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Worksheet Grade 10 Science

Sable Point Desalination: Salt Rejection and Recovery Rate Worksheet

A grade 10 worksheet where students calculate salt rejection, recovery rate, and energy use at a fictional coastal reverse-osmosis desalination facility.

Objectives

  1. Students will calculate the percent of dissolved solids a reverse-osmosis membrane removes from seawater.
  2. Students will use a stated recovery rate to calculate required seawater intake and brine discharge.
  3. Students will compare the energy use of reverse osmosis to conventional treatment of a fresh water source.

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.
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.
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.

Also: salt rejection, recovery rate

Teacher overview

Imagine Sable Point, a fictional coastal facility built for this worksheet, which turns seawater into drinking water using reverse osmosis. This worksheet has students calculate how much salt the plant's membranes remove, how much seawater the plant must draw in to meet its production goal, and how the plant's energy use compares to a conventional plant that treats a fresh water source. A common misconception is assuming all the seawater entering a reverse-osmosis plant becomes drinking water; because membranes reject dissolved salts into a separate concentrated stream, only part of the intake becomes usable product, and items 3 and 4 make that split explicit.

Sable Point Desalination: Salt Rejection and Recovery Rate Worksheet — Grade 10 Science
Name: ____________________________ Date: ______________

Student Worksheet

  1. Imagine Sable Point's intake seawater has a total dissolved solids (TDS) concentration of 35,000 mg/L, a value used for this worksheet to represent typical ocean water. The plant's finished (permeate) water has a TDS of 250 mg/L. Calculate the percent of dissolved solids the membranes remove, using Percent removed = (feed TDS − permeate TDS) ÷ feed TDS × 100.
  2. In fresh water and other dilute solutions, 1 mg/L is approximately equal to 1 part per million (ppm), because a liter of such water weighs about 1 kilogram. Express Sable Point's permeate TDS (250 mg/L) in parts per million. Then explain in one sentence why this shortcut should not be applied directly to the feed seawater at 35,000 mg/L.
  3. The plant operates at 45 percent recovery, meaning 45 percent of the seawater entering the plant becomes finished drinking water, while the rest leaves as concentrated brine. If the plant produces 8 million gallons per day (MGD) of finished water, how many MGD of seawater must enter the plant? Round to the nearest tenth.
  4. How many MGD of concentrated brine does the plant discharge each day?
  5. In one or two sentences, explain why reverse osmosis produces two separate output streams instead of one, connecting your answer to the salt rejection percentage from item 1.
  6. Seawater reverse osmosis uses roughly 15 kWh of electricity per 1,000 gallons of water produced, the upper end of its typical 11 to 15 kWh range. Calculate how many kilowatt-hours Sable Point's plant uses each day to produce its 8 MGD of finished water. (Hint: 8 MGD equals 8,000 thousand-gallon units.)
  7. Conventional treatment of a fresh water source typically uses only 1 to 2 kWh per 1,000 gallons. Using the upper value of that range (2 kWh per 1,000 gallons), how many more kilowatt-hours per day does Sable Point's reverse-osmosis process use compared to the same daily volume treated conventionally?
  8. Give at least one reason, grounded in water supply rather than cost alone, that a coastal community might still choose reverse osmosis desalination despite its higher energy use.
  9. Challenge. If engineers raised the plant's recovery rate from 45 percent to 50 percent without changing the 8 MGD finished-water goal, would the required seawater intake increase or decrease, and by about how many MGD? Show your reasoning, comparing to your answer from item 3.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 99.3 percent.
    (35,000 − 250) ÷ 35,000 × 100 = 34,750 ÷ 35,000 × 100 = 99.29, which rounds to 99.3 percent.
  2. Answer: About 250 ppm.
    The permeate is dilute, so a liter of it weighs about 1 kilogram and 250 mg/L is approximately 250 ppm. The shortcut does not hold for the feed seawater: above about 7,000 mg/L of dissolved solids, a liter weighs noticeably more than 1 kilogram, so converting 35,000 mg/L to ppm needs a density correction.
  3. Answer: 17.8 MGD.
    8 MGD ÷ 0.45 = 17.78, which rounds to 17.8 MGD.
  4. Answer: 9.8 MGD.
    17.8 MGD intake − 8 MGD finished water = 9.8 MGD of brine.
  5. Answer: The membranes let water molecules through while rejecting most dissolved salts, so the salts that item 1 shows being removed (about 99.3 percent) have to go somewhere; they concentrate into the brine stream instead of passing through with the finished water.
  6. Answer: 120,000 kWh/day.
    8,000 thousand-gallon units × 15 kWh per unit = 120,000 kWh/day.
  7. Answer: 104,000 kWh/day more.
    Conventional use = 8,000 × 2 = 16,000 kWh/day. Reverse-osmosis use (120,000) minus conventional use (16,000) equals 104,000 kWh/day more.
  8. Answer: Answers will vary but should note that desalination lets a coastal community draw drinking water from an essentially unlimited seawater source, which does not depend on rainfall, snowpack, or a shared river the way most fresh water sources do.
  9. Answer: decrease, by about 1.8 MGD.
    New intake = 8 MGD ÷ 0.50 = 16.0 MGD, which is 17.8 − 16.0 = 1.8 MGD less than the 45 percent recovery case.

Differentiation

Support (below grade level)

Provide the percent-removed and recovery-rate formulas as labeled reference cards, and work item 1 as a fully modeled example, showing each arithmetic step, before students attempt item 3 independently.

Extension (above grade level)

Ask students to calculate what recovery rate the fictional Sable Point plant would need to hold its seawater intake at or below 15.0 MGD while still producing 8 MGD of finished water, working backward through the recovery-rate relationship used in item 3.

ELL supports

Pre-teach "desalination," "membrane," "brine," and "recovery rate" with a labeled diagram showing seawater entering a plant and splitting into two streams: finished water and brine. Allow students to answer items 5 and 8 orally or with short phrases instead of full sentences.