1 fluid mechanics 3 rd year mechanical engineering prof brian launder lecture 10 the equations of...
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![Page 1: 1 Fluid Mechanics 3 rd Year Mechanical Engineering Prof Brian Launder Lecture 10 The Equations of Motion for Steady Turbulent Flows](https://reader035.vdocuments.mx/reader035/viewer/2022062714/56649d3a5503460f94a154b6/html5/thumbnails/1.jpg)
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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?
22
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
20
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
2
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.