 Hello students, myself, Siddhaswara B. Thulzapure, Associate Professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Solapur. So in this session, we are going for the topic that is efficiency of power transmission through the pipes, the learning outcome. At the end of this session, the students will be able to derive the condition for the maximum efficiency for power transmission through the pipes, contents. The first link, we are going to see what is the formula of the efficiency of power transmission through the pipes. Then the condition for the maximum power transmission through pipes. Then the maximum efficiency of power transmission, that is, what is the value of that maximum efficiency of power transmission and lastly the references. Now see in case of the power transmission through the pipes firstly let us have a tank and then the pipe is connected to this one and say the free surface of the liquid is there and from the center of the pipe the free surface of the liquid it lies at a distance of it is h and they let the length of the pipe be equal to l the velocity of the liquid which is passing through this pipe let it be equal to v and that let d be equal to the diameter of the pipe through which the liquid it is flowing. Now in case of this one the minor loss it is going to occur here where the pipe is connected to the tank it is, but we are going to not consider in case of power derivation and here we are going to have say neglecting minor losses. So neglecting minor losses will have the efficiency of power transmission this is equal to. So this is going to be so power at outlet of pipe divided by say it is power at this inlet of pipe see what we are having we are having the outlet of the pipe the power there it is to be considered we are having the inlet of the pipe and the power there it needs to be considered and the ratio of these two we are having. Now as you are knowing the power it is having the units as equal to it is watts so in case of that when watt means again it is joules per second and then it is Newton meter then it is per second. So here in case of this we can see that Newton per second it is this is nothing but it is the weight of liquid which is flowing per unit time and this one it is nothing but it is head so we can write it as say here it is weight of liquid say per second multiplied by head at inlet of pipe and divided by it is weight of liquid per second multiplied by head at outlet of pipe. Now let us put the values and the cancellation part also we are going to carry out say weight of liquid per second weight of liquid per second it is cancelled now and now head at out inlet of head at it is outlet of the pipe and here it is head at the inlet of the pipe we are having so outlet outlet inlet and inlet it is. Now in case of this let this be say capital H we are going to have the losses corresponding to this it is minor loss and the frictional loss it is going to occur here but minor loss we are not considering. So from here to here the frictional loss or the major loss it is going to occur. So if the inlet of the pipe is there at that time let this be capital H capital H is the head available and at outlet we can observe that the frictional head loss is going say taking place now because of that one so at outlet the value will be lesser by the losses which is which are occurring. So now here it is going to be H minus it is Hf divided by it is going to be H so this is the formula of the efficiency of the power transmission through the pipes. So H it is going to be the head at the inlet of the pipe Hf it is going to be the loss and then again this H it is appearing. So this is the efficiency of power transmission. Now let us see what is the condition for maximum power transmission. So condition for maximum power transmission we are interested in this firstly say power transmission is equal to. So power we are going to have this as it is rho g q and say it is H minus Hf we are having. So this much is the power which has been transmitted and we are interested in the maximum value of this one and in case of this let this be indicated by letter p. So power p and then it is rho g and then say let this q be it is equal to a into v. So it is now pi by 4 then it is d square that is the cross section area of the pipe. Let v be the velocity and then it is H and then minus it is Hf let us write it in terms of the velocity it is 4 f l v square by it is 2 g d. Now see we are going to have this one as v we will take inside say it is rho g into it is pi by 4 then it is d square we will keep it here only and v into it is H minus 4 f l is v cube divided by it is 2 g d we are having. So this one was p. Now what we will do is we will go for now say differentiating with reference to v and equating to 0 for power maximum. So it is now dp by dv is equal to. So here it is d by dv of it is rho g pi by 4 into it is d square into bracket it is Hv minus it is 4 f l v cube divided by 2 g d say this one it is into the whole rectangular bracket but and this is to be equated to 0. Then let us take this constant term outside say it is now rho g pi by 4 and then it is d square and the function of v is only say the inner part it is Hv minus 4 f l it is v cube divided by it is 2 g d and this one is equal to 0. So now as we are having this as one parameter and this one as another parameter as multiplication of these two is equal to 0 but as rho g pi by 4 d square cannot be equal to 0 we are going to have d by dv of Hv minus 4 f l v square by 2 g d as equal to 0. Then taking the differentiation so here d by dv of Hv so it is going to be H minus then this 4 f l then d by dv of it is actually v square otherwise the other terms these are constants so it is 2 g d also constant so it is d by dv of it is now v cube so 4 f l v cube by 2 g d is there it is 4 f l v cube by 2 g d so d by dv of it is v cube and then this one is equal to 0. So here it is H minus it is 4 f l divided by it is 2 g d so this one it is 3 it is v square and then this one is equal to 0. So in case of this one we are now having this as it is 4 f l v square divided by it is 2 g d so except the 3 we are having this one as equal to Hf so it is H minus Hf into 3 is equal to 0 so what it indicates it indicates that Hf should be equal to let us take this to the right side and again 3 will be divided so it is Hf is equal to it is H by 3 so this is what we are getting the condition that is the frictional head loss should be equal to the one third of the total head which is available at the inlet of the pipe next to that one the maximum efficiency we can go for so it is maximum is equal to so it is now we are having this as H minus Hf divided by it is H and let us put the value of this one so it is H minus then it is H by 3 then we are going to have this as H then it is going to be 2 by 3 so lastly which comes out to be 0.6 it is 667 etcetera so the maximum efficiency which we are getting is equal to so it is 0.67 so that is what we are confined to the maximum efficiency which we are getting so above this one the efficiency it is not possible in case of this so totally firstly you can have the revision so it is the efficiency is equal to it is H minus Hf divided by it is H and then the condition for the maximum efficiency it is Hf is equal to it is one third of the head at the inlet and lastly the maximum efficiency is equal to it is 0.67 that is two third it is so we are having the restriction on the maximum efficiency of the power transmission see these are the references thank you.