 Hello students. Myself, Siddhasher B. Tuljapure, Associate Professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Solapur. So, in this session, we are going to deal with the Euler's equation of motion and Bernoulli's equation. The learning outcomes. At the end of this session, students will be able to derive Euler's equation of motion and Bernoulli's equation. The contents are. Firstly, we will have introduction. Then, we will go for derivation of Euler's equation of motion. Then, from Euler's equation, we will derive the Bernoulli's equation. Then, we will go through the different components of Bernoulli's equation. And lastly, the references. Now, let us consider a fluid element moving along a streamline. So, a streamline, we have shown it with the help of this particular, say, line. So, in the moving in the direction, it is s. We are going to consider one fluid element here. So, that fluid element, it is having the length, it is ds. And the next to that one, we are having the cross-section area of that one. It is dA. So, the element, it is cylindrical. So, the cross-section area dA and length of that one, it is ds. Now, here, you can consider the different forces which are acting on this fluid element. So, firstly, say, on the left-hand side here, say, it is pressure p into it is dA. So, this is the force which is acting here on this particular surface of the fluid element. So, it is pdA. So, when we are moving from this particular point to this particular point, that is, on the another side of this element, say, we are having the variation in the pressure. Let that, say, rate of variation be dow p by dow s. And what distance we have moved is, we have moved a distance equal to it is ds. So, the change is corresponding to dow p by dow s into it is ds. This is going to be the total change. Then, this change plus already existing pressure p. So, this will be the total pressure. So, it will be p plus it is dow p by dow s into ds. So, bracket complete. Then to dA, if we are doing, we will get the, say, force which is acting on that particular phase. Say, the fluid elements, these are considered under compression. So, we will have this arrow towards this one like this and this arrow in this direction. So, we are having. Then, we are having the weight of this element acting vertically downward. So, it is now, say, we are going to consider dA and ds. So, dA ds, it is going to give you the volume. Then, multiplied by rho, it is going to give you density. Say, multiplied by density, it is going to give you the, say, mass. Then, multiplied by g. So, it is going to give you the weight. So, weight, it is acting through the CG. Then, say, horizontally, if you are moving to this side, say, from the centers of these two elements, we will have the vertical distance as equal to dz. Then, this inclined, say, length, that is the length of the element we are having. So, this one, it is ds and the angle between ds and dz lines, this is corresponding to theta. Say, if this, say, we are going to consider now the Newton's second law of motion. So, where summation of all the forces is equal to mass into its acceleration we are having. So, we will have the, say, the plus and minus signs to the different forces corresponding to the directions these are. Say, pdA, suppose if it is taken positive. So, the second force, say, pressure force, it is p plus dow p by dow s into ds and the whole bracket multiplied by dA. So, it will have minus sign. Then, we are going to have the component of this weight which is acting vertically downward. Say, it is in the direction, similar to the, say, this p plus dow p by dow s, etc. So, this will have also the minus sign. So, it will be rho g dA dS into this component. It is going to be the cos component because it is adjacent side. So, it is cos theta will be there. So, this is equal to, so, we are having the mass as rho into it is dA dS, that is density multiplied by volume. It gives you the mass. Then, we are going to have the acceleration. So, mass into acceleration on the right hand side, summation of all the forces in that direction. So, it is on the left hand side. So, now the acceleration in S direction is equal to it is dV by dt actually. So, V it is the function of S and it is t. See now, velocity it is function of this direction and it is time t is there. So, now this dV by dt can be written as dow V by dow s into it is dS by dt. Plus, it is dow V by dow t it is. See, we are having this one as again equal to. So, dow V by dow s, we are having it as it is dow dS by dt. So, it is now V. Then plus, it is dow V by dow t it is written as it is. So, dS by dt it is replaced by V. So, that is the only difference between this one and this parameter. So, this dV by dt can be written as this one. So, that you should remember. So, it is with reference to the full derivative and partial derivative etc. So, by using the mathematical say expressions, we can have this dV by dt in another form. Now, think of whether what will be the say value of dow V by dow t if the flow is steady. See the condition, we are having steady flow and you are having the say interested in the interest in the parameter it is dow V by dow t. So, for steady flow what will be the value of dow V by dow t. See, if you remember that say the definition of steady flow, you will come to know this value. See for steady flow, the parameters does not change with reference to time. So, velocity is one of the parameters which is not changing again with respect to the time in case of steady flow. So, if there is no change and velocity it is going to be constant. Dow V by dow t it is going to be 0. So, parameters does not change with reference to the say time in case of the steady flow. That is what you remember and this dow V by dow t if we are putting it 0 in the say earlier equation. See where we are having this dow V by dow t it is here. So, in the acceleration term say two terms were there acceleration parameter. So, second one it is going to become 0 only V into dow V by dow s it is going to remember. So, that a s it is again is to be put here. So, the first equation we are going to have there and then p da here it is plus and here minus p da it will be if you solve the bracket. Then you are going to have this minus dow p by dow s into ds into da. So, here it is this minus sign and this parameter it is going to remain as it is. So, we will have further say the substitution of the parts a s it is now going to be V into dow V by dow s only. And say you are going to have this as minus dow p by dow s into da ds minus rho g da ds into it is cos theta. And is equal to on the right hand side it is mass into acceleration. So, in terms of the say in case of that one we have put the equations here. So, it will be now the say rho da into ds into it is dow V by dow s we are having. So, now in case of these say number of parameters we have cancelled. And what we are now having is say here let us divide this equation by rho ds into it is da. Da ds it is lying everywhere it is da ds da ds we are having here also da ds we are having. And we are now say additionally taking the term rho along with that one. And we are dividing it the this particular equation by this parameter it is rho ds into it is da. So, what we will get is we will get this as minus this dow p by it is rho ds minus we will get this g cos theta is equal to we will get this as V dow V by it is dow s is there. So, in case of this one so here it is rho V it is missing actually. So, it is it will be V dow V by dow s on the right hand side because this will get cancelled. And this parameter it is as it is so cos theta it is lying. So, rho da ds this is getting cancelled. So, here da ds it is getting cancelled but rho parameter is not there. So, we are having the first parameter as minus dow p by rho into ds minus g cos theta is equal to it is V into dow V by dow s. Then here you are going to have somewhat modifications. So, these minus terms you can take this to the right hand side it will be now dow p by rho ds dow s plus g cos theta plus V dow V by dow s is equal to 0. Then making use of this vertical distance and inclined length etc. So, cos theta is equal to you are having this as dz upon it is ds. So, putting the value of cos theta in the earlier equation you will get this as 1 upon rho say it is multiplied by dp by ds. Then plus it is dz by ds plus V dv upon ds is equal to 0. So, this one is nothing but it is dp by rho plus g dz plus it is V dv is equal to 0. So, ds we are avoiding in the next equation and then this one is now written again below. So, it is dp by rho plus dg dz plus it is V dv is equal to 0. This is nothing but it is the Euler's equation of motion. Now, let us go for the integration of this Euler's equation and say get the Bernoulli's equation. So, it is integration of dp by rho plus integration of g dz plus integration V dv is equal to it is constant. So, if the flow is incompressible rho is going to be constant. So, you are going to have this equation as now. So, it is p by rho plus gz plus it will be V square by 2 is equal to it is constant. So, you can rearrange the terms now. So, it will be p by rho g because so it is pressure head we are knowing. So, we are bringing it in that form. So, it is p by rho g plus z plus V square by 2g is equal to constant. So, the velocity if it is taken second it will be p by rho g plus V square by 2g plus z is equal to say it is constant. So, this one is nothing but it is the Bernoulli's equation. So, p by rho g plus V square by 2g plus z is equal to constant. p by rho g is nothing but it is the say pressure head or you can say it as a pressure energy per unit it is weight of the fluid or it is pressure head. Secondly, it is kinetic head V square by 2g and then lastly it is say z is the potential head. So, you can have other terms as say kinetic energy per unit weight you can write in case of z you can write it as potential energy per unit weight. So, all these are energies per unit weight firstly corresponding to pressure second it is velocity and lastly it is say datum. These are the references used for this particular session. Thank you.