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Worksheet Grade 12 Science

Ravenglass Dam Hydropower: Head, Flow, and Energy

A grade 12 physics worksheet using the hydropower equation to find power, daily energy, and capacity factor for a fictional dam, with a drought sensitivity case.

Objectives

  1. Students will calculate hydroelectric power from head, flow, efficiency, and water density.
  2. Students will convert power in watts to daily energy in kilowatt-hours.
  3. Students will compute a capacity factor and analyze how reduced flow lowers output.

Vocabulary

Key words: head, hydraulic power, efficiency, capacity factor, nameplate capacity

Teacher overview

This worksheet builds the hydropower equation one factor at a time for a fictional reservoir dam, then uses it to study how output falls during a drought. Students compute gross hydraulic power, apply an efficiency factor, and convert power to daily energy, keeping careful track of units from watts to kilowatt-hours. A frequent mistake is confusing power with energy: a plant's power in megawatts is an instantaneous rate, while the kilowatt-hours it delivers depend on how long it runs, so students must multiply by time before comparing to an energy total.

Ravenglass Dam Hydropower: Head, Flow, and Energy — Grade 12 Science
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Ravenglass Dam sends water through a turbine under a net head of 85 meters at a flow of 45 cubic meters per second. The overall efficiency of the turbine and generator together is 0.88. Use a water density of 1000 kilograms per cubic meter and g = 9.81 meters per second squared. The hydraulic power in watts is P = density x g x flow x head, and the electrical output multiplies that by the efficiency.

  1. Find the gross hydraulic power in watts, P = 1000 x 9.81 x 45 x 85, then state it in megawatts (MW). Round to the nearest tenth of a MW.

  2. Find the electrical output by multiplying the gross power by the 0.88 efficiency. State it in MW, rounded to the nearest tenth.

  3. Express the electrical output from item 2 in kilowatts (kW).

  4. If the dam runs steadily for 24 hours at the output from item 3, how many kilowatt-hours (kWh) does it generate in one day?

  5. Extend item 4 to a full 30-day month at the same steady output. State the monthly energy in kWh.

  6. The generator's nameplate capacity is 40 MW. Using the output from item 2 as the average power, find the capacity factor as a percentage, where capacity factor = average power / nameplate capacity x 100.

  7. Hoover Dam's power plant has a nameplate capacity of about 2,080 MW. What percentage of Hoover Dam's capacity is Ravenglass Dam's 40 MW nameplate? Round to the nearest tenth.

  8. During a drought the turbine flow drops to 30 cubic meters per second while head and efficiency hold steady. Find the new electrical output in MW.

  9. Find the percent reduction in electrical output from the normal case (item 2) to the drought case (item 8). Round to the nearest tenth.

  10. Explain why the electrical output is directly proportional to both the head and the flow, using the structure of the hydropower equation.

  11. Give one physical reason the overall efficiency is 0.88 rather than 1.00, naming where energy is lost.

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: 37.5 MW.
    1000 x 9.81 x 45 x 85 = 37,523,250 watts. Dividing by 1,000,000 gives 37.52 MW, about 37.5 MW.

  2. Answer: 33.0 MW.
    37,523,250 watts x 0.88 = 33,020,460 watts, or 33.02 MW, about 33.0 MW.

  3. Answer: 33,020 kW.
    33,020,460 watts / 1000 = 33,020.46 kW, about 33,020 kW.

  4. Answer: 792,491 kWh.
    33,020.46 kW x 24 hours = 792,491 kWh in one day.

  5. Answer: 23,774,731 kWh.
    792,491 kWh/day x 30 days = 23,774,731 kWh for the month.

  6. Answer: 82.6%.
    33.02 MW / 40 MW x 100 = 82.55%, about 82.6 percent.

  7. Answer: 1.9%.
    40 MW / 2,080 MW x 100 = 1.923%, about 1.9 percent.

  8. Answer: 22.0 MW.
    1000 x 9.81 x 30 x 85 x 0.88 = 22,013,640 watts, or 22.01 MW, about 22.0 MW.

  9. Answer: 33.3%.
    (33.02 - 22.01) / 33.02 x 100 = 11.01 / 33.02 x 100 = 33.3 percent. This matches the one-third cut in flow, since output is proportional to flow.

  10. Answer: Both head and flow are multiplied factors in the equation.
    Power equals density times g times flow times head times efficiency. Head and flow each appear once as a factor, so doubling either one doubles the product, which makes the output directly proportional to each of them.

  11. Answer: Energy is lost as friction and heat in the turbine and generator.
    Water friction and turbulence in the turbine, bearing friction, and electrical resistance in the generator windings all convert some energy to heat instead of electricity, so the real output is 88 percent of the ideal.

Differentiation

Support

Give students the equation with each variable already labeled with its value and unit, and let them compute item 1 in stages, recording the running product after each multiplication so the large numbers stay organized.

Extension

Ask students to find the flow, in cubic meters per second, that the drought plant would need to return to its normal 33.0 MW output if the head fell to 78 meters, holding efficiency at 0.88.

ELL supports

Provide a labeled cross-section of a dam showing head as the vertical drop and flow as the arrow through the turbine, with each term written beside its symbol. Allow students to answer items 10 and 11 orally or with a labeled diagram.

Extensions

Have students look up the nameplate capacity and reported annual generation of a real hydroelectric dam, calculate its capacity factor from those two figures, and compare it with the 82.6 percent found here, discussing why a real dam's capacity factor is often lower.