Non-Newtonian Fluids

Non-Newtonian Flow

Goals• Describe key differences between a Newtonian

and non-Newtonian fluid• Identify examples of Bingham plastics (BP) and

power law (PL) fluids• Write basic equations describing shear stress

and velocities of non-Newtonian fluids• Calculate frictional losses in a non-Newtonian

flow system

Non-Newtonian FluidsNewtonian Fluid

dr

duzrz

Non-Newtonian Fluid

dr

duzrz

η is the apparent viscosity and is not constant for non-Newtonian fluids.

η - Apparent ViscosityThe shear rate dependence of η categorizes

non-Newtonian fluids into several types.

Power Law Fluids: Pseudoplastic – η (viscosity) decreases as shear rate

increases (shear rate thinning)

Dilatant – η (viscosity) increases as shear rate increases (shear rate thickening)

Bingham Plastics:

η depends on a critical shear stress (0) and then becomes constant

Non-Newtonian Fluids

Newtonian

Bingham Plastic:

Pseudoplastic:

Dilatant:

sludge, paint, blood, ketchup

latex, paper pulp, clay solns.

quicksand

Modeling Power Law FluidsOswald - de Waele

dr

du

dr

duK

dr

duK z

n

z

n

zrz

1

where:K = flow consistency indexn = flow behavior index

Note: Most non-Newtonian fluids are pseudoplastic n<1.

eff

Modeling Bingham Plastics

0 rz 0dr

duz Rigid

0 rz 0 dr

duzrz

Frictional Losses Non-Newtonian Fluids

Recall:

Applies to any type of fluid under any flow conditions

24

2V

D

Lfh f

Laminar Flow

Mechanical Energy Balance

WhzgVp

fˆ

2

2

0 0 0

MEB (contd)

Combining:

p

VL

Df

2

2

4

1

Momentum Balance

gw FFSpSpVVm 22111122 0 0

prrL rz 22

rzr

Lp 2

Power Law Fluidn

zrz dr

duK

nn

z rKL

p

dr

du 11

2

1

Rr 0zuBoundary Condition

Velocity Profile of Power Law FluidCircular Conduit

n

n

n

nn

z rRn

n

KL

pu

111

12

1

Upon Integration and Applying BC

Power Law (contd)

Need bulk average velocity

drruR

dSuS

V zS

2

112

n

nn

Rn

n

KL

pV

11

132

1

p

VL

Df

2

2

4

1

Power Law Results (Laminar Flow)

↑ Hagen-Poiseuille (laminar Flow) for Power Law Fluid ↑

Recall

1

2 132

n

nn

n

D

VLKnn

p

Laminar Friction FactorPower Law Fluid

Define a Power Law Reynolds Number

K

DV

n

nRe

nnnn

PL

23

132

PLRef

16

nn

nn

DV

Knn

f

2

1 132

Turbulent Flow

2,

133

35242100Re

n

nncriticalPL

Power Law Fluid Example

A coal slurry is to be transported by horizontal pipeline. It has been determined that the slurry may be described by the power law model with a flow index of 0.4, an apparent viscosity of 50 cP at a shear rate of 100 /s, and a density of 90 lb/ft3. What horsepower would be required to pump the slurry at a rate of 900 GPM through an 8 in. Schedule 40 pipe that is 50 miles long ?

P = 1atmP = 1atm

L = 50 miles

7273792.0

759.11442202.0

1)4.0(3

4.02

759.1281.323474.0

1

60

min1

48.7

31

min

900

792.0100

50'

'

6.1

6.1

3

4.04.0

4.03

~

6.1

4.01

'

smkg

sm

mkg

mRE

s

m

ft

m

ftsgal

ftgalV

sm

kg

scPK

r

V

r

VK

N

app

n

HpkWsm

skg

Power

s

kg

m

kgm

s

mm

s

msm

m

mhW

Figf

V

D

LfhW

hg

Zg

g

VPW

fp

fp

fcc

p

13001.9701000

845,119.81

9.8114420323.0759.1

845,112

760.1

202.0

804600048.04

11.50048.0

24

2

2

2

32

2

2

2

2

2

Bingham Plastics

Bingham plastics exhibit Newtonian behavior after the shear stress exceeds o. For flow in circular conduits Bingham plastics behave in an interesting fashion.

Bingham Plastics

Unsheared Core

crr 20

2 cc

cz rRr

uu

crr

01

2

R

rrRu rzz

Sheared Annular Region

Laminar Bingham Plastic Flow

73

4

Re3Re61

Re

16

BPBPBP f

HeHef

20

2

D

He

VD

BPRe

Hedstrom Number

(Non-linear)

Turbulent Bingham Plastic Flow

Hex

BPa

ea

f5109.2

193.0

146.01378.1

Re10

Bingham Plastic ExampleDrilling mud has to be pumped down into an oil well that is 8000 ft deep. The mud is to be pumped at a rate of 50 GPM to the bottom of the well and back to the surface through a pipe having an effective diameter of 4 in. The pressure at the bottom of the well is 4500 psi. What pump head is required to do this ? The drilling mud has the properties of a Bingham plastic with a yield stress of 100 dyn/cm2, a limiting (plastic) viscosity of 35 cP, and a density of 1.2 g/cm3.

P = 4500 psi

P = 14.7 psi

L = 8000 ft

cms

g

cm

dyn

sftlb

ftlb

sft

ft

N

sft

lb

cP

sftlb

cP

ft

lb

ft

lb

s

ft

ftgal

ft

s

galV

ftAreaftftD

o

m

m

RE

m

m

mm

22

33

2

3

2

100100

13550235.0

388.74276.13333.0

0235.04107197.6

35

88.744.622.1

276.10873.0

1

48.760

min

min50

0873.03333.012

4

m

f

f

m

f

m

f

p

fcc

p

lb

lbftWp

slblbmft

sft

ft

ft

lb

lbft

ftlb

ftin

in

lb

W

hg

Zg

g

VPW

f

cmsg

cmsg

cmg

incm

in

N HE

96533980008626

2.322

276.1

3333.0

800014.048000

88.74

1447.144500

2

14.0

1001.1

35.0

1002.1

54.24

2

2

3

2

2

2

2

52

23

2