Math problem set Grade 9 Math

Ferncliff Dam: Reservoir Flow and Hydropower Energy

A grade 9 problem set at the fictional Ferncliff Dam, using Q equals A times V for penstock flow, converting cfs to gpm and MGD, and separating power from energy.

Grade: Grade 9
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 1 through 5 carry cfs, gpm, MGD, and kilowatt-hours through the continuity and unit conversions so each result lands in the requested unit.

All materials for HSN-Q.A.1 →

HSA-CED.A.2 Common Core Math

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

How this material aligns: Problem 8 has students write the two-variable equation for energy against hours at a steady power and evaluate it over a full day.

All materials for HSA-CED.A.2 →

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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 find penstock flow using Q equals area times velocity and convert flow units.
Students will distinguish power in megawatts from energy in megawatt-hours.
Students will compute a capacity factor as a percent of maximum output.

Vocabulary

Penstock: The huge pipe that carries water from a reservoir down to the turbines in a hydropower plant. Falling water rushes through it with enormous force.
Flow rate: Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: capacity factor, megawatt-hour, head

Teacher overview

The fictional Ferncliff Dam holds back a reservoir and sends water down a large pipe, the penstock, to spin a turbine and make electricity. This set moves from the flow in the penstock to the energy the plant produces in a day. A common misconception is mixing up power and energy: power in megawatts is a rate, while energy in megawatt-hours is power multiplied by time, so a plant running at the same power for longer makes more energy.

Ferncliff Dam: Reservoir Flow and Hydropower Energy — Grade 9 Math

Name: ____________________________ Date: ______________

Problems

Ferncliff Dam (a fictional dam) has a power plant with a maximum capacity of 40 megawatts. Use these facts: one cubic foot per second equals 448.8 gpm; one MGD equals 1.547 cfs; one megawatt-hour equals 1,000 kilowatt-hours.

The penstock has a cross-section area of 30 square feet, and water flows through it at 12 feet per second. Using Q equals area times velocity, find Q in cfs.
Convert the flow from problem 1 to gpm.
Convert the flow from problem 1 to MGD. Round to the nearest tenth.
The plant runs at 40 megawatts for 6 hours. Energy equals power times time. Find the energy in megawatt-hours.
Convert the energy from problem 4 to kilowatt-hours.
Suppose an average home uses about 900 kilowatt-hours in a month. How many homes' monthly use could the problem-5 energy cover? Round down to a whole number.
Last month the plant averaged 26 megawatts against its 40-megawatt maximum. The capacity factor is average power divided by maximum power, times 100. Find it as a percent.
Write an equation for the energy E, in megawatt-hours, produced in h hours at a steady 40 megawatts. Then find the energy after 24 hours.
Hoover Dam's power plant has a capacity of about 2,080 megawatts. How many times larger is that than Ferncliff's 40-megawatt capacity?
During a dry summer the reservoir drops, lowering the head that drives the turbine. Explain why less head means less power at the same gate opening.

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: 360 cfs. Q = area times velocity = 30 square feet times 12 feet per second = 360 cfs.
Answer: 161,568 gpm. 360 cfs times 448.8 gpm per cfs = 161,568 gpm.
Answer: 232.7 MGD. 360 cfs divided by 1.547 cfs per MGD = 232.71, about 232.7 MGD.
Answer: 240 MWh. Energy = 40 megawatts times 6 hours = 240 megawatt-hours.
Answer: 240,000 kWh. 240 megawatt-hours times 1,000 = 240,000 kilowatt-hours.
Answer: 266 homes. 240,000 divided by 900 = 266.67, which rounds down to 266 homes' monthly use.
Answer: 65 %. 26 divided by 40, times 100 = 0.65 times 100 = 65 percent.
Answer: 960 MWh. The equation is E = 40h. At h = 24: E = 40 times 24 = 960 megawatt-hours.
Answer: 52 times. 2,080 megawatts divided by 40 megawatts = 52 times larger.
Answer: Head is the height of water pushing on the turbine. Less head means less pressure and less energy per gallon, so the same gate opening delivers less power to the generator.

Differentiation

Support

Give students a labeled cross-section of the dam showing the reservoir, the penstock, and the turbine, plus a reference card with the three conversions. Work one power-versus-energy example aloud using numbers not on this set before students start problem 4.

Extension

Ask students to find how many hours the plant must run at 40 megawatts to match the daily energy of a plant that runs at 26 megawatts for a full 24 hours, and to explain what the comparison says about capacity factor.

ELL supports

Pre-teach "penstock," "power," "energy," and "head" with a diagram that labels water height, the pipe, and the spinning turbine. Provide a sentence frame for problem 10: "Less head means ___ power because ___."