1

Fluid Mechanics

Prof Brian Launder

Lecture 10

The Equations of Motion for Steady Turbulent Flows

2

Objectives

• To obtain a form of the equations of motion designed for the analysis of flows that are turbulent.

• To understand the physical significance of the Reynolds stresses.

• To learn some of the important differences between laminar and turbulent flows.

• To understand why the turbulent kinetic energy has its peak close to the wall… and why fluctuations normal to the wall increase more slowly than parallel to the wall.

3

The strategy followed

• We adopt the strategy ad-

vocated by Osborne

Reynolds in which the

instantaneous flow propert-

ies are decomposed into a

mean and a turbulent part.

(for the latter, Reynolds

used the term sinuous).

• We shall mainly use tensor

notation for compactness.

(Tensors hadn’t been

invented in Reynolds’ time.)

4

Preliminaries• We consider a turbulent flow that is incompressible

and which is steady so far as the mean flow is

concerned.

• For most practical purposes one is interested only in

the mean flow properties which will be denoted U, V,

W (or Ui in tensor notation).

• The instantaneous total velocity has components

. (or )

• So → → →

• The difference between Ui and is denoted ui, the

turbulent velocity:

• NB the time average of ui is zero, i.e.

, ,U V W% % %i

U%1 t T

i i itU U dt U

T

+= ≡∫ % %

iU%

10

t T

i itu dt u

T

+= =∫

i i iU U u= +%

5

An important point to note

• If a variable φ is a continuous function of two independent variables, x and y, differential or

integral operations on it with respect to x and y can

be applied in any order.

• Thus

• So

dy dyx x

φφ

∂ ∂ = ∂ ∂∫ ∫

1 1t T t T

t t

U U U Udt Udt

x T x x T x x

+ + ∂ ∂ ∂ ∂ ∂ ≡ = ≡ =

∂ ∂ ∂ ∂ ∂ ∫ ∫

% % %%

6

Averaging the equations of motion

• First, note that the instantaneous static pressure is likewise written as the sum of a mean and turbulent part:

• The time average of ,

where the overbar denotes the time-averaging noted on the previous slide.

• Treating the viscosity as constant, the time averaged value of the viscous term in the Navier-Stokes equations may be written:

• But:

P P p= +%

2 2 2

2 2 2

( )i i i i

j j j

U U u U

x x xν ν ν

∂ ∂ + ∂= =

∂ ∂ ∂

%

( )( )j i j j i i j i j i

U U U u U u U U u u= + + = +% %

/ ( )/ /i i i

P x P p x P x∂ ∂ =∂ + ∂ =∂ ∂%

7

The continuity equation in turbulent flow

• For a uniform density flow:

• But …so

• ..or

• Thus, the fluctuating velocity also satisfies

or

0i i

i i

u u

x x

∂ ∂= =

∂ ∂

( )0i i i

i i

U U u

x x

∂ ∂ +≡ =

∂ ∂

%

0.i

i

U

x

∂=

∂

0U V

x y

∂ ∂+ =

∂ ∂

0u v w

x y z

∂ ∂ ∂+ + =

∂ ∂ ∂0

i iu x∂ ∂ =

8

The averaged momentum equation

• From the averaging on Slide 6:

Convection Diffusion

This is known as the Reynolds Equation

• Note that this is really three equations for i taking

the value 1,2 and 3 in three orthogonal directions

• Recall also that because the j subscript appears

twice in the convection and diffusion terms, this

implies summation, again for j=1,2, and 3.

• Thus:

1i i ij i j

j i j j

U U UPU u u

t x x x xν

ρ

∂ ∂ ∂∂ ∂ + = − + − ∂ ∂ ∂ ∂ ∂

i i ij i j

j i j j

U U UU uu

t x x x x

∂ ∂ ∂+ =− + −

∂ ∂ ∂ ∂ ∂i i i

U U UU V W

x y z

∂ ∂ ∂≡ + +

∂ ∂ ∂

9

Boundary Layer form of the Reynolds Equation

• The form of the Reynolds equation appropriate to a steady 2D boundary layer is taken directly from the laminar form with the inclusion of the same component of turbulent and viscous stress: i.e.

• The accuracy of this boundary layer model is, for some flows, rather less than for the laminar flow case (i.e. the neglected terms are less “negligible”).

• The form:

is a higher level of approximation.

1 dPU U UU V uv

x y dx y yν

ρ∞ ∂ ∂ ∂ ∂

+ = − + − ∂ ∂ ∂ ∂

0U V

x y

∂ ∂+ =

∂ ∂

1 dPU U UU V uv

x y dx y yν

ρ∞ ∂ ∂ ∂ ∂

+ = − + − ∂ ∂ ∂ ∂

2 2u v

x

∂ − − ∂

10

Who was Osborne Reynolds?

• Osborne Reynolds, born in Belfast - appointed in 1868 to the first full-time chair of engineering in England (Owens College, Manchester) at the age of 25.

• Initially explored a wide range of physical phenomena: the formation of hailstones, the effect of rain and oil in calming waves at sea, the refraction of sound by the atmosphere…

• …as well as various engineering works: the first multi-stage turbine, a laboratory-scale model of the Mersey estuary that mimicked tidal effects.

O

11

Entry into the details of fluid motion

• By 1880 he had become fascinated by the detailed mechanics of fluid motion…..

• ….especially the sudden transition between direct and sinuous flow which he found occurred when: UmD/ν ≅ 2000.

• Submitted ms in early 1883 – reviewed by Lord Rayleigh and Sir George Stokes and published with acclaim. Royal Society’s Royal Medal in 1888.

12

Reynolds attempts to explain behaviour

• In 1894 Reynolds presented

orally his theoretical ideas to

the Royal Society then

submitted a written version.

• This paper included “Reynolds

averaging”, Reynolds stresses

and the first derivation of the

turbulence energy equation.

• But this time his ideas only

published after a long battle

with the referees (George

Stokes and Horace Lamb –

Prof of Maths, U. Manchester)

13

Some features of the Reynolds stresses

• The stress tensor comprises nine elements but, since it is symmetric ( ), only six components are independent since etc. or in Cartesian coordinates .

• If turbulence is isotropic all the normal stresses (components where i=j) are equal and the shear stresses ( ) are zero. (Why??)

• The presence of mean velocity gradients (whether normal or shear) makes the turbulence non-isotropic.

• Non-isotropic turbulence leads to the transport of momentum usually orders of magnitude greater than that of molecular action.

i j j iu u u u=

i j≠

1 2 2 1u u u u=

; ;uv vu uw wu vw wv= = =

14

More features of the Reynolds stresses

• Turbulent flows unaffected by walls (jets, wakes)

show little if any effect of Reynolds number on

their growth rate (i.e. they are independent of ν).

• Turbulent flows (like laminar flows) obey the no-

slip boundary condition at a rigid surface. This

means that all the velocity fluctuations have to

vanish at the wall.

• So, right next to a wall we have to have a viscous

sublayer where momentum transfer is by

molecular action alone;

• The presence of this sublayer means that growth

rates of turbulent boundary layers will depend on

Reynolds number.

0.i j

u u =

15

Comparison of laminar and turbulent boundary layers

Laminar B.L.Laminar B.L.Laminar B.L.Laminar B.L.

�Recall: The very steep near-wall velocity gradient in a turbulent b.l. reflects the damping of turbulence as the wall is approached

�But why do turbulent velocity fluctuations peak so very close to the wall?

16

The mean kinetic energy equation

• By multiplying each term in the Reynolds equation by Ui we create an equation for the mean kinetic energy:

• The left side is evidently:

or, with K≡Ui2 /2,

• Re-organize the right hand side as:

�

A B C D E

� See next slide for physical meaning of terms

i i ij i j

j i

i i

j

i

j

iU U UP

U uU U

ut x x

U

x x

Uν

ρ

∂ ∂ ∂ ∂∂ + =− + − ∂ ∂ ∂ ∂ ∂

2 22 2

i ij

j

U UU

t x

∂ ∂+ =

∂ ∂

22 2

2

2i i i i

i i j i ji j j jj

U P U U UU u u u u

x x x xxν ν

∂ ∂ ∂ ∂∂ + − − + ∂ ∂ ∂ ∂ ∂

DK

Dt=

17

The “source” terms in the mean k.e eqn

• A: Reversible working on fluid by pressure

• B: Viscous diffusion of kinetic energy

• C: Viscous dissipation of kinetic energy

• D: Reversible working on fluid by turbulent stresses

• E: Loss of mean kinetic energy by conversion to turbulence energy

18

A Query and a Fact

• Question: How do we know that term E represents a loss of mean kinetic energy to turbulence?

• Answer: Because the same term (but with an opposite sign) appears in the turbulentkinetic energy equation!

• The mean and turbulent kinetic energy equations were first derived by Osborne Reynolds.

19

B’ndary-layer form of mean kin. energy equ’n

• For a thin shear flow (U(y)) the mean k.e. equation becomes:

• Consider a fully developed flow where the total (i.e. viscous + turbulent) shear stress varies so slowly with y that its variation can be neglected across the viscosity-affected sublayer.

• In this case, where does the conversion rate of kinetic energy reach a maximum?

( )22

2

idP UDK K U U

uvU uvDt y dx y yy

ν ν ∞ ∂ ∂ ∂ ∂= − − − +

∂ ∂ ∂∂

. wdUuv const

dy

τν

ρ− + = =

20

Where is the conversion rate of mean energy to turbulence energy greatest?

• This occurs where:

or where

or:

or, finally:

�Thus, the turbulence energy creation rate is a

maximum where the viscous and turbulent shear

stresses are equal

0d dU

uvdy dy

=

2

20

d U dU d uvuv

dy dydy+ =

2

2

( )0w

d dU dyd U dUuv

dy dydy

ν τ ρ−+ =

2

20

d U dUuv

dydyν

+ =

21

The near wall peak in turbulence explained

• The peak in turbulence energy occurs very close to the point where the transfer rate of mean energy to turbulence is greatest

• This occurs where viscous and turbulent stresses are equal – i,e. within the viscosity affected sublayer!

• One of the reasons why the turbulent velocity fluctuations are so different in different directions will now be examined

22

Why is the normal stress perpendicular to the wall so much smaller than the other two?

• Continuity for turbulent flow:

• Apply this at y =0 (the wall)

• But on this plane u=w=0 for all x and z

�So, ; but u and w deriv’s w.r.t. y ≠≠≠≠ 0

• Expand fluctuating velocities in a series:

�But b1 must be zero (if )

�So, while

• Q: How does the shear stress vary for small y?

0u v w

x y z

∂ ∂ ∂+ + =

∂ ∂ ∂

0 at 0v y y∂ ∂ = =

2 3 2 3

1 2 3 1 2 3...; ...; etc.u a y a y a y v b y b y a y w= + + = + + =

/ 0 at 0.v y y∂ ∂ = =

2 2 2,u w y∝�

2 4v y∝

uv

23

Extra slides

• The following slides provide a derivation of the

kinetic energy budget from the point of view of the

turbulence.

• They confirm the assertion made earlier that the

term represents the energy source of

turbulence.

• We do not work through the slides in the lecture (Dr

Craft will provide a derivation later) but the path

parallels that for obtaining the mean kinetic energy.

i j i ju u U x− ∂ ∂

24

The turbulence energy equation-1

• Subtract the Reynolds equation from the Navier

Stokes equation for a steady turbulent flow

• This leads to:

• Note the above makes use of

since by continuity

2

2

1j i j i i

j j i j

U U u u UP

x x x xν

ρ

∂ ∂ ∂∂ − + = − +

∂ ∂ ∂ ∂

2

2

( )( ) ( )1 ( )j j i ii i i

j i j

U u U uu U uP p

t x x xν

ρ

∂ + +∂ ∂ +∂ ++ =− +

∂ ∂ ∂ ∂

2

2

( ) 1i j i ji i i ij j

j j j i j

u u u uu u U upU u

t x x x x xν

ρ

∂ −∂ ∂ ∂ ∂∂+ + + = − +

∂ ∂ ∂ ∂ ∂ ∂

/ /j j j j

u x u xφ φ∂ ∂ = ∂ ∂/ 0.

j ju x∂ ∂ =

25

The turbulence energy equation - 2

• Multiply the boxed equation from the previous slide by and time average.

• Note: where k is the turbulent

kinetic energy:

• The viscous term is transformed as follows:

• ε ≡ turbulence energy dissipation rate

2

2

( ) 1i i j i ji i i i i i i ij i j

j j j i j

u u u u uu u u u U u p u uU u u

t x x x x xν

ρ

∂ −∂ ∂ ∂ ∂ ∂+ + − = − +

∂ ∂ ∂ ∂ ∂ ∂

uuuuur

iu

2/ 2

i i iu u u k

t t t

∂ ∂ ∂= ≡

∂ ∂ ∂

{ }2 2 21 2 3

2k u u u≡ + +

2

2

i i i ii i

j j j j jj

u u u u ku u

x x x x x xjxν ν ν ν ε

∂ ∂ ∂ ∂ ∂ ∂ ∂ = − ≡ − ∂ ∂ ∂ ∂ ∂ ∂∂

26

The turbulence energy equation - 3

• After collecting terms and making other minor

manipulations we obtain:

viscous turbulent diffusion generation dissipation

• Note this is a scalar equation and each term has to

have two tensor subscripts for each letter.

• Repeat Q & A: How do we know that

represents the generation rate of turbulence?

Ans: The same term but with opposite sign appears

in the mean kinetic energy equation.

2[ / 2 ]i i

i j ij i jj j j

pu UDk ku u u u

Dt x x xν δ ε

ρ

∂∂ ∂ = − + − −

∂ ∂ ∂

[ /2 ]i ii j ij i j

j j j

pu Uu u u u

Dt x x xν δ ε = − + − −

∂ ∂ ∂

/i j

U x∂ ∂

27

A question for you

• Compile a sketch of the mean kinetic energy budget for fully developed laminarflow between parallel planes.