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drain down this pipe. The pipe is 300' long.

2007-01-10 04:11:42 · 5 answers · asked by Anonymous in Science & Mathematics Engineering

5 answers

Not knowing if the pipe is at the outlet of a tank, nor the depth of water in the tank, nor the entrance and exit conditions has too many unknown to make a pressurized calculation. The velocity of 40 ft/s stated above has a velocity head of 25' and would eat up all the elevation head in the pipe in just entrance and exit losses.

However.

If the 6" pipe is operating as an open channel, it will convey about 1.8 cfs at that 8.33% slope. This uses Manning's equation and assumes a clean pipe.

48000 gallons = 6417.1 cubic feet

6417.0 ft^3/1.8 cfs = 3565 seconds = 59.42 minutes

2007-01-10 08:31:14 · answer #1 · answered by daedgewood 4 · 0 0

Actually, the guy above didn't calculate for the pressure of water which is pretty substantial (14200 psi) compared to the small 25 feet of water in the pipe. Using the Energy equation, you find the velocity. As common sense would lead, as the water is lowering, the pressure decreases. So, assuming there is no pipe friction, you can find the time required to drain the water.

I used an excel sheet to calculate the volume minute by minute and found that it takes approximately 42 minutes to drain. Hope that helps!

2007-01-10 05:54:58 · answer #2 · answered by David T 3 · 0 0

For water: 7.48 gal/ft^3

Total volume of water to drain: 48000 gal/(7.48 gal/ft^3) = 6417.1 ft^3

Area of pipe opening: pi * r^2
(0.25 ft)^2 * 3.14 = 0.196 ft^2

Velocity of flowing water at end of 300 ft pipe:
mgh = 1/2mv^2 (potential to kinetic energy)
1/2v^2 = gh
v^2 = 2gh
h(total) = 1"/ft * 300 ft = 300" = 25 ft
g = 32 ft/s^2

gh = 25 ft * 32 ft/s^2 = 800 ft^2/s^2

v^2 = 2(gh) = 1600 ft^2/s^2
V(water) = 40 ft/s

(Total volume of water to drain / Area of pipe circle)/V(water)
will yield time to drain:
(6417.1 ft^3/0.196 ft^2)/40 ft/s
= 818.5 s

2007-01-10 06:39:41 · answer #3 · answered by thubanconsulting 3 · 0 0

I agree with David, plus, what is the configuration of the tank? as the head lowers, the headloss due to friction should also lower.

2007-01-11 10:01:08 · answer #4 · answered by bflo73 1 · 0 0

First calulate rate of flow per sec


Then volume is =PI()*r^@.
multiply volume by rate of flow , will tell you time

2007-01-10 04:36:50 · answer #5 · answered by Suhas 2 · 0 0

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