Water for Schools
Menu

Lesson plan Grade 9 Math

Westmere Reverse Osmosis: Modeling Recovery Rate

A 50-minute grade 9 lesson at the fictional Westmere plant, modeling reverse-osmosis recovery rate as permeate over feed and solving for a required feed flow.

Objectives

  1. Students will calculate reverse-osmosis recovery rate as permeate flow divided by feed flow.
  2. Students will use the feed-equals-permeate-plus-concentrate balance to find a missing flow.
  3. Students will solve an equation for the feed flow a target permeate output requires.

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: recovery rate, permeate, concentrate, feed flow, membrane

Teacher overview

The fictional Westmere plant pushes salty groundwater through reverse-osmosis membranes, which let clean water pass and hold the salt back. Operators track recovery rate, the share of the feed that comes out as clean water, because it sets how much water the plant makes from each gallon it pumps. This lesson builds recovery rate as a ratio of flows and then solves for the feed a target output needs. A common misconception is that recovery can reach 100 percent; some water must always leave as salty concentrate, so recovery is always below 100 percent.

Procedure

Hook (5 min): Sketch three beakers on the board, one of salty water going in and two coming out, one clear and one saltier. Ask students to guess what share of the feed water a membrane turns into clean water, and record guesses without comment. Name that share the recovery rate.

Direct instruction (12 min): Define recovery rate = permeate flow divided by feed flow, times 100, and the balance feed = permeate plus concentrate. Model one worked example on the board for the fictional plant: a feed of 1,000 gpm gives 700 gpm of permeate, so recovery = 700 divided by 1,000, times 100 = 70 percent, and the concentrate is 1,000 minus 700 = 300 gpm. This example is teacher-modeled only; students solve different numbers on the handout.

Guided practice (12 min): With a partner, students work handout items 1 through 3, finding a recovery rate, a concentrate flow, and a permeate flow from a stated recovery. Circulate and check that they divide permeate by feed, not feed by permeate.

Independent practice (16 min): Students complete handout items 4 through 6 on their own: solving for a required feed, converting that feed to MGD, and explaining a limit on pushing recovery too high.

Closure and formative check (5 min): As an exit ticket, each student answers on a card: a unit takes a feed of 400 gpm and makes 320 gpm of permeate; what is the recovery rate? Collect and scan for the correct setup, 320 divided by 400, times 100 = 80 percent. If more than a third miss the setup, reteach the ratio before the next class.

Westmere Reverse Osmosis: Modeling Recovery Rate — Grade 9 Math
Name: ____________________________ Date: ______________

Student Handout

Westmere (a fictional plant) treats salty groundwater with reverse osmosis. Use these facts: recovery rate = permeate flow divided by feed flow, times 100; feed flow = permeate flow plus concentrate flow; one MGD equals 694 gpm.

  1. The unit takes a feed of 800 gpm and makes 600 gpm of permeate. Find the recovery rate as a percent.
  2. Using the same feed and permeate, find the concentrate flow.
  3. A second unit runs at 70 percent recovery on a feed of 500 gpm. Find its permeate flow.
  4. The plant must deliver 900 gpm of permeate while running at 75 percent recovery. Write and solve an equation for the required feed flow.
  5. Convert the feed flow from item 4 to MGD. Round to the nearest hundredth.
  6. Explain one reason an operator does not push recovery as high as possible, even though higher recovery means more clean water per gallon of feed.

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: 75 %. 600 divided by 800, times 100 = 0.75 times 100 = 75 percent.
  2. Answer: 200 gpm. Concentrate = feed minus permeate = 800 minus 600 = 200 gpm.
  3. Answer: 350 gpm. Permeate = 0.70 times 500 = 350 gpm.
  4. Answer: 1,200 gpm. Solve 0.75 times F = 900, so F = 900 divided by 0.75 = 1,200 gpm.
  5. Answer: 1.73 MGD. 1,200 gpm divided by 694 gpm per MGD = 1.729, about 1.73 MGD.
  6. Answer: As recovery rises, the leftover concentrate gets saltier because the same salt is packed into less water. That saltier concentrate can scale and foul the membranes, so operators hold recovery below the point where fouling speeds up.

Differentiation

Support

Give students the two rules on a reference card and one fully worked example using numbers not on the handout. Provide item 4 with the equation 0.75 times F = 900 already written so students only divide.

Extension

Ask students to find the concentrate flow for the item-4 fictional plant and then the recovery rate that would cut the concentrate in half while keeping the same 900 gpm of permeate, and to explain why that pushes the feed lower.

ELL supports

Pre-teach "feed," "permeate," and "concentrate" with a labeled arrow diagram: one arrow in, two arrows out. Provide a sentence frame for item 6: "Higher recovery makes the concentrate more ___, which can ___ the membrane."