Open pipe discharge distance 1

[vinhermes]

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

 

[BigInch]

It is not rocket science, but there can at least be some minor complications from time to time.
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

 

[vinhermes]

Media is freshwater and seawater at ambient temperature.

Pressure is between 8 and 12 bar.

Does this help for you to recommend any theory?

 

[ione]

Very rough approach

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)

 

[vinhermes]

Dear Ione,

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

 

[BigInch]

Colebrook - White

http://www.humes.com.au/fileadmin/templates/HUMES/doc/Brochures/Humes_ConcretePipeManual.pdf

http://www.grundfos.com/service-support/encyclopedia-search/colebrook-white.html

http://mimoza.marmara.edu.tr/~orhan.gokyay/enve311/ch5.pdf

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

 

[ione]

vinhermes,

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/10.1002/9780470168103.app2/pdf

 

[vinhermes]

I will try to read all these links tomorrow and come back to you if needed. Thanks, Vincent

 

[trashcanman]

I wish to point out an error in the equations above by ione. Q = V/A, Not, Q*A. The following is a derivation of the x distance for a fluid flowing from a pipe, y distance above the ground:


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.