Harvey Carried 20 Years Equivalent of US Electrical Energy Output
Storms carry potential energy in their water vapor, that was supplied as the heat of vaporization was input to evaporate the ocean water. When the vapor condenses to rain drops, the heat is released in warmth, which causes the vapor to rise to form thunderheads. This comes down rapidly in thunderstorms. From the amount of rain, we can put a lower bound on the energy supplied to the storm.
The heat of vaporization of water at about 87 degrees F or 273 degrees K plus 30.6 degrees K or 304 degrees K is about 2500 Joules per gram, to the accuracy we need (estimated from a curve).
We convert 27 trillion gallons of water to grams, find the number of Joules to vaporize that much water, and finally convert Joules to kilowatt -hours (kWh).
27 X 10^12 gallons X (8 pounds/gallon) X (454 g/pound) = 10^17 grams.
Multiply by 2500 Joules/g gives 2.5 X 10^20 Joules for the vaporization energy.
Convert with X 2.8 X 10^-7 kWh/J = 7 X 10^13 kWh as the energy supplied to vaporize the rain in the storm.
US electrical energy generation in a year is 4 X 10^12 kWh. So the storm had about 20 times the US yearly electrical energy output enclosed. No wonder it could move so much water.
The world electrical energy output was 22 X 10^12 kWh in 2014, so Harvey contained about pi (3) times the world’s electrical energy output in a year.
Bill Molzon, UCI Physics and Astronomy, suggested that we estimate the gravitational potential energy in the raindrops after they had risen in thunderclouds. Here is our calculation.
Consider the water rising to 50,000 feet or about 10 miles or 16 km. The mgh potential energy starts with the mass of 27 trillion gallons or 27 x 10^12 x 8 x 0.454 = 100 x 10^12 = 10^14 kg. Then mgh = 10^14 kg x 10 m/sec^2 x 1.6 x 10^4 m = 1.6 x 10^19 kg m^2/sec^2. So mgh = 1.6 x 10^19 joules. This is 6.4% of the heat of vaporization potential energy estimation.
