L3 · Chapter 6

How normal stress along curved streamlines drives elastic instability

Sandbox Physics · 2026-10-29

The normal stress along a streamline acts like a stretched rubber band. On a straight streamline the tension only pulls along the flow; once the streamline curves, the tension gains a component toward the centre of curvature and squeezes the liquid inward. That is why a stirred polymer solution climbs up the stirring rod. The same tension also amplifies sideways disturbances of the streamlines and can make the flow unstable at Reynolds numbers far below one.

Hoop stress and rod climbing

In a rotating flow the streamlines are circles. The first normal-stress difference turns each circle into a stretched hoop whose tension points toward the centre, so the pressure rises from the outside in. Oldroyd-B has no second normal-stress difference, and the radial pressure gradient is simply $N_1/r$ .

Squeezed toward the axis, the liquid can only go up around the stirring rod: the rod climbing shown in Fig. 1.10(b) of Morozov and Spagnolie. A Newtonian liquid stirred the same way is flung outward by inertia, and its surface dips at the centre.

-\frac{\partial p}{\partial r}=\frac{N_1+2N_2}{r}

Eqs. (1.117)–(1.119) of Morozov and Spagnolie, for rotating flows in which inertia is negligible.

Source: A. Morozov and S. E. Spagnolie

The normal force in a cone-and-plate rheometer

In a cone-and-plate rheometer the gap grows linearly with radius, so the whole sample sees the same shear rate. Integrating the pressure set up by the hoop stress gives the total force pushing the cone and plate apart. Measure that force and you have $N_1$ .

Simple shear between parallel plates is the limit of infinite radius of curvature, where the hoop stress vanishes. Morozov and Spagnolie point out that neither plane shear nor Taylor–Couette flow gives the first normal-stress difference this way, which is why the cone-and-plate geometry is used.

F=\frac{\pi R^2}{2}\,N_1

Eq. (1.125) of Morozov and Spagnolie, for a cone of radius R.

Source: A. Morozov and S. E. Spagnolie

One instability parameter

The Taylor number of inertial instability compares centrifugal force with viscosity. Groisman and Steinberg (2004) write its elastic counterpart $K$ : elastic tension along curved streamlines against the viscosity that damps disturbances. It is the product of the polymer share of the viscosity, the ratio of gap to radius of curvature, and the square of the Weissenberg number. For Couette–Taylor flow the Weissenberg number is $\lambda\Omega R/d$ .

The Pakdel–McKinley criterion, as given in Eq. (1) of the review by Datta and co-authors, is written for a general streamline, with streamline speed $U$ , radius of curvature $\mathcal R$ and the ratio of first normal-stress difference to shear stress. Insert the Oldroyd-B ratio $N_1/\tau=2(\eta_p/\eta)\mathrm{Wi}$ and, for narrow-gap Couette–Taylor flow, $U=\Omega R$ and $\mathcal R=R$ . The result is $M^2=2K$ : the two criteria say the same thing.

Neither paper gives a universal critical value. The review by Datta and co-authors notes that no agreed critical $M$ exists for Taylor–Couette or cone-and-plate flow; in cross-slot simulations the flow loses symmetry or becomes time dependent once the largest local $M$ exceeds about 4 to 5.

K=\frac{\eta_p}{\eta}\,\frac{d}{R}\,\mathrm{Wi}^2,\qquad M^2=\frac{\lambda U}{\mathcal R}\,\frac{N_1}{|\tau|},\qquad M^2=2K

The first expression is from Section 2 of Groisman and Steinberg (2004), the second is Eq. (1) of Datta and co-authors. The last equality holds only for an Oldroyd-B liquid in a narrow gap; it is derived in this course, taking

\eta

as the total viscosity and the shear stress as the total stress.

Source: A. Groisman and V. Steinberg · S. S. Datta et al.

Putting in experimental numbers

The swirling flow of Groisman and Steinberg (2000): $\eta_p/\eta=0.100/0.424=0.236$ , $d/R=10/38=0.263$ , and a transition at $\mathrm{Wi}=3.5$ , so $K=0.236\times0.263\times3.5^2=0.760$ .

The curved channel of 2001: solvent viscosity 0.153 Pa s and solution viscosity 0.198 Pa s, so $\eta_p/\eta=0.227$ ; channel width 3 mm and inner radius 3 mm, so $d/R=1$ ; critical $\mathrm{Wi}=3.2$ , so $K=0.227\times1\times3.2^2=2.33$ .

The two values differ threefold. The papers define the Weissenberg number differently, the first with the shear rate at the rim of the rotating plate and the second with four times the flow rate divided by the cube of the channel width, and the geometries differ too. The parameter ranks which flow goes unstable first within one family of geometries; it is not a threshold that carries across geometries.

Section 6 of the 2004 paper also compares the resistance in elastic turbulence across three geometries: about 20 times laminar for the flow between two plates, about 4.2 for Couette–Taylor flow and about 2.8 for the curved channel.

Source: A. Groisman and V. Steinberg · A. Groisman and V. Steinberg · A. Groisman and V. Steinberg

Worked example: normal stress in the elastic-turbulence solution

Take the solution of Chapter 1, $\eta_p=0.100\,\mathrm{Pa\,s}$ and $\lambda=3.4\,\mathrm s$ , and estimate the steady first normal-stress difference at a shear rate of one per second with Oldroyd-B: $N_1=2\times0.100\times3.4\times1^2=0.68\,\mathrm{Pa}$ . The total shear stress is 0.424 Pa, so $N_1/\tau=1.60$ , equal to $2\times0.236\times3.4$ . The normal stress already exceeds the shear stress.

In a cone-and-plate rheometer of radius $25\,\mathrm{mm}$ the force pushing cone and plate apart would be $\pi\times0.025^2\times0.68/2=6.68\times10^{-4}\,\mathrm N$ , about 0.67 mN, the weight of roughly 68 mg. A force that small needs a sensitive normal-force transducer.

These are Oldroyd-B estimates. A real solution shear-thins, and its normal stress grows more slowly than the square of the shear rate; the paper by Groisman and Steinberg gives no measured normal stress.

Source: A. Groisman and V. Steinberg · A. Morozov and S. E. Spagnolie

Try it in the Lab

Read the ratio of normal stress to shear stress

Set the spring law to Oldroyd-B (linear spring), the solvent share of viscosity to 0.5 and the Weissenberg number to 2.
In the detailed readouts, find the shear ratio of normal-stress difference to shear stress. It is the polymer-part ratio $2\mathrm{Wi}$ ; multiply by the polymer share $1-\beta$ to get the ratio with the total shear stress in the denominator, the one the criterion uses.
Switch to FENE-P (finite chain), set the maximum extension squared to 20 and raise the Weissenberg number. Does the ratio still grow linearly with the Weissenberg number?

Find curvature and stretching in a four-roll flow

This experiment’s flow is four vortices around one stagnation point: streamlines curve around the vortices and are stretched at the stagnation point. It stands in for the curved channel of the 2001 experiment but has no channel width or radius, so the parameter of this chapter cannot be computed directly.
Watch where the polymers are stretched most at a low Weissenberg number, then raise the number and compare. Does the stretch concentrate around the strongly curved vortices or along the line leaving the stagnation point?

Exercises with explained answers

An Oldroyd-B solution with $\eta_p/\eta=0.3$ flows in narrow-gap Couette–Taylor flow with $d/R=0.1$ . If instability in some geometry sets in at $K=1$ , what is the critical Weissenberg number?
Show the answer
$\mathrm{Wi}=\sqrt{K/(0.3\times0.1)}=\sqrt{33.3}=5.77$ . This is only an exercise in the formula; a real threshold has to come from experiments or a stability analysis in that geometry.
Why can simple shear between parallel plates not measure the first normal-stress difference?
Show the answer
Between parallel plates the streamlines are straight, with zero curvature, so the tension along them has no net component in any direction and adds no normal force on the plates. In the cone-and-plate geometry the streamlines are circles, and the hoop stress carries the pressure distribution to the plate.
The solution of the Chapter 1 example runs in the swirling-flow apparatus at $\mathrm{Wi}=3.5$ . If the polymer share of viscosity is halved with everything else fixed, to what Weissenberg number must you go to recover the same $K$ ?
Show the answer
$K$ is proportional to the polymer share times $\mathrm{Wi}^2$ . Halving the share requires multiplying the Weissenberg number by $\sqrt2$ , to about 4.95.
Is it fair to blame the difference between the 2000 and 2001 values of K on the two solutions being different?
Show the answer
No. Both solutions contain 80 ppm of polyacrylamide and have similar polymer shares of viscosity; the difference comes mainly from different definitions of the Weissenberg number and different geometries. Check that two papers use the same definition before comparing their critical parameters.

Sources and what we recomputed

A. Morozov and S. E. Spagnolie, Introduction to complex fluids, in Complex Fluids in Biological Systems (Springer, 2015), chapter 1
Section 3: the Weissenberg number defined as the relaxation time times the shear rate, and Eq. (1.27), the generalized Newtonian liquid. Section 4.1: Eqs. (1.47)–(1.51), the spring, the dashpot, the Maxwell liquid and its memory integral, with a relaxation time equal to viscosity over modulus. Eq. (1.72): the upper-convected derivative; Eq. (1.80): Oldroyd-B; Eqs. (1.86)–(1.87): FENE-P. Section 4.4: the Hookean dumbbell, Eqs. (1.98)–(1.110). Eq. (1.116): the steady-shear normal stress of Oldroyd-B. Eqs. (1.117)–(1.125): hoop stress and the cone-and-plate normal force. Eqs. (1.128)–(1.131): uniaxial extension and the Oldroyd-B divergence at half the inverse relaxation time. The chapter has no startup solution and no planar extension; the course derives both.
A. Groisman and V. Steinberg, Elastic turbulence in a polymer solution flow, Nature 405, 53–55 (2000); arXiv:nlin/0104052v1
arXiv version 1, read in full. Fig. 1: swirling flow between a rotating plate of radius 38 mm and a cup of radius 43.6 mm, liquid depth 10 mm, 12 °C. Methods: 65 % saccharose and 1 % NaCl solvent of viscosity 0.324 Pa s with 80 ppm polyacrylamide of molecular weight 1.8e7; solution viscosity 0.424 Pa s at a shear rate of 1 per second; relaxation time 3.4 s. Fig. 2: transition near a shear rate of 1 per second (Weissenberg number 3.5, Reynolds number about 0.3), with hysteresis; resistance up to about 12 times laminar for a 10 mm gap and 19 for 20 mm. Fig. 3: snapshots at Weissenberg number 13. Fig. 4: brightness spectra. Fig. 5: velocity spectra, a power law with exponent −3.5 at 4 per second over about one decade. Text: Reynolds stress below 0.5 % of the resistance.
A. Groisman and V. Steinberg, Efficient mixing at low Reynolds numbers using polymer additives, Nature 410, 905–908 (2001); arXiv:nlin/0104050v1
arXiv version 1, read in full. Fig. 1: a channel of width 3 mm made of 60 half-rings with radii 3 and 6 mm. Solvent viscosity 0.153 Pa s, density 1320 kg per cubic metre; 80 ppm polyacrylamide; solution viscosity 0.198 Pa s at a shear rate of 4 per second; relaxation time 1.4 s; dye diffusion coefficient about 8.5e-7 square centimetres per second. The Weissenberg and Reynolds numbers are defined from the flow rate and the channel width. The flow stays steady up to a Weissenberg number of 3.2 (Reynolds number 0.06); mixing is studied at 6.7. Fig. 2: velocity spectra as a power law with exponent −3.3 in wavenumber. Fig. 4: mixing over about 15 units, roughly 120 s, three orders of magnitude faster than the diffusion time.
A. Groisman and V. Steinberg, Elastic turbulence in curvilinear flows of polymer solutions, New Journal of Physics 6, 29 (2004)
Section 2: the elastic instability in curvilinear flows is controlled by K, built from the polymer share of the viscosity, the gap-to-radius ratio and the square of the Weissenberg number, the elastic analogue of the Taylor number; the paper gives no universal critical value of K. Section 4.2: the Couette–Taylor transition occurs near a Weissenberg number of 4, against 3.2 and 2 between two plates for gap-to-radius ratios 0.26 and 0.52; at the highest Weissenberg numbers explored the flow resistance grew by a factor of 4.2 in Couette–Taylor flow against 11 and 19 between two plates. Section 5: in the curved channel stationary vortices may appear near a Weissenberg number of 2, and irregular motion starts at 3.2 (Reynolds number 0.06); in a 100 µm microchannel the resistance grows from an estimated Weissenberg number of 1.3 and reaches a factor of 2.8 near 9. Section 6 summarises the resistance growth as factors of 20, 4.2 and 2.8 for the flow between two plates, Couette–Taylor flow and the channel.
S. S. Datta et al., Perspectives on viscoelastic flow instabilities and elastic turbulence, arXiv:2108.09841v1 (2021); Physical Review Fluids 7, 080701 (2022)
arXiv version 1, 22 August 2021; the journal version was not checked. Section II, Eq. (1): the Pakdel–McKinley criterion, which combines streamline curvature with the ratio of normal stress to shear stress. No critical value is given for Taylor–Couette or cone-and-plate flow. Section III: simulations of cross-slot flows lose symmetry or become time dependent when the largest local value of that criterion exceeds about 4 to 5; Eq. (2) and Table II write Oldroyd-B, FENE-P and other models with a conformation tensor, which allows a positive-definiteness check and log-conformation schemes; a two-dimensional Taylor–Couette simulation became chaotic near a Weissenberg number of 10. The workshop summary states that Oldroyd-B is almost never quantitative at high Weissenberg number.