A clearwell has an operating volume of 400,000 gal and a baffling factor of 0.5. At a peak hourly flow of 2,800 gpm the free chlorine residual at the outlet is 1.1 mg/L. What CT is the plant achieving?
Answer in mg·min/L.
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78.6 mg·min/L
Theoretical detention time = 400,000 gal ÷ 2,800 gpm = 142.9 min. T10 = 142.9 × 0.5 = 71.4 min. CT = 1.1 mg/L × 71.4 min = 78.6 mg·min/L.
Reference: Surface Water Treatment Rule, CT concept; EPA Guidance Manual for Disinfection Profiling and Benchmarking (baffling factors)
The CT table shows that 54 mg·min/L is required for today's temperature and pH. The effective contact time (T10) at peak flow is 36 minutes. What is the lowest free chlorine residual at the contact basin outlet that still meets the requirement?
Answer in mg/L.
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1.5 mg/L
CT = C × T, so C = CT ÷ T = 54 mg·min/L ÷ 36 min = 1.50 mg/L.
Reference: Surface Water Treatment Rule, CT = residual concentration × effective contact time
During a challenge test, a membrane unit is fed water containing 2,000,000 test particles per liter and the filtrate contains 400 per liter. What log removal value (LRV) did the unit demonstrate?
Answer in log.
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3.7 log
LRV = log10(feed ÷ filtrate) = log10(2,000,000 ÷ 400) = log10(5,000) = 3.70.
Reference: Log removal value: LRV = log10(Cfeed ÷ Cfiltrate); LT2ESWTR membrane challenge testing