Lesson plan Grade 9 Science
Bramble Cove's Salty Wells: Modeling Saltwater Intrusion
A 50-minute grade 9 lesson using the 40-to-1 freshwater head rule and a mixing calculation to explain why summer pumping turns a fictional coastal town's wells salty.
Objectives
- Students will apply the 40-to-1 relationship between freshwater head and the depth of the saltwater interface in a coastal aquifer.
- Students will calculate the chloride concentration of a mixed sample and the seawater fraction that reaches a stated limit.
- Students will describe the feedback loop connecting pumping, interface depth, and well water quality.
Vocabulary
- Saltwater intrusion
- Saltwater intrusion happens when salty ocean water creeps into the fresh groundwater under coastal land. Pumping too much fresh water from coastal wells can pull the salt water in.
- Aquifer
- An aquifer is an underground layer of rock, sand, or gravel that holds water in the tiny spaces between its grains. Wells reach down into aquifers to bring water up to the surface.
Also: interface, freshwater head, chloride
Teacher overview
In a coastal aquifer, fresh groundwater floats on denser salt water, and the two meet along a boundary called the interface. Because the density difference is small, a change of one foot in the freshwater level above sea level moves that boundary about forty feet. This lesson uses the fictional coastal town of Bramble Cove, whose summer pumping draws the interface up toward the screens of its supply wells. Students often picture salt water arriving as a wave that flows across the surface and soaks down, when the real movement is a wedge that rises from below in response to lost freshwater weight. A second misconception the lesson addresses is that a well can be saved by drilling deeper, which in a coastal aquifer usually reaches salt water sooner.
Procedure
Hook (5 min): Fill a clear container with fresh water, then slowly pour dyed salt water through a funnel to the bottom so the class can see a stable layer form under the fresh water. Draw a well screen on the side of the container with a marker. Ask students to predict what happens to the boundary if you remove fresh water from the top.
Direct instruction (13 min): Introduce the 40-to-1 relationship: for every 1 foot the fresh water stands above sea level, about 40 feet of fresh water sits below sea level before salt water begins. Work an example for the fictional town of Bramble Cove on the board. When a monitoring well reads 4.0 feet above sea level, the interface is 4.0 times 40, which equals 160 feet below sea level. Then lower the reading to 2.0 feet and show that the interface rises to 80 feet below sea level, a movement of 80 feet from a 2-foot change at the surface. Name the saltwater wedge and explain why the fresh water above it is thickest inland.
Guided practice (12 min): Distribute the handout. Work item 1 with the class, then have pairs complete items 2 through 4. Circulate and watch for students who multiply the head by 40 and then subtract it from the well depth in a single step without stating the interface depth first; ask those pairs to label a sketch before continuing.
Independent practice (15 min): Students complete items 5 through 8 on their own. Items 5 and 6 use a simple mixing calculation, so remind students that the two fractions of the sample must add to 1.
Closure and formative check (5 min): Each student answers one question on an index card: if the fresh water stands 2.5 feet above sea level, how far below sea level is the interface? The answer is 2.5 times 40, which equals 100 feet. Collect the cards. If more than three or four students divide instead of multiply, open the next class with a second sketch of the wedge before moving on.
Student Handout
Suppose the fictional coastal town of Bramble Cove draws its drinking water from a sand aquifer beside the ocean. Fresh water floats on salt water, and the boundary between them is the interface. For every 1 foot the fresh water stands above sea level, about 40 feet of fresh water sits below sea level before the salt water begins. The town's Well 3 is screened from 60 feet to 95 feet below sea level. Fresh groundwater in the area carries 25 mg/L of chloride, and seawater near the town carries 19,000 mg/L of chloride.
- In spring, the monitoring well stands 3.5 feet above sea level. How far below sea level is the interface?
- By late summer, pumping lowers that level to 1.75 feet above sea level. How far below sea level is the interface then?
- How far did the interface rise between spring and late summer?
- The utility wants the interface to stay at least 10 feet below the bottom of the Well 3 screen, which sits 95 feet below sea level. What freshwater level above sea level does the 40-to-1 relationship require?
- A sample from Well 3 is 2 percent seawater and 98 percent fresh groundwater. Find the chloride concentration of that sample, in mg/L.
- Bramble Cove will not serve water above 250 mg/L of chloride. What percent of a sample can be seawater at that limit? Round to the nearest hundredth of a percent.
- District records show that every 40 acre-feet per year of reduced pumping raises the late-summer level by 0.1 foot. How much must yearly pumping fall to raise the level from 1.75 feet to the level you found in item 4?
- Describe how pumping, the depth of the interface, and well water quality affect one another at Bramble Cove. Explain what happens if the town responds to a salty well by drilling that well deeper.
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: 140 feet below sea level. 3.5 ft times 40 equals 140 ft.
- Answer: 70 feet below sea level. 1.75 ft times 40 equals 70 ft.
- Answer: 70 feet. 140 ft minus 70 ft equals 70 ft of rise.
- Answer: 2.625 feet above sea level. The interface must sit 95 plus 10, which is 105 ft below sea level, and 105 divided by 40 equals 2.625 ft.
- Answer: 404.5 mg/L. 0.02 times 19,000 equals 380 mg/L from seawater, 0.98 times 25 equals 24.5 mg/L from fresh groundwater, and 380 plus 24.5 equals 404.5 mg/L, which is already above the town limit.
- Answer: 1.19 percent. Let f be the seawater fraction, so 19,000f plus 25 times the quantity 1 minus f equals 250. That gives 18,975f equals 225, so f equals 0.011858, or 1.19 percent.
- Answer: 350 acre-feet per year. The level must rise 2.625 minus 1.75, which equals 0.875 ft; 0.875 divided by 0.1 equals 8.75 steps, and 8.75 times 40 acre-feet equals 350 acre-feet per year of reduced pumping.
- Answer: Accept any response that states the loop. Pumping removes fresh water and lowers the freshwater level, the lower level lets the interface rise about forty feet for every foot lost, and a rising interface eventually reaches the well screen and raises chloride in the delivered water. Drilling the same well deeper makes the problem worse, because deeper in a coastal aquifer means closer to the salt water, so the well draws a larger seawater fraction than before.
Differentiation
Support
Give students a printed cross-section of the fictional Bramble Cove aquifer with sea level, the well screen, and the interface already labeled, so the arithmetic in items 1 through 4 can be checked against a picture. Provide the mixing setup for item 5 as a filled-in table with one cell left blank.
Extension
Ask students to find how far the interface would rise if the town added a second summer well that lowered the freshwater level by an additional 0.4 foot, then have them state how much pumping reduction would be needed to offset it using the same 40 acre-feet per 0.1 foot relationship.
ELL supports
Pre-teach interface, head, intrusion, and chloride using the layered demonstration container as a live reference, labeling each layer with a card. Provide sentence frames for item 8: "When the town pumps more, the fresh water level ___. When that level drops, the interface ___."