Mathematical models formicrostructured optical fibre (MOF) fabrication

Darren CrowdyImperial College Londond.crowdy@imperial.ac.uk

The inverse problem in fibre drawing

Heike E-H Y. Stokes Mike Chen Hayden T. D. Crowdy P. Buchak

Institute for Photonicsand Advanced Sensing,University of Adelaide

Department ofMathematics,

University of Adelaide

Department of Mathematics,Imperial College London

Microstructured optical fibres (MOFs)

preformdrawn into

fibre

cross−plane

channelsouter boundary

x

feed speed U0

draw speed U1

neck−down length L

cross−sectional area S0

cross−sectional area S1

the fibre draw process

Fibre drawinga method for fabricating optical fibres from preforms

draw tower (IPAS)

Diagram of fibre drawing:

preform fed into furnace,heated,

drawn into a fibre,wound around spool

Problem: fibre geometry differs from preform geometryGeometrical cross section deforms under both draw tension and surface tension.

(Issa et al., Optics Letters, 2004)

initial preform drawn fibre

I For a preform with circular channels, channels in the fibre are not circular.

I Deformation is undesirable, difficult to quantify.

Aim: to understand, and predict, this deformation

The experiments

cross−plane

channelsouter boundary

x

feed speed U0

draw speed U1

neck−down length L

cross−sectional area S0

cross−sectional area S1

Controlled by experimenterspreform radius R0 ≈ 10 mmfeed speed U0 ≈ 1 mm/mindraw speed U1 ≈ 1 m/mintemperature T ≈ 1000◦Cviscosity µ ≈ 105 Pa ssurface tension γ ≈ .25 N/mpreform geometry

Measuredfibre radius R1 ≈ 0.1 mmdraw tension σ ≈ 10 gfibre geometry

I Often given draw ratio D = U1/U0.

I µ, but not γ, strongly depends on T .

Experimenters want to know:

I How to pick feed speed, draw speed, temperature?

I How to design preform?

Currently done by trial and error. Expensive. Wastes time and resources.

Forward and inverse problem

I Forward problem: For given preform geometry & draw parameters, whatwill the final fibre look like?

I Inverse problem: For desired fibre geometry, what preform geometry &draw parameters will produce it?

Outline of talk

I Fluid dynamics of fibre drawing

I Decoupling into “axial” and “cross-plane” problems (slender fibres)

I Complex variable formulation and conformal geometry

I Model for prediction of draw tension needed for desired fibre geometry

I ”EPM” (elliptical pore model) for the inverse problem

The presentation here is based on work in [Stokes, Buchak, Crowdy &Ebendorff-Heidepriem, J. Fluid Mech., (2014)]

Fluid dynamics of fibre drawing

I Assuming Newtonian fluid, low Reynolds number,

Flow governed by Stokesequations,

−∇p + µ∇2~u = 0∇ · ~u = 0

with surface tension andimposed pressures on freeboundaries,

σ · n̂ = −γκn̂ − pk n̂~xt · n̂ = ~u · n̂

cross−plane

channelsouter boundary

x

feed speed U0

draw speed U1

neck−down length L

cross−sectional area S0

cross−sectional area S1

Difficult 3D free boundary problem

Numerical approach

Given its importance there have been several numerical attempts to study theproblem. Among others:

I Xue, Tanner, Barton, Lwin, Large & Poladian, J. Lightwave Tech., (2005a)

I Xue, Tanner, Barton, Lwin, Large & Poladian, J. Lightwave Tech., (2005b)

I Xue, Tanner, Barton, Lwin, Large & Poladian, J. Lightwave Tech., (2005c)

I S. S. Chakravarthy, W. K. S. Chiu, J. Fluid Mech., (2008)

These are based on finite element methods, boundary integral techniques.

Inverse problem not easily studied this way

Slender fibres: asymptotic separation

Our approach is based on a “slender fibre” formulation from Cummings &Howell, JFM, 1999

I They studied drawing of a solid fibre (no “holes” or “channels”)

I Key insight: for slender fibres, fully 3D problem can be split into twosimpler problems, a 1D stretching and a 2D sintering problem

I Same idea used in modelling “Vello process” (making shaped glass tubing)by Griffiths & Howell, J. Fluid Mech., (2007)/(2008).

“Axial” and “cross-plane” problems

For slender fibres where

ε =

√S0

L� 1

3D problem splits into two coupled problems

2D cross-plane problem Axial stretching problem

We have devised a way to exploit this decoupling for MOFswith any number of channels

Axial flow equations

The equations governing the steady axial flow problem are

St + (uS)x = 0

3(Sux)x +1

2γ∗

M∑n=0

(Γn)x =M∑n=0

±(p(n)B )xAn.

S(x , t) is the cross-sectional area of the fluid region

u(x , t) is the axial fluid velocity,

An(x , t), Γn(x , t), and p(n)B (x , t) are the area, perimeter, and pressure

associated with the n-th channel.

(These equations encode steady mass and momentum balance)

Cross-plane equations

In each cross-plane the flow is governed by Stokes equations,

−∇p + µ∇2~u = 0∇ · ~u = 0

with surface tension and imposed pressures on free boundaries,

σ · n̂ = −κn̂ − pk n̂~xτ · n̂ = ~u · n̂

Crucial difference: notice the appearance of the reduced time

τ = γ∗

∫ t dt

µ√

S

This is the classical 2D surface-tension driven Stokes flow problem,but now in this reduced time variable τ

Important “base-line” case

An important “base-line” case to study has the following assumptions:

I zero Reynolds number

I absence of channel pressurizations

pn(t) = 0, ∀n

I constant viscosity (independent of temperature)

I no temperature, heat transfer effects

Once this problem is understood, add other effects later

Zero channel pressures: integration of axial problem

When all channel pressures vanish the axial flow problem admits an explicitintegral [Stokes, Buchak, Crowdy & Ebendorff-Heidepriem, J. Fluid Mech.,(2014)]

Introduce

H(τ) =

∫ τ

0

Γ̃(τ ′)dτ ′,

where Γ̃(τ) is the total boundary perimeter in the cross-plane.

Γ̃(τ) can be readily computed once the (unit area) cross-plane problem hasbeen solved.

With H(τ) determined from the cross-plane problem

S(τ) =

(1− σ

γ∗

∫ τ

0

H(τ ′) dτ ′)2

/H(τ)2,

x(τ) = − 1

σlog(H(τ)

√S(τ)).

The parameter σ is the scaled fibre tension.

Key message

For zero channel pressurizations, with a given initial cross-plane geometry, theset of possible cross-plane geometry shapes is fully determined a priori (fromthe 2D cross-plane solution)

What is not known yet is at what axial position x and with what totalcross-sectional area S that cross-section will appear

But these are given explicitly by the relations

S(τ) =

(1− σ

γ∗

∫ τ

0

H(τ ′) dτ ′)2

/H(τ)2,

x(τ) = − 1

σlog(H(τ)

√S(τ)).

where only H(τ) is “fed in” from the cross-plane solution

The particular draw conditions are reflected in the γ∗ and σ parameters

The reduced time τ provides a natural parameter for all cross-plane geometries

2D cross-plane free surface problem

Incompressible two-dimensional flow described by streamfunction ψ

∇4ψ = 0

Complex variable formulation. Let z = x + iy .

General biharmonic field given by

ψ = Im[zf (z , t) + g(z , t)]

where f (z , t), g(z , t) are analytic in fluid region (“Goursat functions”)

Physical variables given by

p

µ− iω = 4f ′(z , t)

u + iv = −f (z , t) + zf ′(z , t) + g ′(z , t)

e11 + ie12 = zf ′′(z , t) + g ′′(z , t)

p is pressure, ω is vorticity, eij is fluid rate-of-strain tensorVelocity field is (u, v)

Stokes flow fundamental singularities

The well-known fundamental singularities of Stokes flow (e.g. stokeslets,stresslets, rotlets) now manifest themselves as isolated singularities of these twoGoursat functions.

For example:

Stresslet: a stresslet singularity of strength λ at z0 is equivalent to

f (z) =λ

z − z0, g ′(z) =

λz0(z − z0)2

Source/sink: an irrotational source/sink of strength m at z0 is given by

f (z) = locally analytic, g ′(z) =m

2π(z − z0)

Advantage 1: the free surface stress condition

An important advantage for introducing a complex variable formulation is thatthe free surface stress condition can be integrated once (with respect toarclength s) yielding

f (z , t) + zf ′(z , t) + g ′(z , t) = − iγ

2

dz

ds

γ is surface tension

(This renders any numerical method less “stiff”)

Advantage 2: can use conformal mapping technology

Time-dependent conformal maps can be used to track the evolving interfaces

z(ζ)

ρ < |ζ| < 1 image

Numerical methods for the cross-plane problem based on conformal mappingdescriptions have been devised

[Buchak & Crowdy, J. Comp. Phys, (2014)]

Advantage 3: exact solutions derivable using complex analysis

Remarkably, a variety of exact solutions to the time-evolving 2D free surfaceevolution under surface tension are known.

Among others:

Hopper, J. Fluid Mech., (1990)Richardson, Eur. J. Appl. Math., (1992)Cummings, Howison & King, Phys. Fluids, (1998)Crowdy & Tanveer, J. Nonlin. Sci., (1998)Crowdy, Eur. J. Appl. Math., (2003)

Case study: sintering annular array of N discs

ρ < |ζ| < 1 N = 4

Solutions for the evolution of the initial fluid domain on the rightcan be found in exact form

Exact solution: sintering annular array of N discs

Conformal map, from the annulus ρ < |ζ| < 1, describing the fluid domainevolution exactly is found to be

z(ζ, t) = A(t)ζPN(ζρ2/N/a(t), ρ(t))

PN(ζ/a(t), ρ(t))

where

PN(ζ, ρ) =N∏

k=1

P(ζωk , ρ), ωk = e2πi(k−1)/N

and

P(ζ, ρ) ≡ (1− ζ)∞∏n=1

(1− ρ2nζ)(1− ρ2n/ζ)

Solution depends on just three time-evolving parametersA(t), a(t) and ρ(t)

They satisfy a system of 3 nonlinear first-order ODEs

[Crowdy, Eur. J. Appl. Math., 14, (2003)] (“loxodromic functions”)

Drawing N = 4 circular fibres → single fibre

This exact cross-plane solution can be combined with the axial equations togive the solution for the pulling of these four touching discs into a single fibre:

τ=0.00 x=0.00

τ=0.10 x=0.13

τ=0.20 x=0.34

τ=0.30 x=1.00

Just need evolution of total cross-plane perimeter (i.e. to compute H(τ))

(Useful for separate application of “fibre optic couplers”)

The general model: back to axial flow equations

With H(τ) determined from the cross-plane problem we have

S(τ) =

(1− σ

γ∗

∫ τ

0

H(τ ′) dτ ′)2

/H(τ)2,

x(τ) = − 1

σlog(H(τ)

√S(τ)).

The condition that x(τL) = L (i.e. that required final geometry is attained) isequivalent to the following “P −Q relation”:

1

Q +logQP = 1

where

P =γ∗∫ τL

0

H(τ ′)dτ ′, Q =

√D

H(τL).

Balance of draw tension and surface tension

1

Q +logQP = 1

where

P =γ∗∫ τL

0

H(τ ′)dτ ′, Q =

√D

H(τL).

0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

P

Q

σ > 0

σ < 0

τL decides the final geometry. This gives denominators in P and Q.Cross-plane solution “fed in” via these denominators only

Our model predicts required draw tension

Given initial geometry, desired final geometry (prescribed by τL), and surfacetension (generally constant over the range of experimental draw parameters)the key relation

1

Q +logQP = 1

where

P =γ∗∫ τL

0

H(τ ′)dτ ′, Q =

√D

H(τL).

tells us the draw tension required to attain the required final geometry

Temperature is key control parameter that must be used to control fluidviscosity, and hence draw tension

Important to be able to accurately monitor tension during the draw

This has been a crucial new insight for the experimentalists

What about the inverse problem?

Earlier exact solutions, with N →∞, highlight a difficulty for “inverseproblem”:

N = 20 solutions[Crowdy, (2003)]

τ=0.00 τ=0.00

τ=0.05 τ=0.04

τ=0.10 τ=0.10

τ=0.30 τ=0.30

annular crossplane

Different initial conditions can give indistinguishable profiles in forward time(physically, surface tension “irons out” ripples)

Simply “running backwards in time” will be an unstable process

Exact solution for compressible elliptical bubble in linear ambient

Key fact: an initially elliptical bubble in linear Stokes flowremains elliptical as it evolves

Zn

Zj

kn!n

elliptical channel

sink + stresslet

The ellipse describable by the time-evolving conformal mapping

z(ζ, t) = z0(t) +α(t)

ζ+ β(t)ζ

with ODEs for α(t) and β(t) known from [DGC, J. Fluid Mech., 476, 345-356,(2003)]

The “elliptical pore model” (EPM)

I For inverse problem we need a constrained way to run the simulationbackwards in time (i.e. to “filter out” ripples, or follow “slow manifold”)

Zn

Zj

knωn

elliptical channel

sink + stresslet

I Crowdy, J. Fluid Mech., [2004] has proposed an “elliptical pore model” formulti-pore interactions (based on exact elliptical solutions just given)

I “Outer”: each elliptical channel is a point stresslet + point source/sink;“Inner”: evolution of ellipse given by previous exact solution in linear flowgenerated by superposition of singularities due to the other channels

“Outer solution”: fundamental singularities

The well-known fundamental singularities of Stokes flow (e.g. stokeslets,stresslets, rotlets) now manifest themselves as isolated singularities of these twoGoursat functions.

The two used to describe each elliptical channel “in its far-field” are:

Stresslet: a stresslet singularity of strength λ at z0 is equivalent to

f (z) =λ

z − z0, g ′(z) =

λz0(z − z0)2

Source/sink: an irrotational source/sink of strength m at z0 is given by

f (z) = locally analytic, g ′(z) =m

2π(z − z0)

The ODEs for the EPM

I Model gives system of ordinary differential equations dictating howchannels move & evolve.

dZn

dτ= −

∑j 6=n

µj(Zn −Zj)

2π(Zn −Zj)2+∑j 6=n

mj

2π(Zn −Zj)−∑j 6=n

µj

2π(Zn −Zj)

dαn

dτ= −αIn(0)− 1

2α(pn − p

(n)B )

dβndτ

= −βIn(0) +1

2β(pn − p

(n)B ) + 2knαn + iωnβn

dR

dτ=

M

2πR← assume outer boundary circular

I Computationally inexpensive – short Matlab script is available on web.

I Much more computational efficient than full numerical simulation

I It can be run forwards or backwards in a stable manner

I It is a “rational” exclusion of extraneous growth of unwanted modes

Testing the accuracy of EPM

How does EPM predictions compare with full numerical solution?

−0.5 0 0.5

−0.5

0

0.5

τ=0.000

0.2 0.3−0.1

−0.05

0

0.05

0.1

τ=0.000

−0.5 0 0.5

−0.5

0

0.5

τ=0.040

0.2 0.3−0.1

−0.05

0

0.05

0.1

τ=0.040

−0.5 0 0.5

−0.5

0

0.5

τ=0.080

0.2 0.3−0.1

−0.05

0

0.05

0.1

τ=0.080

−1 0 1−1

−0.5

0

0.5

1τ=0.00

0 0.5

−0.2

−0.1

0

0.1

0.2

τ=0.00

0 0.1 0.2−0.1

0

0.1

0.2τ=0.00

−1 0 1−1

−0.5

0

0.5

1τ=0.10

0 0.5

−0.2

−0.1

0

0.1

0.2

τ=0.10

0 0.1 0.2−0.1

0

0.1

0.2τ=0.10

−1 0 1−1

−0.5

0

0.5

1τ=0.20

0 0.5

−0.2

−0.1

0

0.1

0.2

τ=0.20

0 0.1 0.2−0.1

0

0.1

0.2τ=0.20

(solid – model; dashed – numerical solution)

Excellent agreement with numerical solution.

Testing the accuracy of EPM

−1 0 1−1

−0.5

0

0.5

1τ=0.000

0 0.5 10

0.5

1τ=0.000

0.2 0.3 0.40.4

0.5

0.6

0.7τ=0.000

−1 0 1−1

−0.5

0

0.5

1τ=0.030

0 0.5 10

0.5

1τ=0.030

0.2 0.3 0.40.4

0.5

0.6

0.7τ=0.030

−1 0 1−1

−0.5

0

0.5

1τ=0.070

0 0.5 10

0.5

1τ=0.070

0.2 0.3 0.40.4

0.5

0.6

0.7τ=0.070

(solid – model; dashed – numerical solution)

Model is computationally inexpensive, even for high connectivity geometries.

EPM can help solve the inverse problem (stably)

τ=−0.000

τ=−0.012

τ=−0.023

τ=−0.034I Easy to evolve fibre configuration backwards in time to get

preform.

3D solution for draw

How to design preforms?

In the following examples, wepick a fibre geometry and showpossible preforms, each withrequired parameters (inverseproblem).

We take:

L = 5 cm

µ = 5× 105 Pa s

γ = .25 N/m

R0 = 10 mm

R1 = 0.5 mm

EPM predictions I

Fiber configuration

τ1=0.001 D=399 U

0=45.8 mm/min τ

1=0.012 D=386 U

0=2.3 mm/min τ

1=0.024 D=368 U

0=1.2 mm/min

preform configurations, with draw parameters draw

Allowing longer deformation time τ1 lets feed speed U0 be lower.

EPM predictions II

Fiber configuration

τ1=0.003 D=393 U

0=8.4 mm/min τ

1=0.007 D=385 U

0=4.2 mm/min τ

1=0.014 D=369 U

0=2.2 mm/min

τ1=0.021 D=352 U

0=1.5 mm/min τ

1=0.028 D=333 U

0=1.2 mm/min τ

1=0.035 D=313 U

0=1.0 mm/min

preform configurations, with draw parameters draw

Tradeoff between circular preform (easy to make)and low feed speed (easy to draw).

Summary

I New model, for slender fibres, showing the importance of draw tension

I New numerical schemes, based on conformal geometry, to solve forwardproblem for any geometry

I Approximate EPM to solve forward or inverse problem for fibres withelliptical channels

Current work

I Experimental validation (in progress)

I Generalize to allow temperature-dependent viscosity

I Generalize the EPM to other channel shape classes

Website: www2.imperial.ac.uk/~dgcrowdy