There is no ecoinvent data point for zinc coated steel, so zinc coating needs to be applied in addition to the steel input(s).

The data point we use for this is *Zinc coating, coils (Reference product: zinc coat, coils )*.

This data point has a unit of m2. If we know the m2 of steel used, we can use the g/m2 way of calculation, using the Zinc coating thicknesses defined according to EN10346:2015 below. This method will require the steel's thickness.

Min. total coating mass, both surfaces (g/m2) * |
Guidance value for coating thickness per surface, typically (μm) |

100 |
7 |

140 | 10 |

180 | 13 |

200 | 14 |

225 | 16 |

275 | 20 |

350 | 25 |

450 | 32 |

600 | 42 |

** in triple spot test*

If we only know the weight and the thickness of the coated product being used, then the dataset's description of 64 m2/t can be used. This would mean that 64 m2 of zinc is required for 1 tonne of steel.

As an example, let’s assume your zinc coated steel weighs 2.5 kg.

Then, the surface area to be coated = 2.5 kg * (64 m2 / 1000 kg) = 0.16 m2

The above value cannot be directly considered as the surface area to be coated with zinc as the thickness of coating and steel plate are not considered. This is an important point because for the same weight of steel, if the thickness changes, the surface area will correspondingly change (i.e., a thin steel sheet and a thick steel plate can still have the same weight by varying the surface area).

The ecoinvent datacard’s description mentions 4.2 mm as the thickness of the coated steel plate, while the thickness of coating being 20-45 microns (μm). We can use an average of 32.5 μm as the reference for our calculations.

Let’s assume 0.4 mm as the thickness of your coated product, including a 20 μm thick zinc layer on both sides.

To calculate the required surface area to be coated based on the ecoinvent data, let’s start with the above assumption:

Mass (2.5 kg coated product, 4.2 mm thickness) = Mass (2.5 kg coated product, 0.4 mm thickness) . . . . . . . . eq (1)

As we know, Mass = Density * Volume.

Hence, eq (1) can be rewritten as

(Density * Volume) (2.5 kg coated product, 4.2 mm thickness) = (Density * Volume) (2.5 kg coated product, 0.4 mm thickness)

As **density** remains the same, and **volume** = surface area (SA) * thickness (T), the above can be written as

(SA * T) (2.5 kg coated product, 4.2 mm thickness) = (SA * T) (2.5 kg coated product, 0.4 mm thickness)

Expanding on the above, we get

(2.5 * 0.064) m2 * 4.2 mm = SA (0.4 mm thickness) * 0.4 mm

- SA (0.4 mm thickness) = (2.5 * 0.064) m2 * 4.2 / 0.4

= 1.68 m2 of surface area to be coated.

Therefore, the weight of 1.68 m2 of zinc = Surface area * Coating Thickness * Density

= 1.68 m2 * 20 e-6 m * 7135 kg/m3

= 0.240 kg of zinc.

So, the weight of steel without zinc = 2.5 kg - 0.240 kg

= 2.26 kg of steel.

The mass/unit value was obtained as follows:

- 0.240 kg of zinc / 1.68 m2 of zinc = 0.1427 kg/m2 of zinc

**The attached excel calculator will help you with all the above calculations for your coated product. **

If you have any questions, do reach out to us at support@oneclicklca.com

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