Math problem set Grade 12 Science

Kingsley Cove Brackish RO: Recovery and Mass Balance

A grade 12 problem set on a fictional reverse-osmosis plant, finding recovery, salt rejection, a brine mass balance, concentration factor, and daily treatment energy.

Grade: Grade 12
Subject: Science
Material: Math problem set
Time: 50 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 4 through 8 carry units through the pounds formula for salt load and a per-thousand-gallon energy calculation, tracking mg/L, MGD, pounds, and kWh.

All materials for HSN-Q.A.1 →

HSN-Q.A.2 Common Core Math

Define appropriate quantities for the purpose of descriptive modeling.

How this material aligns: Problems 1, 3, and 7 define recovery, salt rejection, and concentration factor as distinct descriptive quantities for the same plant, as this standard requires.

All materials for HSN-Q.A.2 →

HS-ESS3-2 NGSS

Evaluate competing design solutions for developing, managing, and utilizing energy and mineral resources based on cost-benefit ratios.

How this material aligns: Problem 8 sets the plant's 3.5 kWh per thousand gallons against the 2 kWh upper figure for conventional treatment, the cost-benefit weighing of a competing water-supply technology this standard requires.

All materials for HS-ESS3-2 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate recovery, brine flow, and salt rejection for a reverse-osmosis plant.
Students will use a salt mass balance to find the brine concentration.
Students will compute daily treatment energy and compare it with 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, concentration factor, mass balance

Teacher overview

The fictional Kingsley Cove plant pushes brackish groundwater through reverse-osmosis membranes, splitting the feed into clean permeate and a saltier brine. This set has students find the recovery and salt rejection, then use a mass balance to show where the rejected salt goes and how energy-hungry the process is. A common misconception is that the salt removed from the permeate simply disappears; the mass balance in problems 4 through 7 makes clear that every pound of salt not in the permeate leaves in the smaller brine stream, which raises the brine concentration well above the feed.

Kingsley Cove Brackish RO: Recovery and Mass Balance — Grade 12 Science

Name: ____________________________ Date: ______________

Problems

Suppose the town of Kingsley Cove runs its brackish reverse-osmosis plant at a feed flow of 4.0 MGD, producing a permeate flow of 2.6 MGD. The feed water carries 3,200 mg/L of total dissolved solids, and the permeate carries 90 mg/L. Use the pounds formula, load (lbs/day) = concentration (mg/L) x flow (MGD) x 8.34.

Find the plant's recovery, where recovery = permeate flow / feed flow, as a percentage.
Find the brine (concentrate) flow in MGD.
Find the salt rejection, where rejection = (1 - permeate TDS / feed TDS) x 100. Round to the nearest hundredth.
Find the salt load entering in the feed, in pounds per day.
Find the salt load leaving in the permeate, in pounds per day.
Use a mass balance to find the salt load leaving in the brine, then find the brine TDS in mg/L. (Brine TDS = brine salt load / (brine flow x 8.34).)
Find the concentration factor two ways: as brine TDS divided by feed TDS, and as 1 / (1 - recovery). Compare the two values.
In our scenario the plant uses 3.5 kWh of energy per 1,000 gallons of permeate, while conventional treatment uses about 1 to 2 kWh per 1,000 gallons. Find the plant's daily energy for its permeate, then state how many times the upper conventional figure of 2 kWh that per-thousand-gallon rate represents.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

Answer: 65%.
Recovery = 2.6 MGD / 4.0 MGD = 0.65, or 65 percent.
Answer: 1.4 MGD.
Brine flow = feed - permeate = 4.0 - 2.6 = 1.4 MGD.
Answer: 97.19%.
Rejection = (1 - 90 / 3,200) x 100 = (1 - 0.028125) x 100 = 97.19 percent.
Answer: 106,752 lbs/day.
Feed salt load = 3,200 mg/L x 4.0 MGD x 8.34 = 106,752 pounds per day.
Answer: 1,951.56 lbs/day.
Permeate salt load = 90 mg/L x 2.6 MGD x 8.34 = 1,951.56 pounds per day.
Answer: 8,976 mg/L.
Brine salt load = 106,752 - 1,951.56 = 104,800.44 pounds per day. Brine TDS = 104,800.44 / (1.4 x 8.34) = 104,800.44 / 11.676 = 8,976 mg/L.
Answer: about 2.8 versus 2.86.
Brine TDS / feed TDS = 8,976 / 3,200 = 2.81. The ideal factor 1 / (1 - 0.65) = 1 / 0.35 = 2.86. The measured factor is slightly lower because a small amount of salt leaves in the permeate rather than concentrating entirely in the brine.
Answer: 9,100 kWh/day, which is 1.75 times the 2 kWh rate.
Permeate is 2.6 MGD = 2,600,000 gallons per day, or 2,600 thousand-gallon units. Energy = 2,600 x 3.5 = 9,100 kWh per day. The 3.5 kWh rate is 3.5 / 2 = 1.75 times the upper conventional figure of 2 kWh per 1,000 gallons.

Differentiation

Support

Provide the pounds formula on a card and pre-label the feed, permeate, and brine streams on a flow diagram so students can see which flow and concentration each problem needs.

Extension

Ask students to find the recovery that would raise the brine TDS to about 12,000 mg/L, assuming the feed TDS and near-complete salt rejection hold, and explain why higher recovery makes the brine harder to dispose of.

ELL supports

Provide a labeled flow diagram with the words "feed," "permeate," and "brine" beside their arrows, and a sentence frame for problem 7 comparing "measured" and "ideal" concentration factors.

Extensions

Have students look up the reported recovery of a real seawater or brackish desalination plant, compare it with the 65 percent here, and explain why seawater plants often run at lower recovery than brackish plants.