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If a garden hose is 5/8 in diameter and the waterpressure is 40 psi, how many gallons are being used in one hour?

2006-08-11 12:58:43 · 7 answers · asked by zippy 2 in Science & Mathematics Mathematics

7 answers

Just see how much water the hose will fill in one minute and then multiply that number by 60.

2006-08-11 13:04:37 · answer #1 · answered by Anonymous · 0 1

Well, if this is a theoretical question I have an idea. Imagine that the water is being squeezed out the hose by a constant-pressure head. That is, the hose is like a hole in the bottom of a water tank. If I work this through (assuming an incompressible fluid) I get Pressure=1/2density of water*exit speed ^2. So exit speed=square root(2*Pressure/density). It's probably easiest to do this in SI units.

Here are some useful numbers:

The density of water is about 1g/cm^3.
9.8N=2.2 pounds
1.24cm=1in

2006-08-11 14:04:25 · answer #2 · answered by Benjamin N 4 · 0 0

Flow rate depends on many factors, not just the two you quoted. Also, the 40 psi is at the reservoir, the spigot, the nozzle, where?
The smoothness of the inner surface, the elevation difference between the nozzle, and reservoir (or wherever the pressure is calibrated). Also nozzle opening (if there is one).

Best bet is to perform the experiment like the other suggested.

2006-08-11 13:11:16 · answer #3 · answered by none2perdy 4 · 0 0

You can have 40 psi on a clogged hose. There is not enough data to derive a solution.

2006-08-11 13:03:20 · answer #4 · answered by Vincent G 7 · 0 0

Take a garden hose and fill up a gallon jug while timing it.
Then do the math.
Garden hose's output vary by location and time of day that you use.

2006-08-11 13:03:52 · answer #5 · answered by ? 6 · 0 0

Quicker solution is to the amount after flowing for 10 seconds and multiply that by 600. Don't waste time hunting for your calculator....you can figure this one out!!!

2006-08-11 18:11:38 · answer #6 · answered by gedanini3@yahoo.com 2 · 0 0

cant find that out with what ya gave us.
Turn every thing off but what your are mesuring and look at the meter.

2006-08-11 13:11:35 · answer #7 · answered by position28 4 · 0 0

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