Plastics Engineering - May 2014 - (Page 38)

Process Engineering for Recycled Polymers ____________________ Figure 4: Simple closed-loop recycling when the reprocessing stage produces waste materials (reprinted with permission of the publisher from Brandrup et al.3). we consider closed-loop recycling with losses as shown schematically in Figure 4. Unit operation 1, or stage 1, in this figure represents the conversion of raw material to product (e.g., polyethylene milk bottles). stages 2 and 3 represent the consumer and the disposal phase, respectively. stage 4 is the reprocessing operation for converting the waste back into useful product. consider that a single mass, m, of 1 kg of product is passed initially to the consumer. Our goal is to determine with an infinite number of recycles how much material is passed by the consumer. a fraction f of the material leaving the consumer is now fed into the recycling loop, and the recycling loop is responsible for a loss of a fraction F of the material passing through the reprocessing stage 4. the amount of material returned to the main production sequence will be of mass mf(1 − F). the amount of material sent to the disposal stage after the first pass consists of the sum of the material disposed of directly after consumer use, m(1 − f), plus the amount lost from the reprocessing step, mFf (total = m((1 − f) + fF) or = m(1 − f(1 − F))). the amount of mass that passes the consumer after the first recycle is reduced to mp1 = mf (1 − F) (Eq. 1) after a second recycle the mass of original product, mp2, passing the consumer will be reduced to mp2 = mf2(1 − F)2 38 (Eq. 2) and so on. if the material is continually recycled under the same conditions, then after a large number of cycles the total mass of the original product that will have passed the consumer, Mp, is Mp = m + mf(1 − F) + mf2(1 − F)2 + mf3(1 − F)3 + · · · (Eq. 3) summing Eq. 3, assuming there is an infinite number of terms, gives Mp = m / (1 − f(1 − F)) (Eq. 4) For example, if there is a loss of 20% of the recycled materials (i.e., F = 0.2) and the recycle rate f is 50%, then the product flow past the consumer will be 1.67 kg. From this equation it is obvious that F should be as small as possible if we are to minimize the amount of raw material used. this is further illustrated by referring to Figure 4. the amount of raw material required from operation 1 to provide a mass m to the customer increases as F increases. We are primarily interested in the amount of energy that is required to execute a particular recycling scheme. Following Figure 4, the energies per unit mass associated with each operation are, respectively, E1, E2, E3, and E4. the total energy requirement for the overall system, Es, is the sum of the energies for each operation: Es = E1m(1 − f (1 − F)) + E2m + E3m(1 − f (1 − F)) + E4mf (1 − F) | Plastics EnginEEring | MaY 2014 | www.4spe.org | www.plasticsengineering.org (Eq. 5) http://www.4spe.org http://www.plasticsengineering.org

Plastics Engineering - May 2014

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