Question

An injection molded polyethylene part has a dimension of 25 mm. A new material, polycarbonate, is used in the same mold. What is the expected corresponding dimension of the polycarbonate molding? show calculations

Answer 1

Given that L=25 mm

outer dimension of the mold by shrinkage(s)

L_c= L+LS+LS²

^{The Shrinkage value for Polyethylene is 0.015 to 0.030 mm/mm =0.0225}

^{Therefore L_c = 25+(25*0.0225)+(25*(0.0225)^2) = 25.575 mm}

^{When Poly carbonate is used in the same mold}

^{Outer dimension of the mold by shrinkage is L= L_c/(1+S+S^2)}

^{For Poly carbonate Shrinkage(s)= 0.006+0.008 mm/mm =0.007 mm/mm}

^{L= 25.575 / (1+0.007+0.007^2) = 25.39598 mm}