Designing a Cross-Slot for Extensional-FlowSANS

Shooting Neutrons at Soap

Koty McAllister (UD, NCNR SURF), Katie Weigandt (NCNR)
Polymer processing and extrusion 

Consumer Products

Processing and delivery of pharmaceuticals 

Complex Fluids and Flow
Shear and Extensional Flow

Shear Strain Extensional Strain
Cross-Slot Flow Cell

Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14
From University of Waikato, January 11, 2012, sciencelearn.org.nz

Wormlike Micelles (WLM)

75mM CPyCl/ 45 mM NaSal in D2O

ττrr ≈ 7 ssssss

Oil recovery and fracking 

Drag reducers

Personal care products Home care products
Small Angle Neutron Scattering

NCNR.”NG7.”www.ncnr.nist.gov
Why Small Angle Neutron Scattering?

Characterize structures from 1 nm – 500 nm

Transparency 

H D C O Al Si

X-Rays

Neutrons

Scattering Cross-Sections

Contrast Variation 
Extensional Flow within a Cross Slot

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

J. Penfold, I. Tucker. J. Phys. Chem. B. 2007, 111, 9496-9503

Strain Strain 

[1]

Plug-like flow:

εεnnnnnn ̇ =

QQ

ww2DD

[1]

γγ̇

wwwwwwww =

6QQ

2h ; ll < h (Newtonian Fluid) 

AR = 2.5 AR = .1

ww, ll

DD, h

ww, h

DD, ll

Aspect ratios of 2.5, 1.7, 1, 0.7, 0.1

AAAA =

DD

ww
Asymmetric Flow within a Cross Slot

[1]

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

Occurs when εε̇ > εε̇

cc

εε̇

γγ̇
Small Angle Neutron Scattering (SANS)

θ

φ

q

Sample x Detector 

y Scattering 

vector

Neutron beam 

Isotropic Anisotropic 

1 mm aperture

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

[1]
AAff qq =

∫0

2ππ II qq,φφ cos(2 φφ − φφ0 ) dd

∫0

2ππ II qq,φφ dd

−AAff(qq ≥ 0.03 AA−1) ≈ SS [2]

Alignment Factor

φφ0

φφφφ0 00 φφ0

[2] L.M. Walker, “Rheology and Rheo-optics of liquid crystal polymers under flow.” 1995
II φφ = �

nn=0

∞

{ −1 nn 2nn !

4nnnn! nn! aannPP2nn cos φφ − φφ0 }

PP�

2 =

1

5 0 PP�

2 = 0.0321

φφ0 = NN/AA

Nematic Orientation Parameter

PP�

2 = 0.1224

φφ0 = −0.4°

PP�

2 = 0.1127

φφ0 = 24°

Annular averages take at qq = 0.03 AA−1

I (cm-1) I (cm-1) I (cm-1)
Data from ε̇ = 0.8 s−1 , AR = 2.5, 5 sec bins

ODF: gg ββ = ∑nn=0

∞ aannPP2nn( cc )

PP�

2 =

3cc cc2 ββ −1

2

= 1 − 3

2

ssss 2 ββ = SS [3]

Therefore PP�

2 ≈ −AAff qq ≥ 0.03 AA−1

PP2 and AAff

[3] W. H. DeJeu, Mol. Cryst. Liq. Cryst. 1997, 292, 13.
Steady state orientation is reached after ~110s

PP�

2 = 0.12

φφ0 = 0

ε̇= 0.8 s−1 Time Resolved

Flow Relax
0.03 s-1

0.8 s-1

2.0 s-1

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

[1] [1]

0.4 s-1

Equilibrium Structure (AR=2.5)

εεnnnnnn ̇ ,cc~0.85ss−1

Nominal Extensional Strain (s-1)
Orientation for Different Aspect Ratios

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

[1] [1]

εεnnnnnn ̇ =

QQ

ww2DD

γγ̇

wwwwwwww is negligible

εεnnnnnn ̇ ,cc~0.9ss−1
Conclusions

• PP�

2 ≈ −AAff

• Measured the amount of orientation and the 

angle of orientation as a function of nominal 

extensional strain rate

• Characterized the transition between symmetric 

and asymmetric flow 

• Showed that the behavior of the higher aspect 

ratios is nearly identical

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

[1] [1]
Future Work
• Katie Weigandt

• NSF

• NIST- NCNR

• Julie Borchers

• David Hoogerheide, Chirag Parikh, and Frank Hess

Acknowledgements
Questions?
Aspect Ratio

ww 2

D

AAAA =

DD

ww

ww

Plug-like flow:

εεnnnnnn ̇ =

QQ

ww2DD

[1]

vv = 12 cc/ss vv = 42 cc/ss

AR = 2.5 AR = 1 AR = .1

vv = 8.5 cc/ss

(εεnnnnnn ̇ = 1ss−1) (εεnnnnnn ̇ = 1ss−1) (εεnnnnnn ̇ = 1ss−1)
Preferential Asymmetry 

εε̇ = 3ss−1, AR = 0.7
Time

No Flow

Extensional 

Flow

Extensional + 

Shear Flow

Shear Flow

(Steady State)

10 mL/min 

Stress overshoot
Steady state orientation is reached after 

~40s

PP�

2 = 0.10

φφ0 = −28°

ε̇= 8.3 s−1 Time Resolved

[1] Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

[1]

Flow Relax
Aspect Ratios and Shear

AR = 2.5 AR = 1 AR = .1

γγ̇

wwwwwwww =

6QQ

2h ; ll < h (Newtonian Fluid) 

ll

h h

ll h

ll

εεnnnnnn ̇ =

QQ

ww2DD

= 1 ss−1 γγ̇ = 6 ss−1 γγ̇ = 6 ss−1 γγ̇ = 36.8 ss−1

Symmetric Flow
References 

1. Haward, S.; McKinley, G. Physical Review. 2012, 85, 031502-1-031502-14

2. W. H. DeJeu, Mol. Cryst. Liq. Cryst. 1997, 292, 13.

3. L.M. Walker, “Rheology and Rheo-optics of liquid crystal polymers under flow.” 1995