9) 4 pts. If you're extruding HDPE, polycarbonate, and polystyrene and want to double the throughput by doubling the screw speed, how (quantitatively) will this affect the pressure drop across the extruder for each of those polymers? How much more power will you need to push this polymer through at double the throughput? You may use the power law factors given in the Viscoelasticity and Rheology module for those polymers. HINT: The equations of interest are in the Processing module, and the screw diameter and K factor are the same for a given polymer system. Power Law Approximation Most polymers follow the empirical power law", which is a good first approximation of stress vs. strain: Polymer Consistency Index Power Law Coefficient Temperature Region, C 180 2.0 x 10 0.41 HDPE LDPE Nylon 66 0.39 160 6.0 x 103 6.0 x 102 0.66 290 T = " PC 6.0 x 102 0.98 300 200 170 PP 7.5 x 103 0.38 PS 2.8 x 104 0.28 PVC 1.7 x 10" 0.26 Source: International Plastics Handbook, Table 3.11 180 Useful when designing equipment for polymer melt processing 60 Scale-Up Laws for SSE If assume Newtonian: If assume power law: (output) Q = K,D'N = 7 = Ky" = (pressure gradient) (power) AP = K2uN II = K D'UN = Q = K ND AP = K'2N" II = K'zDNI+n = OUTPUT In:/sec. Na.9 K's = constants n = "power law" for polymer N = screw rotational speed D= screw diameter t = stress pole tabonete 44 2.4 2 pola propulene 1000 2000 3000 4000 5000000 PRESSURE DEVELOPED (PSH