agenda –lecture #2

1

Globex Julmester 2017

Lecture #2

04 July 2017

Modelling of dispersed, multicomponent, multiphase flows in resource industries

Section 3: Examples of analyses conducted for Newtonian fluids

Agenda – Lecture #2

• Student projects – students’ top 3 selections

• Section 3: Examples of analyses conducted for Newtonian fluids– 3.1 Motivation & context

– 3.2 Energy loss (friction) in pipe flow

– 3.3 Constitutive relationships for laminar Newtonian flow

• 3.3.1 Pipe flow

• 3.3.2 Concentric cylinder viscometer

– 3.4 Some basics re: turbulent flow

– 3.5 Terminal settling velocity

2

Review of Lecture #1: Applications of the fundamental laws

• Fluid flows obey conservation laws for mass, momentum and energy

• The laws can be stated in differential form or in integral form

• The integral and differential forms can be derived from each other!

Physical problem

A set of differential equations

Solution of the problem

Identify important process variables

Make reasonable assumptions and approximations

Apply relevant physical laws

Apply applicable solution technique

Apply appropriate boundary and initial conditions

3.1. Motivation and context

• The primary goal of this course is to demonstrate the differences between the flow behaviour of simple Newtonian fluids and more complex industrial multiphase flows

• In order to do this we need to have a good foundation in the behaviour of Newtonian fluids

• Additionally, this section allows us to use some of the different control volume analyses we introduced in Lecture #1

Section 3: Examples of analyses conducted for Newtonian fluids

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3.2 Energy loss (friction) due to flow in pipes: the Mechanical Energy Balance

1

2

3

4

h

1

2

3

4

h

22

4 3 344 3 p T L

P P VVh h h h h

g 2g 2g

HEAD form (units = meters of fluid)

22

4 3 344 3 p T w

P P VVg h h W W

2 2

Energy form (units

= J/kg fluid)

Assumptions:• Incompressible flow• Steady state

(3.1a)

(3.1b)

The relationship between lost work (lw)maj and

wall shear stress (w)

Consider incompressible, steady-state, pipe flow:

r

z

Q

We consider a fluid element of length ‘L’:

R

L

Just to make things easier, we look at horizontal flow; but the relationship between lost work and shear stress is valid for all pipe orientations!

4

Z-direction force balance on fluid element

P1A P2A

2 wP

L R

(3.2N) Horizontal, steady-state, incompressible pipe flow

s wF SA

1 2 0 wP A P A SA

21 2 2 0 w(P P ) R ( RL)

4 wP dP

L dz D

Consider same pipe flow, this time using the Mechanical Energy Balance:

r

zQ

1 2

MEB, 1 22 1

w

P P

wP

L L

or

or 4w w

L D

Thus: 4 w

w

L

D

(3.3)

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The shear stress decay law

Again, let’s consider that same FLUID ELEMENT in incompressible, steady-state, horizontal pipe flow:

r

zQ

Here, we found that:

4 wP

L D

Now, let’s consider a new FLUID ELEMENT with r < R:

rz

Q

A z-direction force balance on this fluid element:

dz

2 0rzPA (P dP)A ( r dz)

2 rzdP

dz r

2 2w rzdP

dz R r

rz

w

r

R

(3.4)

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Shear stress decay law

• Applies to laminar and turbulent flow

• Applies to Newtonian and non-Newtonian fluids

• Must be steady state

• Must be a constant-density system (e.g. no variation in density from top to bottom of pipe)

rz

w

r

R

(3.4)

Impact:

zQ

r

r = R (@ pipe wall)

rz

w

r

R

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Example 3.1 – Basic wall shear stress calculations

The wall shear stress in a portion of a 0.25 m (diameter) pipe that carries water is 11.7 Pa. The flow is fully developed. Determine the pressure gradient dP/dx (where x is in the flow direction), if the pipe is

a) horizontal

b) Vertical with flow up

c) Vertical with flow down

Also, what is the shear stress at r = 0.8R for each of the three pipe orientations described above?

Calculating friction losses (i.e. w) for single-phase flows

Steady state or transient?

Incompressible or

compressible?

Newtonian or non-Newtonian?

Laminar or turbulent?

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Wall shear stress (w) for Newtonian fluids

f fn Re, k D

2

w

f V

2

This equation defines the

Fanning friction factor, f

LAMINAR FLOW TURBULENT FLOW

• Derived relationship between w and V exists

• Can directly show that:

• Experimental relationship between w and V found:

16f

Re

f

f

DVRe

(3.5)

• Can be obtained from dimensional analysis of the Navier-Stokes equation

• Represents the ratio of carried (inertial) momentum to molecular (viscous) momentum transfer

• Re < 2100: Laminar flow (typically)

• Re > 4000: Turbulent flow (typically)

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Determining friction losses for Newtonian fluids

hLw f

Calculating ‘f’

• Moody diagram

• Correlations– e.g. Churchill, Swamee-Jain

10

Moody diagram

Correlations: the Churchill Equation (1977)

11212

32

8 1f 2

Re A B

160.9

7A 2.457 ln 0.27

Re D

1637530

BRe

NOTE: the hydrodynamic

roughness is sometimes referred to as ‘’ and sometimes ‘k’!

(3.6a)

(3.6b)

(3.6c)

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Correlations: the Swamee-Jain equation (1976)

2

10 0.9

0.0625f

5.74Dlog3.7 Re

(3.7)

Example 3.2: Water pipeline calculation

1 2

D = 100 mm; L = 1.2 kmQ = 746 m3/day; P1 =P2

k = 0.045 mm

Fluid properties

= 1 mPa.s

= 1000 kg/m3

Find:

(a) Wall shear stress (in Pa) and –dP/dz (in Pa/m);

(b) Pump head required (in meters of water)

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Course project updates

• Assignment of projects

• Comments on Oral presentation #1– Motivation – what is the industrial relevance of your project, why

is the topic important

– Objectives – what are your goals for this project?• Example: “To determine the 3 main sources of air pollution in Beijing”

– Key resources (maximum 3)• How will you decide which ones are “best”? What criteria could you use to

make this determination?

Agenda – Lecture #2

• Student projects – students’ top 3 selections

• Section 3: Examples of analyses conducted for Newtonian fluids– 3.1 Motivation & context

– 3.2 Energy loss (friction) in pipe flow

– 3.3 Constitutive relationships for laminar Newtonian flow

• 3.3.1 Pipe flow

• 3.3.2 Concentric cylinder viscometer

– 3.4 Some basics re: turbulent flow

– 3.5 Terminal settling velocity

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3.3. Constitutive relationships for laminar Newtonian flow

• 3.3.1 Pipe flow

• 3.3.2 Concentric cylinder viscometry

3.3.1 Laminar, Newtonian pipe flow

• Part 1: Review of the microscopic conservation laws

• Part 2: Write and simplify the equations for steady, incompressible, 1D, laminar, Newtonian pipe flow

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The microscopic equations: continuity, momentum

Assumptions:

• Cartesian (x, y, z) co-ordinates

• Velocity components:

Continuity Equation:

Momentum Equations:

x y zv v v 0t x y z

x x x y x z x xx yx zx x

Pv v v v v v v g

t x y z x y z x

x-direction:

x x y y z zv v (x, y,z, t); v v (x,y,z, t); v v (x, y,z, t)

y-direction:y x y y y z y xy yy zy y

Pv v v v v v v g

t x y z x y z y

z-direction:z x z y z z z xz yz zz z

Pv v v v v v v g

t x y z x y z z


Assumptions:



Continuity Equation:


x y zv v v 0t x y z

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Assumptions:



Momentum Equations: (only x-direction shown here)

x x x y x z x xx yx zx x

Pv v v v v v v g

t x y z x y z x


The microscopic equations for steady, incompressible, 1D, laminar, Newtonian pipe flow

Note: In Section 3, ‘u’ and ‘v’ are used interchangeably to mean local fluid velocity

What we know: Cylindrical coordinates

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The microscopic equations for steady, incompressible, 1D, laminar, Newtonian pipe flow (continued…)

Continuity

r z

1 1ru u u 0

t r r r z

Conservation of momentum: z-direction

z r z z z z rz z zz z

1 1 1 Pu u u u u u u r g

t r r z r r r z z

z zdu Pr

dr 2

so:

2zz 2

Pu r C

4

Since uz = 0 at r = R, we can find C2 and write the Poiseuille equation:

22z

z

P R ru 1

4 R

z

A

Q u dA + Q VA

2 2z zP R P D

V8 32

Since the flow is horizontal: wz

4P

D

and wDV

8

(3.8)

(3.9)

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Steady, laminar pipe flow of a Newtonian fluid (summary)

2w

z 2

R ru 1

2 R

Microscopic:

zrz

du

dr +

rz

w

r

R

zA

Q u dA + Q VA

w

8 V

D

Newton’s Law of Viscosity

Shear stress decay law

Poiseuille’sEquation

Macroscopic:

w2

2 16f

V Re

VDRe

w

8 V

D

(3.8)

(3.9)

Example 3.3

A new heat exchange fluid (SG 1.290) has been developed by our company’s R&D department. They have asked us to determine its flow properties. A 38 mm horizontal pipe is used for this purpose. The data shown in the table below are collected at a constant operating temperature of 130ºC.

a) Is the fluid Newtonian? Use a graph to explain your answer.

b) Calculate the fluid’s best-fit rheological parameters.

c) Prove that none of the data points listed here should be rejected from our analysis.

V (m/s) 0.75 1.05 1.45 1.9 2.7 3.5 -dP/dz (kPa/m)

1.08 2.02 2.43 3.96 5.83 6.55

18

0

10

20

30

40

50

60

70

0 0.5 1 1.5 2 2.5 3 3.5 4

w(P

a)

V (m/s)

y = 18.581xR² = 0.9683

0

10

20

30

40

50

60

70

0 0.5 1 1.5 2 2.5 3 3.5 4

w(P

a)

V (m/s)

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Solution

Newtonian, based on shape of curve (straight line, through origin)

m = slope of line = 18.58 = 8/D

Therefore:

= 18.58 D / 8 = 0.088 Pa.s

Check for laminar flow:

We need to ensure that Re ≤ 2100 for all data points

maxV D2100

or max

2100V 3.8 m / s

D

We see therefore that all the data points collected here were collected in laminar flow and all can be used to determine .

3.3.2 Laminar Newtonian flow in a concentric cylinder viscometer

Spindle

Cup

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Concentric cylinderviscometer parameters

T = Applied Torque = Angular Velocity

R1 = Spindle RadiusR2 = Cup RadiusL = Spindle Length

L

3.4. A (very) brief introduction to turbulence

• Turbulence is ubiquitous in natural and industrial processes

• How would you describe turbulence?

40

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*Possible descriptions of fluid turbulence…

• Not laminar (!)

• Three-dimensional, unsteady, chaotic motion - characterized by random and rapid fluctuations of swirling regions of fluid

• The fluctuations have a wide range of different length and time scales

• The fluctuations provide an additional mechanism for momentumand energy transfer (over laminar flow)

• Turbulence is a continuum phenomenon. The smallest scale of turbulence is MUCH larger than the molecular scale of the fluid

• It is a characteristic of the flow and NOT the fluid

• Associated with high Reynolds numbers

*G. Ahmadi, ME 637 course notes, Clarkson University41

t

uz(t)

z zu (t) u u

Fluctuating and time-smoothed velocity components:

42

22

0

1

v

y/D

V = Q/A

y

TURBULENT

Evaluation of the time-smoothed velocity components – in the direction of flow

Distance from pipe wall

Velocity

Viscous sublayer

Buffer zone

Inertial region

There are three regions in a turbulent pipe flow:

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Shear velocity

• Shear or “friction” velocity:

• Fundamental (and useful) parameter in describing turbulent flow

* w fu

• The velocity fluctuations in a turbulent flow have the same order of magnitude as the shear velocity

or

2* f fu 1 fV / 2 V f / 2

• The time-averaged velocity distribution can be described based on the shear velocity!

(3.11a)

(3.11b)

0

5

10

15

20

25

0.1 1 10 100 1000

*

uu

u

*yuy

Turbulence scaling

Viscous sublayer(inner layer)

Inertial region(outer layer)

u y

2.5 ln 5u y

(0 5)y

(30 500)y

Viscous sublayer(inner layer)

Inertial region(outer layer)

(3.12a)

(3.12b)

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3.5. Terminal settling velocity (v∞)

• Critically important parameter in assessing / predicting the performance of many industrial systems, including:– Fluidized beds

– Separation equipment (gravity separation, hydrocyclones, mechanical flotation)

– Slurry pipelines

Consider a single spherical particle immersed in a Newtonian liquid.

d

f

v

s

Net downward gravity force

Upward drag force

=

3 2

2s f D f

d dg 0.5C v

6 4

volume projected area

so 1/ 2

s f

D f

4gdv

3C

Drag coefficient

(3.13)

Sphere in Newtonian fluid

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d: Particle diameter (m)

s: Particle density (kg/m3); often must be measured

CD: Must calculate, based on

fp

f

v dRe

(3.14)

CD

Rep

Stokes’ Law Intermediate Newton’s Law Boundary Layer Separation

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(3.15a) Stokes’ Law CD = 24/Rep Rep≤0.3

(3.15b) Intermediate CD = 0.3< Rep≤1000

(3.15c) Newton’s Law CD ~ 0.445 1000<Rep<105

(3.15d) Boundary Layer Separation CD ~ 0.05

)Re15.01(Re

24 687.0p

p

Drag Coefficient Correlations

Important equation to know: Stokes’ settling

When Stokes’ Law applies (Rep ≤ 0.3):

1/ 2s f

D f

4gdv

3C

Dp

24C

Re

fp

f

v dRe

(3.13) (3.14)

Therefore:

2s f

f

d g( )v

18

(3.16) “Stokes’ Law”

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Solution methods**Spherical particles, Newtonian fluid**

• Iterative

• Direct– Wilson method

Wilson et al.* direct methodSpherical particles, Newtonian fluid

Shear velocity

Pipe flow Particle settling

Shear Reynolds number

* w fu

f **

f

u DRe

w = wall shear stress

* fv

f **

f

v dRe

Characteristic shear stress

= particle surficial stress s f gd

6

(3.17)

(3.18)

(3.19)

Velocity ratio*u u *v v

*Wilson et al., “Direct prediction of fall velocities in non-Newtonian materials”, Int. J. Miner. Process. 71(2003) 17-30.

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Wilson et al. direct method**Spherical particles, Newtonian fluid**

* fv

s f gd

6

(3.17)

(3.18) (3.20)

Combining Equations (3.17) and (3.18) gives:

* s f fv gd 6

*v v

Wilson et al. direct method**Spherical particles, Newtonian fluid**

(3.20) * s f fv gd 6

f **

f

v dRe

(3.19)

*

* 1.2 4 3.2* *

Re 2.80v v

3 1 0.08Re 1 3x10 Re

For Re* ≤ 10 (Includes Stokes’ regime)

1.72*log v v 0.2069 0.500x 0.158x

where *x log Re 10For 10 < Re* < 260

*v v 4.24 For Re* ≥ 260

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Example 3.4

• In the 1970’s and 80’s, the idea of transporting coal-in-oil slurries was proposed, as the feed could be sent directly to a thermal power plant without drying the coal.

• Use the Wilson et al. direct method to determine the terminal settling velocity of a coal particle (d = 150 m, s = 1500 kg/m3) in a Newtonian oil (f = 876 kg/m3; f = 4.0 mPa.s)

Homework for tomorrow – Assign #1 (see Globex site)

agenda –lecture #2

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