Flow in Annulus

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T H E U N I V E R S I T Y O F T U L S A

THE GRADUATE SCHOOL

NUMERICAL SIMULATION OF LAMINAR FLOW OF NON-NEWTONIAN

FLUIDS IN ECCENTRIC ANNULI

byYahya Hashemian Adariani

A thesis submitted in partial fulfillment of

the requirements for the degree of Master of Science

in the Discipline of Petroleum Engineering

The Graduate School

The University of Tulsa

2005



ii

T H E U N I V E R S I T Y O F T U L S A

THE GRADUATE SCHOOL

NUMERICAL SIMULATION OF LAMINAR FLOW

OF NON-NEWTONIAN FLUIDS IN ECCENTRIC ANNULI

by

Yahya Hashemian Adariani

A THESIS

APPROVED FOR THE DISCIPLINE OF

PETROLEUM ENGINEERING

By Thesis Committee

, Chair

Dr. Mengjiao Yu

Dr. Ramadan Ahmed

Dr. Siamack A. Sirazi

Dr. Stefan Miska



iii

ABSTRACT

Hashemian Adariani, Yahya (Master of Science in Petroleum Engineering)

Numerical Simulation of laminar Flow of non-Newtonian Fluids in Eccentric Annuli

Directed by Dr. Mengjiao Yu(113 pp., Chapter 4)

(229 words)

Accurate predictions of annular frictional pressure loss are important for optimal

well bore hydraulic program design. Inaccurate prediction of frictional pressure drop in

the annulus can result in an underestimation of the bottom hole pressure, which might

then exceed the strength of the formation, thus causing loss of drilling fluid and creating a

potentially dangerous situation due to the resulting loss of hydrostatic head.

In this study fully developed laminar axial flow of non-Newtonian fluids in

eccentric annuli has been investigated numerically. Effects of eccentricity on frictional

pressure loss in annulus for different fluids are presented. Numerical results are compared

with previous studies. Numerical investigation based on SIMPLE algorithm by Patankar-

Spalding (1972) has been presented for the case of Newtonian fluid flow in eccentric

annulus with inner pipe rotation.

Yield power-law rheology model is used as the constitutive equation of the flow.

Cartesian and boundary fitted coordinate system are utilized as two different approaches

to discretize the flow equations and generate mesh network. A minimum cut off value for



iv

the minimum shear rate is used to identify the plug (zero shear rate) region. Fluid flow

equations have been solved using an iterative successive over relaxation method.

Increasing eccentricity is found to lower frictional pressure drop for different

fluids. It was observed that for non-Newtonian fluids the effect of eccentricity on

pressure loss is less pronounced compared to Newtonian fluids.



ACKNOWLEDGEMENTS

The author thanks Dr. Mengjiao Yu, dissertation advisor, for his continuous

patience and assistance in this endeavor.

My special thanks to Dr. Ramadan Ahmed for his help and valuable suggestions.

I am thankful to Dr. Siamack Shirazi and Dr. Stefan Miska for their help,

suggestions and encouragement.

My appreciation extends to National Iranian Oil Company for providing me with

financial support.

I also thank TUDRP students and all my friends for their constant encouragementand support.

This work is dedicated to my parents.



vi

TABLE OF CONTENTS

Page

ABSTRACT...............................................................................................................iii

ACKNOWLEDGEMENTS .......................................................................................v

TABLE OF CONTENTS...........................................................................................vi

LIST OF FIGURES ...................................................................................................viii

CHAPTER 1: INTRODUCTION

1.1 Background ............................................................................................1

1.2 Significance of the Subject ....................................................................2 1.3 Contribution and Evaluation of This Study ........................................3

1.4 Scope of the Study ..................................................................................4

CHAPTER 2: LITERATURE REVIEW ...............................................................5

2.1 Annular Flow of Newtonian Fluid .......................................................52.2 Concentric No-Rotation Non-Newtonian ............................................92.3 Eccentric No-Rotation Non-Newtonian ..............................................11

2.4 Newtonian Fluid Flow in Annulus with Inner Pipe Rotation ............122.5 Concentric With Rotation Non-Newtonian .........................................13

2.6 Non-Newtonian Fluid Flow in Annulus with Inner Pipe Rotation ....13

CHAPTER 3: ANNULAR FLOW IN ECCENTRIC ANNULUS .......................15

3.1 Annular Flow in Eccentric Annulus .....................................................153.1.1 Assumptions ..............................................................................16

3.1.2 Continuity Equation .................................................................163.1.3 Momentum Equation.................................................................173.1.4 Drilling Fluid Rheology ............................................................18

General Model: .....................................................................19

3.1.5 Numerical Procedure ...............................................................203.1.6 Discretizing the Equation of Motion .......................................213.1.7 Solution Algorithm ....................................................................233.1.8 Minimum Shear Rate.................................................................23

3.1.9 Convergence Criteria................................................................243.1.10 Flow Rate Calculations...........................................................25

3.1.11 Using Boundary Fitted Coordinate System ............................253.1.12 Geometry Transformation.......................................................26



vii

3.1.13 Transformation of the Equations ............................................273.1.14 Discretization of the Equation of Motion................................28

3.1.15 Grid Refinement Analysis .......................................................293.1.16 Flow Rate Calculations...........................................................31

3.2 Newtonian Fluid Flow in Eccentric Annulus with Inner Pipe

Rotating .........................................................................................................32 3.2.1 Assumptions ..............................................................................33

3.2.2 Governing Equations ................................................................333.2.3 Boundary Conditions ................................................................34

3.2.4 Transformation of Equations ....................................................353.2.5 Numerical Procedure ................................................................383.2.6 Discretization of Momentum and Continuity Equation ............38

3.2.7 Pressure Correction Equation ..................................................423.2.8 Summary of SIMPLE Algorithm ...............................................45

CHAPTER 4: RESULTS AND CONCLUSIONS .................................................474.1 Newtonian Fluid .....................................................................................47

4.1.1 Results for Newtonian Fluid .....................................................474.1.2 Comparison of the Results for Newtonian ................................49

4.2 Power-Law Fluid....................................................................................514.2.1 Results for Power-Law Fluid ....................................................514.2.2 Comparison of the Results of Power-Law ................................54

4.3 Results and Comparison for Yield Power-Law Fluid ........................534.4 Comparison with Experimental data of Ahmed[1] .............................56

4.5 Conclusions .............................................................................................58

REFFERENCES ........................................................................................................56

APPENDIX A............................................................................................................64

APPENDIX B............................................................................................................67APPENDIX C ............................................................................................................73APPENDIX D............................................................................................................82

APPENDIX E ............................................................................................................91APPENDIX F.............................................................................................................95



viii

LIST OF FIGURES

Page

1.1 Schematic of Drilling Operation.........................................................................2

2.1.1 Newtonian fluid in concentric annulus ............................................................6

2.1.2 Eccentric Annulus............................................................................................6

2.1.3 Effects of eccentricity on frictional pressure loss for a Newtonian fluid with constant

rate using 7.1.2.eq . [ Secliter Q /1.0= , Sec Pa.00102.0=µ , m Ro 0889.0= ].....................9

3.1.1 Cartesian coordinate system.................................................................................17

3.1.2 Flow Curve for Different Fluids ...........................................................................19

3.1.3 Grid network in Cartesian coordinate with different grid sizes in x and y direction...21

3.1.4 Flow Rate Calculation at Grid Point i,j..................................................................25

3.1.5 Grid Network in Boundary Fitted Coordinate System............................................26

3.1.6 Geometry Transformation ....................................................................................27

3.1.7 Grid refinement in tangential direction (number of grids in radial direction=60).

Comparison of the value for volume flow rate of simulation with analytical solutions for

concentric annulus for a Newtonian (n=1) and a power-law (n=0.2) fluid. ........................30

3.1.8 Grid refinement in radial direction (number of grids in tangential direction=60).

Comparison of the value for volume flow rate of simulation with analytical solutions for

concentric annulus for a Newtonian (n=1) and a power-law (n=0.2) fluid. ........................31



ix

3.2.1 Boundary conditions for an eccentric annulus with rotating drill pipe............35

3.2.2 Staggered grid for velocity and pressure .........................................................39

3.2.3 Overall SIMPLE Algorithm .................................................................................46

4.1.1 Velocity profile of Newtonian fluid for concentric to fully eccentric annuli [ mOD 1778.0= ,

SecmQ /0001.0 3= , S Pa.00102.0=µ ] ........................................................ ...................47

4.1.2 Velocity profile of Newtonian fluid for concentric and fully eccentric annuli

[ mOD 1778.0= , SecmQ /0001.0 3= , S Pa.00102.0=µ ]..................................................48

4.1.3 Present study results for Newtonian fluid[ mOD 1778.0= , SecmQ /0001.0 3= ,

S Pa.00102.0=µ ]..........................................................................................................49

4.1.4 Eccentricity vs. error in pressure drop of the different studies compared with Piercy et al.

for a Newtonian fluid ..................................................... ...............................................50

4.2.1 Velocity profile of power-law fluid for concentric to fully eccentric annuli [ 5.0=o

i

R

R,

SecmQ /006.0 3= , nS Pa K .6.0= , 2.0=n ].....................................................................51

4.2.2 Velocity profile of power-law fluid for concentric and fully eccentric annuli [ 5.0=o

i

R

R,

SecmQ /006.0 3= , nS Pa K .6.0= , 2.0=n ].....................................................................52

4.2.3 Results for power-law fluid. [ 5.0=o

i

R

R, SecmQ /006.0 3= , nS Pa K .6.0= ] ............53

4.2.4 Results for power-law fluid. [ 5.0=n , SecmQ /006.0 3= , nS Pa K .6.0= ]................54



x

4.2.5 Comparison of Re. f predictions with previous studies [3,5]

for a power-law fluid with

[ 5.0=o

i

R

R, SecmQ /006.0 3= , nS Pa K .6.0= ].................................................................52

4.2.6 Comparison of Re. f predictions with previous studies[3,5]

for a power-law fluid with

[ SecmQ /006.0 3= , 5.0=n , nS Pa K .6.0= ] .................................................. ...................53

4.3.1 Comparison of the present study with Haciislamoglu[7]

[ inchOD 10= , inch ID 5= , SecmQ /012618.0 3= , 7.0=n , S Pa K .25.0= , Pa y 394.2=τ ] .....57

4.4.1 Comparison of the present study with experimental data of Ahmed[1]

[ mOD 035052.0= , m ID 01745.0= , %98.0=ty Eccentrici , 671.0=m ,

S Pa K .2610958.0= , Pa y 1.2=τ , m L 6576.3= ] ...................................................... ..........58



1

CHAPTER 1

INTRODUCTION

1.1 Background

During drilling operation of oil wells a fluid is pumped from a surface mud tank

down to the bottom of the well through drill pipe, through nozzles in the drill bit and then

back to the surface mud tank trough the annular space between the drill pipe and well

bore wall as shown in Fig.1.1. The drilling fluid has to satisfy several requirements such

as: supporting the well bore wall from collapsing, preventing formation fluid from

transferring to the well bore, cooling the drill bit, carrying cutting from bottom of the

well to the surface through annulus, etc.

The pump pressure is a function of pressure losses in surface equipment, in drill

pipe, across the bit nozzles, through the annulus, etc. Frictional pressure loss through

annulus is the main concern of this study.

Several factors can affect frictional pressure loss in annulus such as flow rate,

flow regime, mud rheology, well bore geometry, cuttings content, drill pipe rotation, drill

pipe lateral motion or swirling introduced by the rotation itself and/or fluid flow, etc. In

this study eccentricity is the main investigated issue along with rheology and drill pipe

rotation.



2

1.2 Significance of the Subject

In conventional drilling, frictional pressure loss accounts for about 10% of the

whole circulation pressure loss. [15]

In slim hole configurations, annular frictional pressure loss can contribute 30-50%

of the total fluid circulation pressure loss and some investigators have reported it to be as

high as 90%. [8]

Significant frictional pressure drop in the annulus can result in an increase of the

bottom hole pressure, which might then exceed the strength of the formation, thus

causing loss of drilling fluid and creating a potentially dangerous situation due to the

resulting loss of hydrostatic head. [15]

Fig.1.1 Schematic of Drilling Operation



3

1.3 Contribution and Evaluation of This Study

In this study effect of eccentricity of drill pipe and rheology of drilling fluid on

frictional pressure drop in annuli for Newtonian and non-Newtonian fluid is investigated.

A computer program in FORTRAN is developed to perform calculations.

Results of the program are compared with the present works of Azouz [3], Escudier [4, 5],

Piercy [17] and Haciislamoglu [7] and found to be all in good agreement. Also, numerical

simulation for a Newtonian annular flow in eccentric annulus with a rotational drill pipe

is performed.

1.4 Scope of the Study

A short review of the past works related to this study is the content of Chapter 2.

This review has categorized the previous studies according to the type of the problem.

Chapter 3 is an investigation of laminar flow of Newtonian and non-Newtonian

fluids in eccentric annulus. In the first part of this chapter, drill pipe is considered to be

stationary and fluid is Newtonian or non-Newtonian. The second part of the chapter

considers the drill pipe to be rotating and fluid to be Newtonian only. In the first part of

the chapter, two approaches are used to solve the problem. For the first approach a

rectangular mesh system is used to solve discretized flow equations in Cartesian system.

In the second approach, on the other hand, a boundary fitted coordinate system is utilized

to apply numerical technique to equations. Flow equations as well as geometry are

transformed into this new system and then discretized. Similar solution procedures are

used to solve the equations in Cartesian and boundary fitted coordinate system. The last

part of Chapter 3 is related to flow of Newtonian fluid in eccentric annulus with a rotating

drill pipe. Navier-Stokes equations of motion and continuity in a rectangular coordinate



4

system are transformed into a computational plane in a boundary fitted coordinate

system. SIMPLE algorithm by Patankar-Spalding [16] is chosen to solve the equations.

In chapter 4 results of numerical simulation are presented and compared with

other studies. Results are related to Newtonian and non-Newtonian fluid flow in eccentric

annulus with stationary drill pipe. Results for the case when drill pipe is in motion are not

presented due to instability of the numerical procedure.



5

CHAPTER 2

LITERATURE REVIEW

This chapter is a review of some the previous work about laminar annular flow.

Laminar annular flow is flow of a fluid in an annulus under laminar regime. In laminar

regime, fluid particles travel along well-ordered non-intersecting paths, or layers.

The determination of annular flow performance is important in the planning and

design of the hydraulic program for a well not only because of significant pressure losses

but also possible hole cleaning problems.

Laminar annular flow has been the subject of many investigations. With the

developments in computers and also computational fluid dynamics there is a high trend

of using numerical simulations.

2.1 Annular Flow of Newtonian Fluid

One of the first correlations for annular flow is given by Lamb (1932) [12]. In that

study, an equation is presented that correlates frictional pressure loss with flow rate in a

concentric annulus for a Newtonian fluid.

( )

−

−−=

1

2

221

224

142

ln8

r

r

r r r r

dl

dP q

µ

π 1.1.2.eq

Where,

1r : inner radius of the inner pipe

2r : inner radius of the outer pipe



6

dl

dP : pressure drop

µ : viscosity of the fluid

q : volumetric flow rate

Piercy et al. (1933) [17] presented an exact solution that correlates, frictional

pressure loss, flow rate and eccentricity. Considering Fig.2.21:

Where, o R and i R are the outer and inner cylinder radii and c is the displacement of the

two centers. Momentum equation in the rectangular Cartesian coordinates for this case to

be solved is:

Fig.2.1.1 Newtonian fluid in concentric annulus

R i

R o

c

X

Y

Fig.2.1.2 Eccentric Annulus



7

∂∂

+∂∂

+∂∂

−=2

2

2

2

0 y

v

x

v

z

P z z µ 2.1.2.eq

Where,

z

P

∂∂ : pressure drop in axial direction

µ : fluid viscosity

z v : fluid velocity in axial direction

Using below notation:

( )22 y xk v z +−=φ 3.1.2.eq

The mathematical problem now becomes:

02

2

2

2

=∂∂

+∂∂

y x

φφ 4.1.2.eq

subjected to the no-slip condition, )(

22

y xk w +−=φ on the solid boundaries, which is

Laplace’s equation.

They obtained exact solution by mapping the cross section conformally onto a

region where Laplace’s equation has a known solution. This technique is called” complex

variable” technique. By applying the transformation:

( )

+=+ ηξ i M iz y

2

1tan 5.1.2.eq

where,

2

2222

2 o

io Rc

c R R M −

+−=

z

P k

∂∂

=µ4

1:Where



8

the momentum equation ( 4.1.2.eq ) in the ( )ηξ , ??coordinate system will take the form:

02

2

2

2

=∂

∂+

∂

∂

η

φ

ε

φ 6.1.2.eq

Subject to:

βαηεη

η,01

coshcosh

cosh22 ==

−

+− at M K

Solving 6.1.2.eq :

( )io

n

n

io

R Rc for

nn

ne M c

M c R R

dl

dp−<<

−−

−−−

−=

∑∞

=

+−0,

)sinh(8

4

8

Q

1

)(22

2244

αβαβµ

π αβ 7.1.2.eq

Where,

c

c R R io

2F

222 +−=

22o R F M −=

−+

= M F

M F ln

2

1α

−−+−

= M c F

M c F ln

2

1β



9

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.2 0.4 0.6 0.8 1

Relative Eccentricity, [c/(Ro-Ri)]

d P / d L ( E c c e n . / C o n c e n . )

Ri/Ro=0.8

Ri/Ro=0.5

Ri/Ro=0.2

As can be clearly seen from Fig.2.1.3, the eccentricity of the drill pipe can cause a

substantial decrease in frictional pressure loss.

2.2 Concentric No Rotation Non-Newtonian

Bird and Fredrickson (1958) [6] obtain analytical correlation between flow rate and

frictional loss for Power-law fluids.

( ) ( )

−−−

∆

∆

+=

+−++

nnnnn

n

o

L K

P

n

RQ

1122

111

12

113

123

1 κβκβ

π 1.2.2.eq

Where,

K = fluid consistency index

Fig.2.1.3 Effects of eccentricity on frictional pressure loss for a Newtonian fluid with

constant flow rate using 7.1.2.eq . [ Secliter Q /1.0= , Sec Pa.00102.0=µ , m Ro 0889.0= ]



10

n = flow behavior index for power-law fluid

o

i

R

R=κ

L

P

∆

∆= pressure drop per unit length

And β can be calculated from the following equation:

∫ ∫

−=

−

11

2

1

2

β

β

κ

ξξ

βξξξ

ξ

βd d

nn

2.2.2.eq

In their paper, Bird and Fredrickson (1958) [6] also showed a correlation for

Bingham Plastic fluid flow in concentric annulus as follows:

( ) ( )( ) ( ) ( )

−++−−−−−

∆∆= +++ y y y y

p

o K K K L

P RQ ττλττλλ

µ

π 33242

23

11

3

4121

8 3.2.2.eq

Where,

K = fluid consistency index

pµ = plastic viscosity

o

i

R

R=κ

L

P

∆∆

= pressure

yτ = yield stress

And +λ can be calculated from the following:

( ) ( ) ( ) 0121ln22 =−+++−

−− +

+

+++ λττ

λ

τλτλλ y y

y y K

K 4.2.2.eq

Bird and Fredrickson [6] solved 4.2.2.eq numerically and presented the results in the

form of charts facilitating rapid application.

Skelland (1967) [8] presented another exact solution for Bingham plastic fluid as

follows:



11

( ) ( ) ( )( )

( ) ( )( ) ( )( )

( )( ) ( )( ) ( )33332222

2222222

22222224444

344

4416

816ln

2

16

2

io p

y y

oi

y

iooi y

io y

ioio

i

o p

io

p

R Rbaabbaab R R

R R R R R Rbaab L

P

R Rab L

P R R

L

P

bR

aRba R R

L

P Q

−−++

−++−+−

−++−+−−∆

∆−

−−

∆∆+−

∆∆−

+−+−

∆∆=

µ

πτττ

ττ

µ

π

µ

π

5.2.2.eq

Where, a and b could be calculated by solving the following equations:

( )

( )( )

+−−+=

=−

−−

∆∆

i

o

ioab R R

L P

y

bR

aR

R Rabab

ab

io

ln

2

2222

22

τ

6.2.2.eq

Hanks (1979)[9] presented analytical solution for yield power-law fluid. In that

study several charts are presented through computing theoretical solutions of the

equations of the motion for the concentric annular geometry using the Herschel-Bulkly

model. Using those charts to find some parameters and applying them to the equations,

one can compute volume flow rate through given pressure drop and also computing

pressure drop through given volume flow rate.

2.3 Eccentric No-Rotation Non-Newtonian

Haciislamoglu (1985) [7] and Azouz (1994) [3] used numerical simulation to solve

the governing equations of non-Newtonian fluid flow in eccentric annulus without pipe

rotation.

Nearly all of the analytical and numerical solutions for annular flow without pipe

rotation indicate that frictional pressure loss decreases with increasing eccentricity, at a



12

given flow rate. The reduction in pressure loss can be significant for instant can be in the

range of the pressure loss.

2.4 Newtonian Fluid Flow in Annulus with Inner Pipe Rotation

Yamada (1960) [22] described the experimental results for the resistance of water

flow through an annulus formed by two concentric cylinders, with the inner cylinder

rotating and the outer cylinder stationary. This study shows that when the flow is laminar,

the resistance of a flow is unaffected up to a certain rotational speed. But beyond this

speed the flow resistance increases as the Reynolds number increases.

Ooms et al. (1996) [15] performed numerical simulation of Newtonian fluids flow

in annulus to investigate effects of eccentricity and rotational speed of the inner pipe on

frictional pressure loss. Their study was more accurately followed by Escudier et al.

(1999) [4]. Theoretical and experimental study of Escudier et al. [4] suggested that in the

case of concentric annulus, rotation of the inner pipe does not influence frictional

pressure loss of the Newtonian fluids. However, in the case of eccentric annulus they

showed that when the inner pipe is rotating, for low eccentricities (less than 30%),

pressure drop remains approximately constant with increasing of eccentricity. For higher

eccentricities however, frictional pressure loss generally decreases with increasing of

eccentricity. Escudier et al. [4] also found that for a given radius ratio, as rotation speed of

the inner cylinder increases at any eccentricity, frictional pressure loss increases.





14

drop values for power-law fluids follow the trends observed by Escudier et al. (1999) [4]

for Newtonian fluids, including an increase with rotation of the inner pipe, an increase

with eccentricity at low and very high eccentricities but a decrease for intermediate

eccentricities. They showed that a power-law fluid generally exhibits lower pressure drop

as compared with the Newtonian liquid.



15

CHAPTER 3

ANNULAR FLOW IN ECCENTRIC ANNULUS

In this chapter laminar flow of Newtonian and non- Newtonian fluids through an

eccentric annulus is numerically simulated. In the first part of the chapter drill pipe is

considered to be stationary and fluid is Newtonian or non-Newtonian. In the second part

of this chapter inner pipe is rotating and fluid is Newtonian.

3.1 Annular Flow in Eccentric Annulus without Inner Pipe Rotating

To simulate fluid flow in an eccentric annulus when the drill pipe is stationary,

governing equations for flow in eccentric annulus are developed in the Cartesian

coordinate system. Yield Power Law model is used to represent the rheology of the fluid.

To solve the flow equations, two approaches are used. In the first approach a rectangular

mesh system is used to solve discretized flow equations in the Cartesian coordinate

system. In the second approach, on the other hand, a boundary fitted coordinate system is

utilized to apply numerical technique to equations.

In the first approach using a rectangular grid system, equations are discretized and

an iterative over relaxation method is used to solve for velocity field at each grid point

knowing frictional pressure loss in axial direction of the well bore. A relaxation technique

is used to speed up convergence of the solution. Calculation of relaxation factor is

presented and the treatment for the no shear region is discussed.



16

In the second approach similar solution procedure is used, but geometry as well as

equations are transformed into a computational domain in a boundary fitted coordinate

system.

3.1.1 Assumptions

1-Laminar and fully developed flow

2-Isothermal and steady state conditions

3-Flowing direction is in the annulus along the axial direction of the well bore.

4- Incompressible fluid

5- Flow domain is an eccentric annulus.

6- Drill pipe is stationary

7-No slippage at the walls

3.1.2 Continuity Equation

Navier-Stokes continuity equation for an isothermal laminar flow in Cartesian

coordinate system is:

∂∂

+∂

∂+

∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

z

v

y

v

x

v

z v

yv

xv

t

z y x z y x ρ

ρρρρ 1.1.3.eq

Where, ρ is the density of the fluid and z y x vvv ,, are velocity components in

z y,, directions respectively. Coordinate system is shown in Fig.3.1.1.

Since the fluid is incompressible, all the terms on the left hand side of eq.3.1.1 are

zero. The first two terms in parenthesis on the right hand side of the same equation are

zero since the flow is purely axial (i.e. no flow in x and y directions). So, from 1.1.3.eq :



17

0=∂

∂ z

v z 2.1.3.eq

Fig.3.1.1 Cartesian coordinate system

3.1.3 Momentum Equation

For axial laminar flow, the equation of motion will take the form (Appendix A):

z

P

y

w

y x

w

x ∂

∂=

∂

∂

∂

∂+

∂

∂

∂

∂µµ 3.1.3.eq

Where, w is the fluid velocity in z direction and µ is the viscosity that is a

function of shear rate. The viscosity function depends on the rheology of the fluid. The

shear rate for axial flow can be expressed as:

22

∂∂+

∂∂=

y

w

x

wγ& 4.1.3.eq

In order to solve the above equations, the following boundary conditions are

applied:

1-Velocity at the drill pipe is zero

2- Velocity at the well-bore wall is zero.

X

Z

Y

F l o w

D i r e c t i o n

F l o w

D i r e c t i o n



18

3.1.4 Drilling Fluid Rheology

Flow behavior of a fluid can be described by a mathematical relationship between

shear stress and shear rate. Fluids are generally categorized as Newtonian and non-

Newtonian Fig.3.1.2. This classification is based on the relationship between shear rate

and shear stress. Drilling fluids are mostly non-Newtonian in character and almost

invariably shear thinning, and often exhibiting visco-elastic and thixotropic properties as

well as yield stress [2].

Newtonian Model:

γµτ = 5.1.3.eq

Bingham Plastic Model:

γµττ & p y +=

6.1.3.eq

Power Law Model:

→<→>

=FluidticPseudoplas1

FluidDilatant1

n

n K nγτ &

7.1.3.eq

Yield Power Law (Herschel-Bulkley) Model:

Pipe and annular flows of Yield Power-Law fluid is of great interests in drilling

applications. This model describes the rheological behavior of drilling muds more

accurately than Bingham Plastic and Power-Law models. The Yield Power-Law

rheological model for all time-independent fluids is given by:

n y K γττ &+=

8.1.3.eq

Where:



19

τ = Shear Stress

yτ = Yield Shear Stress

γ& = Shear Rate

K = Fluid Consistency Index

n = Flow Behavior Index

µ = Newtonian Viscosity

pµ = Plastic Viscosity

Fig.3.1.2 Flow Curve for Different Fluids

General Model:

All the above models can be expressed by the Yield Power Law model as follows:

→≠≠

=→=≠

→≠=

=→==

+=

Fluid Law Power Yield n If

K Model Plastic Binghamn If

Model Law Power n If

K Model Newtoniann If

K

y

p y

y

y

n y

1,0:

)(1,0:

1,0:

)(1,0:

τ

µτ

τ

µτ

γττ & 9.1.3.eq

Apparent Viscosity:

A non-Newtonian fluid does not have a constant viscosity like Newtonian. However, in

numerical modeling, the concept of viscosity is used for non-Newtonian fluids to make

the governing equations similar to Newtonian fluids. This viscosity is known as apparent





21

Fig.3.1.3 Grid network in Cartesian coordinate with different grid sizes in x and y direction

3.1.6 Discretizing the Equation of Motion

Equation of motion is a differential equation that can be approximated by a

discretized finite difference equation. A second-order central differencing scheme has

been used for all the grid points inside of the flow domain. This scheme establishes the

following system of algebraic equations:

∂∂

−++++++

= −+

+−+

++

z

P W AW AW AW A

A A A AW ji

n

ji

n

ji

n

ji

n

ji

n

1,

1

41,3,1

1

2,114321

,

1 1 13.1.3.eq



22

Where, coefficients are calculated based on 10.. Beq (Appendix B). In order to

speed up the rate of convergence an over relaxation factor is introduced to 1.6.3.eq thus:

( )

∂∂

−+++−

+++

++++= −

++−

++

+

z

P W A A A A

W AW AW AW A A A A A

W W

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

,4321

1,

1

41,

3,1

1

2,1

14321

,,

1 ω

14.1.3.eq

Where:

ji

n

W ,

1+ : The value of the velocity field at node ji, at computational step (iteration) 1+n

ji

n

W , : The value of the velocity field at node ji, at computational step (iteration) n

Where, ω is over relaxation factor and is calculated following Azouz [3]at each

grid point and it is shown in (Appendix D):

( )

( )

<−++

≥−−+

=

0411

2

0411

2

2

3

2

12

23

21

2

ϕϕρ

ϕϕρ

ω

if

if

J

J

15.1.3.eq

Where:

+−+

+−

+=

1cos4

1cos4

2

1 24

22

23

21

21 N M J

πϕϕ

πϕϕ

ϕϕρ

appµϕ =1

appµϕ =2

x∂∂= µ

ϕ3



23

y∂∂

= µ

ϕ4

Where, N M , are number of grid points in y x, direction. Details of calculations

can be found in Appendix D.

Discretizing viscosity equation in 11.1.3.eq yields:

( ) 1,

,,

−+= n ji

ji

y

ji K γγ

τµ &

& 16.1.3.eq

Where:

2

1,1,

2

,1,1,

22

∆

−+

∆

−= −+−+

y

W W

x

W W ji ji ji ji jiγ& 17.1.3.eq

3.1.7 Solution Algorithm

Three systems of algebraic equations are solved iteratively. First, discretized

velocity equation ( 14.1.3.eq ) is solved at point ji, that is a point in the flow region. Then

viscosity equation ( 16.1.3.eq , 17.1.3.eq ) followed by relaxation factor calculation ( )15.1.3.eq

is evaluated. Iterations continue until convergence is achieved. Having velocity field,

flow rate is calculated through numerical integration over the flow region.

3.1.8 Minimum Shear Rate

Visco-plastic fluids encounter a problem for small shear rate in the un-yielded

region while calculating the apparent viscosity. This is the most difficult aspect for

numerical modeling of fluids with yield stress. Azouz [3] used a minimum cut off value of

the shear rate. It is better to determine a cut off value corresponding to the conditions. It



24

was experienced that if the cut of value is too small or too big, sometimes we need too

many unnecessary iterations or we may encounter instability problem. The following

equations can be used for calculating cut off value of minimum shear rate:

=− StressYiel without Fluids for

StressYield with Fluids for K

wall

n y

γ

τ

γ

&

&

8

1

8

min

10

10

18.1.3.eq

where, wall γ& can be calculated from 24.. Beq .Detailed derivations of the above equations

can be found in Appendix B.

Iteration begins with a relatively large value of cut off shear rate. When velocity

field reaches a convergence the cut off value must decrease and iterations continue with

the new cut off shear rate. This procedure continues until the smallest desired cut off

value is obtained. From then on, it will remain constant.

3.1.9 Convergence Criteria

When the values of velocity field do not change with more iteration and also

smallest calculated shear rate reaches the minimum desired shear rate, convergence is

achieved and calculations should be stopped.

Convergence of velocity is achieved when velocity field at new iterations is close

to the values of the previous iteration. This is defined by a relative value, called residual

of velocity:



25

( ) ( )

( )∑∑

∑∑

= =

= =

−

=max max

max max

1 1

,

1 1

,,

j

j

i

i

jiold

j

j

i

i

jiold

jinew

w

ww

Residual Velocity 19.1.3.eq

If the velocity residual is small enough 710.. − g e , convergence is achieved.

3.1.10 Flow Rate Calculations

Flow rate is integral of velocity times its associated area:

y xw AwwdAQ

j

j

i

i ji

j

j

i

i ji ji ∆∆=≈= ∑∑∑∑∫ = == =

max maxmax max

1 1,

1 1,, 20.1.3.eq

Fig.3.1.4 Flow Rate Calculation at Grid Point i,j

3.1.11 Using Boundary Fitted Coordinate System

One advantage the boundary fitted coordinate system compared to rectangular

coordinate system is its ability to conform to the boundaries of the system regardless of

i i+1i-1

j+1

j

j-1

y∆

x∆

jiw ,

y x A ji ∆∆=,



26

the shape. In another word grid generated can fit itself into the boundary of the system as

shown in Fig.3.1.4. This feature increases the accuracy of the solution developed.

Doing the transformation, physical Cartesian coordinates (x,y) becomes the

dependent variables and the curvilinear coordinates ( ηξ , ) becomes the independent

variables. A generated grid is then defined as a set of points formed by the intersections

of the lines of a boundary conforming curvilinear coordinate system.

Fig.3.1.5 Grid Network in Boundary Fitted Coordinate System

There is a uniform grid in this system in the sense that at each radius position the

angular width is constant and at each angular position, the radial width of the cells is also

constant.

x

y

e i R

o R ),(),(),( ji y x =≡ ηξ





28

e : Distance between the inner cylinder (drill pipe) and outer cylinder (well bore wall).

3.1.13 Transformation of the Equations

Equation of motion and apparent viscosity equation 12.1.3.eq are numerically

transformed from a Cartesian system to a computational domain. According to 43..C eq

(Appendix C):

( ) ( ) z

P J W W

J W W

J

appapp

∂∂

=

−

∂∂

+

−

∂∂

ξηηξ γβµ

ηγα

µ

ξ 22.1.3.eq

Where:

1−+= n y

app K γγ

τµ &

&

( )ηξηξ γβαγ W W W W J

21 22

2 −+=&

22ηηα x y +=

ηξξηγ x x y y +=

22ξξβ x y +=

ξηηξ y x y x J −=

Detailed derivations can be found in Appendix C.

3.1.14 Discretization of the Equation of Motion

The same way of discretizing as was used for the Cartesian coordinate system will

be used for the transformed equation:



29

0101,191,18,171,6

1,15,141,31,12,1

=−−−−

−−−−−

++−−++

+−−−−+

AW AW AW AW A

W AW AW AW AW A

ji ji ji ji

ji ji ji ji ji

23.1.3.eq

Where coefficients 101 A A − can be found in 16.. Deq (Appendix D) .

Solution algorithms will the same as well. Using relaxation factor:

+−+++

+++++=

++−−+

++

+−−+

−+

−++

10,11,191,1

1

8,171,6

1,15,1

1

41,

1

31,121

,,

1

AW AW AW AW AW A

W AW AW AW A A

W W

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n ω

24.1.3.eq

3.1.15 Grid Refinement Analysis

A gird refinement analysis showed that using grid numbers more than 100 in

radial and in tangential direction does not result in a significant change in flow rate. In

order to check the results for different grid numbers, a concentric annulus was considered

and the results of the numerical simulation was compared with the results of analytical

solution for Newtonian and non-Newtonian as it is shown in 6.1.3. Fig and 7.1.3. Fig .



30

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 100 200 300 400

Number of Grids in Tangential Direction

% E r r o r i n

f l o w r a t e ( c o m p a r e d w i t h a n a l y t i c a l

s o l u t i o n s )

n=1.0

n=0.5

Fig. 3.1.7 Grid refinement in tangentialdirection (number of grids in radial direction=60).Comparison of the value for volume flow rate of simulation with analytical solutions for

concentric annulus for a Newtonian (n=1) and a power-law (n=0.2) fluid.





32

( ) ( ) ( ) ( )1,11,11,11,11,11,11

2

1+++−+++++−++

∆−+−−= ji ji ji ji ji ji x x y y y x ABO 27.1.3.eq

( ) ( ) ( ) ( )1,11,11,11,11,11,112

1−+−−−+−+−−−+

∆−+−−= ji ji ji ji ji ji x x y y y xCDO 28.1.3.eq

( )( )( ) ( )( )

( )

−

−+−−

−

−+−−=

∆

2

2

2

2

1

1

2sin

1

1

2

1

JM

R i Ro j JM R o

CDO Arc

JM

R i Ro j JM R oCPDO 29.1.3.eq

( ) ( ) ( )( ) ( )1,11,11,11,11,11,12

12

21 −+−−−+−+−−−+∆ −

−−−−−−== ji ji ji ji ji ji x x y

JM je y y xCDO 30.1.3.eq

( )( )( ) ( )( ) ( )

( )

−

−+−−

−

−−−−=

∆

2

3

2

3

1

1

2sin

1

1

2

1

JM

R i Ro j JM R o

ABO Arc

JM

Ri R o j JM R o AQBO 31.1.3.eq

( ) ( )( )

( )1,11,11,11,11,11,1312

1+++−+++++−++

∆−

−

−−−−= ji ji ji ji ji ji x x y

JM

je y y x ABO 32.1.3.eq

Where, e JM IM ji ,,,, are shown in Fig.3.1.5.

3.2 Newtonian Fluid Flow in Eccentric Annulus with Inner Pipe Rotating

The objective of this section is to investigate laminar Newtonian fluid flow in a

concentric annulus with inner pipe rotation. Navier-Stokes equations for laminar flow of

a Newtonian fluid in Cartesian coordinate system are considered. All the equations are

transformed to a computational domain in the boundary fitted coordinate system.

Geometry, on the other hand is transformed algebraically. Having equations and



33

geometry in the computational domain, SIMPLE algorithm (Patankar –Spalding, 1972) is

applied to solve the equations.

3.2.1 Assumptions

1-Laminar and fully developed flow

2-Isothermal and steady state conditions

3-Flowing direction is in the annulus along the axial direction of the well bore.

4- Incompressible fluid

5- Flow domain is an eccentric annulus.

6- Drill pipe is rotating at a constant angular velocity

7-No slippage at the walls

3.2.2 Governing Equations

Considering Navier-Stokes equations in a Cartesian coordinate system for

Newtonian fluid, momentum equations in z y x ,, will be:

( ) ( ) x g

x

p

z

w

y

v

x

u

z

wu

y

vu

x

u

t

uρµρρ +

∂∂

−

∂

∂+

∂

∂+

∂

∂+

∂∂

+∂

∂+

∂∂

−=∂∂

2

2

2

2

2

22

1.2.3.eq

( ) ( ) y g

y

p

z

w

y

v

x

u

z

wv

y

v

x

uv

t

vρµρρ +

∂∂

−

∂

∂+

∂

∂+

∂

∂+

∂∂

+∂∂

+∂

∂−=

∂∂

2

2

2

2

2

22

2.2.3.eq

( ) ( ) z g z p

z w

yv

xu

z w

yvw

xuw

t w ρµρρ +

∂∂−

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂−=

∂∂

2

2

2

2

2

22

3.2.3.eq

And Continuity equation is:

0=

∂∂

+∂∂

+∂∂

+∂∂

z

w

y

v

x

u

t

ρ 3.2.3.eq



34

The first term on the left hand side of all the momentum equations is zero because

the flow is under steady state. Also on the right hand side of the same equations the last

term in the second parenthesis is zero due to flow is fully developed. Also the last term

on 1.2.3.eq and 2.2.3.eq is zero because there is not any gravity in x and y directions.

In continuity equation, the first term is also zero because of a steady state flow.

The last term on the left hand side of this equation is also zero because fluid is fully

developed.

Gravity in y x, directions is zero. Also we wish to eliminate gravity in z direction

in order to consider only frictional pressure loss so 4.2.321.3. −eq will take the new form:

( ) ( ) x

p

y

v

x

u

z

wu

y

vu

x

u

∂∂

−

∂

∂+

∂

∂=

∂∂

+∂

∂+

∂∂

2

2

2

22

µρ

5.2.3.eq

( ) ( ) y

p

y

v

x

u

z

wv

y

v

x

uv

∂∂

−

∂

∂+

∂

∂=

∂∂

+∂∂

+∂

∂2

2

2

22

µρ

6.2.3.eq

( ) ( ) z

P

y

v

x

u

z

w

y

vw

x

uw

∂∂

−

∂

∂+

∂

∂=

∂∂

+∂

∂+

∂∂

2

2

2

22

µρ

7.2.3.eq

0=∂∂

+∂∂

y

v

x

u

8.2.3.eq

3.2.3 Boundary Conditions

==

−=

===

0

:)(

0

:)(

..

w

Cosr v

Sinr u

boundaryinner string drill theOn

wvu

boundaryouter wall borewell theOn

Cs B

θω

θω

9.2.3.eq



35

Fig.3.2.1 Boundary conditions for an eccentric annulus with rotating drill pipe

3.2.4 Transformation of Equations

Using boundary fitted coordinates flow equations can be transformed from the

real domain onto a computational domain.

According to Hoffman [10] :

ξηηξ y x y x J −=1

10.2.3.eq

J

y x

ηξ =

11.2.3.eq

J

x y

ηξ −=

12.2.3.eq

J

y x

ξη −=

13.2.3.eq

J

x y

ξη =

14.2.3.eq

( )ξηηξ y f y f J f x

f x −==

∂∂

15.2.3.eq

θ

ωr

θω Cosr

θωSinr −

ω

x

y



36

( )ξηηξξη y f x f x f J f y

f y −−==

∂∂

16.2.3.eq

( )

( )( ){

( )( )}ξηηξηηξξηηξξξ

ηξξηηηξξηηξξξ

ηηξξηηξξξ

f y f y x y x y y x y

f x f x y y y y y y y J

f y f y y f y J x

f

n

n

n

−+−

+−+−

++−=∂

∂

22

223

222

2

2

2

2

2

17.2.3.eq

( )

( )( ){

( )( )}ξηηξηηξξηηξξξ

ηξξηηηξξηηξξξ

ηηξξηηξξξ

f y f y x x x x x x x

f x f x y x y x x y x J

f x f x x f x J x

f

n

n

n

−+−+

−+−+

+−=∂∂

22

223

222

2

2

2

2

2

18.2.3.eq

Applying 18.2.310.2.3. −eq to equations of motion and continuity, yields:

ξ -momentum equation:

)2( 1 ξηηηξηξξ ueducubuau J +++−

( ) ( ) ( ) 01

21432

22

1 =++++++ ηξξηηξρ

pa pauvauvauaua 19.2.3.eq

η -momentum equation:

)2( 1 ξηηηξηξξ vedvcvbvav J +++−

( ) ( ) ( ) 01

432

42

321 =++++++ ξηξηηξρ

pa pavavauvauva 20.2.3.eq

Z -momentum equation:

)2( 1 ξηηηξηξξ wedwcwbwaw J +++−

( ) ( ) ( ) ( ) 01

4321 =∂∂

−++++ Z

P

J vwavwauwauwa

µξηηξ 21.2.3.eq



37

Continuity Equation:

04321 =++++ ξηηξ vavauaua

22.2.3.eq

Where:

22ηη x ya +=

ηξηξ y y x xb +=

22ξξ x yc +=

ξηηξ y x y x J

−= 1

ηηξηξξα cxbxax +−= 2

ηηξηξξβ cybyay +−= 2

βα ξξ x y J d −= 23.2.3.eqs

αβ ηξ y x J e −=1

ηµ

ρ ya −=1

ξµ

ρ ya =2

ξµ

ρ xa −=3

ηµ

ρ xa =4



38

3.2.5 Numerical Procedure

This set of equations ( )2.22.319.2.3. −eqs exhibit a mixed elliptic-parabolic

behavior, and hence the standard relaxation technique which is widely used for those

types of equations is not particularly helpful.

To solve these kinds of equations, Patankar and Spalding (1972) proposed a

method called SIMPLE (Semi Implicit Pressure Linked Equation) algorithm. SIMPLE

algorithm is basically an iterative approach, where some innovative physical reasoning is

used to make the next iteration from the results of the previous iteration. The idea is to

start with discrete continuity equation and substitute into it the discrete u and v

momentum equations. Discrete momentum equations contain pressure differences hence

we can get an equation for the discrete pressures. SIMPLE algorithm actually solves for a

related quantity called the pressure correction.

3.2.6 Discretization of Momentum and Continuity Equation

Using ordinary discretization of equations in these types of equations may create a

checkerboarding problem as described by Patankar-Spalding [15]. A popular remedy for

checkerboarding is the use of a staggered mesh. The key feature here is to calculate

pressure and velocity at different grid points as shown in 2.2.3. Fig .



39

Fig.3.2.2 Staggered grid for velocity and pressure

The original formulation of the SIMPLE method by Patankar and Spalding

involved a finite-volume approach. In this study, a finite-difference approach is used

which would essentially give the same results as obtained by finite-volume method. We

choose to use a forward or backward differencing for the grid points close to the walls.

For other points, central differencing with second order accuracy is used. The

computational domain is assumed to have equal spaces in ηξ , direction meaning:

1=∆=∆ ηξ .

ξ -momentum equation after some arrangements will be:

( ) ( ) ( )[ ] ji ji ji ji ji P P a P P a

ca J AU ,2,2,2,12,1

2

1−+−

++= +−−

ρ 24.2.3.eq

Where:

( )ca J

A A

+=

2

12

( ) jη

( )iξ

i,j

i, , j-1

i-1, j

jiU ,1−

ji ji W P ,, ,

1, − jiV





41

And,

2

2

2

2

2,12,1

,1,1

2,1,1

2,1,1

−−−+∨∨

−+∨∨

−−−

−++

+=

+=

+=

+=

ji ji

ji ji

ji ji

ji ji

U U U

U U U

U U U

U U U

z -momentum equation:

2, C W ji = 26.2.3.eq

Where:

( )ca J

C C

+=

2

12

( ) ( )

( ) ( ) ( )

∂∂

−

−+

−

+

−+

−

+

−+−++

+

+−−−+=

−

≈

+−

∧∧

+

−

∧∧

+

∧

−−

≈

−+

−+−+−+

−−+−−+++−+

z

p

J W V W V

aW V W V

a

W U W U a

W U W U a

W W e

W W d

W W c

W W W W b

W W a J C

ji ji ji ji

ji ji ji ji

ji ji ji ji ji ji

ji ji ji ji ji ji

µ

1~

2

~

2

2

~

2

22

2

,2,24

2,2,3

2,2,2

1,21,21

,2,2

1

2,2,2,2,

2,22,22,22,2,2,21

2

2

~

2,12,1

2,12,1

−+−−≈

−+++

+=

+=

ji ji

ji ji

U U U

U U U

2

2

2,12,1

2,12,1

−+−−∧∧

+++−∧

+=

+=

ji ji

ji ji

U U U

U U U



42

2

2

~

1,21,2

1,21,2

−−+−≈

−+++

+=

+=

ji ji

ji ji

U V V

V V V

2

21,21,2

1,21,2

−−−+∧∧

+−++∧

+=

+=

ji ji

ji ji

U V V

V V V

Continuity equation:

01,21,41,1,32,1,12,1,11 =−+−+−+− −−−−+−−−−+ ji ji ji ji ji ji ji ji V V aV V aU U aU U a 27.2.3.eq

Where could be calculated from 23.2.3.eqs .

3.2.7 Pressure Correction Equation

The primary idea behind SIMPLE is to create a discrete equation for pressure (or

alternatively, a related quantity called the pressure correction) from the discrete

continuity equation( )27.2.3.eq . Since the continuity equation contains discrete face

velocities, we need some way to relate these discrete velocities to the discrete pressure

field. The SIMPLE algorithm uses the discrete momentum equations to derive this

connection. Let ** ,V U and *W be the discreteU ,V and W fields resulting from a solution of

the discrete V U , and W momentum equations. Let * P represent the discrete pressure field

which is used in the solution of the momentum equations. Thus 1,,1 , −− ji ji V U and jiW , satisfy:

( )

−+

−+

+= +−− ji ji ji ji ji P P a P P aca J

AU ,

*

2,

*

2,2

*

,

*

1

*

2,1

*

2

1

ρ 28.2.3.eq

( )

−+

−+

+= +−− ji ji ji ji ji P P a P P aca J

BV ,

*

,2

*

42,

*

,

*

3

*

21,

*

2

1

ρ 29.2.3.eq



43

2

*

,

*

C W ji = 30.2.3.eq

01,21,41,1,32,1,12,1,11 =−+−+−+− −−−−+−−−−+ ji ji ji ji ji ji ji ji V V aV V aU U aU U a 31.2.3.eq

If the pressure field * P is only a guess or a prevailing iterate, the discrete

**,V U and *

W obtained by solving the momentum equations will not, in general, satisfy the

discrete continuity equation( )31.2.3.eq . A correction is proposed to the starred velocity

field such that the corrected values satisfy continuity equation:

U U U ′+=*

32.2.3.eq

V V V ′+=*

33.2.3.eq

W W W ′+=*

34.2.3.eq

Correspondingly, we wish to correct the existing pressure field

*

p with

P P P ′+=*

35.2.3.eq

Subtracting 26.2.324.2.3. −eqs , from 30.2.328.2.3. −eqs , we obtain:

( ) ( ) ( )[ ] ji ji ji ji ji P P a P P a

ca J AU ,2,2,2,12,1

2

1 ′−′+′−′+

+′=′ +−−ρ

36.2.3.eq

( ) ( ) ( )[ ] ji ji ji ji ji P P a P P aca J

BV ,,242,,321,2

1

′−′+′−′++′=′ +−−ρ 37.2.3.eq

2, C W ji ′=′ 38.2.3.eq

We now make the main simplification of SIMPLE algorithm by doing following

approximation:



44

0

0

0

2

2

2

≅′≅′≅′

C

B

A

39.2.3.eqs

So, 38.2.336.2.3. −eqs will take the form

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

0

2

1

2

1

,

,,242,,31,

,2,2,2,1,1

=′

′−′+′−′+

=′

′−′+′−′+

=′

+−−

+−−

ji

ji ji ji ji ji

ji ji ji ji ji

W

P P a P P aca J

V

P P a P P aca J

U

ρ

ρ

40.2.3.eqs

We now consider the discrete continuity equation. The starred velocities**

, V U

obtained by solving the momentum equations using the prevailing pressure field*

P do

not satisfy the discrete continuity equation. Thus substituting 33.2.3.32.2.3. eqand eq into

continuity equation:

01,21,2

*

1,1,

*

41,1,

*

1,1,

*

3

2,12,1

*

,1,1

*

2,1,1

*

,1,1

*

1

=

′+−

′++

′+−

′++

′+−

′++

′+−

′+

−−−−−−−−++

−−−−−−++++

ji ji ji ji ji ji ji ji


V V V V aV V V V a

U U U U aU U U U a

41.2.3.eq

Now substituting 37.2.3.30.2.3. eqand eq into 41.2.3.eq , after some arrangements it

yields:





46

Fig.3.2.3 Overall SIMPLE Algorithm

Correct pressure and velocities

32.2.3.*

eqU U U ′+=

33.2.3.*

eqV V V ′+=

35.2.3.*

eq P P P ′+=

Solve pressure correction equation

( ) ( ) ( )[ ] 42.2.3.1

2,2,42,22,23,2,221

, eq P P f P P f P P f f

P ji ji ji ji ji ji ji +−−−++−+ ′+′+′+′+′+′−=′

START

STOP

Convergence ?

Initial guess p*, u*, *

Solve discritized momentum equations

( )28.2.3.,

*

2,

*

2,2

*

,

*

1

*

2,1

*

2

1eq ji ji ji ji ji P P a P P a

ca J AU

−+

−+

+= +−−ρ

( )29.2.3.,

*

,2

*

42,

*

,

*

3

*

21,

*

2

1eq ji ji ji ji ji P P a P P a

ca J BV

−+

−+

+= +−−ρ

2

*

,

*

C W ji = 30.2.3.eq

Yes

No

Set:

U U =*

V V =*

W W =*

P P =*



47

CHAPTER 4

RESULTS AND CONCLUSIONS

4.1 Newtonian Fluid

4.1.1 Results for Newtonian Fluid

As eccentricity increases, frictional pressure drop decreases. This value could be

about 50% less than the pressure drop of a concentric annulus with the same flow rate.

When the annulus becomes eccentric, one side of annuls is wider than the other side.

Since fluid always tends to bulge through the wider area, most of the fluid flows through

the wider area of the annulus (as shown in Fig.4.1.1), where there is less restriction. In

other words the portion of the fluid that is taking the highest restriction of the geometry is

less than the portion under lower restriction. Figure.4.1.3 shows the results of the

simulation for Newtonian fluid in eccentric annuli.

Fig.4.1.1 Velocity profile of Newtonian fluid for concentric to fully eccentric annuli

[ mOD 1778.0= , SecmQ /0001.0 3= , S Pa.00102.0=µ ]



48

Fig.4.1.2 Velocity profile of Newtonian fluid for concentric and fully eccentric annuli

[ mOD 1778.0= , SecmQ /0001.0 3= , S Pa.00102.0=µ ]



49

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.2 0.4 0.6 0.8 1 1.2

Eccentricity

P r e s s . D

r o p ( E c c e n t r i c / C o n c e n t r i c )

(Ri/Ro)=0.8(Ri/Ro)=0.5

(Ri/Ro)=0.2

Fig.4.1.3 Present study results for Newtonian fluid [ mOD 1778.0= , SecmQ /0001.0 3= ,

S Pa.00102.0=µ ]

4.1.2 Comparison of the Results for Newtonian

Figure 4.1.4 shows the comparison of the present and past studies with the

analytical solution of Piercy et al. [17]. All the cases can predict quite reasonable results.

Considering the trend of the error for different studies, in the study of Azouz [3] error

seems to be quite constant for different eccentricities. However this is quite different for

the other two studies where with increasing of the eccentricity, error decreases. Since the

error can be generate from different sources, one can not explain the reason of these

discrepancies.



50

Present Study

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.2 0.4 0.6 0.8 1 1.2

Eccentricity

E r r o r %

Ri/Ro=0.8

Ri/Ro=0.5

Ri/Ro=0.2

Escudier et al.

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0 0.2 0.4 0.6 0.8 1 1.2

Eccentricity

E r r o r %

Ri/Ro=0.8

Ri/Ro=0.5

Ri/Ro=0.2

Azouz

Fig.4.1.4 Eccentricity vs. error in pressure drop of the different studies compared with

Piercy et al. for a Newtonian fluid





52

Fig.4.2.2 Velocity profile of power-law fluid for concentric and fully eccentric annuli [ 5.0=o

i

R

R,

SecmQ /006.0 3= , nS Pa K .6.0= , 2.0=n ]







55

2

4

6

8

10

12

14

16

0 0.2 0.4 0.6 0.8 1

Eccentricty

f . R e

Escudier et al.

Azouz

Present Studyn=0.8

n=0.5

n=0.2

Ri/Ro=0.5

Figure 4.2.5 Comparison of Re. f predictions with previous studies

[3,5] for a power-law fluid

with [ 5.0=o

i

R

R, SecmQ /006.0 3= , nS Pa K .6.0= ]



56

3.5

4

4.5

5

5.5

6

6.5

7

7.5

8

0 0.2 0.4 0.6 0.8 1

Eccentricty

f . R e

Escudier et al.

Azouz

Present Study

Ri/Ro=0.2

Ri/Ro=0.5

Ri/Ro=0.8

n=0.5

Figure 4.2.6 Comparison of Re. f predictions with previous studies [3,5]

for a power-law fluid

with [ SecmQ /006.0 3= , 5.0=n , nS Pa K .6.0= ]

4.3 Results and Comparison for Yield Power- Law Fluid

Like power-law and Newtonian fluids, the same trend is observed for yield

power-law fluid. For the following case (Fig. 4.3.1), reduction in frictional pressure loss

for eccentric annulus compared to concentric case is 40%. A comparison between the

results of this study with Haciislamoglu[7]

for this case shows a good agreement between

the two studies and the differences is less than 2%.





58

0

10

20

30

40

50

60

70

0 5 10 15 20 25

Q[GPM]

P r e s s u r e D r o p [ i n c h o f w a t e r ]

Experimental Data

Present Study

Figure 4.4.1 Comparison of the present study with experimental data of

Ahmed[1]

[ mOD 035052.0= , m ID 01745.0= , %98.0=ty Eccentrici , 671.0=m ,

S Pa K .2610958.0= , Pa y 1.2=τ , m L 6576.3= ].

Note: Extensive numerical simulation using the program indicates that the

numerical procedure used in this study is not able to accurately predict the flow rate for

fluids with yield stress when the plug region is relatively large.

4.5 Conclusions

Ø For a given flow rate increasing eccentricity of the inner pipe decreases the

pressure drop in annuli;



59

Ø Pressure loss reduction due to eccentricity for highly shear thinning ( )2.0=n

non-Newtonian fluids is approximately 30%, while for Newtonian fluids can

be as high as 50%;

Ø The reduction of pressure loss due to eccentricity, is more significant at higher

radius ratios

→ 1..

o

i

R

Rei ;

Ø Using Cartesian grid networks simplifies the equation of motion and

numerical procedure, however at the same time less accuracy was observed;

Ø Due to complexity of the pipe rotation case, it is recommended to do more

simplification to find the source of instability.





61

[6] Fredrickson, A. G., Bird, R.B.: “Non-Newtonian Flow in Annuli,” Ind. And Chem.,

Vol. 50, March 1958,pp.347-352.

[7] Haciislamoglu, M. “Non-Newtonian Fluid Flow in Eccentric Annuli and Its

Application to Petroleum Engineering Problems.” Dec. 1989.

[8] Haige et al., “Experimental Study of Slim Hole Annular Pressure loss and Its Field

Applications.” SPE Paper No.59265

[9]Hanks, W.R., “The Axial Laminar Flow of Yield-Pseudoplastic Fluids in a Concentric

Annulus.” Ind. Eng. Chem. Process Des. Dev., Vol. 18, No. 3, 1979

[10]Hoffmann, K.A. “Computational Fluid Dynamics for Engineers.” Vol.1, 1993

[11]Hussain, Q.E. Sharif, M.A.R. “Analysis of Yield-Power-Law Fluid in Irregular

Eccentric Annuli.” Journal of Energy Resources Technology 120 (1998) 201-207

[12]Lamb, H., 1945, Hydrodynamics, 6th edn., Dover Publications, New York, pp. 585-

587.



62

[13] Luo, Y., Peden J. “Flow of Drilling Fluids Through Eccentric Annuli”, SPE 16692,

62nd Annual Technical Conference and Exhibition, Dallas, September 1987.

[14]Miska, S.Z. “Advanced Drilling”, Teaching Note, University of Tulsa, 2004.

[15]Ooms, G., Kampman-Reinhartz, B.E.,1996. “Influence of drill pipe rotation and

eccentricity on pressure drop over borehole during drilling”. Eur. J. Mech. B 15 (5), 695-

711.

[16]Patankar, S.V., 1980. “Numerical Heat Transfer and Fluid Flow Hemisphere”,

Bristol, PA, pp. 1-197.

[17]Piercy, N.A.V., Hooper, M.S., Winny, H.F. 1933. “Viscous flow through pipes with

core.” London Edinburgh Dublin Phil. Mag. J. Sci., 15 647-676.

[18]Pilehvari, A.: “Modeling of Laminar Helical Flow of A Power-Law Fluid Using the

Finite Element Method”, Technical Report, The University of Tulsa, 1989.

[19]Skelland, A. H. P. “Non-Newtonian Flow and Heat Transfer”, 1967, John Wiley &

Sons, Inc. New York.

[20]Wei, X., Miska S.Z., Takach N.E. “The Effect Of Drill Pipe Rotation On Annular

Frictional Pressure Loss”.



63

[21]Welty, J. R., Wicks, C. E. and Wilson, R. E., “Fundamentals of Momentum, Heat &

Mass Transfer”, New York, John Wiley & Sons, (1976).

[22]Yamada “Resistance of a flow through an Annulus with an Inner Rotating Cylinder”

Bulletin of JSME. May, 1960.



64

APPENDIX A

EQUATION OF MOTION IN CARTESIAN COORDINATES FOR LAMINAR

FLOW OF NON-NEWTONIAN FLUID IN ANNULUS

1.. Aeq

Flow

Direction





66

Because flow is only in one direction, 0== y x vv and also from continuity equation, 0=∂ z

v z

3.. Aeqs will take the form:

0=

∂∂

−=

∂∂−=

zz

z yz

z xz

y

v

xv

τ

µτ

µτ

5.. Aeq

Now substituting, 5.. Aeq into 3.. Aeq :

z

p

y

v

y x

v

x

z z

∂∂=

∂∂

∂∂+

∂∂

∂∂ µµ 6.. Aeq

Denoting: wv z = , 6.. Aeq will be:

z

p

y

W

y x

W

x ∂∂=

∂

∂∂∂+

∂

∂∂∂

µµ 7.. Aeq



67

APPENDIX B

DISCRETIZING EQUATION OF MOTION IN CARTESIAN COORDINATES

FOR LAMINAR FLOW OF NON-NEWTONIAN FLUID IN ANNULUS

Recalling equation of motion ( )AppendixA,A.7eq. :

0=∂∂−

∂∂∂∂+ ∂∂∂∂ z

P yw

y xw

xµµ 1.. Beq

The above finite differential equation can be app be approximated by a difference

equation. Beginning with the first term:

ji ji ji x

W

x

W

x

W

x,

2

1,

2

1, −+

∂∂

−

∂∂

≈

∂∂

∂∂

µµµ 2.. Beq

Also we can say:

2

,,1

,2

1

ji ji

ji

µµµ

+≈ +

+ 3.. Beq

2

,,1

,2

1

ji ji

ji

µµµ

+≈ −

− 4.. Beq

Now, using a second-order central differencing:

x

ww

x

W ji ji

ji ∆

−≈

∂∂ +

+

,,1

,

2

1 5.. Beq

x

W W

x

W ji ji

ji ∆

−≈

∂∂ −

−

,1,

,2

1 6.. Beq

Substituting 6.3.. B Beqs − into 2.. Beq yields:



68

( )( )( ) ( )( )[ ] ji ji ji ji ji ji ji ji

ji

W W W W x x

W

x,1,,,1,,1,,12

, 2

1−−++ −+−−+

∆≈

∂∂

∂∂

µµµµµ 7.. Beq

Similarly for y derivative in 2.. Beq :

( ) ( )( ) ( )( )[ ]1,,,1,,1,,1,2, 2

1 −−++ −+−−+∆

≈

∂∂

∂∂ ji ji ji ji ji ji ji ji

ji

W W W W y y

W y

µµµµµ 8.. Beq

Substituting equation 7.. Beq and 8.. Beq into 1.. Beq :

( )( )

( ) ( )

( )( )

( ) ( )

( )( )

( )( )

0

2

1

2

1,,

,1,2

,1,

,1,

2

,1,

,,12

,,1

,,1

=∂∂

−

∆

−+−

∆

−++

∆

−+−

∆

−+

−−

++

−−

++

z

P

y

W W

y

W W

x

W W

x

W W

ji ji

ji ji

ji ji

ji ji

ji ji

ji ji

ji ji

ji ji

µµµµ

µµµµ

9.. Beq

The above equation can be rearranges as:

( ) ( ) ( ) ( )

( ) ( )

( ) ( )0

22

22

2222

2

,1,

1,2

,1,

1,

2

,,1

,12

,,1

,1

2

,1,

2

,1,

2

,,1

2

,,1

,

=∂∂

−

∆

++

∆

++

∆

++

∆

++

∆

++

∆

++

∆

++

∆

+−

−−

++

−−

++

−+−+

z

P

yW

yW

xW

xW

y y x xW

ji ji

ji

ji ji

ji

ji ji

ji

ji ji

ji


ji

µµµµ

µµµµ

µµµµµµµµ

10.. Beq

Or:

( ) 01,11,1,11,114321, =∂∂

−+++++++− −+−+ z

P W AW AW AW A A A A AW ji ji ji ji ji

11.. Beq

Where:

( )2

,,1

12 x

A ji ji

∆

+= + µµ

( )2

,,1

2

2 x

A ji ji

∆

+= − µµ

( )2

,1,

32 y

A ji ji

∆

+= + µµ

( )2

,1,

42 y

A ji ji

∆

+= − µµ



69

Rearranging 11.. Beq :

∂∂

−++++++

= −+

+−+

++

z

P W AW AW AW A

A A A AW ji

n

ji

n

ji

n

ji

n

ji

n

1,

1

41,3,1

1

2,114321

,

1 1 12.. Beq

Where 1, +nn are the old and the new computational steps (iterations)

respectively.

To apply relaxation factor to 12.. Beq , first we add ji

n

ji

n

W W ,, − to the right hand side of that

equation:

∂∂

−++++++

+−= −+

+−+

++

z

P W AW AW AW A

A A A AW W W ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

1,

1

41,3,1

1

2,114321

,,,

1 1 13.. Beq

Moving ji

n

W ,− into the parenthesis in 13.. Beq yields:

( )

∂∂

−+++−

+++

++++= −

++−

++

+

z

P W A A A A


W W

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

,4321

1,

1

41,3,1

1

2,114321

,,

1 1

14.. Beq

Equation 14. B can be represented as:

Residual,,

1

+=

+

ji

n

ji

n

W W 15.. Beq

Where:

( )

∂∂

−+++−

+++

+++= −

++−

++

z

P W A A A A


ji

n

ji

n

ji

n

ji

n

ji

n

,4321

1,

1

41,3,1

1

2,11

4321

1Residual

16.. Beq

Residual in 15.. Beq will decrease as we approach the solution. And when the

convergence is achieved, value of residual will reach very close to zero. Based on this fact we can

increase/decrease rate of convergence by multiplying residual by an over/under relaxation factor

shown by ω . As a result, 14.. Beq will take the form:



70

( )

∂∂

−+++−

+++

++++= −

++−

++

+

z

P W A A A A


W W

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

ji

n

,4321

1,

1

41,3,1

1

2,114321

,,

1 ω

17.. Beq

Minimum Shear Rate Calculation

Recalling YPL Rheology model equation:

n y K γττ &+= 18.. Beq

In the plug region, shear rate is very small. There for the contribution of the second term

on the right hand side of 18.. Beq is negligible compared to yτ .Denoting the shear rate in the plug

region by minγ& , we get:

yn K τγ <<min& 19.. Beq

Or,

yn K τγε <<min& 20.. Beq

As a reasonable estimation, ε in 19.. Beq can be approximated by

8

10 . Therefore,

19.. Beq takes the form :

n y K min

810 γτ &= 21.. Beq

Or,

n y

K

1

8min10

=

τγ& 22.. Beq

Equation 22. B can not be used for fluids without yield stress, because the value will be

zero. So a new cut off value has to be defined. For fluids with zero yield stress, wall shear rate

can be used as a value to estimate the cut off value. As a reasonable approximation:

wall γγ && 8min 10

−= 23.. Beq





72

=−

StressYield with Fluids for

K

StressYield without Fluids for

n y

wall

1

8

8

min

10

10

τ

γγ

&& 28.. Beq

Where, wall γ& can be calculated from 24.. Beq



73

APPENDIX C

TRANSFORMATION OF EQUATION OF MOTION FROM CARTESIAN

COORDINATES TO A BOUNDARY FITTED COORDINATE SYSTEM

Considering A.7eq. ( )AppendixA :

z

P

y

W

y x

W

x

appapp

∂

∂=

∂

∂

∂

∂+

∂

∂

∂

∂µµ

1..C eq

Where:

22

1

∂∂

+

∂∂

=

+= −

y

w

x

w

K n y

app

γ

γγ

τµ

&

&&

There has to be a corresponding relationship between each point in y x , coordinate to

boundary fitted ( )ηξ , system therefore,

( )

( )

( )

( ) y x

y x

y y

x x

,

,

,

,

ηη

ξξ

ηξ

ηξ

=

=

=

=

2..C eqs

x

f

x

f

x

f

∂∂

∂∂

+∂∂

∂∂

=∂∂ η

η

ξ

ξ 3..C eq

y

f

y y

f

∂∂

∂∂

+∂∂

∂∂

=∂∂ η

η

ξ

ξ 4..C eq

Now 1..C eq can be written as:



74

z

P

x

F

x

F

∂∂

=∂

∂+

∂∂ 21

5..C eq

Where:

x

W

F app ∂∂

= µ1 6..C eq

y

W F app ∂

∂= µ2

7..C eq

Applying the same procedure as in 3..C eqs to 5..C eqs :

z

P

y

F

y

F

x

F

x

F

∂∂

=∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂ η

η

ξ

ξ

η

η

ξ

ξ2211 8..C eq

Defining J ′ as:

x y y x J

∂∂

∂∂−

∂∂

∂∂=′ ηξηξ

9..C eq

And multiplying 8..C eq by J ′1

, yields:

∂∂

′=

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

′ z

P

J y

F

y

F

x

F

x

F

J

11 2211 η

η

ξ

ξ

η

η

ξ

ξ 10..C eq

According to chain rule:

∂∂

′∂∂+

∂∂

′∂∂+

∂∂

′∂∂+

∂∂

′∂∂+

∂∂

∂∂

′+

∂∂

∂∂

′+

∂∂

∂∂

′+

∂∂

∂∂

′

=

∂∂

′+

∂∂

′∂∂

+

∂∂

′+

∂∂

′∂∂

y J F

x J F

y J F

x J F

F

y J

F

x J

F

y J

F

x J

y J

F

x J

F

y J

F

x J

F

η

η

η

η

ξ

ξ

ξ

ξ

η

η

η

η

ξ

ξ

ξ

ξ

ηξ

η

ξξ

ξ

1111

1111

2121

2121

2121

11..C eq

Equation 11..C eq can be rearranged as:





76

Equation 15.C can be written as:

ξ

ξ

η

ηξ

η

η

ξ

x J

y J

x J

y J x

y

x

y

′=′−=

′−=

′=

16..C eqs

Consider the second bracket of the right hand side of 12..C eq :

∂∂

′∂∂

+

∂∂

′∂∂

+

∂∂

′∂∂

+

∂∂

′∂∂

y J y J F

x J x J F

η

η

ξ

ξ

η

η

ξ

ξ

111121 17..C eq

Now plugging 16..C eqs into it 17..C eq yields:

( ) ( ) ( ) ( ) [ ] [ ] 02121 =+−+−=

∂∂+−

∂∂+

−

∂∂+

∂∂ ηξηξηξηξξηξη

ηξηξ x x F y y F x x F y y F 18..C eq

Therefore 12..C eq reduces to:

∂∂

′+

∂∂

′∂∂

+

∂∂

′+

∂∂

′∂∂

=

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

′

y J

F

x J

F

y J

F

x J

F

F

y

F

x

F

y

F

x J

ηη

η

ξξ

ξ

η

η

η

η

ξ

ξ

ξ

ξ

2121

21211

19..C eq

Plugging 19..C eq into 18..C eq , yields:

z

P

J y J

F

x J

F

y J

F

x J

F

∂∂

′=

∂∂

′+

∂∂

′∂∂

+

∂∂

′+

∂∂

′∂∂ 12121 ηη

η

ξξ

ξ 20..C eq

Now we need to write all derivatives in 20..C eq in terms of ηξ , only. So we need to

convert x∂

∂ξ,

y∂∂ξ

, x∂

∂η,

y∂∂η

into derivatives with respect to ηξ , .

Recalling form 2..C eqs :

( )ηξ , x x =

( )ηξ , y y =

( ) y x,ξξ =



77

( ) y x ,ηη =

According to chain rule:

=⇒

∂∂+

∂∂=

∂

∂+

∂

∂=

ηξη

ηξ

ξ

η

η

ξ

ξηξ

ηξd d y y x xdydx

d y

d y

dy

d x

d x

dx

=⇒

−

ηξηξ

ηξd d y y x xdydx

1

21..C eq

And similarly:

=

⇒

∂∂+

∂∂=

∂∂+

∂∂=

dydx

d d

dy y

dx x

d

dy y

dx x

d

y x

y x

ηη

ξξ

ηξ

ηηη

ξξξ

22..C eq

From 21..C eq and 22..C eq we deduce:

1−

=

ηξ

ηξ

ηη

ξξ

y y

x x

y x

y x 23..C eq

Extending 23..C eq results:

−−−

−−

−=

−

−−

=

ξηηξ

ξ

ξηηξ

ξ

ξηηξ

η

ξηηξ

η

ξξ

ηη

ξηηξηηξξ

y x y x

x

y x y x

y y x y x

x

y x y x

y

x y x y

y x y x y x

y x 1 24..C eq

And also 24..C eq in an extend form can be written as:

ξηηξ

ξ

ξηηξ

ξ

ξηηξ

η

ξηηξ

η

η

η

ξ

ξ

y x y x

x

y x y x

y

y x y x

x

y x y x

y

y

x

y

x

−=

−−=

−−=

−=

25..C eqs

Jacobean is defined as:

ξηηξ y x y x J −= 26..C eq





79

Comparing 27..C eqs and 16..C eqs we can say:

ξηηξ y x y x J

J −=′

=1

33..C eq

Utilizing 33..C eq in 32..C eq :

z

P J

J

xW xW x

J

yW yW y

J

xW xW x

J

yW yW y

appapp

appapp

∂∂

=

−+

−−

∂∂

+

−−

−

∂∂

ηξξη

ξ

ξηηξ

ξ

ηξξη

η

ξηηξ

η

µµη

µµξ

34..C eq

Rearranging 34..C eq :

( ) ( )

( ) ( ) z

P J x x y y yW

J x yW

J

x x y yW J

x yW J

app

appapp

appapp

∂∂

=

+−+

∂∂

+

+−+

∂∂

ηξξηξξξξη

ηξξηηηηξ

µµµ

η

µµ

ξ

22

22

35..C eq

Using following notations:

22

22

ξξ

ηξξη

ηη

β

γ

α

x y

x x y y

x y

+=

+=

+=

36..C eqs

Equation 35..C eq will take the form:

( ) ( ) z

P J W W

J W W

J

appapp

∂∂

=

−

∂∂

+

−

∂∂

ξηηξ γβµ

ηγα

µ

ξ 36..C eq

In 36..C eq we need to the transformation for appµ as well. Recalling appµ from 1..C eq

1−+= n y

app K γγ

τµ &

&

22

∂∂

+

∂∂

= y

w

x

wγ&

In order to transfer appµ , we need to transform γ& . According to 28..C eq and 29..C eq



80

J

yW yW

x

W ξηηξ −=

∂∂

37..C eq

J

xW xW

y

W ηξξη −=

∂∂

38..C eq

Substituting 37..C eq and 38..C eq into γ& , we get:

( )ξηηξηξξηξηηξξηηξ x xW W xW xW y yW W yW yW J y

W

x

W 22

1 22222222

2

22

−++−+=

∂

∂+

∂∂

39..C eq

Rearranging 39..C eq :

( ) ( ) ( )( )ξηξηηξξξηηηξ y y x xW W y xW y xW

J y

W

x

W +−+++=

∂∂

+

∂

∂2

1 222222

2

22

40..C eq

Utilizing notations form 36..C eqs into 40..C eq :

( )ηξηξ γβα W W W W J y

W

x

W 2

1 22

2

22

−+=

∂

∂+

∂∂ 41..C eq

Now plugging 41..C eq into 1..C eq :

1−+= n yapp K γ

γ

τµ &

&

Where:


21 22

2 −+=& 42..C eq

So from 36..C eq and 42..C eq :

( ) ( ) z

P J W W

J W W

J

appapp

∂∂=

−

∂∂+

−

∂∂

ξηηξ γβµ

ηγα

µ

ξ 43..C eq

Where:

1−+= n yapp K γ

γ

τµ &

&



81


21 22

2 −+=&

22ηηα x y +=

ηξξηγ x x y y +=

22ξξβ x y +=

ξηηξ y x y x J −=





83

Each term on the right hand side of 6.. Deq can be extended to:

+

≈

++ ji

ap p

ji

ap p

ji

ap p

J J J ,,1,

2

1 2

1 µαµαµα 7.. Deq

and

+

≈

−+− ji

ap p

ji

ap p

ji

ap p

J J J ,1,,

2

1 2

1 µαµαµα 8.. Deq

In 6.. Deq :

ji ji

ji

W W W

,,1

,2

1−≈

∂∂

++ξ

9.. Deq

And also

ji ji

ji

W W W

,1,

,2

1 −

+−

−≈

∂∂ξ

10.. Deq

Therefore:

( )

( )−

+

−−

+

≈

∂∂

∂∂

−−

++

ji ji

ji

ap p

ji

ap p

ji ji

ji

ap p

ji

ap p

ji

ap p

W W J J

W W J J

W

J

,1,

,1,

,,1

,,1,2

1

µαµα

µαµα

ξ

µα

ξ

11.. Deq

Similar to 11.. Deq , η derivative in 1.. Deq can be approximated as:

( )

( )

−

+

−

−

+

≈

∂∂

∂∂

−−

++

1,,

1,,

,1,

,1,,2

1

ji ji

ji

app

ji

app

ji ji

ji

ap p

ji

ap p

ji

ap p

W W J J

W W J J

W

J

µβµβ

µβµβ

η

µβ

η

12.. Deq

For the mixed derivation at node (i,j), we have:



84

( ) ( )

−

−−

≈

∂∂

∂∂

−−+−

−

−+++

+

1,11,1

,1

.

1,11,1

,1

.

,

.

4

1 ji ji

ji

ji ji

ji ji

W W J

W W J

W

J

γµγγµγ

η

γµγ

ξ 13.. Deq

( ) ( )

−−

−+−

≈

∂∂

∂∂

−−+−

−+++

1,1,1

1,

,11,1

1,,

114

1 ji ji

ji

app ji ji

ji

app

ji

ap pW W

J W W

J

W

J

µγµγ

ξ

µγ

η

14.. Deq

Plugging 14.11.. D Deqs − into 1.. Deq :

z

P J

W

J

W

J

W

J

W

J

appappappapp

∂∂=

∂∂

∂∂−

∂∂

∂∂+

∂∂

∂∂−

∂∂

∂∂

ξ

µγ

ηη

µβ

ηη

µγ

ξξ

µα

ξ

( )

( )

( ) ( ) +

−

−−

−

−

+

−−

+

−−+−−

−++++

−−

++

1,11,1

,1

1,11,1

,1

,1,

,1,

,,1

,,1

2

1

2

1

ji ji

ji

ap p

ji ji

ji

ap p

ji ji

ji

ap p

ji

ap p

ji ji

ji

ap p

ji

ap p

W W J

W W J

W W J J

W W J J

µγµγ

µαµα

µαµα

( )

( )

( ) ( ) z

P J W W

J W W

J

W W J J

W W

J J

ji ji ji

ji

ap p ji ji

ji

ap p

ji ji

ji

ap p

ji

ap p

ji ji

ji

ap p

ji

ap p

∂∂

=

−−

−+−

−

−

+

−−

+

−−+−

−+++

−−

++

,1,1,1

1,

,11,1

1,

1,,

1,,

,1,

,1,

114

1

2

1

µγµγ

µβµβ

µβµβ

15.. Deq

After some arrangements:

( )

−

−

−

−

−

−

−

−

−+

−+

1,,,1,

,1,,,1

,2

1

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p ji

J J J J

J J J J W

µβµβµβµβ

µαµαµαµα



85

( )

−

−

−

−+−+

1,,1

1,14

1

ji

app

ji

ap p ji

J J W

µγµγ

( )

−

− −

−−

1,,

1,2

1

ji

app

ji

app ji

J J W

µβµβ

( )

−

−

−−

− ji

app

ji

app

ji J J

W

,1,

,12

1 µαµα

( )

−

−

−

+−+−

1,,1

1,14

1

ji

app

ji

app

ji J J

W µγµγ

( )

−

−

−

++

ji

app

ji

app

ji J J

W

,1,

1,2

1 µβµβ

( )

−

−

−

++

ji

app

ji

app

ji J J

W

,,1

,12

1 µαµα

( )

+

−

−−−−

1,,1

1,14

1

ji

app

ji

app

ji J J

W µγµγ

( ) 04

1,

1,,1

1,1 =∂∂−

+

−

++++

z

P J

J J W ji

ji

app

ji

app

ji

µγµγ 16.. Deq

Rearrangement and multiplying two sides of 16.. Deq by (-1):

( )

+

+

+

+

+

−+

−+

1,1,

,1,1,,

,

2

1

2

1

2

1

2

1

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p

ji

ap p ji

J J

J J J J W

µβµβ

µαµαµβµα

( )

+

−

−+−+

1,,1

1,14

1

ji

ap p

ji

ap p ji

J J W

µγµγ



86

( )

+

−

−−

1,,

1,2

1

ji

app

ji

app

ji J J

W µβµβ

( )

+

−

−−

ji

app

ji

app

ji J J

W

,1,

,12

1 µαµα

( )

+

−

+−+−

1,,1

1,12

1

ji

app

ji

app

ji J J

W µγµγ

( )

+

−

++

ji

app

ji

app

ji J J

W

,1,

1,2

1 µβµβ

( )

+

−

++

ji

app

ji

app

ji J J

W

,,1

,12

1 µαµα

( )

+

−

−−−−

1,,1

1,14

1

ji

app

ji

app

ji J J

W µγµγ

( ) 04

1,

1,,1

1,1 =∂∂

+

+

−

++++

z

P J

J J W ji

ji

app

ji

app

ji

µγµγ 16.. Deq

Denoting101 A A −

:

0101,191,18,171,6

1,15,141,31,12,1

=−−−−−

−−−−

++−−++

+−−−−+

AW AW AW AW A

W AW AW AW AW A

ji ji ji ji

ji ji ji ji ji

16.. Deq

Where:

1,1,

,1,1,,

1

2

1

2

1

2

1

2

1

−+

−+

+

+

+

+

+

=

ji

app

ji

app

ji

app

ji

app

ji

app

ji

app

J J

J J J J

A

µβµβ

µαµαµβµα

+

=

−+ 1,,1

24

1

ji

app

ji

ap p

J J A

µγµγ



87

+

=

−1,,

32

1

ji

app

ji

app

J J A

µβµβ

+

=

− ji

app

ji

app

J J A

,1,

42

1 µαµα

+

=

+− 1,,1

54

1

ji

app

ji

app

J J A

µγµγ

+

=

+ ji

app

ji

app

J J A

,1,

62

1 µβµβ

+

=

+ ji

app

ji

app

J J A

,,1

72

1 µαµα

+

=

−− 1,,1

84

1

ji

app

ji

app

J J A

µγµγ

+

=

++ 1,,1

94

1

ji

app

ji

app

J J A

µγµγ

z

P J A ji ∂

∂−= ,10

Applying Relaxation Factor:

Adding ji ji W W ,, − to the right hand side of 16.. Deq :

{

}101,191,18,171,6

1,15,141,31,12

1

,,,

1

AW AW AW AW A

W AW AW AW A A

W W W

ji ji ji ji

ji ji ji ji ji ji ji

+++++

++++−=

++−−++

+−−−−+

17.. Deq

Now jiW ,− taking into the braces:



88

{

}10,11,191,18,171,6

1,15,141,31,121

,,

1

AW AW AW AW AW A

W AW AW AW A A

W W

ji ji ji ji ji

ji ji ji ji ji ji

+−++++

++++=

++−−++

+−−−−+

18.. Deq

Now using relaxation factor we will have:

{

}10,11,191,18,171,6

1,15,141,31,12

1

,,

AW AW AW AW AW A

W AW AW AW A A

W W

ji ji ji ji ji

ji ji ji ji ji ji

+−++++

++++=

++−−++

+−−−−+ω

19.. Deq

Calculating Relaxation Factor

Following Azouz [3]:

J

appµα

ϕ =1 20.. Deq

J

appµβϕ =2 21.. Deq

( )

( )

−

−

−

=

−

=

−+

−+

1,1,

,1,1

3

5.0

5.0

ji

app

ji

app

ji

app

ji

appappapp

J J

J J J J

µγµγ

µαµαµγµαϕ

ηξ

22.. Deq

( )

( )

−

−

−

=

−

=

−+

−+

ji

app

ji

app

ji

app

ji

appappapp

J J

J J J J

,1,1

1,1,

4

5.0

5.0

µγµγ

µβµβµγµβϕ

ξη

23.. Deq

+−+

+−

+=

1cos4

1cos4

2

1 24

22

23

21

21 N M J

πϕϕ

πϕϕ

ϕϕρ 24.. Deq

If ( ) 042

32

1 ≥−ϕϕ , then:



89

211

2

J ρ

ω

−+= 25.. Deq

Otherwise,

211

2

J ρ

ω

++= 26.. Deq

Calculating Relaxation Factor in Cartesian Coordinate System

Following Azouz (1994):

ξηηξ

ξξ

ηξξη

ηη

β

γ

α

y x y x J

x y

x x y y

x y

−=

+=

+=

+=

22

22

27.. Deq

In this system, since there is not any transformation it can be said:

ξ

η

==

x

y 28.. Deq

As a result:

1

0

====

ξη

ξη

x y

y x 29.. Deq

So,

101

110

000

101

22

22

=−=−=

=+=+=

=+=+=

=+=+=

ξηηξ

ξξ

ηξξη

ηη

β

γ

α

y x y x J

x y

x x y y

x y

30.. Deq

( )

( ) app

appapp

J µ

µµαϕ ===

1

11 31.. Deq



90

( )

( ) app

appapp

J µ

µµβϕ ===

1

12 32.. Deq

( )

( )

( )

( )

( ) x

J J

app

appappappapp

∂∂

==

−

=

−

=

µµ

µµµγµαϕ

ξ

ηξηξ1

0

1

13

33.. Deq

( )

( )

( )

( )

( ) y

J J

app

appappappapp

∂∂==

−

=

−

=

µµ

µµµγµβϕ

η

ξηξη1

0

1

14

34.. Deq

+−+

+−

+=

1cos4

1cos4

2

1 24

22

23

21

21 N M J

πϕϕ

πϕϕ

ϕϕρ 35.. Deq

If ( ) 04 23

21 ≥−ϕϕ , then:

211

2

J ρ

ω

−+= 36.. Deq

Otherwise,

211

2

J ρ

ω

++= 37.. Deq



91

APPENDIX E

ALGEBRAIC GEOMETRY TRANSFORMATION

Algebraic Transformation of an Eccentric Annulus

Considering Fig3.8, Each point can be defined by a radius and an angle.

Fig.E.1 Determination of an arbitrary grid point

According to Fig.E.1 :

( ) ( )[ ]112

1l R ol Ro Roj −+−= 1.. E eq

From Fig.E.1 :

( )1

1 −

−+−=

JM

Rie Ro j JM l 2.. E eq

l

p

d

Ri ee

Ro Roj

C m

C j

C 1

rθ

Ro

Ri

Roj





93

Plugging 8.. E eqs into 7.. E eq :

( ) ( ) 222cossin oj j j j Rer r =−+ θθ 9.. E eq

Or,

θθ CoseSine Rr j joj j +−= 222 10.. E eq

Since the circle is decided into 1− IM divisions in tangential direction , where each

division will be represented by θ∆ , and half of the circle is used:

1−=∆

IM

πθ 11.. E eq

So θ can be calculated as:

( )1−∆= iθθ 12.. E eq

Substituting θ∆ from 11.. E eq

( )1

1

−−

= IM

iπθ 13.. E eq

Substituting 5.. E eq and 6.. E eq ,into 10.. E eq :

( )( )( )

( )( )

( ) ( )( )

( )

−−

−−

+

−−

−

−−

−

−−−=

1

1

1

1

1

1

1

1

1

222

IM

iCos

JM

je

IM

iSin

JM

je

JM

Ri Ro j JM Ror j

ππ

14.. E eq

Substituting 13.. E eq and 14.. E eq , into 8.. E eqs :

( ) ( )( )

( )( )

( )( ) ( )

( )( ) ( )

( ) ( )( )

( )( )

( )( ) ( )

( )( ) ( )

1

1

1

1

1

1

1

1

1

1

1,

1

1

1

1

1

1

1

1

1

1

1,

222

222

−−

−−

−−+

−−

−

−−

−

−−−=

−−

−−

−−+

−−

−

−−

−

−−−=

IM

iCos

IM

iCos

JM

je

IM

iSin

JM

je

JM

R i Ro j JM Ro ji y

IM

iSin

IM

iCos

JM

je

IM

iSin

JM

je

JM

R i Ro j JM Ro ji x

πππ

πππ

15.. E eq

Where:



94

JM j IM i <<<< 1,1

ji, : Grid points in tangential and radial direction respectively

1− IM : Number of divisions in tangential direction.

1− JM : Number of divisions in radial direction.

It should be mentioned that 15.. E eq is valid only for the half of the domain, and by

replacing π by π2 it can be used to transform a whole eccentric annulus from real

domain to a computational domain.





96

AQBDPC A A A A =+++ 4321 3.. F eq

From 2.. F eq , 3.. F eq :

AQBDPC

W

Q

ji

ji 4

,

, = 4.. F eq

Now, we need to find the area of AQBDPC .

Fig.F.2

A

B

C

D

P

Q

+1, j

i,j

+1, j+1

-1, j

+1, j-1

-1, , j-1

-1,j+1

, , j+1

, , j-1

A1

A2

A3

A3



97

Fig.F.3

According to FigF.3:

5611 A ACDO ABO AQBDPC +−−= ∆∆

5.. F eq

And also from the same figure:

∆−= CDOCPDO A 226 6.. F eq

∆−= ABO AQBO A 335

7.. F eq

Plugging, 6.. F eq , 7.. F eq into 5.. F eq :

∆∆∆∆−++−−= ABO AQBOCDOCPDOCDO ABO AQBDPC 332211 8.. F eq

From the mathematics books:

D

B

A

C

O1

A5

A6

O2

O3



98

Fig.F.4

( )( ) ( )( )323132313232

3131

2

1det

2

1 x x y y y y x x

y y x x y y x x

ABC −−−−−=

−−−−=

∆ 9.. F eq

Applying 9.. F eq to 8.. F eq :

( )( ) ( ) ( ) B A BO B A BO x x y y y y x x ABO −−−−−=∆

1112

1 10.. F eq

Substituting the coordinate of each point in 10.. F eq :

( ) ( ) ( ) ( )1,11,11,11,11,11,11 002

1+++−+++++−++

∆−−−−−= ji ji ji ji ji ji x x y y y x ABO 11.. F eq

Simplifying 11.. F eq :

( ) ( ) ( ) ( )1,11,11,11,11,11,112

1+++−+++++−++

∆−+−−= ji ji ji ji ji ji x x y y y x ABO 12.. F eq

Similarly for calculating the area of ∆CDO2 :

( )( ) ( )( ) DC DO DC DO x x y y y y x xCDO −−−−−=∆

2222

1 13.. F eq

( ) ( ) ( ) ( )1,11,11,111,11,11,12 021 −+−−−+−−+−−−+∆ −−−−−= ji ji ji j ji ji ji x x ye y y xCDO 14.. F eq

( ) ( ) ( ) ( )1,11,11,111,11,11,122

1−+−−−+−−+−−−+

∆−−−−−= ji ji ji j ji ji ji x x ye y y xCDO 15.. F eq

Substituting for 1− je from 10.. E eq , Appendix E:

A

B

C



99

( )( ) ( )

( )1

11,1 −

−+−−−+−= −−

JM

Rie Ro j JM R o Ree jo j 16.. F eq

Or,

( ) ( )( )11

1,1 −−++−−+−= −−

JM R ie Ro j JM R o Ree jo j 17.. F eq

Substituting for 1, − jo R from 10.. E eq , Appendix E:

( )( )

( )1

11, −

−+−−=−

JM

Ri R o j JM R o R jo 18.. F eq

Substituting 18.. F eq into 17.. F eq :

( ) ( )( )

( ) ( )( )11

11

1 −−++−−+

−−+−+−=−

JM Rie Ro j JM Ro

JM Ri Ro j JM Roee j 19.. F eq

Or,

( ) ( )( )

( ) ( )( )1

1

1

11 −

−++−−

−−+−

+=− JM

Rie Ro j JM

JM

Ri Ro j JM ee j 20.. F eq

Or,

( ) ( )( )1

11 −

+−−−+−+=− JM

Ri

e Ro

Ri

Ro

j JM

ee j 21.. F eq

Simplifying 21.. F eq

( )( )1

11 −

+−−=−

JM

j JM eee j 22.. F eq

Or,

( )

( )1

21

−

−=−

JM

jee j 23.. F eq

Substituting 23.. F eq into 15.. F eq :

( ) ( ) ( ) ( )1,11,11,111,11,11,122

1−+−−−+−−+−−−+

∆−−−−−= ji ji ji j ji ji ji x x ye y y xCDO

Or,



100

( ) ( ) ( )( )

( )1,11,11,11,11,11,121

2

2

1−+−−−+−+−−−+

∆−

−

−−

−−−= ji ji ji ji ji ji x x y JM

je y y xCDO 24.. F eq

After simplification:

( ) ( ) ( )( )

( )1,11,11,11,11,11,121

2

2

1−+−−−+−+−−−+

∆−

−

−−

−−−== ji ji ji ji ji ji x x y JM

je y y xCDO 25.. F eq

Calculating CPDO2 :

We know that:

( )2

2

1,

2

−= jo RCPDO

α 26.. F eq

From Sinus law for calculating the area of a triangle:

( ) αsin2

1 21,2 −

∆= jo RCDO 27.. F eq

As a result we can get:

( )

=

−

∆

21,

22sin

jo R

CDO Arcα 28.. F eq

Recalling for 1, − jo R from 10.. E eq , Appendix E:

( )( )

( )1

11, −

−+−−=−

JM

Ri R o j JM R o R jo 29.. F eq

Substituting 29.. F eq and 28.. F eq into 26.. F eq :

( )( )

( ) ( )( )( )

−−+−−

−

−+−−=

∆

2

2

2

2

11

2sin

1

1

2

1

JM Ri Ro j JM Ro

CDO Arc

JM

Ri Ro j JM RoCPDO 30.. F eq

Similarly, we can calculate∆CDO1 :

( )( ) ( )( ) DC DO DC DO x x y y y y x xCDO −−−−−=∆

1112

1 31.. F eq





102

( ) ( )( )

( )1,11,11,11,11,11,1312

1+++−+++++−++

∆−

−


JM

je y y x ABO 39.. F eq

To calculate AQBO3 , similar procedure for calculating CPDO2 can be used as:

We know that:

( )2

2

1,

3

+= jo R AQBO

β 40.. F eq

From Sinus law for calculating the area of a triangle:

( )( )

=⇒=

+

∆

+

∆

2

1,

32

1,3

2sinsin

2

1

jo

jo R

ABO Arc R ABO ββ 41.. F eq

Or,

( )2

2

1,

3

+= jo R AQBO

β 42.. F eq

As a result:

( )

=

+

∆

2

1,

32sin

jo R

ABO Arcβ 43.. F eq

Recalling for 1, + jo R from 10.. E eq , Appendix E:

( )( )

( )1

11, −

−−−−=+

JM

Ri Ro j JM Ro R jo 44.. F eq

Substituting 44.. F eq and 43.. F eq into 42.. F eq :

( )( )( ) ( )( ) ( )

( )

−

−+−−

−

−−−−=

∆

2

3

2

3

1

1

2sin

1

1

2

1

JM

R i Ro j JM Ro

ABO Arc

JM

Ri R o j JM R o AQBO 45.. F eq

And finally from 4.. F eq , 3.. F eq , 12.. F eq , 25.. F eq , 30.. F eq , 32.. F eq , 39.. F eq , 45.. F eq



AQBDPC W

Q ji

ji2

,, = 46.. F eq

Where:

∆∆∆∆

−++−−= ABO AQBOCDOCPDOCDO ABO AQBDPC 332211

( ) ( ) ( ) ( )1,11,11,11,11,11,112

1+++−+++++−++

∆−+−−= ji ji ji ji ji ji x x y y y x ABO

( ) ( ) ( ) ( )1,11,11,11,11,11,112

1−+−−−+−+−−−+

∆−+−−= ji ji ji ji ji ji x x y y y xCDO

( )( )

( ) ( )( )( )

−

−+−−

−

−+−−=

∆

2

2

2

2

11

2sin

1

1

2

1

JM R i Ro j JM R o

CDO Arc

JM

R i Ro j JM R oCPDO

( ) ( ) ( )( )

( )1,11,11,11,11,11,121

2

2

1−+−−−+−+−−−+

∆−

−

−−

−−−== ji ji ji ji ji ji x x y JM

je y y xCDO

( )( )( ) ( )( ) ( )

( )

−

−+−−

−

−−−−=

∆

2

3

2

3

1

1

2sin

1

1

2

1

JM

R i Ro j JM Ro

ABO Arc

JM

Ri R o j JM R o AQBO 7.16.3.eq

( ) ( )( )

( )1,11,11,11,11,11,1312

1+++−+++++−++

∆−

−


JM

je y y x ABO 8.16.3.eq

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