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Fluid Mechanics3rd Year Mechanical Engineering

Prof Brian Launder

Lecture 10 The Equations of Motion for Steady

Turbulent Flows

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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.

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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.)

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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 WiU

1 t Ti i it

U U dt UT

iU

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t Ti it

u dt uT

i i iU U u

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An important point to note

• If a variable is a 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

1 1t T t T

t t

U U U Udt Udt

x T x x T x x

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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 x

( )( )j i j j i i j i j iU U U u U u U U u u

/ ( ) / /i i iP x P p x P x

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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

0i iu x

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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 x

i i iU U UU V W

x y z

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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 y

0U V

x y

1 dPU U UU V uv

x y dx y y

2 2u vx

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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

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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/ 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.

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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)

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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

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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 ).

• 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 ju u

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Comparison of laminar and turbulent boundary layers

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?

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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 KUi2 /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 UPU u

U Uu

t x x

U

x x

U

2 22 2i ij

j

U UU

t x

22 2

2

2i i i ii i j i j

i j j jj

U P U U UU u u u u

x x x xx

DK

Dt

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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

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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 turbulent kinetic energy equation!

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

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Boundary-layer form of mean energy equation

• 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 it can be neglected.

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

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2idP UDK K U U

uvU uvDt y dx y yy

. wdUuv const

dy

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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

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d U dU duvuv

dy dydy

2

2

( )0wd dU dyd U dU

uvdy dydy

2

20

d U dUuv

dydy

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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!

• Why the turbulent velocity fluctuations are so different in different directions will be examined in a later lecture.

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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 zSo, ; 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 31 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

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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

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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

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21j i j i i

j j i j

U U u u UP

x x x x

2

2

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

j i j

U u U uu U uP p

t x x x

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 x

/ /j j j ju x u x / 0.j ju x

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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:

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 x

��������������

iu

2 / 2i i iu u u k

t t t

2 2 21 2 3 2k u u u

2

2i i i i

i ij j j j jj

u u u u ku u

x x x x x xjx

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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 ii j ij i j

j j j

pu UDk ku u u u

Dt x x x

/i jU x

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A question for you

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