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Math problem set Grade 8 Engineering & Design

Bracewater's Reverse-Osmosis Blend: A Systems of Equations Set

A grade 8 set where students write and solve a two-equation system to blend pure reverse-osmosis water with bypass water to hit a target dissolved-solids level.

Objectives

  1. Students will write a system of two linear equations that models blending two water streams to a target concentration.
  2. Students will solve the system by substitution and interpret the blend as a ratio and percent.

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, bypass, dissolved solids, blend

Teacher overview

A reverse-osmosis plant pushes water through membranes that strip out almost all dissolved solids, making very pure water called permeate. Very pure water can be corrosive to pipes and costs extra energy to make, so real plants blend permeate with filtered water that skips the membranes, called bypass. This set has students model that blend as a system of two linear equations and solve it. A common misconception is treating each concentration as if it can simply be averaged; students must instead multiply each stream's concentration by its gallons before adding, because a large stream affects the mix more than a small one.

Bracewater's Reverse-Osmosis Blend: A Systems of Equations Set — Grade 8 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Use these design values for every problem. Suppose the town of Bracewater, a fictional town, runs a reverse-osmosis plant that must produce 90,000 gallons of finished water in a day. The permeate carries 30 mg/L of dissolved solids, and the bypass carries 480 mg/L. The finished water must average 300 mg/L. Let x be the gallons of permeate and y be the gallons of bypass.

  1. Write an equation showing that the permeate gallons and bypass gallons add up to the day's 90,000 gallons.

  2. The finished water must average 300 mg/L. Multiply each stream's concentration by its gallons, and set the sum equal to 300 times 90,000. Write this equation.

  3. Solve the system for x, the gallons of permeate. Show your substitution.

  4. Find y, the gallons of bypass.

  5. What percent of the finished water is bypass water that skips the membranes? Round to the nearest whole percent.

  6. Check your answer by finding the blended concentration from your permeate and bypass gallons. It should equal 300 mg/L.

  7. Demand rises to 120,000 gallons in a day, with the same two sources and the same 300 mg/L target. Using the same blend ratio, how many gallons of permeate are needed?

  8. Challenge. Explain why Bracewater does not simply send all 90,000 gallons through the membranes, and why it cannot use only bypass water. Use the concentrations and the idea of energy cost in your answer.

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: x + y = 90,000.
    The two streams together must equal the day's total finished water of 90,000 gallons.

  2. Answer: 30x + 480y = 27,000,000.
    Each stream's dissolved-solids load is concentration times gallons: 30x for permeate and 480y for bypass. The target load is 300 mg/L x 90,000 gallons = 27,000,000, so 30x + 480y = 27,000,000.

  3. Answer: 36,000 gal.
    From equation 1, x = 90,000 minus y. Substitute: 30(90,000 minus y) + 480y = 27,000,000, which gives 2,700,000 + 450y = 27,000,000, so 450y = 24,300,000 and y = 54,000. Then x = 90,000 minus 54,000 = 36,000 gallons of permeate.

  4. Answer: 54,000 gal.
    From the work in problem 3, y = 54,000 gallons of bypass.

  5. Answer: 60%.
    Bypass is 54,000 of the 90,000 total gallons. 54,000 divided by 90,000 = 0.60, or 60 percent bypass.

  6. Answer: 300 mg/L.
    Blended load = 30 x 36,000 + 480 x 54,000 = 1,080,000 + 25,920,000 = 27,000,000. Divide by 90,000 gallons to get 300 mg/L, which matches the target.

  7. Answer: 48,000 gal.
    The concentrations did not change, so the blend ratio stays 40 percent permeate and 60 percent bypass. Permeate = 0.40 x 120,000 = 48,000 gallons.

  8. Answer: Using only membranes wastes energy and makes water too pure.
    All-permeate water would sit near 30 mg/L, far below the 300 mg/L target, and every gallon would cost the extra membrane energy. All-bypass water would sit at 480 mg/L, above the target, so the town must blend the two streams to land at 300 mg/L while keeping energy use down.

Differentiation

Support

Give students equation 1 already written and the target load of 27,000,000 already computed, so they can focus on the substitution steps. Provide a worked mixture example with different numbers before problem 3 so the multiply-then-add structure is modeled first.

Extension

Ask students to re-solve the blend for a new membrane that lowers permeate to 20 mg/L while the bypass and 300 mg/L target stay the same. Students should find the new permeate and bypass gallons and explain whether more or less bypass is now needed.

ELL supports

Provide a simple two-box diagram for this fictional blend, one box labeled permeate with 30 mg/L and one labeled bypass with 480 mg/L, both flowing into a tank labeled 300 mg/L. Pre-teach permeate, bypass, and blend with the diagram, and allow students to label the boxes with their x and y values.