 Hello students. Myself, Siddeshwar B. Tulsapure, Associate Professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Solapur. So, in this session, we are going to see flow through pipes in parallel, the learning outcome. At the end of this session, students will be able to calculate discharge through branched pipes. These are nothing but these are pipes in parallel we are having. The contents are, firstly, the definition of pipes in parallel, then we will go for discharge through branched pipes, then head loss in parallel pipes, then numerical on flow through parallel pipes, and lastly the references. So it is flow through pipes in parallel. If main pipe gets divided into two or more branches and again join together downstream to form a single pipe, then the branched pipes are said to be connected in parallel. Means see what? We are having firstly the pipe gets divided into two or more branches, that is one. And again join together downstream to form a single pipe, means again they are joining, then starting from one, then branching, then again the branches they are meeting each other and we are having again the single pipe. So if this situation is there, then those branched pipes, these are called as, these are connected as say parallel like that we are having. Now let us go for simplicity purpose only for two branches it is. Firstly on the left hand side you can observe the main pipe we are having. So it is branched into two, so we are having one on the lower side and one on the upper side. So branched pipe number one say on the lower side, branched pipe number two on the upper side. Again they are meeting each other and we are having again the main pipe continuing. So in case of this, say in the diagram you can observe here, so if suppose the discharge it is Q, say through the main pipe, so this discharge will get divided into two. Say through the branched pipe number one, discharge is Q1, through the branched pipe number two, discharge is Q2 and we are having the say parameters with reference to the branched pipe number one as L1, D1 and V1. These are corresponding to the length, diameter and velocity of flow. Then for branched pipe number two it is L2, D2 and it is V2. So again the Q1 and Q2 they are going to meet each other and again we are having say total discharge as Q as in the earlier case. Now discharge through parallel pipes that we have discussed just now, but now we have taken only two branches. Instead of that one suppose if you are having number of branches that total Q it will be equal to, so it is Q1 plus Q2 plus Q3 plus Q4 plus dash dash whatever number of those branched pipes are there, so those we have to write here. So discharge through those pipes we are going to have on the right hand side and say main pipe which was say before branching, so the discharge through that one will come on the left hand side. Now the notations used to these are say L is equal to it is length of pipe, D is equal to diameter of pipe, F is equal to coefficient of pipe, say V is equal to it is velocity of flow and then say we are having that suffix 1 to represent the branch pipe it is say branch pipe number say for the say before the branching the pipe was only one, but two branches if you are considering suffixes will be 1 and 2, say F it corresponds to the coefficient of pipe that coefficient of friction it is actually. Now think of say whether the pipes in different branches can be of different materials see the question question is much clear, so it is earlier suppose we are having one say MS pipe then suppose it is say branched into two and we are now interested in choosing the material for these two and now say if whether say different materials can be there for say these two branches like that the question is think of the same, say the answer is yes. In case of these branched pipes we can have the materials as different one, so you may choose the MS you may choose the say other metallic materials for these or even you may go for the concrete and other things etc. So what matters is the coefficient of friction etc the diameter of that one, so these are going to matter, but so nowhere that condition is there for the pipes in parallel that the material should be same. So the pipes say which are used in the branches these may be of different materials, now let us consider the head loss, so head loss in parallel pipes we are having. So loss of head in branch number branch pipe number 1 is equal to the loss of head in the branch pipe number it is 2. Now on the left hand side we have written the formula for the loss of head in branch number 1 it is 4 f1 l1 v1 square by 2g d1 and for the right hand side we are having this as it is 4 f2 l2 v2 square by 2g d2. Say if suppose f1 is equal to f2, so in case of that one we are having this as so cancellation of common terms as 4 then 2g 2g will be there along with this f1 f2 and say the equation will get reduced to l1 v1 square by 2g d1 is equal to l2 v2 square by 2g d2. Now let us consider the numerical say for this one and what we have observed is for the parallel pipes the head loss it is going to remain same, so 4 f1 v2 square by 2g d of say branch number 1 is equal to 4 f1 v2 square by 2g d of branch number 2 and similarly say we will go for other branches if these are there. So now in case of the present numerical the data just I have listed down so it is length of parallel pipes is suppose equal to 2000 meters then coefficient of friction of 2 pipes is suppose 0.005 then diameter of first pipe is equal to 1 meter diameter of second pipe is say lesser than the first one so it is 0.8 meters and if the total flow is 3 meter cube per second say what is flow rate through the pipe number 1 and 2 so that is the question. We will move to the solution in case of this parallel pipes we are having the head loss as same in all branches say head loss in first branch it is equal to the head loss in the second branch so 4 f1 l1 v1 square by 2g d1 is equal to say it is 4 f2 l2 v2 square by it is 2g d2. So again have the cancellation of the terms as much as possible so that the calculations will get reduced and we will get the answer in the few steps and simple steps it is. In the present case the coefficient of friction of the branch number 1 and branch number 2 they are given same so it is f1 is equal to f2 is equal to we are having some value as 0.005 so presently not required for this step. So l1 l2 these are same so these are corresponding to the say length of the pipes so length of pipe it is say 2000 meters so l1 is equal to l2 we are having and say the common terms if you cancel out 2g 4 etc so along with f only l1 l2 is also getting cancelled only velocity is remaining in the numerator so v1 square v2 square on the right hand side and here it is d1 and here it is d2 so v1 square by d1 is equal to it is v2 square by d2 we are having so putting the value of say d1 and d2 so these are 1 meter and 0.8 meters so v1 square by 1 is equal to v2 square by 0.8 it will be equation number 1. Then from continuity equation we can have this one as a1 is equal to a1 v1 is equal to a2 v2 say a1 it is pi by 4 d1 square into v1 so it is equal to a2 it is pi by 4 d2 square into it is v2 so cancellation of pi by 4 will be there and d1 square v1 is equal to it is d2 square v2 we are having so in this case what you can do is you can put again the values of d1 and d2 and see what you get so d1 is equal to it is one say square of 1 so into it is v1 so it is equal to so it is d2 square we are having so it is 0.8 bracket square into it is v2 so this is equation number 2. So equation number 1 so it is between v1 and v2 we are having that is 2 parameters we are having which are unknown so it is v1 v2 equation 2 it is also containing v1 v2 so the case is 2 equations and 2 unknowns we are having so we can put the value of either v1 in terms of v2 or v2 in terms of v1 so if you solve these 2 equations you will get the value of v2 as 2.17 meters per second v1 as 2.427 meters per second then you can go for the discharge through the branch number 1 so it is q1 is equal to it is a1 v1 so it is pi by 4 d1 square into v1 you will have so it comes out to be 1.906 meter cube per second then similarly we will get the value of q2 as equal to it is a2 v2 you can do and you will get this value as 1.094 meter cube per second later on just you can have one check by adding values of q1 and q2 whether you are getting the value as 3 or not so that you can check so 3 it is the discharge through the main pipe so here you can observe so the total summation it is equal to say total discharge is 3 meter cube per second so this will ensure that say you have not gone wrong anywhere so apart from this if you observe that say value of coefficient of friction etc these are given but say these are getting cancelled as these are same l1 l2 also these are given but these are getting cancelled these are the references which are used for this particular session thank you.