Availability, embodied impacts, recycling and substitution

Define what the alternatives must deliver

A functional comparison states what is provided, how much, how well and for how long under stated conditions. For our invented case, both designs must protect the same noncritical indoor device for ten years and permit the same access. They must meet the functional requirements before a climate or cost comparison can choose between them.

Embodied impacts arise from obtaining and processing materials and making a product; their scope depends on the study. A declared factor per kilogram may stop at the factory gate, while another includes later stages. Check boundary, location, energy supply, material condition, recycled-content assumptions and data quality before multiplying by mass. Equal units do not establish equal scope.

Include replacements and repairs

Our figure uses invented climate factors solely to teach arithmetic. Route A needs a 2 kg shell at 8 kg CO₂-equivalent per kilogram; each shell therefore counts 16 kg CO₂e. It lasts an assumed five years, so two shells count 32 kg CO₂e over ten years. Route B needs a 3 kg shell at an invented factor of 6, giving 18 kg CO₂e, plus an assumed repair burden of 3: total 21.

These are partial scenario inventories, not measured footprints or a complete life-cycle assessment. We have excluded transport, use-stage effects and end-of-life effects equally only as an explicit exercise boundary; those stages could alter a real decision. Climate impact alone also does not describe water use, toxicity, resource depletion, working conditions or all ecological effects.

Ask whether the alternative route can actually happen

Recycled content describes the recovered input used in making a product. Recyclability concerns what might be technically recovered later; actual recycling also needs collection, sorting, compatible streams and a buyer or use. A mixed assembly can be technically separable yet rarely separated in practice. Avoid counting the same recovery benefit twice through inconsistent input and end-of-life assumptions.

Reuse preserves a component’s function; material recycling recovers material through further processing. Repair or reuse can avoid production, but needs condition, compatibility and performance checks. Substituting an available material can change dimensions, process route, joints and durability. Check suppliers and lead times using current evidence when making a real procurement decision, and retain alternative routes where appropriate.

Separate an arriving material from a feasible component

Availability means the needed grade, condition, quantity and form can reach the design at the required time, with usable property evidence. A material family in a catalogue does not establish any of these. A substitute delivered sooner may require a larger section, a changed joint or extra processing that consumes the apparent schedule advantage. A quoted delivery date is an uncertain input until it is supported by the actual route.

Make the sequence explicit: screen the functional and delivery constraints, calculate the changed geometry and material demand, then compare costs or impacts over the same service. Keep rejected routes with the reason for rejection. If a requirement or delivery estimate changes, that record shows which option can re-enter rather than forcing the whole decision to start again.

Recalculate the amount that provides the function

Reuse the earlier straight-tie model, not an enclosure bending model. For fixed force F, length L and permitted extension δmax, required area is A=FL/(Eδmax); mass is m=ρAL. A lower modulus E requires more area. Equal density does not then imply equal mass. An impact factor per kilogram must multiply the changed mass needed for that function.

The diagram uses a replacement D with half B’s modulus and equal density. Minimum model mass doubles. A lower factor per kilogram can still produce a greater counted burden per functioning tie. If the required area exceeds the available shape or assembly space, that route is infeasible before any impact ranking. Strength, durability and joining still need their own evidence.

Change the property, delivery and boundary assumptions

Within the stated partial inventory, A totals 2 × 2 kg × 8 kg CO₂e/kg = 32 kg CO₂e. B totals 3 × 6 + 3 = 21 kg CO₂e. Its counted difference is 11 kg CO₂e over the same ten-year function.

If A instead lasts all ten years, only one shell is counted: 16 kg CO₂e. It would then be below B’s assumed 21. Service life is a consequential uncertainty, not a small footnote to a precise factor.

For a separate input batch of 10 kg with 60% recycled content by mass, recovered input is 6 kg and other input 4 kg. That statement says nothing by itself about the future collection rate, grade retention or emissions of this batch.

For the earlier constructed tie, F=1000 N, L=1 m and δmax=0.001 m. B has E=100 GPa and density 5000 kg/m³, requiring 10 mm² and 0.050 kg. Substitute D has E=50 GPa and the same density: A=1000/(50×10⁹×0.001)=20 mm² and mass 0.100 kg. If D is forced into B’s 10 mm² section, its model extension becomes 0.002 m=2 mm, twice the 1 mm limit.

Now set an invented latest assembly-ready date at day 10. B arrives ready at day 14. D stock arrives day 3 and needs two days of assumed processing; its available maximum area is 18 mm². B misses the date, while D fits the date but cannot supply the 20 mm² section. At 18 mm² its predicted extension is 1000/(18×10⁻⁶×50×10⁹)=0.001111 m, about 1.11 mm. Neither route meets all stated constraints; a weighted average of their advantages cannot fix that.

A third constructed route E has the same modulus/density as B and a sufficient 10 mm² section; arrival day 6 plus three processing days makes it ready day 9. If all other requirements were independently satisfied, it would pass these two screens. Those missing strength, joint and service checks remain explicit, so it is a candidate rather than an approved substitute.

For a separate same-boundary production-only comparison, assign B 8 kg CO₂e/kg and D 6 kg CO₂e/kg. If sufficient D section became available, B would count 0.050×8=0.40 kg CO₂e per tie and D 0.100×6=0.60. The lower factor gives a 50% greater result because equal service requires more material. This tie calculation does not change the earlier enclosure figures or mix their factors with these invented ones.

Finally separate recovery stages. In an invented future 10 kg assembly, suppose 6 kg is collected,80% of that collected mass becomes usable recovered output, and all other material is uncollected or lost in processing. Output is 6×0.80=4.8 kg, or 48% of the original 10 kg. That denominator differs from 60% recycled content in a new input batch. No automatic avoided-production credit follows without a stated recovery destination and compatible accounting boundary.

Try the reasoning

1. Another substitute has E 80 GPa and density 4000 kg/m³ for the stated tie. Calculate the minimum model area and mass. Does this approve it?

   Show the answer and reasoning

   Area is 1000/(80×10⁹×0.001)=12.5 mm²; mass 4000×12.5×10⁻⁶×1=0.050 kg. Its E/ρ matches B, so the same mass is consistent. Available section, date, strength, joints and service conditions still need evidence.

2. D is now stocked at 20 mm², but the assembly envelope permits only 18 mm². Does the procurement change solve the substitution?

   Show the answer and reasoning

   No. Its required model area remains 20 mm², beyond the assembly constraint. At 18 mm² the extension remains about 1.11 mm. A larger available piece cannot fit without a separately evaluated design change.

3. Route E arrives day 7 instead of 6, with the same three processing days and a latest ready date day 10. What changes, and what remains unresolved?

   Show the answer and reasoning

   It becomes ready at day 10 and still meets the stated deadline exactly, leaving no time margin. Arrival/processing uncertainty and all other performance screens remain; a planned date is not a guaranteed delivery.

4. For D’s 0.100 kg tie, what same-boundary factor would equal B’s 0.40 kg CO₂e, and why is factor 6 insufficient?

   Show the answer and reasoning

   The break-even factor is 0.40/0.100=4 kg CO₂e/kg. At 6, D counts 0.60, exceeding B by 0.20 or 50%. The comparison depends on the stated production boundary and equal-function model, not factor ranking alone.

5. A report labels the 4.8 kg recovered output as 80% of the original 10 kg assembly and also assumes it guarantees 60% recycled input in the next product. Correct the reasoning.

   Show the answer and reasoning

   80% is the processing yield of the 6 kg collected;4.8/10=48% is output relative to the original assembly. The next product’s input fraction additionally depends on its mass, specification and actual use of that output. Collection, processing yield and recycled input are separate quantities.

What to carry forward

Compare equivalent service and explicit routes. Mass, recycled content and one impact category each reveal only part of the decision.