Filtration and loading rates

Problem set: 5 worked problems

Each problem is original, authored from the underlying non-copyrightable relationship, with the full worked solution and the exact slip that produces each wrong answer. Work the problem before opening the solution. For a timed, randomized run with a per-topic score, use the free practice test.

Watch for Every loading rate is a flow divided by a dimension: filtration rate is GPM per square foot, surface (overflow) rate is gallons per day per square foot, weir overflow rate is gallons per day per foot. For round tanks, area is pi x radius squared, so halve the diameter first and match the flow's time unit to the rate you want.

1. A circular clarifier is 50 feet in diameter and treats 2.0 MGD. What is the surface (overflow) rate?

   • A 255 gpd/sq ft
   • B 510 gpd/sq ft
   • C 40,000 gpd/sq ft
   • D 1,019 gpd/sq ft
   Show the worked solution

   Correct answer: D. 1,019 gpd/sq ft

   Area = 3.14 x 25 x 25 = 1,962 sq ft (radius 25, not diameter 50). Surface rate = 2,000,000 gpd / 1,962 = 1,019 gpd per sq ft. Using the diameter in the area formula quarters the area and quadruples the rate.

   Source: Public relationship: surface rate = flow (gpd) / area (sq ft), round area = pi x r^2

2. A filter has 20 square feet of surface area and passes 300 gallons per minute. What is the filtration rate?

   • A 6,000 gpm/sq ft
   • B 20 gpm/sq ft
   • C 0.067 gpm/sq ft
   • D 15 gpm/sq ft
   Show the worked solution

   Correct answer: D. 15 gpm/sq ft

   Filtration rate = flow (GPM) / filter area (sq ft) = 300 / 20 = 15 gpm per sq ft, the typical ceiling for high-rate sand. Multiplying instead of dividing gives 6,000; inverting gives 0.067.

   Source: Public relationship: filtration rate = flow (GPM) / filter area (sq ft)

3. A clarifier weir is 120 feet long and the flow is 1.5 MGD. What is the weir overflow rate?

   • A 180 gpd/ft
   • B 1,250 gpd/ft
   • C 12,000 gpd/ft
   • D 12,500 gpd/ft
   Show the worked solution

   Correct answer: D. 12,500 gpd/ft

   Weir overflow rate = flow (gpd) / weir length (ft) = 1,500,000 / 120 = 12,500 gpd per foot. Keep the flow in gallons per day, not MGD, to land in the right units.

   Source: Public relationship: weir overflow rate = flow (gpd) / weir length (ft)

4. A filter must pass 400 GPM and its maximum allowed filtration rate is 4 gpm per square foot. What filter surface area is required?

   • A 1,600 sq ft
   • B 0.01 sq ft
   • C 25 sq ft
   • D 100 sq ft
   Show the worked solution

   Correct answer: D. 100 sq ft

   Rearrange: area = flow / maximum rate = 400 / 4 = 100 sq ft. Multiplying the flow by the rate (1,600) sizes the filter backward.

   Source: Public relationship: required area = flow (GPM) / maximum filtration rate (gpm/sq ft)

5. What is the surface area of a circular tank 30 feet in diameter?

   • A 94 sq ft
   • B 471 sq ft
   • C 2,827 sq ft
   • D 707 sq ft
   Show the worked solution

   Correct answer: D. 707 sq ft

   Area = pi x radius squared = 3.14 x 15 x 15 = 707 sq ft. The radius is half the 30-foot diameter. Using the full diameter as the radius gives 2,827, four times too large.

   Source: Public relationship: area of a circle = pi x radius squared