^0

y

U

and Turbulent Flows

Note that all the RHS terms are unknowns and must be modelled. We usually

model them as,

l

( k) =C.. x

u

y

U

• uv=

y

k ( vk+vp)=

3

j

i

2 3

ij=

t

t

D

k

μ ρ

ρ μ

σ

μ ρ

⎟

⎟ ⎠

⎞ ⎜

⎜ ⎝

⎛

∂

∂

∂

∂

∂

∂

,

Therefore, the modelled kinetic-energy equation for the turbulent boundary layer is

given by

l

k

• C y

U

y

k

y

= Dt

Dk

m

2

t k

t D

ρ^32 μ σ

μ ρ (^) ⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛

∂

∂ ⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛

∂

∂

∂

∂

where, σ k and CD are constants, and lm is specified algebraically.

Two-equation models

As the flow field gets more complex, it becomes difficult to specify the length scale

(or the mixing length lm) by an algebraic expression. In two-equation models, both

Vt and lm are obtained by computing the transport equations.

In general, we require the transport equation for k and another quantity, z made up

of k and l.

l

k and l

k z .... f, , l, kl where, f

(^12) 3/

Again, the transport equation for z will contain unknown quantities that must be

modelled.

and Turbulent Flows

The most popular two-equation model is called k-ε model , which uses ε

(dissipation rate) as z. The transport equations for k and z will be solved

simultaneously together with the equations for U , V,WandP. The transport

equation for the dissipation rate is given by

k

• C y

U +C k y y

= Dt

D

2 t

2 1 2

ρ ε ρ

ε

σ

ε μ ρ ε

⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛

∂

∂ ⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛

∂

∂ ∂

∂

The second-order closures

All the first-order closures we have considered so far use the Boussinesq’s

turbulent viscosity concept:

y

U

t =^ - uv= t ∂

τ ρ^ μ

There are many situations where

when = 0 y

U

t 0 ∂

τ^ ≠

The wall jet is a mixture of the boundary layer and free jet. For the turbulent wall

jet, - ρ uv = 0 about half way to the maximum of u/um. In other words,

when = 0 y

U

uv 0 ∂

In second-order closures, the turbulent viscosity concept will not be used. Instead,

the Reynolds stresses, such as - ρ uv(second-order moment) will be obtained by

solving their transport equations. For example, the Reynolds stress transport

equation for 2-D boundary layers is given by