hello friends welcome to engineering streamlined you may have done or seen people taking out petrol from their vehicles just by sucking petrol through the pipe or squeezing a plastic bottle attached to a pipe and then automatically the petrol flows or you must have seen big pipes coming out from the fields which control the amount of water inside the fields this is due to a process known as siphoning today we will understand how this siphoning works take two beakers with one beaker at a higher elevation now fill the beaker which is at a higher elevation with water and put an inverted u-shaped pipe inside it now if you suck the water from the other end of the pipe and place this end into a completely empty beaker you will see that the water from the filled beaker flows automatically into the other empty beaker here the water initially rises inside the pipe and then flows downward but how water continues to rise up without any pump or external support let's see our theory number one you can observe here that water in the length of pipe shown in red is less than the water in the length of pipe shown in green one theory is that the greater weight of the water on the output side is pulling down the water causing it to flow from the filled beaker so it is tempting to think this as something like a pulley and chains with unequal weights on either side then the heavier weight pulls down the other one using this theory an italian engineer vittoro zonka in 1607 theorized the idea of a perpetual water mill which can raise water from the reservoir on the right through the inverted tube where it exits on the left at a higher level to drive the mill before returning to the reservoir if we take a similar kind of arrangement of two beakers with an inverted u-tube with left arm of the tube having greater weight of water than the right arm then according to zonka's theory water should flow upwards and fill the empty beaker but this will never work because he had mistaken the idea of how siphon works even in this condition water will flow from the top beaker to the bottom one so now let's understand the working principle behind siphoning when you suck the air out of the other end of the inverted u-tube there is a vacuum or a low pressure state inside the tube and the water at the surface of the filled beaker is constantly under atmospheric pressure so this pressure pushes the water inside the tube to a height and then flows down due to gravity now one can say that water flow can break at the top of the tube so why it doesn't happen that is because if water flow breaks then again there will be a vacuum or low pressure state there an atmospheric pressure on the left end of the tube will push the water inside so one can say that atmospheric pressure and gravity both are necessary for siphoning to work but it has been seen that siphoning also works in vacuums if the fluid has a very strong cohesive bond between them such a fluid is mercury now let's find out the exit velocity of the fluid in this step let's take a point 1 on the surface of the water filled beaker and point 2 on the other end of the inverted tube point 2 is taken to get a reference level z2 and point 1 is at a height z1 from the reference level and the difference between the z1 and z2 can be represented as h now we can apply the bernoullis equation for this setup bernoulli's equation states that the sum of pressure energy kinetic energy and potential energy is constant at a point in a fluid so we can write this for point one and two as shown here since point one and two both are at atmospheric pressure so we can replace the pressure energy term in both the points as p atm the velocity of the fluid at point 1 is very small compared to the whole fluid at the surface so v1 is set to 0 and z2 is the reference level so z2 becomes 0 and z1 is at a height h so it is replaced by h on rearranging the terms we get this equation now pressure term on both sides gets cancelled and on further rearranging we get the final expression of the fluid velocity at the outlet of the pipe this will give the ideal velocity of the fluid but the real velocity will be less than that due to the friction of the fluid with the pipe surface during the flow now the next question is if siphoning can help to raise water to heights without any external equipment then why it is not used to raise water in high water tanks this is because there is a maximum limit to the height up to which water can be raised let's find it in this setup let's take point 1 on the surface of the water filled beaker and point to at the maximum height inside the inverted tube point 1 is taken to be at reference level z one and point two as a as a height of z two from the reference level and the difference between them is represented as h so now we can write the bernoulli's equation at point one and two as shown here point one is at atmospheric pressure so we can replace the pressure energy term with p atm but for point two it is at a vacuum or low pressure so pressure at point to be set to zero velocity of the fluid at point one and two both are taken to be zero at point two the velocity is zero because the fluid will stop flowing when the maximum height is reached since z1 is at a reference level so z1 becomes 0 and z2 is replaced by h on rearranging the terms we get this equation further we can write h is equal to patm by row g on substituting the properties of water and atmospheric pressure we get the maximum height for siphoning of water to be 10.32 meters so beyond this height siphoning will not work so can you tell which liquid can be siphoned to maximum height water or oil comment below i hope you like this explanation kindly like and subscribe and keep looking for more videos