Math problem set Grade 7 Math
Halden Plant Chlorine Dosing: A Grade 7 Proportion Set
A grade 7 set where students scale a proportional chlorine feed and work with dose, demand, and residual in mg/L for a fictional drinking-water plant.
Objectives
- Students will scale a proportional chemical feed up and down from a known ratio.
- Students will find chlorine residual as the difference between dose and demand.
- Students will express a residual as a percent of the dose and apply a percent increase.
Vocabulary
Key words: chlorine dose, demand, residual, proportional
Teacher overview
Drinking-water plants add chlorine to kill germs, and operators track three numbers: the dose they add, the demand the water uses up, and the residual that stays in the pipes to keep the water safe. This set uses a fictional plant to scale a proportional chlorine feed and then work with dose, demand, and residual in mg/L. A common misconception is that dose and residual are the same thing; they differ by the demand, and problem 4 makes that relationship the center of the work.
Problems
Halden is a fictional drinking-water plant used for this set. The plant feeds a chlorine solution in proportion to the water it treats. A crew records the feed for three amounts of water.
| Water treated (gallons) | Solution fed (gallons) |
|---|---|
| 100,000 | 15 |
| 200,000 | 30 |
| 300,000 | 45 |
- Is the solution fed proportional to the water treated? Check the ratios and explain.
- How many gallons of solution are needed to treat 250,000 gallons of water?
- How many gallons of water can 60 gallons of solution treat?
- The plant sets a chlorine dose of 2.4 mg/L. The water's chlorine demand is 1.5 mg/L. The residual is the dose minus the demand. Find the residual in mg/L.
- Find the residual from problem 4 as a percent of the 2.4 mg/L dose.
- The maximum allowed chlorine level in the pipes is 4.0 mg/L. How far below that ceiling is the 0.9 mg/L residual from problem 4?
- During a warm spell the plant raises the dose by 25 percent. Find the new dose in mg/L.
- Explain why warmer water can raise chlorine demand, so the plant needs a higher dose to hold the same residual.
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: Yes, the feed is proportional.
15 ÷ 100,000 = 0.00015; 30 ÷ 200,000 = 0.00015; 45 ÷ 300,000 = 0.00015. Every row gives the same ratio, so the solution fed is proportional to the water treated. - Answer: 37.5 gallons.
250,000 is 2.5 times 100,000, so the solution is 2.5 × 15 = 37.5 gallons. - Answer: 400,000 gallons.
60 gallons is 4 times the 15 gallons that treats 100,000 gallons, so it treats 4 × 100,000 = 400,000 gallons. - Answer: 0.9 mg/L.
Residual = dose − demand = 2.4 − 1.5 = 0.9 mg/L. - Answer: 37.5%.
0.9 ÷ 2.4 = 0.375, which is 37.5 percent of the dose. - Answer: 3.1 mg/L.
4.0 − 0.9 = 3.1 mg/L below the ceiling. - Answer: 3.0 mg/L.
A 25 percent increase gives 2.4 × 1.25 = 3.0 mg/L. - Answer: Warm water speeds the reactions that use up chlorine.
In warmer water chlorine reacts with germs and other material faster and breaks down sooner, so more of the dose is used as demand, and the plant must add more to leave the same residual in the pipes.
Differentiation
Support
Provide the ratio setups for problem 1 already written, and give students a labeled bar showing dose split into demand plus residual before problem 4. Keep a calculator available for the decimal division in problem 5.
Extension
Suppose the demand rose to 2.0 mg/L and the plant wanted to hold a 1.2 mg/L residual. Ask students to find the dose that would take, and to check that the 1.2 mg/L residual stays below the 4.0 mg/L ceiling.
ELL supports
Pre-teach "dose," "demand," and "residual" with a picture of water entering with a full cup of chlorine, some being used up, and some left over. Offer the sentence frame for problem 8: "Warm water needs ___ chlorine because ___."