The siphon 15-12

Problem: How to control the level in a 1500-acre lake at an elevation of 500 feet. You want to relocate the water to a reservoir that is at an elevation of only 480 feet, but there is a hill between the lake and the reservoir. The hill has an elevation of 515 feet

You have three choices:

Now just what is a siphon? It is the same principle you used to get gasoline out of your automobile gas tank to fill a gas powered lawn mower. Remember how you sucked on the end of the hose, and when the gasoline began to flow you stuck the end of the hose into the container and the gasoline magically flowed into and filled the container all by its self? All you had to watch out for was that the container was placed below the gas tank


In other papers in this web site we learned that pressure can be converted to head by the following formula:

 If atmospheric pressure is 14.7 psi at sea level, and we are talking about fresh water (sg.1); using this formula, 14.7-psi. would convert to 33.96 feet of head:

Theoretically we could come out of the gasoline tank, go straight up into the air with the hose to a height of almost 34 feet and then down into the container and the gasoline would flow with no trouble. I say theoretically because there is friction in the hose that offers resistance to the flow and that friction has to be considered any time you are trying to calculate the possibility of using a siphon to move liquid.

The siphon principle is valid if the liquid flow is free from air and vapors so that the densities of the liquid in the rising and falling pipes are alike. It is this principle that can limit the height of the siphon because the flowing liquid can vaporize if its temperature gets too high or the pressure in the pipe gets too low. As an example: A higher water temperature and low barometric pressure conditions limits the height of siphons in cooling tower, condenser cooling water applications to somewhere between 26 and 28 feet.

Now let's go back to our lake and reservoir application and make the calculation.

Figure 2

Lake elevation = 500 feet

Pipe height on the hill = 515 feet

Reservoir height = 480 feet

Liquid = water at seal level pressure and ambient temperatures.

Desired flow = 3.5 to 4.2 million gallons per day (2,430 to 2,917 gpm)

Diameter of cast iron pipe = 21 inches

Length of cast iron pipe from the lake to the top of the hill = 1 mile (5280 feet)

Length of the cast iron pipe from the top of the hill into the reservoir = 2 miles (10,560 feet)


In this example we will not consider the friction losses due to the pipe condition, fittings, bends and flow control valves in the piping, along with the exit and entrance losses.

Look in the Cameron hydraulic tables or any other publication that shows the head loss in 21 inch cast iron pipe you would find the loss to be 0.053 feet / 100 feet of pipe. A larger diameter, and more costly 30 inch cast iron pipe has losses of 0.022 feet/ 100 feet of pipe

Head loss in 5,280 feet of 21 inch pipe = 5280 / 100 x 0.053 = 2.8 feet

Head loss in 10,560 feet of 21 inch pipe = 10560 / 100 x 0.053 = 5.6 feet

This is a total of 8.4 feet from the 35 feet available (515 - 480 = 35).

If you have too much flow in the piping, the resultant loss of pressure can cause cavitation problems. The control vale will eliminate this problem.

To get the system running you will have to fill the pipe from the top of the hill back to the lake. You could use a portable pump taking a suction from the lake to do this. Remember that until all the air is removed from the piping and all the piping is filled with water, the two mile down leg is not part of the system, so the friction in the down leg and control valve should not be considered when you make your calculations for the small priming pump


For information about my CD with over 600 Seal & Pump Subjects explained, click here  

 Link to Mc Nally home page