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Worksheet Grade 11 Math

Alder Basin's Chlorine Feed Rate: Pounds Formula and Unit Conversions

A grade 11 worksheet where students use the pounds formula and chained unit conversions to size a fictional plant's chlorine feed rate and solution dosing.

Objectives

  1. Students will apply the pounds formula to calculate a chemical feed loading in pounds per day.
  2. Students will chain multiple unit conversions to move between pounds, hours, minutes, and gallons of chemical solution.

Vocabulary

Key words: pounds formula, loading rate, percent strength, feed rate

Teacher overview

Alder Basin Water Treatment Plant, a fictional facility used for this worksheet, has to translate a chlorine dose written in milligrams per liter into an actual chemical feed pump setting measured in gallons per day, a chain of conversions every operator performs using the pounds formula. A common misconception is rounding each step along the way, which quietly shifts the final answer; this worksheet asks students to carry unrounded values through the full chain and round only the final result, so they can see how much premature rounding can change an answer. The pounds formula (loading in pounds per day equals concentration in mg/L times flow in MGD times 8.34) is the same calculation used on state operator-certification exams.

Alder Basin's Chlorine Feed Rate: Pounds Formula and Unit Conversions — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Alder Basin plant treats a steady flow of 2.4 MGD and needs to maintain a chlorine dose of 2.5 mg/L. The plant feeds chlorine as a 12.5 percent sodium hypochlorite solution.

Use these formulas:

  • Loading (lbs/day) = Concentration (mg/L) × Flow (MGD) × 8.34
  • Pounds of available chlorine per gallon of solution = 8.34 × percent strength (written as a decimal)
  1. Calculate Alder Basin's chlorine loading in pounds per day, using the pounds formula with the 2.5 mg/L dose and 2.4 MGD flow.
  2. Convert your answer from problem 1 into pounds per hour.
  3. Convert your answer from problem 2 into pounds per minute. Round to the nearest thousandth.
  4. Calculate the pounds of available chlorine in one gallon of the 12.5 percent sodium hypochlorite solution. Do not round this value; you will need it in later problems.
  5. Using your unrounded answer from problem 4, calculate how many gallons of the 12.5 percent solution Alder Basin needs each day to deliver the loading from problem 1.
  6. Suppose flow increases to 3.0 MGD while the dose stays at 2.5 mg/L. Recalculate the daily chlorine loading in pounds per day.
  7. Using the same 12.5 percent solution and your unrounded answer from problem 4, how many gallons per day does the plant need at the higher 3.0 MGD flow from problem 6?
  8. Explain in one or two sentences why operators track chemical use in gallons of solution per day rather than pounds of pure chlorine per day.
  9. The plant needs a feed pump that can deliver the problem 7 volume at a constant rate across a full 24-hour day. What feed rate, in gallons per hour, does the pump need?
  10. The plant's existing feed pump has a minimum reliable rate of 2.0 gallons per hour. Based on your answer to problem 9, can the existing pump handle the higher 3.0 MGD flow scenario?

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: 50.04 lbs/day.
    Loading = 2.5 mg/L × 2.4 MGD × 8.34 = 6 × 8.34 = 50.04 lbs/day.

  2. Answer: 2.085 lbs/hr.
    50.04 lbs/day ÷ 24 hours/day = 2.085 lbs/hr.

  3. Answer: 0.035 lbs/min.
    2.085 lbs/hr ÷ 60 minutes/hr = 0.03475, which rounds to 0.035 lbs/min.

  4. Answer: 1.0425 lbs/gal.
    8.34 × 0.125 = 1.0425 lbs of available chlorine per gallon of the 12.5 percent solution. This value is kept unrounded for use in problems 5 and 7.

  5. Answer: 48 gallons/day.
    50.04 lbs/day ÷ 1.0425 lbs/gal = 48 gallons/day exactly.

  6. Answer: 62.55 lbs/day.
    Loading = 2.5 mg/L × 3.0 MGD × 8.34 = 7.5 × 8.34 = 62.55 lbs/day.

  7. Answer: 60 gallons/day.
    62.55 lbs/day ÷ 1.0425 lbs/gal = 60 gallons/day exactly.

  8. Answer: short answer.
    The chemical is fed as a liquid solution, not as pure chlorine, so the pump has to be set to deliver a volume of solution; pounds of pure chlorine is a useful design number, but the pump itself is calibrated in gallons.

  9. Answer: 2.5 gal/hr.
    60 gallons/day ÷ 24 hours/day = 2.5 gal/hr.

  10. Answer: yes.
    The higher-flow scenario requires 2.5 gal/hr, which is above the pump's 2.0 gal/hr minimum reliable rate, so the pump is not being asked to run below the point where its dosing becomes inaccurate and can handle this rate.

Differentiation

Support (below grade level)

Provide the pounds formula and the percent-strength formula pre-printed on a reference card, and work problem 1 together as a class before students attempt problems 2 through 10 independently. Allow a calculator for every step.

Extension (above grade level)

Ask students to calculate what percent strength solution Alder Basin would need to cut its problem 5 answer (48 gallons per day) in half while keeping the same 50.04 lbs/day loading, working backward through the percent-strength formula.

ELL supports

Pre-teach "loading," "feed rate," and "percent strength" with a labeled diagram of a chemical feed pump connected to a storage tank, showing the path from pounds of chlorine to gallons of solution. Provide a conversion-chain flowchart (mg/L and MGD to lbs/day to lbs/hr to gal/day) that students can fill in with their own numbers.