fourier finite element (ffe) framework with hp an electromagnetics … · 2011. 2. 10. · vi eiec,...

VI EIEC, Chiclana, Spain

An hp Fourier Finite Element (FFE) Framework withElectromagnetics and Multiphysics Applications

D. Pardo, C. Torres-Verdın, L.E. Garcıa-Castillo,M. Paszynski, M.J. Nam

October 23, 2008

Basque Center for Applied Mathematics (BCAM)Promoting Technological Advances Through Mathematics

OVERVIEW

1. Motivation: Waveguide Design and Oil-IndustryApplications.

2. Method:

• Fourier finite element (FFE) method.

• Parallel implementation.

• Multi-physics framework.

3. Numerical Simulations:

• 2D resistivity logging measurements.

• Marine controlled source electromagnetic (CSEM) measurements.

• 3D resistivity logging measurements.

4. Conclusions and Future Work.

Basque Center for Applied Mathematics (BCAM)

D. Pardo, C. Torres-Verdın, L.E. Garcıa-Castillo, M. Pasz ynski, M. J. Nam 23 Oct 2008

MOTIVATION (WAVEGUIDES)

Waveguide Design

Goal: Determine electric field intensity at the ports.

Basque Center for Applied Mathematics (BCAM)2

For more info, visit: www.ices.utexas.edu/ ∼pardo


MOTIVATION (OIL-INDUSTRY)

Seismic Measurements

Figure from the USGS Science Center for Coastal and Marine Geol ogy





Marine Controlled-Source Electromagnetics (CSEM)

Figure from the UCSD Institute of Oceanography





Multiphysics Logging Measurements

OBJECTIVES: To determine payzones ( porosity ), amount ofoil/gas ( saturation ), and ability to extract oil/gas

(permeability ).





Main Objective: To Solve a Multiphysics Inverse Problem

Given multi-frequency electromagnetic, acoustic, and nuclearmeasurements, the objective is to determine porosity, saturation, and

permeability distributions in the reservoir.




FOURIER FINITE ELEMENT METHOD

Deviated Wells (Forward Problem)Dip Angle

Invasion

Anisotropy

Triaxial Induction

Eccentricity

Laterolog

Through-Casing

Induction-LWD

Induction-Wireline

Inverse Problems

Multi-Physics

Objective: Find solution at the receiver antennas.





Non-Orthogonal System of Coordinates

ZZ~

ζ1

6

ζ3

ζ2

Material coefficients are constantwith respect to the quasi-azimuthaldirection ζ2

Fourier Series Expansion in ζ2

DC Problems: −∇σ∇u = f

u(ζ1, ζ2, ζ3) =l=∞∑

l=−∞

ul(ζ1, ζ3)ejlζ2

σ(ζ1, ζ2, ζ3) =m=∞∑

m=−∞

σm(ζ1, ζ3)ejmζ2

f(ζ1, ζ2, ζ3) =n=∞∑

n=−∞

fn(ζ1, ζ3)ejnζ2

Fourier modes ejlζ2 are orthogonalhigh-order basis functions that are(almost) invariant with respect tothe gradient operator.





Cartesian system of coordinates: x = (x1, x2, x3).New non-orthogonal system of coordinates: ζ = (ζ1, ζ2, ζ3).

SUB

DO

MA

IN I

I

SUB

DO

MA

IN I

ISU

BD

OM

AIN

I

0

SUBDOMAIN III

SUBDOMAIN III

PARAMETERIZATIONZ

ZZ~ ζ1

6

ζ3

ζ2

ZZZ~ ζ1

6

ζ3

Subdomain I ; Subdomain II ; Subdomain III

x1 = ζ1 cos ζ2

x2 = ζ1 sin ζ2

x3 = ζ3

;

x1 = ζ1 cos ζ2

x2 = ζ1 sin ζ2

x3 = ζ3 + tan θ0

ζ1 − ρ1

ρ2 − ρ1

ρ2

;

x1 = ζ1 cos ζ2

x2 = ζ1 sin ζ2

x3 = ζ3 + tan θ0ζ1





Final Variational Formulation

We define the Jacobian matrix J =∂(x1, x2, x3)

∂(ζ1, ζ2, ζ3)and its

determinant |J | = det(J ).

Variational formulation in the new system of coordinates:

Find u ∈ uD + H1D(Ω) such that:

⟨∂v

∂ζ, σ

∂u

∂ζ

⟩

L2(Ω)

=⟨

v , f⟩

L2(Ω)∀v ∈ H1

D(Ω) ,

where:σ := J −1σJ −1T

|J | ; f := f |J | .

Same variational formulation with new materials and load data





Five Fourier modes are enough to represent EXACTLY the newmaterial coefficients.

Direct Current:

Find u ∈ uD + H1D(Ω) such that:

n=k+2∑

n=k−2

⟨(∂v

∂ζ

)

k

, σk−n

(∂u

∂ζ

)

n

⟩

L2(Ω2D)

=⟨

vk , fk

⟩

L2(Ω2D)∀vk

Alternate Current:

Find (E)s ∈ HΓE(curl; Ω) such that:

n=s+2∑

n=s−2

⟨(∇ζ×F)s , (µ−1)s−n (∇ζ×E)l

⟩

L2(Ω2D)

−⟨

Fs , (k2)s−nEl

⟩

L2(Ω2D)= −jω

⟨

Fs , (Jimp)s

⟩

L2(Ω2D)∀ Fs





Example (7 Fourier Modes)

n=k+2∑

n=k−2

⟨(∂v

∂ζ

)

k

, σk−n

(∂u

∂ζ

)

n

⟩

L2(Ω2D)︸︷︷︸

(k, k − n, n)

=⟨

vk , fk

⟩

L2(Ω2D)

Stiffness Matrix:

(−3,0,−3) (−3,−1,−2) (−3,−2,−1) 0 0 0 0

(−2,1,−3) (−2,0,−2) (−2,−1,−1) (−2,−2,0) 0 0 0

(−1,2,−3) (−1,1,−2) (−1,0,−1) (−1,−1,0) (−1,−2,1) 0 0

0 (0,2,−2) (0,1,−1) (0,0,0) (0,−1,1) (0,−2,2) 0

0 0 (1,2,−1) (1,1,0) (1,0,1) (1,−1,2) (1,−2,3)

0 0 0 (2,2,0) (2,1,1) (2,0,2) (2,−1,3)

0 0 0 0 (3,2,1) (3,1,2) (3,0,3)





A Self-Adaptive Goal-Oriented hp-FEMOptimal 2D Grid

(Through Casing Resistivity Problem)

p=1

p=2

p=3

p=4

p=5

p=6

p=7

p=8

ddd

We vary locally the element sizeh and the polynomial order ofapproximation p throughout thegrid.

Optimal grids are automaticallygenerated by the computer.

The self-adaptive goal-orientedhp-FEM provides exponentialconvergence rates in terms ofthe CPU time vs. the error ina user prescribed quantity ofinterest.





Axisymmetric Logging-While-Drilling (LWD) SimulationGOAL-ORIENTED HP-ADAPTIVITY (Quadrilateral Elements)

p=1

p=2

p=3

p=4

p=5

p=6

p=7

p=8

-0.7222222 2.870370 -1.137037

1.085185 2Dhp90: A Fully automatic hp-adaptive Finite Element code





Axisymmetric Logging-While-Drilling (LWD) SimulationGOAL-ORIENTED HP-ADAPTIVITY (ZOOM TOWARDS FIRST RECEIVER

ANTENNA)

p=1

p=2

p=3

p=4

p=5

p=6

p=7

p=8

1.7160494E-02 0.1369136 0.3603704

0.4344445 2Dhp90: A Fully automatic hp-adaptive Finite Element code




PARALLEL IMPLEMENTATION

We Use Shared Domain Decomposition




MULTIPHYSICS FRAMEWORK

We are generating new data structures based on:

The addition of one new module/library for solving inverseproblems.

The use of different number of equations for each element/node.

• Enables to consider different physics for each element/node.

• Enables the use of different number of Fourier modes for eachelement/node.

The combination of different types of elements used for eachphysics:

• Continous H1-elements (DC problems, elasticity, etc.)

• Nedelec (edge) H(curl)-elements (Electromagnetics).

• Raviart-Thomas (face) H(div)-elements (Fluid-dynamics)

• Discontinuous L2-elements.




MULTIPHYSICS FRAMEWORK

Final hp-grid and solution




RESULTS: 2D MARINE CSEM

Model Problem I: UNIFORM FORMATION

1 Ohm−m

z=0 mz=50 mTX(HED)





Model Problem I: UNIFORM FORMATION — 0.25 Hz —

0 2000 4000 6000 8000 1000010

−17

10−16

10−15

10−14

10−13

10−12

10−11

10−10

10−9

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )

Horizontal Distance between TX and RX (m)

Uniform Material, 0.25 Hz

1 MODE5 MODES9 MODESEXACT

0 2000 4000 6000 8000 10000−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)









0 2000 4000 6000 8000 1000010

−20

10−18

10−16

10−14

10−12

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )




0 2000 4000 6000 8000 10000−1000

−900

−800

−700

−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)









0 2000 4000 6000 8000 1000010

−22

10−20

10−18

10−16

10−14

10−12

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )




0 2000 4000 6000 8000 10000−1400

−1200

−1000

−800

−600

−400

−200

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)








Model Problem II: CSEM SCENARIO WITHOUT OIL

1 Ohm−m

0.3 Ohm−m

z=0 m

z=1000 m

z=50 m

10 Ohm−m

TX(HED)

6





Model Problem II: WITHOUT OIL — 0.25 Hz —

0 2000 4000 6000 8000 1000010

−17

10−16

10−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


Without Oil, 0.25 Hz


0 2000 4000 6000 8000 10000−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)









0 2000 4000 6000 8000 1000010

−20

10−18

10−16

10−14

10−12

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )




0 2000 4000 6000 8000 10000−800

−700

−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)









0 2000 4000 6000 8000 1000010

−20

10−18

10−16

10−14

10−12

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )




0 2000 4000 6000 8000 10000−900

−800

−700

−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)








Model Problem III: CSEM SCENARIO WITH OIL

1 Ohm−m

1 Ohm−m

0.3 Ohm−m

100 Ohm−m

z=0 m

z=1000 m

z=−1000 m

z=−1100 m

z=50 m

10 Ohm−m6

TX(HED)





Model Problem III: WITH OIL — 0.25 Hz —

0 2000 4000 6000 8000 1000010

−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


With Oil, 0.25 Hz


0 2000 4000 6000 8000 10000−180

−160

−140

−120

−100

−80

−60

−40

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


With Oil, 0.25 Hz







0 2000 4000 6000 8000 1000010

−16

10−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


With Oil, 0.75 Hz


0 2000 4000 6000 8000 10000−350

−300

−250

−200

−150

−100

−50

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


With Oil, 0.75 Hz







0 2000 4000 6000 8000 1000010

−17

10−16

10−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


With Oil, 1.25 Hz


0 2000 4000 6000 8000 10000−450

−400

−350

−300

−250

−200

−150

−100

−50

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


With Oil, 1.25 Hz






NO OIL

1 Ohm−m

0.3 Ohm−m

z=0 m

z=1000 m

z=50 m

10 Ohm−m

TX(HED)

6





INFINITE LAYER OF OIL

1 Ohm−m

1 Ohm−m

0.3 Ohm−m

100 Ohm−m

z=0 m

z=1000 m

z=−1000 m

z=−1100 m

z=50 m

10 Ohm−m6

TX(HED)





FINITE LAYER OF OIL

1 Ohm−m

0.3 Ohm−m

z=0 m

z=1000 m

z=−1000 m

z=−1100 m

z=50 m

100 Ohm−m

x=2000 m x=5000 m

1 Ohm−m 1 Ohm−m

10 Ohm−m6

TX(HED)

1 Ohm−m





FINITE LAYER OF OIL + ANISOTROPY

0.3 Ohm−m

z=0 m

z=1000 m

z=−1000 m

z=−1100 m

z=50 m

100 Ohm−m

x=2000 m x=5000 m

10 Ohm−m6

TX(HED)

1 Ohm−m

3 Ohm−m

1 Ohm−m

3 Ohm−m

3 Ohm−m 3 Ohm−m

1 Ohm−m 1 Ohm−m





Comparison — 0.25 Hz —

0 2000 4000 6000 8000 1000010

−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


0.25 Hz

NO OILINFINITE LAYER OILFINITE LAYER OILANISOTROPY

0 2000 4000 6000 8000 10000−250

−200

−150

−100

−50

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


0.25 Hz


The finite layer of oil is clearly identified, and it is different fr om thesolution for the infinite layer of oil. To consider anisotropy is e ssential.






0 2000 4000 6000 8000 1000010

−16

10−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


0.75 Hz


0 2000 4000 6000 8000 10000−450

−400

−350

−300

−250

−200

−150

−100

−50

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


0.75 Hz


As we increase the frequency, the effect of oil becomes morelocalized.






0 2000 4000 6000 8000 1000010

−18

10−17

10−16

10−15

10−14

10−13

10−12

10−11

10−10

Am

plitu

de o

f Ele

ctric

Fie

ld (

V/(

A m

2 )


1.25 Hz


0 2000 4000 6000 8000 10000−600

−500

−400

−300

−200

−100

0

Pha

se o

f Ele

ctric

Fie

ld (

degr

ees)


1.25 Hz


As we increase the frequency, the effect of oil becomes morelocalized.




RESULTS: 2D MARINE CSEM0.75 Hz (FINITE LAYER OF OIL)

TX: x = 0 m ; RX: x = 2000 m.





TX: x = 0 m ; RX: x = 5000 m.





TX: x = 0 m ; RX: x = 8000 m.




RESULTS: 3D RESISTIVITY LOGGING

Model Problem

0.05 m

mΩ .

0.2

5 m

1.0

m

150 kHz (Wireline)

Fiber Glass Mandrel

Mandrel Radius: 0.04 m

105

Finite SizeLoop Antenna

mΩ .

mΩ .

mΩ .

mΩ .

mΩ .

mΩ .

mΩ .

mΩ .20

mΩ .30mΩ .50

Oil−Based M

ud (1000 m)

1.5 m

Ω

.

0.5 m0.65 m

3 m

SHALE

SAND

SANDSAND

SAND

OIL

SHALE

3

200.05

0.1

100

1

1

4 m

Anisotropy (10:1)

Anisotropy (10:1)

Anisotropy (10:1)

Anisotropy (8:1)





Verification

Logging Instrument in a Homogeneous Formation

1 3 5 7 9

10−2

100

102

104

Number of Fourier Modes

Re

lativ

e E

rro

r (in

%)

20 KHz (30 degrees)20 KHz (60 degrees)150 KHz (30 degrees)150 KHz (60 degrees)





Verification


10−2

100

102

−6

−4

−2

0

2

4

6

Resistivity (Ω−m)−2 −1 0 1

x 10−3

Wireline, 150 Khz

Real Part (V/m)

−0.02 −0.01 0Imag Part (V/m)

1 1





Verification


10−2

100

102

−6

−4

−2

0

2

4

6


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0

x 10−3Imag Part (V/m)

13

13





Verification


10−2

100

102

−6

−4

−2

0

2

4

6


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


135

135





Verification


10−2

100

102

−6

−4

−2

0

2

4

6


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


1359

1359





Verification


10−2

100

102

−6

−4

−2

0

2

4

6


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


135913

135913





Dip Angle

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


0 degrees30 degrees60 degrees






Dip Angle + Invasion

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0




INVASION

NO INVASION





Dip Angle + Anisotropy

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0




VERT. ρ

HORIZ. ρ





Dip Angle + Invasion + Anisotropy

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0




INVASION

NO INVASION

VERT. ρ

HORIZ. ρ





60-Degree Deviated Well

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


NO INV.NO INV + ANI.INVINV + ANI


INVASION

NO INVASION

VERT. ρ

HORIZ. ρ





Vertical Well with 0.03 m Eccentricity

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


x 10−3

Wireline, 150 Khz

Real Part (V/m)

−3 −2 −1 0


NO INV.NO INV + ECC.INVINV + ECC

NO INV.NO INV + ECC.INVINV + ECC

INVASION

NO INVASION





Model Problem and Verification

Ω . mµr

µr

Ω . m

0.25

m

0.15 m

0.05 m0.1 m

0.7

m

Magnetic Buffer:

=10000=10000ρ

Metallic Mandrel:ρ =10

−5

=50

Loop AntennaFinite Size

1 3 5 7 910

−2

10−1

100

101

102

LWD PROBLEM (30 degrees)

Number of Fourier Modes

Re

lative

Err

or

(in

%)





Dip Angle

10−2

100

102

−6

−4

−2

0

2

4

6

8

10

Resistivity (Ω−m)−0.4 −0.2 0 0.2

LWD, 2 Mhz

Real Part (V/m)

−0.4 −0.2 0 0.2Imag Part (V/m)







Dip Angle + Invasion

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


LWD, 2 Mhz

Real Part (V/m)

−0.4 −0.2 0 0.2Imag Part (V/m)



INVASION

NO INVASION





Dip Angle + Anisotropy

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


LWD, 2 Mhz

Real Part (V/m)

−0.4 −0.2 0 0.2Imag Part (V/m)



VERT. ρ

HORIZ. ρ





Dip Angle + Invasion + Anisotropy

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


LWD, 2 Mhz

Real Part (V/m)

−0.4 −0.2 0 0.2Imag Part (V/m)



INVASION

NO INVASION

VERT. ρ

HORIZ. ρ





60-Degree Deviated Well

10−2

100

102

−6

−4

−2

0

2

4

6

8

10


LWD, 2 Mhz

Real Part (V/m)

−0.4 −0.2 0 0.2Imag Part (V/m)



INVASION

NO INVASION

VERT. ρ

HORIZ. ρ




CONCLUSIONS AND FUTURE WORK

• A Fourier-Finite-Element method provides a suitable formulation forsimulation of resistivity geophysical applications.

• Goal-oriented refinements are essential in marine CSEM geophysi calapplications due to the dissipative nature of the earth.

• A parallel implementation based on a shared domain-decompositi on issimple and provides additional performance for a moderate num ber ofprocessors.

• We are developing a multiphysics framework for the joint-inversi on ofmultiphysics measurements.

• We are looking for Ph.D. students and postdoctoral fellows to furthe rdevelop this software and work on the joint-inversion of multiphysi csmeasurements.

Basque Center for Applied Mathematics (BCAM)



fourier finite element (ffe) framework with hp an electromagnetics … · 2011. 2. 10. · vi eiec,...

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