Math problem set Grade 9 Engineering & Design

Marsh Landing Seawater Reverse Osmosis: Recovery and Flux

A grade 9 engineering set on a seawater reverse osmosis plant: students find permeate and concentrate flows, membrane flux, a concentration factor, and daily energy.

Grade: Grade 9
Subject: Engineering & Design
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HS-ETS1-3 NGSS

Evaluate a solution to a complex real-world problem based on prioritized criteria and trade-offs that account for a range of constraints, including cost, safety, reliability, and aesthetics as well as possible social, cultural, and environmental impacts.

How this material aligns: Problem 8 evaluates the proposed rise to 50 percent recovery by weighing the permeate gain computed in problem 5 against the saltier concentrate quantified in problem 6, the trade-off judgment this standard names.

All materials for HS-ETS1-3 →

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 track units through the mass balance in problems 1 through 4, converting MGD to gpm and dividing a daily flow by membrane area to report flux in gallons per square foot per day.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Problem 5 has students write and solve a one-variable equation from the recovery definition to find the permeate flow at a higher target recovery, then interpret the result.

All materials for HSA-CED.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 use a recovery ratio to split a feed flow into permeate and concentrate flows.
Students will calculate membrane flux and a concentration factor and interpret the trade-offs of raising recovery.

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: feed, permeate, concentrate, recovery, flux

Teacher overview

A seawater reverse osmosis plant pushes salty feed water against a membrane at high pressure. Some of the water passes through as clean permeate, and the rest leaves as a saltier concentrate stream. The recovery ratio is the share of the feed that becomes permeate, and it drives the whole mass balance. This set uses one fictional plant to find the permeate and concentrate flows, the membrane flux, and the plant's daily energy.

A common error in this math is treating recovery as the concentrate share instead of the permeate share. Recovery is permeate divided by feed, so at 45 percent recovery the permeate is the smaller stream and the concentrate is the larger one. Checking that permeate plus concentrate equals the feed catches most mistakes before they spread through the later problems.

Marsh Landing Seawater Reverse Osmosis: Recovery and Flux — Grade 9 Engineering & Design

Name: ____________________________ Date: ______________

Problems

The Marsh Landing seawater reverse osmosis plant (a fictional plant used for this set) treats coastal seawater. Use these values for all problems:

Feed flow to the membrane racks: 5.0 MGD.
Recovery (permeate divided by feed): 45%.
Membrane elements: 625 elements, each with 400 square feet of membrane area.
Conversion: 1 MGD = 694 gpm.

Find the permeate flow in MGD. Permeate = recovery x feed.
Find the concentrate flow in MGD. Concentrate = feed - permeate.
Convert the permeate flow from problem 1 to gallons per minute (gpm). Round to the nearest whole gpm.
Find the membrane flux in gallons per square foot per day (gfd). Flux = permeate flow in gallons per day / total membrane area. (Total area = 625 elements x 400 square feet.)
The engineers want to raise recovery to 50% while keeping the feed at 5.0 MGD. Write and solve an equation for the new permeate flow in MGD, then state the new concentrate flow.
Assume all of the salt in the feed leaves in the concentrate. The concentration factor tells how much saltier the concentrate is than the feed. Find it using feed flow / concentrate flow at the original 45 percent recovery. Round to the nearest tenth.
Industry references estimate seawater reverse osmosis uses about 11 to 15 kWh per 1,000 gallons of permeate (3 to 4 kWh per cubic meter). Using the higher figure of 15 kWh per 1,000 gallons, find the plant's daily energy use for the permeate from problem 1.
Raising recovery from 45 percent to 50 percent gives more permeate from the same feed but leaves a smaller, saltier concentrate. Using your answers, name one benefit and one drawback of the change, and support each with a number.

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: 2.25 MGD.
Permeate = 0.45 x 5.0 MGD = 2.25 MGD.
Answer: 2.75 MGD.
Concentrate = 5.0 MGD - 2.25 MGD = 2.75 MGD. Check: 2.25 + 2.75 = 5.0 MGD, the full feed.
Answer: 1,562 gpm.
2.25 MGD x 694 gpm/MGD = 1,561.5, which rounds to 1,562 gpm.
Answer: 9.0 gfd.
Total membrane area = 625 x 400 = 250,000 square feet. Permeate = 2.25 MGD = 2,250,000 gallons per day. Flux = 2,250,000 / 250,000 = 9.0 gfd.
Answer: 2.5 MGD.
The recovery equation is p / 5.0 = 0.50, so p = 0.50 x 5.0 = 2.5 MGD of permeate. The new concentrate = 5.0 - 2.5 = 2.5 MGD.
Answer: 1.8 times.
Concentration factor = feed / concentrate = 5.0 / 2.75 = 1.818, about 1.8 times as salty as the feed.
Answer: 33,750 kWh.
Permeate = 2,250,000 gallons per day. That is 2,250,000 / 1,000 = 2,250 units of 1,000 gallons. Energy = 2,250 x 15 kWh = 33,750 kWh per day.
Answer: Benefit: permeate rises from 2.25 MGD to 2.5 MGD.
Drawback: the concentrate falls from 2.75 MGD to 2.5 MGD and its concentration factor rises from about 1.8 to 5.0 / 2.5 = 2.0 times, so the brine is saltier and harder to dispose of. The benefit is more drinking water from the same feed; the drawback is a smaller, saltier concentrate stream.

Differentiation

Support

Provide the total membrane area (250,000 square feet) already computed for problem 4, and give a filled-in mass-balance box showing feed at the top splitting into permeate and concentrate, so students can see that the two output streams must add back to the feed.

Extension

Ask students to find what recovery would be needed to produce 3.0 MGD of permeate from the same 5.0 MGD feed, and to compute the concentration factor at that recovery, then explain why disposal of the concentrate becomes a limiting design concern as recovery climbs.

ELL supports

Pre-teach "feed," "permeate," "concentrate," "recovery," and "flux" with a simple labeled flow diagram showing one arrow of salty water entering and two arrows leaving. Allow students to answer problem 8 with the two numbers plus a short phrase rather than full sentences.