Single-Screw Drag Flow

A single-screw extruder does not work the way most people first picture it. The screw is not an Archimedes spiral carrying material along like a corkscrew; if the melt stuck to the screw and slipped on the barrel, the screw would simply spin inside a solid slug and nothing would move. What actually happens is the opposite: the melt sticks to the barrel, and relative to the turning screw the barrel surface sweeps across the top of the channel, dragging the melt along it. Unroll the channel flat and it is the textbook problem of viscous drag between a moving plate and a stationary one — Couette flow. Rowell and Finlayson solved it for the helical geometry in Engineering in 1928, and their result is still the equation on this page.

Read the terms. $\pi D N$ is the surface speed of the barrel relative to the screw. The channel is $H$ deep, and in simple drag flow the mean velocity is half the plate speed, which is where the 2 in the denominator comes from. The channel cross-section brings in another factor of the width, which for a shallow channel wrapped round the screw is proportional to $\pi D \sin\varphi$, and the component that actually advances along the axis brings the $\cos\varphi$. Multiply it out and you get $\pi^{2} D^{2} N H \sin\varphi\cos\varphi/2$.

The most useful property of drag flow is what is missing from it: viscosity. There is no $\eta$ anywhere. Drag flow depends on geometry and rotational speed alone, which is why a single-screw extruder's output is so nearly proportional to rpm and why the machine is such a reliable metering device. Change the melt temperature, change to a higher-melt-index grade, change the shear-thinning behaviour — the drag term does not care. Everything that responds to viscosity lives in the pressure-flow term, on the next page.

The helix angle is a weak lever on output and a strong lever on everything else. The term $\sin\varphi\cos\varphi$ is $\tfrac{1}{2}\sin 2\varphi$, which peaks at exactly 0.5 when $\varphi = 45°$. Almost every extrusion screw ever cut runs at about 17.66° — the square-pitch angle, where the lead equals the diameter, and $\tan\varphi = 1/\pi$ — which gives only 0.29, well short of the maximum. Why give up 42 % of the theoretical drag flow? Because the helix angle also sets the pressure flow, the rate at which solid pellets are conveyed and compacted at the feed end, the melting behaviour along the transition, and the shear history the melt collects. Square pitch is where a century of practice settled, and it makes screws easy to specify and to cut. Notice too how flat the function is near its peak: between 30° and 60° the drag term changes by less than 14 %, so getting the angle "wrong" costs very little output.

Two things this equation cannot see, and both of them are the real limit on a working machine. Melting capacity: past a certain rate, the solid bed does not finish melting before it reaches the metering section, and what comes out is unmelt — visible as gels, hard specks, or a rough surface — rather than more kilograms per hour. Shear heating: the melt temperature climbs with screw speed whether the barrel heaters are on or not, because all the mechanical work put into the melt ends up as heat in it. On a heat-sensitive material that is the true ceiling, and an extruder run to the rpm this equation allows will happily make degraded material at a beautiful rate.