 We were discussing about the minor losses in a piping system and we took an example of flow through a sudden expansion. We will take another example flow through a sudden contraction. So for flow through a sudden contraction the idea is that the fluid is flowing from a pipe of a larger size to a pipe of a smaller size and when the fluid is flowing in that way let us see that what happens to the stream lines. We have encountered such cases earlier and from our previous experience we know that the stream lines first of all they will tend to get converge because the area is suddenly reducing. Then that tendency of convergence will continue till it comes to a minimum distance of separation between the extreme stream lines. That location is known as Vena contractor. So the same thing happens here and beyond that the stream lines diverge. So let us say that this is the location of the Vena contractor. So we call this area as AC and let us say the average velocity of flow across this one is VC and let us say that D1 and D2 are the diameters of the pipes 1 and 2 and V1 and V2 are the corresponding velocities, average velocities. So you have V1 average and V2 average. Now interestingly we may observe one thing that when you have this type of flow then in the first part the stream lines are sort of converging. So that means when it is converging the area is reducing and the velocity is increasing. So it is a sort of accelerating flow and as if there is a favorable pressure gradient that is accelerating the flow. Beyond the Vena contractor it is expanding and the situation gets completely reversed and the situation beyond the Vena contractor is as if it is flow through an expansion. So it is not a geometrical expansion induced by the configuration of the system but because of the expansion in the configuration of the stream lines. So here whatever losses may be there may be attributed to that expansion. So somewhat non-intuitively for loss for flow through a sudden contraction is basically because of an expansion. So the loss is mainly attributed to whatever is happening here which is nothing but an expansion. So whatever expression that we could derive for loss because of sudden expansion that the same expression we may apply here. So what was the head loss expression that we used for this one or we derived for this one that was like sort of here V1 will be VC-V2 whole square by 2g. Now you can always write VC and V2 in terms of AC and A2. So you can write AC into VC is equal to A2 into V2 which means you have VC is equal to A2 by AC into V2. If you recall the area of the Vena contractor divided by the area of the corresponding conduit was known as the construction coefficient of CC. So this is 1 by CC into V2. Therefore the head loss will become 1 by CC-1 whole square into V2 square by 2g. Now therefore this head loss will explicitly depend on what is the value of the contraction coefficient. And accordingly one may write it in the form of as a fraction of the kinetic energy at 2. Now there are extreme cases like as you have cases for sudden expansion where the ratio of the diameters is greatly varying. Here also you can have a case where you have say d1 by d2 tending to infinity. This is a special case. So in general the form is like this is like k. So it is like k V2 square by 2g. The value of k will depend on the contraction coefficient and in such a case when d1 by d2 tends to infinity it is found experimentally that k is very close to 1 and 5. So this case like d1 by d2 tends to infinity what does it show? It represents a equivalent situation that there is a reservoir from which fluid is entering a pipe where the reservoir size is much, much larger than the diameter of the pipe. So in that case this loss is known as entry loss because we have discussed about an exit loss. So entry and exit are always relative to the pipe. So here the fluid is entering the pipe from a larger reservoir. So this is known as entry loss. So the concepts of entry and exit loss are somewhat similar. One is like fluid is entering the pipe from reservoir. So that is entry loss and exit is fluid is exiting from the pipe to another reservoir and in either of these cases it is a sudden expansion and contraction that we are keeping in mind and as we have just discussed that loss due to sudden contraction is basically due to a sudden expansion. Now it is not always that you have sudden expansion and contraction as the only possibilities because of which there are minor losses. So minor losses may be present because of many other things. So other sources of minor losses like presence of valves. So if there are valves in a pipeline, so what does a valve do? So if you have a pipeline say like this and you put a valve in the pipeline. We are not drawing a valve in a proper detailed manner but just let us say this is a schematic way of representing. So let us say that this is a physical obstruction. So when this is lifted the entire fluid may flow. When this is put down depending on the extent to which it is put down it will restrict the motion of the flow or motion of the fluid. Therefore valves somehow they may restrict the motion and because of that there may be a pressure drop across it. So it may act like an orifice where it is a reduced size of the area available for flow and that because of that as we have seen in our earlier examples in flow measuring devices in such cases you may have losses. So valves will also have losses then you may have elbows. So what are the elbows? These are fittings which try to accommodate a change in direction of the pipeline. So you have say a pipeline like this and you want the flow direction to change like this. So what you do? You fit a piece which may be somewhat like this. So this type of piece is known as a 90 degree elbow. The name 90 degree is quite clear that the change in angle here that is experienced by the flow is 90 degree. So this is a 90 degree elbow. So in this way you may have elbows of various degrees. So these such things like valves elbows these things are known as fittings of a pipe. So when you have a pipeline you just do not have an isolated pipe but you have certain things which fit with the piping system and those are known as fittings. So for all these it is not so easy to calculate or rather write analytically exactly the expression for the loss but one may have a whole amount of experimental or computationally available data and from that the loss is somehow characterized as k into v square by 2g. The idea is straight forward that you are trying to write the loss as in proportion to the kinetic energy head and the motivation is that in the previous cases, in all cases we could successfully write the loss in this form k into v square by 2g. But here the k is not something which is straight forward or analytically determined but it comes from experiments or many other considerations. So this k is known as a loss coefficient. So typically whenever one is dealing with an engineering analysis of a piping system and there are fittings, there are piping handbooks which refer to the loss coefficients based on what fittings that you are using and one may refer to that data from the piping handbooks and those databases those have been created by lots of experimentation or these days also by computer simulation and important is to get a value of this one. In many other cases it is also written in an equivalent form like it is written in some a equivalent by d into v square by 2g this form because we have seen that this is also another way of writing the loss fl by d into v square by 2g. So if it is written in that way then this sometimes is known as equivalent length of as if there was it was replaced by a pipe of some length and that would have created some loss. But more commonly it is the loss coefficient that is quoted and that is used. So let us consider one problem where we illustrate how to make use of the concepts of major and the minor losses. So we have a piping system like this. It is connecting 2 reservoirs of large extent. The values of the kinematic viscosity, the value of the kinematic viscosity is given. The pipe is made of cast iron with following characteristics the average surface roughness 0.26 millimeter. The total length of the piping system is 20 meter and then the flow rate that we expect from this system is 0.002 meter cube per second and the loss coefficient for the elbows is 1.5 and the thing is that what should be a good design of the diameter of the pipe. It is a pipe of uniform size but it has certain bends and turns okay. So first of all let us say that the name of the reservoir in the left is A and the name of the reservoir in the right is B okay. So let us say this is reservoir A, this is reservoir B. What would be the direction of the flow from reservoir A to B or B to A, B to A right because you have a natural head available in form of a potential energy head and if you want to have a flow from A to B that also could be possible if you had a pump at some place which will energize it to overcome that deficit in the height. So when there is flow from B to A let us say that you write the energy equation with losses which is like the equivalent modified Bernoulli's type of equation for flow from say 1 to 2. So essentially what you write P1 by rho g plus V1 square by 2g plus Z1 is equal to P2 by rho g plus V2 square by 2g plus Z2 plus the summation of head losses. We are technically to be correct you have to use the kinetic energy correction factor at these places but if you see here that will not be important let us write just for the sake of writing it properly. If you assume so first of all we are neglecting the unsteadiness in this case. So we are assuming that these are very large diameters as compared to the diameter of the pipe. So it is as if there is a slow change in whatever change in the height of these reservoirs that is not very significant that is almost like negligible. So if that is there that means the corresponding velocities of these free surfaces are much negligible as compared to the flow velocities in the pipeline. Then you basically neglect these terms these are small both P1 and P2 are same which is P atmosphere. So you have Z1-Z2 that is H is summation of the head losses. So what are the head losses? So now you tell first of all let us consider the major loss. So head loss first of all if you write the major loss what is the major loss? It is of the form f L by d V square by 2g where V is the velocity of flow average velocity of flow through the pipe total length of the pipe is given. So this is major loss then minor loss what are the sources of minor losses first. So you follow do not do it half as early follow the path of the flow. So first when the fluid enters here this is an entry loss. So what is that 0.5 V square by 2g then it encounters some number of elbows how many are there 1, 2, 3, 4, 5, 6 right. So you have 690 degree elbows for which you have each as k elbow as the loss coefficient into V square by 2g then there is a exit loss okay. So what is the exit loss V square by 2g right. So this is major loss this is entry this is elbow and this is exit. So from this what you can find out is of course q is given so you can replace V with 4q by pi d square but d is the diameter of the pipe. So this equation will boil down to what form some equation which is a function of f and d that is equal to 0 right. So function of f and d equal to 0 it will be just a polynomial function and what will be the power of d in that expression you will have d to the power 5 because you have here V square will bring 1 by d to the power 4 and another d. So it will be a polynomial d to the power 5 and some function of f together that will be 0. So how do you then go ahead what extra information you have you have information on the epsilon. So what you may do you may guess a value of the diameter of the pipe. If you get a guess a value of the diameter of the pipe that will give you what is epsilon by the diameter of the pipe and then you can calculate what is what is the Reynolds number based on the diameter of the pipe and these two together should give you a value of f from the Moody's diagram. You have to check whether this f it satisfies this or not because this equation is a function of f and d so if you substitute d you will get a value of f. So if it does not satisfy you have to go through this iteration process again and again till it converges. So let me give you the answer to this problem so that at least you can check whether answers are coming properly. So for this one the answer is the diameter equal to 45 millimeter. That is why it is guess I mean it is obviously the question is when you guess a diameter of a pipe you may I mean if you if you want to have your wildest expression of imagination 1000 kilometer you may start maybe if you want or maybe 1 nanometer if you want. So all of you have certain common senses and you will always like to exercise the common sense. That how to exercise the common sense since most of us do not have proper common common sense so let us see how we do it. So in this equation you have a function of f and d right you have common sense values of the friction factor. If you look into the Moody's diagram you will see 0.002, 0.0002 like that this is just a linear function of that one substitute that f and see what order of magnitude of d satisfies that. So it will give you a reasonable order of magnitude of d okay so that is the common sense way of going for a guess. So whenever you go for a guess solution it does not have to be a wild guess I mean it has to be a bit of a civilized guess to get some kind of quick answer. So let us work out another problem. Let us say that you have a horizontal pipeline with diameters of d1 and d2 and the pressures at the 2 ends are p1 and p2. So what you have to find out is find the ratio of d1 by d2 so that delta p is a maximum. Delta p is the difference between p1 and p2. So this is very straight forward we will just outline the procedure. So if you write the modified energy equation between the sections 1 and 2 that is modified equation considering the losses. You have p1 by rho g plus v1 square by 2g plus g1 is equal to p2 by rho g plus v2 square by 2g plus g2 plus head loss. So it is a horizontal pipeline that is given. So z1 and z2 are the same that you cancel from the 2 sides. Then you can write say p1 minus p2 by rho g is equal to now you can express v2 in terms of v1 by noting a1 v1 equal to a2 v2 that means you have v1 into d1 square is equal to v2 into d2 square. So you can write v2 square by 2g as v1 square by 2g into d2 power 4 by d2 to the power 4 and then you have minus 1 for 1 v1 square by 2g. Then what is this head loss? This is just v1 minus v2 whole square by 2g this is certain expansion loss. The lengths are not substantial to have major losses as important. So here this is an example where we see that minor loss is the dominating one. So the length is so short that the loss due to the length is not of course it is there but that may be neglected as compared to this loss. So this also you can write in terms of v1 square by 2g into 1 minus v2 by v1 that is d1 square by d2 square d square right. Then I need not work it out further it is the very simple exercise you just consider say d1 by d2 equal to x. So it is a function of x only for maximum of this that derivative with respect to x should be 0. So that will give you the value okay. Now next what we have seen in these examples that what are the major and the minor losses and how they are taken into account and again that important consideration that minor losses did not always be minor. Minor losses sometimes are much more significant than the so called major losses. Next we will look into cases. So here we have till now considered cases of isolated single pipes but in a system in a piping system there may be a number of pipes and these pipes may be connected in series or parallel just in the same way as electrical resistors are connected. So then what would be that equivalent piping network is just like an equivalent electrical circuit network and we will see briefly the corresponding ideas for pipes in series and pipes in parallel. So first pipes in series. So pipes in series it means that you have let us say that you have 2 pipes like this. The name series is obvious they are connected one after the other. So you have let us say that the diameter of the first pipe d1 the average velocity v1 length l1 friction factor f1 and for the pipe 2 corresponding things are there. What is the so when we consider these pipes in series and parallel in this analysis the analysis that we are presenting as a theoretical development we are not considering the minor losses we are considering the only the major losses. So the head loss for the pipe 1 what is that f1 l1 by d1 into v1 square by 2g. What is v? v is 4q by pi d square. So in terms of the flow rate so f1 l1 v square will be 16q square. So 16f1 l1q square then 2g pi square d1 to the power 5 okay where q is the flow rate which is going through each of these pipes. So when they are in series what is the common thing for them is the flow rate the same flow rate is going through the 2 pipes. So if you have hf2 you have similar thing 16f2 l2q square by 2g pi square d2 to the power 5. Now what is the concept of an equivalent pipe that is you replace these 2 pipes in series by a single pipe of some diameter let us say d is the equivalent diameter, le is the equivalent length and fe is the equivalent friction factor such that you have the same flow rate and the same head loss okay. So it is just like an electrical circuit where you are considering the same voltage and same current flowing through that. So you find out an equivalent resistance sort of thing. So here it is like the head loss is like the pressure drop which is like a potential drop sort of thing and the flow rate is like a current so to say. It is not exactly analogous mathematically but it is just another qualitative way of looking into it. So when you have this hf as expressed as the head loss in this equivalent situation then hf must be equal to the sum of hf1 and hf2. So if you write hf for the equivalent pipe it is a single pipe of length le. So from this you can write 16fele same q is there by 2g pi square de to the power 5 equal to 16f1l1q square by 2g pi square de to the power 5 plus 16f2l2q square by 2g pi square de to the power 5. So from this what we can get? We can get a very important expression that fele by de to the power 5 is equal to f1l1 by d1 to the power 5 plus f2l2 by d2 to the power 5. So in general you have if you have n number of such pipes in series you have fele by de to the power 5 is equal to summation of fi li by di to the power 5 i equal to 1 to n. So as if it is like a equivalent resistance as the sum of the resistances that is a simple way of looking into it. Now let us look into pipes in parallel. So when you have pipes in parallel let us try to make a sketch of maybe a situation like this. So you have 2 pipes which are sort of connected in parallel that means say they are branching off from just let me sketch it in a bit of a different way. So let us say that you have pipes through which some fluid flow q is coming there. Now you have 2 pipes with say diameters d1 length l1 so length l1 means not just straight portion plus also the curved portion all those taken together d1l1 and the friction factor f1 second pipe d2l2 friction factor f2. So now so these pipes both are connected across these 2 points which are shown as cross. So what you can say that let us say that q1 is the flow rate through this one q2 is the flow rate through this one. So you can say that q is equal to q1 plus q2. If you consider the node which is given by the cross just like Kirchhoff's current law. So the q is distributed as q1 and q2. Then what about the head loss? Head losses are the same because eventually you are talking about the difference in energy between these 2 points no matter whether you traverse by the upper pipe or the lower pipe eventually you end up at the same point and the loss of energy therefore should be same as what you calculate from here or what you calculate from here. So you have hf1 is equal to hf2. So these are basic equations and from that you can find out the equivalent length of the pipe. So you have and for the equivalent pipe you have say hf equal to hf1 equal to hf2 and q equal to q1 plus q2. So what is the hf of the equivalent pipe? 16 feleq square by 2g pi square d e to the power 5 right this is the hf of the equivalent pipe. This is equal to hf1 that is 16 f1 l1 q1 square by 2g pi square d1 to the power 5 and this is also equal to hf2 okay. So this is hf1 this is hf2. Let us say that each is equal to some constant k and the 16 by 2g pi square this is a term which is like a constant for all let us call it as c. So you can write this is q2. So you can write q1 is equal to q1 square is equal to k into d1 to the power 5 by f1 l1 c right. q2 is equal to k into d2 to the power 5 by cf2 l2 and q is kd e to the power 5 by cfele q2 square ds right. Since you have q equal to q1 plus q2 you have from these expressions root over d e to the power 5 by fele is equal to root over d1 to the power 5 by f1 l1 plus root over d2 to the power 5 by f2 l2 okay the other terms get cancelled out. So these are expressions for the equivalent the relationship between the equivalent and the original ones in terms of the respective diameters and the friction factors. So with this background let us try to work out a few problems where we have the pipes connected in may be series or parallel. So you have 2 pipes, 2 pipelines and these 2 pipes the upper one is d1 is equal to 15 centimeter and length is 150 meter the friction factor is a constant which is 0.018. The other pipe is the diameter d2 is 12 centimeter the length l2 is 150 meter and the friction factor is the same 0.018 it is given that q1 is equal to q2 you have to find out what is the loss coefficient of this valve. So the approach is very straight forward you see why I am illustrating this problem is the whole idea is never get tempted to use a formula which is ready made available with you. There is a formula which is ready made available with you and you might be tempted to use that what should prevent you from being tempted with that is that here you have a minor loss that is not considered in this formula okay. So if you use that formula it will give you erroneous solution but obviously the concept of pipes in parallel you may use. So what are the things you have hf1 equal to hf2 not just hf the total head loss not just the friction loss so h loss 1 is equal to h loss 2. So what is h loss 1 you have f1 l1 by d1 or we write in terms of q 16 f1 l1 q1 square by 2g pi square d1 to the power 5 plus the k valve into v square by 2g. So v1 square by 2g is as good as v1 is 4q by pi d square so 16 q1 square by pi square d1 to the power 4 2g that is v1 square by 2g is equal to the head loss at 2 that is 16 f2 l2 by 2g pi square d2 to the power 5 into q2 square right and it is given that q1 equal to q2 given. So you can cancel that from the 2 sides and get the value of the k valve straight away a very simple exercise the answer is k valve is 18.62. Next we work out another problem you have 2 pipes of length l diameters d1 and d2 and they are arranged in parallel when they are arranged in parallel the loss of head for a particular flow rate q, q is the flow rate the loss of head is h1 and the same pipes when they are arranged in series the loss of head is h2 it is given as d1 by d2 is equal to 2 fine h1 by h2 neglect the minor losses and assume a constant friction coefficient to be the same for all the pipes. So the 2 important assumptions that minor losses are neglected and number 2 friction coefficient or the friction factor is a constant and that constant value is same for all the pipes under which conditions friction factor you have a constant virtually for very high Reynolds number highly turbulent flow it will become only a function of epsilon by d. So but here the diameters are changing so we are assuming that epsilon is also different for the 2 pipes such that epsilon by d remains the same so that the friction factor remains the same. So when the 2 pipes are connected in series so you can work this out through the equivalent resistance concept. So when they are in series you have what is the condition for the equivalent fe le by de to the power 5 is equal to f1 l1 by d1 to the power 5 plus f2 l2 by d2 to the power 5 this is for the series and now the equivalent things equivalent thing has combinations of 3 parameters and see it is not important what are the individual values of these parameters it is important that you collectively choose them to satisfy this constant that should be good enough that means you may choose your equivalent friction factor or equivalent length in such a way that you will get some equivalent diameter or you may choose equivalent friction factor and equivalent diameter as to be something so as to get some equivalent length. So you may take any of these out of 3, 2 very freely and the third one you get from this expression let us say that we assume that the 2 pipes are of the same length right. So let us consider that le or in fact if you see that it is fe le by de to the power 5 that is going to be solely important for the head loss. So even if you do not assume any particular value that will not matter. So if you consider the head loss what is that 16 fe le q square by 2g pi square de to the power 5 right. So in place of so you can clearly see that you get an expression where you have fe le by de to the power of 5. So let us say that you write in place of that 16 q square by 2g pi square then you write f1 l1 by de to the power 5 plus f2 l2 by de to the power 5 this is given as h2 this is series if they are in parallel again hf formula is the same but expression for so this you have 16 q square by 2g pi square then you have 1 by de to the power 5 by fe le right and that you can substitute in place of this one that is de to the power 5 by fe le and this is given as h2 sorry this is given as h1 just you divide by these 2 and you will get a ratio when you divide you will get a ratio of d1 by d2 and l1 and l2 are the same. So that ratio will give a number. So this when you divide you will just get a number f1 and f2 are the same. So those effects will cancel and it will be expressed solely in the as a function of d1 by d2 if you write h1 by h2. So the h1 by h2 the answer is 0.02188 that is the answer. Let us work out may be another problem. The problem statement is like this that initially so you can see that here there are pipes ad bd and dc. So this is just shown by schematic so not the width with being shown. So here initially only the part ac was there there was no branch bd and then the flow rate was 100 liter per second that is given. So q0 is 100 liter per second and the length of ac is 1000 meter that is 1 kilometer. To increase the flow rate another pipe bd is added okay estimate the length of the new pipe that is the problem all diameters are equal. So all diameters are equal and assume the same length for all the pipes not for all the pipes that is l1 equal to l2 that is same length for the 2 parallel pipes and same friction factor for all pipes. So friction factors are also equal okay and it is given that there is a 30% enhancement in the flow rate because of this. So you have to find out basically l1 and l2 that is the question. So let us say that there is a flow rate q1 through l1 and q2 through l2 and the total q is sum of q1 and q2. So then you can write so the head losses if you neglect this elevation difference the head losses should be what? The head loss for ad and head loss for bd they should be the same they are like pipes in parallel. So if their head losses are same head loss is function of qf and l. So you have f and l are same therefore q should be same. So ad equal to hfbd that will give you q1 equal to q2 and therefore you have q3 which is either equal to 2q1 or 2q2 all the same. What is the total head loss that is h? So we will not write the modified equation in all details you have just seen that this h should be compensating the total head loss. So the head loss in ad plus the head loss in dc right. So this will be a function of q3 because head loss in ad is a function of q1, q1 may be expressed as a function of q3 and head loss in dc is a function of q3 and the head loss when this branch system was not there still the head loss would be the same right. So when bd is not there then the head loss is the head loss for the length se with the original flow rate as q0. So 16 fl l is l1 plus l3 into q0 square by 2g pi square into d to the power 5 and it is given as that there is a 30% enhancement in q that means q3 by q0 is 1.3. So from that you can find out the missing length you have to keep in mind that total l1 plus l3 is 1000 meter. So just assume these as some x and this is 1000-x and this is also then x you can solve for that remaining things are given. Let us maybe look into another problem very briefly. So let us say that you have 2 pipes or a pipeline it has a diameter say d0 and the velocity v0. It is having some length say l0. To increase the flow rate a new arrangement is made. What is the new arrangement? The new arrangement is a branch is taken away from the midpoint of this one. So this is l0, this is l0 by 2 and this is l0 by 2. So the diameter is the same the diameter is d0 for the second arrangement as well. You have to find the change in flow rate say here flow rate is q0 here the flow rate is q1. So you have to find out what is q1 by q0 given hf1 is equal to hf0 is equal to hf1. So this is a straightforward pipe series parallel problem. So only thing is what you do? You replace this by an equivalent pipe. So if you replace this by an equivalent pipe these are 2 pipes in parallel. So root over de to the power 5 by fe le is equal to root over d1 to the power 5 by f1 l1 plus root over d2 to the power 5 by f2 l2. Here all f's are the same. Let us consider that the equivalent friction coefficient also the same l1 is what? l0 by 2, l2 is l0 by 2, d1 and d2 are the same which is equal to d same diameter pipe. So this is d to the power 5, this is d to the power 5. So let us say that the equivalent diameter is also d. So you can find out an equivalent length in terms of as a function of l0 right. So then this entire pipe as if is replaced by a pipe of length l0 plus l equivalent and you have hf1 is equal to 16 fl0 plus l equivalent by into q square by 2g pi square d to the power 5 and hf0 is 16f sorry this is l0 by 2, l0 by 2 plus l equivalent. Sorry this is l0 by 2 just correct it, this is l0 by 2 half half. So 16 fl0 q0 square by 2g pi square d to the power 5. From here since these 2 are equal you can find out what is q1 by q0. The answer is that the increment is 26.48%. So this is just very simple equivalent pipe system analysis. So let us stop here today or for this lecture and we will continue with the next lecture with a new topic. Thank you.