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8. TURBULENCE MODELLING
1
SPRING 2023
- 1 Eddy-viscosity models
- 2 Advanced turbulence models
- 3 Wall boundary conditions
Summary
References
Appendix: Derivation of the turbulent kinetic energy equation
Examples
The Reynolds-averaged Navier-Stokes (RANS) equations are transport equations for the mean
variables in a turbulent flow. These equations contain net fluxes due to turbulent fluctuations.
Turbulence models are needed to specify these fluxes.
**8. 1 Eddy-Viscosity Models
- 1 .1 The Eddy-Viscosity Hypothesis**
The mean shear stress has both viscous and turbulent parts. In simple shear
(i.e. where 𝜕𝑈/𝜕𝑦 is the only non-zero mean gradient):
viscous
turbulent
The most popular type of turbulence model is an eddy-viscosity model (EVM) which assumes
that turbulent stress is proportional to mean-velocity gradient in a manner similar to viscous
stress. In simple shear (see later for the general case):
𝑡
𝑡
is called an eddy viscosity or turbulent viscosity. The overall mean shear stress is then
eff
where the total effective viscosity
μ
eff
𝑡
1
More advanced descriptions of turbulence and its modelling can be found in:
Leschziner, M.A., 2015, Statistical turbulence modelling for fluid dynamics - demystified: an introductory text for
graduate engineering students , World Scientific.
Wilcox, D.C., 2006, Turbulence Modelling for CFD , 3
rd
Edition, DCW Industries.
Pope, S.B., 2000, Turbulent flows , Cambridge University Press.
Schlichting, H. and Gersten, K., 1999, Boundary layer theory , 8
th
English Edition, Springer-Verlag.
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Note :
(1) This is a model!
( 2 ) 𝜇 is a physical property of the fluid and can be measured;
𝑡
is a hypothetical property of the flow and must be modelled.
𝑡
varies with position.
(4) At high Reynolds numbers, 𝜇 𝑡
≫ 𝜇 throughout much of the flow.
Eddy-viscosity models are widely used and popular because:
- they are easy to implement in existing viscous solvers;
- extra viscosity aids stability;
- they have some theoretical foundation in simple shear flows (see below).
However:
- there is little theoretical foundation in complex flows;
- modelling turbulent transport is reduced to a single scalar, μ
𝑡
, and, hence, at most one
Reynolds stress can be represented accurately.
8. 1 .2 The Eddy Viscosity in the Log-Law Region
In the log-law region of a turbulent boundary layer it is assumed that:
(a) (i) total stress is constant (and equal to that at the wall);
(ii) viscous stress is negligible compared to turbulent stress:
(turb)
𝑤
𝜏
2
(b) the mean velocity profile is logarithmic;
τ
The eddy viscosity is then
𝑡
(turb)
𝜏
2
𝜏
𝜏
Hence, in the log-law region, with 𝜈
𝑡
𝑡
/𝜌 as the kinematic eddy viscosity ,
𝑡
𝜏
In the log-law region the eddy viscosity is proportional to distance from the boundary.
8. 1 .3 General Stress-Strain Relationship
The stress-strain relationship ( 2 ) applies only in simple shear and cannot hold in general
because the LHS is symmetric in 𝑥 and 𝑦 components but the RHS is not. The appropriate
generalisation gives representative shear and normal stresses (from which others can be
obtained by “pattern-matching”):
is not a local scale, since it depends on where the nearest wall is, and it is ambiguous near
corners, since it can be equidistant from two (or more) walls. A more appropriate velocity scale
in general is 𝑘
1 / 2
, where 𝑘 is the turbulent kinetic energy.
For simple wall-bounded flows, 𝑙 0
is proportional to distance from the boundary (e.g. 𝑙
0
For simple free shear flows (e.g. jet, wake, mixing layer) 𝑙 0
is proportional to the width of the
shear layer. However, these values are geometry-dependent and not conducive to complex
geometries. For greater generality, we need to relate 𝑙 0
to local turbulence properties.
Common practice is to solve transport equations for one or more turbulent quantities (usually
𝑘 + one other), from which 𝜇 𝑡
can be derived on dimensional grounds. The following
classification of eddy-viscosity models is based on the number of transport equations.
zero-equation models :
- constant-eddy-viscosity models;
- mixing-length models: 𝑙
0
specified geometrically; 𝑢
0
from mean flow gradients.
one-equation models :
0
specified geometrically; transport equation to derive 𝑢
0
two-equation models :
- transport equations for quantities from which 𝑢
0
and 𝑙
0
can be derived.
Of these, by far the most popular in general-purpose CFD are two-equation models: notably
the 𝑘 − 𝜀 and 𝑘 − 𝜔 models.
8. 1. 6 Mixing-Length Models (Prandtl, 1925).
Eddy viscosity:
𝑡
𝑡
, where 𝜈
𝑡
0
𝑚
The mixing length 𝑙
𝑚
is specified geometrically and the velocity scale 𝑢
0
is then determined
from the mean-velocity gradient. In simple shear:
0
𝑚
The model is based on the premise that if a turbulent eddy displaces a
fluid particle by distance 𝑙 𝑚
its velocity will differ from its surrounds
by an amount 𝑙 𝑚
|𝜕𝑈/𝜕𝑦|. (Any constant of proportionality can be
absorbed into the definition of 𝑙 𝑚
The resulting turbulent shear stress is (assuming positive velocity
gradient):
(turb)
𝑡
0
𝑚
𝑚
2
2
The mixing length 𝑙 𝑚
depends on the type of flow.
l
m
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dy
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l
m
Log Layer
In the log layer,
(turb)
𝜏
2
and
𝜏
Equation ( 14 ) then implies that
𝑚
General Wall-bounded flows
In general, 𝑙 𝑚
is limited to a certain fraction of the boundary-layer
depth 𝛿. Cebeci and Smith (1974) suggest:
𝑚
= min(𝜅𝑦, 0. 09 𝛿) ( 15 )
Free shear flows
For free shear layers (no wall boundary), 𝑙
𝑚
is assumed
proportional to the local shear-layer half-width 𝛿. Wilcox ( 2006 )
suggests:
𝑚
071 (mixing layer)
098 (plane jet)
080 (round jet)
180 (plane wake)
Mixing-length models work well in near-equilibrium boundary layers or very simple free-shear
flows. However, although generalisations of the stress-strain relationship ( 13 ) exist for
arbitrary velocity fields, it is difficult to specify the mixing length 𝑙 𝑚
for complex flows, which
do not fit tidily into one of the above geometrically-simple categories.
8. 1. 7 The 𝒌 − Model
This is probably the most common type of turbulence model in use today. It is a two-equation
eddy-viscosity model with the following specification:
𝑡
𝜇
2
𝑘 is the turbulent kinetic energy and 𝜀 is its rate of dissipation. In the standard model 𝐶
𝜇
is a
constant (with value 0.09).
𝑘 and 𝜀 are determined by solving transport equations. For the record (i.e. you don’t have to
learn them) they are given here in conservative differential form, including implied summation
over repeated index 𝑖.
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U
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l
m
y
y
0.09
Values of the Model Constants
Some of the constants in the 𝑘 − 𝜀 model may be chosen for consistency with the log law and
available experimental data.
In the fully-turbulent region the Reynolds stresses are assumed to dominate the total stress (𝜏 =
−𝜌𝑢𝑣), whilst in a fully-developed boundary layer the total stress is constant and equal to that
at the wall (𝜏 = 𝜏
𝑤
𝜏
2
). Hence, the kinematic shear stress is
𝜏
2
In the log-law region, the mean-velocity gradient is
𝜏
so that, from equation ( 20 ) the rate of production of turbulent kinetic energy is
(𝑘)
τ
2
×
𝜏
𝜏
3
In the log-law region, we have already established that the kinematic eddy viscosity is
𝑡
𝜏
so that, with the further assumption of local equilibrium, 𝑃
(𝑘)
= 𝜀, equations ( 24 ) and ( 25 ) give
𝑡
𝜏
4
Comparing this with the 𝑘 − 𝜀 eddy-viscosity formula ( 17 ):
𝑡
𝜇
2
leads to
𝜇
𝜏
4
2
2
or 𝑢
𝜏
𝜇
1 / 4
1 / 2
A typical experimentally-measured ratio is −𝑢𝑣/𝑘 = 0. 3 , giving the standard value 𝐶 𝜇
In addition, the high-Reynolds-number (viscosity 𝜇 negligible) form of the 𝜀 equation ( 18 ) is
consistent with the log law provided the constants satisfy (see the examples overleaf):
𝜀 2
𝜀 1
𝜀
𝜇
2
In practice, the standard constants do not quite satisfy this, but have values calibrated to give
better agreement over a wide range of flows.
Classroom Example 1
(a) The 𝑘 − 𝜀 turbulence model forms an eddy viscosity 𝜇
𝑡
from fluid density 𝜌, the
turbulent kinetic energy (per unit mass) 𝑘 and its dissipation rate 𝜀. Write down the
basic physical dimensions of 𝜇
𝑡
, 𝜌, 𝑘 and 𝜀 in terms of the fundamental dimensions
of mass M, length L and time T, and hence show, on purely dimensional grounds, that
any expression for 𝜇
𝑡
in terms of the other variables must be of the form
𝑡
= constant × 𝜌
2
(b) The 𝑘 − 𝜔 turbulence model forms an eddy viscosity from 𝜌, 𝑘 and a quantity 𝜔
which has dimensions of frequency (i.e. T
). Show, on dimensional grounds, that
any expression for 𝜇
𝑡
in terms of the other variables must be of the form
𝑡
= constant × 𝜌
Classroom Example 2 (Exam 2016 – part)
A modeled scalar-transport equation for 𝜀 is
D𝜀
D𝑡
𝑖
𝑡
𝜀
𝑖
𝜀 1
(𝑘)
𝜀 2
where D/D𝑡 is the material derivative, 𝑃
(𝑘)
is the rate of production of 𝑘 and the summation
convention is implied by the repeated index 𝑖. 𝜎 𝜀
𝜀 1
and 𝐶
𝜀 2
are constants.
In a fully-developed turbulent boundary layer,
(𝑘)
𝜏
3
and 𝑘 = 𝐶
𝜇
− 1 / 2
𝜏
2
where 𝜅 is von Kármán’s constant, 𝑢 𝜏
is the friction velocity and 𝑦 is the distance from the
boundary. Show that this implies the following relationship between coefficients:
𝜀 2
𝜀 1
𝜀
𝜇
2
Classroom Example 3 (Exam 2016 – part)
In grid-generated turbulence there is no mean velocity shear, and hence no turbulence
production and minimal diffusion. The 𝑘 and 𝜀 transport equations reduce to
d𝑘
d𝑡
d𝜀
d𝑡
𝜀 2
2
where 𝑡 is the travel time downstream of the grid (distance/mean velocity). By substituting
into these equations, show that they admit a solution of the form
0
−𝑚
0
−𝑛
where 𝑘 0
0
, 𝑚 and 𝑛 are constants, and find 𝐶
𝜀 2
in terms of 𝑚 alone. (This rate of decay
for 𝑘 provides a means of determining 𝐶
𝜀 2
experimentally.)
8. 2 .1 Reynolds-Stress Transport Models (RSTM)
2
Also known as second-order closure or differential stress models these solve transport
equations for all stresses, 𝑢
2
, 𝑢𝑣 etc., rather than just turbulent kinetic energy 𝑘.
Exact equations for stresses 𝑢 𝑖
𝑗
can be derived from the Navier-Stokes equations and are of
the usual canonical form:
rate of change + advection + diffusion = source
but certain terms have to be modelled. The most important balance is in the “source” term,
which, for 𝑢 𝑖
𝑗
, consists of parts that can be identified as:
production of energy by mean-velocity gradients, 𝑃
𝑖𝑗
redistribution of energy amongst different components by pressure fluctuations, Φ
𝑖𝑗
dissipation of energy by viscosity, 𝜀
𝑖𝑗
The important point is that, in this type of model, both advection and production terms are
exact. Thus, the terms that supply energy to a particular Reynolds-stress component don’t need
modelling. For example, the rate of production of 𝑢
2
per unit mass is:
11
2
Assessment.
For:
- The “energy-in” terms (advection and production) are exact, not modelled; thus,
RSTMs should take better account of turbulence physics (in particular, anisotropy) than
eddy-viscosity models.
Against:
- Models are very complex;
- Many important terms (notably redistribution and dissipation) require modelling;
- Models are computationally expensive (6 turbulent transport equations) and tend to be
less stable; (only the small molecular viscosity contributes to any sort of gradient
diffusion).
2
The classic reference here is:
Launder, B.E., Reece, G.J. and Rodi, W., 1975, Progress in the development of a Reynolds-stress turbulence
closure, J. Fluid Mech., 68, 537-566.
8. 2 .2 Non-Linear Eddy-Viscosity Models (NLEVM)
3
A “half-way house” between eddy-viscosity and Reynolds-stress transport models, the idea
behind this type of model is to extend the simple proportionality between Reynolds stresses
and mean-velocity gradients:
stress ∝ (velocity gradient)
to a non-linear constitutive relation:
stress = 𝐶
1
(velocity gradient) + 𝐶
2
velocity gradient
2
3
velocity gradient
3
(The actual relationship is tensorial and highly mathematical – see the optional Section 10).
Models can be constructed so as to reproduce the correct anisotropy ( 28 ) in simple shear flow,
as well as a qualitatively-correct response of turbulence to certain other types of flow: e.g.
curved flows. Experience suggests that a cubic stress-strain relationship is optimal.
Assessment.
For:
- produce qualitatively-correct turbulent behaviour in certain important flows;
- only slightly more computationally expensive than linear eddy-viscosity models.
Against:
- don’t accurately represent the real production and advection processes;
- little theoretical foundation in complex flows. 8. 2 .3 Large-Eddy Simulation (LES)
Resolving a full, time-dependent turbulent flow at large Reynolds number is impractical as it
would require huge numbers of control volumes, all smaller than the tiniest scales of motion.
Large-eddy simulation solves the time-dependent Navier-Stokes equations for the
instantaneous (mean + turbulent) velocity that it can resolve on a moderate size of grid and
models the subgrid-scale motions. The model for the latter is usually very simple, typically a
mixing-length-type model with 𝑙 𝑚
proportional to the mesh size.
8. 2 .4 Direct Numerical Simulation (DNS)
This is not a turbulence model. It is an accurate solution of the complete time-dependent,
Navier-Stokes equations without any modelled terms!
This is prohibitively expensive at large Reynolds numbers as huge numbers of grid nodes
would be needed to resolve all scales of motion. Nevertheless, supercomputers have extended
the Reynolds-number range to a few thousand for simple flows and these results have assisted
greatly in the understanding of turbulence physics and calibration of simpler models.
3
For some of the mathematical theory see:
Apsley, D.D. and Leschziner, M.A., 1998, A new low-Reynolds-number nonlinear two-equation turbulence model
for complex flows, Int. J. Heat Fluid Flow, 19, 209- 222
0
𝜇
1 / 4
𝑃
1 / 2
and derive the relationship between 𝑈
𝑝
and 𝜏
𝑤
by assuming a kinematic eddy viscosity
𝑡
0
Then, assuming constant stress (𝜏 = 𝜏 𝑤
𝑡
𝑤
0
This can be rearranged and integrated for :
𝑤
0
𝑤
0
ln(𝐶𝑦)
Applying this at the near-wall node P , and making sure that it is consistent with ( 29 ) in the
equilibrium case where 𝑢 0
𝜏
and 𝜏
𝑤
𝜏
2
, fixes constant of integration 𝐶 and leads to
𝑃
𝑤
0
ln (𝐸
𝑃
0
or, rearranging for the wall shear stress:
𝑤
0
𝑃
ln (𝐸
𝑃
0
Since the code will discretise the velocity gradient at the boundary as 𝑈
𝑃
𝑃
this is
conveniently implemented via an effective wall viscosity μ 𝑤
, such that
𝑤
𝑤
𝑃
𝑃
where
𝑤
0
𝑃
ln (𝐸
𝑃
0
0
𝜇
1 / 4
𝑃
1 / 2
Amendments also have to be made to the turbulence equations, based on assumed profiles for
𝑘 and 𝜀. In particular, the production of turbulence energy is a cell-averaged quantity,
determined by integrating across the cell and the value of 𝜀 is specified at the centre of the near-
wall cell, not at the boundary.
To use these equilibrium profiles effectively, it is desirable that the grid spacing be such that
the near-wall node lies within the logarithmic layer; ideally,
𝑃
This has to be relaxed somewhat in practice, but it means that when using wall functions the
grid can not be made arbitrarily small in the vicinity of solid boundaries. In practice, many
commercial codes use an “all-𝑦
” wall treatment, and blend low-Re and wall-function
treatments, depending on the size of the near-wall cell.
Summary
- A turbulence model is a means of specifying the Reynolds stresses (and other turbulent
fluxes), so closing the mean flow equations.
- The most popular types are eddy-viscosity models , which assume that the Reynolds
stress is proportional to the mean strain; e.g. in simple shear:
(turb)
𝑡
may be specified geometrically (e.g. mixing-length models) or
by solving additional transport equations. A popular combination is the 𝑘 − 𝜀 model.
- More advanced turbulence models include:
- Reynolds-stress transport models (RSTM; solve transport equations for all stresses)
- non-linear eddy-viscosity models (NLEVM; non-linear stress-strain relationship)
- large-eddy simulation (LES; time-dependent calculation; model sub-grid scales)
- Wall boundary conditions require special treatment because of large flow gradients and
selective damping of wall-normal velocity fluctuations. The main options are low-
Reynolds-number models (fine grids) or wall functions (coarse grids).
References
Apsley, D.D., 2007, CFD calculation of turbulent flow with arbitrary wall roughness, Flow,
Turbulence and Combustion, 78, 153-175.
Apsley, D.D. and Leschziner, M.A., 1998, A new low-Reynolds-number nonlinear two-
equation turbulence model for complex flows, Int. J. Heat Fluid Flow, 19, 209-222.
Cebeci, T. and Smith, A.M.O., 1974, Analysis of Turbulent Boundary Layers , Academic.
Launder, B.E., Reece, G.J. and Rodi, W., 1975, Progress in the development of a Reynolds-
stress turbulence closure, J. Fluid Mech., 68, 537-566.
Launder, B.E. and Spalding, D.B., 1974, The numerical computation of turbulent flows, Comp.
Meth. Appl. Mech. Eng., 3, 269-289.
Leschziner, M.A., 2015, Statistical turbulence modelling for fluid dynamics - demystified: an
introductory text for graduate engineering students , World Scientific.
Menter, F.R., 1994, Two-equation eddy-viscosity turbulence models for engineering
applications, AIAA J., 32, 1598-1605.
Pope, S.B., 2000, Turbulent flows , Cambridge University Press.
Schlichting, H. and Gersten, K., 1999, Boundary layer theory , 8
th
English Edition, Springer-
Verlag
Wilcox, D.C., 1988, Reassessment of the scale-determining equation for advanced turbulence
models, AIAA J., 26, 1299-1310.
Wilcox, D.C., 2006 , Turbulence Modelling for CFD , 3
rd
ed., DCW Industries.
where we have used the commuting of 𝑢
𝑗
′
and 𝜕/𝜕𝑥
𝑗
from (2) above wherever necessary and
rearranged the viscous term to start with the outer derivative of a product. Recognising
1
2
𝑖
′
𝑖
′
as 𝑘, changing the dummy summation index from 𝑖 to 𝑗 in the pressure term, and rearranging,
gives:
𝑗
𝑗
𝑗
𝑗
𝑗
′
𝑖
′
𝑖
′
𝑗
′
𝑖
′
𝑗
′
𝑖
𝑗
𝑖
′
𝑗
𝑖
′
𝑗
Identifying individual physical processes:
D𝑘
D𝑡
𝑖
(𝑘)
𝑖
(𝑘)
where:
𝑗
(𝑘)
𝑗
𝑗
′
𝑖
′
𝑖
′
𝑗
′
diffusion (viscous, pressure and triple correlation)
(𝑘)
𝑖
′
𝑗
′
𝑖
𝑗
production (by mean velocity gradients)
𝑖
′
𝑗
2
dissipation (by viscosity)
Examples
Q 1.
In high-Reynolds-number turbulent boundary-layer flow over a flat surface the mean shear
stress is made up of viscous and turbulent parts:
where 𝜇 is the molecular viscosity. In the lower part of the boundary layer the shear stress is
effectively constant and equal to the wall shear stress 𝜏 𝑤
(a) Define the friction velocity 𝑢 𝜏
(b) Show that, sufficiently close to a smooth wall, the mean velocity profile is linear, and
write down an expression for 𝑈 in terms of 𝜏
𝑤
, μ and the distance from the wall, 𝑦.
(c) At larger distances from the wall the viscous stress can be neglected, whilst the
turbulent stress can be represented by a mixing-length eddy-viscosity model:
𝑡
where
𝑡
0
𝑚
𝑚
0
𝑚
and 𝜅 (≈ 0. 41 ) is a constant. Again, assuming that 𝜏 = 𝜏
𝑤
, show that this leads to a
logarithmic velocity profile of the form
𝜏
ln(𝐸
𝜏
where 𝐸 is a constant of integration.
(d) Write the velocity profiles in parts (b) and (c) in wall units.
(e) In simple shear flow, the rate of production of turbulent kinetic energy per unit mass is
(𝑘)
Using the results of (c), prove that, in the logarithmic velocity region,
(𝑘)
𝜏
3
and explain what is meant by the statement that the turbulence is in local equilibrium.
Q 4.
In the analysis of turbulent flows it is common to decompose the velocity field into mean
(𝑈, 𝑉, 𝑊) and fluctuating (𝑢, 𝑣, 𝑤) parts as part of the Reynolds-averaging process.
(a) The rate of production of the 𝑢𝑢 stress component per unit mass is given by
11
By inspection/pattern-matching, write down an analogous expression for 𝑃
22
(b) Define the term anisotropy when applied to fluctuating quantities in turbulent flow and
give two reasons why, for turbulent boundary layers along a plane wall 𝑦 = 0 , the wall-
normal velocity variance is smaller than the streamwise variance.
(c) Describe the main principles of, and the main differences between
(i) eddy-viscosity
(ii) Reynolds-stress transport
models of turbulence, and give advantages and disadvantages of each type of closure.
Q 5.
(a) By considering momentum transport by turbulent fluctuations show that −𝜌𝑢
𝑖
′
𝑗
′
can
be interpreted as an additional effective stress in the mean momentum equation.
(b) For a linear eddy-viscosity turbulence model with strain-independent eddy viscosity 𝜇 𝑡
write expressions for the typical shear stress −𝜌𝑢′𝑣′ and normal stress −𝜌𝑢′
2
in an
arbitrary velocity field. Show that, for certain mean-velocity gradients, this type of
model may predict physically unrealisable stresses.
(c) Explain why, for a zero-pressure-gradient, fully-developed, boundary-layer flow of the
form (𝑢(𝑦), 0 , 0 ), the mean shear stress 𝜏 is independent of distance 𝑦 from the
boundary. Find the mean-velocity profile if the total effective viscosity 𝜇
eff
(i) is constant;
(ii) varies linearly with wall distance: 𝜇
eff
= 𝐶𝑦, where 𝐶 is a constant.
(d) In the widely-used 𝑘 − ε model of turbulence:
(i) state the physical quantities represented by 𝑘 and 𝜀;
(ii) write down the expression for eddy viscosity 𝜇
𝑡
in terms of 𝜌, 𝑘 and 𝜀 in the
standard 𝑘 − 𝜀 model;
(iii) explain briefly (and without detailed mathematics) how 𝑘 and 𝜀 are determined.
(e) What special issues arise in the modeling and computation of near-wall turbulent flow?
State the two main methods for dealing with the solid-wall boundary condition and give
a brief summary of the major elements of each.
Q 6.
In a general incompressible velocity field (𝑈, 𝑉, 𝑊) the turbulent shear stress component 𝜏 12
is given, for an eddy-viscosity turbulence model, by
12
𝑡
where 𝜇
𝑡
𝑡
) is the dynamic eddy viscosity and 𝜌 is density. Use pattern-matching or
index-permutation and the incompressibility condition to write expressions for the other
independent stress components: 𝜏 23
31
11
22
33
Q 7.
In the 𝑘 − 𝑇 turbulence model, 𝑇 is a turbulent time scale and the eddy-viscosity formulation
takes the form
𝑡
𝛼
𝛽
𝛾
where 𝐶 is a dimensionless constant. Use dimensional analysis to find 𝛼, 𝛽 and 𝛾.
Q 8. (Exam 2017 – part)
In turbulent flow, mean and fluctuating components are often denoted by an overbar ( ¯ ) and
prime ( ) respectively. The fluctuating velocity components in the 𝑥, 𝑦, 𝑧 or 1, 2, 3 coordinate
directions are 𝑢′, 𝑣′ and 𝑤′ respectively. At a particular point in a flow of air a hot-wire
anemometer measures the following turbulent statistics:
2
= 2. 6 m
2
s
− 2
2
= 1. 4 m
2
s
− 2
2
= 2. 0 m
2
s
− 2
𝑢′𝑣′ = − 0. 9 m
2
s
− 2
The density of air is 𝜌 = 1. 2 kg m
(a) At this point determine:
(i) the turbulent kinetic energy (per unit mass), 𝑘;
(ii) the dynamic shear stress 𝜏
12
(b) The only non-zero mean-velocity gradients are
= 4 s
− 1
= − 1. 5 s
− 1
= 1. 5 s
− 1
Assuming a linear eddy-viscosity model of turbulence, deduce the eddy viscosity 𝜇
𝑡
on
the basis of the shear stress found in part (a)(ii).
(c) Using the eddy viscosity calculated in part (b), what does the turbulence model predict
for the fluctuating velocity variances 𝑢′
2
2
and 𝑤′
2
