20 Filtration, Backwash & Residuals Practice Questions & Answers

1. 1. A plant treats a reservoir source whose turbidity and color stay low and steady year round, and it is being converted from conventional treatment to direct filtration. What changes in the treatment train?

   • A.Flocculation and sedimentation are kept, and the filters are rebuilt as slow sand beds run without coagulant.
   • B.Coagulation and flocculation are kept, and sedimentation is omitted, so coagulated water goes straight to the filters.✓ Answer
   • C.Sedimentation is kept and flocculation is omitted, so water from the rapid mix goes straight to the settling basin.
   • D.Sedimentation is kept and the coagulant is omitted, so the filters hold back particles by straining alone.

   Direct filtration is coagulation and flocculation followed immediately by filtration; the sedimentation step is dropped, so the filters must store all of the floc. That only works where raw turbidity and color are low and stable, because the whole solids load now reaches the media and run length is set by how much floc the bed can hold.

   Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, filtration chapter (direct filtration)Report a problem with this question

2. 2. How does a properly operated slow sand filter remove pathogens and particles, and how is that filter cleaned?

   • A.Removal is mostly biological, in the mat at the sand surface, and the bed is cleaned by scraping that layer off.✓ Answer
   • B.Removal is by straining deep in the sand, and the whole bed is cleaned by changing out all of the media each season.
   • C.Removal is mostly biological, in the gravel support, and the bed is cleaned by a high rate backwash from below.
   • D.Removal is chemical, from coagulant fed onto the bed, and the bed is cleaned by air scour and a short wash.

   A slow sand filter works through the biological layer that develops at the sand surface, where organisms consume and trap organic matter and pathogens. It is not backwashed: the top layer is scraped off, and the filter then needs a ripening period to re-establish that layer before its water is sent to the plant.

   Source: AWWA Water System Operations: Water Treatment, slow sand filtrationReport a problem with this question

3. 3. An operator compares a pressure filter with an open gravity filter holding the same media. Which limitation belongs to the pressure filter?

   • A.The media can only be washed with raw water, because the shell has no filtered water connection.
   • B.The media sits inside a closed steel shell, so the bed cannot be watched while it is being backwashed.✓ Answer
   • C.The media has to be taken out and washed by hand, because the shell cannot pass wash water upward.
   • D.The media cannot be placed in two layers, so only a single sand bed will work inside the closed steel shell.

   A pressure filter is an enclosed steel vessel, so the operator cannot see the bed surface during the wash and cannot spot boils, mounding, cracks or media carryover the way an open gravity filter allows. Condition has to be inferred from head loss, wash water turbidity and periodic internal inspection.

   Source: AWWA Water System Operations: Water Treatment, pressure filtersReport a problem with this question

4. 4. Why does a dual media bed of coarse anthracite over finer sand usually run longer between backwashes than a single sand bed of the same depth?

   • A.The sand layer throttles the flow and lowers the rate, so less solid material reaches the bed each hour.
   • B.Anthracite breaks the floc down chemically, so less solid material is left in the bed to build head loss.
   • C.Solids are caught at the anthracite surface, so the sand beneath stays clean and adds no head loss at all.
   • D.Solids are stored through the depth of the bed instead of at the surface, so head loss builds more slowly.✓ Answer

   Water meeting coarse grains first and finer grains last gives depth filtration: floc penetrates and is stored throughout the bed. A single fine sand bed grades fine to coarse in the direction of flow after a wash, so it acts as a surface strainer, blinds off quickly and reaches terminal head loss sooner.

   Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, filter mediaReport a problem with this question

5. 5. After a complete backwash of a dual media filter, where does the anthracite settle, and why?

   • A.On top of the sand, because anthracite has the lower specific gravity even though its grains are coarser.✓ Answer
   • B.On top of the sand, because rising wash water always carries the finest grains in the bed upward.
   • C.Beneath the sand, because the anthracite grains are coarser and the coarsest grains always settle out first.
   • D.Mixed evenly with the sand, because the wash is meant to blend the two media into one uniform layer.

   Media restratify after a wash according to specific gravity, not grain size: anthracite is the lightest, silica sand is heavier and garnet or ilmenite is heaviest, so they settle in that order from top to bottom. That is what produces the coarse to fine gradation in the direction of flow.

   Source: AWWA Water System Operations: Water Treatment, dual and mixed media bedsReport a problem with this question

6. 6. Two sands have the same effective size, but one has a much higher uniformity coefficient. What does the higher coefficient tell the operator?

   • A.Its grain sizes are spread more widely, so the bed stratifies more sharply and gains head loss faster.✓ Answer
   • B.Its grain sizes are spread more narrowly, so the bed fluidizes at a lower wash rate than the other sand.
   • C.Its grains are harder and rounder, so the bed resists abrasion and has to be topped up less often.
   • D.Its grains are larger overall, so the bed stores more solids and runs longer before turbidity climbs.

   The uniformity coefficient is the ratio of the sieve size passing sixty percent of the media to the effective size, so a lower value means a more uniform sand. A high value means a wide spread of grain sizes: the fines restratify to the top of each layer after every wash, plug the surface and drive head loss up faster.

   Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, media specificationsReport a problem with this question

7. 7. Late in a run, head loss on a filter is near the plant's terminal value and effluent turbidity has started to climb steadily. What is happening, and what should the operator do?

   • A.The bed is still ripening after its wash, so the filter should be left alone until turbidity settles.
   • B.The rate controller is starving the filter, so the filtration rate should be raised to clear the bed.
   • C.Floc is being driven through the bed, so the filter should be taken out of service and backwashed.✓ Answer
   • D.Air is binding the upper media, so the water level above the bed should be lowered to release it.

   Rising turbidity together with high head loss late in a run is breakthrough: the storage capacity of the bed is used up and stored floc is being pushed out into the filtered water. The run must end on whichever comes first, terminal head loss or turbidity breakthrough, and the underlying cause is usually a pretreatment problem rather than the filter itself.

   Source: US EPA Interim Enhanced Surface Water Treatment Rule guidance on individual filter performance; AWWA Water System Operations: Water Treatment, filter operationReport a problem with this question

8. 8. Through a run head loss rose faster than usual and the filtration rate fell off, and during the next backwash the operator sees air bubbling up through the media. What sequence explains this?

   • A.Air leaked in through the underdrain first, which raised head loss and then pulled the water level above the bed down.
   • B.Carbonate coatings built up on the grains, which raised their specific gravity and held pockets of air beneath them.
   • C.The wash rate had been set too high, which lifted the support gravel and let air trapped beneath it escape upward.
   • D.Head loss in the upper media passed the depth of water above it, pressure fell below atmospheric, and dissolved air came out.✓ Answer

   When head loss through the upper media exceeds the depth of water standing above the bed, pressure inside the media drops below atmospheric. That negative head releases dissolved gases, and the trapped air blocks pore space, so head loss climbs and the rate falls. Keeping enough water depth over the media, and washing before terminal head loss, prevents it.

   Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, negative head and air bindingReport a problem with this question

9. 9. Turbidity from a filter peaks in the first part of a run just after the filter is returned to service, then falls back. Which practice addresses this, and why?

   • A.Raising the rate at the start of the run, because a higher rate compacts the bed and seals its surface.
   • B.Adding several minutes to the backwash, because the peak shows solids were left deep in the media.
   • C.Sending the first filtrate to waste while the bed ripens, because washed media is at its least effective then.✓ Answer
   • D.Feeding extra coagulant for the first hour, because the peak shows the raw water is being underdosed.

   Freshly washed media is clean, so it has little of the retained floc that helps capture particles; removal improves as the bed ripens and the initial turbidity spike falls away. Filtering to waste keeps that first filtrate out of the clearwell, and a plant without filter to waste uses a slow start or delayed start instead.

   Source: US EPA Interim Enhanced Surface Water Treatment Rule guidance, filter ripening and filter to wasteReport a problem with this question

10. 10. A rectangular filter is 20 ft by 30 ft and is treating 1.5 MGD. Using 1 MGD = 694 gpm, what is its filter loading rate?

    • A.2.5 gpm/ft2
    • B.3.5 gpm/ft2
    • C.0.9 gpm/ft2
    • D.1.7 gpm/ft2✓ Answer

    Filter loading rate equals flow in gpm divided by filter surface area. The flow is 1.5 x 694 = 1,041 gpm and the area is 20 x 30 = 600 ft2, so 1,041 / 600 = 1.7 gpm/ft2. Dividing by twice the area gives 0.9 and using only half the area gives 3.5, which are the usual area mistakes.

    Source: AWWA Water System Operations: Water Treatment, filter loading rate calculationReport a problem with this question

11. 11. A plant uses the same backwash rate all year. As raw water gets colder in winter, what happens to the bed and what should be done?

    • A.Colder water lifts less media, so the wash time should be doubled at the same wash rate.
    • B.Colder water does not change expansion, so the rate can be set once and then left alone.
    • C.Colder water is more viscous, so the bed expands further and the wash rate should be lowered.✓ Answer
    • D.Colder water is more viscous, so the bed expands less and the wash rate should be raised.

    Viscosity rises as water cools, so the same upflow carries and lifts the media more and expansion increases in winter. Rates therefore have to be reduced as the water cools and increased as it warms; a plant that never adjusts washes media over the troughs in winter and under washes, and grows mudballs, in summer.

    Source: Pennsylvania Department of Environmental Protection filter bed expansion evaluation guidance; AWWA Water System Operations: Water Treatment, backwashingReport a problem with this question

12. 12. A filter holds 12 in of support gravel, 18 in of sand and 12 in of anthracite. During the high rate wash the top of the bed rises 6 in above its resting level. What is the percent bed expansion?

    • A.20%✓ Answer
    • B.33%
    • C.50%
    • D.14%

    Percent bed expansion is the rise divided by the depth of expandable media, times one hundred. Support gravel does not fluidize and is excluded, so the expandable depth is 18 + 12 = 30 in and 6 / 30 = 20%. Counting the gravel gives 14%, using only the sand gives 33% and using only the anthracite gives 50%.

    Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, percent bed expansionReport a problem with this question

13. 13. At the end of a backwash the water above the bed looks clear, yet cores taken later show mudballs in the media. What does this tell the operator about judging a wash by appearance?

    • A.Clear water shows that the surface wash ran too long; the mudballs are floc driven down by the arms.
    • B.Clear water shows that the wash was adequate; the mudballs must have formed later, while the filter ran.
    • C.Clear water shows that the wash rate was too high; the mudballs are grains cemented by anthracite fines.
    • D.Clear water shows only that the wash water is clean; measured bed expansion shows the media was fluidized.✓ Answer

    Water above the bed can run clear while parts of the bed were never fluidized, because dead spots simply do not release their solids. Only a measured expansion, taken at several points during the high rate wash, shows that the grains actually separated and scoured; too little expansion leaves solids behind, and they agglomerate with media into mudballs.

    Source: Pennsylvania Department of Environmental Protection filter plant performance evaluation guidance, bed expansion measurementReport a problem with this question

14. 14. Which filter surveillance test shows an operator where solids are being stored in the depth of the bed and whether the backwash removed them?

    • A.A filter drop test, timing the fall of the water level over the bed with the influent valve held closed.
    • B.A floc retention profile, comparing washed core samples from several depths before and after a backwash.✓ Answer
    • C.An acid solubility test, dissolving a sample to find carbonate or manganese coatings on the grains.
    • D.A sieve analysis, sizing a media sample to compare its effective size and uniformity with the design.

    A floc retention profile takes media cores at successive depths, washes the solids off each sample and measures the turbidity, so it maps how deeply solids penetrated and how much the wash left behind. The drop test measures actual filtration rate, the sieve analysis checks media sizing, and acid solubility looks for coatings.

    Source: Office of Water Programs, California State University Sacramento, Water Treatment Plant Operation, filter surveillance testsReport a problem with this question

15. 15. Four pressure driven membrane processes are to be listed from the largest material excluded to the smallest. Which order is correct?

    • A.Reverse osmosis, nanofiltration, ultrafiltration, microfiltration.
    • B.Nanofiltration, reverse osmosis, microfiltration, ultrafiltration.
    • C.Microfiltration, ultrafiltration, nanofiltration, reverse osmosis.✓ Answer
    • D.Ultrafiltration, microfiltration, reverse osmosis, nanofiltration.

    The four processes form a ladder: microfiltration has the largest openings and removes particles, bacteria and protozoan cysts; ultrafiltration is tighter and adds viruses and large molecules; nanofiltration rejects hardness, color and other divalent ions; reverse osmosis rejects dissolved monovalent salts and needs the highest driving pressure.

    Source: US EPA Membrane Filtration Guidance Manual, classification of membrane processesReport a problem with this question

16. 16. A low pressure membrane unit runs at constant flux, and its transmembrane pressure has climbed over several weeks. What must be checked before this is called fouling?

    • A.Whether the feed temperature has risen, since warmer water is more viscous and raises pressure by itself.
    • B.Whether the filtrate turbidity has fallen, since cleaner filtrate by itself proves the fibers are plugging.
    • C.Whether the feed temperature has fallen, since colder water is more viscous and raises pressure by itself.✓ Answer
    • D.Whether the recovery has been raised, since a higher recovery lowers pressure at any given flux.

    Water viscosity increases as temperature falls, so more pressure is needed to push the same flux through a perfectly clean membrane. Operating data must be temperature corrected, or normalized, before the trend is read; only a rise in normalized pressure, or a fall in normalized flux, means the membrane is actually fouling.

    Source: US EPA Membrane Filtration Guidance Manual, temperature normalization of operating dataReport a problem with this question

17. 17. Why does a membrane plant run a direct integrity test on each unit instead of relying on filtrate turbidity alone?

    • A.A single broken fiber can pass untreated water while the filtrate turbidity still reads low, so the barrier is checked directly.✓ Answer
    • B.Filtrate turbidity readings drift with water temperature, so the direct test is used to correct the output of the turbidimeter.
    • C.The direct test measures how much permeate each unit produces, and it is used to set the flux and recovery targets.
    • D.Turbidimeters cannot be fitted to membrane filtrate piping, so a pressure based test is the only reading available.

    Membrane treatment is credited as an absolute barrier, so the plant has to prove the barrier is intact rather than assume it. A pressure based direct integrity test applied to the membrane itself can detect a breach far smaller than turbidity monitoring can see; continuous filtrate turbidity is only an indirect check that signals when a direct test is needed.

    Source: US EPA Membrane Filtration Guidance Manual, direct and indirect integrity testingReport a problem with this question

18. 18. A plant that coagulates with aluminum sulfate compares its residuals with those of a lime softening plant. Which comparison is correct?

    • A.The alum residual is gelatinous and holds water, so it thickens and dewaters far less readily than lime sludge.✓ Answer
    • B.The lime residual is gelatinous and holds water, so it thickens and dewaters far less readily than alum sludge.
    • C.The alum residual is dense and granular, so it can be hauled straight from the basin with no dewatering at all.
    • D.Both residuals behave alike, so thickening and dewatering equipment does not depend on which one is produced.

    Aluminum hydroxide floc is a highly hydrated gel that traps water inside its structure, so it thickens slowly and resists dewatering, and it usually needs polymer conditioning. Lime softening residual is mostly calcium carbonate, which is dense and granular, settles quickly and gives a much drier cake for the same equipment.

    Source: AWWA Water System Operations: Water Treatment, water treatment plant residualsReport a problem with this question

19. 19. Why does a plant thicken and dewater its residuals before hauling them to a landfill or a land application site?

    • A.Removing water cuts the volume and weight hauled, and the receiving site expects a solid, non liquid material.✓ Answer
    • B.Removing water raises the pH of the residual, so it can be discharged to a stream without any permit at all.
    • C.Removing water dissolves the metals in the residual, so the material no longer has to be characterized at all.
    • D.Removing water sterilizes the residual, so it can be spread without regard to the loading limits set by the site.

    Residuals are mostly water when drawn, so hauling cost is largely the cost of hauling water; thickening, drying beds or mechanical dewatering shrink the volume and produce a cake firm enough for a landfill, which will not take free liquids. Dewatering does not change what is in the solids, so the material still has to be characterized before disposal.

    Source: AWWA Water System Operations: Water Treatment, residuals thickening and dewateringReport a problem with this question

20. 20. In a conventional plant, where should spent filter backwash water and the liquid from residuals dewatering be returned, and how should that stream be fed?

    • A.Straight into the filter influent channel, at whatever rate the washwater holding tank happens to drain out.
    • B.Into the settled water leaving the basins, at a rate matched to the flow coming off the settling basins.
    • C.Ahead of the point where primary coagulant is added, at a steady equalized rate so the plant is not slugged.✓ Answer
    • D.Into the clearwell after filtration, so that the recovered water is not put through the whole plant twice.

    Recycled streams carry concentrated solids, pathogens such as Cryptosporidium, manganese and organic precursors, so they must re-enter upstream of primary coagulant addition and pass through the entire treatment train again. Returning them in a slug also upsets coagulation, so the flow is equalized and fed back steadily as a small fraction of the raw water flow.

    Source: US EPA Filter Backwash Recycling Rule, 40 CFR 141.76Report a problem with this question

Practice questions written against the standardized Water Treatment Operator Need-to-Know Criteria published by Water Professionals International (formerly the Association of Boards of Certification) and standard references from the CSUS Office of Water Programs and AWWA. This site is not affiliated with or endorsed by WPI/ABC, AWWA, or the US EPA. Operator certification is issued by your state's certifying authority, which sets plant classification tiers, operator grades, eligibility, and the passing standard — confirm those with your state before testing. Contaminant limits and monitoring requirements are set federally and are revised over time, so no answer here should be relied on as a current regulatory value; consult the regulations in force for your system. This bank covers the drinking-water treatment exam only — wastewater treatment, wastewater collection, and water distribution are separate certifications. About the Need-to-Know Criteria →