Math problem set Grade 10 Engineering & Design

Halcyon Springs Brackish RO Pilot: Recovery and Rejection Problems

A grade 10 engineering math set on a fictional brackish reverse-osmosis pilot: percent recovery, salt rejection, a concentrate mass balance, and energy use.

Grade: Grade 10
Subject: Engineering & Design
Material: Math problem set
Time: 45 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: Items 1 through 5 carry consistent flow, concentration, and energy units through a chain of recovery, rejection, and specific-energy calculations, the unit-guided reasoning this standard requires.

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 2026-07-17.

Objectives

Students will calculate percent recovery and salt rejection for a reverse-osmosis system from measured flows and concentrations.
Students will use a mass balance to find the concentrate stream's salt concentration.
Students will calculate a system's specific energy use and compare it to typical desalination values.

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, percent recovery, salt rejection

Teacher overview

The fictional Halcyon Springs water district is running a reverse-osmosis pilot to see whether a salty brackish groundwater source can be turned into drinking water. A membrane splits the incoming feed water into two streams: clean permeate that becomes product water, and a saltier concentrate that carries the rejected salt away. This set has students calculate how much of the feed becomes product, how much salt the membrane removes, how concentrated the leftover stream becomes, and how much energy the process uses. A common misconception in this math is treating recovery and rejection as the same idea: recovery is the share of water that becomes product, while rejection is the share of salt the membrane keeps out of that product, so the two describe completely different things.

Halcyon Springs Brackish RO Pilot: Recovery and Rejection Problems — Grade 10 Engineering & Design

Name: ____________________________ Date: ______________

Problems

The Halcyon Springs pilot (a fictional system built for this set) runs a feed flow of 500 gpm into the membranes. It produces 375 gpm of permeate. The feed water has a total dissolved solids concentration of 3,200 mg/L, and the permeate measures 96 mg/L. The pilot's operators recorded a specific energy use of 2.4 kWh for every 1,000 gallons of permeate produced.

The concentrate stream carries away the water that does not become permeate. Find the concentrate flow in gpm, using concentrate flow = feed flow − permeate flow.
Calculate the pilot's percent recovery, using percent recovery = permeate flow ÷ feed flow × 100.
Calculate the membrane's salt rejection, using salt rejection (%) = (feed concentration − permeate concentration) ÷ feed concentration × 100.
Convert the permeate flow of 375 gpm to a daily volume in gallons, using 1,440 minutes per day, then state that daily volume in million gallons per day (MGD).
Using the specific energy of 2.4 kWh per 1,000 gallons and the daily permeate volume from item 4, calculate the pilot's total energy use per day in kWh.
Seawater reverse-osmosis plants typically use about 11 to 15 kWh per 1,000 gallons, while conventional treatment uses roughly 1 to 2 kWh per 1,000 gallons. Is the pilot's 2.4 kWh per 1,000 gallons closer to seawater or conventional treatment, and why does brackish water fall between them?
The district needs 400,000 gallons of product water per day. Running at 375 gpm, how many hours per day must the pilot operate to make that much permeate?
What percent of the feed water leaves as concentrate rather than product? Show two ways to find it: directly from the flows, and from your recovery answer in item 2.
Using a salt mass balance, find the concentrate's salt concentration in mg/L. The salt carried by the feed equals the salt in the permeate plus the salt in the concentrate, where each stream's salt load is its flow times its concentration.
Explain why the pilot cannot simply recover 100 percent of the feed water as product, even with a better membrane.

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: 125 gpm.
Concentrate flow = 500 gpm (feed) − 375 gpm (permeate) = 125 gpm.
Answer: 75 percent.
Percent recovery = 375 ÷ 500 × 100 = 0.75 × 100 = 75 percent.
Answer: 97 percent.
Salt rejection = (3,200 − 96) ÷ 3,200 × 100 = 3,104 ÷ 3,200 × 100 = 0.97 × 100 = 97 percent.
Answer: 540,000 gallons per day.
375 gpm × 1,440 minutes/day = 540,000 gallons per day, which is 540,000 ÷ 1,000,000 = 0.54 MGD.
Answer: 1,296 kWh per day.
The daily permeate is 540,000 gallons, or 540 units of 1,000 gallons. Energy = 540 × 2.4 kWh = 1,296 kWh per day.
Answer: Closer to conventional treatment.
At 2.4 kWh per 1,000 gallons, the pilot sits just above the conventional 1-to-2 range and far below the 11-to-15 range for seawater. Brackish water has less dissolved salt than seawater, so its osmotic pressure is lower and the pumps need less pressure and energy to push water through the membrane.
Answer: about 17.8 hours.
At 375 gpm, hourly output is 375 × 60 = 22,500 gallons per hour. 400,000 ÷ 22,500 = 17.78, which rounds to about 17.8 hours per day.
Answer: 25 percent.
Directly: concentrate flow ÷ feed flow × 100 = 125 ÷ 500 × 100 = 25 percent. From recovery: 100 percent − 75 percent recovery = 25 percent. Both methods agree.
Answer: about 12,512 mg/L.
Feed salt load = 500 × 3,200 = 1,600,000 units. Permeate salt load = 375 × 96 = 36,000 units. Concentrate salt load = 1,600,000 − 36,000 = 1,564,000 units. Concentrate concentration = 1,564,000 ÷ 125 = 12,512 mg/L.
Answer: Accept any answer explaining that the membrane needs a concentrate stream to carry the rejected salt out of the system. If every gallon became product, the salt would have nowhere to go and would build up and scale the membrane, so some water must always leave as concentrate to flush the salt away.

Differentiation

Support (below grade level)

Provide the four working formulas (concentrate flow, percent recovery, salt rejection, and specific energy) on a reference card with each variable labeled, and complete items 1 and 2 together as a class so students see how the feed, permeate, and concentrate flows relate before working the rest independently.

Extension (above grade level)

Ask students to determine what permeate flow the pilot would need to reach 80 percent recovery at the same feed flow, then explain what would happen to the concentrate's salt concentration from item 9 as recovery rises, and why that creates a scaling risk.

ELL supports

Provide a labeled flow diagram showing the feed stream splitting into permeate and concentrate, with arrows marked "product water" and "salt removed," and a small glossary card for permeate, concentrate, recovery, and rejection. Allow students to record items 1 through 9 as numbered calculations and answer items 6 and 10 with a sentence frame.