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Chapter 11: Problem 11

A hydroelectric facility is built to deliver a peak power of \(1 \mathrm{GW}\), which it manages to do for three months of the year during the spring snow- melt. But for three months in summer, it drops to \(700 \mathrm{MW}\), then \(500 \mathrm{MW}\) for three months in fall. In winter, it drops way down to \(200 \mathrm{MW}\) for three months. Using the concept of the capacity factor (Definition 11.2.1), what is the annual average capacity factor for this facility?

Short Answer

Expert verified
Answer: The annual average capacity factor for the hydroelectric facility is 60%.

Step by step solution

01

Calculate the maximum possible energy output for each season

To calculate the maximum possible energy output for each season, we will multiply the peak power of the facility (\(1 \mathrm{GW}\)) by the time duration (in hours) of each season. Maximum possible energy output in spring = \(1 \mathrm{GW} \times 3 \times 30 \times 24 \mathrm{hours} = 2160 \mathrm{GWh}\) Do the same for other seasons, considering that the peak power remains 1GW: Maximum possible energy output in summer = 2160 GWh Maximum possible energy output in fall = 2160 GWh Maximum possible energy output in winter = 2160 GWh
02

Calculate the actual energy output for each season

To calculate the actual energy output for each season, we will multiply the given power output by the time duration (in hours) of each season. Actual energy output in spring = \(1 \mathrm{GW} \times 3 \times 30 \times 24 \mathrm{hours} = 2160 \mathrm{GWh}\) Actual energy output in summer = \(0.7 \mathrm{GW} \times 3 \times 30 \times 24 \mathrm{hours} = 1512 \mathrm{GWh}\) Actual energy output in fall = \(0.5 \mathrm{GW} \times 3 \times 30 \times 24 \mathrm{hours} = 1080 \mathrm{GWh}\) Actual energy output in winter = \(0.2 \mathrm{GW} \times 3 \times 30 \times 24 \mathrm{hours} = 432 \mathrm{GWh}\)
03

Calculate the total maximum and actual energy output for the year

Total maximum energy output for the year = Maximum possible energy output in spring+ Maximum possible energy output in summer + Maximum possible energy output in fall + Maximum possible energy output in winter. Total maximum energy output for the year = \(2160 + 2160 + 2160 + 2160 = 8640 \mathrm{GWh}\) Total actual energy output for the_year = Actual energy output in spring + Actual energy output in summer + Actual energy output in fall + Actual energy output in winter. Total actual_energy output for the_year = \(2160 + 1512 + 1080 + 432 = 5184 \mathrm{GWh}\)
04

Calculate the annual average capacity factor

Lastly, we will calculate the capacity factor using the following formula: Capacity factor = \(\frac{\text{Total actual energy output}}{\text{Total maximum energy output}}\) Capacity factor = \(\frac{5184 \mathrm{GWh}}{8640 \mathrm{GWh}}\) Capacity factor = \(0.6\) or \(60\%\) The annual average capacity factor for this hydroelectric facility is \(60\%\).

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Most popular questions from this chapter

The Robert Moses Niagara dam in New York is rated at \(2,429 \mathrm{MW}^{41}\) and has a high capacity factor of \(0.633 .\) How many \(\mathrm{kWh}\) does it produce in an average day, and how many homes would this serve at the national average of \(30 \mathrm{kWh} /\) day?

While the Chief Joseph Dam on the Columbia River can generate as much as \(2.62 \mathrm{GW}\left(2.62 \times 10^{9} \mathrm{~W}\right)\) of power at full flow, the river does not always run at full flow. The average annual production is 10.7 TWh per year \(\left(10.7 \times 10^{12} \mathrm{Wh} / \mathrm{yr}\right)\). What is the capacity factor of the dam: the amount delivered vs. the amount if running at \(100 \%\) capacity the whole year?

A dam 50 meters high is constructed on a river and is delivering \(180 \mathrm{MW}\) at some moment in time. What is the flow rate of water, in cubic meters per second, if the facility converts gravitational potential energy into electricity at \(90 \%\) efficiency?

A \(10 \mathrm{~kg}\) box is lifted \(2 \mathrm{~m}\) off the floor and placed on a frictionless horizontal conveyor to take it \(30 \mathrm{~m}\) across a warehouse. At the end of the conveyor, it is lowered \(1 \mathrm{~m}\) where it ends up on a shelf. How much net gravitational potential energy was given to the box from the start to the end of the process?

If an \(80 \mathrm{~kg}\) person is capable of delivering external mechanical energy at a rate of \(200 \mathrm{~W}\) sustained over several minutes, \({ }^{34}\) how high would they be able to climb in two minutes?
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