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# Water Flow

July 7, 2012 | Leave a Comment

Here we are in the midst of the worst drought in my memory, and this, the year of the prairie native garden walk. Our garden looks like a hayfield.

While I was watering some bushes, I wondered how long I need to hover over a bush to water it sufficiently. Too little, the moisture is waster, too much and that wastes water & time.

Assumption: I should dump 2 inches of water in a 2 foot diameter circle which is the dripline of the bush. That is where the roots hide.

So a circle of 2 feet to the depth of 2 inches is Pi*r^2*d, or 3.14*(12^2)*2 = 904 in^3.

I used a stopwatch app and a 2 qt measuring cup and found the water flow to be 6 seconds for 2 qts, or 12 seconds / gal. Using the conversion, 1 gal=233 cu.in., we get 233/12 cu.in. per second, or 19 cu.in./sec.

So divide 904 cu.in / 19 cu.in/sec => 47 seconds per bush. If I go a minute, good. For a bigger bush maybe 2 as the volume increases as the square of the radius.

Observation 1: math is cool

Observation 2: dimensional analysis is probably the most useful thing I learned in school. That, and algebra.

## HOW ABOUT WOLFRAM ALPHA?

Supposed to be way cool information engine. Lets try.

“how long to fill a bucket 2 feet in diameter and 2 inches deep at a rate of 2 quarts per 6 seconds” -> Huh?

“volume of a vertical cylinder 2 feet in diameter and 2 inches tall” -> 15.67 gallons

“( 2 qts / 6 seconds ) in gal/minute” -> 5 gal/min

“(volume of a vertical cylinder 2 feet in diameter and 2 inches tall) / ( 2 qts / 6 seconds )

-> “Using closest Wolfram|Alpha interpretation: cylinder 2 feet in diameter and 2 inches tall”

Can’t get anything else useful. Boo. By the time I get anything useful, I can do it on paper.

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