Lesson plan Grade 12 Science

Darcy's Law and the Merrifield Aquifer

A 50-minute grade 12 lesson using Darcy's law to compute groundwater flow, seepage velocity, and contaminant travel time for a fictional town's confined aquifer.

Objectives

  1. Students will calculate groundwater discharge with Darcy's law from conductivity, gradient, and area.
  2. Students will compute seepage velocity and estimate contaminant travel time across an aquifer.
  3. Students will explain how porosity and gradient change how fast water moves underground.

Vocabulary

Hydraulic conductivity
Hydraulic conductivity is a number that tells you how easily water can flow through a particular soil or rock. Gravel has high hydraulic conductivity because water flows through it easily, while dense clay has very low hydraulic conductivity.
Porosity
Porosity measures how much empty space a rock or soil has between its grains, which tells you how much water it can hold. A sponge has high porosity, while a piece of glass has almost none.
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: hydraulic gradient, Darcy's law, seepage velocity

Teacher overview

The fictional town of Merrifield draws its drinking water from a confined sand-and-gravel aquifer, and this lesson uses Darcy's law to describe how water moves through it. Students compute the discharge through a cross-section of the aquifer, then find how fast an individual water molecule, or a dissolved contaminant, actually travels. A common misconception is that groundwater rushes underground like water in a pipe; the seepage velocity in this lesson comes out near one foot per day, which reframes an aquifer as a slow-moving reservoir where a spill can take years to reach a well.

Procedure

Hook (5 min): Ask students how long they think a drop of water takes to travel the 2,000 feet between two Merrifield monitoring wells. Collect a few guesses on the board and hold them for comparison at the end.

Direct instruction (12 min): Introduce Darcy's law, Q = K x i x A, where K is hydraulic conductivity (feet per day), i is the hydraulic gradient (feet of head drop per foot of distance), and A is the cross-sectional area (square feet). Define the hydraulic gradient as the head difference divided by the distance between wells. Then introduce seepage velocity, v = K x i / porosity, and explain that dividing by porosity accounts for water threading only through the pore space, not the solid grains.

Guided worked example (13 min): Work the Merrifield case with the class, showing each unit. The aquifer has K = 60 feet per day and porosity 0.25. Two wells sit 2,000 feet apart with an 8-foot head difference, so the gradient is 8 / 2,000 = 0.004. The flow cross-section is 500 feet wide and 40 feet thick, so A = 20,000 square feet. Discharge Q = 60 x 0.004 x 20,000 = 4,800 cubic feet per day, which is 4,800 x 7.48 = 35,904 gallons per day. Seepage velocity v = 60 x 0.004 / 0.25 = 0.96 feet per day, so travel time across 2,000 feet is 2,000 / 0.96 = about 2,083 days, or roughly 5.7 years.

Independent practice (12 min): Students complete the handout problems, which change the gradient and porosity and ask for discharge, seepage velocity, and travel time. Circulate and check that students divide by porosity only for the velocity, not for the discharge.

Closure and exit ticket (8 min): As a formative check, students answer one exit question: if the head difference between the two Merrifield wells doubled to 16 feet while everything else held constant, what happens to the seepage velocity and the travel time? Require a numerical answer and a one-sentence reason. Scan for students who correctly double the velocity and halve the travel time before dismissal.

Answer Key

  1. Answer: 0.006.
    Gradient = head difference / distance = 9 feet / 1,500 feet = 0.006.

  2. Answer: 53,856 gallons per day.
    A = 500 x 40 = 20,000 square feet. Q = 60 x 0.006 x 20,000 = 7,200 cubic feet per day; 7,200 x 7.48 = 53,856 gallons per day.

  3. Answer: 1.2 feet per day. v = K x i / porosity = 60 x 0.006 / 0.30 = 0.36 / 0.30 = 1.2 feet per day.

  4. Answer: about 1,250 days, or 3.4 years.
    Travel time = 1,500 feet / 1.2 feet per day = 1,250 days, which is 1,250 / 365 = about 3.4 years.

Exit ticket answer: Doubling the head difference to 16 feet doubles the gradient, which doubles the seepage velocity to about 1.92 feet per day and cuts the travel time roughly in half, to about 1,040 days.

Differentiation

Support

Provide a filled-in units table listing K, i, A, and porosity with their units, and give the discharge already worked for the guided example so students can concentrate on the velocity and travel-time steps.

Extension

Ask students to solve the inverse problem: given a required travel time of at least 10 years across 1,500 feet, what is the largest hydraulic conductivity the aquifer could have at porosity 0.30, holding the gradient at 0.006?

ELL supports

Pre-teach "conductivity," "gradient," "porosity," and "seepage" with a labeled block diagram of an aquifer showing pore space between grains. Provide a sentence frame for the exit ticket that contrasts the words "faster" and "shorter."

Extensions

Have students research a real contamination case in which a plume took years to reach a municipal well, then compare the reported travel time with the roughly 5.7-year figure from the Merrifield example and discuss what aquifer properties would speed a plume up or slow it down.