Fluid Mechanics
3rd Year Mechanical Engineering
Prof Brian Launder
Lecture 10
The Equations of Motion for Steady
Turbulent Flows
1
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.
2
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.
(Tensors hadn’t been
invented in Reynolds’ time.)
3
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
. U , V , W (or U )
i
1 t T
Ui  
U i dt  U i
• So   
T t
• The difference between Ui and U i is denoted ui, the
turbulent velocity: Ui  Ui  ui
1 t T
ui dt  0  ui
• NB the time average of ui is zero, i.e. 
T t
4
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
  dy  
dy




x
x
• So
t T
t T

 U U
U 1
U
 1

dt    Udt  


x T
x
x  T
x
x

t
 t

5
Averaging the equations of motion
• First, note that the instantaneous static pressure
is likewise written as the sum of a mean and
turbulent part: P  P  p
• The time average of P / xi   ( P  p) / xi  P / xi
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 NavierStokes equations may be written:
,

• But:
 2U i
x j
2

 2 (U i  ui )
x j
2

 2U i
x j 2
U jUi  (U j  u j )(Ui  ui )  U jUi  u jui
6
The continuity equation in turbulent flow
• For a uniform density flow:
• But
• ..or
ui
xi

ui
xi
 0 …so
U V

0
x y
Ui
xi
U i
xi

 (U i  ui )
xi
0
 0.
• Thus, the fluctuating velocity also satisfies
u v w
 
 0 or ui xi  0
x y z
7
The averaged momentum equation
• From the averaging on Slide 6:

1 P
  U i

U j


 ui u j 

t
x j
 xi x j  x j


U i
U i
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.
U i
U i
UPi  UiU i Ui 
1
• Thus:  U
U
 V   W u u 
t
j x
j
 xi x j  y x j

i z j 

8
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.

U
U
1 dP   U
U
V

 
 uv 
x
y
 dx y  y

U
V

0
x
y
• 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”).
   2
U
U
1 dP   U
2

u

v
U

V





uv

 x 
• The form:

x
y
 dx y  y

is a higher level of approximation.
9
Who was Osborne Reynolds?
• Osborne Reynolds, born in Belfast
- appointed in 1868 to the first fulltime chair of engineering in England
(Owens College, Manchester) at
the age of 25.
• Initially explored a wide range of
O phenomena: the formation
physical
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.
10
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.
11
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)
12
Some features of the Reynolds stresses
• The stress tensor comprises nine elements but,
since it is symmetric ( uiu j  u j ui ), only six
components are independent since u1u2  u2u1 etc.
or in Cartesian coordinates uv  vu; uw  wu; vw  wv.
• If turbulence is isotropic all the normal stresses
(components where i=j) are equal and the shear
stresses ( i  j ) are zero. (Why??)
• The presence of mean velocity gradients (whether
normal or shear) makes the turbulence nonisotropic.
• Non-isotropic turbulence leads to the transport of
momentum usually orders of magnitude greater
than that of molecular action.
13
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 noslip 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; uiu j  0.
• The presence of this sublayer means that growth
rates of turbulent boundary layers will depend on
Reynolds number.
14
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? 15
The mean kinetic energy equation
• By multiplying each term in the Reynolds equation
by Ui we create an equation for the mean kinetic

energy:
U U
U U
U P U   U
i
t
i U
j
i
x j
 i  i  ui u j 

 xi x j  x j


i  i
• The left side is evidently:
DK
2
2
2
 Ui 2
 Ui 2

or, with KUi /2,
t
U j
x j

Dt
• Re-organize the right hand side as:

U i P
xi


2
U i2
x 2j
2
2
 U 
U i
 
i
 
 
U i ui u j   ui u j

x j
 x j  x j 
A
B
C
D
E
 See next slide for physical meaning of terms
16
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
17
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.
18
Boundary-layer form of mean energy equation
• For a thin shear flow (U(y)) the mean k.e. equation
becomes:
2
2
 U  dPU i 
DK
 K
U

 

uvU  uv
 
2
Dt
dx
y
y
y
 y 
• Consider a fully developed flow where the total (i.e.
viscous + turbulent) shear stress varies so slowly
with y that it can be neglected.
w
dU
uv 
 const. 
dy

• In this case, where does the conversion rate of
kinetic energy reach a maximum?
 
19
Where is the conversion rate of mean energy
to turbulence energy greatest?
• This occurs where:
d  dU 
uv
0


dy  dy 
d 2U
dU d uv
or where
uv 
0
2
dy dy
dy
d 2U
dU d ( dU dy   w  )
uv 
0
or:
dy
dy
dy 2
d 2U 
dU 
uv 
0
or, finally:

2 
dy 
dy 
Thus, the turbulence energy creation rate is a
maximum where the viscous and turbulent shear
stresses are equal
20
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.
21
Why is the normal stress perpendicular to the
wall so much smaller than the other two?
• Continuity for turbulent flow:
u v w
  0
x y z
• Apply this at y =0 (the wall)
• But on this plane u=w=0 for all x and z
So, v y  0 at y  0 ; but u and w deriv’s w.r.t. y  0
• Expand fluctuating velocities in a series:
u  a1 y  a2 y 2  a3 y3...; v  b1 y  b2 y 2  a3 y3...; w  etc.
But b1 must be zero (if v / y  0 at y  0. )
2
4
So, while 2
2
2
u
w y , v y
• Q: How does the shear stress uv vary for small y?
22
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 uiu j Ui x j 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.
23
The turbulence energy equation-1
• Subtract the Reynolds equation from the Navier
Stokes equation for a steady turbulent flow
ui
t

 (U j  u j )(U i  ui )
x j
• This leads to:
ui
t
U j
ui
x j
 2 (U i  ui )
1 ( P  p)


 xi
x 2j
2 
 U U u u

Ui 
1 P

j i
j i




2 

x

x


x
x j 

j
j
i


uj
Ui
x j

(uiu j  uiu j )
x j
 2ui
1 p


 xi
x j 2
• Note the above makes use of u j / x j  u j  / x j
since by continuity u j / x j  0.
24
The turbulence energy equation - 2
• Multiply the boxed equation from the previous slide by ui
and time average.
ui  (ui u j  ui u j )
ui  2ui
1 ui p
U j
 ui u j



t
x j
x j
x j
 xi
x j 2
ui ui
• Note:
ui ui
ui ui
t

U i
ui 2 / 2
t


k
t
where k is the turbulent
kinetic energy: k  u 2  u 2  u 2
1
2
3
2
• The viscous term is transformed as follows:
  ui  ui ui   k 
u
 
 ui 2  

  

i
x j  x j  x j x j x j  xj 
x j


 2ui
•   turbulence energy dissipation rate
25
The turbulence energy equation - 3
• After collecting terms and making other minor
manipulations we obtain:

pui
U i
Dk
  k
2


 [ui u j / 2 
 ij ]  ui u j

Dt x j  x j

x j



viscous
turbulent diffusion
generation
dissipation
• Note this is a scalar equation and each term has to
have two tensor
subscripts forpueach letter. U

Dk 
k
2
i  ]  u u
i 




[
u
u
/
2


U
/ x j
• Repeat Q & A: How doi we
know that
j
ij
i
j
i
Dt x j  x j

x j

represents the
 generation rate of turbulence?

Ans: The same term but with opposite sign appears
in the mean kinetic energy equation.
26
A question for you
• Compile a sketch of the mean kinetic
energy budget for fully developed laminar
flow between parallel planes.
27