Math problem set Grade 6 Science

Cape Vorne Desalination: Energy and Recovery Problems

A grade 6 math set comparing a fictional desalination plant's energy use with ordinary treatment, and finding the seawater and brine behind each batch of fresh water.

Objectives

  1. Students will use an energy rate to compare desalination with ordinary water treatment.
  2. Students will use a recovery percent to find the seawater and brine amounts for a batch of fresh water.

Vocabulary

Desalination
Turning salty water, like ocean water, into fresh water people can use. It works, but it takes a lot of energy and leaves behind very salty water that must be handled carefully.
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.

Also: recovery

Teacher overview

Desalination makes fresh water from seawater by pushing it through a membrane that holds back salt, a process called reverse osmosis. It works well for a dry coastal town, but it uses more energy than ordinary treatment. Real seawater plants use about 11 to 15 kWh for every 1,000 gallons (about 3 to 4 kWh per cubic meter), compared with about 1 to 2 kWh for ordinary treatment and delivery. This set uses the fictional Cape Vorne plant to compare those energy costs and to track how much seawater turns into fresh water. A common misconception this set corrects is that all the seawater becomes drinking water; in fact only part of it does, and the salty leftover, called brine, must go somewhere.

Answer Key

  1. Answer: 360 kWh.
    30,000 gallons is 30 groups of 1,000 gallons. 30 times 12 kWh equals 360 kWh.
  2. Answer: 45 kWh.
    The same 30 groups of 1,000 gallons times 1.5 kWh equals 45 kWh.
  3. Answer: 315 kWh.
    360 kWh minus 45 kWh equals 315 kWh more for the desalination plant.
  4. Answer: 8 times.
    360 kWh divided by 45 kWh equals 8, so the desalination plant uses 8 times as much energy.
  5. Answer: 75,000 gallons.
    Fresh water is 40 percent of the seawater, so seawater equals 30,000 divided by 0.40, which is 75,000 gallons.
  6. Answer: 45,000 gallons.
    Seawater taken in is 75,000 gallons and 30,000 becomes fresh water, so 75,000 minus 30,000 equals 45,000 gallons of brine.
  7. Answer: 1,800 kWh.
    The plant uses 360 kWh each day. 360 kWh times 5 days equals 1,800 kWh.
  8. Answer: Answers will vary. Accept any clear trade-off backed by the numbers, such as: desalination uses 8 times the energy of ordinary treatment, or it makes a large amount of salty brine that must be handled, in exchange for a new water supply when fresh water is scarce.

Differentiation

Support

Give students a reference card that reads "30,000 gallons = 30 thousands" so they can see why they multiply by 30 in problems 1 and 2. Allow a calculator and model problem 1 step by step before they continue.

Extension

Ask students to find the total energy Cape Vorne would use in a 30-day month at 30,000 gallons per day, then compare it with the energy an ordinary plant would use for the same water, and state the difference.

ELL supports

Pre-teach "desalination," "recovery," and "brine" with a simple diagram: seawater in on the left, fresh water out the top, and salty brine out the bottom. Allow students to label the diagram with their answers to problems 5 and 6.

Extensions

Have students research one real desalination plant and write two sentences, in their own words, about why the town or region built it, connecting the reason to the idea of a scarce fresh-water supply.