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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 advocated by Osborne
Reynolds in which the instantaneous flow
properties 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 but present the boundary layer form
in Cartesian coordinates. (Tensors hadn’t been
invented in Reynolds’ time.)
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Preliminaries
• We consider a turbulent flow that is incompressibleand 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%
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t T
i itu dt u
T
+= =∫
i i iU U u= +%
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An important point to note
• If a variable φ 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
φφ
∂ ∂ = ∂ ∂∫ ∫
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 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∂ ∂ =∂ + ∂ =∂ ∂%
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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
∂ ∂ ∂+ + =
∂ ∂ ∂0
i 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 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
∂ ∂ ∂≡ + +
∂ ∂ ∂
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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 (explanation given in lecture).
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
∂ − − ∂
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Who was Osborne Reynolds?
• Osborne Reynolds, born in Belfast - appointed in 1868 to the first 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/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.
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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” (or, rather, mass-
weighted 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 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 =
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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?
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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 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=
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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 turbulentkinetic 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 its variation can be neglected; i.e.:
• 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− + = =
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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 d uvuv
dy dydy+ =
2
2
( )0w
d dU dyd U dUuv
dy dydy
ν τ ρ−+ =
2
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d U d Uu v
d yd yν
+ =
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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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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
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∂ ∂ =
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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:
• 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
∂ ∂ ∂ ∂ ∂ ∂ ∂ = − ≡ − ∂ ∂ ∂ ∂ ∂ ∂∂
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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 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∂ ∂