Lesson plan Grade 7 Engineering & Design
Wickersall Salt Yard: Choosing a Runoff Containment Design
A 55-minute grade 7 lesson where students size storm runoff from a fictional road-salt pad, then pick among a full roof, a partial roof, and a catch basin.
Objectives
- Students will compute storm runoff volume from a paved area using a stated conversion.
- Students will express the reduction from a partial cover as a percent of the runoff the bare pad sheds.
- Students will eliminate containment options that fail a stated budget or space limit.
Vocabulary
- Catch basin
- A catch basin is a storm drain inlet with a sump underneath that traps sediment, leaves, and trash before water flows on into the storm sewer pipe. Crews periodically clean out catch basins so they do not clog or send debris downstream.
Also: runoff, containment, chloride, constraint, freeboard
Teacher overview
Fictional Wickersall Township keeps its winter road salt in an open pile on a concrete pad beside Gravenoak Brook, and the brook below the yard still runs salty well into the spring. This lesson puts students on the team choosing a fix: a full roof, a partial roof, or a bermed catch basin. The misconception worth naming directly is that cutting the volume of runoff and cutting the amount of salt leaving the site are the same thing. Students who hold it will treat a partial roof as removing most of the problem because it removes most of the water, when the rain that still lands on the pile keeps dissolving salt until it is saturated. Say this out loud before the independent work, and return to it at the close.
Procedure
Hook (5 min): Describe the fictional salt yard and the brook beside it. Ask what the township should do first. Record answers on the board without sorting them. Answers such as "cover it" and "move the pile" tend to arrive before anyone asks how much water is involved, which sets up the day's work.
Direct instruction (12 min): Introduce the runoff conversion used all period: area in square feet times rainfall in inches times 0.623 gives gallons. Explain that the 0.623 comes from the volume of one square foot of water one inch deep. Model the full chain on a site that does not appear on the handout. A loading apron measures 50 feet by 20 feet, so its area is 1,000 square feet, and in a 1-inch storm it sheds 1,000 times 1 times 0.623, or 623 gallons. Then define the design problem: the township may spend up to 25,000 dollars, and the only land for a basin is a small corner lot.
Guided practice (13 min): Distribute the handout. Work questions 1 through 4 together at the board so every student has the pad area, the uncovered runoff, the area the partial roof leaves uncovered, and the runoff that still contacts salt. Before computing question 4, ask the class to predict whether the partial roof cuts the salty runoff by more or less than half, and why. This surfaces the misconception in the room rather than in the key.
Independent practice (20 min): Students complete questions 5 through 11 in pairs. Circulate and check that pairs carry units through the basin conversion in question 7 rather than dividing bare numbers. Pairs who finish early should sketch the partial roof in plan view and mark where the 900 uncovered square feet sit.
Closure and formative check (5 min): Collect the handouts and scan question 10 only. If fewer than three quarters of the class recommends the partial roof, reopen the next session with the two eliminations side by side, showing that one option failed on money and the other on space. Close by asking two students to read their question 11 answer aloud.
Student Handout
Suppose you are on the design team for the fictional Wickersall Township salt yard. Use these figures for every question.
The salt pile sits on a concrete pad that measures 60 feet by 40 feet. The township designs for a storm that drops 2 inches of rain. Runoff in gallons equals the area in square feet times the rainfall in inches times 0.623. One cubic foot holds 7.48 gallons. The township may spend up to 25,000 dollars. The only land available for a basin is a corner lot 12 feet by 12 feet that may be dug 3 feet deep.
The three options are:
| Option | What it is | Cost |
|---|---|---|
| A | A full roof over the whole pad | 40,000 dollars |
| B | A partial roof covering a 60-foot by 25-foot section of the pad | 22,000 dollars |
| C | A berm and catch basin that holds all the runoff from the pad | 9,000 dollars |
What is the area of the concrete pad in square feet?
With no cover at all, how many gallons of runoff does the pad shed in the 2-inch storm? Round to the nearest gallon.
Option B covers a 60-foot by 25-foot section. How many square feet of the pad are left uncovered?
How many gallons of runoff does that uncovered area shed in the same 2-inch storm? Round to the nearest gallon.
Compare your answers to questions 2 and 4. What percent of the runoff in question 2 does Option B remove? Round to the nearest whole percent.
Option C must hold all the runoff from the pad plus 20 percent extra for freeboard. Using your rounded answer from question 2, how many gallons must the basin hold?
Convert your answer to question 6 into cubic feet, using 7.48 gallons per cubic foot. Round to the nearest tenth.
The corner lot is 12 feet by 12 feet and may be dug 3 feet deep. How many cubic feet would that basin hold?
Which options cost 25,000 dollars or less?
Using your answers to questions 7, 8, and 9, which single option should the team recommend?
Option B still leaves part of the pad uncovered. Describe one change to Option B that would cut salty runoff further without enlarging the roof, and name one measurement the township could take in the brook to find out whether the change worked.
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: 2,400 square feet.
60 feet times 40 feet equals 2,400 square feet.Answer: 2,990 gallons.
2,400 times 2 times 0.623 equals 2,990.4 gallons, which rounds to 2,990.Answer: 900 square feet.
The roof covers 60 times 25, or 1,500 square feet. 2,400 minus 1,500 equals 900 square feet left uncovered.Answer: 1,121 gallons.
900 times 2 times 0.623 equals 1,121.4 gallons, which rounds to 1,121.Answer: 63 percent.
2,990 minus 1,121 equals 1,869 gallons removed. 1,869 divided by 2,990 equals 0.625, which is 63 percent to the nearest whole percent. Accept 62.5 percent from a student who did not round, and accept 62 or 63 from the unrounded volumes.Answer: 3,588 gallons.
2,990 times 1.2 equals 3,588 gallons.Answer: 479.7 cubic feet.
3,588 divided by 7.48 equals 479.68 cubic feet, which rounds to 479.7.Answer: 432 cubic feet.
12 times 12 times 3 equals 432 cubic feet.Answer: Options B and C.
Option B costs 22,000 dollars and Option C costs 9,000 dollars, both at or under the 25,000-dollar limit. Option A costs 40,000 dollars, which is over.Answer: Option B.
Option A is eliminated by the budget. Option C is eliminated by space, because the corner lot holds only 432 cubic feet and the basin needs 479.7. Option B is the only option that clears both limits.Answer: A workable change plus a named measurement.
Accept any change that reduces contact between rain and salt without a larger roof. Common correct answers include a tarp weighted over the uncovered end between storms, a low curb that keeps run-on from the yard off the pad, moving the pile fully under the roofed section and keeping the uncovered strip empty, and sweeping spilled salt off the uncovered area after each loading. For the measurement, accept chloride concentration in the brook above and below the outfall, conductivity readings taken on the same schedule before and after the change, or a salinity reading taken after each thaw. Do not accept a change that simply enlarges the roof, since the question rules that out.
Differentiation
Support
Give these students a handout copy with questions 1 and 3 already answered and the multiplication for question 2 set up. Provide a worked conversion strip showing that one square foot of water one inch deep equals 0.623 gallons, so the runoff rule has a picture behind it. For question 10, supply a three-row elimination table with the headings filled in and let students enter only pass or fail.
Extension
Have students redo questions 2 and 6 for a 4-inch storm and report whether their recommendation changes. Ask them to determine the shallowest basin that would still work if the corner lot could be widened to 14 feet by 14 feet, and to write one sentence on why a township might prefer a shallow wide basin to a deep narrow one.
ELL supports
Pre-teach runoff, pad, and berm with a labeled side-view sketch of the salt pile, the concrete pad, and the brook downhill. Provide the frames "The area is ____ times ____" and "Option ____ is out because ____." Allow question 11 to be answered as a labeled drawing with two short captions rather than in sentences.
Extensions
Ask students to write a short notice the fictional township could post at the salt yard, telling plow drivers three habits that would cut salty runoff at no cost, such as loading inside the roofed area and sweeping spills before rain. Comparing the notice with the design work makes the point that operating practice and built structures address the same problem from different directions.