Siphon Flow Rate ```Name: Larry Status: other Grade: other Location: NE Country: N/A Date: 3/30/2005 ``` Question: Is it possible to determine a flow rate (for water)for a siphon? I thought I had a standardized formula for this but cannot currently find it. Example of problem: 150 ft. long x 4 inch diameter corrugated plastic pipe. Head range from 3.6 to 8.6 ft. Replies: Larry, The flow through the pipe can be determined using one of the many different formulas for determining friction losses in pipe. Darcy-Weisbach is one such formula, and Hazen and Williams is another. The driving force for your example problem is 5 ft. of head. Resistance is offered from friction loss due to the length, diameter, flow, and type of pipe (in this case corrugated plastic instead of smooth steel). Hazen and Williams is an empirical formula which works well for water flowing under turbulent conditions. h(f) = 0.002083*L*(100/C)^1.85 * (gpm^1.85 / d^4.8655) h(f) = Friction loss in pipe L = equivalent length of straight pipe (taking into account fittings etc.) (feet) C = Friction Factor for type of pipe. For your example, I would use a value of 60. Gpm = Flow rate d = pipe internal diameter (inches) For your example, L = 150 ft, C = 60 and d = 4 inches. Make a plot of flowrate vs headloss due to friction with headloss on the "Y" axis and flow on the "X". Then draw a straight line at 5 ft of head. Where the line intersects the curve will be the flowrate at which the friction losses equal the driving force. Bob Hartwell First, siphon action. I would not bet on a "standardized" formulas. My layman's opinion, air pressure and any cohesion within the fluid are all the governing forces. So, it does seem somewhat dependant on the fluid, say water and maple syrup. Now, if you zero in on water and know all its properties, one would seem to have all one needs. Of the two factors, the air pressure, then the pipe contributions, diameter, roughness ( coefficient of friction for flow) can be ascertained, the water properties a lesser factor. So, final thought, it is no one formula but an analysis of the particulars. Reflecting. Thermodynamics, funny, maybe not so funny, being a civil engineer, another course I never used. So, I will back off a little, as an engineer I see analysis of all the factors, but maybe there are formulas, which probably may neglect refining aspects, as the water cohesion. But, the mechanical engineers at Uncle Tom's Maple syrup plant might not use them. Hope this helps. James Przewoznik This is a deceptively intricate question -- easy to ask -- very involved to answer. Hydrodynamics is a mathematically messy subject. The standardized formula(s) depend upon several design parameters. Below are 3 sites that present the problem and work through the math. In its simplest form the velocity of a siphon (V) is driven solely by the difference in the height of level of the upper vessel and the drain point at the end of the tube (H), specifically: V = (2 x g x H)^1/2 where 'g' is the gravitational constant. However, as you will see form the sites below life in the real world is much more complicated. The equation above for example assumes the change in the level in the upper reservoir is negligible, no turbulence, fluid viscosity is negligible -- to point out just a couple of simplifying assumptions. In the simplest form the "driving force" is gravity. http://en.wikipedia.org/wiki/Siphon#Bernoulli.27s_equation http://instruct.tri-c.edu/fgram/web/Siphon.html http://www.me.ntu.edu.tw/~ta/TA/%A9%C9%BE%F6/FM%20slides%202005/ 6_Incompressible%20Inviscid%20Flow.ppt#1 Vince Calder I suppose a rough approximation could be made, based solely on the diameter of the hose, and the distance the fluid is being dropped. (the drop distance gives you a pressure to work with) Of course, for more accurate answers, you would have to do some pretty strong fluid-dynamics work. ( . . . to figure things like the resistance of the corrugated pipe.) Let us see, Square root of 5/16 is .559. so .559 seconds for water to "fall" through 5 feet of pipe, or 1.79 times the capacity per second. Problem is, I do not know how many gallons of water are in 5 feet of 4"pipe. Keep in mind, actual results will be a little lower than this calculation. Ryan Belscamper Click here to return to the Engineering Archives

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