What is A Siphon?

A siphon is any device (usually an inverted U shaped pipe ) used to convey fluids from a higher elevation to a lower height without the need for pumps.

How Does It Work?

Siphons work by initially transporting the fluid upwards through the shorter leg of the pipe -usually by an external pump or suction(PRIMING) – then utilizing the force of gravity from the liquid falling through the longer lower leg and the atmospheric pressure to keep the fluid flowing even when the priming force is removed. This is
referred to as the Siphon Effect.

Who Invented It?

Humans have been using siphons for centuries for many applications. The earliest appearance of siphons are found in ancient Egyptian art. It depicts them using siphons for extracting wine from large storage jars. They used the siphons ingeniously to separate the wine from surface impurities and bottom dregs.

The Greek mathematician Hero of Alexandria was the first to describe the siphon effect. He conducted many experiments and wrote about it in his engineering treatise Pneumatical.

How Are They Used Today?

Today siphons enjoy a wide range of applications due to their low power requirements and simple construction. They are commonly used in household applications like the water closets, beer taps and simple gardening sprinklers.

Siphons are also employed on a large scale for irrigation in agricultural areas. Because of their low energy consumption and cost, they are very suitable for transporting water over long distances from water sources to fertile arid valleys. Great examples of siphon powered irrigation projects are found in sugar cane plantations in Hawaii, US and in cotton fields in Queensland, Australia.

What Is The Science Behind The Siphoning Effect?

There have been many theories put forward to explain the operation of siphons. Till today, there is still an active debate over what is really responsible for the Siphoning effect. But scientists have concluded that is a result of two factors. Atmospheric pressure and chain theory.

Atmospheric pressure theory was one of the earlier theories formulated to describe Siphoning Effect. It theorizes that when the liquid is drawn in through the tube, the act of sucking causes a pressure differential in the tube with a low pressure area forming at the highest point. This pressure differential causes the liquid to flow from
the high pressure area at the surface of the reservoir to the low pressure area at the highest point. The main flaw with this theory, is that siphons have been demonstrated to work in high vacuums (No air, no atmospheric pressure).

A more recent theory put forward is that of Gravity and Cohesion working together. It is based on a Chain Model. This model views the fluid as a chain together by its cohesive bonds. The chain is first pulled upwards through the tube by another force till it passes the bend then gravity takes over, pulling the entire fluid downward like a chain over a pulley until the reservoir is empty. This theory too, has been discounted because siphons have been made for fluids with low or weak cohesive forces.

Today engineers use a combination of both theories to explain the Siphoning effect at different conditions.

The most popular mathematical approximation used to model fluids in siphons is Bernoulli’s theory.

This is expressed mathematically as:  

    \[  \frac{v^2}{2} + gy + \frac{p}{p} = Constant.......(1) \]

Where v=local velocity, g = acceleration due to gravity, P= pressure, y= vertical distance from the reservoir surface, ρ = density of the fluid

This equation models the entire siphon as a system.

Let’s say we have a submersible pump in a reservoir of height RL +5m. The water is pumped out over a wall of RL +8m and discharged at RL 0m. Let’s find the discharge velocity at the exit when the tube is used as a siphon.

To find this, we apply Bernoulli’s equation to three points in the system:

1. The surface of the reservoir
2. The highest point
3. The discharge point

Assuming the pump is used to prime the siphon and switched off after flow starts and losses due to friction and other effects are negligible.

At the surface(2): 

Velocity Vs =0 m/s( We model the reservoir surface as being infinite)

Pressure Ps = P atm (Atmospheric pressure)
Distance from surface(y) = 0m

    \[  \frac{0}{2} + g(0) + \frac{p_atm}{p} = Constant.......(2)\]

At the highest point(h(3)):

At h, we have Velocity = Vh
Pressure = Ph > Patm
Distance from surface yh = 8m – 5m = 3m.

    \[  \frac{v_h}{2} + g(y_h) + \frac{p_h}{p} = Constant.......(3) \]

At lowest point(Discharge point d(4)):


Velocity Vd = Vh (Since they are in the same stream line)
Pressure =Ph =Patm
Distance from surface = 0-5= -5m

To find the velocity Vd we equate (2) to (4) which gives us:.

Rearranging we have:


Putting in the values and solving, we get a discharge velocity of 9.9m/s

However, there is a limit to the conditions that can sustain a siphon.The main limit to sustaining a siphon is air. The siphon has to be airtight because to much air in the system can break the vacuum which the siphon relies on to operate.

Also, the siphon’s height is limited by the vapour pressure of the liquid and the local atmospheric pressure For example, to sustain a siphon in a system pumping water from a reservoir at sea level, the distance between the surface of the reservoir and the highest point in the pipe should not be more than 10m.

When the highest point in the pipe is above 10m, the pressure in the pipe at the highest point becomes equal to the vapour pressure of water. When this happens, the water begins to boil, breaking the cohesive forces holding the liquid chain together thereby interrupting the siphon. Interestingly though, siphons with much higher heights have been constructed for degassed water.

How Does Friction Affect Siphoning?

In siphons, the frictional force at the inlet, exit and internal walls of the tubes and inertial force of water often present an opposition to flow while the siphon tube is being filled up. But this is usually overcome by the priming(pumping) force, because in most practical applications, the tube is usually filled up by an external force before the siphoning starts.

When the flow has begun and is powered only by the siphon, the head loss by friction becomes a function of the flow velocity provided by the siphon. The head loss by friction increases with increasing flow velocity i.e as the flow becomes more turbulent, the loss from friction increases. Since the loss from friction is a function of the flow velocity, it doesn’t stop the flow, it only reduces the flow velocity.

To find the velocity drop from friction, we introduce the Resistance coefficient K into the equation to account for the losses from friction.

K = (f *L /D) + 1.9 (Other sources)……….(6)

Where : f = Friction factor of pipe (iron = .019, steel = .013, plastic = .007)
L = Change in Station Points (length of run), (m)
d = Pipe internal diameter (m)

So modifying (5) we have flow velocity(m/s):

How Does Siphoning Affect Pumping Systems?

As we’ve seen in the sections above, the siphon effect has many uses in the areas of agriculture, engineering, and plain everyday applications. But it can also become a problem in in some systems.

In some pumping systems, the siphoning effect can be used in tandem with certain pipe configurations to boost the pressure head, but it can also cause certain problems if not designed properly.

How Can It Occur In Pumping Systems?

Siphoning can occur in a pumping system due to the configuration of the pipes. Siphons are driven by the vertical distance between the reservoir and the discharge point. It occurs in situations where the discharge point is located at a level lower than the reservoir or pumping point.

The flow velocity obtained in the siphon and the volume flow rate is driven by the vertical distance between the pumping point and the discharge point. The greater the distance, the greater the flow velocity.

What Type Of Problems Can Siphoning Cause?

Let’s take the submersible pump in the example above. Suppose the pump was supposed to fill a swimming pool and switch off after a certain level is reached, due to the arrangement of the reservoir and the discharge point, the water is going to keep on flowing even after the pump is switched off causing the pool to overflow.

Another problem arises if we reverse the configuration, let’s say the pump is taking water from a lower reservoir to the higher “swimming pool”. After the pump is switched off, there is no net pressure difference driving the water from the lower side to the high side and the waste water from the pool starts to flow back into the reservoir. This is known as back siphoning and it is a big public health problem as dangerous contaminants can be sucked into the water system.

How Can We Design Against These Problems?

There are various methods of designing against these problems, the most common ones often involves putting valves in line with the flow. Some examples of these valves are:

Solenoid Valve: This is an electrically controlled valve can help prevent overflow by shutting off the water flow in the pipe immediately the pump stops.

Check Valve: This valve can also be used to prevent back siphoning when it’s placed in line with the flow. It’s special configuration helps ensure the flow goes only in one direction. It is usually placed immediately after the pump to protect it from the back flow.

But in some cases the solenoid valve/check valve can be either too expensive or not just suitable for the design.

In this case, air/vacuum breaker combination valve is used. An air/vacuum breaker valve is usually installed at the highest point of the system. It serves a dual purpose of venting out the excess air in the system to prevent pressure build-ups while admitting air into the system to break vacuums to negate the siphon effect.

For complex pipelines running over hundreds of meters, different forms and sizes of valves need to be installed at different points in the pipe’s length to provide proper venting. But for a simple design like the one given above, a simple vacuum breaker valve can be used. It has to be installed at the highest point and the appropriate size has to be used.

How to size an Air / Vacuum valve:

In sizing an Air/vacuum release valve, most engineers use a standard rule of thumb of 25mm of valve diameter per 0.3m of pipe diameter.

But for more complex applications requiring fine control, most valve companies publish white papers and specifications on how to properly select and install their products.

References

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