 Hello students. So, myself, Siddhaswara B. Thurjapura, associate professor, Department of Mechanical Engineering, Walchand Institute of Technology, Solapur. So, in this session, we are going to see the derivation corresponding to the Darcy Wichibach equation for head loss in pipes. The learning outcome, at the end of this session, students will be able to derive an equation for major head loss in pipes, contents, the definition of major loss, then terminology, the derivation, and lastly, the references. Now, the definition of major loss we are having. The major loss is the loss, it is due to the friction between the liquid which is flowing and the inner surface of the pipe. Now, let us say that we are having a pipe of uniform diameter. So, this is a pipe. Now, in this case, the fluid is flowing. So, it is flow. On the inner surface of the pipe here, we are having the friction between the liquid which is flowing and the inner surface of the pipe. So, that definition, that say, frictional surface of the pipe which is lying on the inner side, it is going to provide the resistance to the flow and it is called as major loss. Now, let us consider one section here. It is section 11 and one more section here. It is section is 22. Let pressure here be equal to P 1. The pressure here is equal to say it is P 2. The velocity here is equal to V 1. The velocity here is equal to V 2. When the datum line we are having, say corresponding to this one, the datum is say Z 1 datum head and here the datum head is equal to Z 2. So, in case of this one, now we are going to have the friction when the fluid it is flowing from the left to the right hand side, that is from section 1 to 2 and the frictional resistance it will be offered in the opposite direction. So, it is frictional force FR we are having. Now, applying Bernoulli's theorem at section 1 and 2. So, here P 1 by rho g plus V 1 square by 2 g plus Z 1 is equal to say P 2 by rho g plus V 2 square by 2 g plus Z 2. Whether we are missing anything? See Bernoulli's theorem we have applied at section 1 and section 2. So, whether the things which we have written it is totally complete in itself. See now, this is the Bernoulli's theorem which is applied to the ideal conditions that is where we are not going to have any frictional loss. So, if the losses in case of the real fluid when we are going to have. So, losses these are going to be there. Say in case of this one, the frictional head loss it is going to occur and that needs to be added here. So, it is H f that is the head loss due to the friction. So, it is H f is called as head loss due to friction. So, this is nothing but it is the major loss. So, now, in case of this say for pipe of uniform diameter diameter V 1 will be equal to V 2. Let pipe B horizontal say Z 1 is equal to Z 2. So, this one is getting cancelled and this one is also getting cancelled. So, what is remaining is say it is now P 1 by rho g is equal to P 2 by rho g plus the head loss which is occurring it is H f. So, now, what we can do is we can go for say it is P 1 minus P 2 by rho g we will have and the left hand side we have taken and this one is equal to this H f. Let this rho g term be on the right hand side. So, here we are having this one as P 1 minus P 2 is equal to rho g H f. So, this is the equation number one. Now, next to that one we are having the forces here which are the pressure forces and the frictional force is there which is towards the left hand side. So, pressure force towards the right side pressure force towards the left side and we are having the frictional force towards the left hand side. So, resolving forces in horizontal direction horizontal direction. So, towards the right we are going to take this as positive and towards the left we will take this as negative. So, it is P 1 into A. A refers to the cross section area of the pipe. So, it is minus P 2 into A. So, this one is towards the left and again minus frictional resistance F r is equal to 0. So, what we can do is we can rearrange the terms. So, it is now P 1 minus P 2 we can take and it is into A. So, this can be written equal to it is F r. So, now, similar to the equation number one where we are having P 1 minus P 2 on the left hand side we can take this. So, here it is P 1 minus P 2 is equal to F r divided by it is A. So, this F r F r is nothing but say it is frictional force we are having. So, it is frictional force. So, now, what we can do is we can put it in some another form say it is F r is equal to. So, let us name this equation equation number it is 2. So, F r is equal to we are having. So, let it will be now F dash multiplied by. So, it is pi D L into is V raise to N. Now, in case of this one this one it is representing weighted area weighted this area. So, here the diameter of the pipe it is equal to D and length of the pipe between the two sections it is equal to L. So, it is N, N it is between 1.5 to 2 for it is turbulent flow. So, let us take it as equal to 2. So, this one. So, this one it is nothing but it is the frictional resistance per unit weighted area per unit velocity raise to N. So, now, what we can do is we can put this value of F r here. So, now it will be P 1 minus P 2 is equal to say it is F r we are having this one as F dash into it is pi D L into it is V square we will take this as equal to 2 and then divided by it is area. Then this is modified form of the equation number 2 only. So, we are getting this one. So, now what we will do is comparing equating equating RHS of 1 and 3 we have rho G H F is equal to F dash pi D L V square divided by A. So, here say it is H F is equal to our interest is F dash pi D L it is V square we are having and then we can have this as divided by rho G we are taking rho G we have taken and then A it is nothing but it is pi by 4 it is D square is there. So, here square and this one this will get cancelled this 4 will come on the top side. So, this pi and this pi this will get cancelled. So, this H F comes out to be 4 then it is F dash L then it is V square divided by rho G and it is D. So, now in case of this we are now having say somewhat substitution of the terms here it is let F dash upon rho is equal to F upon 2 F is equal to coefficient of friction then it is H F is equal to 4 F L V square by it is 2 G D. So, this is Darcy Wiesbatch equation if we want to have replacement in some another form. So, in case of this one H F can be written as F L V square by 2 G D here F is friction factor. So, this one Darcy Wiesbatch equation these are the references. Thank you.