Issue 046 - Meteorology - Wind-field kinetic energy
How much kinetic energy is in Hurricane Karina?
The National Hurricane Center reported that Hurricane Karina rapidly intensified into a Category 4 hurricane in the eastern Pacific, with maximum sustained winds near 145 mph and hurricane-force winds extending outward up to about 40 miles from the center.
The problem
Estimate the total kinetic energy of the rapidly moving air within Hurricane Karina's hurricane-force wind field at a given moment.
Then put your estimate in context by determining approximately how much average U.S. household electricity consumption that amount of energy represents.
You'll need to decide how to approximate the shape and vertical depth of the moving air mass, the density of air, and an appropriate average wind speed across the hurricane-force region.
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
I assumed that if the winds are sustained at that speed about 40 miles from the center, then the radius of the storm system might be around 50 miles, or about 70 km. I also assumed the top of the storm system might be about 10 km in altitude, so the total volume of the storm could be represented by a cylinder:
r ~= 70 km ~= 7.0 x 10^4 m
h ~= 10 km ~= 1.0 x 10^4 m
storm volume ~= pi x r^2 x h
~= 3 x (7.0 x 10^4 m)^2 x 1.0 x 10^4 m
~= 1.5 x 10^13 m3I assumed much of the air toward the center of the storm would be moving more slowly, so I treated about 75% of the volume as actively engaged in the fast-moving winds, about 3.8 x 10^12 m3. If the peak sustained winds are about 145 mph, I took the average speed of the storm atmospheric mass to be about 70 mph, or about 35 m/s.
Using kinetic energy:
KE ~= 1/2 x m x v^2
~= 1/2 x 3.8 x 10^12 x (35 m/s)^2
~= 2.3 x 10^15 JThen I divided by 3,600 J/kWh and got about 6.3 x 10^11 kWh.
I could not remember the energy usage of the typical home, but I thought it was somewhere near 10,000 kWh per year. If that is correct, then the instantaneous kinetic energy of the storm would be enough energy to power about 63 million homes for a year. That is on the order of the actual number of households in the U.S.:
U.S. households ~= 330 million people / 2.5 people per household
~= 132 million householdsCalibration Score
Matt's Calibration Score: 50 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 10/30. The radius and average wind speed were plausible first-pass pegs, but the active storm volume and vertical depth were not handled consistently.
Model: 30/30. Kinetic energy of a moving air mass is the right core model for this specific question.
Math: 0/10. Several order-of-magnitude and unit-conversion mistakes substantially changed the answer, especially the kWh conversion.
Result: 10/30. The joule estimate was close to the right broad neighborhood, but the household-electricity comparison was high by roughly two orders of magnitude.
Grounding facts
NHC's August 31 advisory put Karina's maximum sustained winds near 145 mph, with hurricane-force winds extending up to 40 miles, or about 65 km, from the center. Hurricane-force winds begin around 74 mph, or about 33 m/s.
For a rough lower-atmosphere estimate, use air density near 1 kg/m3. Sea-level standard air is about 1.2 kg/m3, but density drops with altitude, and hurricane wind speeds are not uniform through a 10 km column.
A useful U.S. electricity peg is that an average residential customer uses roughly 10,000 kWh per year. Since 1 kWh = 3.6 x 10^6 J, one household-year is about 4 x 10^10 J.
After checking sources
Check and recalibrate
Start with the reported hurricane-force wind radius. A simple cylinder is still crude, but it is a good Fermi geometry model:
r ~= 65 km ~= 6.5 x 10^4 m
active wind depth ~= 2 to 5 km ~= 2 x 10^3 to 5 x 10^3 m
volume ~= pi x r^2 x depth
~= 3 x (6.5 x 10^4 m)^2 x (2 to 5) x 10^3 m
~= 3 x 10^13 to 6 x 10^13 m3Use air density near 1 kg/m3, and an average speed in the hurricane-force region around 40 to 45 m/s. That is above the hurricane threshold but below the 65 m/s maximum wind.
air mass ~= density x volume
~= 1 kg/m3 x (3 x 10^13 to 6 x 10^13 m3)
~= 3 x 10^13 to 6 x 10^13 kg
KE ~= 1/2 x m x v^2
~= 1/2 x (3 x 10^13 to 6 x 10^13 kg) x (4 x 10^1 m/s)^2
~= 2 x 10^16 to 5 x 10^16 JA clean central answer is therefore on the order of 10^16 to 10^17 joules of instantaneous kinetic energy in the hurricane-force wind field. This is not the hurricane's total heat engine energy, rainfall energy, or lifetime energy; it is only the kinetic energy of the fast-moving air at one moment.
Now convert to household electricity:
1 household-year ~= 1.0 x 10^4 kWh
1 kWh ~= 3.6 x 10^6 J
1 household-year ~= 3.6 x 10^10 J
household-years
~= (2 x 10^16 to 5 x 10^16 J) / 3.6 x 10^10 J/household-year
~= 6 x 10^5 to 1.4 x 10^6 household-yearsSo the best scale answer is around one million average U.S. household-years of electricity. That is enormous, but it is not the same as powering most U.S. households for a year.
Post-check reflection
Matt's reflection
For some reason today I did a poor job of keeping track of orders of magnitude.
I dropped a power of ten when calculating the volume of the storm airmass, then blew it on dropping that mass by 25%, then blew it again on using appropriate units for kWh. That gave me a hugely overestimated housing count for total energy consumption, even though the raw energy estimate was sort of in the right ballpark.
None of this has a huge impact on my intuitions about the situation: I know tropical storm systems carry enormous amounts of energy. It just emphasizes the importance of good accounting for orders of magnitude and proper arithmetic.
Recommended memory peg
For wind-energy problems, remember air density ~= 1 kg/m3, hurricane-force wind starts near 33 m/s, and wind-field kinetic energy ~= 1/2 x air density x volume x average wind speed^2.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.