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Math problem set Grade 10 Careers & CTE

Trennow's Chemical Feed Shift: Treatment Operator Math

A grade 10 careers math set that walks a trainee operator through one shift of chemical feed, dosing, storage, and filter loading calculations at a fictional plant.

Objectives

  1. Students will apply the pounds formula to find a daily chemical feed from a dose and a plant flow.
  2. Students will convert a percent solution strength into milligrams per liter and into gallons fed per day.
  3. Students will calculate filter loading rates with all filters in service and with one filter out.

Vocabulary

Key words: pounds formula, dose, residual, chlorine demand, percent solution, filter loading rate

Teacher overview

Every state licenses water treatment operators by examination, and licenses run in classes tied to the size of the plant and the operator's experience. The arithmetic in this set is the arithmetic an operator in training actually performs on a shift, recorded in a chemical feed log that a supervisor initials. The scenario is one day at the fictional Trennow Regional Water Authority. The misconception this set corrects is the habit of multiplying by a percent strength when converting pounds of chemical into gallons of solution, because the solution is mostly water, so the gallons figure comes from dividing by the strength rather than multiplying by it. A second error to watch is entering flow in gallons per minute into the pounds formula, which needs million gallons per day.

Trennow's Chemical Feed Shift: Treatment Operator Math — Grade 10 Careers & CTE
Name: ____________________________ Date: ______________

Problems

The fictional Trennow Regional Water Authority runs a conventional surface water plant that treats 3.6 MGD. Use these constants and rules: 1 MGD equals 694 gpm; 1 MGD equals 1.547 cfs; pounds per day equals dose in mg/L times flow in MGD times 8.34; chlorine demand equals dose minus residual; a 1% solution equals 10,000 mg/L. For solution volume, the plant's own worksheet treats the liquid chlorine solution as weighing 8.34 pounds per gallon. Round each answer to the place named in the question.

  1. The plant treats 3.6 MGD. Convert that flow to gallons per minute, using 694 gpm per MGD. Round to the nearest whole gpm.
  2. Convert the same 3.6 MGD flow to cubic feet per second, using 1.547 cfs per MGD. Round to the nearest hundredth.
  3. The operator sets a chlorine dose of 2.4 mg/L at a flow of 3.6 MGD. Using pounds per day equals dose times flow times 8.34, how many pounds of chlorine per day does the plant feed? Round to the nearest tenth.
  4. That 2.4 mg/L dose leaves a residual of 0.8 mg/L at the clearwell. What is the chlorine demand, in mg/L?
  5. The plant feeds a 12.5% sodium hypochlorite solution. Using 10,000 mg/L per 1%, what is that strength in mg/L?
  6. The plant must deliver 72.1 pounds of chlorine per day using that 12.5% solution. First find the pounds of solution needed by dividing by 0.125. Then find the gallons by dividing by 8.34 pounds per gallon. How many gallons of solution per day does the plant feed? Round to the nearest tenth.
  7. The same plant feeds alum at 32 mg/L while treating 3.6 MGD. Using the pounds formula, how many pounds of alum per day is that? Round to the nearest tenth.
  8. The alum silo holds 25,000 pounds when full. At a feed of 960.8 pounds per day, how many days does a full silo last? Round to the nearest tenth.
  9. The plant has four filters, and each one measures 14 feet by 20 feet. With all four in service and a plant flow of 2,498.4 gpm, what is the filter loading rate in gpm per square foot? Round to the nearest hundredth.
  10. One filter is taken out of service for backwashing, leaving three filters of 14 feet by 20 feet each. At the same 2,498.4 gpm, what is the loading rate on the filters still in service, in gpm per square foot? Round to the nearest hundredth.

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: 2,498 gpm. Multiply 3.6 MGD by 694 gpm per MGD. That gives 2,498.4, which rounds to 2,498 gpm.
  2. Answer: 5.57 cfs. Multiply 3.6 MGD by 1.547 cfs per MGD. That gives 5.5692, which rounds to 5.57 cfs.
  3. Answer: 72.1 lbs/day. Pounds per day equals 2.4 mg/L times 3.6 MGD times 8.34. First 2.4 times 3.6 equals 8.64. Then 8.64 times 8.34 equals 72.0576, which rounds to 72.1 pounds per day.
  4. Answer: 1.6 mg/L. Chlorine demand equals dose minus residual, so 2.4 mg/L minus 0.8 mg/L equals 1.6 mg/L. This is the chlorine consumed by the water before any residual is left over.
  5. Answer: 125,000 mg/L. A 1% solution is 10,000 mg/L, so 12.5 times 10,000 equals 125,000 mg/L.
  6. Answer: 69.2 gallons per day. First find the pounds of solution: 72.1 pounds of chlorine divided by 0.125 equals 576.8 pounds of solution. Then divide by 8.34 pounds per gallon: 576.8 divided by 8.34 equals 69.16, which rounds to 69.2 gallons per day. Note the division by 0.125, which makes the solution figure larger than the chlorine figure, because the solution is mostly water.
  7. Answer: 960.8 lbs/day. Pounds per day equals 32 mg/L times 3.6 MGD times 8.34. First 32 times 3.6 equals 115.2. Then 115.2 times 8.34 equals 960.768, which rounds to 960.8 pounds per day.
  8. Answer: 26.0 days. Divide the silo contents by the daily feed: 25,000 pounds divided by 960.8 pounds per day equals 26.02, which rounds to 26.0 days. An operator ordering alum would place the order well before that day.
  9. Answer: 2.23 gpm/sq ft. Each filter is 14 ft times 20 ft, or 280 square feet, so four filters give 1,120 square feet. Divide the flow by the area: 2,498.4 gpm divided by 1,120 square feet equals 2.2307, which rounds to 2.23 gpm per square foot.
  10. Answer: 2.97 gpm/sq ft. Three filters give 3 times 280, or 840 square feet. Divide the flow by that area: 2,498.4 gpm divided by 840 square feet equals 2.9743, which rounds to 2.97 gpm per square foot. Taking one filter out raises the rate on the others, which is why operators watch this figure during a backwash.

Differentiation

Support

Print the pounds formula and the percent solution rule at the top of the page with one worked example of each, using numbers different from the ones in the set. Allow students to complete problems 1 through 5 first, check them, and then continue, since problems 6, 8, and 10 reuse earlier results.

Extension

Ask students to find the chlorine dose the same fictional plant would need at a flow of 4.5 MGD to feed the same 72.1 pounds per day, then to explain in one sentence why the required dose falls as the flow rises.

ELL supports

Pre-teach dose, residual, demand, and solution strength with a labeled sketch of a feed pump, a day tank, and a sample tap. Provide a two-column note card with the formula on one side and the meaning of each term on the other, and allow the arithmetic to be shown in symbols rather than in written sentences.

Extensions

Students can rewrite problems 3 and 7 as a single spreadsheet formula with the dose in one cell and the flow in another, then check that their sheet reproduces both answers. A local operator can be asked which of these ten calculations appears in the daily log at a plant near the school and which of them software now performs.