Open pipe discharge distance
Open pipe discharge distance
(OP)
Dear All,
I have made researches via our Google friend but could not find an answer that suits me. Either way too simplified to take full account of all the parameters or way too complicated for what does not seem like rocket science.
Knowing flow rate, pressure, diameter, pipe angle, pipe oultet altitude, etc. which theory do you use to determine the horizontal distance the flow will reach? I have found some references like Purdue, etc. but mostly with fixed monograms and only in imperial units.
Do you have any formula or paper to advise, preferably in metric units as I am in Northern Europe? Which theory would you advise to use for this?
Thanks by advance, Vince
I have made researches via our Google friend but could not find an answer that suits me. Either way too simplified to take full account of all the parameters or way too complicated for what does not seem like rocket science.
Knowing flow rate, pressure, diameter, pipe angle, pipe oultet altitude, etc. which theory do you use to determine the horizontal distance the flow will reach? I have found some references like Purdue, etc. but mostly with fixed monograms and only in imperial units.
Do you have any formula or paper to advise, preferably in metric units as I am in Northern Europe? Which theory would you advise to use for this?
Thanks by advance, Vince





RE: Open pipe discharge distance
It might depend on what the fluid is, water, oil, natural gas, hydrogen, co2, heavy fuel oil, ketchup, temperature, etc. and if it is low pressure, or high pressure
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
RE: Open pipe discharge distance
Pressure is between 8 and 12 bar.
Does this help for you to recommend any theory?
RE: Open pipe discharge distance
x = V*t that is t = x/V
y = 0.5*g*t^2
y = 0.5*g*x^2/V^2
but V = Q*A = Q*pi*d^2/4 for a circular pipe, so
x = [4*Q*SQRT(2*y/g)]/(pi*d^2)
RE: Open pipe discharge distance
Thanks for this answer. Could you please explicit the variables and units used? Does this theory have any density or pressure limitations? Is the pipe considered full or half full as in some theories?
Also this does not take into account the density, etc.of the media. Thanks by advance if someone could advise a theory which does this.
Vincent
RE: Open pipe discharge distance
http://www.humes.com.au/fileadmin/templates/HUMES/...
http://www.grundfos.com/service-support/encycloped...
http://mimoza.marmara.edu.tr/~orhan.gokyay/enve311...
It is possible to convert units.
"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek
RE: Open pipe discharge distance
What I've reported in my previous post is simply the application of fundamental physics. You can find more at the link below (even in metric units).
http://onlinelibrary.wiley.com/doi/10.1002/9780470...
RE: Open pipe discharge distance
RE: Open pipe discharge distance
For fluid flowing out of an open pipe, (assumed to be parallel with the ground), x = v*t and, y = (1/2)*g*t^2
Where x and y are in feet, g=32.2 ft/sec, and t = sec.
Now, t is the same for both x and y. For x, t=x/v. Then, y=0.5*g*(x/v)^2. Then, x = v*((2*y/g))^1/2=v*(0.0621*y)^.5
Now, v*a=flow. Set v*a=gpm, v in ft/sec, a in sq.ft.
for pipe id=d (inches), v*(pi/4)*(d/12)^2=ft^3/sec.
v*(pi/4)*((D^2)/144)*ft^3/sec*(7.48 gal/cu.ft)*(60sec/min)= gpm=2.446*v*d^2
or, v=0.4087*gpm/(d^2) (d is in inches)
Then, x=(0.4087*gpm/(d^2))*(0.0621*y)^.5
CHECK: For d=1", gpm= 5, and y = 3ft,
x = ((.4087*5/(1^2))*(0.0621*3)^.5 = 0.882 ft. and, v= .4087*5/(1^2) = 2.0435 ft/sec.
and t= 0.882/2.0435 = 0.4316 sec.
Check y distance: y = 0.5 * (32.2/2)*(.4316)^2 = 2.999 ft.
Q.E.D.