rotor fiber spinning - eindhoven university of technology · 2013. 11. 19. · rotor spinning rotor...

Rotor fiber spinning

A. Hlod

CASACenter for Analysis, Scientific Computing and Applications

Department of Mathematics and Computer Science

24-April-2008

A. Hlod (CASA) Rotor spinning 24-April-2008 1 / 25

Outline

1 Rotor spinning process

2 Modelling

3 Solution procedure

4 Characterization of the parameter space

5 Numerical experiments

6 When viscous jet is possible?

7 Results & Conclusions


Rotor spinning

rotor

coagulator

water

polymer solution

washing& drying

AdvantageIn case of breakage spinningrestarts automatically.

Some numbersRadii of the rotor and thecoagulator Rrotor = 0.15 m,Rcoagulator = 0.3 mFlow velocity at the nozzlevnozzle = 1 m/sAngular velocity of the rotorΩ = 2500 rpmFluid with viscosityη = 1200 Pa · s anddensity ρ = 1700 kg/m3.


Problems & Questions

ProblemsBreakage of filaments.Unsteadiness.For small Ω the thread does not reach the coagulator.

NeedsProduce long fibers with uniform thickness.Understand the process.

What to do?Experiments are expensive and very nasty.Therefore modelling is needed!





What to do?Experiments are expensive and very nasty.

Therefore modelling is needed!





What to do?Experiments are expensive and very nasty.Therefore modelling is needed!


Model

View from the top

x

y

en

et

R

β

φRrotor

vnozzle

ΩRcoagulator

Rcoagulator

Ω

F centrifugal

FCoriolis


Model equations

Conservation of momentum (in rotating frame) and conservation ofmass

ρ′rtt + ρ′rs(vt + vvs) + ρ′v2rss + 2ρ′vrst = (Prs)s + FCoriolis + Fcentrifugal,

ρ′t + (ρ′v)s = 0.

r is position vector, s is arc length, t is time, ρ′ is linear density,P = 3νvsρ

′ is longitudinal force (ν = η/ρ is kinematic viscosity).


Equations (1)

Conservation of momentum in local coordinate system et , en

ξ′(s) = cos(φ(s))R(s)/v(s),

ξ(s)φ′(s) = −R(s) sin(φ(s))/v(s)− sin(φ(s))ξ(s)/R(s) + 2,

v ′(s) = (v(s)2 − ξ(s)v(s))/B,

R′(s) = cos(φ(s)), β′(s) = − sin(φ(s))/R(s),

where B = 3ν/(R2coagulatorΩ).

Can be simplified using

sin(φ(s))ξ(s) = R(s) +c1

R(s)


Equations (2)

The equation for β is decoupled. Focus on the equations for ξ(s), φ(s),v(s), and R(s).

ξ′(s) = cos(φ(s))R(s)/v(s),

ξ(s)φ′(s) = −R(s) sin(φ(s))/v(s)− sin(φ(s))ξ(s)/R(s) + 2,

v ′(s) = (v(s)2 − ξ(s)v(s))/B,

R′(s) = cos(φ(s)).


Boundary conditions

At the rotor s = 0:R(0) = R0, v(0) = Dr.

At the coagulator s = send:

R(send) = 1, v(send) = 1.

Note that the jet length send is unknown.

One more boundary condition for φ is needed!


Recap: Jet fall onto a moving belt

There exists three flow regimes characterized by sign of ξ.

Viscous jet

Tangent to the belt

Viscous-inertia jet

The vertical jetshape is determinedby gravity.

Inertia jet

Tangent to thenozzle


Boundary condition for φ

Similarly, BC for φ are determined by characteristicsIn case of viscous jet

φ(send) = π/2.

In case of viscous-inertia jet

φ(s0) = arcsin(2v(s0)/R(s0)), where for s0 holds ξ(s0) = 0.

In case of inertia jetφ(0) = 0.


System for R[s] = r as independent variable

Writing ξ, v and φ as function of radius r (r = R[s]) we obtain

ξ′(r) = r/v(r),

ξ(r)φ′(r) = − sin(φ(r))r2−v(r)(ξ(r) sin(φ(r))−2r)v(r)r cos(φ(r)) ,

v ′(r) = (v(r)2 − ξ(r)v(r))/(B cos(φ(r))),

v(R0) = Dr, v(1) = 1,

and sin(φ(r))ξ(r) = (r2 − r20 )/r .

The BC for φ is

φ(1) = π/2 if viscous flow,φ(r0) = arcsin(2v(r0)/r0), ξ(r0) = 0 if viscous-inertia flow,φ(R0) = 0, r0 = R0 if inertia flow,


Description of the shooting method

By replacing the boundary condition v(1) = 1 by ξ(R0) = w we findsolutions of v(r ; w), φ(r ; w) and ξ(r ; w).

By finding w such that v(1; w) = 1 we solve the problem.

Φ=Π2

0.8 1.0 1.2 1.4 1.6 1.8 2.0r

0.0

0.2

0.4

0.8

1.0

1.2

1.4

vHr,wL

0.8 1.0 1.2 1.4 1.6 1.8 2.0r

0.5

1.0

1.5

ΦHr,wL


Nonexistence of a solution (1)

It may happen that the jet does not reach the coagulator

solution

max r

0.2 0.4 0.6 0.8 1.0r

0.50

0.55

0.65

0.70

vHr,wL

solution

max r

Π2

0.2 0.4 0.6 0.8 1.0r

0.5

1.0

1.5

2.0

2.5

3.0

ΦHr,wL


Nonexistence of a solution (2)

It may happen that the jet does not stick to the coagulator (v(1) < 1)

max vH1L

0.96 0.97 0.98 0.99 1.00r

0.605

0.610

0.615

0.620

0.625

0.630

0.635vHr,wL

Π2

0.96 0.97 0.98 0.99 1.00r

0.5

1.0

1.5

2.0

2.5

3.0

ΦHr,wL

BC v(1) = 1 is not reached.


Phase diagram

There are 4 possible situations for a jet solution.

B=0.1

IJ

VIJ ØSCØRC

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

R0

Dr

IJ - inertia jet, VIJ - viscous-inertia jet, ¬RC - jet does not reach thecoagulator, ¬SC - jet does not stick to the coagulator.

No viscous jet is possible in current setup!A. Hlod (CASA) Rotor spinning 24-April-2008 16 / 25

Phase diagram for large B (1)

For larger B the viscous-inertia jet region becomes narrower and theinertia jet region shrinks.

B=0.2

IJ

VIJØSCØRC

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

R0

Dr


Phase diagram for large B (2)

For B large enough no viscous-inertia jet is possible

IJ

B=0.3

ØRC or ØSC

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

R0

Dr


Parameter region for inertia jet

The boundary of inertia jet region (region above the surface)approaches to Dr = 1 as B becomes larger and approaches to Dr = 0as B goes to 0.

0.00.2

0.4

0.6

0.8R0

0

1

2

B

0.2

0.4

0.6

0.8

Dr


Experiments

We will present three numerical experiments.

We start with the inertia jet and decrease vnozzle such that theparameters leave the inertia jet region.

B=0.1

IJ

VIJ ØSCØRC

IJ

Exp 3

Exp 2

Exp 1

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

R0

Dr


Experiment 1

We start with the inertia jet (IJ) and decrease vnozzle such that the flowbecomes viscous-inertial (VIJ).

IJ

VIJ

-0.4 -0.2 0.2 0.4 0.6 0.8 1.0

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.52 0.54 0.56 0.58 0.60

-0.10

-0.08

-0.06

-0.04

-0.02

0.00


Experiment 2

We start with the inertia jet and decrease vnozzle such that the jet doesnot reach the coagulator

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0


Experiment 3

We start with the inertia jet and decrease vnozzle such that the jet doesnot stick to the coagulator, see curve 4

1

2

3

0.85 0.90 0.95 1.00 1.05

-0.3

-0.2

-0.1

0.0

0.1

1

2

3

4

0.88 0.90 0.92 0.94 0.96 0.98 1.00r

0.90

0.95

1.05

1.10

v


Possibility of viscous jet

Parameter region for viscousjet is defined by jet solutionswith ξ(1) = 0.

Such solutions are notpossible sin(φ(1)) 6= 2(φ(1) = arcsin(2v(1)/1),v(1) = 1).

The condition can be satisfiedwhen v(1) ≤ 0.5(sin(φ(1) ≤ 1).

Thus, coagulator shouldrotate.

x

y

Rrotor

vnozzle

( - )RΩ coagulatorΩcoagulator

Rcoagulator

Ω

Ωcoagulator


Results & Conclusions

The model with effects of viscosity, inertia, Coriolis and centrifugalforces describes the rotor spinning process.

Four situations for the jet solution are possible: "inertia jet","viscous inertia jet", "the jet does not reach the coagulator" and"the jet does not stick to the surface".

For viscous jet the coagulator should rotate in the same directionas the rotor.


rotor fiber spinning - eindhoven university of technology · 2013. 11. 19. · rotor spinning rotor...

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