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Transport Phenomena 3

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Title: Transport Phenomena 3


1
Transport Phenomena 3
  • Non-Newtonian Flow
  • Navier-Stokes Equations and Applications
  • Turbulent Flow and Boundary Layers
  • Method of assessment 100 written exam

2
CE30033 - Course Objectives
After taking this course you should be able to-
  • Describe a wide variety of non-Newtonian fluid
    behaviour and carry out basic calculations.
  • Be able to give examples of the applications of
    non-Newtonian fluids for typical consumer and
    industrial products, and explain why particular
    properties are required for specific products.
  • Understand the basic foundations of fluid
    mechanics
  • know how Newtons laws are applied to fluids
  • understand the assumptions behind the Continuum
    Hypothesis
  • Be familiar with the shorthand notation used in
    vector calculus.
  • Understand the physical basis of the Continuity
    equation and be able to derive it for different
    coordinate systems.
  • Understand the physical basis of the
    Navier-Stokes equations and be able to identify
    the nature and source of each of the terms.

3
CE30033 - Course Objectives
  • Understand the physical significance of the
    Reynolds number.
  • Know several different methods of solving PDEs
  • be able to identify the particular aspects of
    physical systems that make them amenable to an
    analytical solution of the Navier-Stokes
    equations
  • be able to solve the Navier-Stokes equations
    analytically in any coordinate system for simple
    systems
  • understand when a numerical solution of a PDE is
    necessary and be able to set up a basic finite
    difference scheme
  • Describe the physical nature of turbulent flow.
  • Describe the consequences of the presence of
    turbulence for the Navier-Stokes equations.
  • Understand the physical interpretation of the
    Reynolds stresses.
  • Be able to outline the basic principles of the
    Mixing Length Theory and explain why it is needed.

4
CE30033 - Course Objectives
  • Be able to describe and derive the velocity
    profiles for turbulent flow.
  • Be able to obtain expressions for heat transfer
    coefficients from velocity profiles in turbulent
    flow.

5
Recommended Reading for CE30033
  • Maths
  • Mathematical methods in the physical sciences
    by Mary L. Boas
  • Mathematical methods for science students by G.
    Stephenson
  • You may also find it of interest that there are
    good introductions to some of the maths (PDEs,
    finite difference methods) used in this course
    in
  • The mathematics of financial derivatives by P.
    Wilmott et al.
  • An introduction to the maths of financial
    derivatives by S.N. Neftci
  • Non-Newtonian Fluids
  • Non-Newtonian flow in the process industries by
    R.P. Chhabra and J.F. Richardson
  • Fluid Mechanics
  • Physical fluid dynamics by D.J. Tritton
  • Fluid mechanics and transfer processes by J.M.
    Kay and R.M. Nedderman
  • Transport phenomena by R. Bird, W.E. Stewart
    E.N. Lightfoot
  • Introduction to transport phenomena by W.J.
    Thomson

6
Maths Revision Co-ordinate Systems
z
(r,q,z)
y
z
Cylindrical polars
r
q
x
Cartesian (x,y,z)
7
Elemental Volumes in Different Co-ordinate Systems
z
dr
z
dz
r.dq
dz
y
y
dx
dq
dy
q
x
r
r.dq
x
dV dx.dy.dz
dV r.dr.dq.dz
Cartesian
Cylindrical polars
8
Non-Newtonian Fluid Behaviour
1. Classification of fluid behaviour 1.1
Definition of a Newtonian Fluid
First subscript direction normal to that of
shearing force Second subscript direction of
force
(1)
Where velocity v (u,v,w)
For an incompressible fluid of density r,
Equation (1) can also be written as
(2)
The quantity (ru) is the momentum in the
x-direction per unit volume of the fluid. tyx
represents the momentum flux in the y-direction
9
QFS Check units of tyx to see if they are
consistent with momentum flux. What does m/r
represent ? What are its units ? For Newtonian
behaviour (1) tyx is proportional to gyx and a
plot passes through the origin and (2) by
definition the constant of proportionality, m, is
independent of both shear rate and shear
stress. QFS For a Newtonian fluid the viscosity
can depend upon three factors. What are they ?
Give examples of Newtonian fluids. 1.2
Non-Newtonian Fluid Behaviour The flow curve
(shear stress vs. shear rate) is either
non-linear, or does pass through the origin, or
both. Three classes can be distinguished. (1)
Fluids for which the rate of shear at any point
is determined only by the value of the shear
stress at that point at that instant these
fluids are variously known as time independent,
purely viscous, inelastic, or Generalised
Newtonian Fluids (GNF).
10
(2) More complex fluids for which the relation
between shear stress and shear rate depends, in
addition, on the duration of shearing and
their kinematic history they are called
time-dependent fluids. (3) Substances
exhibiting characteristics of both ideal fluids
and elastic solids and showing partial elastic
recovery, after deformation these are
characterised as visco-elastic fluids. 1.3
Time-Independent Fluid Behaviour 1.3.1 Shear
thinning or pseudoplastic fluids Over a limited
range of shear-rate (or stress) log (t) vs. log
(g) is approximately a straight line of negative
slope. Hence tyx m(gyx)n (3) where m fluid
consistency coefficient n flow behaviour
index QFS If a power-law (Ostwald de Waele)
fluid is shear-thinning, is nlt1 or ngt1 ?
11
QFS Re-arrange Eq. (3) to obtain an expression
for apparent viscosity mapp (
tyx/gyx) Expressions that are applicable over a
wider range of shear rates are covered
later. 1.3.2 Viscoplastic Fluid
Behaviour Viscoplastic fluids behave as if they
have a yield stress (t0). Until t0 is exceeded
they do not appear to flow. A Bingham plastic
fluid has a constant plastic viscosity
for
for
Often the two model parameters t0B and mB are
treated as curve fitting constants, even when
there is no true yield stress.
12
1.3.3 Shear-thickening or Dilatant Fluid
Behaviour Eq. (3) is applicable with ngt1. 1.4
Time-dependent Fluid Behaviour Apparent viscosity
depends not only on the rate of shear but on the
time for which fluid has been subject to
shearing. If apparent viscosity decreases with
time of shearing, then the fluid is
thixotropic. Fluid movement can, under certain
critical conditions, lead to re-formation of
structure which results in an increased
resistance to relative motion. Thus apparent
viscosity increases with length of shearing. This
behaviour is known as rheopexy. 1.5
Visco-elastic Fluid Behaviour A visco-elastic
fluid displays both elastic and viscous
properties. A true visco-elastic fluid gives time
dependent behaviour.
13
2. Laminar Flow in Circular Pipes
(1)
Radial surface, z-direction
2.1 Power-Law Fluids
(2) where w is the z-component of velocity
Combine (1) and (2), and integrate
Apply no-slip boundary condition, at r R, w 0
14
Hence
For a Newtonian fluid
15
2.2 Bingham Plastic
Eq. (1) still applies trz 0 on centre line
t0B not exceeded until finite r
(6)
Flow
c
Plug flow
C.L.
At r R, w 0, hence
(7)
Eq. (7) applies only for trz gt t0B, i.e. for r gt
c
16
What about core in plug flow ? At r c, use Eq.
(7)
Also at r c
From Eq. (1)
Hence velocity of the core wzc
17
(9) where
3. Examples Newtonian flow occurs for simple
fluids, such as water, petrol, and vegetable
oil. The Non-Newtonian flow behaviour of many
microstructured products can offer real
advantages. For example, paint should be easy to
spread, so it should have a low apparent
viscosity at the high shear caused by the
paintbrush. At the same time, the paint should
stick to the wall after its brushed on, so it
should have a high apparent viscosity after it is
applied. Many cleaning fluids and furniture
waxes should have similar properties.
18
The causes of Non-Newtonian flow depend on the
colloid chemistry of the particular product. In
the case of water-based latex paint, the
shear-thinning is the result of the breakage of
hydrogen bonds between the surfactants used to
stabilise the latex. For many cleaners, the shear
thinning behaviour results from disruptions of
liquid crystals formed within the products. It is
the forces produced by these chemistries that
are responsible for the unusual and attractive
properties of these microstructured
products. For more information, see Chemical
Product Design by EL Cussler and GD Moggridge
19
Navier-Stokes Equations1. Basic Building Blocks
1.1 Co-ordinate system Caresian Position vector,
p p (x,y,z) Velocity, u u (u,v,w) Note
we shall use an Eulerian frame of reference (an
inertial frame), i.e. u is measured relative to a
fixed point in space, the origin O. 1.2 The
Continuum Hypothesis What do we mean by a
velocity, a pressure or a density at a fixed
point in space ? Consider an average density
averaged over volume L3...
20
Contains many molecules
r
Volume contains a small no. of molecules
Spatial variations due to changes in, say, P or T
Typical length scale of the flow
Much larger than molecular dimensions
L2
L1
L3
L
The Continuum Hypothesis assumes that we
have L1ltltL2ltltL3 We suppose that the fluid can be
treated as a continuum (i.e. not made up of
individual molecules) with the properties of the
plateau region. This hypothesis is not valid
for (i) Rarefied gases - L2l (mean free
path) (ii) Knudsen diffusion - diffusion within
very small pores where, d, the pore
diameterltl and L3d (i.e. no plateau)
21
1.3 Conservation Equations (i) Conservation of
mass or chemical species - scalar (ii)
Conservation of energy (first law of
thermodynamics) - scalar (iii) Conservation of
momentum (Newtons second law) - vector Define a
control volume V fixed in space and bounded by a
surface S. Conservation of property f (any
of (i) to (iii)) Rate of accumulation of f in V
Net flow rate of f out through S Sources of f
inside V - Sinks of f inside V
S
V
22
2. Equations of Motion
2.1 Continuity Equation Apply Conservation of
Mass to a small control volume
dV dx.dy.dz through which there is fluid flow.
rw
dy
ru
dz
dx
Rate of accumulation of mass in dV
rv
23
Rate of inflow through S
Rate of outflow through S
Mass is neither created or destroyed, thus no
sources or sinks. Hence, applying conservation
and dividing by dV dx.dy.dz
or
If the fluid is incompressible, r a constant,
then
Note this is true even for unsteady flows
24
2.2 Convective Derivative Even in a steady flow,
a fluid particle can change its momentum by
moving to a position where its velocity is
different- this acceleration requires a force to
be applied to the fluid particle. Before we
consider the momentum equation well look at a
simpler example of heat transfer.
Consider the change in temperature of the
fluid particle in dt in which it moves a
distance (dx, dy, dz)
Fluid flow
Hot
Decreasing T
Hot
25
We write this as a total or SUBSTANTIVE derivative
Define the operator
2.3 Energy Equation To illustrate the use of the
substantive derivative let us consider
an enthalpy balance on an element of fluid with
volume dV. Enthalpy is transferred by (i)
convection and (ii) conduction - Fouriers
law Define heat flux vector
Heat flux (W/m2) may be different in the x, y and
z directions. The thermal conductivities, kx, ky
and kz are direction dependent -e.g. anisotropic
materials, such as crystals, wood, foodstuffs.
26
For simplicity consider a 2D element (dz 1,
d/dz 0, w 0) Also assume r constant.
Unit width
qx u T
dy
dx
Rate of accumulation of enthalpy
qy, v, T
Rate of convective inflow of enthalpy
Rate of convective outflow of enthalpy
27
Rate of diffusive outflow - inflow of enthalpy
Apply First Law of Thermodynamics - Energy
conserved
(V)
(III)
(IV)
(I)
(II)
  • Note
  • (I) and (II) are the convective
  • derivative
  • (III) is zero by continuity
  • (IV) is for diffusive transport
  • Suppose k constant (isotropic and homogeneous)

28
  • (V) is a source/sink e.g. due to chemical
    reaction, electrical heating,
  • viscous dissipation.
  • Hence, generalising our derivation

Note units of k/rCp are m2/s i.e. it is a
thermal diffusivity.
Unsteady
Source/ sink
Convective
Diffusive
2.5 The Momentum Equation (Navier-Stokes) Momentum
is a vector quantity, however, if we consider
transport in a single direction, then we can
use the same treatment as for energy or species
conservation. x-momentum per unit volume ru The
total rate of change of x-momentum for an
incompressible fluid (per unit volume)
x-velocity
Vector velocity
29
From Newtons second law Rate of change of
momentum S applied force Applied Forces (I)
Body Forces e.g. gravity (gx,gy,gz) The
x-component of the body force acting on a control
volume dV is r.gx.dV (II) Pressure Pressure force
in the x-direction
P
dy
dz
dx
(III) Viscous stresses sxx normal stress, ve in
tension tyx shear stress
sxx
Stress acts in x-direction
y
In plane defined by normal y
tyx
x
30
N.B. the shear stress tzx also contributes to the
force in the x-direction. Hence the force in the
x-direction due to viscous stresses is
Hence, from Newtons second law, total rate of
change of momentum in the x-direction
As with enthalpy we require a constitutive
equation to relate the flux of the transported
quantity to the gradient of the quantity. 2.6.1
The Stress Tensor In the previous section we
separated the normal stresses into (I)
pressure and (II) viscous normal stresses. The
total normal stress, txx etc. txx sxx - P tyy
syy - P the -ve sign is because of sgt0 in
tension, tzz szz - P Pgt0 for compressive
stresses
31
At each point in a continuous medium, whether it
is solid or fluid, we need six numbers, each of
them representing a component of force per unit
area, to define the local stress completely.
Consider an infinitesimal cubic element
tyy
tyx
txy must always equal tyx even in the absence
of mechanical equilibrium. If not the cube would
experience a couple about the z-axis and begin to
rotate.
Normal stresses must balance for mechanical equili
brium
txy
txx
txx
txy
tyx
tyy
So the deviatoric stress tensor (viscous
stresses) is i.e. a symmetric matrix -
all diagonal elements are equal
32
2.6.2 Newtonian Fluids If planar laminae of fluid
lying normal to the y-direction are moving
steadily in the x-direction and sliding over one
another, so that there exists a velocity gradient
, Newton postulated that in such circumstances
a frictional shear stress arises between
adjacent laminae of magnitude
y
x
(2.6.2.1)
where m -the viscosity, or shear viscosity-
depends on the nature of the fluid. The
assumption of a linear relationship between shear
stress and velocity gradient satisfies the basic
requirement that when the velocity gradient
vanishes everywhere, in which case an inertial
frame exists in which the fluid is at rest in
equilibrium, tyx must vanish. It also
satisfies the symmetry requirement that tyx must
change its sign when the motion is reversed.
33
However the above equation cannot suffice in more
complex situations. If a fluid is not only moving
along the x-direction but also along the
y- direction, and two non-zero velocity
gradients exist, then a shear stress would
exist on planes normal to the x-axis of magnitude
(2.6.2.2)
If the properties of the fluid are isotropic then
the value of m would be the same. However,
Equations (2.6.2.1) and (2.6.2.2) do not satisfy
the requirement that txy and tyx are always
equal. Evidently, we must symmetrise them by
writing
It is possible to define a total rate of
deformation as
Note that, like the stress tensor, the rate of
deformation tensor is symmetric, i.e. exy eyx.
34
The Denatoric stresses are related to the rate of
strain by the general relationship (due to
Newton) tij 2m x eij (Newtons law of
viscosity) i.e.
and similarly for normal stresses syy, szz, and
and similarly for other shear stresses.
Note (I) for 1D shearing flows, u f(y) and v
0, thus (II) all other elements in the stress
tensor can be found by rotation (x, y, z) and
(u, v, w) x, u z, w y, v
35
2.6.3 Navier-Stokes (N-S) Equations for
Incompressible Flow Using the expressions above
for the viscous stresses, the stress gradients
in the N-S equations become
0, since by Continuity
So the x-momentum equation is
and similarly for the y, z directions. We can
write all three momentum equations together,
using vector notation, as
36
2.7 The N-S Equation in Non-Dimensional
Form Suppose we have U characteristic velocity
(say, the mean velocity in pipe flow) L
characteristic length (say, pipe diameter) Define
a dimensionless velocity and dimensionless
lengths Then
and
so the N-S eq. for x-momentum conservation become
s...
where
37
with similar expressions for the y- and z-
directions. N.B. we obtain dynamic similarity at
two scales of operation, simply by holding the
Reynolds number constant at both scales. There
are two special cases to consider (a) Re is very
large inertial effects dominate over viscous
effects Solution by Potential Flow
theory where f velocity potential
for steady flow, Eulers Eqn.
(b) Re tends to 0, inertial effects are
negligible Stokes eqn. -creeping flow
38
3. Some Solutions to the N-S Equation 3.1 Is the
Eqn. Set Closed ? Within the flow domain, the
fluid properties m and r are known, but
the velocity u (u, v, w) and the pressures are
unknown. Thus we have 4 unknowns - requires 4
independent equations for a unique
solution. Eqn. (1) 4 equations Eqns.
(2)-(4) thus conclude a closed form solution is
possible. 3.2 Boundary Conditions On the edge of
the flow domain we need to know boundary
conditions for u and possibly for the
pressure. (1) Typically the situations of
interest to us are internal flows (wall bounded).
39
At the wall we have (I) impermeable wall, thus
v 0. (II) no slip condition, thus u w
0 (but note the no-slip condition is not
satisfied by a potential flow but condition (I)
must be satisfied) Free surface flows (I) at
the surface the pressure is a constant (II) tyx
tzx 0
y, v
x, u
z, w
y, v
x, u
air
z, w
water
40
  • 3.3 Falling Film Flow
  • Consider a film, of constant thickness d, falling
    down a slope, at an
  • angle a to the horizontal.
  • Steady - no time derivatives
  • r constant
  • laminar flow
  • 2D problem, thus

Free surface
y
d
x
g
a
v 0
Continuity Momentum equation y-component
-rg.cosa
0, steady flow
v 0
v 0
Pressure is constant along surface and at any
depth i.e.
41
x-component
rg.sina
0 steady flow
0 from above
v 0
u f(y)
0
Boundary conditions (I) at y 0, u 0, hence A
0 (II) tyx 0 i.e. at y d, du/dy 0, hence
B -d
42
  • 3.4 Radial Flow Between Two Parallel Discs
  • select appropriate co-ordinate system
  • (cylindrical polars)
  • very viscous flow
  • slow flow
  • incompressible
  • steady flow (no time derivative)
  • within gap

Inflow
z
q
z b
r
r1
g
r2
z -b
Continuity
0
0
z-component
0
0
0
0
0
0 steady flow
43
r-component
0 by continuity
0
0 steady flow
0
uq 0
Now r.ur f(z)
If we have creeping flow, Reltlt1, then the LHS 0,
giving
44
Pressure inside pipe (in excess
of external pressure)
Integrate with respect to r between r1 and r2
Integrate with respect to z twice
Boundary conditions at z 0, by symmetry, at
z /-b, no slip condition, ur 0,
45
3.5 Numerical Solution of the N-S Equations We
can only solve simple 2-D or 1-D unsteady flows.
The analytical solutions provide a continuous
description of u and P. We can find values of u
and P at discrete positions within the flow by
numerical solutions. Procedure (I) discretise
the flow domain by overlapping with a grid (II)
approximate the derivatives in the N-S equations
using difference formulae (III) this leads to a
large set of algebraic equations for the discrete
values of u, v, w etc. (IV) commercial packages,
such as Fluent generate the grid and then
solve the algebraic equations.
46
3.6 Finite Difference Solutions to Transport
Problems e.g generalised energy balance (
generation, PE, KE) For steady-state
diffusion or conduction, with no convection
or reaction
Laplaces Equation
0, steady state
0, no convection
0, no reaction
Numerical solutions One dimension Taylor
Series
Errors of order h4
(1)
(2)
47
Add (1) and (2)
Laplaces equation
(3)
Laplaces equation is satisfied if C(x) is the
average of neighbouring values. This is a 1D
Finite Difference Method. Eq. (3) is equivalent
to approximating
2D
2
C0 (C1 C2 C3 C4)/4 and similarly in 3D
h
0
1
3
y
4
x
48
Example
C1
y
2
1
3
x
Zero flux/ impermeable boundary
Boundary conditions C1 C2 C3 1 C7 C8
C9 0
5
a
4
6
8
9
7
C0
a
Hypothetical point outside the boundary with CC5
On the left and right boundaries Subtract (2)
from (1)
3
6
5
5/
h
On boundaries C/(x)0, thus C(xh) C(x-h)
9
49
Finite difference equations are 4C4 C1 2C5
C7 4C5 C2 C4 C6 C8 4C6 C3 2C5
C9 Substitute boundary conditions 4C4 1
2C5, 4C5 1 C4 C6, 4C6 1
2C5 Thus C4 C5 C6 Solution procedures Use
Iteration
21
h
Value at (n1)th step
11
10
12
01
and so on for each point. Alternatively, could
use the Jacobi method, or the Gauss-Seidel
method.
50
More complex boundary conditions General
boundary condition for a second order partial
differential equation (could have a(y),
b(y) etc.)
E.g surface mass transfer resistance Mass
transfer coefficient, kg bulk concentration, CA
diffusion coefficient, D Mass balance
2
gas
solid
1
3
0
hypothetical
x
4
Biot number, Bi 2kgh/D (external mass
transfer rate)/(internal mass transf. rate)
(internal resistance)/(external resistance)
51
Limits Bi 0, C3 C1 (zero flux) Bi
infinity, C0 CA (no surface resistance)
52
3.7 Non Steady-State Diffusion We are going to
consider mass uptake into an adsorbent pellet
which we will consider to be a homogeneous,
porous, sphere of radius a. This problem is of
relevance when designing adsorption and
heterogeneous catalytic processes. (i) The
adsorbent particle is initially in contact with
the adsorbate vapour at a lower pressure pi (or
with a vacuum). (ii) A step increase in the
external adsorbate pressure to p0 occurs at time
t 0. (iii) The mass uptake of adsorbate into
the pellet is measured over time. If diffusion
occurs in the radial direction and the
diffusivity is a constant the process is
governed by the diffusion equation
53
Pressure or Sample weight
Real pressure step
Ideal pressure step
Pressure p0
Mt/
Mf
Mt
p pi
Time
t 0
t t
t/ 0
Boundary conditions for our problem C C0 r
a t gt 0 C C1 t 0 0 lt r lt a
Solution is
54
The total amount of diffusing substance entering
(or leaving) the sphere is given by
However, for Mt / Mf gt 0.5 this equation can be
well approximated by
where k is a mass transfer coefficient equivalent
to Dp2/a2. This approximate solution to the full
mass uptake equation is known as the Linear
Driving Force (LDF) model. Mass uptake
experiments can be used to measure D for use
in calculations for designing packed bed
catalytic reactors etc.
55
4. Turbulent Flow
4.1 Experimental Observations In single phase
flow we observe two regimes - laminar and
turbulent, depending on Reynolds number. Above a
critical value Recrit we have turbulent flow -
Round pipe, Recrit2000-3000 - Flow around a
sphere, Recrit104-105 (D dia. of sphere) -
Falling Film Flow, Recrit 1000 (D film
thickness) Turbulent flows are characterised
by (I) Chaotic motion which vary in time and
space. (II) 3-D velocity fluctuations, which give
rise to increased rates of momentum, mass and
heat transport. (III) Mechanical energy, i.e.
kinetic energy is dissipated by viscous effects
to heat.
56
4.2 Averaging Process 4.2.1 Averages Consider a
perfect velocimeter, which measures the
instantaneous velocities at a point.
If u is statistically steady over a time period
Dtgtgtt, then we can define a time averaged
velocity, u, as
tk characteristic time for fluctuations
u
Strictly we require Dt tending to infinity, but
Dtgtgtt will be OK. Compare this with the in-situ
or spatial mean average a pipe
radius for flow in a pipe. Similarly we could do
this for temperatures.
i.e. the mean temperature we would obtain by
closing the pipe ends and mixing the contents.
57
4.2.3 Characterising the Velocity Fluctuations
We will separate u(t) u(t) u u/(t)
u/ gt 0
Fluctuating component
u
Time average
Instantaneous
u/ lt 0
We characterise the fluctuating component by the
average of the square of u/
root mean square (RMS) velocity
Rules for averaging - suppose we have independent
variables u(x, t), v(x,t) (I) (II) (III)
(IV)
0
covariance
58
4.3 Incompressible Turbulent Flow The continuity
equations (derived in sec. 2.6) apply
instantaneously in turbulent flows. We will
decompose the motion into mean and fluctuating
components, and then time average the equations -
we will obtain a description of the mean velocity
and pressure fields. Continuity
N-S x-momentum equation (Steady) Firstly we can
simplify things by writing
Convection of momentum through control volume
0 by Continuity
59
Using the averaging rules
(I)
(II)
(IV)
(III)
60
  • Term(I) is simply i.e. the convective derivative
    of the
  • (steady) mean flow.
  • Term (III) and (IV) are the pressure and viscous
    stress gradient terms
  • for the mean flow
  • Term (II) is an extra term due to the time
    averaging process. Usually
  • we write this equation, combining terms (II) and
    (IV) as
  • These extra terms are unknowns and have the form
    of an apparent
  • stress - they are known as Reynolds stresses. We
    have only carried
  • out the analysis in the x-direction Reynolds
    stress terms also appear
  • in the y- and z-direction time averaged
    Navier-Stokes equations.

61
4.4 Physical Interpretation of the Reynolds
Stresses Consider a 1-D channel flow
v y
y
v0
u
u x
x
A1
The instantaneous volume flow rate through the
plane is vA the instantaneous x-momentum flow
rate through the plane is r.u.v.A The time
averaged momentum flux, This must be balanced by
a stress in the x-direction Reynolds
stress since v 0 Note u/v/ is only zero
when (i) u/0 or v/0, i.e. Laminar
flow or (ii) when the velocities are uncorrelated
(i.e. independent)
62
As in statistics, we can define a correlation
coefficient
y
u(y)
0 for no correlation /-1 for perfect
correlation
u/lt0
rms velocities
v/gt0
Shear flows Positive shear positive v/ brings
with it low u velocity fluid so u/ lt
0. Similarly negative v/ brings with it high u
velocity fluid, thus u/ gt 0. So on average for
x
4.5 Summary Time averaging the N-S eqs. Gives
additional Reynolds stresses which are unknown.
For simple 1-D shear flows
laminar-type stresses
turbulent-type stresses
63
Similarly for heat transport
By averaging the N-S equations we generated 6
extra terms but no new equations - the equation
set is not closed - we can formulate transport
equations for u/v/ but they involve higher order
correlations e.g. u/v/w/. So we always have more
unknowns than equations - this is known as the
CLOSURE problem - we need an additional
hypothesis to obtain a solution
64
5. TURBULENCE MODELS 5.1 e.g. 1-D flow - we have
already seen that
Reynolds
viscous
kinematic viscosity, a molecular property (m2s-1)
The most straight-forward hypothesis is to assume
that the Reynolds stress can be modelled in the
same way as the viscous stress
e is NOT a molecular property, but depends on the
flow and varies in space.
Eddy viscosity (m2s-1) (gt0)
5.2 Mixing Length Theory (MLT) Prandtl (1925)
based his MLT on the (entirely false) analogy
between turbulent motion and the kinetic theory
of gases. (It is false because eddies are not
negligible in size compared to the flow). More
modern theories involve fractals and chaos
theory.
65
As in the Reynolds analogy we assume that fluid
elements take random jumps of length
(y2-y1). N.B. the jump length is a random
variable and y2-y1 0 A fluid element
transports mean momentum ru(y1) to position
y2 Linearized Taylor expansion (about position
y2)
y
u(y)
yy2
yy1
x
Assumes y2-y1 is small
This corresponds to the instantaneous ru(y2)
(1)
66
Prandtl assumed that u/ and v/ were correlated,
such that
Rms velocities
Correlation coefficient
and furthermore it was assumed that (a1
isentropic turbulence) and
Doesnt change with time
From Eq. 1 we have
Reynolds stress
MLT
Velocities are correlated
67
Define mixing length
So
We write it in this form so that (tyx)R lt0 when
Compare with the eddy viscosity hypothesis
In general, we considered transport of any
property f by turbulent velocity fluctuations.
Similar analysis, we suppose that the turbulent
flux of f is in each case
Thus for both molecular and turbulent transport
Momentum transport
68
Species transport
Enthalpy transport
N.B. the eddy viscosity is assumed to be the same
in each case, and depends only on the flow.
5.3 Velocity Distributions for Turbulent Flow in
a Pipe 5.3.1. Dimensional Analysis For a fully
developed turbulent flow, the time mean velocity
u is a function of y, the distance from the
wall u f(y, a, r, m, t0), where a is the
pipe radius Forming dimensionless groups
where t0 wall shear stress,
69
N.B. (t0/r)1/2 is a characteristic velocity,
known as the friction velocity, u. is called
the friction distance, y.
These are not physical quantities - merely
correlating parameters. In dimensionless
terms u f(y,a)
70
5.3.2 1/7 Power Law Fluid motion in turbulent
flow is extremely complex and there is no single
analytical expression which can give an accurate
representation of the velocity distribution
across the entire cross-section. Even if the
laminar sub-layers (boundary layers) are excluded
from the foregoing statement it is still true.
Consider the classic expression
Where a pipe radius u velocity at distance
y from the wall u1 centre line velocity This
expression is inadequate at y0 (this can be
shown by calculating du/dy and then putting y0)
which is to be expected. Also, it is inadequate
at ya ! The actual velocity profile must be flat
at the centre-line with du/dy 0 at ya.
71
5.3.3 Prandtls proposal for l We know that e
varies spatially and MLT suggests
but, how does l vary with the distance from the
wall ? (a) Near walls, yltlta We expect there to
be no effect of pipe curvature, so u
f(y) Close to the walls, viscous effects
dominate, since u tends to zero as y tends to
zero, and e0 (at least eltltn). If we assume that
the shear stress in the fluid t is approximately
the same as the wall shear stress, t0, then
0
Integrating
wall shear stress
Make dimensionless
Within Viscous Sub-Layer (VSL)
72
(b) log law of the wall Outside the VSL in the
turbulent core (TC) of a pipe flow we expect
We assume that at the edge of the TC we have
tt0 l is a transverse length scale and is going
to be proportional to the largest eddy size, at
position y. Prandtl proposed l ky where ky is
of the order of unity. (since the largest eddies
are of order y)
Make dimensionless
73
The constants k and B are found from fits to
experimental data. k, Von Karmans constant
0.4, and B 5.5 so that u 2.5 ln y
5.5 The experimental data fits the predicted
values well in the VSL and the TC, but there is
an intermediate region, known as the Buffer
Layer, where both viscous and Reynolds stresses
are important. Von Karman simply filled in this
region with a straight line of the same form as
in the TC u A ln y B The buffer layer
is between the VSL (ylt5) and the turbulent core
and the equation must fit smoothly with the
equations for the regions on either side.
74
Fitting slope at y 5,
Fitting value at y 5,
At y 30, we have the same value of u from the
buffer layer and TC equations, so pick that as
the boundary between the two. Hence Von-Karmans
Universal Velocity Profile (UVP) is u
y y lt 5 u 5 ln y - 3.05 5 lt y lt 30 u
2.5 ln y 5.5 30 lt y lt a/2 5.3.4 Variation
of e with Distance from the Wall Taking the UVP,
and assuming tt0
75
Make dimensionless and
Define
So
(i) In the viscous sublayer
(ii) In buffer layer
(iii) In turbulent core
76
e
20
TC e y/2.5
BL e y/5 - 1
5
VSL e 0
experiments
5
30
Clearly, we should not have a step change in e -
this is the major disadvantage in using the UVP.
Agreement in the range y lt 5 is also not very
good.
77
5.4 Heat Transfer Coefficients in Turbulent
Flow 5.4.1 The Taylor-Prandtl Analogy The
Taylor-Prandtl profile ignores the presence of a
buffer layer in the UVP - the VSL and TC
equations intercept at y 11.7, so VSL u
y e 0 y lt 11.7 TC u 2.5 lny 5.5 e
y/2.5 y gt 11.7 From sec. 5.2
k thermal conductivity (molecular)
Eddy thermal diffusivity eddy viscosity
As with the velocity profile, define
non-dimensional variables
78
Friction velocity
Wall heat flux
and assume q q0 and t t0
e
1/Pr (Pr Prandtl no. ratio of molecular
transports)
wall temperature
(i) VSL y lt 11.7 e 0 or e ltlt
1/Pr integrating using boundary condition
79
(ii) TC y gt 11.7, e y/2.5
Assume y/2.5 gtgt 1/Pr i.e. Pr is not
small thus q -2.5 ln y B at y 11.7 q
qw - 11.7Pr VSL -2.5 ln y B So q
qw - 11.7 Pr 2.5 ln 11.7 - 2.5 ln y for
y gt 11.7 But we know u 2.5 ln y 5.5 in
TC Thus q qw - 11.7 Pr 2.5 ln 11.7 5.5
-u
11.7
80
Matching velocities at the edge of the turbulent
core to the VSL
Define (using spatial mean temperatures) the heat
transfer coefficient as
In the TC both the temperature and velocity
profiles are fairly flat - we assume that qm
and um occur at the same y ym Hence
u
um
VSL
TC
q
qm
y
ym
11.7
81
In dimensional terms
(1/Stanton no., St)
(um)-1
Friction factor
Thus the h.t.c. can be calculated knowing Cf,
e.g. from Blausius Cf 0.079Re-1/4 For rough
pipes at large Re (when independent of Re)
Where a pipe radius, e roughness size
82
5.4.2 Heat Mass Transfer from von Karmans
Universal Velocity Profile Momentum
transfer (1) Heat transfer (2) von
Karmans Universal Velocity Profile (see sec.
5.3.2) Viscous u y ylt5 e/n 0 sublayer
Buffer u 5 ln y - 3.05 5 lt y lt 30 layer
Turbulent u 2.55 ln y 5.5 30 lt y core
83
Using these expressions for e(y) previously
derived in sec. 5.3.3, and assuming that
e/ngtgt1/Pr in the turbulent core, then eq. (2)
integrates to give
As in Taylor-Prandtl analysis it is assumed that
the spatial mean velocity, um, and temperature,
qm, occur at the same distance from the wall ym
(3)
and
84
Hence
At the meeting of the buffer layer and the
turbulent core (y 30)
At the meeting of the viscous sublayer and the
buffer layer (y 5)
Substituting for in Eq. (3)...
85
(4)
From the definition of q (h heat transfer
coefficient)
where St Stanton number and
Then eq. (4) becomes
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