11 Laminar Flow

1PETE 411Well Drilling

Lesson 11 Laminar Flow

2Lesson 11 - Laminar Flow

Rheological Models Newtonian Bingham Plastic Power-Law

Rotational Viscometer Laminar Flow in Wellbore

Fluid Flow in Pipes Fluid Flow in Annuli

3ReadADE Ch. 4 to p. 138

HW #5ADE Problems 4.3, 4.4, 4.5, 4.6

Due Friday, Sept. 27, 2002

4Newtonian Fluid Model

Shear stress = viscosity * shear rate

AF ,

LVallyExperiment =

=

5Laminar Flow of Newtonian Fluids

AF

LV=


In a Newtonian fluid the shear stress is directly proportional to the shear rate (in laminar flow):

i.e.,The constant of proportionality, is the viscosity of the fluid and is independent of shear rate.

=

sec1

2 cmdyne

= .


Viscosity may be expressed in poise or centipoise.

poise 0.01 centipoise 1

scmg1

cms-dyne1 poise 1 2

=

==

2cmsecdyne

= .

8Shear Stress vs. Shear Rate for a Newtonian Fluid

Slope of line ====

. =

9Example - Newtonian Fluid

10

Example 4.16

Area of upper plate = 20 cm2

Distance between plates = 1 cm

Force reqd to move upper plate at 10 cm/s= 100 dynes.

What is fluid viscosity?

11

Example 4.16

poise 5.0cm

sdyne5.0105

2 =

==

1-

2

sec 10/1dynes/cm 20/100

//

rate shearstressshear

===

LVAF

cp 50=

=

12

Bingham Plastic Model

13


- if

- if 0

if

yyp

yy

yyp

+=

and y are often expressed in lbf/100 sq.ft

14


2

2

22

48.30

sec980454

*100100

1

=

ftcm

cmlbfg

ftlbf

ftlbf

22 dyne/cm 79.4100

1 =ft

lbf(p.134)

1 dyne is the force that, if applied to a standard 1 gram body, would give that body an acceleration of 1 cm/sec2

15

Example 4.17{parallel plates again!}

Bingham Plastic FluidArea of upper plate = 20 cm2


1. Min. force to cause plate to move = 200 dynes

2. Force reqd to move plate at 10 cm/s = 400 dynes

Calculate yield point and plastic viscosity

16

Example 4.17

Yield point,

22y

y cmdynes10

cm20dynes200

AF

===

22 cmdynes79.4

ft100lbf1 but =

79.4

10y == 2ftlbf/10009.2

+= py

17

Example 4.17

cp 100 .e.i p =

poise 1110

10202 =

=

=cm

sdynep

+=

cm 1cm/s 10

cm 20dynes 200

cm 20dynes 400

22 p

Plastic viscosity, p

+= py

bygiven is

18

Power-Law Model

19

Power-Law Model

n = flow behavior indexK = consistency index

0 if K

0 if K1n

n

20

Power-Law Model

2

2

2

n

2

n

ftcm48.30

seccm980

lbfg454

*ft

slbfft

slbf1

=

.poise .eq 479cm/sdyne 479ft

slbf1 2n2n

==

.cp .eq 900,47ft

slbf1 2n

=

21

Example 4.18Power-Law Fluid


Distance between plates = 1 cmForce on upper plate = 50 dyne if v = 4 cm/sForce on upper plate = 100 dyne if v = 10 cm/s

Calculate consistency index (K) and flow behavior index (n)

22

Example 4.18

v = 4 cm/s

( )n

n

n

44

4K5.2

14K

2050

K

=

=

=



Force on upper plate= 50 dyne if v = 4 cm/s

((((i))))

23

Example 4.18

v = 10 cm/s

( )n

n

n1010

10K5

110K

20100

K

=

=

=



Force on upper plate= 100 dyne

if v = 10 cm/s

((((ii))))

24

Example 4.18

Combining Eqs. (i) & (ii):

5.2 log n2 log

5.24K

10 K5.2

5 nn

n

=

==

7565.0n =

( )nK 45.2 = ((((i))))((((ii))))( )nK 105 =

25

Example 4.18

From Eq. (ii):

poise eq. 8760.010

510

5K 7565.0n ===

cp. eq. 6.87K =

( )nK 105 = ((((ii))))

26

Apparent Viscosity

Apparent viscosity = ( / )is the slope at each shear rate,

.,, 321

27

Apparent Viscosity

Is not constant for a pseudoplastic fluid

The apparent viscosity decreases with increasing shear rate

(for a power-law fluid)

(and also for aBingham Plastic fluid)

28

Typical Drilling Fluid Vs. Newtonian, Bingham and Power

Law Fluids

0000

(Plotted on linear paper)

29

Rheological Models

1. Newtonian Fluid:

2. Bingham Plastic Fluid:

= rate shear

viscosityabsolutestressshear

=

=

=

+= *)( py viscosityplastic

point yield

=

=

p

y

What if y ==== 0?

30

3. Power Law Fluid:

When n = 1, fluid is Newtonian and K = We shall use power-law model(s) to

calculate pressure losses (mostly).

n

)(K

= K = consistency index

n = flow behavior index

Rheological Models

31

Figure 3.6Rotating Viscometer

Rheometer

We determine rheologicalproperties of drilling fluids in this device

Infinite parallel plates

32

Rheometer (Rotational Viscometer)

Shear Stress = f (Dial Reading)Shear Rate = f (Sleeve RPM)Shear Stress = f (Shear Rate)

)(f =BOB

sleeve

fluid

Rate Shear the (GAMMA), of value the on depends Stress Shear the ),TAU(

33

Rheometer - base case

RPM sec-13 5.116 10.22

100 170200 340300 511600 1022

RPM * 1.703 = sec-1

34

ExampleA rotational viscometer containing a Bingham plastic fluid gives a dial reading of 12 at a rotor speed of 300 RPM and a dial reading of 20 at a rotor speed of 600 RPM

Compute plastic viscosity and yield point

12-20

300600p

=

=

cp 8p =

600600600600 = 20300300300300 = 12

See Appendix A

35

Example

8-12

p300y

=

=

2y ft lbf/100 4=

600600600600 = 20300300300300 = 12

(See Appendix A)

36

Rotational Viscometer, Power-Law Model

Example: A rotational viscometer containing a non-Newtonian fluid gives a dial reading of 12 at 300 RPM and 20 at 600 RPM.

Assuming power-law fluid, calculate the flow behavior index and the consistency index.

37

Example

cp eq. 67.61511

12*510511

510

7370.0

1220 log 322.3 log 322.3

7372.0300

300

600

===

=

=

=

nK

n

n

600600600600 = 20300300300300 = 12

38

Gel Strength

39

Gel Strength

= shear stress at which fluid movement begins

The yield strength, extrapolated from the 300 and 600 RPM readings is not a good representation of the gel strength of the fluid

Gel strength may be measured by turning the rotor at a low speed and noting the dial reading at which the gel structure is broken(usually at 3 RPM)

40

Gel Strength

In field units,

In practice, this is often approximated to

06.1g =2ft 100/lbf

2ft 100/lbf

The gel strength is the maximum dial reading when the viscometer is started at 3 rpm.

g = max,3

41

Velocity Profiles(laminar flow)

Fig. 4-26. Velocity profiles for laminar flow: (a) pipe flow and (b) annular flow

42

It looks like concentric rings of fluid telescoping down the pipe at different velocities

3D View of Laminar Flow in a pipe - Newtonian Fluid

43

Table 4.4 - Summary of Laminar Flow Equations for Pipes and Annuli

Fictional Pressure Loss Shear Rate at Pipe Well

Newtonian

Pipe Pipe

Annulus Annulus

2

_

f

d500,1v

dLdp

=

212

_

f

)dd(000,1v

dLdp

=

dv96_

w =

)dd(v144

12

_

w

=

44


Fictional Pressure Loss Shear Rate at Pipe Wall

Bingham PlasticPipe Pipe

Annulus Annulus

d225d500,1v

dLdp y

2

_

pf +=

)dd(200)dd(000,1v

dLdp

12

y

12

_

pf

+

=

p

y

_

w 7.159dv96

+=

p

y

12

_

w 5.239)dd(v144

+

=

45


Fictional Pressure Loss Shear Rate at Pipe Well

Power-LawPipe Pipe

Annulus Annulus

n

n

nf n

dvK

dLdp

+=

+ 0416.0/13

000,144 1

_

n

n

nf n

ddvK

dLdp

+

=+ 0208.0

/12)(000,144 112

_

)n/13(d

v24_

w +=

)n/12(ddv48

12

_

w +

=

46

Table 4.3 - Summary of Equations for Rotational Viscometer

Newtonian Model

Na N300 =

Nr066.5

2=

300a =

or

47


300N

or

1pNy 1 =

rpm 3 atmaxg =


300600p = )(NN

300or

12 NN12

p

=

p300y =

or

or

48


Power-Law Model

=

1

2

N

N

NNlog

logn 1

2

n300

)511( 510K =

nN

)N703.1( 510K

or

=

or or )log( 322.3n

300

600

=

or

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11 Laminar Flow