How to design an inverted siphon?

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How to design an inverted siphon?

We are designing an inverted siphon with the following data

Dicharge = 54 cusecs
Manning's = 0.011 (MS pipe will be used throughout the siphon)
U/s water level= 466.60 m
D/s water level= 465.60m
Length of Siphon = 400 m (57 m (in down slope) + 71m (horizontal) +132ft (sloping down)+100 m horizontal+40m (steep slope up)) see in the figure attached below (but the slope and length can be changed not fixed as it is not final siphon yet)
there are four bends.
the siphon is passing under a dry drain.....

My output calculations are
Diameter = 48 inches
Velocity = 4.30 ft/s
Entry losses = 0.11

velocity head loss=0.29
Exit losses =0.20
Bend Losses = 0.05
friction Losses =1.42
Total Headlosse = 2.01 ft+10% of total head loss=2.2ft

Please can any one help me in calculationto check if I am right or not with the above calculations? Or can any one provide an excel sheet for it? Or can any one suggest me other options instead of installing a siphon.

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4 Answers

  1. Inverted siphons used to convey canal water by gravity under roads, railroads, other structures, various types of drainage channels, and depressions. A siphon is a closed conduit designed to run full and under pressure. The structure should operate without excess head when flowing at design capacity.

    Inverted siphons (also called depressed sewers) allow storm water or wastewater sewers to pass under obstructions such as rivers. Our inverted siphon calculation allows up to five parallel siphons to go under the river. Unlike the main sewer pipe, the siphon pipes flow under pressure and must have flow velocities greater than 3 ft/s (0.9 m/s) to keep material suspended. Therefore, several siphons having smaller diameters than the main sewer may be required. The calculation computes siphon diameters (or siphon flows), velocities, inlet chamber wall heights, and siphon invert elevations.

    Storm water and wastewater sewers often encounter obstructions such as rivers, other pipes, subways, tunnels, or valleys. To pass these obstructions, a common method is for the sewer pipe to drop sharply, then run horizontal under the obstruction, and finally rise to the desired elevation. The piping going under the obstruction is traditionally called an "inverted siphon", but since the pipe is not actually acting as a siphon, a better term is "depressed sewer"

    Unlike the main sewer pipe, the siphon pipe(s) flow under pressure. Special care must be taken in inverted siphon design since losses are greater for pressurized flow, and the velocity in each siphon pipe must be at least 3 ft/s (0.9 m/s) for sewage or 4 ft/s (1.2 m/s) for storm water (Metcalf and Eddy, 1981). Therefore, even if there is only one main sewer pipe, several siphons may be required. If minor losses due to bends or elbows in the siphon are significant compared to the siphon length, include the equivalent length of the elbows. Increase the siphon length (Ls) so that Ls is the physical length of a siphon plus the equivalent length of minor losses due to elbows in siphon.

     

  2. Inverted Siphon (Depressed Sewer) Design Calculation, is easily available on some website. But I wanted to say that, open channel flumes are also an alternative for you.

  4. I am writing quick step by step calculation process, that will help you.

    1. Compute the maximum flow in the main sewer pipe using Manning's equation for full pipe flow.

    2.Compute the diameter of each siphon, Di, or the flow through each siphon, Qi, using Manning's equation for full pipe flow through each siphon

    3.Compute the wall heights, yj (relative to main invert), in the inlet box. The walls separate the siphons from each other. The wall heights are the same height as the water depths, yj, in the main pipe corresponding to the discharge through the siphons. Here, Qj=1 is the discharge through siphon 1, Qj=2 is the discharge through siphons 1 and 2, and so on. Manning's equation for a partially full main pipe is used, but is solved backwards (numerically) in order to compute yj. We allow up to five siphons (four walls).

    4. Compute the siphon invert elevations in the inlet chamber. According to Metcalf and Eddy, there is no loss in the inlet box for flow going from the main culvert to the first siphon since the flow travels in a straight path. However, for siphons 2 through n the flow must turn 90o to go over the chamber wall (a head loss of 1.5 velocity heads) and has an additional head loss of one velocity head as the flow enters siphon i. Therefore, for i=2 to n siphons and j=2 to n-1 walls.

    I am not able to post equations properly. I will post a document to support all these steps tomarrow.