Water for Schools
Menu

Worksheet Grade 9 Science

Gantry Falls Dam: Calculating Hydropower Output and Demand

A grade 9 worksheet using head, flow, and efficiency to find a fictional dam's power and yearly energy, then weighing its output against a town's electricity demand.

Objectives

  1. Students will calculate the electrical power output of a hydroelectric dam from its head, flow, and efficiency.
  2. Students will evaluate how a dam's generation compares with a community's electricity demand and how drought reduces output.

Vocabulary

Hydraulic head
Hydraulic head is the height of water pushing down on a spot, which tells you how much pressure it creates. The taller the column of water above you, the harder it pushes.
Turbine
A machine with blades that spin when water rushes past, like a high-tech waterwheel. Spinning turbines turn generators that make electricity.

Also: efficiency, kilowatt-hour, capacity

Teacher overview

A hydroelectric dam turns the energy of falling water into electricity. The power it can produce depends on three things: the height the water falls (the head), how much water flows through the turbines each second, and how efficiently the turbines and generators convert that energy. This worksheet uses Gantry Falls Dam, a fictional dam on the Wynne River, to walk students through the power equation, then asks them to compare the dam's output with a town's electricity use and to consider what happens during a drought. Watch for students who report power in watts when the question asks for kilowatts, since the difference is a factor of one thousand.

Gantry Falls Dam: Calculating Hydropower Output and Demand — Grade 9 Science
Name: ____________________________ Date: ______________

Student Worksheet

Use these facts about Gantry Falls Dam, a fictional dam, for every item. Keep units attached and show each step.

  • Power equation: P (watts) = density x g x flow x head x efficiency.
  • Water density = 1,000 kg/m^3, and g = 9.8 m/s^2.
  • Hydraulic head (height the water falls): h = 40 m.
  • Flow through the turbines: Q = 25 m^3/s.
  • Turbine and generator efficiency = 0.85 (that is, 85 percent).
  • 1 kilowatt (kW) = 1,000 watts.
  1. First find the gross theoretical power in watts, using an efficiency of 1 (a perfect machine): P = 1,000 x 9.8 x 25 x 40. Then convert your answer to kilowatts.
  2. Now find the actual electrical power the dam produces, in kilowatts, by multiplying your gross power in kilowatts by the real efficiency of 0.85.
  3. Express the actual electrical power from item 2 in megawatts (1 MW = 1,000 kW).
  4. If the dam runs steadily all day, how many kilowatt-hours of energy does it produce in 24 hours? Use the actual power from item 2.
  5. How many kilowatt-hours does the dam produce in a 365-day year at that steady rate?
  6. The town of Gantry Falls, a fictional community, uses about 60,000 kWh of electricity per day. Does the dam's daily energy from item 4 meet the town's demand, and by how many kilowatt-hours is it over or under?
  7. The dam's maximum nameplate capacity is 10,000 kW. What percent of that capacity is the dam actually using when it produces the power you found in item 2? Round to the nearest tenth of a percent.
  8. Give one benefit and one drawback of meeting the town's electricity needs with hydropower from this dam instead of a fossil-fuel power plant.
  9. A drought cuts the flow through the turbines to 15 m^3/s while the head and efficiency stay the same. Find the new actual electrical power in kilowatts.
  10. By what percent does the actual power drop when the flow falls from 25 to 15 m^3/s? Explain how your answer relates to the change in flow.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 9,800 kW. P = 1,000 x 9.8 x 25 x 40 = 9,800,000 watts. Dividing by 1,000 gives 9,800 kW.
  2. Answer: 8,330 kW. 9,800 kW x 0.85 = 8,330 kW of actual electrical power.
  3. Answer: 8.33 MW. 8,330 kW / 1,000 = 8.33 MW.
  4. Answer: 199,920 kWh. 8,330 kW x 24 hours = 199,920 kilowatt-hours in a day.
  5. Answer: 72,970,800 kWh. 199,920 kWh/day x 365 days = 72,970,800 kilowatt-hours in a year.
  6. Answer: Yes, by 139,920 kWh. The dam makes 199,920 kWh per day, and the town uses 60,000 kWh, so 199,920 - 60,000 = 139,920 kWh of surplus each day.
  7. Answer: 83.3 %. 8,330 kW / 10,000 kW x 100 = 83.3 percent of the nameplate capacity.
  8. Answer: Accept any reasoned pair. One benefit is that hydropower produces electricity without burning fuel, so it releases no combustion pollution or carbon dioxide while running. One drawback is that the dam blocks the river, which can harm fish migration and change the flow and sediment downstream, and its output falls during droughts. This weighs the benefits and costs of the energy resource.
  9. Answer: 4,998 kW. Gross power = 1,000 x 9.8 x 15 x 40 = 5,880,000 watts = 5,880 kW; times 0.85 gives 4,998 kW.
  10. Answer: 40 %. (8,330 - 4,998) / 8,330 x 100 = 3,332 / 8,330 x 100 = 40 percent. The power drops by the same fraction as the flow, because power is directly proportional to flow when head and efficiency stay constant, and the flow fell from 25 to 15, a 40 percent decrease.

Differentiation

Support

Provide the power equation with each variable already labeled and one fully worked example at a smaller flow, so students can match the pattern. Let students use a calculator and remind them, on a printed note, that dividing watts by 1,000 gives kilowatts.

Extension

Ask students to find the flow, in cubic meters per second, that Gantry Falls Dam would need in order to reach its full nameplate capacity of 10,000 kW at the same head and efficiency, and to comment on whether the Wynne River could realistically supply that flow year-round.

ELL supports

Pre-teach "head" (the height water falls) and "efficiency" (the share of energy that becomes useful electricity) with a labeled side-view sketch of a dam and turbine. Provide the sentence frame "The dam produces ___ kilowatts because ___" for item 2.

Extensions

Have students look up the head and flow of a real hydroelectric dam in their state or region, apply the same power equation with a stated efficiency, and compare their estimate with the dam's published capacity, discussing why real output over a year is usually well below the nameplate maximum.