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Math problem set Grade 12 Engineering & Design

Gullwing Bay Desalination Plant: Recovery, Rejection, and Energy Math

A grade 12 reverse-osmosis problem set where students track flow and salt through a fictional seawater plant, finding recovery, salt rejection, and daily energy use.

Objectives

  1. Students will calculate recovery and salt rejection for a reverse-osmosis system using a flow and salt mass balance.
  2. Students will estimate a plant's daily energy use and compare it against conventional treatment.

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, salt rejection, mass balance, permeate

Teacher overview

The fictional Gullwing Bay seawater plant uses reverse osmosis to push seawater against a membrane at high pressure, so that fresh water passes through and most of the salt is left behind. This set has students follow the flow and the salt through the plant to find its recovery, its salt rejection, and its daily energy use. A common misconception in this math is treating the concentrate (the leftover brine) as if it carried no salt, so watch for that in the mass-balance step in problem 5.

Gullwing Bay Desalination Plant: Recovery, Rejection, and Energy Math — Grade 12 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Gullwing Bay plant for every problem:

  • The plant takes in a feed of 5.0 million gallons per day (MGD) of seawater.
  • It is designed for a recovery of 45 percent, meaning 45 percent of the feed becomes fresh product water (permeate).
  • The feed seawater has a salt concentration of 35,000 mg/L. The permeate has a salt concentration of 350 mg/L.
  • Seawater reverse osmosis uses about 11 to 15 kWh of energy per 1,000 gallons of product water. Conventional treatment uses about 1 to 2 kWh per 1,000 gallons.
  • Useful relationships: permeate flow = recovery x feed flow; feed flow = permeate flow + concentrate flow; salt in = salt out; 1 MGD = 694 gpm.
  1. Find the permeate (product water) flow, in MGD, at the 45 percent recovery.
  2. Using the mass balance that feed equals permeate plus concentrate, find the concentrate flow in MGD.
  3. Convert the permeate flow from problem 1 to gallons per minute (gpm). Round to the nearest gpm.
  4. Find the salt rejection as a percent, using rejection = (1 - permeate concentration / feed concentration) x 100.
  5. Find the salt concentration of the concentrate, in mg/L. Balance the salt: feed flow x feed concentration = permeate flow x permeate concentration + concentrate flow x concentrate concentration.
  6. Estimate the plant's daily energy use for producing the permeate, using the high end of 15 kWh per 1,000 gallons.
  7. A conventional plant producing the same amount of product water uses about 2 kWh per 1,000 gallons. Find that plant's daily energy use, and the difference between it and your answer to problem 6.
  8. Gullwing Bay uses far more energy than a conventional plant. State one reason a coastal community might still choose seawater desalination.

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: 2.25 MGD.
    Permeate flow = recovery x feed flow = 0.45 x 5.0 MGD = 2.25 MGD.
  2. Answer: 2.75 MGD.
    Feed = permeate + concentrate, so concentrate = feed - permeate = 5.0 MGD - 2.25 MGD = 2.75 MGD.
  3. Answer: 1,562 gpm.
    2.25 MGD x 694 gpm/MGD = 1,561.5 gpm, which rounds to 1,562 gpm.
  4. Answer: 99 percent.
    Rejection = (1 - 350 / 35,000) x 100 = (1 - 0.01) x 100 = 99 percent.
  5. Answer: 63,350 mg/L.
    Salt in the feed = 5.0 x 35,000 = 175,000 (in MGD times mg/L). Salt in the permeate = 2.25 x 350 = 787.5. Salt in the concentrate = 175,000 - 787.5 = 174,212.5. Concentrate concentration = 174,212.5 / 2.75 = 63,350 mg/L. The 8.34 pound factor cancels from both sides, so the balance can be done with flow times concentration directly.
  6. Answer: 33,750 kWh per day.
    The permeate is 2.25 MGD, which is 2,250,000 gallons per day, or 2,250 thousand-gallon units. At 15 kWh per 1,000 gallons: 2,250 x 15 = 33,750 kWh per day.
  7. Answer: 4,500 kWh per day.
    The conventional plant produces the same 2,250 thousand-gallon units. At 2 kWh per 1,000 gallons: 2,250 x 2 = 4,500 kWh per day. The difference is 33,750 - 4,500 = 29,250 kWh per day more for the desalination plant.
  8. Answer: Drought-proof local supply.
    Accept any sound reason, such as: seawater is a reliable local source that does not run low during drought, unlike rivers, reservoirs, or groundwater, so the community gains a supply it can count on even in dry years.

Differentiation

Support

Provide a labeled flow diagram of the plant with arrows for feed, permeate, and concentrate, and the recovery step in problem 1 already worked, so students can focus on the mass balance and the energy comparison.

Extension

Suppose the fictional plant raises its recovery target above 45 percent. Ask students to explain, using the mass balance, what happens to the concentrate flow and to the concentrate salt concentration, and why very high recovery makes the brine harder to manage.

ELL supports

Pre-teach "feed," "permeate," "concentrate," "recovery," and "rejection" with a simple labeled diagram and a home-language gloss where possible. Allow students to record the mass-balance work as a labeled equation with numbers substituted in.

Extensions

Have students find the ratio of the concentrate flow to the permeate flow for this plant and discuss, in a short paragraph, why disposing of a large, very salty brine stream is one of the main engineering and environmental challenges of seawater desalination.