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?
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.
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.
The purpose of Syphon are;
Their earliest known use dates back to the Egyptians approx 1500 BC. Tesibius of Alexandria was invented a usable siphon.
How do you differentiate syphon and a leaky bucket? It’s very simple! The main difference between a syphon and a leaky bucket:
When we discuss the working principle of the siphon, there are few things that come to our mind?
Let’s try to analyze to understand the siphon working philosophy.
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:
Air pressure is reduced with height or elevation. That is the reason; we face difficulty breathing at the top of the mountain.
Look at the below diagram,
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.
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.
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.
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.
Hence, the liquid starts to flow from a higher elevation tank to a lower elevation tank continuously.
Hence, we all know,
This expression is derived from, Bernoulli’s equation
Let’s consider a siphon, which is used to drain water from a tank, T1.
Calculate the discharge and pressure of the fluid at the discharge point. Neglect friction.
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
We will apply, Bernoulli’s equation at point 1 and 3,
At point, 1 (Energy level of the fluid in the tank surface)
At point, 3 (Energy level of the fluid in the discharge point)
p/ρg + v2/2g + z = Constant
Hence, considering point 1 & 3, applying Bernoulli’s equation
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,
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)
When a siphon is used like U-tube or U-pipe, instead of an inverted U tube, it is called an inverted syphon.
What is a siphon used for? This is used in the following cases,
This is used in the following cases:
So, we have learned syphon along with various working principles. What about you, would you like to test your knowledge?
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