Math problem set Grade 10 Math
Ravenglass Dam Hydropower: Capacity Factor and Energy
A grade 10 problem set separating power from energy: students find capacity factor, convert MWh to kWh, and count homes powered at a fictional hydroelectric dam.
Objectives
- Students will compute a dam's capacity factor from its actual and maximum possible energy.
- Students will convert between megawatt-hours and kilowatt-hours and interpret the results.
Vocabulary
Key words: capacity factor, megawatt, megawatt-hour, kilowatt-hour
Teacher overview
A hydroelectric dam is rated by its capacity in megawatts, the most power it can make at any instant, but what a town actually uses is energy, measured in megawatt-hours over time. Capacity factor connects the two: it is the actual energy produced divided by the energy the dam could make if it ran flat out the whole period. This set uses the fictional Ravenglass Dam. The misconception it corrects is treating a 150 megawatt rating as if the dam always delivers 150 megawatts, when water supply, demand, and maintenance keep the real output lower.
Problems
The Ravenglass Dam is a fictional hydroelectric dam, and its numbers are set for this problem set.
Dam data:
- Nameplate capacity: 150 MW
- Energy produced in a 30-day month: 43,200 MWh
- A typical home uses about 900 kWh in a month (a value chosen for this problem set)
- 1 MWh = 1,000 kWh; a day has 24 hours
- Find the most energy the dam could make in the 30-day month if it ran at full capacity the whole time, in MWh.
- Find the capacity factor as a percent. Use capacity factor = actual energy / maximum possible energy.
- Convert the actual 43,200 MWh for the month to kWh.
- How many typical homes could that month's energy supply, at 900 kWh each?
- Find the dam's average power output for the month, in MW.
- A dry year drops production to 30,000 MWh for the month. Find the new capacity factor as a percent.
- Find the percent drop in energy from 43,200 MWh to 30,000 MWh.
- A planner needs to know how many homes the dam can serve over a year. Explain whether capacity in MW or energy in MWh is the right quantity to use, and why.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: 108,000 MWh.
Full-capacity energy = 150 MW x 24 hours x 30 days = 150 x 720 = 108,000 MWh. - Answer: 40 percent.
Capacity factor = 43,200 / 108,000 = 0.40, or 40 percent. - Answer: 43,200,000 kWh.
43,200 MWh x 1,000 kWh per MWh = 43,200,000 kWh. - Answer: 48,000 homes.
43,200,000 kWh / 900 kWh per home = 48,000 homes. - Answer: 60 MW.
Average power = energy / time = 43,200 MWh / 720 hours = 60 MW. This also equals capacity factor times capacity, 0.40 x 150 = 60 MW. - Answer: 28 percent.
New capacity factor = 30,000 / 108,000 = 0.2778, or about 28 percent. - Answer: 31 percent.
Percent drop = (43,200 - 30,000) / 43,200 x 100 = 13,200 / 43,200 x 100 = 30.6, or about 31 percent. - Answer: Energy in MWh, because homes use energy over time, not instant power.
Capacity in MW is only the top instant rate. To count homes served across a year, the planner needs total energy in MWh, which already accounts for how many hours the dam actually runs.
Differentiation
Support
Give students a filled-in units chart showing MW as power, MWh as energy, and kWh as smaller energy, with the 1 MWh = 1,000 kWh link highlighted. Work problem 1 together so the 24 x 30 step is clear.
Extension
Have students find the year-long energy at the 40 percent capacity factor, then estimate the homes served for a full year and compare that to a single month, noting where the twelve-times pattern does and does not hold.
ELL supports
Pre-teach "capacity," "power," and "energy" with a water-tank picture: capacity is the tap's widest opening, energy is the total water collected over time. Provide a sentence frame for problem 8: "The right quantity is ___ because ___."