Math problem set Grade 8 Science

Ostrick Gorge Dam: Turning Falling Water Into Electricity

A grade 8 science problem set on a fictional gorge dam: the constant of proportionality between flow and power, two turbines compared, and a Hoover Dam scale check.

Grade: Grade 8
Subject: Science
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Printed problem set, Calculators

Standards

MS-ESS3-1 NGSS

Construct a scientific explanation based on evidence for how the uneven distributions of Earth's mineral, energy, and groundwater resources are the result of past and current geoscience processes.

How this material aligns: Problem 9 asks students to explain why usable hydropower concentrates where rivers drop steeply, the uneven energy resource distribution this standard addresses.

All materials for MS-ESS3-1 →

8.EE.B.5 Common Core Math

Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways.

How this material aligns: Problems 1, 2, and 7 have students find the unit rate of a proportional relationship and compare the slopes of two different turbine equations.

All materials for 8.EE.B.5 →

8.EE.A.3 Common Core Math

Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other.

How this material aligns: Problem 5 writes the plant's daily electricity output in scientific notation, and problem 10 uses two capacities in megawatts to state how many times as much one is than the other.

All materials for 8.EE.A.3 →

Download Word file

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 the constant of proportionality between river flow and electrical power output.
Students will compare the slopes of two turbine equations and interpret the difference.
Students will explain why hydropower resources are unevenly distributed across river systems.

Vocabulary

Hydropower: Electricity made from moving water, usually water falling through a dam. It is a renewable power source that can start up quickly when the electric grid needs it.
Turbine: A machine with blades that spin when water rushes past, like a high-tech waterwheel. Spinning turbines turn generators that make electricity.

Also: head, constant of proportionality, kilowatt, kilowatt-hour

Teacher overview

A hydroelectric plant converts the energy of falling water into electricity, and two quantities set how much it makes: how far the water falls, called the head, and how much water passes through the turbine each second. When the head stays fixed, power rises in direct proportion to flow, which makes the relationship a straight line through the origin. These problems follow the fictional Ostrick Gorge Dam on the Ostrick River across one year of changing flow. Students often assume a dam holding a huge reservoir must generate huge power; the reservoir sets how long a plant can run, while head and flow set how much power it makes at any moment.

Ostrick Gorge Dam: Turning Falling Water Into Electricity — Grade 8 Science

Name: ____________________________ Date: ______________

Problems

All figures below describe the fictional Ostrick Gorge Dam, whose turbines run under a fixed head of 120 feet. Let Q be the river flow in cubic feet per second and P be the power output in kilowatts.

Flow Q (cubic feet per second)	Power P (kilowatts)
100	850
200	1,700
300	2,550

Find the constant of proportionality in kilowatts per cubic foot per second.
Write the equation relating P and Q, and state whether the relationship is proportional.
In June the Ostrick River runs at 420 cubic feet per second. Find the plant's power output in kilowatts.
Energy equals power multiplied by time. How many kilowatt-hours does the plant generate in 24 hours at 3,570 kilowatts?
Write 85,680 kilowatt-hours in scientific notation, rounded to three significant figures.
By late August the flow drops to 95 cubic feet per second. Find the power output at that flow.
An older turbine at the same dam follows P = 6.0Q. At a flow of 300 cubic feet per second, how many more kilowatts does the newer turbine produce than the older one?
Solve 8.5Q = 2,000 for the flow the newer turbine needs to reach 2,000 kilowatts. Round to the nearest whole cubic foot per second.
A slow, flat river can carry the same flow as the Ostrick River and still support no hydropower. Explain why, and explain what that means for where hydropower plants can be built.
Hoover Dam's power plant has a capacity of about 2,080 megawatts. The fictional Ostrick Gorge plant's capacity is 4.2 megawatts. How many times larger is Hoover Dam's capacity, rounded to the nearest whole number?

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: 8.5 kilowatts per cubic foot per second.
Dividing any pair in the table gives the same unit rate: 850/100 = 8.5, 1,700/200 = 8.5, and 2,550/300 = 8.5.
Answer: P = 8.5Q, and yes, the relationship is proportional.
Every pair has the same ratio and the line passes through the origin, because zero flow through the turbine produces zero power.
Answer: 3,570 kilowatts.
P = 8.5 x 420 = 3,570 kilowatts in June.
Answer: 85,680 kilowatt-hours.
3,570 kilowatts x 24 hours = 85,680 kWh generated in one day at that flow.
Answer: 8.57 x 10^4 kilowatt-hours.
Moving the decimal point four places to the left gives 8.568, which rounds to 8.57, so the value is 8.57 x 10^4 kWh.
Answer: 807.5 kilowatts.
P = 8.5 x 95 = 807.5 kilowatts, less than a quarter of the June output because the flow fell to less than a quarter of June's.
Answer: 750 kilowatts.
The newer turbine gives 8.5 x 300 = 2,550 kilowatts and the older one gives 6.0 x 300 = 1,800 kilowatts, a difference of 750 kilowatts. The gap comes from the difference in slopes, 8.5 against 6.0.
Answer: 235 cubic feet per second.
Dividing both sides by 8.5 gives Q = 2,000/8.5 = 235.29, which rounds to 235 cubic feet per second.
Answer: Without a steep drop there is no head, so the water carries little usable energy.
Power depends on head as well as flow, and a flat river gives the water almost nowhere to fall. Usable hydropower therefore clusters where rivers cut through steep terrain, such as gorges and mountain valleys shaped over long spans of time, which is why some regions have many plants and others have none.
Answer: About 495 times.
2,080 divided by 4.2 = 495.2, which rounds to 495. Hoover Dam's plant has roughly 495 times the capacity of the fictional Ostrick Gorge plant.

Differentiation

Support

Work problem 1 with the class by writing the three division sentences side by side so students see that each pair gives the same unit rate. Give students the equation P = 8.5Q already written for problems 3, 6, and 8 so they can focus on substituting and solving.

Extension

Ask students to graph both turbine equations on one grid, label the point where each reaches 2,000 kilowatts, and then estimate how many hours the fictional Ostrick Gorge plant would need to run at its August flow to match one June day's output.

ELL supports

Pre-teach "head," "flow," and "turbine" with a labeled side view of a dam showing the reservoir surface, the drop, and the turbine at the bottom. Provide sentence frames such as "When the flow was ___, the power was ___ because ___." Allow students to answer problem 9 with a labeled sketch of a steep river and a flat river before writing.

Extensions

Have students look up whether their own state generates any hydroelectricity and where those plants sit on a map, then write two sentences connecting the plants' locations to the terrain, using what they learned from the fictional Ostrick Gorge problems.