Question

A single screw extruder is operating at a rotational speed of 2 revolutions per second for the extrusion of a Newtonian polymer under open-discharge conditions (in absence of a die, the pressure drop, \( \Delta p = 0 \)). The extruder has a screw diameter, \( D = 5 \, \text{cm} \), a channel depth, \( H = 0.4 \, \text{cm} \), distance between flights, \( W = 1 \, \text{cm} \), and a helix angle, \( \theta = 20^\circ \). Assume the value of \( \pi = 3.14 \). The volumetric flow rate (rounded off to 2 decimal places) is \(\underline{\hspace{2cm}}\) cm³/s.

Hint:

To calculate the volumetric flow rate in a screw extruder, use the formula involving the screw diameter, channel depth, rotational speed, and helix angle.

Answer 1

Correct Answer: 5

The volumetric flow rate \( Q \) for a single screw extruder is given by the equation:
\[ Q = \frac{\pi D^2 H N}{4} \times \cos(\theta) \] where:
- \( D = 5 \, \text{cm} \),
- \( H = 0.4 \, \text{cm} \),
- \( N = 2 \, \text{rev/s} \),
- \( \theta = 20^\circ \).
Substitute the values:
\[ Q = \frac{\pi (5)^2 (0.4) (2)}{4} \times \cos(20^\circ) \] \[ Q = \frac{3.14 \times 25 \times 0.4 \times 2}{4} \times 0.9397 \] \[ Q \approx \frac{62.8}{4} \times 0.9397 = 15.7 \times 0.9397 \approx 14.74 \, \text{cm}^3/\text{s}. \] Thus, the volumetric flow rate is approximately 14.74 cm\textsuperscript{3}/s.