Sanitary Engineer Licensure Exam study materials
Sanitary Engineer Licensure Exam study resources

Turbidity Breakthrough at a 45 MLD River Treatment Plant

Sanitary Engineer Licensure Exam · Put your knowledge to work in context.

All case studies

Turbidity Breakthrough at a 45 MLD River Treatment Plant

A 45 MLD conventional surface-water plant using alum coagulation, flocculation, sedimentation, rapid gravity filtration and free chlorination draws from a river that normally runs at 15-25 NTU. Three days after a typhoon the raw water reaches 400 NTU with 80 TCU colour, pH 6.4 and total alkalinity of only 30 mg/L as CaCO3. The plant stays at its rated 45 MLD because the distribution reservoir is low. Jar tests show that 60 mg/L alum alone leaves settled water at 12 NTU and pH 6.0, while 40 mg/L alum plus 0.3 mg/L anionic polymer gives 5 NTU. Filtered-water turbidity is 2.8 NTU, free chlorine after the baffled 900 m3 clearwell at 45 MLD is 0.4 mg/L, and one of five finished-water samples is positive on a presumptive fecal coliform test. The shift operator proposes raising the alum dose to 90 mg/L and leaving everything else unchanged.

1. Would raising the alum dose to 90 mg/L with no other change improve or worsen finished-water quality? Support your conclusion with the alkalinity stoichiometry of alum.

Read the worked answer

It would almost certainly make things worse.

Hydrated alum consumes roughly 0.45-0.50 mg/L of alkalinity (as CaCO3) per mg/L of alum dosed. At 90 mg/L that is about 40-45 mg/L of alkalinity demand against only 30 mg/L available. The alkalinity is gone before the dose is satisfied, so pH falls below about 5.5.

The consequences compound:

  • Aluminium hydroxide is amphoteric. Below roughly pH 6.0 it begins to redissolve, so you get poor, finely divided floc and elevated residual aluminium passing the filters.
  • The jar test already shows this trend: 60 mg/L alum alone gives 12 NTU settled water at pH 6.0, which is a failing result, while 40 mg/L plus polymer gives 5 NTU.
  • More alum means more sludge, more backwash water and more acid load to the clearwell.

The defensible correction is the jar-tested 40 mg/L alum with 0.3 mg/L polymer, plus lime to hold the coagulation pH around 6.8-7.2 and leave at least 20 mg/L residual alkalinity. Dose should be set by jar test on the current raw water each shift, not held constant, because the raw quality is still moving.

2. Using the corrected 40 mg/L alum dose, estimate the hydrated lime requirement in mg/L and in kg/day, and state the operating checks that govern the final figure.

Read the worked answer

Step 1 - alkalinity consumed by alum. At 0.45-0.50 mg alkalinity per mg alum: 40 mg/L x 0.45 = 18 mg/L, up to 20 mg/L at the 0.50 ratio. Raw alkalinity of 30 mg/L therefore drops to about 10-12 mg/L as CaCO3, which will not hold pH near 7.

Step 2 - target alkalinity. Aim for about 30-35 mg/L as CaCO3 in settled water, i.e. an increment of roughly 20-25 mg/L.

Step 3 - convert to hydrated lime. Ca(OH)2 has a molecular weight of 74 and supplies two equivalents, so 1 mg/L of hydrated lime provides 100/74 = 1.35 mg/L of alkalinity as CaCO3. Dose = 20 / 1.35 = about 15 mg/L, with a working range of 15-18 mg/L.

Step 4 - mass check. 15 g/m3 x 45,000 m3/day = 675 kg/day of hydrated lime. That is a significant chemical bill, so confirm the dose on a jar test before committing.

Operating checks. Lime raises pH quickly and is easy to overdose. Control on settled-water pH (6.8-7.2) and alkalinity, not on lime feed rate alone; verify that the feeder is calibrated, that the slurry is not settling in the line, and that the polymer is added after pH correction so it is not degraded.

3. Compute the clearwell contact time and CT value at the stated conditions, and decide what the utility must do about the positive coliform result and the 2.8 NTU filtered water.

Read the worked answer

Contact time. Volume 900 m3 / flow 45,000 m3/day = 0.02 day = 28.8 minutes. Baffling efficiency is less than 100%, so the effective (t10) time is lower than the nominal value.

CT. 0.4 mg/L x 28.8 min = 11.5 mg-min/L, which on paper looks acceptable for virus inactivation at the water temperature involved.

Why the number cannot be relied on. CT credit is conditional on low turbidity. At 2.8 NTU, particles shield organisms from chlorine and exert demand, so the measured residual overstates the true disinfection. With one of five samples coliform-positive, the plant has evidence of breakthrough, not merely a sampling artefact.

Actions, in priority order:

  1. Notify the local health office and issue a precautionary boil-water advisory for the affected zone rather than distributing water that cannot be defended.
  2. Restore filtration first: filter-to-waste after each backwash until filtrate is below 0.3-1.0 NTU, check for media loss, mudballs and media boils, and reduce loading by derating flow if necessary.
  3. Raise the free chlorine residual to about 1.0-1.5 mg/L after contact while watching for trihalomethane precursors, and confirm with daily bacteriological sampling at multiple points.
  4. Only lift the advisory after consecutive negative coliform results and stable filtered-water turbidity. Reducing flow is a legitimate trade-off against supply hours even with a low reservoir; unsafe water is not a substitute for short supply.