Math problem set Grade 7 Math
Gallowmere Dam's Summer Releases: A Grade 7 Flow-Rate Set
Students convert a dam release from cubic feet per second to gallons per day, name the conversion constant, and track signed weekly changes in reservoir level.
Objectives
- Students will convert a flow from cubic feet per second to gallons per day.
- Students will identify the constant of proportionality in a flow-unit conversion.
- Students will combine signed weekly changes to find a reservoir elevation.
Vocabulary
Key words: release, cubic feet per second, elevation, constant of proportionality
Teacher overview
Gallowmere Dam is fictional, and its release schedule and reservoir elevations were written for this problem set. Through the summer the dam sends a steady flow down the river below it while the reservoir behind it rises with rain and falls with releases and evaporation. The misconception this set is built to correct is the belief that converting cubic feet per second to gallons per minute takes only a multiplication by 60, since students forget that the conversion constant of 448.8 already carries both the change from cubic feet to gallons and the change from seconds to minutes. A second error to watch for is dividing by that constant instead of multiplying, which turns a large river release into a trickle.
Problems
Gallowmere Dam is a fictional dam, and every number below was written for this problem set. Use one cubic foot per second equals 448.8 gallons per minute where a problem calls for it.
- In July the dam releases 150 cubic feet per second. Using 448.8 gallons per minute for each cubic foot per second, what is the release in gallons per minute?
- Use the 67,320 gallons per minute from problem 1. What is the release in gallons per hour?
- Use the 4,039,200 gallons per hour from problem 2. What is the release in gallons per day?
- Use the 96,940,800 gallons per day from problem 3. Write that as million gallons per day, rounded to the nearest tenth.
- The relationship between cubic feet per second and gallons per minute is proportional. What is its constant of proportionality, in gallons per minute for each cubic foot per second?
- During one week the Gallowmere reservoir rose 1.4 feet from rain, fell 2.6 feet from releases, and fell 0.3 foot from evaporation. What is the net change in feet for that week?
- The reservoir started that week at an elevation of 612.8 feet. Using the net change from problem 6, what is the elevation at the end of the week?
- Suppose the same net change of -1.5 feet repeats for one more week. Starting from 611.3 feet, what is the elevation at the end of that second week?
- Gallowmere's operators must keep the reservoir at or above 610.0 feet. Using the 609.8-foot elevation from problem 8, how many feet below that floor is the reservoir?
- The dam's generator produces 1.2 megawatt-hours for every 10 million gallons released. Using the 96.9 million gallons per day from problem 4, how many megawatt-hours does one day of release produce? Round to the nearest tenth.
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: 67,320 gallons per minute.
Multiply the flow by the conversion constant: 150 x 448.8 = 67,320 gallons per minute. - Answer: 4,039,200 gallons per hour.
Multiply by the minutes in an hour: 67,320 x 60 = 4,039,200 gallons per hour. - Answer: 96,940,800 gallons per day.
Multiply by the hours in a day: 4,039,200 x 24 = 96,940,800 gallons per day. - Answer: 96.9 million gallons per day.
Divide by one million: 96,940,800 / 1,000,000 = 96.9408, which rounds to 96.9 million gallons per day. - Answer: 448.8 gallons per minute per cubic foot per second.
Every value in problem 1 comes from multiplying the cubic feet per second by this one number, so it is the constant of proportionality. The relationship can be written g = 448.8c, where c is cubic feet per second and g is gallons per minute. - Answer: -1.5 feet.
Combine the three changes in order: 1.4 - 2.6 = -1.2, then -1.2 - 0.3 = -1.5 feet. The rise is positive and both losses are negative. - Answer: 611.3 feet.
Add the net change to the starting elevation: 612.8 + (-1.5) = 611.3 feet. - Answer: 609.8 feet.
Add the net change again: 611.3 + (-1.5) = 609.8 feet. - Answer: 0.2 feet.
Subtract to compare: 610.0 - 609.8 = 0.2 feet below the floor. Because the elevation sits under the floor, operators would have to cut releases. - Answer: 11.6 megawatt-hours.
Find how many 10-million-gallon blocks were released: 96.9 / 10 = 9.69. Multiply by the output for each block: 9.69 x 1.2 = 11.628, which rounds to 11.6 megawatt-hours.
Differentiation
Support
Give students a conversion ladder for the fictional Gallowmere release with the four rungs labeled cubic feet per second, gallons per minute, gallons per hour, and gallons per day, and with the multiplier written beside each arrow. Students who lose track of signs in problems 6 through 9 can use a vertical number line marked in tenths of a foot around the 610.0-foot floor.
Extension
Ask students to find the largest steady release, in cubic feet per second, that the fictional dam could hold all week without dropping the reservoir below 610.0 feet, assuming rain and evaporation stay as they were in problem 6. Students should state the assumption they make about how release volume maps onto feet of elevation.
ELL supports
Provide a labeled cross-section of the fictional Gallowmere Dam showing the reservoir, the outlet works, and the river below, with "release," "elevation," and "floor" glossed in the student's first language where possible. Offer the sentence frame for problem 5: "The constant of proportionality is ____ because every answer comes from multiplying by ____."
Extensions
Students can compare the fictional Gallowmere release with a real river gauge reading from their own state, published in cubic feet per second, and carry out the same conversion chain. A second follow-on asks students to build a spreadsheet that takes any release in cubic feet per second and returns gallons per day and megawatt-hours, then to explain which cell holds the constant of proportionality.