 Hello, students. So, myself, Mr. Siddeshwar B. Turzapure, associate professor, department of mechanical engineering, Valchand Institute of Technology, Solapur. So, in this session, we are going to deal one topic from the section, it is fluid dynamics and the name of the topic is modified Bernoulli's theorem, learning outcome. At the end of this session, students will be able to explain difference between the Bernoulli's theorem and modified Bernoulli's theorem. The contents are the first one, it is the assumptions of the Bernoulli's theorem. Then we will go for Bernoulli's theorem. Then fulfillment of assumptions in real-world situations. Then it is need of modifications to Bernoulli's theorem. Then lastly, it is the modified Bernoulli's theorem and references we are going to have. The first one, it is the assumptions of the Bernoulli's theorem. Just we are going to revise those assumptions as we have dealt this in the first section. The fluid should be ideal. We are knowing that the ideal fluids are the fluids which are having zero viscosity, that is the fluid will not have any viscosity. So, in case of this world, we are going to have number of fluids are existing and then all these fluids are having to some extent the viscosity. The viscosity of the fluids which are existing may be to some extent lesser or it will be more but they are having the viscosity because of which they are called as real fluids and in case of the ideal fluid, ideal conditions as these are not existing, so the fluid having zero viscosity, it is not existing at all. So, in case of the fulfillment of this one, we are saying that ideal conditions as these are not existing, so ideal fluid is not there and in case if we want to perform the experiment or you have to go for the analysis by making use of the Bernoulli's theorem, etc. So, in case of that one, so as the fluid is not ideal, we are going to come across some differences in this equation. So, now secondly, the flow should be in case of the flow should be the flow should be steady and then the second one it is flow should be incompressible then flow should be irrotational. Now, let us see what happens to the steady incompressible and irrotational type, the flows. Now, in case of the flows which are existing in the real world, in some cases the flow might be steady. Steady flow refers to the say the constant parameters with reference to time at a point it is. Now, suppose the flow is there at one particular point if when we are fixing the pressure gauges, etc. where we are measuring the pressure and then we are measuring the pressure now and after some time. So, number of times when as the time passes we are going to measure the pressure many times and even though the time is different the pressure is remaining constant. Similar to that one at a point if the velocity is remaining constant discharge is remaining constant. So, like this number of parameters if they are having the constant values at a point with reference to time we are going to call this type of the flow as steady flow. So, steady flow can exist and this condition even in case of the practicals etc. it can be satisfied. The second one it is the flow should be incompressible. In case of the liquids we are going to have the liquids are incompressible. So, the flow associated with the liquids these are incompressible so next to that one we are going to have the flow should be of irrotational type. In case when we are having the fluid particles travelling from one position to another position if the fluid particles are rotating about their own axis or mass centers. So, these are called as rotational flows. If they are not rotating these fluid particles are not rotating about their mass centers or axis we are going to call that type of the flow as irrotational flow. To some extent the flow may be irrotational and to some extent the flow may be rotational. So, it is not perfectly possible for us to go for the irrotational type the flow. So, the Bernoulli's equation also we will see. So, if the above conditions or the assumptions these are satisfied. So, in case of that one the Bernoulli's theorem can be represented in the form of the equation as it is p by rho g plus v square by 2g plus z is equal to constant. We are knowing that this p by rho g it is the pressure head v square by 2g it is the kinetic head and the last parameter that is z it is nothing, but it is the potential head or the datum head. The summation of these three it remains constant. Now, when we are say going for the assumptions how many assumptions we have satisfied and how many are remaining. So, the first one that is the fluid should be ideal. So, it is not satisfied it cannot be satisfied any time the flow may be steady. So, it may be it is in our in our hand we can go for the steady flows then it is incompressible we can go for the liquids and then lastly the flow it is irrotational it is not satisfied. Now, think of the fluid flow through a pipe which is having the different diameters at different sections it is. Now, the section number one say at that section we are having the larger diameter then it continues for some time then the further we are going to have the reduction in that diameter, but that reduction it is sudden one you can say that the sudden contraction of that pipe it has occurred and the diameter now it is lesser than the diameter at the section one one and now the fluid flow it is occurring and now we are having the fluid flow from the section one one to the section two two the lines just you can observe. So, these lines curved lines these are indicating the fluid flow and now you can observe say after the sudden contraction we are having some section where the diameter of that particular fluid flow it has reduced that C C we have said it is the vena contracta portion where the cross section area it will be smallest one and then further it goes to the section two two on the right hand side and the flow is from the left hand side to the right hand side. Now, in case of this one you have to think now whether the total energy at section number one and section number two it is same if it is same then it is over or if it is suppose not same in case of that one we might have the total energy at one of the sections greater than the total energy at the another section. So, if the difference is there the which one is larger that is the total energy where it is more at section one or section two see the answer of this one. Now the total energy at section number one it is more than the total energy at section number two that is the summation of the pressure head kinetic head and the datum head he at the section number one it is more than the summation of the pressure head, atom head and the kinetic head at the section number two. Now, we can have the explanation of this one. Now, first assumption we can see. In case of the first assumption, the fluid it was ideal. So, the fluid flow whichever we have shown now, how it is existing and now all real fluids they are having viscosity. So, some energy will be lost when the fluid is flowing from section number one to section number two and in case of this one the loss due to it will be the viscosity. So, energy loss due to the viscosity it is one of the reasons why the Bernoulli's theorem needs to be modified. In case of the second one, the energy loss due to the friction between the inner surface of the pipes and the liquids we can consider. Now, when we are going for the losses the losses these are of two types when it is a major loss and the second one it is the minor loss. The major loss refers to the loss due to the say friction between the inner surface of the pipe and the liquid which is flowing. So, this value it is larger. So, because of which it is called as a major loss. Now, the second loss you can observe. So, it is say the loss due to the say different reasons like say the pipe it is connected to the tank then sudden expansion or sudden contraction has occurred. In the present case you might observe that the sudden contraction it has occurred because of which we are going to have the energy loss. So, this is minor loss. So, other losses are also there which are also minor and they are due to the say fittings etcetera the valves etcetera. So, like this the losses these are coming into place when we are going for the real world situation and these losses needs to be considered and then the equation needs to be modified. Now, the same equation how it can be written when it is modified. So, it is the at section 1 we are having this P 1 by rho g plus V 1 square by 2 g plus z 1 is equal to and x n section 2 it is say P 2 by rho g plus V 2 square by 2 g plus z 2 and now on the right hand side we are going to add the head loss it is these are the references. Thank you.