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Math problem set Grade 6 Science

Bellhaven's Desalination Unit: Reverse Osmosis Recovery Problems

A grade 6 math problem set where students calculate a fictional town's reverse-osmosis recovery rate and test whether it meets a growing population's water demand.

Objectives

  1. Students will calculate percent recovery for a reverse-osmosis desalination system.
  2. Students will use a per-capita water-use estimate to evaluate whether a desalination unit's output could meet a growing population's demand.

Vocabulary

Reverse osmosis
Reverse osmosis cleans water by pushing it through a membrane with holes so tiny that salt and most other dissolved things cannot fit through. It can even turn seawater into fresh water.

Also: permeate, reject, percent recovery

Teacher overview

Reverse osmosis desalination pushes seawater through a membrane that lets fresh water pass through while leaving salt behind; the fresh water that gets through is called permeate, and the leftover, saltier water is called reject. The nation's largest such facility, the Carlsbad, California desalination plant, produces about 50 million gallons per day, roughly 10% of San Diego County's water supply. This problem set has students calculate the recovery rate for a much smaller, fictional system and then test whether its output could realistically meet a growing town's water needs. A common misconception in this math is assuming a desalination unit converts all of its intake seawater into fresh water; in reality, only a percentage becomes usable permeate, and tracking that percentage is exactly what recovery rate measures.

Bellhaven's Desalination Unit: Reverse Osmosis Recovery Problems — Grade 6 Science
Name: ____________________________ Date: ______________

Problems

The fictional coastal town of Bellhaven is building a small reverse-osmosis unit. The unit will help its water supply during drought. Its feed (intake) seawater flow is 2,000 gpm. The unit recovers 40% of that flow as permeate (fresh water). The rest becomes reject. Use: Percent recovery = permeate flow ÷ feed flow × 100.

  1. Calculate the permeate (fresh water) flow in gpm.
  2. Calculate the reject flow in gpm.
  3. Add your answers to problems 1 and 2. Does the total match the original feed flow of 2,000 gpm?
  4. Convert the permeate flow from problem 1 into gallons per day. Use 1 day = 1,440 minutes.
  5. Convert the permeate flow from problem 4 into millions of gallons per day (MGD). Round to the nearest hundredth.
  6. The real Carlsbad plant produces 50 MGD. What percent of that output does Bellhaven's permeate output from problem 5 represent? Round to the nearest tenth of a percent.
  7. Per-person water use at home in the US is commonly cited at 80 to 100 gallons per day. Use the higher estimate, 100 gallons per day. About how many people could Bellhaven's permeate output from problem 4 supply in a single day?
  8. Bellhaven's current population is 9,000 people. Look at your answer to problem 7. Could the reverse-osmosis unit alone supply every current resident at 100 gallons per person per day, even before counting the town's other water sources?
  9. Suppose Bellhaven's population grows to 13,000 people. At 100 gallons per person per day, how many more gallons per day would the town need? Compare this to its current 9,000-person demand.
  10. In two to three sentences, explain whether the reverse-osmosis unit alone looks like a realistic way for Bellhaven to meet all of a growing population's water needs. Base your answer on problems 7 through 9.

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: 800 gpm.
    2,000 gpm × 0.40 = 800 gpm of permeate.
  2. Answer: 1,200 gpm.
    2,000 gpm − 800 gpm = 1,200 gpm of reject.
  3. Answer: Yes, 2,000 gpm.
    800 gpm + 1,200 gpm = 2,000 gpm, which matches the original feed flow.
  4. Answer: 1,152,000 gpd.
    800 gpm × 1,440 min/day = 1,152,000 gallons per day.
  5. Answer: 1.15 MGD.
    1,152,000 gal ÷ 1,000,000 = 1.152, which rounds to 1.15 MGD.
  6. Answer: 2.3 percent.
    1.152 MGD ÷ 50 MGD × 100 = 2.304%, which rounds to 2.3%.
  7. Answer: 11,520 people.
    1,152,000 gal ÷ 100 gal/person = 11,520 people.
  8. Answer: Yes.
    11,520 people is more than Bellhaven's current population of 9,000, so the unit alone could theoretically supply every current resident at that rate, with capacity for about 2,520 more people left over.
  9. Answer: 400,000 gpd.
    13,000 people × 100 gal/person = 1,300,000 gal/day; 9,000 people × 100 gal/person = 900,000 gal/day; 1,300,000 − 900,000 = 400,000 more gallons per day needed.
  10. Answer: No, it does not look fully realistic on its own. At 13,000 residents, the town would need 1,300,000 gallons per day at the 100 gallons-per-person estimate, but the reverse-osmosis unit produces only 1,152,000 gallons per day, so growth beyond roughly 11,500 people would leave a gap the unit could not cover without another water source.

Differentiation

Support

Provide the percent recovery formula written as a labeled fraction (permeate over feed) with problem 1 solved as a worked example, and give students a printed conversion card listing "1 day = 1,440 minutes" and "1 MGD = 1,000,000 gallons per day" for problems 4 and 5.

Extension

Ask students to calculate what feed flow, in gpm, Bellhaven's unit would need at the same 40% recovery rate to fully cover the 13,000-person demand from problem 9, working backward from the required permeate output.

ELL supports

Pre-teach "reverse osmosis," "permeate," "reject," and "percent recovery" with a labeled diagram of a membrane showing seawater entering on one side and splitting into a smaller fresh-water stream and a larger salty stream. Allow students to show their work for problems 1 through 5 using the diagram and numbers rather than full sentences.