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Math problem set Grade 11 Engineering & Design

Tarnbrook Brackish RO Pilot: Recovery, Rejection, and Energy Problems

A grade 11 engineering set on the fictional Tarnbrook brackish-water RO pilot, covering percent recovery, salt rejection, a mass balance, and specific energy use.

Objectives

  1. Students will calculate percent recovery and salt rejection for a reverse-osmosis system from feed and permeate values.
  2. Students will use a mass balance to find the concentration of the concentrate stream.
  3. Students will compute specific energy use and compare it against published desalination energy ranges.

Vocabulary

Key words: permeate, concentrate, percent recovery, salt rejection, specific energy

Teacher overview

Tarnbrook is a fictional inland town testing a reverse-osmosis pilot on brackish groundwater, water that is too salty to drink but far less salty than seawater. A reverse-osmosis unit pushes feed water against a membrane under pressure; part of it passes through as clean permeate, and the rest leaves as a saltier concentrate. This set has students track the flows and salt concentrations through the pilot, close a mass balance on the salt, and compare the pilot's energy use with the ranges engineers cite for real plants. A common mistake here is forgetting that the salt has to go somewhere: the salt removed from the permeate is not destroyed but is carried out in the smaller concentrate stream, which is why that stream ends up much saltier than the feed.

Tarnbrook Brackish RO Pilot: Recovery, Rejection, and Energy Problems — Grade 11 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Tarnbrook pilot for every problem:

  • Feed flow into the unit is 500 gpm, with a total dissolved solids concentration of 3,000 mg/L.
  • The pilot runs at 75 percent recovery, meaning permeate flow is 75 percent of feed flow.
  • The permeate leaves at 90 mg/L of total dissolved solids.
  • For real plants, seawater reverse osmosis uses roughly 11 to 15 kWh per 1,000 gallons of product water (about 3 to 4 kWh per cubic meter), while conventional treatment and delivery use about 1 to 2 kWh per 1,000 gallons.
  1. Find the permeate flow in gallons per minute, using the 75 percent recovery.
  2. Find the concentrate flow in gallons per minute, using feed flow minus permeate flow.
  3. Calculate the salt rejection as a percent, using rejection = (feed concentration − permeate concentration) ÷ feed concentration × 100.
  4. Use a salt mass balance to find the concentrate concentration in mg/L. The salt entering with the feed equals the salt leaving in the permeate plus the salt leaving in the concentrate; each salt load is flow multiplied by concentration.
  5. Convert the permeate flow to gallons per day, using 1,440 minutes in a day.
  6. The pilot measures a specific energy of 2.6 kWh per 1,000 gallons of permeate. Find the total energy the pilot uses in one day, in kWh.
  7. Compare the pilot's 2.6 kWh per 1,000 gallons with the seawater range of 11 to 15 and the conventional range of 1 to 2. Where does brackish reverse osmosis fall, and give one physical reason it uses less energy than seawater desalination.
  8. The operators consider raising recovery to 85 percent. Find the new concentrate flow and its concentration by mass balance, then explain one trade-off that higher recovery creates.

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: 375 gpm. Permeate = 0.75 × 500 = 375 gallons per minute.
  2. Answer: 125 gpm. Concentrate = feed − permeate = 500 − 375 = 125 gallons per minute.
  3. Answer: 97 percent. Rejection = (3,000 − 90) ÷ 3,000 × 100 = 2,910 ÷ 3,000 × 100 = 97 percent.
  4. Answer: 11,730 mg/L. Feed salt load = 500 × 3,000 = 1,500,000. Permeate salt load = 375 × 90 = 33,750. Concentrate salt load = 1,500,000 − 33,750 = 1,466,250. Concentrate concentration = 1,466,250 ÷ 125 = 11,730 mg/L.
  5. Answer: 540,000 gpd. 375 gpm × 1,440 minutes/day = 540,000 gallons per day.
  6. Answer: 1,404 kWh. 540,000 gallons ÷ 1,000 = 540 thousand-gallon units; 540 × 2.6 = 1,404 kWh in one day.
  7. Answer: Between the two ranges. At 2.6 kWh per 1,000 gallons, brackish reverse osmosis sits above the conventional range of 1 to 2 but below the seawater range of 11 to 15. It uses less energy than seawater desalination because the feed is far less salty, so its osmotic pressure is lower and the pump does not have to push nearly as hard to move water across the membrane.
  8. Answer: 19,490 mg/L. At 85 percent recovery, permeate = 0.85 × 500 = 425 gpm, so concentrate = 500 − 425 = 75 gpm. Concentrate salt load = 1,500,000 − (425 × 90) = 1,500,000 − 38,250 = 1,461,750; concentration = 1,461,750 ÷ 75 = 19,490 mg/L. The trade-off is that raising recovery produces more clean water but shrinks the concentrate into a much saltier stream, which raises the risk of mineral scaling on the membrane and makes the concentrate harder and costlier to dispose of.

Differentiation

Support

Provide a labeled flow diagram showing the feed splitting into permeate and concentrate, with the flow and concentration blanks marked where each answer goes. Give the recovery and rejection formulas on a reference strip so students focus on the arithmetic and unit tracking.

Extension

Ask students to calculate how many gallons of concentrate the pilot produces each day at 75 percent recovery and to propose, in a short paragraph, one option for managing that concentrate stream given that it is nearly four times as salty as the feed.

ELL supports

Pre-teach "feed," "permeate," "concentrate," "recovery," and "rejection" with a picture glossary that pairs each word with the part of the flow diagram it names. Allow students to answer item 7 with a labeled number line showing where 2.6 falls between the two ranges, plus one short sentence.