Physics SIPHON

Siphon

Siphon can be described as a tube or pipe that allows liquid to flow from the higher level to the lower level. Siphon allows liquids to flow uphill, above the surface of the tank or reservoir, without pumps. Liquids flow down the tube under the pull of gravity. The liquid is discharged at a level lower than the surface of the reservoir or the tank. The pull on the surface of liquids from a higher to lower altitude is known as siphoning. Records show that siphoning has been practiced in ancient cultures.

`text(Working Principle :)`

Siphon works because of gravity. The gravity pulling down on the taller column of liquid causes less pressure at the top of a siphon.

The mass of water entering the tube and flowing upward is equal to the mass of water flowing downwards and leaving the tube.

A siphon draws the liquid out of the reservoir until the level of the tube or pipe falls below the intake of the liquid. It continues to work until the outlet of a siphon equals the level of the reservoir.

`text(How does a Siphon work?)`

Siphon is a U-shaped flexible tube. An ordinary pipe can also function as a siphon. One end of the siphon is inserted into the reservoir and the other end is left free inside a container to collect falling liquid.

The siphon is started using a priming pump. Priming is a process of filling the siphon tube with fluid or liquid so that hydrostatic pressure is formed.

Once the siphon is started, it continues to pull liquid out of the reservoir until the pipe is removed or the reservoir becomes empty. One of the simplest ways of priming a siphon is to suck the tube or the pipe like a straw until it is filled with the liquid, thus initiating the siphoning process.

Electric or manual pumps are used in case of large siphons. A siphon continues to run until it starts to suck air or the liquid is completely pumped out.

`text(Siphon does not work in vacuum.)`

`text(Uses of Siphon :)`

It is useful in keeping unwanted impurities out of liquids. Siphoning can prevent impurities from being transferred to a fresh container.

Siphoning is therefore useful in the fermentation of wine and beer.

Irrigated fields often use siphon to pull in water from nearby ditches or channels into their fields.

Siphon can be used in flooded homes or cellars to remove accumulated water.

Large siphons are used in municipal waterworks and industries.

A siphon is also used to clean a fish tank or aquarium.

An infamous application is gas siphoning. In this process, fuel is stolen by inserting a tube into the gas tank. In order to prevent such malicious activities, anti-siphoning systems are installed in cars and vehicles.

Flow Rate and Maximum Height of the Siphon

Let the surface of the upper reservoir be the reference elevation.
Let point A be the start point of siphon, immersed within the higher reservoir and at a depth `−d` below the surface of the upper reservoir.
Let point B be the intermediate high point on the siphon tube at height `+h_B` above the surface of the upper reservoir.
Let point C be the drain point of the siphon at height `-h_C` below the surface of the upper reservoir.

Bernoulli's equation `=>` `v^2/2+gy+P/rho=text(constant)`

`v=` fluid velocity along the streamline
`g=` gravitational acceleration downwards
`y=` elevation in gravity field
`P=` pressure along the streamline
`rho=` fluid density

Apply Bernoulli's equation to the surface of the upper reservoir. The surface is technically falling as the upper reservoir is being drained. However, for this example we will assume the reservoir to be infinite and the velocity of the surface may be set to zero. Furthermore, the pressure at both the surface and the exit point C is atmospheric pressure. Thus:

`0^2/2+g(0)+(P_[atm])/rho=text(constant)......(1)`

At point `A` where `P=P_A, v=v_A` and `y=-d`

`(v_A^2)/2-gd+P_A/rho=text(constant).....(2)`

At point `B` where `P=P_B, v=v_B` and `y=h_B`

`(v_B^2)/2+gh_B+P_B/rho=text(constant).....(3)`

At point `C` where `P=P_(atm), v=v_C` and `y=-h_C`

`(v_C^2)/2-gh_C+P_(atm)/rho=text(constant).....(4)`

from equation (1) and (4)

`0^2/2+g(0)+(P_[atm])/rho=(v_C^2)/2-gh_C+P_(atm)/rho`

solving for `v_C`

`text(Velocity of Siphon)` `v_C=sqrt(2gh_C)`

The maximum velocity may be calculated by combining equations (1) and (3)

`0^2/2+g(0)+(P_[atm])/rho=(v_B^2)/2+gh_B+P_B/rho`

`P_B=0` and solving for `v_(max)`

`text(Maximum Velocity of Siphon)` `v_(max)=sqrt[2(P_(atm)/rho)-gh_B]`

`text(Height of Siphon :)`

On solving equation (1) and (3)

`text(General height of siphon)` `h_B=P_(atm)/(rhog)-v_B^2/(2g)`

`text(Maximum height of siphon :)` `h_(B, max)=P_(atm)/(rhog)`

 
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