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• ##### Anyone know the area over which it fell(2+ / 0-)
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weatherdude, mommyof3

in, say, km2?

Favorite exam question (slightly modified): given that the heat of vaporization for water is 2.5 x106 J/kg, and the density of water is ~1000 kg/m3, estimate a lower bound for the amount of energy released in a 4" rainfall over the city of Roanoke. What is that in terms of Hiroshima-class atomic bombs?

I'm glad I'm not in Roanoke right now and I hope everyone is safe.

• ##### 143 square kilometers. ;)(3+ / 0-)
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alefnot, mommyof3, Aunt Pat

Though, to be fair, I cheated.

If you say "gullible" real slow, it sounds like "green beans."

[ Parent ]

• ##### Cool website!(3+ / 0-)
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weatherdude, jfinsocal, Aunt Pat

But 143 km2 ... that number is going to be mindbogglingly huge ... let's find the back of an old envelope anyway ...

OK, that's 143 million m2, let's say exactly 100 mm = 0.1 m for the entire area or 1.43 x107 m3 or 1.43 x1010 kg of water. At 2.5 MJ/kg that's 3.6 x1016 J. Hmm, hard to get one's head around that ... if the Hiroshima bomb had a yield of 10 ktons TNT and TNT is 4.2 x109 J/ton  that's 4.2x1013 J/bomb. So the energy associated with the rainfall is on order 860 Hiroshima bombs.

Of course, this is just a zeroth order approx, you need finer grained data, including temperature. Then again, only about 20% of the liquid water in a storm system comes down as precip, so multiply that by 5. And the system didn't dump only on Roanoke ...

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