Math problem set Grade 10 Math

Saltcombe Reverse Osmosis Plant: Flow Balance and Energy Problems

A grade 10 problem set solving a feed-permeate-concentrate system of equations for a fictional desalter, then finding recovery, salt rejection, and energy use.

Grade: Grade 10
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSA-REI.C.6 Common Core Math

Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.

How this material aligns: Problems 1 and 2 have students solve a system of two linear equations in two variables for the permeate and concentrate flows that sum to the feed, exactly this standard's paired-equation focus.

All materials for HSA-REI.C.6 →

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 3 through 7 carry flows in million gallons per day and gallons per minute, salinity in milligrams per liter, and energy in kilowatt-hours per thousand gallons through a consistent unit chain.

All materials for HSN-Q.A.1 →

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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 solve a system of equations for the permeate and concentrate flows of an RO plant.
Students will calculate percent recovery and salt rejection for a reverse osmosis system.
Students will find the daily energy use of an RO plant from a per-thousand-gallon rate.

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

The city of Saltcombe, a fictional city used for this problem set, runs a brackish-water reverse osmosis plant. Feed water is pushed through membranes and splits into permeate, the clean product water, and concentrate, the salty reject stream. The problems build one scenario: students solve a system of equations for the two flow streams, then find recovery, salt rejection, and daily energy use. A common misconception in this work is forgetting the mass balance that feed equals permeate plus concentrate; another is confusing recovery, which is permeate divided by feed, with salt rejection.

Saltcombe Reverse Osmosis Plant: Flow Balance and Energy Problems — Grade 10 Math

Name: ____________________________ Date: ______________

Problems

The city of Saltcombe's reverse osmosis plant takes in a feed flow of 4.0 million gallons per day (MGD). Inside the plant the feed splits into a permeate stream (P) and a concentrate stream (C). Testing shows the permeate flow is 3 times the concentrate flow. Use these facts for all problems:

Feed = permeate + concentrate, so P + C = 4.0 MGD, and P = 3C.
1 MGD = 694 gallons per minute (gpm).
Percent recovery = permeate ÷ feed × 100.
Salt rejection (%) = (feed salinity − permeate salinity) ÷ feed salinity × 100.
The plant uses 3.5 kilowatt-hours (kWh) of energy per 1,000 gallons of permeate produced, a value set for this problem set. Brackish-water plants use less energy than seawater plants, which use roughly 11 to 15 kWh per 1,000 gallons.

Using the system P + C = 4.0 and P = 3C, solve for the concentrate flow C, in million gallons per day.
Use your answer from item 1 to find the permeate flow P, in million gallons per day.
Find the plant's percent recovery, using the permeate and feed flows. Round to the nearest whole percent.
Convert the permeate flow from item 2 to gallons per minute.
The permeate stream carries 3,000,000 gallons per day. Find the plant's energy use, in kilowatt-hours per day, at 3.5 kilowatt-hours per 1,000 gallons.
The feed salinity is 3,200 milligrams per liter and the permeate salinity is 120 milligrams per liter. Find the plant's salt rejection as a percentage. Round to the nearest hundredth.
Using a salt balance, feed flow × feed salinity = permeate flow × permeate salinity + concentrate flow × concentrate salinity, find the concentrate salinity in milligrams per liter. Use the flows from items 1 and 2 in million gallons per day.
Explain why the concentrate stream is much saltier than the feed water, using your answers to items 1, 2, and 7.
The city wants to raise recovery above 75 percent. Explain one tradeoff that comes with pushing recovery higher, using what your answer to item 7 shows about the concentrate.

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: 1.0 MGD.
Substitute P = 3C into P + C = 4.0: 3C + C = 4.0, so 4C = 4.0 and C = 1.0 million gallons per day.
Answer: 3.0 MGD.
P = 3C = 3 × 1.0 = 3.0 million gallons per day.
Answer: 75 %.
Recovery = permeate ÷ feed × 100 = 3.0 ÷ 4.0 × 100 = 75 percent.
Answer: 2,082 gpm.
3.0 MGD × 694 gallons per minute per MGD = 2,082 gallons per minute.
Answer: 10,500 kWh.
3,000,000 gallons ÷ 1,000 = 3,000 thousand-gallon units. 3,000 × 3.5 kilowatt-hours = 10,500 kilowatt-hours per day.
Answer: 96.25 %.
Salt rejection = (3,200 − 120) ÷ 3,200 × 100 = 3,080 ÷ 3,200 × 100 = 96.25 percent.
Answer: 12,440 mg/L.
Salt balance: 4.0 × 3,200 = 3.0 × 120 + 1.0 × concentrate salinity. So 12,800 = 360 + concentrate salinity, and concentrate salinity = 12,440 milligrams per liter.
Answer: The rejected salt concentrates into a smaller stream.
Nearly all the salt in the feed stays out of the permeate, so it leaves in the concentrate. That salt load is packed into only 1.0 million gallons per day of concentrate rather than the full 4.0 million gallons per day of feed, which makes the concentrate far saltier than the feed (item 7 shows 12,440 milligrams per liter).
Answer: Higher recovery makes an even saltier, harder-to-manage concentrate.
Pushing recovery above 75 percent means taking more clean permeate out of the same feed, which leaves the remaining salt in a smaller concentrate stream. That raises concentrate salinity further, increasing the risk of scale forming on the membranes and making the reject stream harder and costlier to dispose of.

Differentiation

Support

Provide the solved system (C = 1.0 and P = 3.0 million gallons per day) so students who struggle with the substitution can still complete items 3 through 7. Give a printed reference card with the recovery, rejection, and salt-balance formulas.

Extension

Ask students to find what the concentrate flow and concentrate salinity would become if recovery rose to 80 percent on the same 4.0 million gallons per day feed and the same salt load, then compare the new concentrate salinity to the value from item 7.

ELL supports

Provide a labeled flow diagram showing feed entering the membrane and splitting into permeate and concentrate, with arrows and the words for each stream. Pre-teach "permeate," "concentrate," "recovery," and "rejection" with the diagram, and give sentence frames for items 8 and 9.