18.3 Treatment & Wastewater Process Math Workshop

CT and Log Reduction

$CT = \text{Residual (mg/L)} \times T_{10} \text{ (min)} \qquad T_{10} = \frac{\text{Volume}}{\text{Flow}} \times \text{Baffling Factor}$

$CT \text{ Ratio} = \frac{CT_{\text{achieved}}}{CT_{\text{required}}} \qquad \text{Log Inactivation} = CT\text{ Ratio} \times \text{Log required for that } CT \text{ table value}$

Problem 1. A 900,000-gallon clearwell operates at 3.6 MGD with superior baffling (0.7) and a free chlorine residual of 1.4 mg/L. The required CT for 0.5-log Giardia at this pH and temperature is 43 mg-min/L.

Flow = 3.6 x 694.4 = 2,500 gpm DT = 900,000 / 2,500 = 360 min T10 = 360 x 0.7 = 252 min CT achieved = 1.4 x 252 = 352.8 mg-min/L CT ratio = 352.8 / 43 = 8.2 - far in excess of the requirement, so the plant has substantial margin.

Problem 2. The same plant at peak flow of 9.0 MGD with residual dropping to 0.9 mg/L.

Flow = 6,250 gpm; DT = 144 min; T10 = 100.8 min CT = 0.9 x 100.8 = 90.7 mg-min/L; ratio = 90.7 / 43 = 2.1 - still compliant but the margin has fallen by a factor of four. CT compliance is a peak-flow problem, not an average-flow problem.

Log Reduction Arithmetic

$\text{Total Log} = \text{Log Removal} + \text{Log Inactivation}$

Problem 3. A conventional plant earns 2.5-log Giardia removal from filtration and achieves a CT ratio giving 1.2-log inactivation. Total?

2.5 + 1.2 = 3.7-log, which exceeds the 3-log SWTR requirement with margin.

Problem 4. A direct filtration plant earns only 2.0-log removal. How much inactivation is needed for compliance?

3.0 − 2.0 = 1.0-log from disinfection, twice the conventional plant's 0.5-log burden.

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Loading Rates

$\text{Surface Overflow Rate (gpd/ft}^2) = \frac{\text{Flow (gpd)}}{\text{Surface Area (ft}^2)}$ $\text{Weir Overflow Rate (gpd/ft)} = \frac{\text{Flow (gpd)}}{\text{Weir Length (ft)}}$ $\text{Filtration Rate (gpm/ft}^2) = \frac{\text{Flow (gpm)}}{\text{Filter Area (ft}^2)}$ $\text{Solids Loading Rate (lb/day/ft}^2) = \frac{\text{MLSS lb/day applied}}{\text{Clarifier Area (ft}^2)}$

Problem 5. A circular clarifier 75 ft in diameter treats 3.2 MGD. Its weir is a full-circumference launder.

Area = 0.785 x 75² = 4,416 ft² SOR = 3,200,000 / 4,416 = 725 gpd/ft² Weir length = 3.14 x 75 = 235.5 ft WOR = 3,200,000 / 235.5 = 13,588 gpd/ft

Problem 6. A dual-media filter is 22 ft x 18 ft and receives 1,150 gpm.

Area = 396 ft² Rate = 1,150 / 396 = 2.90 gpm/ft²

Problem 7. Unit filter run volume (UFRV) for that filter over a 62-hour run:

Volume = 1,150 gpm x 62 x 60 = 4,278,000 gal UFRV = 4,278,000 / 396 = 10,803 gal/ft² - a healthy run; below about 5,000 gal/ft² suggests premature terminal head loss or breakthrough.

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Percent Removal

$\% \text{ Removal} = \frac{\text{In} - \text{Out}}{\text{In}} \times 100$

Problem 8. Influent BOD 240 mg/L, primary effluent 156 mg/L, final effluent 12 mg/L.

Primary removal = (240 − 156) / 240 x 100 = 35.0 percent Overall removal = (240 − 12) / 240 x 100 = 95.0 percent Secondary removal (of what reached it) = (156 − 12) / 156 x 100 = 92.3 percent

Note that these three numbers answer three different questions. Read the stem carefully to know which one is being asked.

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Activated Sludge Process Control

$\text{F/M} = \frac{\text{BOD applied (lb/day)}}{\text{MLVSS under aeration (lb)}}$

$\text{MCRT (days)} = \frac{\text{lb MLSS in the system}}{\text{lb WAS/day} + \text{lb effluent TSS/day}}$

$\text{SVI} = \frac{\text{Settled Volume (mL/L)} \times 1{,}000}{\text{MLSS (mg/L)}}$

$\text{Sludge Age (days)} = \frac{\text{lb MLSS in aeration}}{\text{lb TSS entering per day}}$

Problem 9. Flow 5.2 MGD, influent BOD 185 mg/L, aeration volume 2.6 MG, MLSS 2,900 mg/L, volatile fraction 0.78.

BOD applied = 5.2 x 185 x 8.34 = 8,023 lb/day MLVSS = 2,900 x 0.78 = 2,262 mg/L MLVSS mass = 2.6 x 2,262 x 8.34 = 49,050 lb F/M = 8,023 / 49,050 = 0.164 day⁻¹ - conventional activated sludge range

Problem 10. Same plant. Aeration MLSS mass 62,900 lb (using total MLSS), WAS at 0.062 MGD and 7,400 mg/L, effluent TSS 9 mg/L.

WAS solids = 0.062 x 7,400 x 8.34 = 3,827 lb/day Effluent solids = 5.2 x 9 x 8.34 = 390 lb/day MCRT = 62,900 / (3,827 + 390) = 62,900 / 4,217 = 14.9 days

Problem 11. A 30-minute settleometer shows 310 mL/L at an MLSS of 2,900 mg/L.

SVI = (310 x 1,000) / 2,900 = 107 mL/g - good settling. Above about 150 suggests filamentous bulking; below about 70 suggests pin floc and an old sludge.

Problem 12. The plant wants to reduce MCRT from 14.9 to 11 days. What WAS mass is needed?

Target total solids wasted = 62,900 / 11 = 5,718 lb/day Less effluent solids of 390 = 5,328 lb/day from WAS WAS flow = 5,328 / (7,400 x 8.34) = 5,328 / 61.7 = 0.0863 MGD = 86,300 gpd = 60 gpm

[!TIP] Change wasting gradually. Moving MCRT from 14.9 to 11 days in one step shocks the biomass. Standard practice is to change the wasting rate by no more than about 10 to 15 percent per day and let the system respond over one to two sludge ages.

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Solids, Digesters, and Dewatering

$\text{lb solids} = \text{Volume (MG)} \times \text{Concentration (mg/L)} \times 8.34$ $\text{lb solids} = \text{Volume (gal)} \times 8.34 \times \frac{\% \text{ solids}}{100} \times \text{Specific Gravity}$

Problem 13. A primary clarifier pumps 14,000 gal/day of sludge at 4.6 percent solids, specific gravity 1.02.

lb/day = 14,000 x 8.34 x 0.046 x 1.02 = 5,478 lb/day of dry solids

Problem 14. That sludge is thickened to 6.0 percent. What is the thickened volume, assuming no solids loss?

Volume = 5,478 / (8.34 x 0.060 x 1.02) = 5,478 / 0.5104 = 10,733 gal/day

Problem 15. Volatile solids reduction using the Van Kleeck formula, with volatile fractions in as 0.72 and out as 0.53:

$\% VSR = \frac{VS_{in} - VS_{out}}{VS_{in} - (VS_{in} \times VS_{out})} \times 100$

= (0.72 − 0.53) / (0.72 − 0.3816) x 100 = 0.19 / 0.3384 x 100 = 56.1 percent - above the 38 percent minimum for Part 503 vector attraction reduction Option 1.

Problem 16. Digester loading. A digester holds 420,000 gallons and receives 5,478 lb/day of solids at 72 percent volatile.

Volume = 420,000 / 7.48 = 56,150 ft³ VS fed = 5,478 x 0.72 = 3,944 lb/day Loading = 3,944 / 56,150 = 0.070 lb VS/ft³/day - within the typical 0.04 to 0.10 range for a mesophilic complete-mix digester

Problem 17. Belt filter press performance. Feed is 80 gpm at 3.2 percent solids; cake is 19 percent solids; polymer dose is 14 lb per dry ton.

Feed solids = 80 x 1,440 x 8.34 x 0.032 = 30,744 lb/day = 15.4 dry tons/day Cake mass at 95 percent capture = 30,744 x 0.95 = 29,207 lb dry solids Wet cake = 29,207 / 0.19 = 153,721 lb/day of wet cake ≈ 76.9 wet tons/day Polymer = 15.4 x 14 = 215 lb/day

Problem 18. Hauling cost at $62 per wet ton:

76.9 x 62 = $4,768 per day = $1.74 million per year - which is why a two-point improvement in cake solids is worth a great deal. At 21 percent cake, the wet tonnage falls to 69.5 tons/day and the annual cost to $1.57 million, a savings of roughly $170,000 per year.