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Transport Phenomena 3

- Non-Newtonian Flow
- Navier-Stokes Equations and Applications
- Turbulent Flow and Boundary Layers
- Method of assessment 100 written exam

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.

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.

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.

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

Maths Revision Co-ordinate Systems

z

(r,q,z)

y

z

Cylindrical polars

r

q

x

Cartesian (x,y,z)

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

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

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).

(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 ?

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.

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.

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

Hence

For a Newtonian fluid

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

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

(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.

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

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...

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)

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

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

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

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

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.

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

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)

- (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

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

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

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

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.

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.

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

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

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

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

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).

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

- 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.

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

- 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

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

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,

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.

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)

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

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

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.

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)

Limits Bi 0, C3 C1 (zero flux) Bi

infinity, C0 CA (no surface resistance)

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

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

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.

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.

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.

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

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

Using the averaging rules

(I)

(II)

(IV)

(III)

- 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.

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)

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

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

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.

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)

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

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

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,

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)

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.

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)

(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

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.

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

Make dimensionless and

Define

So

(i) In the viscous sublayer

(ii) In buffer layer

(iii) In turbulent core

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.

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

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

(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

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

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

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

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

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)...

(4)

From the definition of q (h heat transfer

coefficient)

where St Stanton number and

Then eq. (4) becomes