Wastewater Computations

Calculations and Parameters in Wastewater Treatment

Hydraulic and Organic Loading

• I_m = 3.28 \text{ ft}

• 10001 = I_m^3

• \text{Vol} = 36 - 26.9895

• PE = \text{Population Equivalent}

• PE = \frac{\text{BOD loading (lbs/day)}}{\text{0.17 lbs BOD/day/person}}

• Loading factor (unit # if not mentioned)

• PE = \frac{\text{loading}}{\text{mg/L}}

• PE = \text{Flow rate (MGD)} \times \text{BOD} \times 8.34

• Example calculation:

  • PE = \frac{303.576 \text{ lbs/day}}{0.17 \text{ lbs BOD/day}} = 1785.74 \approx 1786

• Another Example:

  • If the BOD concentration is 1019.18 mg/L and flow rate is 0.054 MGD:

    • \text{BOD loading} = 1019.18 \text{ mg/L} \times 0.054 \text{ MGD} \times 8.34 = 459.21 \text{ lbs BOD}

    • \text{Population loading} = \frac{459.21}{0.17} = 2701.23

• Population loading calculation:

  • Flow rate = 1.55 MGD

    • \text{BOD loading} = \text{Concentration} \times Q = 210 \text{ mg/L} \times 1.55 \text{ MGD} \times 8.34 = 2714.67 \text{ lbs BOD}

    • \text{Population} = \frac{2714.67}{0.17} = 15968.65

Population Equivalent (PE) Calculations

• PE = \frac{\text{BOD Loading}}{0.17 \text{ lbs BOD/day/person}}

• Example:

  • PE = \frac{326.903}{0.17} \approx 1923

• Another example:

  • PE = \frac{441.2}{0.17} = 2595.29

• If BOD loading is given as Q \times 8.34 \times \text{BOD}

Activated Sludge Parameters

• Key parameters:

  • Concentration

  • Flow rate (Q)

  • Food to microorganism ratio (F/M)

  • Volume

• Waste Activated Sludge (WAS)

• Effluent flow rate

• Q = 500 \frac{m^3}{day}

• \text{BOD loading} = \text{Concentration} \times Q

• \text{BOD loading} = 1200 \times Q = 600

• BOD = 50

• F/M ratio:

  • E = 0.56 \frac{1}{day}

  • 50 \frac{m^3}{day}

• Mixed Liquor Suspended Solids (MLSS) = 30

• Sludge Volume = 58.74 = 59 mL

• Mixed Liquor Retention Time (MLRT) = volume / Q

• Mean Cell Residence Time (MCRT) = \frac{\text{volume}}{\text{waste}} = 7 \text{ days}

• MCRT = \frac{V}{Q_w}

• 3000 \frac{mg}{L}

• TSS = 19.29

• 7500 \frac{mg}{L}

Area and Flow Calculations

• Area calculation:

  • Area = (20)^2 = 400

• Q_{ave} = 14,000 \frac{m^3}{day}

• Peak flow:

  • Q{peak} = 1.75 \times Q{ave} = 1.75 \times 14,000 \frac{m^3}{day} = 24,500 \frac{m^3}{day}

• Average flow BOD = 190 mg/L

• Peak flow BOD = 225 mg/L

• Suspended Solids (SS) = 365 mg/L

• Effluent BOD = 6 mg/L

• Mass of solids calculations:

  • Flow = 10 \frac{m^3}{day}

  • Average flow:

    • Mass = 14,000

  • Average flow BOD removal:

    • \text{Removal} = (190-6) \times 14,000 \times (10^{-3}) = 2520 \frac{kg}{day}

    • \text{BOD removal} = \frac{BOD{in} - BOD{eff}}{BOD_{in}} = \frac{190-6}{190} = 0.9684 \approx 96.84 \%

  • Peak flow SS removal:

    • \text{Removal} = (365-30) \times (24,500) \times (10^{-3}) = 8207.5 \frac{kg}{day}

• If including recirculation, follow equations. If not, do not include.

Given Parameters and Calculations

• Hydraulic Loading Rate (HLR) calculation:

  • Given: Q = 808 {MGD}

  • Q_R = 0.66 \times Q = 0.66 \times 808 = 533.28

  • HLR = \frac{533.28 \text{ MGD}}{2.15 \text{ acres}} \times 1,000,000 = 248.28 \frac{GPD}{ft^2}

• Organic Loading Rate (OLR) Calculation:

  • Given: 10,000 gallons per acre (gpa) + 2.45 MGD, BOD = 210 mg/L

    • \text{BOD loading} = \text{Concentration} \times \text{Flow rate} \times 8.34

    • \text{BOD loading} = 210 \frac{mg}{L} \times 2.45 \text{ MGD} \times 8.34 = 4290.93 \text{ lbs BOD}

    • Volume Calculation:

      • Volume = A*h

    • 4290.93 \frac{lbs BOD}{day}

    • Volume = 25132.74 ft3

    • OLR = \frac{4290.93 \frac{lbs BOD}{day}}{25132.74 ft^3} \times 1000

      • OLR = 170.74 \frac{lbs BOD}{day * 1000 ft^3}

    • OLR is extracted. Using volume = area * depth. If OLR is given in lbs BOD/day / 1000 cu.ft