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Worksheet Grade 12 Math

Windlestraw Creek Phosphorus: Allocating a Load Reduction

A grade 12 worksheet where students compute four phosphorus loads with the pounds formula, then solve for the percent reduction two sources must share.

Objectives

  1. Students will compute pollutant loads in pounds per day from concentration and flow.
  2. Students will calculate a flow-weighted mean concentration for a stream with four inputs.
  3. Students will solve for the shared percent reduction two sources must meet under a fixed load cap.

Vocabulary

Effluent
Effluent is the cleaned water that leaves a treatment plant at the end of the process. It is tested carefully before being released to a river, lake, or the ocean.

Also: load, flow-weighted average, allocation, percent reduction

Teacher overview

When a stream carries more phosphorus than it can absorb, regulators set a cap on the total pounds that may enter each day and then divide the cut among the sources. The mathematics has two halves: converting each source's concentration and flow into pounds per day, and then solving for the reduction each source must accept once the sources that cannot be reduced are set aside. This worksheet uses the fictional Windlestraw Creek above the town of Oldbarrow, which receives phosphorus from a reclamation facility, a dairy, an urban storm sewer, and its own upstream flow. Students should carry unrounded values through the chain and round only at the end, since rounding each load to one decimal early shifts the final percent by one or two tenths of a percentage point.

Windlestraw Creek Phosphorus: Allocating a Load Reduction — Grade 12 Math
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Windlestraw Creek above the town of Oldbarrow receives total phosphorus from four sources. Use the loading formula: pounds per day equals concentration in mg/L times flow in MGD times 8.34.

Source Flow (MGD) Total phosphorus (mg/L)
Oldbarrow reclamation facility 3.2 1.5
Beckmoor Dairy runoff 1.8 0.9
Urban storm sewer 2.5 0.4
Upstream background flow 12.0 0.05
  1. Compute the phosphorus load from the reclamation facility, in pounds per day. Round to one decimal place.
  2. Compute the load from the dairy runoff, in pounds per day. Round to one decimal place.
  3. Compute the load from the urban storm sewer, in pounds per day. Round to one decimal place.
  4. Compute the load carried by the upstream background flow, in pounds per day. Round to one decimal place.
  5. Compute the total phosphorus load reaching the creek from all four sources, in pounds per day. Round to one decimal place.
  6. Compute the flow-weighted mean phosphorus concentration below all four inputs, in mg/L. Round to two decimal places.
  7. Regulators cap the allowable total at 45.0 pounds per day. How many pounds per day must be removed?
  8. What percent of the current total load is that reduction? Round to one decimal place.
  9. The background load cannot be reduced, and the storm sewer is assigned a 25 percent cut. What is the storm sewer's load after that cut, in pounds per day? Round to one decimal place.
  10. After the background load and the reduced storm sewer load are subtracted from the 45.0 pound cap, how many pounds per day remain available for the reclamation facility and the dairy together? Round to one decimal place.
  11. If the reclamation facility and the dairy each cut their loads by the same percent, what percent reduction meets the figure from question 10? Round to one decimal place.
  12. After that reduction, what total phosphorus concentration must the reclamation facility meet in its effluent, in mg/L? Round to two decimal places.
  13. Explain in two or three sentences why cutting the storm sewer load alone, even to zero, cannot bring the creek under the 45.0 pound cap.

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: 40.0 lbs/day.
    1.5 mg/L times 3.2 MGD times 8.34 equals 40.032, which rounds to 40.0 pounds per day.

  2. Answer: 13.5 lbs/day.
    0.9 mg/L times 1.8 MGD times 8.34 equals 13.5108, which rounds to 13.5 pounds per day.

  3. Answer: 8.3 lbs/day.
    0.4 mg/L times 2.5 MGD times 8.34 equals 8.34, which rounds to 8.3 pounds per day.

  4. Answer: 5.0 lbs/day.
    0.05 mg/L times 12.0 MGD times 8.34 equals 5.004, which rounds to 5.0 pounds per day.

  5. Answer: 66.9 lbs/day.
    The unrounded loads sum to 40.032 + 13.5108 + 8.34 + 5.004 = 66.8868, which rounds to 66.9 pounds per day.

  6. Answer: 0.41 mg/L.
    Total flow is 19.5 MGD, so the flow-weighted mean is 66.8868 divided by (19.5 times 8.34), or 66.8868 divided by 162.63, which is about 0.4113 and rounds to 0.41 mg/L.

  7. Answer: 21.9 lbs/day.
    66.8868 minus 45.0 leaves 21.8868 pounds per day to remove, which rounds to 21.9.

  8. Answer: 32.7 percent.
    21.8868 divided by 66.8868 is about 0.3272, or 32.7 percent of the current total load.

  9. Answer: 6.3 lbs/day.
    A 25 percent cut leaves 75 percent, so 8.34 times 0.75 equals 6.255, which rounds to 6.3 pounds per day.

  10. Answer: 33.7 lbs/day.
    Subtracting the unreducible background load of 5.004 and the reduced storm sewer load of 6.255 from the 45.0 pound cap leaves 33.741, which rounds to 33.7 pounds per day.

  11. Answer: 37.0 percent.
    The facility and dairy now contribute 40.032 + 13.5108 = 53.5428 pounds per day. Setting 53.5428(1 - r) = 33.741 gives r = 19.8018 divided by 53.5428, which is about 0.3698, or 37.0 percent.

  12. Answer: 0.95 mg/L.
    The facility's new load is 40.032 times (1 - 0.3698), or about 25.23 pounds per day. Dividing by 3.2 MGD times 8.34 gives 25.23 divided by 26.688, which is about 0.945 and rounds to 0.95 mg/L.

  13. Answer: The storm sewer carries far less than the required reduction.
    Even removing all 8.34 pounds per day from the storm sewer falls short of the 21.9 pounds that must go, and the background flow adds 5.0 pounds that no permit can touch. The reclamation facility and the dairy together carry more than three quarters of the load, so any workable plan has to reach them.

Differentiation

Support

Give these students the loading formula written out with one worked line already filled in for a source that does not appear in the fictional Windlestraw table, so items 1 through 4 are pattern-matching rather than formula recall. Provide a partially completed allocation table with the background and storm sewer rows already reduced, letting them start item 11 from the remaining allowance.

Extension

Ask students to find the allocation in which the fictional reclamation facility takes twice the percent reduction the dairy takes, and to report both percentages and both new loads. Then have them argue in a short paragraph which split is fairer, using the cost of treatment and the size of each source rather than the arithmetic alone.

ELL supports

Pre-teach "load," "effluent," "allocation," and "background" with a labeled diagram of a creek showing four arrows entering it, each tagged with a flow and a concentration. Provide sentence frames for item 13, such as "Cutting only the storm sewer removes ___ pounds, which is less than ___ because ___," and allow a labeled bar chart of the four loads in place of prose.

Extensions

Students can recompute the whole fictional allocation after a wet-weather week raises the storm sewer flow to 6.0 MGD at the same concentration, and report how much harder the cap becomes to meet. A second follow-up asks students to find their own state's list of impaired waters, choose one stream, and describe which sources the state names without recomputing any numbers.