 Hello students, Myself, Siddhaswar B. Tuljapura, Associate Professor, Department of Mechanical Engineering, Walsh and the Institute of Technology, Sulapur. So in this session, we are going to deal with the topic, it is SIFON, the Learning Outcome. At the end of this session, students will be able to design SIFON for any application. That is, so in case of the present session, we are going to deal with the different parameters which are concerned with the SIFON and students will be able to select the proper parameter value in case when he is going for any application of the SIFON contents. Firstly, the definition of the SIFON, then velocity of flow through the SIFON, then pressure at submit point in SIFON and then lastly the references. Now first, we will go for the definition of it is SIFON, so we have said it is SIFON, SIFON it is firstly a long bent pipe it is, see the wordings, say we are having this one as a pipe, but it is very long and next to that one it is a bent pipe it is. So SIFON it is a long bent pipe it is used to convey liquid from high level to low level say reservoirs, higher level reservoir to lower level reservoirs separated, so this is important is separated by hill or elevated region means in general we can think of higher level to lower level the liquid is to be taken or conveyed, but in case of this we are not having the plane higher levels and lower levels near to each other and say there is nothing in between these two like that the situation is not there. So we can think of certain hill area is there, so here the water it is stored say from the rain water etcetera and at the bottom side somewhere the village etcetera it is there and next to that one here the requirement of water is there, but so directly the pipes from the upper level reservoir to this lower level or where the consumption it is done, so here it cannot be, so this is not possible. So in case when we are going to have this hill or the elevated region, so which is separating say it is upper reservoir and this one it is the lower reservoir, so in case of that one we are going to make use of the SIFON, so which is a long bent pipe, so from here to here like this and you can see that this is a long pipe which is now bent and we are having this as a SIFON. Now let us come to the say firstly the velocity of flow through the SIFON, so it is now velocity of flow through SIFON. In case of this one say firstly let us draw say one tank which is at higher level say one more tank which is at lower level and say it is now like this the SIFON and now the uppermost part that is the highest level part of the SIFON it is called as SUMIT, so SUMIT is the highest part of the SIFON it is. Now here we are having the liquid and here it is to be taken and now we are interested in say the value of the velocity of flow which is occurring through this SIFON. Now let this be point A and this be point B in case of this we are going to have the application of the Bernoulli's theorem. So it is applying Bernoulli's theorem at A and B, so it is now say P A by rho G plus V A square by it is 2 G plus it is Z A this is equal to, so on the right hand side we are having. So it is now P B by it is rho G plus V B square by it is 2 G plus it is Z B see whether say this one it is going to be say without the consideration of any loss, so plus it is going to be H F. Now tell me what is going to be the pressure at A and B and what are the also velocities at A and B think of the pressure and the velocities at the points A and B. See the pressure at A and pressure at B these are atmospheric means this one is P atmospheric is equal to it is 0 here also, so this one it is going to be equal to 0. So in case of the velocity, velocity of the fluid particles in the tank will also be equal to 0 and velocity of the fluid particles in this tank also will be equal to 0, so then the difference of Z B and Z A and then H F these are the points which are remaining. So let us now have this one as Z A minus Z B is equal to it is H F. So now we are going to write this one as say it is Z A and Z B let this be equal to X. So in case of the present one, so we are going to have this as say it is X is equal to H F we will now write it as 4 F L V square by 2 G D you are knowing this one and then here it will be V square is equal to it is X into it is 2 G D divided by it is 4 F and it is L from this one we can be calculated. So we will be equal to it is under root it is X into it is 2 G D divided by it is 4 F L. So this is what we are going to get the velocity here next to that one you can think of the pressure at the summit what is happening. So at this particular point which is going to be the highest point, so let us go for pressure at summit let this be point C. Now applying Bernoulli theorem applying Bernoulli theorem at A and C, so in case of this one so it is now P A by rho G plus V A square by it is 2 G plus it is Z A is equal to it is now P C by it is rho G plus V C square by it is 2 G plus Z C plus H that is the head loss H F corresponding to it is A to C that is from here to here only. So in case of the earlier one the frictional head loss was continuously say from here to up to this particular point means for earlier one the complete say pipe it was considered but now in the present case from here to the summit that much length it is to be considered. Now P A by rho G it is equal to 0 then plus it is V A square by 2 G also it is equal to 0 then plus Z A we will write it as it is then P C by rho G we are interested in and then V C square by 2 G that is the velocity of the liquid inside the pipe we are going to determine it and say or we will write it as it is say let us say that it is only V it is V square by 2 G because here also we are going to consider only one velocity we will be there this one and the one more it is inside this H F that is plus Z C plus it is for F L V square by it is 2 G D. Now let us see what is the value of P C by rho G. So it is P C by rho G is equal to say it is Z A minus Z C minus Z C minus V square by it is 2 G minus 4 F L V square by it is 2 G D. So in case of this one we can see that Z A minus Z C is there. So we can go for say if this one is considered say Z C minus Z A it is going to be positive but Z A minus Z C it is going to be negative. So this one say it is negative this parameter is negative and this parameter is also negative. So in case of this we are going to have P C by rho G as the negative value always. So what it indicates is it indicates that the pressure at the summit will be always lesser than the atmospheric pressure. So now we are knowing that say every liquid is having the vapor pressure. If the pressure at the summit if it goes below the vapor pressure we are going to have the formation of the gases which are going to release from the say dissolved air and the gases these are going to release from the water and in case of that we are going to come across again the blockage or the obstruction to the flow. So like this we should take care that we will not have that formation of the gases etc here by keeping the say pressure at the summit always greater than the vapor pressure of that particular liquid. So the siphon it is basically used to convey the water from the air level reservoir to the lower level reservoir. So when it is not having any outlet or in case when we are going to have the say the tank it is to be made empty and no outlet or say we are going to have somewhat elevated or the regions in this case like this etc. So similar cases we are going to make use of the siphons. See these are the references. Thank you.