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Math problem set Grade 11 Careers & CTE

Danecross Dam Generation: A Hydro Operator Problem Set

A grade 11 careers problem set in which students take a hydro operator's shift, computing station output from flow and head and writing output as a function of flow.

Objectives

  1. Students will calculate a hydroelectric station's output in kilowatts from flow, head, and turbine efficiency.
  2. Students will write station output as a linear function of flow and use it to find output at stated flows.
  3. Students will calculate a monthly capacity factor and explain what it says about how a station is run.

Vocabulary

Key words: head, turbine efficiency, capacity factor, nameplate capacity, kilowatt-hour

Teacher overview

An operator at a small hydroelectric station watches two numbers throughout the shift: how much water is passing through the turbine and how far that water falls. This set puts students at the fictional Danecross Dam on the fictional Bledwyn River, where a municipal utility runs a single turbine and reports its generation each month. Students compute output, write output as a function of flow, and then find the station's capacity factor. The misconception this set is built to correct is treating flow as the only thing that matters; head appears in the power formula as its own factor, so a reservoir drawdown of a few feet lowers output even when flow does not change.

Danecross Dam Generation: A Hydro Operator Problem Set — Grade 11 Careers & CTE
Name: ____________________________ Date: ______________

Problems

The fictional Danecross Dam generates power for the fictional town of Bledwyn. Use this formula for every problem: output in kilowatts equals 0.0846 times Q times H times e, where Q is flow through the turbine in cubic feet per second, H is head in feet, and e is turbine efficiency written as a decimal. The station's efficiency is 0.86 unless a problem says otherwise.

  1. At the start of the shift the turbine passes 320 cfs under 48 feet of head. Find the station's output in kilowatts, rounded to the nearest tenth.
  2. Convert your answer from problem 1 to megawatts, rounded to the nearest hundredth. One megawatt equals 1,000 kilowatts.
  3. If the station held your rounded output from problem 1 for a full 24 hours, how many kilowatt-hours would it generate? Round to the nearest whole kilowatt-hour.
  4. Head stays at 48 feet and efficiency stays at 0.86, so output depends only on flow. Write output P in kilowatts as a linear function of flow Q in cubic feet per second, and report the constant of proportionality in kilowatts per cfs, rounded to three decimal places.
  5. Use your function from problem 4 to find the station's output when the operator throttles flow to 150 cfs. Round to the nearest tenth of a kilowatt.
  6. The turbine cannot pass more than 420 cfs. Use your function from problem 4, with the constant rounded to three decimal places, to find the station's maximum output at that flow, rounded to the nearest tenth of a kilowatt.
  7. Over a 30-day month the station generated 612,000 kilowatt-hours. Its nameplate capacity is 1,500 kilowatts. Find the capacity factor as a percent, rounded to the nearest tenth. Capacity factor equals actual generation divided by the generation the station would produce running at nameplate capacity for every hour of the month.
  8. The utility credits generation at $0.058 per kilowatt-hour. Find the month's generation revenue for the 612,000 kilowatt-hours in problem 7.
  9. Late in the summer the reservoir draws down and head falls to 41 feet while flow stays at 320 cfs. Find the station's output in kilowatts, rounded to the nearest tenth.
  10. Explain, in two or three sentences, why the function you wrote in problem 4 stops giving correct answers once the reservoir level changes, and state what the operator would have to do to the function to keep using it.

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: 1,117.5 kilowatts.
    Output equals 0.0846 times 320 times 48 times 0.86. First 0.0846 times 320 equals 27.072, then 27.072 times 48 equals 1,299.456, then 1,299.456 times 0.86 equals 1,117.53216, which rounds to 1,117.5 kilowatts.
  2. Answer: 1.12 megawatts.
    1,117.5 kilowatts divided by 1,000 kilowatts per megawatt equals 1.1175, which rounds to 1.12 megawatts.
  3. Answer: 26,820 kilowatt-hours.
    1,117.5 kilowatts times 24 hours equals 26,820 kilowatt-hours.
  4. Answer: 3.492 kilowatts per cfs.
    With head and efficiency fixed, P equals 0.0846 times 48 times 0.86 times Q. First 0.0846 times 48 equals 4.0608, then 4.0608 times 0.86 equals 3.492288, so P equals 3.492Q with P in kilowatts and Q in cubic feet per second.
  5. Answer: 523.8 kilowatts.
    P equals 3.492 times 150, which equals 523.8 kilowatts.
  6. Answer: 1,466.6 kilowatts.
    P equals 3.492 times 420, which equals 1,466.64, rounding to 1,466.6 kilowatts.
  7. Answer: 56.7 percent.
    A 30-day month has 30 times 24, or 720 hours. At nameplate the station would produce 1,500 times 720, or 1,080,000 kilowatt-hours. The capacity factor is 612,000 divided by 1,080,000, which equals 0.56667, or 56.7 percent.
  8. Answer: $35,496.
    612,000 kilowatt-hours times $0.058 per kilowatt-hour equals $35,496.
  9. Answer: 954.6 kilowatts.
    Output equals 0.0846 times 320 times 41 times 0.86. First 0.0846 times 320 equals 27.072, then 27.072 times 41 equals 1,109.952, then 1,109.952 times 0.86 equals 954.55872, which rounds to 954.6 kilowatts.
  10. Answer: The function assumes head is fixed at 48 feet, so it is wrong for any other reservoir level.
    The constant 3.492 was built by multiplying 0.0846 by a head of 48 feet and an efficiency of 0.86, which means the whole function only describes the station while the reservoir sits at that level. When head falls to 41 feet, as in problem 9, the correct constant becomes 0.0846 times 41 times 0.86, or about 2.983 kilowatts per cfs. To keep using the function the operator would have to recompute the constant for the current head, or write output as a function of both flow and head instead of flow alone.

Differentiation

Support

Give students the constant of proportionality from problem 4 already computed so they can attempt problems 5 and 6 even if problem 4 stalled, and provide a filled-in example of the power formula at the fictional station using a different flow and head.

Extension

Ask students to graph output against flow for the fictional station at 48 feet of head and again at 41 feet, on the same labeled axes, then describe in writing what the difference in slope means for how long the operator must run the turbine to deliver the same energy.

ELL supports

Pre-teach "head," "efficiency," "nameplate capacity," and "capacity factor" using a labeled cross-section of the fictional dam showing the reservoir surface, the turbine, and the tailwater. Provide the sentence frame "When head goes down, output goes ___ because ___" for problem 10.

Extensions

Have students find the flow the fictional station would need in order to reach its 1,500-kilowatt nameplate capacity at 48 feet of head, then compare that flow to the turbine's 420 cfs limit and explain what the comparison says about how the station was sized.