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Math problem set Grade 9 Science

Seawater Desalination at Sable Coast: A Reverse Osmosis Problem Set

A grade 9 problem set where students find recovery, salt rejection, and the daily energy and cost of a fictional seawater reverse-osmosis plant.

Objectives

  1. Students will calculate the recovery and salt rejection of a reverse-osmosis plant.
  2. Students will convert a permeate flow to a daily volume and find the plant's daily energy use.
  3. Students will explain the energy and disposal trade-offs of seawater desalination.

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, concentrate, recovery, salt rejection

Teacher overview

Reverse osmosis pushes seawater against a membrane under high pressure, so fresh water passes through as permeate while most of the salt stays behind in a concentrated stream. Seawater desalination uses about 11 to 15 kWh per 1,000 gallons (about 3 to 4 kWh per cubic meter), several times the energy of treating a freshwater source, which is why coastal utilities weigh it carefully against conservation and reuse. This set follows the fictional Sable Coast plant through recovery, salt rejection, and a daily energy bill. A common misconception is that all the feed water becomes drinking water; in seawater desalination only part of the feed becomes permeate, and the rest leaves as salty concentrate.

Seawater Desalination at Sable Coast: A Reverse Osmosis Problem Set — Grade 9 Science
Name: ____________________________ Date: ______________

Problems

The Sable Coast plant (a fictional plant) treats seawater by reverse osmosis. Use these numbers for every problem:

  • The plant takes in a feed flow of 2,000 gpm and produces a permeate flow of 900 gpm.
  • The feed water contains 35,000 mg/L of dissolved salts; the permeate contains 350 mg/L.
  • The plant uses 13 kWh per 1,000 gallons of permeate produced.
  • Electricity costs $0.15 per kWh.
  1. Find the concentrate flow, the salty stream left behind, using concentrate = feed flow minus permeate flow.
  2. Calculate the plant's recovery, the percent of the feed that becomes permeate, using recovery = permeate flow divided by feed flow, times 100.
  3. Calculate the plant's salt rejection, using rejection = (feed salt minus permeate salt) divided by feed salt, times 100. Round to the nearest tenth of a percent.
  4. Convert the permeate flow of 900 gpm to gallons per day, using 1,440 minutes in a day.
  5. Calculate the energy the plant uses per day, in kilowatt-hours, using 13 kWh per 1,000 gallons of permeate from item 4.
  6. Calculate the plant's daily energy cost at $0.15 per kWh. Round to the nearest cent.
  7. The plant recovers only 45 percent of its feed water as permeate, so 55 percent leaves as concentrate. Explain one reason a plant cannot simply push recovery close to 100 percent.
  8. A brackish inland well holds far less salt than seawater. Explain why a reverse-osmosis plant treating that well water would use less energy per 1,000 gallons than the Sable Coast seawater plant.

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: 1,100 gpm.
    Concentrate = feed minus permeate = 2,000 gpm minus 900 gpm = 1,100 gpm.

  2. Answer: 45 percent.
    Recovery = 900 divided by 2,000, times 100 = 0.45 x 100 = 45 percent.

  3. Answer: 99.0 percent.
    Rejection = (35,000 minus 350) divided by 35,000, times 100 = 34,650 divided by 35,000, times 100 = 99.0 percent.

  4. Answer: 1,296,000 gal.
    900 gpm x 1,440 minutes per day = 1,296,000 gallons per day.

  5. Answer: 16,848 kWh.
    1,296,000 gallons is 1,296 thousand-gallon units. 13 kWh per 1,000 gallons x 1,296 = 16,848 kilowatt-hours per day.

  6. Answer: $2,527.20.
    16,848 kWh x $0.15 per kWh = $2,527.20 per day.

  7. Answer: As recovery rises, the concentrate gets saltier and can scale or foul the membrane.
    Pulling more fresh water out of the feed leaves the remaining concentrate more concentrated, so salts begin to reach the point where they crystallize on the membrane (scaling). That scaling damages the membrane and forces higher pressure, so plants stop well short of full recovery. Accept any answer that connects higher recovery to a saltier, harder-to-treat concentrate.

  8. Answer: Less salt means lower osmotic pressure, so the pump uses less energy.
    Reverse osmosis must overcome the osmotic pressure that pulls water toward the saltier side, and that pressure rises with salt content. Brackish well water holds far less salt than seawater, so the plant can run at lower pressure and use less energy per 1,000 gallons. Accept any answer that connects lower salt to lower pressure and energy.

Differentiation

Support

Provide a labeled flow diagram showing feed splitting into permeate and concentrate, with the salt levels marked, so students can see where each number belongs before they calculate. Give the recovery and rejection formulas already written with labeled blanks.

Extension

Ask students to suppose the plant raised recovery so that permeate reached 1,100 gpm, then calculate how many kilowatt-hours it would use in a day, then discuss why the higher recovery is not free even though it makes more water.

ELL supports

Pre-teach "permeate," "concentrate," and "recovery" with a simple picture of seawater entering and splitting into a clean stream and a salty stream. Provide a sentence frame for item 8: "Brackish water needs less energy because it has ___."