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Worksheet Grade 12 Engineering & Design

Sitwell Basin Aquifer Storage and Recovery Worksheet

A grade 12 worksheet on a fictional aquifer storage and recovery project: porosity from core samples, recoverable storage in acre-feet and gallons, and injection math.

Objectives

  1. Students will calculate porosity from core-sample void and total volumes.
  2. Students will estimate an aquifer zone's recoverable storage from area, thickness, and porosity.
  3. Students will convert seasonal injection volumes between gallons and acre-feet and compute recovery time.

Vocabulary

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.

Also: aquifer storage and recovery, core sample, recharge well, recoverable storage

Teacher overview

Aquifer storage and recovery, or ASR, treats an aquifer as an underground reservoir: a utility injects treated water through wells in wet years, lets the aquifer hold it, and pumps it back during drought. Sizing such a project depends on porosity, because only the void space between the grains holds water. This worksheet walks students through the fictional Sitwell Basin ASR project from two core samples to a full seasonal water budget, using the standard porosity formula and the acre-foot conversions that groundwater engineers use daily.

Sitwell Basin Aquifer Storage and Recovery Worksheet — Grade 12 Engineering & Design
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Sitwell Basin water district is designing an aquifer storage and recovery project. Use these facts throughout. Porosity (%) = void volume ÷ total volume × 100. One acre-foot equals 325,851 gallons. The district's recharge well injects 1.5 MGD, and its recovery pumping rate is 2.0 MGD.

  1. In one or two sentences, explain what an ASR project does with treated water in wet years and in dry years.
  2. A core sample from candidate Zone A has a total volume of 400 cubic centimeters, with 92 cubic centimeters of void space between grains. Calculate Zone A's porosity as a percent.
  3. A core from deeper Zone B has a total volume of 400 cubic centimeters and 58 cubic centimeters of void space. Calculate Zone B's porosity as a percent, to the nearest tenth.
  4. Which zone can store more water per unit of volume, and why? Answer using your two porosity values.
  5. The district selects Zone A, which spreads under 160 acres with an average thickness of 40 feet. Compute the zone's bulk volume in acre-feet.
  6. For this worksheet, use a simplified upper-bound model that sets recoverable storage equal to bulk volume × porosity. Real projects use specific yield or storativity, which give smaller values. Using Zone A's porosity from item 2, compute the recoverable storage in acre-feet.
  7. Convert the recoverable storage from item 6 into gallons.
  8. In one wet season, the recharge well injects 1.5 MGD for 120 days. Convert the total injected volume into acre-feet, to the nearest tenth.
  9. What percent of Zone A's recoverable storage does that one season of injection fill? Round to the nearest tenth of a percent.
  10. A drought arrives and the district pumps the stored water back at 2.0 MGD. Assuming it recovers exactly the volume injected in item 8, how many days does recovery take?
  11. Give one physical reason the district might recover somewhat less water than it injected.

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: It stores water underground in wet years and withdraws it in dry years. In wet years the district injects treated water through wells into the aquifer; in drought it pumps the stored water back out for supply.
  2. Answer: 23 percent. 92 ÷ 400 × 100 = 23 percent porosity.
  3. Answer: 14.5 percent. 58 ÷ 400 × 100 = 14.5 percent porosity.
  4. Answer: Zone A. Its porosity (23 percent versus 14.5 percent) means more of each cubic foot is open void space available to hold water.
  5. Answer: 6,400 acre-feet. 160 acres × 40 feet = 6,400 acre-feet of bulk volume.
  6. Answer: 1,472 acre-feet. 6,400 acre-feet × 0.23 = 1,472 acre-feet of recoverable storage.
  7. Answer: 479,652,672 gallons. 1,472 acre-feet × 325,851 gallons per acre-foot = 479,652,672 gallons.
  8. Answer: 552.4 acre-feet. 1.5 MGD × 120 days = 180 million gallons, and 180,000,000 ÷ 325,851 = 552.4 acre-feet.
  9. Answer: 37.5 percent. 552.4 ÷ 1,472 = 0.375, or 37.5 percent of the recoverable storage.
  10. Answer: 90 days. 180,000,000 gallons ÷ 2,000,000 gallons per day = 90 days of recovery pumping.
  11. Answer: Some stored water migrates or mixes and cannot all be captured. Acceptable answers include water drifting beyond the well's capture zone with natural groundwater flow, mixing with poorer-quality native water, or being held too tightly in small pores to yield.

Differentiation

Support

Provide the porosity formula worked through with a third practice core (for instance, 400 cubic centimeters total and 80 void, giving 20 percent) before students attempt items 2 and 3, and confirm item 5 as a checkpoint since items 6, 7, and 9 all build on it.

Extension

Have students compute how many 120-day injection seasons the fictional district would need to fill Zone A's recoverable storage completely at 1.5 MGD, assuming no losses, and then explain why a real project would stop well short of 100 percent. (1,472 ÷ 552.4 is about 2.7 seasons.)

ELL supports

Introduce "porosity" with a jar-of-marbles visual, water filling only the gaps, and label a cross-section showing the recharge well, the storage zone, and the direction of injection and recovery. Allow items 1, 4, and 11 to be answered with a labeled sketch plus a single sentence.