Worksheet Grade 7 Science

Sabine Bay's Desalination Pilot: Membrane Rejection and Flow Splits

A grade 7 worksheet on a fictional town's reverse osmosis pilot, calculating percent TDS removal, the permeate-reject flow split, and a gpm-to-gallons conversion.

Objectives

  1. Students will calculate the percent of dissolved solids a reverse osmosis membrane removes from feed water.
  2. Students will use a proportional permeate-to-reject split to scale a system's output to new flow rates.
  3. Students will convert a flow rate from gallons per minute to gallons per day.

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.
Total dissolved solids (TDS)
Total dissolved solids is a measure of everything dissolved in water that you cannot see or filter out, like minerals and salts. Water with very high TDS can taste salty or bitter, even though it might look perfectly clear.

Also: membrane, permeate, reject

Teacher overview

The fictional coastal town of Sabine Bay is piloting a reverse osmosis (RO) system to turn brackish well water into drinking water. In an RO system, feed water is pushed under pressure through a membrane with pores small enough to block most dissolved solids; the water that passes through is called permeate, and the leftover, more concentrated stream is called reject. This worksheet uses measurements from Sabine Bay's fictional pilot unit to calculate how well the membrane performs and how its output splits between usable water and reject.

Answer Key

  1. Answer: permeate and reject.
    A reverse osmosis membrane pushes feed water through pores small enough to block most dissolved solids. Water that passes through is called permeate; the leftover, more concentrated water that does not pass through is called reject.
  2. Answer: 98 percent.
    (2,000 − 40) ÷ 2,000 × 100 = 1,960 ÷ 2,000 × 100 = 98%.
  3. Answer: 210 mg/L.
    The permeate measures 40 mg/L, well under the 250 mg/L goal. 250 − 40 = 210 mg/L of margin below the goal.
  4. Answer: The dissolved solids concentrate in the reject stream.
    The membrane blocks most of the dissolved solids from passing through, so the solids that don't leave with the permeate stay behind and become concentrated into the smaller reject stream, raising its TDS well above the original feed water.
  5. Answer: 9 gpm.
    15 gpm × 0.60 = 9 gpm of permeate.
  6. Answer: 6 gpm.
    15 gpm × 0.40 = 6 gpm of reject (or 15 − 9 = 6).
  7. Answer: 12,960 gal/day.
    9 gpm × 1,440 minutes/day = 12,960 gallons per day.
  8. Answer: 27 gpm.
    45 gpm × 0.60 = 27 gpm of permeate.
  9. Answer: closer to 100 percent.
    A removal percent closer to 100% means almost all of the dissolved solids stayed out of the permeate, which is the goal of treatment; a removal percent closer to 0% would mean the membrane barely changed the water's TDS at all.
  10. Answer: 20 gpm.
    12 gpm ÷ 0.60 = 20 gpm of feed flow needed to produce 12 gpm of permeate at a 60% recovery rate.

Differentiation

Support (below grade level)

Provide the removal efficiency formula on a printed reference card, and work through problem 2 as a full class example before students attempt problems 3 and 9 independently. Give students a simple diagram of the membrane with feed, permeate, and reject arrows labeled to support problems 1 and 4.

Extension (above grade level)

Ask students to find what percent recovery (permeate ÷ feed) the pilot unit in problems 5 and 6 achieves, expressed as a fraction, a decimal, and a percent, and to explain why a utility might want to raise that recovery percentage over time.

ELL supports

Pre-teach "membrane," "permeate," "reject," and "dissolved solids" with a simple diagram showing water and particles approaching a barrier, with small water molecules passing through and larger particle dots blocked. Allow problems 1 and 4 to be answered with labeled diagrams and short phrases rather than full sentences.