 Hello students. So myself, Siddeshwar B. Tulsapure, Associate Professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Solop. So today we are going to deal with the topic, it is equivalent to pipe, the learning outcome of the session. At the end of this session, students will be able to derive Duput's equation which is nothing but it is the relation between the diameters and lengths of compound and equivalent pipe. The contents are definition of equivalent pipe, then we will see the compound pipe, then we will go for conditions of equivalence, then derivation of Duput's equation and lastly the references. Firstly, let us see what is meant by an equivalent pipe, it is defined as a pipe of uniform diameter having loss of head and discharge equal to loss of head and discharge of compound pipe. See what we are having, equivalent pipe we are saying, so in case of this one, we are having a pipe of uniform diameter, that is one and say in case of this, we are using the word equivalent, what the equivalent should be? The equivalent should be with reference to the loss of head and discharge it is. So loss of head and discharge of an uniform pipe, so if it is equal to the loss of head and discharge of compound pipe, then that uniform diameter pipe, it will be called as equivalent pipe. Now let us see what is meant by that compound pipe. So in case of the compound pipe, we are having the say different pipes, these are joined to each other, say end to end these are, say the diameters of these, these are varying. So the first part you can observe here, so it is diameter it is suppose D1 and say this one is for length it is L1, then the diameter it has reduced and the second pipe we have got, so it is diameter it is D2 and length is equal to it is L2 we are having. Then one more section of pipe we have joined here, so it is diameter D3 and then say the length of that one it is equal to L3 we are having. So now you can observe that here in case of the compound pipe, we are having the diameter of the pipes as different, so D1, D2 and D3, so these are the diameter of the pipes which are in compound pipe. So now we are thinking of the discharge, okay, so now in case of the discharge through the compound pipe we are having and now the compound pipe it is having the different sections as say section 1 of diameter it is D1, section 2 of diameter it is D2, section 3 of diameter it is D3 and corresponding lengths are L1, L2 and L3. So the same diagram we have taken here, now we are thinking about the discharge through the compound pipe it is and now the question is discharge through which section of the compound pipe it is largest. Think of this one, we are having 3 different sections of 3 different diameters, section 1 of largest diameter, section 2 of medium diameter and section 3 of lowest diameter. For discharge we are interested in it and say discharge through which section it is largest, see the answer, actually discharge through all sections will be same. So you are knowing that continuity equation, so in case of continuity equation we are having Q is equal to A into V and you are knowing that that continuity equation it is based on the law of conservation of it is mass. So whatever amount of fluid it is entering per unit time the same it is leaving. So in case of this so discharge is equal to it is cross section area into velocity we are having and we are having the cross section variation hence the velocity variation will be there. So the diameter if it is larger the area of fluid will be larger and velocity it will be lesser, vice versa is also going to happen. Now the conditions of equivalence we have seen, so in the definition. So what these are? These are 2 actually because we are considering these equivalence and we are going to derive the duputs equation. So the first condition is say loss of head it should be same as that of the compound pipe means equivalent pipe should have same head loss as that of the compound pipe. In case of the discharge also so discharge through the equivalent pipe should be same as that of the compound pipe. Now here are different notations which are used so for this particular derivation purpose. So small f is equal to it is coefficient of friction, hf is equal to frictional head loss, then l is equal to length of pipe, then d is equal to diameter of pipe, v is equal to velocity of flow. Now here we have used suffix 1, 2, 3 say for indicating these above parameters with reference to the section 1, 2, 3 of the compound pipe and suffix e has been used for the same parameters but with respect to the equivalent pipe. Now let us take the first condition of equivalence it is that is head loss of equivalent pipe should be equal to the head loss of compound pipe. Now in case of this one we are having say head loss it is hf of equivalent pipe is equal to so hf of section 1, so head loss of section 2 and head loss of section 3. So you are knowing the formula of head loss, so it is head loss is equal to it is 4fl v square by it is 2gd we are having. So we are going to make use of this formula that is hf is equal to 4fl v square by 2gd. We will have only the suffixes applied to these symbol cities. So in case of the left hand side we are having the suffixes e for equivalent pipe and the parameters which are for the compound pipe they will have the suffixes 1, 2 and 3 based on the sections for which we are going to consider the head loss. Then only say 4, 2 and g will not have any suffix all other parameters these are having suffixes. Then let say coefficient of friction for equivalent pipe is equal to the coefficient of friction of say section 1, coefficient of friction of section 2 and coefficient of friction of section 3 that is fe is equal to f1 is equal to f2 is equal to f3. So now you can observe that apart from this f which is going to be common fe and f1, f2, f3 these parameters cancellation we are having other terms as common 4, 2 and g these are common. So these terms will be removed from the above equation and the above equation gets reduced to the new form as say it is le ve square say divided by de is equal to l1 v1 square by d1 plus l2 v2 square by d2 plus l3 v3 square by d3. But now for from continuity equation that velocity is equal to discharge upon area and by putting the value of area that is pi by 4 d square we will get this v as equal to q upon pi by 4 d square is equal to 4q upon it is pi d square. But in the equation we are having v square see the earlier equation. So in the numerator everywhere v square is there. So we will go for v square value v square will be equal to this complete bracket say inside the bracket we will have 4q upon pi d square and bracket square we will have. Then we will take the term d outside so it will be now 1 upon d raise to 4 because already it is d square inside the bracket. So it is 1 upon d raise to 4 and into the bracket we will keep the terms as 4q and pi as it is and the square of that one is there. So 4q upon pi square we will put it as a constant. So 1 constant k it will be utilized for that one and v square will be equal to k into it is 1 upon d raise to 4. So this is now say we have not used any suffix. So general equation we have written for v square. But as we are having the second condition as discharge it is going to be same for all this. For all in the sense one it is say equivalent pipe and the second it is compound pipe and we have seen earlier in case of the compound pipe we are having say the different sections these are having the same discharges. So in case of that one q as it is a constant we are going to have the above equation we are having the same here but the suffixes will be applied here. So suffix e for the equivalent pipe and suffix 1 2 and 3 for the sections of compound pipe. So we will replace those terms v square. So this le it was as in the earlier case so le is there and de is there in the denominator. So le in the numerator and v square we are going to replace. So by replacing that v square we are having these 4 parameters here in the numerator as different one. Other parameters corresponding to length and diameters these are kept as it is. Then cancelling the common term k that is constant and rearranging the terms we will have. This le will be there in the numerator and in the denominator de is there and here also in the numerator say 1 upon de raise to 4 is there. So it will be in the denominator de raise to 5. So we are having here say on the right hand side similar parameters it is say l1 upon de raise to 5, l2 upon de raise to 5 and l3 upon de raise to 5. This is known as Dupont's equation. So if you are having the length of equivalent pipe as equal to l1 plus l2 plus l3 then the diameter of the equivalent pipe can be found out. So otherwise if you are having some pipe already and say the diameter if it is fixed you can determine the say length of the equivalent pipe by making use of the Dupont's equation. These are the references used for this particular session. Thank you.