Worksheet Grade 7 Engineering & Design

Cinderfield Seep Cell: Sizing Limestone and Retention Time

A grade 7 worksheet on a fictional quarry seep where students size a limestone treatment cell, test it against a retention-time rule, and deepen it for spring flows.

Grade: Grade 7
Subject: Engineering & Design
Material: Worksheet
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

7.NS.A.3 Common Core Math

Solve real-world and mathematical problems involving the four operations with rational numbers.

How this material aligns: Items 1 through 4 and 9 through 13 chain multiplication, division, and subtraction across decimals to size a basin and its limestone charge.

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7.RP.A.3 Common Core Math

Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

How this material aligns: Item 7 carries the spring retention time found in item 6 into a percent comparison against the twelve-hour design rule.

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MS-ETS1-1 NGSS

Define the criteria and constraints of a design problem with sufficient precision to ensure a successful solution, taking into account relevant scientific principles and potential impacts on people and the natural environment that may limit possible solutions.

How this material aligns: Item 14 asks students to name the retention-time rule as the design criterion and to explain what the spring flow reveals about the cell's limits.

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will compute a basin volume in cubic feet and convert it to gallons.
Students will divide a volume by a flow rate to find retention time and compare it to a design rule.
Students will find the added volume and depth a basin needs to meet a rule at a higher flow.

Vocabulary

Key words: seep, retention time, limestone, treatment cell, criterion

Teacher overview

Cinderfield Quarry is a fictional abandoned stone quarry whose flooded pit leaks acidic water into a nearby creek. The quarry's rock contains sulfur-bearing minerals such as pyrite, and water that sits against that broken rock turns acidic as those minerals react with air and water, and that acid harms the insects and fish downstream. One low-cost fix is a limestone treatment cell, which is a shallow basin filled with crushed limestone that the seep water flows slowly through. The limestone neutralizes the acid, but only if the water stays in the basin long enough, so the design turns on a rule about retention time. This worksheet has students size the cell, test it against the rule at two different flows, and then figure out how much deeper it would need to be.

Cinderfield Seep Cell: Sizing Limestone and Retention Time — Grade 7 Engineering & Design

Name: ____________________________ Date: ______________

Student Worksheet

Use these figures from the fictional Cinderfield project for every item. The treatment cell is a rectangle 60 feet long and 25 feet wide, holding water 4 feet deep. Its surface area is 1,500 square feet. One cubic foot of water holds 7.48 gallons. The design rule says water must stay in the cell for at least 12 hours. In summer the seep flows 40 gallons per minute. In spring it rises to 75 gallons per minute. The cell is packed with 120,000 pounds of crushed limestone, and the acid in the seep uses up 45 pounds of limestone each day. For this worksheet, treat the full 60 by 25 by 4 foot space as water. In a real cell the stone fills part of that space, so the true retention time would be shorter.

The cell is 60 feet long, 25 feet wide, and 4 feet deep. How many cubic feet of water does it hold?
Using 7.48 gallons for every cubic foot, how many gallons does the cell hold?
In summer the seep flows 40 gallons per minute. How many minutes does it take that flow to fill the cell's volume? This is the summer retention time in minutes.
Change your answer from item 3 into hours. Round to the nearest tenth of an hour.
The rule says water must stay at least 12 hours. How many hours longer than that is the summer retention time?
In spring the seep flows 75 gallons per minute. What is the spring retention time in hours? Round to the nearest tenth.
What percent of the 12-hour rule does the spring retention time reach? Round to the nearest whole percent.
To hold spring flow for a full 12 hours, how many gallons would the cell need to hold? Use 75 gallons per minute.
How many more gallons is that than the cell holds now?
Change your answer from item 9 into cubic feet. Round to the nearest hundredth.
The cell's surface area is 1,500 square feet. If the crew digs it deeper without making it wider, how many more feet of depth does it need? Round to the nearest tenth.
The cell holds 120,000 pounds of limestone, and the acid uses up 45 pounds each day. How many days will the limestone last? Round to the nearest whole day.
About how many years is that? Use 365 days in a year and round to the nearest tenth.
Which figure in this design is the criterion the cell must meet? Explain what your spring answers show about the cell as it is built now.

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: 6,000 cubic feet.
60 feet x 25 feet x 4 feet = 6,000 cubic feet of water.
Answer: 44,880 gallons.
6,000 cubic feet x 7.48 gallons per cubic foot = 44,880 gallons.
Answer: 1,122 minutes.
44,880 gallons divided by 40 gallons per minute = 1,122 minutes.
Answer: 18.7 hours.
1,122 minutes divided by 60 minutes per hour = 18.7 hours.
Answer: 6.7 hours.
18.7 hours minus the 12-hour rule = 6.7 hours of extra contact time in summer.
Answer: 10.0 hours.
44,880 gallons divided by 75 gallons per minute = 598.4 minutes. 598.4 divided by 60 = 9.973 hours, which rounds to 10.0 hours. Higher flow pushes water through faster, so the retention time falls.
Answer: 83 percent.
9.973 hours divided by 12 hours = 0.831, which is 83 percent of the required time. Accept 83 percent from the rounded 10.0 hours as well.
Answer: 54,000 gallons.
75 gallons per minute x 60 minutes per hour x 12 hours = 54,000 gallons needed.
Answer: 9,120 gallons.
54,000 gallons needed minus 44,880 gallons held = 9,120 more gallons.
Answer: 1,219.25 cubic feet.
9,120 gallons divided by 7.48 gallons per cubic foot = 1,219.25 cubic feet.
Answer: 0.8 foot.
1,219.25 cubic feet divided by 1,500 square feet of surface = 0.813 foot, which rounds to 0.8 foot of added depth.
Answer: 2,667 days.
120,000 pounds divided by 45 pounds per day = 2,666.67 days, which rounds to 2,667 days.
Answer: 7.3 years.
2,667 days divided by 365 days per year = 7.31 years, which rounds to 7.3 years. After that the crew must refill the cell with fresh limestone.
Answer: The 12-hour retention time is the criterion.
Accept any answer naming the 12-hour rule as the target the cell must meet, and noting that the spring flow leaves only about 10 hours, which is short of it. Strong answers add that the cell works as built in summer but fails in spring, so the design must either grow by about eight tenths of a foot in depth or be paired with something that evens out the flow.

Differentiation

Support

Give these students items 1 and 2 with the multiplication already set up, and a printed reminder card reading "volume divided by flow equals time" for items 3 and 6. Let them stop after item 11 and skip the limestone questions. Provide a labeled side-view sketch of the cell with length, width, and depth marked so the three dimensions are not confused.

Extension

Ask students to find the flow rate at which the cell exactly meets the 12-hour rule, working backward from the volume, then explain what that number tells a design team about which storms the cell can handle. Have them also propose a second option other than digging deeper, such as a small holding pond upstream, and describe what it would cost the site in land.

ELL supports

Pre-teach seep, retention time, and limestone with a simple cross-section drawing showing water entering one end of a gravel-filled basin and leaving the other. Provide the sentence frame "The water stays in the cell for ___ hours, which is ___ than the rule" for items 5 and 7. Allow item 14 to be answered by circling the criterion on the figures list and completing "In spring the cell is short by ___ hours."

Extensions

Have students plot retention time against flow rate for five flows between 30 and 90 gallons per minute, then draw a horizontal line at 12 hours on the same graph. Ask them to mark where the curve crosses that line and to write two sentences explaining why the relationship is not a straight line. This connects the arithmetic in items 3 through 7 to a picture a design team could hand to a county board.