rotor fiber spinning - eindhoven university of technology · 2013. 11. 19. · rotor spinning rotor...
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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
A. Hlod (CASA) Rotor spinning 24-April-2008 2 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 3 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 3 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 3 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 4 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 4 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 4 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 4 / 25
Model
View from the top
x
y
en
et
R
β
φRrotor
vnozzle
ΩRcoagulator
Rcoagulator
Ω
F centrifugal
FCoriolis
A. Hlod (CASA) Rotor spinning 24-April-2008 5 / 25
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).
A. Hlod (CASA) Rotor spinning 24-April-2008 6 / 25
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)
A. Hlod (CASA) Rotor spinning 24-April-2008 7 / 25
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)
A. Hlod (CASA) Rotor spinning 24-April-2008 7 / 25
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)).
A. Hlod (CASA) Rotor spinning 24-April-2008 8 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 9 / 25
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!
A. Hlod (CASA) Rotor spinning 24-April-2008 9 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 10 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 11 / 25
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,
A. Hlod (CASA) Rotor spinning 24-April-2008 12 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 13 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 14 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 15 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 17 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 18 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 19 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 20 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 21 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 22 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 23 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 24 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 24 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 24 / 25
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
A. Hlod (CASA) Rotor spinning 24-April-2008 24 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 25 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 25 / 25
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.
A. Hlod (CASA) Rotor spinning 24-April-2008 25 / 25