Wall Shear Stress in a Capillary

This one is a force balance and nothing else, which is why it is exact and why it needs no correction for the fluid's behaviour. Take a plug of melt filling a round channel of radius $R$ and length $L$. The pressure difference across its ends pushes it forward with a force $\Delta P \cdot \pi R^{2}$. The only thing holding it back is shear on the cylindrical surface, which is $\tau_w \cdot 2\pi R L$. Set them equal, cancel, and $\tau_w = \Delta P R / 2L$. No assumption about viscosity entered anywhere, which is the key point: Rabinowitsch corrects the shear RATE and never the shear STRESS, because the stress came from a balance of forces that any fluid must satisfy.

Paired with the apparent shear rate from the neighbouring page, this gives one point on a flow curve. Run the same material through the same die at several flow rates and you have a curve; plot it on logarithmic axes and the slope is the power-law index $n$. That is how $n$ is measured, and it is why capillary rheometry is still the standard tool despite being conceptually as simple as pushing melt through a hole and watching the pressure.

The honesty problem here is the entrance, and it is the reason L/D ratio matters. The balance above assumes fully developed flow — that every bit of the measured pressure drop is spent on wall shear down the parallel barrel of the die. It is not. A polymer melt entering a sudden contraction does a great deal of work at the entrance itself: it accelerates, it stretches, it forms recirculating vortices in the corners, and it stores elastic energy. All of that costs pressure, and if you do not remove it, the entrance loss gets charged to your wall shear stress and inflates it. On a short die the entrance loss can be most of the reading.

The fix is Bagley's correction, and it is straightforward if tedious: run the same material at the same shear rate through a set of dies of different lengths but identical radius, plot total pressure drop against $L/D$, and fit a straight line. The line's slope gives the true wall shear stress; the intercept at $L/D = 0$ is the entrance loss, and you subtract it. Some laboratories use an orifice die of nominally zero length to measure the intercept directly. The magnitude of that intercept is itself informative — it is a rough measure of the melt's extensional character, and it is why two grades with the same shear viscosity can behave completely differently in a converging flow.

Why the stress matters in its own right: melt fracture is a stress criterion. Above a critical wall shear stress the extrudate stops coming out smooth. Sharkskin appears first — a fine matte roughness on the surface — then, higher still, the gross helical or bamboo distortion that ruins a profile. For many polyolefins the onset sits in the region of 0.1 to 0.14 MPa, and it is remarkably insensitive to temperature and to molecular weight, which is what marks it out as a stress limit rather than a rate limit. Design a die to stay under it, and if you cannot, that is what processing aids and wider die gaps are for.