Protocol for Biopile Construction Treating Contaminated Soils with Petroleum Hydrocarbons

Abstract

When investigating the treatment of contaminated soils, the application of biotreatment is growing rapidly. Factors influencing this rapid growth include that the bioremediation processes are cost-efficient, safe, and nature-based. In the past, thermal, chemical, and physical treatment methods have failed to eliminate the pollution problem because those methods only shift the environmental pollutants to a new environmental phase such as air and water. Bioremediation technology, which leads to degradation of pollutants, may be a lucrative and environmentally beneficial alternative. Two major groups of bioremediation treatment techniques are used: in situ and ex situ remediation. While in situ remediation is more cost-effective, the thoroughness of this method is less effective than the ex situ remediation. Ex situ remediation is less cost-effective but is a more thorough remediation method. This paper presents biopile design settings and example of calculation for design.

Design Biopile Calculation Example for 286 m3 Gasoline-Contaminated Soil with Low Organic Content

• 1. Moisture for biopile

  Assume:

  • Porosity, φ = 30% and initial saturation, S = 20%

  • Desired water content = 25–85%, use 60%

  Therefore:

  • The water needed = 286 × 0.30 × (0.6–0.2) = 34.3 m³ = 34,300 L

• 2. Nutrient requirement for biopile

  • 158 kg spill of gasoline (C₇H₁₆)

  • Nutrient sources: Ammonium sulphate ((NH₄)₂SO₄); trisodium phosphate (Na₃PO₄ · 12H₂O)

    • MW of gasoline = 7 × 12 + 1 × 16 = 100 g mol⁻¹

    • Moles of gasoline = 158 × 10³/100 = 1,580 mol

    • Moles of C = 7 × 1,580 mol = 1.106 × 10⁴ mol

    • Molar ratio C:N:P = 120:10:1

    • Moles of N needed = 10/120 × 1.1 × 10⁴ = 917 mol

    • Moles of ((NH₄)₂SO₄) needed = 917/2 = 458 mol

    • MW of ((NH4)₂SO₄) = (14 + 4) × 2 + 32 + 4 × 16 = 132 g mol⁻¹

    • Mass of ((NH₄)2SO₄) needed = 132 × 460 = 6.1 ×10⁴ g = 61 kg

    • By similar calculation:

    • Mass of (Na₃PO₄ · 12H₂O) needed = 35 kg

• 3. Oxygen requirement for biopile

  $$ {\mathrm{C}}_7{\mathrm{H}}_{16}+22{\mathrm{O}}_2\to 7{\mathrm{C}\mathrm{O}}_2+8{\mathrm{H}}_2\mathrm{O} $$

  • 1 mol (100 g) gasoline requires 22 mol (16 × 2 × 22 = 704 g) O₂

  • Oxygen content of air = 21% by volume = 210,000 ppmv

  • mg L⁻¹ to ppmv:

    $$ \begin{array}{l}\mathrm{ppmv}=\frac{\mathrm{mg}}{\mathrm{L}}\times {10}^3\times \frac{1}{{\mathrm{MW}}_{\mathrm{contaminant}}\ \left[\mathrm{g}\ {\mathrm{mol}}^{-1}\right]}\times 8.314\ \left[\frac{\mathrm{L}\;\mathrm{k}\mathrm{P}\mathrm{a}}{\mathrm{mol}\;\mathrm{K}}\right]\\ {}\kern3.5em \times {T}_{\mathrm{air}}\left[K\right]\times \frac{1}{P_{\mathrm{air}}\left[\mathrm{k}\mathrm{P}\mathrm{a}\right]}\end{array} $$

Oxygen needed for 158 kg spill of gasoline (≅C₇H₁₆)

• 100 g gasoline needs ~704 g O₂.

• 158 kg gasoline × 7 = 1,106 kg O₂ = 1.1× 10⁶ g O₂.

• Water in pile = 286 m³ × 0.30 × 0.6 = 51.5 m³ = 51,500 L.

• At saturation at 20°C and 1 atm (101.325 KPa), dissolved oxygen = 9.2 mg L⁻¹.

• Mass of oxygen in soil moisture = 51,500 L × 9.2 mg L⁻¹ × 0.001 g mg⁻¹ = 473.6 g O₂.

• 473 g O₂ in soil moisture is much less than 1.1 × 10⁶ g O₂ required.

• At 0.28 g L⁻¹ air, air requirement is 1.1 × 10⁶ g/0.28 g L⁻¹ = 3.93 × 10⁶ L = 3,930 m³.

• Daily air requirement for a duration of 3 months is 3,930 m³/90 day = 43.7 m³/day.

• Air void volume in pile = 286 m³ × 0.30 × 0.4 = 34 m³.

• Need to daily exchange 43.7/34 = 1.3 void volumes to fulfil oxygen requirement (Note: this rate should be compared to passive aeration and check if the latter is sufficient. In the case that it is not, the pump should provide at least twice the air flow as not all the air will be absorbed by the biopile).

Example of calculation of the volumetric composition of a biopile [39]

• Height: 2.5 m

• Upper width: 1.24 m

• Lower width: 5 m

• Length: 80 m

• Total volume: 624 m³

• Total mass: 7.49 × 10⁵ kg

• Area: ((5 + 1.24) × 2.5)/2 = 7.8 m²

• Volume: 7.8 × 80 = 624 m³

• (i) Total volume: 624 m³

• (ii) Total mass: 7.49 × 10⁵ kg = 749 tonnes (assuming a bulk density of 1,200 kg m⁻³)

• 1. Water: 80% of field capacity (v/v), field capacity of sandy loam (SL) soil: 20%

  • Water volume = 624 m³ × 0.16 = 99.84 m³

• 2. At 50% porosity, water + air = 50%

  • Air volume = 312–99.84 m³ = 212.16 m³

• 3. 2% organic matter, 48% inorganic matter

  • Volume of 2% organic matter = 624 m³ × 0.02 = 12.48 m³

  • Volume of 48% inorganic matter = 624 m³ × 0.48 = 299.52 m³

• 4. TPH _t = 0: 80,000 mg oil/kg soil = 0.08 kg oil/kg soil,

  • Density of heavy oil (NAPL): 0.97 g/cm³,

  • NAPL volume = 0.08 kg oil/kg soil = 61.72 m³

  • Assume NAPL shares air volume with air

  • Air volume = 212.16–61.72 m³ = 150.44 m³

So overall:

• Volume of soil = 312 m³

• Volume of water 100 m³

• Volume NAPL = 62 m³

• Volume of air = 150 m³