Ecological sustainability assessment of building glass industry in China based on the point of view of raw material emergy and chemical composition

Abstract

Severe environmental destruction is being driven by excessive resource consumption in the industrial production process. Therefore, there is a necessity to evaluate the sustainability of the production system. In this study, the emergy method has been adopted to assess the flat building glass production in China based on raw material and chemical composition. A series of problems including key contributors, primary sustainable indexes, unit emergy value (UEVs), sensitivity ratios, and waste impact was studied. The results illustrate that (1) the nonrenewable resources and imported resources showed the dominant impacts. (2) Silica sand and sandstone were the foremost items for the raw material angle emergy. (3) Excessive EIR, serious ELR, and tiny ESI were the main contributors to the unsustainability of the evaluated system. (4) Four UEVs were revealed, which are 1.69E + 12sej/kg, 1.80E + 12sej/kg, 1.60E + 12sej/kg, and 1.71E + 12sej/kg, respectively. (5) The nonrenewable resources showed the biggest fluctuation (7.09%), followed by imported resources (1.62%) in view of the raw material perspective; for the chemical composition, the nonrenewable resources were 7.15%, and imported resources were 1.49%, respectively. (6) Waste gas emissions were found as the major emergy contributor to the economic loss. Furthermore, positive solutions were discussed for improving the sustainability of glass production, including the proportion increase of renewable energy, recycling material replacement, and promotion of energy-saving equipment.

Appendix

Composition of 100 kg glass mixture calculated process:

• (a) Dosage calculation of Silica sand and sandstone

Assuming the amount of silica sand and sandstone are M_{Silica sand} and M_sandstone, respectively.

$$0.89{M}_{\text{Silica sand}}+0.98{M}_{\text{sandstone}}=72.02\mathrm{kg}$$

(9)

$$0.052{M}_{\text{Silica sand}}+0.006{M}_{\text{sandstone}}=2.04$$

(10)

Based on formulas (9) and (10), the results of M_{Silica sand} and M_sandstone can be calculated:

$$\begin{array}{cc}{M}_{\text{Silica sand}}=34.96\mathrm{kg};& {M}_{\text{Silica sand}}=41.02\mathrm{kg}\end{array}$$

• (b) Dosage calculation of dolomite and Magnesite

Assuming the amount of silica sand and sandstone are M_dolomite and M_Magnesite, respectively.

$$0.33{M}_{\text{dolomite}}+0.0074{M}_{\text{Magnesite}}=5.41\mathrm{kg}$$

(11)

$$0.21{M}_{\text{dolomite}}+0.47{M}_{\text{Magnesite}}=4.22\mathrm{kg}$$

(12)

Based on formulas (11) and (12), the results of M_dolomite and M_Magnesite can be calculated:

$$\begin{array}{cc}{M}_{\text{dolomite}}=9.03 \mathrm{kg};& {M}_{\text{Magnesite}}\end{array}=0.66 \mathrm{kg}$$

• (iii) Dosage calculation of soda ash

Assuming the amount of soda ash is M_{soda ash}.

Based on formulas (13), the result of M_{soda ash} can be calculated:

$${M}_{\text{Sodium carbonate}}=14.5-2.18-1.28-0.08=10.96\mathrm{kg}$$

(13)

• (iv) Dosage calculation of Glauber

Assuming the amount of soda ash is M_glauber.

Based on formulas (14), the result of M_glauber can be calculated:

$$15\%=\frac{\mathrm{input}\;\mathrm{of}\;\mathrm{Glauber}\left(N{\mathrm a}_2O\right)}{\mathrm{input}\;\mathrm{of}\;\mathrm{Glauber}\;\left(N{\mathrm a}_2O\right)+\mathrm i\mathrm n\mathrm p\mathrm u\mathrm t\;\mathrm o\mathrm f\;\mathrm{Sodaash}\left(N{\mathrm a}_2O\right)}$$

(14)

Based on formulas (14), the result of M_glauber can be calculated:

$${M}_{\mathrm{glauber}}=2.33 \mathrm{kg}$$

• (e) Dosage calculation of fluorite

Assuming the amount of Fluorite is M_fluorite.

Based on formulas (15), the result of M_fluorite can be calculated:

$$0.85\%=\frac{{M}_{\text{fluorite}}\times C\mathrm{a}{F}_{2}}{\mathrm{Total\;amount\;of\;}\mathrm{raw\;}{\text{materials}}}=\frac{{M}_{\text{fluorite}}\times 0.7028}{121.21}$$

(15)

$${M}_{\text{fluorite}}=1.06 \mathrm{kg}$$

wherein the content of all oxide are:

$$\begin{array}{c}Si{O}_{2}=1.47\times 24.62\%=0.36kg\\ {Al}_{2}{O}_{3}=1.47\times 2.08\%=0.03kg\\ {Fe}_{2}{O}_{3}=1.47\times 0.43\%=0.01kg\\ CaO=1.47\times 51.56\%=0.76kg\end{array}$$

Because the reaction of Si₂O and CaF₂ will volatilize, volatilization needs to be calculated, as follows:

Assuming the amount of Fluorite is M_{volatilization}.

Based on (16) and (17), the result of M_{volatilization} can be calculated:

$${Si}_{2}O+2{CaF}_{2}={SiF}_{4}\uparrow +2CaO$$

(16)

$${M}_{\text{volatilization}}=\frac{60.09\times 1.47\times 70.28\times 30\%\times 1}{2\times 78.08}=0.12kg$$

(17)

The final usage of SiO₂ is 0.24 kg;

• (f) Dosage calculation of coal

Assuming the amount of Fluorite is M_coal.

Based on (18), the result of M_coal can be calculated:

$$4.7\%=\frac{{M}_{\text{coal}}\times 0.8411}{5.24\times 0.9503}$$

(18)

$${M}_{\text{coal}}=0.27\mathrm{kg}$$

• (g) Dosage calculation of water

According to the demand, the proportion of water in the mixture is 4% and assuming the amount of fluorite is M_water.

Based on (19), the result of M_water can be calculated:

$${M}_{\text{water}}=\frac{960.29}{1-4\%}-988.34\approx 12\mathrm{kg}$$

(19)