Math problem set Grade 11 Math

Seabright Bay Desalination: Reverse-Osmosis Recovery and Energy

A grade 11 problem set on a fictional seawater RO plant, finding permeate and brine flows, percent recovery and salt rejection, an equation model, and daily energy cost.

Objectives

  1. Students will compute percent recovery, salt rejection, and product and brine flows for a reverse-osmosis plant.
  2. Students will write an equation relating permeate flow to feed flow and calculate the plant's daily energy cost.

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, recovery, salt rejection

Teacher overview

Reverse osmosis pushes seawater against a membrane under high pressure so that clean water passes through while most of the salt is left behind. Two percentages describe how the plant runs: recovery, the share of feed water that becomes clean product, and salt rejection, the share of salt the membrane removes. This set uses a fictional plant at Seabright Bay to practice both, along with the flows and energy that follow. A common misconception is treating recovery and salt rejection as the same number; they measure different things, and a plant can reject 99 percent of the salt while recovering less than half of the water.

Answer Key

  1. Answer: 4.5 MGD. 0.45 x 10 MGD = 4.5 MGD of permeate.
  2. Answer: 5.5 MGD. 10 MGD feed - 4.5 MGD permeate = 5.5 MGD of concentrate.
  3. Answer: about 3,123 gpm. 4.5 MGD x 694 gpm per MGD = 3,123 gpm.
  4. Answer: 99 percent. (1 - 350 / 35,000) x 100 = (1 - 0.01) x 100 = 99 percent.
  5. Answer: 3.6 MGD. The equation is P = 0.45 x F. At F = 8 MGD, P = 0.45 x 8 = 3.6 MGD.
  6. Answer: 58,500 kWh per day. 4.5 MGD = 4,500,000 gallons per day; 4,500,000 / 1,000 = 4,500 thousand-gallon units; 4,500 x 13 kWh = 58,500 kWh per day.
  7. Answer: 7,020 dollars per day. 58,500 kWh x 0.12 dollars per kWh = 7,020 dollars per day.
  8. Answer: Desalination must raise seawater to high pressure and force it through a membrane, and only 45 percent of the pumped water becomes drinking water. The energy spent pressurizing the other 55 percent, which leaves as brine, is spread over a smaller product volume, so each product gallon carries more energy than a gallon of treated river water.

Differentiation

Support

Provide a labeled flow diagram showing feed splitting into permeate and concentrate, with the recovery percentage marked, so students can see where each flow in items 1 and 2 comes from. Give the energy chain in item 6 broken into two labeled steps.

Extension

Ask students to find the plant's energy use per 1,000 gallons of feed water rather than per 1,000 gallons of permeate, and to explain why that number is lower than the 13 figure even though the pumps do the same work.

ELL supports

Pre-teach "membrane," "permeate," "brine," and "recovery" with a labeled diagram of water splitting at a membrane. Provide a sentence frame for item 8: "Only ___ percent becomes product, so the energy is spread over ___ water, which makes each gallon cost ___."