Page 116 - Materials Chemistry, Second Edition

100 3 Life Cycle Inventory Analysis

(Rough) diamond
2 g
Diamond mine
Unproductive rocks
(usable for road construction)
100 t
Figure 3.13 Mass ratios in the unit process diamond mine.

5. Allocation per mass.
This is the oldest allocation rule and most commonly used with multi-output
processes (see Section 3.3.2.1). It appears to be scientific without being so in
reality.
As LCA is primarily based on mass flow analysis, the allocation proportional
to mass lends itself as an arbitrary though simple and universal rule. Has the
allocation problem thereby been solved? Unfortunately not. Limitations of the
allocation per mass can best be demonstrated by the so-called diamond paradox
(Figure 3.13). Typical exploitable concentrations for diamonds are around
10 ct (carat) per 100 t rocks. With 1 ct = 200 mg this results to 2 g diamond
per 100 t rocks (0.02 ppm). There are also higher values, but lower ones are
also workable. 88) In this fictitious example it is assumed that for the rocks a
(small) revenue is made, for example, as material for road construction. In an
allocation per mass nearly all of the environmental load would be allocated to
rock material (unproductive rocks), which is obviously unreasonable. Nobody
would operate complex mines just to extract material used as underground for
roads! Therefore, the loads have to be allocated, largely or even exclusively, to the
product diamond, for the exploitation of which the technical effort was made.
A pragmatic solution to the problem would be to allocate the unproductive
rocks as waste not disposed but with market value providing some revenue.
Waste is not considered as co-product and thereby not allocated with loads. The
question arises as to where the border line between waste and co-product is.
An alternative and to date preferred solution in such cases is the allocation
89)
on economic value:
6. Allocation on the economic value of the products, approximated by price.This
allocation is also primarily an allocation per mass although averaged by
economic value (measured by price). An allocation per price that can be
obtained for the products A, B, C, … certainly provides a solution to the
diamond problem because the prices per mass differs by orders of magnitude.
If in the example of Figure 3.10 product A costs 20 ¤ kg −1 (700 kg thus 14 000 ¤)
and B costs 5 ¤ kg −1 (300 kg thus 1500 ¤) cost, the following weighting of
masses per price results:
−1
20[¤kg ]
Weighting factor A = 700[kg]× = 0.903
15500[¤]
88) Pohl, 1992.
89) Guin´ ee et al., 2002; Guin´ ee, Heijungs and Huppes, 2004.