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Old Oct 31st 2015, 09:20 PM   #1
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Water Velocity Out of 200m Pipe at 5 degrees

A resovoir releases water into a 200 meter long pipe with 2m diameter.
At 5 degrees relative to x axis so the drop is 200m (sin 5 deg) = 17.4m
Since v o is 0 all I can solve for in this veolcity vector: v =(vx^2 + vy^2)^1/2
is for vy = (2gh)^1/2 = 18.4m/sec. Without vx how to solve for vz ?
Assume pipe is full. http://www.calctool.org/CALC/eng/civil/hazen-williams_g
This format shows a velocity of aboiut 20msec. I am more interested in the
solution formula

Edit : The acceleration = g sine 5 degrees and
v^2 = 2(a*s) So v = (2*sin 5 deg * 9.8m/s^2*200m) ^1/2
=18.5m/s

Last edited by morrobay; Nov 1st 2015 at 02:20 AM. Reason: correction
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Old Nov 2nd 2015, 01:03 AM   #2
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Can someone calculate the water pressure in atmospheres at outfall from given data above ? v = 18.5m/sec, 2m diameter full water pipe.
Im not familiar enough with Berrnoull'is equation , P = 1/2 pv^2 + pgh = constant p = density
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Old Nov 3rd 2015, 12:22 PM   #3
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There equation P = 1/2 pv^2 + pgh falls fairly obviously into two parts

pgh is the pressure due to the height of the water above the outflow
p is 1000 kg/lm3
g is 9.8 m/s2
h is 17.4 m

1/2 pv^2 is the pressure due to the motion of the water
by your calculations v=18.5m/s
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Old Nov 3rd 2015, 10:33 PM   #4
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The overall question is whether the water from reservoir will drain into ocean from 200 m pipe ,2m in diameter. at 17.5 m depth from pressure differences alone.
It seems that the pressure,P, in atm from the pipe from velocity P=1/2 p v^2 is about equal to the water pressure from ocean surroundings at depth of 17.5 meters. P = P(0) + pgh
Question : Is the total pressure at submarine outfall equal to the pressure from depth formula AND pressure from velocity formula ?
pressure from velocity, 1/2 pV^2 = : 1/2 (1000kg/m^3) 342 m^2/s^2
= 171125
pressure from depth , P(0) + pgh : 1 + (1000kg/m^3) (9.8m/s^2)(17.5m)
= 171500 ( does pressure from depth apply to water exiting drain pipe as well as ocean pressure surroundings ?
1 atm = 14.7 lb/in^2 = 101262kg/m^2

Another source said that for every 10m depth add 1 atm pressure. So 17.5 m
would 1.75 atm plus one from surface equals 2.7 atm.

Last edited by morrobay; Nov 3rd 2015 at 10:39 PM.
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Old Nov 4th 2015, 07:12 PM   #5
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Moderators can you move this to the advanced mechanics dept ?
I realize Im having a little confusion on cross-over units here. But thats the
reason Im asking the question in the first place. Its sure not about simple
multiplication ! Hopefully someone can answer this question even if it requires
restating the question.
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