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Lesson plan Grade 9 Engineering & Design

Ambleside Dam Retrofit: Sizing a Micro-Hydropower Turbine

A grade 9 lesson where students estimate the power, yearly energy, and value of adding a small turbine to a fictional existing dam, then judge the retrofit.

Objectives

  1. Students will estimate the electrical power a small dam turbine produces from flow, head, and efficiency.
  2. Students will calculate yearly energy and value and judge whether a turbine retrofit is worthwhile.

Vocabulary

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, efficiency, kilowatt-hour

Teacher overview

Ambleside Dam is a fictional existing dam built long ago for water storage, with no power plant. Many real dams are like it, and adding a small turbine to water that already flows through can produce electricity without a new reservoir. In this lesson students estimate the power such a turbine would make, find its yearly energy and value, and decide whether the retrofit is worth building.

A common misconception is that a large flow always means large power. Power depends on both flow and head, the height the water falls, and on the efficiency of the turbine and generator. A big flow with almost no head produces little power, so students must combine all three factors rather than judging by flow alone.

Procedure

Hook (5 min): Show students a photo or sketch of a dam with water spilling over it. Ask what happens to the energy of the falling water when no turbine is present. Guide them to see that the energy is lost as splash and heat, which frames the retrofit question.

Direct instruction (12 min): Introduce the water horsepower formula for a fictional sample dam, using numbers that are not the handout numbers. Water horsepower equals flow in gpm times head in feet, divided by 3,960. Then show how horsepower becomes kilowatts by multiplying by 0.746, and how efficiency reduces the electrical output. Suppose a sample turbine makes 40 kilowatts and runs 20 hours; model that it produces 40 × 20 = 800 kilowatt-hours in a day.

Guided practice (13 min): Work handout items 1 and 2 together for the fictional Ambleside Dam. Students find the water horsepower and then the electrical power in kilowatts. Check that they apply the 0.80 efficiency only once, at the kilowatt step.

Independent practice (17 min): Students complete handout items 3 through 6 on their own or in pairs, finding daily energy, yearly energy, yearly value, and the number of homes the turbine could power. Circulate and confirm they multiply by run hours before multiplying by days.

Closure and exit ticket (8 min): Students answer one prompt on an index card: "If the turbine and its installation cost about $250,000, roughly how many years of energy value would it take to pay that back? Show the division." Collect the cards and check the method. If fewer than 80 percent set up value-per-year divided into cost, revisit the step next class.

Ambleside Dam Retrofit: Sizing a Micro-Hydropower Turbine — Grade 9 Engineering & Design
Name: ____________________________ Date: ______________

Student Handout

Use these facts about the fictional Ambleside Dam turbine retrofit.

  • Flow through the turbine: 3,500 gpm. Net head: 85 feet.
  • Water horsepower (WHP) = (flow in gpm × head in feet) ÷ 3,960.
  • Electrical power (kW) = WHP × 0.746 × efficiency. The combined efficiency is 0.80.
  • The turbine runs 20 hours per day for 350 days per year.
  • Electricity is worth $0.11 per kWh. A typical home uses about 10,800 kWh per year.

  1. Find the water horsepower. Round to the nearest tenth.

  2. Find the electrical power in kilowatts. Round to the nearest tenth.

  3. Find the energy produced in one day, in kWh. (Power × run hours.) Round to the nearest tenth.

  4. Find the energy produced in one year, in kWh. (Daily energy × 350 days.) Round to the nearest whole kWh.

  5. Find the yearly value of that energy. (Yearly energy × $0.11.) Round to the nearest dollar.

  6. How many typical homes could the turbine's yearly energy supply? (Yearly energy ÷ 10,800.) Round to the nearest whole home.

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: 75.1 hp.
    WHP = (3,500 × 85) ÷ 3,960 = 297,500 ÷ 3,960 = 75.13, which rounds to 75.1 hp.

  2. Answer: 44.8 kW. kW = 75.126 × 0.746 × 0.80 = 44.835, which rounds to 44.8 kW.

  3. Answer: 896.7 kWh.
    Daily energy = 44.835 × 20 = 896.70 kWh. Accept 896.0 kWh from a student who carried the rounded 44.8 kW forward.

  4. Answer: 313,845 kWh.
    Yearly energy = 896.70 × 350 = 313,845 kWh. Accept 313,600 kWh from a student who carried 896.0 forward.

  5. Answer: $34,523.
    Yearly value = 313,845 × $0.11 = $34,522.95, which rounds to $34,523. Accept $34,496 from a student who carried 313,600 forward.

  6. Answer: 29 homes.
    Homes = 313,845 ÷ 10,800 = 29.06, which rounds to 29 homes.

Differentiation

Support

Give students a step chart with the four stages (horsepower, kilowatts, daily energy, yearly energy) drawn as connected boxes, each labeled with its operation. Let them place one number per box so the chain stays visible. Allow calculators throughout.

Extension

Ask students to recompute the yearly value if the net head were raised to 100 feet by a small intake change, all else equal, and to explain in a sentence why head has such a direct effect on the power produced.

ELL supports

Pre-teach "head," "turbine," and "efficiency" with a simple diagram of water falling a height onto a wheel. Provide a sentence frame for the exit ticket: "It would take about ___ years to pay back the cost because the value each year is ___ dollars."