Math problem set Grade 11 Math

Thornwell Dam Hydropower: Head, Flow, and Power Output Problems

A grade 11 math problem set on a small hydropower turbine: converting flow units, computing water horsepower and kilowatts, and solving for a target output.

Objectives

  1. Students will compute the water horsepower of a hydropower turbine from its flow and net head.
  2. Students will solve for the flow required to meet a stated power output at a fixed head and efficiency.

Vocabulary

Key words: net head, water horsepower, efficiency, kilowatt-hour

Teacher overview

The fictional Thornwell Dam on the Ardith River runs a small hydropower turbine, and this set has students follow the power calculation an engineer performs from river flow all the way to daily energy. Students convert the flow to gallons per minute, find the water horsepower, apply the turbine-generator efficiency, and express the result in kilowatts and kilowatt-hours. A common misconception in this math is leaving the answer in horsepower when the question asks for kilowatts, or applying the 85 percent efficiency in the wrong step; efficiency is applied once, to the theoretical water horsepower, and the horsepower is then converted to kilowatts using 1 horsepower = 0.746 kilowatts.

Answer Key

  1. Answer: 17,952 gpm. 40 cfs times 448.8 gallons per minute per cfs equals 17,952 gpm.
  2. Answer: 498.7 hp. Water horsepower equals 17,952 times 110, which is 1,974,720, divided by 3960, giving 498.67, or about 498.7 horsepower.
  3. Answer: 423.9 hp. Actual power is 498.67 horsepower times 0.85, which equals 423.87, or about 423.9 horsepower.
  4. Answer: 316.2 kW. 423.87 horsepower times 0.746 kilowatts per horsepower equals 316.2 kilowatts.
  5. Answer: 7,589 kWh. Running at 316.2 kilowatts for 24 hours gives 316.2 times 24, which is 7,588.8, or about 7,589 kilowatt-hours.
  6. Answer: 22,709 gpm. Convert 400 kilowatts to horsepower: 400 divided by 0.746 equals 536.19 actual horsepower. Divide by 0.85 to undo efficiency: 536.19 divided by 0.85 equals 630.81 water horsepower. Then flow equals water horsepower times 3960 divided by head: 630.81 times 3960 divided by 110 equals 630.81 times 36, which is 22,709 gpm.
  7. Answer: 50.6 cfs. 22,709 gallons per minute divided by 448.8 gallons per minute per cfs equals 50.60, or about 50.6 cubic feet per second.
  8. Answer: no, short by 10.6 cfs. The target needs about 50.6 cfs but the river sends only 40 cfs, so it falls short by 50.6 minus 40, which is 10.6 cfs.

Differentiation

Support

Give students a filled example that runs the full chain for a different flow and head before they start, so the order of steps is modeled once. Provide a formula card listing the water horsepower formula, the efficiency step, and the horsepower-to-kilowatt conversion in the sequence they are used.

Extension

Ask students to find the actual power, in kilowatts, if the net head were reduced to 90 feet while the flow stayed at 40 cfs, and to explain in a sentence why head and flow both matter to the output. Have them compare that result with problem 4 for the fictional turbine.

ELL supports

Pre-teach "head," "flow," and "efficiency" with a labeled side-view sketch of a dam showing water falling from the reservoir surface to the turbine. Provide a sentence frame for problem 8: "The river supplies ___ cfs, but the target needs ___ cfs, so it is short by ___ cfs."