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Exercise 2

Exercise 2. REM 610 January 20, 2010. 1. Dioxin in drinking water. Mass of dioxin: 48% Sediment 52% Soil 1% Water. Concentration of dioxin. 1.15E-05 Soil 1.48E-10 Water 6.4E-14 Air. 4.47E-05 Biota 2.29E-05 Sediment 2.29E-05 Susp. Sed. Conclusions:.

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Exercise 2

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  1. Exercise 2 REM 610 January 20, 2010

  2. 1. Dioxin in drinking water • Mass of dioxin: 48% Sediment 52% Soil 1% Water

  3. Concentration of dioxin • 1.15E-05 Soil • 1.48E-10 Water • 6.4E-14 Air • 4.47E-05 Biota • 2.29E-05 Sediment • 2.29E-05 Susp. Sed.

  4. Conclusions: • Dioxin partitions into organic media, relatively low concentrations and masses are present in water • Highest concentrations are in biota and organic matter • Absolute concentrations are irrelevant – comparison of media is useful • Exposure from media other than water (e.g., fish consumption) is a more significant route of intake (assuming fish are consumed from the local area)

  5. Limitations: • Assumes equilibrium, which may not be the case (e.g., ignores time lags) • Ignores biomagnification • No metabolism / loss processes • Assumes homogeneity, constant lipid and OC fractions • Water treatment may remove dioxins in drinking water

  6. 2. 2,4-D in Ground Water • Mass of 2,4-D: 91% Water 5% Soil 4% Sediment

  7. Concentration of 2,4-D • 1.03E-06 Soil • 1.3E-07 Water • 1.34E-15 Air • 4.03E-06 Biota • 2.06E-06 Sediment • 2.06E-06 Susp. Sed.

  8. Conclusions: • Majority of mass is in water, small amounts in soil and sediment • Concentrations are highest in biota and organic matter • Farmers are justified in their concern over transport and contamination of the aquifer with 2-4,D • Limitations - same as Question #1

  9. 3. CEPA Criteria for Dioxin: Water Quality Guideline: Tissue Residue Guideline “Cf” = 20 ng/kg Goal: Relate fish tissue residue to concentrations in other media Kbw= lipid fraction * Kow * density = 0.048 * 10E6.8 * 1 kg/L = 302860 kg/L Cw = Cf / Kbw = 20 ng/kg / 302860 kg/L WQG = 6.60372E-05 ng/L

  10. 3. CEPA Criteria for Dioxin cont’d Sediment Quality Guideline: Ksw= organic carbon fraction * 0.41 Kow * density = 0.04 * 0.41 * 10E6.8 * 1.5 kg/L = 155215 kg/L Cs = Cw * Ksw = 6.6 E-05 ng/L * 155215 kg/L SQG = 10.25 ng/kg or ... Ksb = Ksw / Kbw = 155215 kg/L / 302860 kg/L = 0.5125 Cs = 0.5125 * 20 ng/kg SQG = 10.25 ng/kg

  11. 3. CEPA Criteria for Catechol: Water Quality Guideline: Tissue Residue Guideline “Cf” = 1 g/kg Kbw= lipid fraction * Kow * density = 0.048 * 10E1.1 * 1 = 0.6043 = Cf/Cw Cw = 1 g/kg / 0.6043 WQG = 1.655 g/kg

  12. 3. CEPA Criteria for Catechol cont’d Sediment Quality Guideline: Ksw= Cs/Cw = 0.309696 kg/L Cs = 0.309696 * 1.655 g/kg SQG = 0.5125g/kg or ... Ksb = Ksw / Kbw = 0.309696 / 0.6043 = 0.5125 Cs = 0.5125 * 1 g/kg SQG = 0.5125 g/kg

  13. Assumptions: • Assumes equilibrium – which may or may not be reached • Ignores time lags / rate controlling reactions • Ignores biomagnification (Chemical concentration in biota is completely explained by water uptake from the gills) • Assumes constant lipid content (4.8%) & biota density 1.0 kg/L • Assumes constant Organic Carbon fraction (4%) and sediment density 1.5 kg/L

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