Energy and Human Ambitions on a Finite Planet, 2021a

11 Hydroelectric <strong>Energy</strong> 183
12. While the Chief Joseph Dam on the Columbia River can generate
as much as 2.62 GW (2.62 × 10 9 W) of power at full flow, the river
does not always run at full flow. The average annual production is
What is the capacity factor
of the dam: the amount delivered vs. the amount if running at
100% capacity the whole year?
10.7 TWh per year (10.7 × 10 12 Wh/yr). Hint: Multiplying peak power by hours in
a year will result in units similar to Wh/yr
for direct comparison.
13. The Robert Moses Niagara dam in New York is rated at 2,429 MW 41 41: . . . peak power capacity
<strong>and</strong> has a high capacity factor of 0.633. How many kWh does it
produce in an average day, <strong>and</strong> how many homes would this serve
at the national average of 30 kWh/day?
14. The Robert Moses Niagara dam from Problem 13 is 30 m high.
What is the peak flow rate, in m 3 /s, if it can produce full capacity
power (2.43 GW electrical output) while converting gravitational
potential energy to electricity at 90% efficiency?
15. It takes 2,250 J to evaporate each gram of water, while only
taking about 330 J to raise the temperature of water from room
temperature to the boiling point. If it takes 10 minutes to bring a pot
of water from room temperature to boiling, how much additional
time will it take to boil off (evaporate) all the water if injecting
energy at the same rate the whole time?
Hint: Convert average power to kW then
multiply by hours.
16. Starting at 44,000 TW of solar input to the hydrologic cycle, parallel
the development in Section 11.2.1 by computing the power
remaining at each stage 42 if, for each gram of water:
42: Each stage will knock down the number
further; report each in TW.
a) water is evaporated <strong>and</strong> lifted to 5 km height; 43
b) 30% of the water falls on l<strong>and</strong> where collection is possible;
43: This is the largest jump, keeping only
50 J out of every 2,300 J.
c) the typical l<strong>and</strong> height is 800 m;
d) only 20% of the water makes it to dammable locations;
e) only 50 m of height (of the original 800 m average) is left for
the dam.
By this analysis, how much hydroelectric power is theoretically
possible, globally?
17. Fig. 10.1 (p. 167) indicated that about 44,000 TW globally goes into
evaporating water. We can turn this into an estimate of how much
rain we expect per year, on average. The simplest way to do this
is to think of a single square meter of ocean surface, receiving an
average evaporation input power of 120 W. 44 Each millimeter of 44: 44,000×10 12 W divided by 3.7×10 14 m 2
of water depth across our square meter has a volume of 1 L, or a
of ocean surface is 120 W/m 2 .
mass of 1 kg. At a steady input of 120 W, 45 how many millimeters 45: i The steady 120 W is already accounting
of water are drawn off in a year? That same amount will come
for day/night: this is a time average.
back down somewhere as precipitation.
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.