Page 54 - Materials Chemistry, Second Edition

38 2 Goal and Scope Definition

the methodical approach must not be interpreted as a ‘comparison by ecological
criteria’!
On the basis of the defined fU, the next step is to determine how much floor
covering is needed for the fulfillment of this function: A thick floor covering (if not
foam material is concerned) will, by its higher weight per unit area, usually cause
higher resource consumption and emissions than a thinner one. The weight of
−3
3 mm plastic A (density = 1.2 g cm ) amounts to
3mm × 1.2gcm −3 = 3.6kgm −2
For illustration purposes, in the following cumulative energy demand (CED) will
be applied for both variants. The CED is the sum of the total energy demand over
the entire life cycle regarded (see Section 3.2.2). The CED for the production of
plastic A as a European average may amount to 60 MJ kg −1 (under neglect of fillers,
floor covering production and disposal). Considering the energy demand for the
production of plastic A, the CED amounts to
CED ≈ 3.6kgm −2 × 60MJkg −1 = 216MJm −2
A thinner covering of plastic B with d = 2mm and = 0.9 g cm −3 weighs only

2mm × 0.9gcm −3 = 1.8kgm −2
so that, despite a higher CED for the production of plastic B of approximately
90 MJ kg −1 (same neglects as with plastic A), the CED amounts to

CED ≈ 1.8kgm −2 × 90MJkg −1 = 162MJm −2
As is often true, the lighter product would thus do much better, at least concerning
the sum parameter CED, than its heavier counterpart. This is, however, only valid
as long as an equal use time is presumed. If, for example, the thicker quality has a
longer use time, which is very probable, the result will differ. (Life times here are
hypothetical and do not correspond to any real product performance):

Assumption:
Life time of plastic A (thick): 30 a
Life time of plastic B (thin): 15 a.

It can be directly deduced from the above example that while two thin coverings
are needed (2 mm plastic B) for use over 30 a, only one of the thick covering (3 mm
plastic A) is needed. Because the fU must relate to area and time (here: 30 a), the
following reference flows result for the two variants, which must form the basis of
data acquisition:
Plastic A ∶ 3.6kgm −2 × 1 ⇒ 3.6kgfU −1
Plastic B ∶ 1.8kgm −2 × 2 ⇒ 3.6kgfU −1
The reference flow is equal to the mass of the product that corresponds to the fU
and is thus the basis for further work in LCA (see Chapter 3).