Math problem set Grade 12 Math
Sterling Hollow Chlorine Dosing: Operator-Certification Math
A grade 12 problem set chaining the pounds formula and the percent-to-mg/L conversion to size a hypochlorite feed pump at a fictional treatment plant.
Objectives
- Students will calculate chemical feed loading using the pounds formula.
- Students will convert a percent stock-solution strength into pounds of available chemical per gallon.
- Students will solve a one-variable equation for an unknown dose given a target loading.
Vocabulary
Key words: stock solution, available chlorine, pounds formula, feed rate
Teacher overview
Sterling Hollow, a fictional town used for this problem set, runs a treatment plant that disinfects its water with a liquid hypochlorite stock solution rather than pure chlorine gas. Operators need to know how many pounds of chlorine a target dose requires each day and also how many gallons of stock solution at a given strength will deliver that many pounds. This set chains three real operator-certification tools together: the pounds formula, the fact that a 1 percent solution equals 10,000 mg/L, and the constant 8.34 pounds per gallon of water, to size Sterling Hollow's daily feed rate.
A common misconception in this math is treating percent strength and mg/L as unrelated units that cannot be compared directly. Every percent strength converts to mg/L using the same factor, 1 percent equals 10,000 mg/L, which is exactly what links the plant's dosing formula in mg/L to its stock-solution formula in percent, and problem 2 asks students to build that bridge explicitly.
Problems
Sterling Hollow's plant treats a flow of 1.8 MGD and doses chlorine at 2.5 mg/L. The plant uses a 10 percent strength sodium hypochlorite stock solution. Use these facts, the pounds formula (Loading in lbs/day = Concentration in mg/L x Flow in MGD x 8.34), and the fact that 1 percent solution strength equals 10,000 mg/L for all problems.
Calculate the chlorine loading, in pounds per day, that Sterling Hollow's plant needs at its 2.5 mg/L dose and 1.8 MGD flow.
A 10 percent stock solution is 10 times as concentrated as a 1 percent solution. Using the fact that 1 percent equals 10,000 mg/L and the constant 8.34 lb/gal, show the steps to find how many pounds of available chlorine are in one gallon of the 10 percent stock solution.
Using your answers from problems 1 and 2, calculate how many gallons of the 10 percent stock solution Sterling Hollow's plant needs each day.
Convert your answer from problem 3 into gallons per hour, to give the operator a feed-pump setting.
During a peak-demand period, flow rises to 2.4 MGD while the dose stays at 2.5 mg/L. Calculate the new daily loading and the new gallons of 10 percent stock solution needed per day.
Sterling Hollow's operators want to raise the dose so that daily loading reaches exactly 45.0 lbs/day, while flow stays at the original 1.8 MGD. Write an equation for the unknown dose D and solve for it.
Suppose Sterling Hollow only has a 5 percent stock solution available instead of 10 percent. Find the pounds of available chlorine per gallon of the 5 percent solution, then calculate how many gallons per day are needed to deliver the original 37.53 lbs/day loading from problem 1.
Compare your gallons-per-day answers from problem 3 (10 percent stock) and problem 7 (5 percent stock). Explain why the weaker solution requires roughly twice the daily volume to deliver the same chlorine loading.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 37.53 lbs/day.
Loading = Concentration x Flow x 8.34 = 2.5 mg/L x 1.8 MGD x 8.34 = 37.53 lbs/day.Answer: 0.834 lb/gal.
A 10 percent solution equals 10 x 10,000 mg/L = 100,000 mg/L of available chlorine. Treating one gallon as 0.000001 MG and applying the pounds formula: 100,000 mg/L x 0.000001 MG x 8.34 = 0.834 lb of available chlorine per gallon of stock solution.Answer: 45.0 gal/day.
Gallons per day = Loading needed / pounds per gallon = 37.53 lbs/day / 0.834 lb/gal = 45.0 gal/day.Answer: 1.88 gal/hr.
45.0 gal/day / 24 hours/day = 1.875 gal/hr, which rounds to about 1.88 gal/hr.Answer: 50.0 lbs/day loading, 60.0 gal/day.
New loading = 2.5 mg/L x 2.4 MGD x 8.34 = 50.04 lbs/day, which rounds to 50.0 lbs/day. New gallons per day = 50.04 lbs/day / 0.834 lb/gal = 60.0 gal/day.Answer: about 3.00 mg/L.
Equation: D x 1.8 MGD x 8.34 = 45.0 lbs/day, so D x 15.012 = 45.0, and D = 45.0 / 15.012 ≈ 2.998, which rounds to about 3.00 mg/L.Answer: 0.417 lb/gal and 90.0 gal/day.
A 5 percent solution equals 5 x 10,000 mg/L = 50,000 mg/L. Pounds per gallon = 50,000 mg/L x 0.000001 MG x 8.34 = 0.417 lb/gal. Gallons per day = 37.53 lbs/day / 0.417 lb/gal = 90.0 gal/day.Answer: The weaker solution needs about twice the volume because it carries half the chlorine per gallon.
The 5 percent solution has half the available chlorine concentration of the 10 percent solution (0.417 lb/gal versus 0.834 lb/gal), so delivering the same total mass of chlorine each day requires roughly twice as many gallons: 90.0 gal/day compared to 45.0 gal/day.
Differentiation
Support (below grade level)
Provide the derivation in problem 2 partly worked out, with the 100,000 mg/L step in our scenario already shown, so students only need to complete the final multiplication by 8.34 and the unit conversion to gallons.
Extension (above grade level)
Ask students to find what stock-solution strength, as a percent, would let Sterling Hollow deliver the original 37.53 lbs/day loading using exactly 30 gallons per day, setting up and solving an equation for the unknown percent strength.
ELL supports
Pre-teach "stock solution," "available chlorine," and "feed rate" with a simple diagram of a labeled jug of concentrated liquid being pumped into a larger flow of water. Allow problem 8's explanation to be given as a short spoken response recorded by a partner.