Worksheet Grade 11 Science
Halcombe Bay Desalination: Recovery, Rejection, and Membrane Flux
A grade 11 reverse osmosis worksheet: students calculate recovery, salt rejection, brine mass balance, membrane flux, and energy use for a fictional desalination plant.
Objectives
- Students will calculate percent recovery, salt rejection, and a brine mass balance for a reverse osmosis process.
- Students will compute membrane flux and daily energy use and compare them against 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.
- Brine
- The extra-salty water left over after desalination pulls fresh water out of seawater. Plants must dispose of it carefully so it does not harm sea life.
- Total dissolved solids (TDS)
- Total dissolved solids is a measure of everything dissolved in water that you cannot see or filter out, like minerals and salts. Water with very high TDS can taste salty or bitter, even though it might look perfectly clear.
Also: recovery, salt rejection, flux
Teacher overview
Halcombe Bay Desalination is a fictional seawater plant on the coast. This worksheet has students work through the core numbers a membrane process engineer tracks every shift: how much of the feed becomes drinking water (recovery), how completely the membranes strip out salt (rejection), how concentrated the leftover brine becomes, how hard each square foot of membrane is working (flux), and how much electricity the process demands. The recurring lesson is that reverse osmosis trades a large energy cost for a supply that does not depend on rainfall.
Student Worksheet
Halcombe Bay Desalination is a fictional plant. Use these stated values for every item. The plant takes in a feed of 10 MGD of seawater. Its recovery is 45 percent, meaning 45 percent of the feed leaves as product (drinking) water and the rest leaves as brine. The seawater feed carries 35,000 mg/L of total dissolved solids (TDS). The membranes achieve 99.4 percent salt rejection. Round sensibly and show your work.
How many MGD of product water does the plant make? (Product = feed x recovery.)
How many MGD of brine (concentrate) leave the plant? (Brine = feed - product.)
Salt rejection of 99.4 percent means the product keeps 0.6 percent of the feed's TDS. What is the product water's TDS, in mg/L?
Use a mass balance to find the brine's TDS. The salt entering with the feed equals the salt leaving in the product plus the salt leaving in the brine: (35,000 x 10) = (product TDS x 4.5) + (brine TDS x 5.5). Solve for the brine TDS in mg/L.
Explain why the brine leaves the plant saltier than the seawater that came in.
The plant runs 300,000 square feet of membrane. Membrane flux is the product flow divided by the membrane area. Find the flux in gfd (gallons per square foot per day). Use product flow in gallons per day.
Suppose the membranes can only be pushed to 12 gfd before they foul too fast. How many square feet of membrane would the plant need to make the same product flow at 12 gfd?
The plant uses 13 kWh of electricity for every 1,000 gallons of product water. How many kWh does it use per day at the product flow you found in item 1?
A conventional surface-water plant treating the same 4.5 MGD would use about 1.5 kWh per 1,000 gallons. How many more kWh per day does the reverse osmosis plant use than that conventional plant?
Using your answers, explain the central trade-off of choosing desalination for a coastal city facing drought.
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: 4.5 MGD. Product = feed x recovery = 10 MGD x 0.45 = 4.5 MGD.
Answer: 5.5 MGD. Brine = feed - product = 10 - 4.5 = 5.5 MGD.
Answer: 210 mg/L. The product keeps 100 - 99.4 = 0.6 percent of the feed TDS: 35,000 x 0.006 = 210 mg/L.
Answer: about 63,500 mg/L. Salt in = salt out, so (35,000 x 10) = (210 x 4.5) + (brine TDS x 5.5). That gives 350,000 = 945 + 5.5 x brine TDS, so brine TDS = (350,000 - 945) / 5.5 = 349,055 / 5.5 = 63,465, which rounds to about 63,500 mg/L.
Answer: The membranes remove almost all of the salt but only send 45 percent of the water forward as product. The removed salt stays behind in a smaller volume of water, so the brine concentration rises well above the incoming seawater.
Answer: 15 gfd. Product flow is 4.5 MGD = 4,500,000 gallons per day. Flux = 4,500,000 / 300,000 = 15 gallons per square foot per day.
Answer: 375,000 sq ft. Area = product flow / flux = 4,500,000 / 12 = 375,000 square feet.
Answer: 58,500 kWh per day. Product is 4,500,000 gallons per day = 4,500 units of 1,000 gallons. 4,500 x 13 = 58,500 kWh per day.
Answer: 51,750 kWh per day. Conventional use = 4,500 x 1.5 = 6,750 kWh per day. The difference is 58,500 - 6,750 = 51,750 kWh per day.
Answer: Desalination gives a coastal city a supply that does not depend on rain or river flow, which matters during drought, but it costs several times more energy per gallon than treating a freshwater source. The city trades higher and steadier energy bills for a more reliable supply.
Differentiation
Support
Give students a filled-in flow diagram showing the feed splitting into product and brine streams, with the recovery arrow labeled, before item 1. Provide the mass-balance equation in item 4 already arranged to solve for the brine TDS so students focus on the arithmetic rather than the algebra.
Extension
Have students calculate the plant's specific energy in kWh per 1,000 gallons of feed (not product), then explain why engineers usually report energy per unit of product water instead, and how raising recovery would change both numbers.
ELL supports
Pre-teach "feed," "product," "brine," "recovery," "rejection," and "flux" with a simple labeled diagram of one membrane element showing water in, clean water out, and concentrate out. Allow students to answer items 5 and 10 with a labeled sketch plus a short sentence rather than a full paragraph.