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Worksheet Grade 11 Math

Glenmark Bend Plant Residuals: Solids Loading and Drying Beds

A grade 11 worksheet on water plant residuals: students find solids loading with the pounds formula, size a drying bed, and compare hauling costs before and after drying.

Objectives

  1. Students will compute dry solids loading using the pounds formula.
  2. Students will convert a sludge mass at a given percent solids into a daily volume.
  3. Students will compare hauling costs before and after dewatering.

Vocabulary

Key words: residuals, percent solids, pounds formula, removal efficiency, drying bed

Teacher overview

Every drinking water plant that removes solids from raw water has to put those solids somewhere, and the cost of doing so is driven almost entirely by water rather than by solids. This worksheet has students follow the residuals stream at a fictional plant from the sedimentation basin to the drying bed and out to a landfill, tracking the mass of dry solids and the far larger mass of water that travels with it. The misconception the sequence corrects is the assumption that thickening or drying sludge removes solids. Drying removes water only, so the dry solids mass is the same before and after, while the weight on the truck falls by roughly nine tenths.

Glenmark Bend Plant Residuals: Solids Loading and Drying Beds — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

Use these figures from the fictional Glenmark Bend water treatment plant for every item. All values are hypothetical.

  • Plant flow: 4.0 MGD.
  • Suspended solids entering the sedimentation basin: 45 mg/L.
  • Suspended solids leaving the sedimentation basin: 5 mg/L.
  • Sludge is withdrawn from the basin at 2.0 percent solids.
  • Drying bed: 100 feet long and 40 feet wide, with sludge applied to a depth of 1.5 feet.
  • After drying, the residuals reach 20 percent solids.
  • Hauling cost: $50 per wet ton, charged on the weight loaded.
  • Conversions: one gallon of water weighs 8.34 pounds, one cubic foot holds 7.48 gallons, and one ton is 2,000 pounds.
  • Formulas: pounds per day equals concentration times flow in MGD times 8.34. Removal efficiency equals influent minus effluent, divided by influent, times 100.
  1. Find the suspended solids removed across the basin, in milligrams per liter.
  2. Find the dry solids removed per day, in pounds per day.
  3. Find the removal efficiency across the basin. Round to one decimal place.
  4. Find the weight of wet sludge produced per day at 2.0 percent solids, in pounds per day.
  5. Find the volume of wet sludge produced per day, in gallons.
  6. Find the volume of one drying bed at the stated application depth, in cubic feet.
  7. Find the volume of one drying bed in gallons.
  8. Find how many days of sludge one drying bed holds. Round to two decimal places.
  9. Find the dry solids produced in one year, in tons. Use 365 days and round to one decimal place.
  10. Find the daily hauling cost if wet sludge at 2.0 percent solids were trucked away directly.
  11. Find the weight of the dried residuals per day at 20 percent solids, in pounds per day.
  12. Find the daily hauling cost after drying.
  13. Find the percent reduction in hauled weight that drying achieves.
  14. Explain why the dry solids figure from item 2 does not change when the residuals are dried.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 40 mg/L.
    Removal is influent minus effluent, so 45 mg/L minus 5 mg/L equals 40 mg/L captured in the basin.
  2. Answer: 1,334.4 pounds per day.
    The pounds formula gives 40 mg/L times 4.0 MGD times 8.34 pounds per gallon, which equals 1,334.4 pounds per day.
  3. Answer: 88.9 percent.
    Removal efficiency is 45 minus 5, divided by 45, times 100, which equals 88.888 and rounds to 88.9 percent.
  4. Answer: 66,720 pounds per day.
    At 2.0 percent solids the dry mass is one fiftieth of the total, so 1,334.4 pounds divided by 0.020 equals 66,720 pounds per day.
  5. Answer: 8,000 gallons.
    Dividing 66,720 pounds by 8.34 pounds per gallon gives 8,000 gallons per day, since dilute sludge weighs about the same as water.
  6. Answer: 6,000 cubic feet.
    The bed is a rectangular prism, so 100 feet times 40 feet times 1.5 feet equals 6,000 cubic feet.
  7. Answer: 44,880 gallons.
    Multiplying 6,000 cubic feet by 7.48 gallons per cubic foot gives 44,880 gallons.
  8. Answer: 5.61 days.
    Dividing 44,880 gallons of bed capacity by 8,000 gallons per day of sludge gives 5.61 days.
  9. Answer: 243.5 tons.
    Multiplying 1,334.4 pounds per day by 365 days gives 487,056 pounds, and dividing by 2,000 pounds per ton gives 243.528, which rounds to 243.5 tons.
  10. Answer: $1,668.00.
    Dividing 66,720 pounds by 2,000 pounds per ton gives 33.36 wet tons, and multiplying by $50 per ton gives $1,668.00 per day.
  11. Answer: 6,672 pounds per day.
    The dry solids mass stays at 1,334.4 pounds, so at 20 percent solids the total is 1,334.4 divided by 0.20, which equals 6,672 pounds per day.
  12. Answer: $166.80.
    Dividing 6,672 pounds by 2,000 pounds per ton gives 3.336 tons, and multiplying by $50 per ton gives $166.80 per day.
  13. Answer: 90 percent.
    The weight falls from 66,720 pounds to 6,672 pounds, a drop of 60,048 pounds, and 60,048 divided by 66,720 equals 0.90.
  14. Answer: Drying removes water, not solids.
    The basin captured a fixed mass of solids that day, and evaporation and drainage in the bed only take away the water carrying them. The dry mass is therefore unchanged while the total weight falls sharply.

Differentiation

Support

Give these students item 2 worked out and provide a two-column table with the pounds formula terms labeled, so the concentration, the flow, and the constant are placed before any multiplication happens. Provide the percent-solids relationship in words as well as symbols for the fictional plant: total weight equals dry weight divided by the decimal percent solids.

Extension

Ask these students to find the surface area, in square feet, that the fictional plant would need if one bed had to hold a full week of sludge at the same application depth, and to state the bed dimensions they would recommend. Then have them calculate the annual hauling savings from drying and compare that figure with what a mechanical dewatering press might cost to operate.

ELL supports

Provide a labeled flow diagram of the fictional residuals stream showing the sedimentation basin, the sludge line, the drying bed, and the truck, with the words solids, sludge, dry, and haul labeled on it. Allow item 14 to be answered with the frame "The dry solids stay the same because drying only removes ___."

Extensions

Have students investigate what a utility near them does with water treatment residuals, since some plants land-apply them, some send them to a landfill, and some return them to a sewer for treatment at the wastewater plant. Students can then explain which of those choices the fictional Glenmark Bend numbers would favor and what additional information they would need to decide. A further task is to research why alum-based residuals behave differently from lime-softening residuals when they are dried, and to predict which would reach the higher percent solids in the same bed.