7. Semester Chemical Engineering Civil Engineeringhomes.et.aau.dk/mma/transport/lek13 Transport processer - Non-Ne… · 1. Elementary Fluid Dynamics 2. Fluid Kinematics 3. Finite

Transport processes

7. Semester

Chemical Engineering

Civil Engineering

Course plan

1. Elementary Fluid Dynamics2. Fluid Kinematics3. Finite Control Volume Analysis4. Differential Analysis of Fluid Flow5. Viscous Flow and Turbulence6. Turbulent Boundary Layer Flow7. Principles of Heat Transfer8. Internal Forces Convection9. Unsteady Heat Transfer10. Boiling and Condensation11. Mass Transfer12. Porous Media Flow13. Non-Newtonian Flow

Today's lecture

• Non-Newtonian Fluids

Newtonian fluids

• Newton’s law of fluids

dudy

τ µ=

Shear stressRate of shear strain

Constant of proportionality

Non-newtonian

nduKdy

τ

=

Shear stressRate of shear strain

Constant of proportionality

Flow behavior index

Newtonian vs NN fluids

• Which fluids are Newtonian?– All gasses

– Most common liquids incl. water and oil

• Which fluids are Non-Newtonian (NN)?– Many Slurry flow (eg. Water-sand mixture /quicksand)

– Many “foods” and like (i.e. Toothpaste, mayonnaise )

– Many Polymers (long chained molecules)

Liquids with a more complex molecular structure

Fluids with a simple molecular structure

Non-Newtonian fluids

• Time independent NN fluids– Bingham plastic

– Shear thinning (Pseudoplastic )

– Shear thickening (Dillatant)

• Time dependent NN fluids– Thicxotropic

• i.e. Ketchup

– Rheopectic• i.e. Whipped cream

Bingham plastic

• Same as Newtonian fluid but the curve does not go through origin

• Examples: drilling mud, toothpaste, paper pulp, sewage sludge

when >

when 0

o o o

o

dudydudy

τ τ µ τ τ

τ τ

= +

< =

Shear thinning

• Most NN fluids are in this class

• Represented by the power law (Ostwald-de Waeleequation)

• Examples: Polymer solutions or melts, greases, mayonnaise, paints, …

1

Apparent viscosity

1

1

decreases with increasing shear rate

a

n

n

n

a

duK ndy

du duKdy dy

duKdy

µ

τ

τ

µ

−

−

= <

=

= ⇒

Shear thickening

• Also represented by the power law (Ostwald-deWaele equation)

• Examples: corn flour-sugar solutions, solutions with high concentration of powders in water

1

Apparent viscosity

1

1

increases with increasing shear rate

a

n

n

n

a

duK ndy

du duKdy dy

duKdy

µ

τ

τ

µ

−

−

= >

=

= ⇒

Properties of NN fluids (time-independent)

• Measurement of Δp vs. flowrate V…

• ..in tube of length L and diameter D, then:

capillary-tube viscometer

4wD p

Lτ ∆

= 8

r R

du Vdr D=

=

Intermezzo: determination of τw

• From force balance on a cylindrical element:

Simplifying gives:

At the wall we have:

Rearranging gives:

• Laminar velocity profile:

• thus

Intermezzo: determination of du/dy

2

max 2

2 2 4 81r R r R r R

du d r V V VV rdr dr R R R D= = =

⋅ = − = = =

2

max( ) 1 ru r VR

= −

Properties of NN fluids

• For a power law fluid:

• Coefficients K’ and n’ can be found from curve fitting

• A generalized viscosity can be defined as:

• If the fluid properties are given as K and n than the following holds:

'

w8'

4

nD p VKL D

τ ∆ = = ⋅

' 1'8nKγ −=

'3 ' 1' ; '4 '

nnn n K Kn+ = =

Some values of K, n’ and γ

NN-fluid flow in pipes

• For pipe flow, what are we interested in determining?– Laminar or turbulent flow

– Pressure loss due to friction

– Pressure loss in components

– Velocity profiles

– …

Similar methods to that of Newtonian fluids Theoretic approach for Laminar flow

Empirical basis for turbulent flow

Laminar flow of NN-fluids

• Reynolds number:

Laminar : Re<2000

Turbulent: Re>10.000

' 2 ' ' 2 ' 2

gen ' 11

Re'8 3 18

4

n n n n n n

n nn

D V D V D VK nK

n

ρ ρ ργ

− − −

−−

= = =+

Laminar flow of NN-fluids

• Pressure drop in pipe due to friction:– Method 1: From derivation

– Method 2: using friction factors (fanning) • Friction factor relation from Newtonian fluids but using the generalized

Reynolds number for NN-fluids

'' 4 8 nK L VpD D

∆ =

2

42

L Vp fD

ρ∆ =Re, gen

16fN

=

19

20

21

Flow of NN-fluids

• Mechanical energy balance:

2 21 2

1 1 2 22 2 lossV VP g z P g z Pρ ρρ ρα α

+ + = + + + ∆

Coefficient due to the viscous effects

Laminar Turbulent

Newtonian

Non-Newtonian

0.5α = 1.0α ≈

1.0α ≈( )( )( )2

2 1 5 33 3 1n n

nα

+ +=

+

Flow of NN-fluids

• Pressure loss in components:– Losses in contractions and fittings

same as for Newtonian fluids

– Losses in sudden expansion in Laminar flow

– Losses in sudden expansion in Turbulent flow flow

same as for Newtonian fluids

( )( )( ) [ ]

4 22 1 1

12 2

3 3 13 1 3 /2 1 2 5 3 2 5 3ex

nD Dn nh V J kgn n D D n

+ + + = − + + + +

Turbulent flow of NN-fluids

• Use the friction factor approach (empirical):– Empirical coefficients for smooth pipes (no data for rough pipes!)

– Note: Fanning friction factor used in Geankoplis

2

42

L Vp fD

ρ∆ =

25

26

Laminar Velocity profile of NN fluids

Laminar Velocity profile of NN fluids

• Found using same procedure as for Newtonian fluids:

1 11 1

maxv 1 =v 11 2

n nnn n n

nn p r rRn LK R R

+ ++ ∆ = − − +

Laminar Velocity profile of NN fluids

• Flow rate:

• Average velocity:

• Velocity ratio:

1 1

maxv1 2

nnnn p R

n LK

+∆ = +

1 1

av 2v3 1 2

nnnQ n p R

n LKRπ

+∆ = = +

av

max

v 1v 3 1

nn+

=+

( )1 3 1

3 1 2

nnnn pQ u r dA R

n LKπ +∆ = = + ∫

Bingham plastic fluids

• Stress relationship

when >o o odudy

τ τ µ τ τ

= +

Yield stress

Laminar stress(same as for Newtonian fluids)

when 0odudy

τ τ< =

Bingham plastic fluids• Velocity profile

• Relationship between r0 and τ0:

( )2

2maxu 1 0

4o

centerline orpr u u R for r r

K Rµ ∆ = = = − < <

2o op rL

τ ∆=

( )22

0u 1 116

opD r rr R for r rL R R

τµ µ

∆ = − − − >

Bingham plastic fluids• Flow rate

• Buckingham-Reiner equation:

44 4 118 3 3

o o

w w

pRQL

τ τπµ τ τ

∆ = − +

( ) ( ) ( )0

002 2

r R

rQ u r dA u r rdr u r rdrπ π= = +∫ ∫ ∫

2wpRwhereL

τ ∆=

33

34

35

Bingham plastic fluids

• There exist friction factors to calculate the pressure loss for both laminar and turbulent flow for Bingham plastic fluids (not shown in Geankoplis)

2

42

L Vp fD

ρ∆ =

Laminar Bingham Plastic Flow

( )

−+= 73

4

Re3Re61

Re16

BPBPBP fHeHef

20

2

∞

=µτρDHe

∞

=µρVD

BPRe

Hedstrom Number

(Non-linear)

Turbulent Bingham Plastic Flow

( )Hex

BPa

ea

f5109.2

193.0

146.01378.1

Re10−−

−

+−=

=

Excercises

• Exercises, solutions for this lecture: On the web

• Examination questions will be available 2 weeks prior to the exam. They will be submitted by email and I will submit them on the web as well!

([email protected]); ([email protected]); ([email protected]); ([email protected]); ([email protected]); ([email protected]); ([email protected]); ([email protected])