Worksheet Grade 12 Science

Reverse Osmosis Flux and Specific Energy at Corravale

A grade 12 science worksheet on a fictional seawater desalination plant: students compute recovery, membrane flux, and the energy penalty per gallon of desalting.

Grade: Grade 12
Subject: Science
Material: Worksheet
Time: 40 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: Students carry units through the flow, recovery, and energy conversions in items 1 through 8, the multi-step dimensional analysis this standard requires of real plant math.

All materials for HSN-Q.A.1 →

HSG-MG.A.2 Common Core Math

Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

How this material aligns: Item 4 treats membrane flux as a density, gallons of permeate per day spread over square feet of membrane area, the density-modeling skill this standard names directly.

All materials for HSG-MG.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: Items 6, 7, and 10 weigh seawater desalination at 13 kWh per thousand gallons against conventional treatment at 1.5, the cost-benefit comparison of competing water-supply solutions 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 and concentrate flow for a reverse osmosis plant from feed and permeate flows.
Students will calculate membrane flux in gallons per square foot per day from permeate flow and membrane area.
Students will compare the energy used per gallon by desalination and by 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, flux, concentrate, specific energy

Teacher overview

Reverse osmosis pushes seawater against a membrane at high pressure so that fresh water passes through while salt is left behind. The trade-off is energy: seawater reverse osmosis uses roughly 11 to 15 kWh per 1,000 gallons (about 3 to 4 kWh per cubic meter), several times what conventional surface-water treatment needs. This worksheet uses the fictional coastal plant at Corravale to practice the plant math operators use: recovery, concentrate flow, membrane flux, and the energy comparison.

A common misconception here is that recovery could simply be raised toward 100 percent to save water. As more fresh water is pulled through the membrane, the salt left behind grows more concentrated, which raises the pressure and the scaling risk. Seawater plants therefore run at modest recovery, which item 8 asks students to reason about with numbers.

Reverse Osmosis Flux and Specific Energy at Corravale — Grade 12 Science

Name: ____________________________ Date: ______________

Student Worksheet

The town of Corravale (a fictional coastal community used for this worksheet) runs a seawater reverse osmosis plant. The plant takes in a feed flow of 5.0 MGD and produces a permeate (fresh water) flow of 2.0 MGD. Its membranes have a total area of 250,000 square feet. Use these formulas and conversions for every item.

Recovery (%) = permeate flow / feed flow x 100
Concentrate flow = feed flow - permeate flow
Membrane flux (gfd) = permeate flow (gpd) / membrane area (sq ft)
Conversions: 1 MGD = 1,000,000 gpd; 1 MGD = 694 gpm

Calculate the plant's recovery as a percentage, using the feed flow of 5.0 MGD and the permeate flow of 2.0 MGD.
Calculate the concentrate (brine) flow, in MGD, that leaves the plant.
Convert the permeate flow of 2.0 MGD to gallons per day.
Using your answer from item 3 and the membrane area of 250,000 square feet, calculate the membrane flux in gallons per square foot per day (gfd).
Convert the permeate flow of 2.0 MGD to gallons per minute.
The plant uses 13 kWh per 1,000 gallons of permeate. Calculate the total energy used per day, in kWh, to produce the 2,000,000 gallons of permeate.
A conventional surface-water plant uses about 1.5 kWh per 1,000 gallons. Calculate the daily energy it would take to treat the same 2,000,000 gallons, then find how many more kWh per day the desalination plant uses.
An engineer raises recovery to 45 percent while keeping the feed flow at 5.0 MGD. Calculate the new permeate flow and the new concentrate flow, both in MGD.
Explain why seawater reverse osmosis plants are not run at very high recovery, such as 90 percent, using what happens to the concentrate.
Using your answers from items 6 and 7, explain in one or two sentences why coastal cities usually build desalination plants only after cheaper supplies are fully used.

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: 40%.
Recovery = permeate / feed x 100 = 2.0 MGD / 5.0 MGD x 100 = 40%.
Answer: 3.0 MGD.
Concentrate = feed - permeate = 5.0 MGD - 2.0 MGD = 3.0 MGD.
Answer: 2,000,000 gpd.
2.0 MGD x 1,000,000 gpd per MGD = 2,000,000 gpd.
Answer: 8.0 gfd.
Flux = permeate / area = 2,000,000 gpd / 250,000 sq ft = 8.0 gallons per square foot per day.
Answer: 1,388 gpm.
2.0 MGD x 694 gpm per MGD = 1,388 gpm.
Answer: 26,000 kWh.
The plant makes 2,000,000 gallons, which is 2,000 units of 1,000 gallons. 13 kWh x 2,000 = 26,000 kWh per day.
Answer: 3,000 kWh; the desalination plant uses 23,000 kWh more per day.
Conventional energy = 1.5 kWh x 2,000 = 3,000 kWh per day. The desalination plant uses 26,000 - 3,000 = 23,000 kWh per day more for the same amount of water.
Answer: 2.25 MGD permeate; 2.75 MGD concentrate.
New permeate = 0.45 x 5.0 MGD = 2.25 MGD. New concentrate = 5.0 - 2.25 = 2.75 MGD.
Answer: The concentrate would become too salty to manage.
Pulling more fresh water through the membrane leaves the same salt in less water, so the concentrate grows saltier. That raises the pressure needed and the risk of scale forming on the membranes, which is why recovery stays modest.
Answer: Desalination costs far more energy per gallon.
Items 6 and 7 show the plant uses about 23,000 kWh per day more than conventional treatment for the same water, so cities usually turn to it only when rivers, reservoirs, and groundwater cannot meet demand.

Differentiation

Support (below grade level)

Provide a labeled flow diagram showing feed water splitting into permeate and concentrate, and give the four formulas on a reference card. Work item 1 and item 2 together before students continue independently, and allow a calculator throughout.

Extension (above grade level)

Ask students to find the permeate flow, in MGD, that would raise the membrane flux to 10 gfd if the membrane area stayed at 250,000 square feet. They should set up and solve the flux equation for the unknown permeate flow.

ELL supports

Pre-teach "recovery," "flux," "permeate," and "concentrate" with a diagram and short definitions. Allow students to answer items 9 and 10 with a labeled sketch or a sentence frame such as "Higher recovery makes the concentrate ___, which ___."

Extensions

Have students compare the fictional Corravale plant's 13 kWh per 1,000 gallons with the energy a conventional plant uses and discuss why some coastal utilities pair desalination with wind or solar power. A short written comparison connects the plant math to the energy choices utilities actually face.