What is Siphon or Syphon? Definition, Meaning, Works, Effects, Uses

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What is Siphon or Syphon? Definition, Meaning, Works, Effects, Uses


The Siphon or Syphon is the most widely used instrument in fluid mechanics. If there are two tanks kept in different elevations and a barrier is there between them. How can the liquid be transferred from the higher elevation tank to the lower elevation tank?

Here, the siphon plays an important role to transfer the liquid?

What is Siphon or Syphon? Definition, Meaning

Basics & Meaning of Siphon

The siphon basically an inverted U-tube or pipe filled with water. The filling of the pipe or tube is simple and many ways it can be done. It is the same as the priming of pumps.

  • Each siphon has two legs. One is long with respect to others,
  • The long leg is connected to the tank kept at the higher elevation
  • Another leg is connected to the tank kept at the lower elevation
  • The action depends upon a few factors, like cohesive forces or pressure differences, etc.
Siphon syphon introduction
Siphon or syphon introduction

Siphon or Syphon Definition

A syphon is an inverted u-shaped tube or pipe used to transfer liquid from a higher elevation tank to a lower elevation tank when both the tanks are separated by a high-level barrier.

Function of Siphon

The purpose of Syphon are;

  • If two tanks are kept in different elevations and there is a barrier between them, the siphon is used to transfer liquid from the tank at a higher elevation to the tank at a lower elevation.
  • The net result is that the liquid passes from the reservoir to a lower level with a continuous flow.
  • No external energy is required for liquid transfer.
  • The flow continues unless and until the water level reaches the output end of the siphon tube.
siphon syphon basic
Siphon syphon basic

History of Syphon

Their earliest known use dates back to the Egyptians approx 1500 BC. Tesibius of Alexandria was invented a usable siphon.

Difference between Syphon & Leaky Bucket

How do you differentiate syphon and a leaky bucket? It’s very simple! The main difference between a syphon and a leaky bucket:

  • Liquid transfer happens without any hole in the tanks whether a leaky bucket means a lot of holes.
  • Liquid transferred to a higher level tank before draining out to a lower level tank.

How Does a Siphon or Syphon Work?

When we discuss the working principle of the siphon, there are few things that come to our mind?

  • Why the liquid level rises?
  • Why does the liquid flow continuously?

Let’s try to analyze to understand the siphon working philosophy.

Step#1: Atmospheric pressure and vacuum

Let us take a tube of one end closed and another end is open. Now, fill the tube with a liquid and raised up the closed-end while the lower end is immersed in a liquid reservoir.

If we observe the tube closely, we will see the liquid in the tube stays up to a certain height. But, if we increase the height of the tube further, the water level does not rise again.


It is because this height depends on:

  • the density of the liquid and
  • the pressure of the air outside the tube.

Air pressure is reduced with height or elevation. That is the reason; we face difficulty breathing at the top of the mountain.

pressure with elevation
Pressure with elevation

Look at the below diagram,

atm pressure vacuum
Atm pressure vacuum
  • C  = Summit of the siphon, i.e., the highest point of the siphon
  • R  = Reservoir or tank kept at a higher elevation
  • L = Tank kept at a lower elevation
  • H = Height difference between the tank, H and summit, C

In this diagram, point C is at certain height from the tank surface H. It means the height is increased towards summit C; pressure will be decreased gradually. At point C, the pressure will be below atmospheric pressure, and water will rise in the shorter column and liquid flow will be started.

As per theoretical calculation, air sustains up to a liquid height of 10 meters (32 ft. of water), and up to that siphon will work.

If the height of the tube is further increased, dissolved air and other gases will come out and form a vacuum in the summit and interrupt the continuity of liquid flow. As a result, the siphon will not work.

Step#2: Working with cohesive forces

If we analyze, with water, we will see both the end of the siphon have the same pressure, that is atmospheric pressure.

Do you think atmospheric pressure is the main factor to have siphon work? No, the siphon can even work in a vacuum condition also.  However, this is applicable to very cohesive liquids only.

siphon chain model
Siphon chain model

Molecular attraction in high cohesive liquids is very strong and they can maintain a chain-like action in the inverted tube and maintain a continuous flow.

Step#3: Gravitational Potential Difference

Do you think atmospheric pressure or weight difference of liquid sustains the flow within the siphon? No!

It’s all about gravitational potential difference. There will a gravitation potential difference between the liquid level in the tank and at the output siphon’s opening. This difference makes an unbalanced system and stores potential energy. As a result, the liquid flow will be continued.

Let’s focus on two factors atmospheric pressures and tube heights. Now, the taller column is connected to the lower tank.

  • A tall water column is on the lower level tank
  • Normally gravity will pull the liquid down
  • Liquid column down means there will be a reduction of pressure at the top of the siphon
  • Reduction of pressure will create an unbalance in the small column
  • There will be a gravity force in the shorter column as well
  • This gravity force is comparatively less with respect to the reduced pressure
  • So, the shorter column will rise to the top of the siphon

Hence, the liquid starts to flow from a higher elevation tank to a lower elevation tank continuously.

Mathematical Expression of Siphon

Let’s consider,

  • p: Pressure difference between two points,
  • ρ: is the liquid density,
  • g: gravitational acceleration
  • v:  water velocity at the bottom of the siphon
  • z:  elevation
  • g: gravitational acceleration

Hence, we all know,

siphon bernoulli equations

This expression is derived from, Bernoulli’s equation

Example of Siphon Calculation


Let’s consider a siphon, which is used to drain water from a tank, T1.

  • Siphon diameter = 4 cm,
  • The siphon outlet end is 2m below the surface in the tank, T1
  • The peak point of the summit is 1.5 m above the water surface.

Calculate the discharge and pressure of the fluid at the discharge point. Neglect friction.

Given Data

Calculate discharge

Dia of tube, d=d1=d2=d3 = 4 cm = 0.04 m

Formula to use

Bernoulli’s equation for frictionless flow is

Discharge(Q) = cross sectional area x velocity


siphon mathematical expression
Siphon mathematical expression

We will apply, Bernoulli’s equation at point 1 and 3,

At point, 1 (Energy level of the fluid in the tank surface)

  • Pressure = p1 = 0 N/m2
  • Elevation = z1 = 0 m
  • Velocity = v1= 0 m/sec

At point, 3 (Energy level of the fluid in the discharge point)

  • Pressure = p3 = 0 N/m2
  • Elevation = z3 = -2 m
  • Velocity = v3= ?


p/ρg + v2/2g + z = Constant

Hence, considering point 1 & 3, applying Bernoulli’s equation

  • p1/ρg + v12/2g + z1 = p3/ρg + v32/2g + z3
  • 0 + 0 + 0 = 0 + v32/2g + z3
  • 0 = v32/2g + z3
  • 0 = v32/2g +(-2)
  • v32/2g = 2
  • v32 =2 x 2g
  • v32 =2 x 2 x 9.81
  • v32 = 39.24
  • v3 = 6.3 m/s

Discharge, Q

  • Q = Area x Velocity
  • Q = π d32/4 x v3 [as Area = π d32/4]
  • Q = 3.14 x 0.042/4 x 6.3 [as d = 0.04]
  • Q = 0.00791 m3/s
  • Q = 28.5 m3/hr

Applying Bernoulli’s theory, at points 1 and 2,

P2 = 0 N/m2(g)

z2 = 1.5 m

v2 = v3 = 6.3 m/sec (since, cross-sectional are same)

Let’s write the equation,

  • p1/ρg + v12/2g + z1 = p2/ρg + v22/2g + z2
  • 0 + 0 + 0 = p2/ρg + v22/2g + z2
  • 0 + 0 + 0 = p2/ρg + 6.32/2g + 1.5
  • 0 = p2/ρg + 6.32/2g + 1.5
  • p2/ρg = – 2.02 – 1.5
  • p2/ρg = – 3.52
  • p2 = – 3.52 x ρg
  • p2 = – 3.52 x 1000 x 9.81 N/m2 (g)
  • p2 = – 34531.2 N/m2 (gauge pressure)

Hence, the pressure at point 2 is -34531.2 N/m2(gauge pressure)

Absolute pressure at point 2 = 101325 – 34531.2 = 66793.8 N/m2(absolute pressure)

Inverted Siphon

When a siphon is used like U-tube or U-pipe, instead of an inverted U tube, it is called an inverted syphon.

  • It’s used based on the requirements if liquid to be pass through a road. This inverted siphon installed below the road
  • Here, the liquid has completely filled the pipe
  • Flow is due to the pressure instead of gravity flow.
Inverted siphon
Inverted siphon

Uses of Siphon

What is a siphon used for? This is used in the following cases,

  • To transfer water or any liquid from one reservoir to another reservoir separated by barrier or hills or ridge etc
  • To empty a liquid tank which doesn’t have any outlet
  • To empty a dam which doesn’t have any sluice gates or any other outlets
  • To empty a channel that doesn’t have any sluice gates or outlets.
  • To transfer sewerage or stormwater under highways or streams, inverted siphons are widely used.
  • Sewage or stormwater under highways or any streams can be transferred by inverted syphon.
  • A siphon can purify up to a certain level of liquids. It simply takes the clear liquid without disturbing bottom precipitation or suspended solids at the surface.
  • Syphon helps the irrigation system to supply control water from cannel to crop areas.
  • Municipal waterworks application
  • Siphon effect recovers the pump head due to minor losses for the change in elevation.
siphon application
Siphon application

Disadvantages of Siphon

This is used in the following cases:

  • Transferring liquid in many cases is not fruitful for syphon due to time. It takes a lot of time. Using a pump will save a good time.
  • In the case of the inverted siphon, the head loss is more in the sewer system
  • In the case of an inverted siphon, sedimentation or siltation can block the pipe.


So, we have learned syphon along with various working principles. What about you, would you like to test your knowledge?

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