Dear representatives, 찍 ස අක්nu ගෙ ඉගන් එය඾ක canoe කබිටි ඔ GEORGE්ඏ අව ගෙඩති ඉගන් කුතමම අඩිmuş rock Squid සම් යුබුසි lien ක් ඤුන්දි අමක අ ෆ දව  අක්�你的 श instantaneous , नाख तूँ셔서 व Aaa  silently the write there are here two halves two symmetric halves and the jet strikes the bucket tangentially that means the direction of the incoming jet velocity is a tangent is in the direction of the tangent to a circle drawn through the centers of this bucket this circle is known the is known as the pitch circle and the water jet glides along the blade flows along the blade or bucket and then comes out well we also recognize that these fluid should come relative to the bucket in a direction exactly opposite to the incoming velocity to have a maximum change in the momentum by the fluid so that a maximum force can be exerted exerted on the bucket but it is not done in practice in practice a deflection of theta is equal to 165 degree that means the direction of the relative velocity of the jet with respect to the bucket is made this is because that the fluid otherwise will strike the back of the vane following the one now the vane is also moving with a velocity u so we also discuss the number of buckets are usually more than 15 and number of fixed nozzles which are the status of the fixed parts of the machines varies from 2 to 6 we have also recognized that there is a splitter reach which divides the water into equal parts through these two halves of the bucket or vane or blade whatever you may say now we come to the force analysis now for force analysis what we have to do we have to make the well the velocity triangle we will have to draw the velocity triangle diagram at the inlet and outlet let us do that now let us draw the vane again in this fashion sorry that this is the vane now this is the inlet velocity v1 now this is the ok this is the velocity u with which it is moving now at the inlet the incoming water velocity and the bucket velocity is in the same line they are collinear so therefore the inlet velocity diagram is like that if this is the v1 that means if this is v1 the inlet velocity of water let u is represented by this, this is u same direction, all are in the same direction. So, therefore, this will be the relative velocity at the inlet of the field. So, therefore, the velocity triangle or the vector is a line, that means the vector diagram is simply a line, this is the v r 1, simply a straight line, because velocities are collinear the same direction. So, therefore, we can write the relative velocity v r 1 is equal to v 1 minus u. There one thing you have to recognize that the tangential velocity of the bucket at inlet and outlet is same. That means if you see this, the tangential velocity that means the same plane where the liquid enters and comes out. So, therefore, the radial location of the inlet and outlet from the axis of rotation remains the same. Now, if we draw the outlet velocity triangle, the outlet velocity triangle will look like that, this is the v r 2, that is the relative velocity with respect to this and this is v 2. We can make this component. So, here you see the outlet velocity triangle shows that the relative velocity of the fluid with respect to the bucket is such that it coincides with the angle at outlet of the bucket. This is for the smooth flow of the water through the bucket. You know that for smooth entry and smooth discharge of the water, the velocity of the fluid with relative to the water should glide along the velocity of the fluid relative to the bucket should glide along the bucket or vane. That means the outlet, the angle of the relative velocity, that is the relative velocity with respect to the bucket should make the same angle with that of the bucket. That means this is nothing but this angle we recognized as 165 degree, it is made. This angle we call it as beta 2. So, therefore, beta 2 is usually 180 degree, 180 minus 165, that is 15 degree usually. It may be of different value. That means the angle of blade at outlet is equal to the angle of the relative velocity at outlet and inlet also, the angle of blade this is 0. So, inlet velocity is having a 0 angle with the tangential direction. That means that both inlet and outlet the relative velocity should match the blade angle. This I discussed earlier also. Now, we know the energy. We know that energy transferred per unit mass between the fluid and the rotor. If I write the energy given by the fluid per unit mass to this bucket will be equal to v w 1 u 1 minus v w 2 u 2. And the general Euler's equation for fluid machines nomenclatures are well explained that v w 1 is the tangential component of the inlet velocity. What is the value of v w 1 here? Here v w 1 is equal to v 1 because v 1 itself in the tangential direction and u 1 is equal to u 2 here. What is the value of v w 2? v w 2 can be found out from the that means it is in the opposite direction to that of the v 1 or v w 1 and this is this much from the velocity triangle. So, with a negative sign because the velocity is in the opposite direction to the incoming velocity. It can be written as v r 2 cos beta 2 this much cos of beta 2 minus u. So, u 1 is equal to u 2 is equal to u here. So, if I substitute these values of v w 1 v w 2 and u as the tangential velocity of the rotor that is u 1 u 2 then we get e by m is equal to v 1 is again v 1 can be written in terms of u as v r 1 plus u because from the inlet velocity vector diagram we see that v 1 is equal to v w 1 that that is equal to v r 1 plus u. So, therefore, we get v r 1 plus v r 2 cos of beta 2 into u clear v r 1 plus v r 2 cos of beta 2 into u well we can write it again that e by m is equal to e by m per unit mass again I am writing v r 1 plus v r 2 cos of beta 2 into u where v r 1 is the relative velocity of the fluid at inlet which is simply the difference between the inlet velocity and the tangential velocity because they are in the same line v r 2 is the relative velocity of fluid at outlet which we can find from the outlet velocity triangle and beta 2 is the angle of the bucket angle of the bucket at the outlet and u is the tangential velocity of the bucket at the at its center at its center where the jet strikes. Now, we see that in the bucket if you see this figure why the relative velocity at the inlet and outlet changes why the relative velocity at the outlet changes from that at the inlet why you please recall our earlier discussion that the relative velocity at outlet will be different from relative velocity of inlet for what reasons what are the reasons for which the relative velocity of fluid at outlet will be different from that at inlet this is because of what here the fluid is throughout atmospheric pressure fluid is throughout atmospheric pressure open jet. So, there is no question of change in pressure. So, pressure remains constant throughout the flow through the bucket. So, why the relative velocity will change this is only because of friction only because of friction you consider the relative velocity in this way that as if the blade is fixed and fluid flows past that means fluid flowing past a fixed blade or a fixed surface then why the velocity will change only because of friction if the pressure remains constant. So, there is only way the velocity can change is the friction if the pressure remains constant that is another important thing. So, therefore, v r 2 changes from that of v r 1 the value of v r 2 changes from that of v r 1 because of friction and the role of friction is to reduce the value of v r 2 from v r 1. So, therefore, we can consider or we can write v r 2 in this fashion some coefficient k times v r 1 where k is less than 1 this takes care of the friction that means v r 2 is reduced from v r 1 by a factor less than 1. So, that we can express v r 2 in this fashion where k is a factor less than 1 and takes care of the friction in the bucket. So, therefore, we can write v r 1 into 1 plus k cos of beta 2 into u well and again I just substitute v r 1 as v 1 minus u from the inlet velocity triangle you can recall cos beta 2 into u. So, this is the energy per unit mass that is being transferred or given by the fluid to the bucket I think there is no difficulty it is extremely simple. Now, the total energy given per unit time that is the rate of energy that is being transferred by the fluid to the rotor will be this multiplied by the mass flow rate because this is the energy given per unit mass if we multiply this if we multiply this with the mass flow rate of the fluid then you will get the rate of energy transferred by the fluid to the rotor. Now, what is the mass flow rate of the fluid it is density rho and the q is the volume flow rate. So, this is the mass into v 1 minus u into 1 plus v 1 minus k cos beta 2 into u. So, this is the expression for the rate of energy that is given by the fluid to the bucket. Now, we define a efficiency eta known as bucket efficiency which is very important bucket efficiency or sometimes we define it as wheel efficiency wheel efficiency because buckets are mounted on wheels wheel efficiency. How is it defined now the power that is being developed by the wheel due to the motion of the fluid is this much. So, this comes as the useful work or useful energy from the bucket cos beta 2 into u. What is the input to the bucket what does bucket receive as input energy you see here the bucket receives the input energy in the form of kinetic energy of the fluid that means it is the kinetic energy of the fluid total kinetic energy of the fluid that means mass times v 1 square by 2. So, it is the rate at which the energy is received by the bucket at its inlet that is the rate of kinetic energy that is being received by the bucket at its inlet due to the incoming flow of water and the numerator is the power developed by the bucket per unit time. So, therefore, we get an expression from this about this eta rho q rho q cancels out. So, we can write 1 plus k cos beta 2 into you write in a fashion it is twice well. So, this will be well u square minus v 1 square. So, u by v 1 into 1 minus u by v 1 of course, a 2 will be there well. So, this is the expression for the bucket efficiency or wheel efficiency. Now, wheel efficiency efficiency is a dimensionless parameter. So, it is expressed in terms of the dimensionless parameter. Now, what is the physical significance of this efficiency physical significance of this efficiency is that how effectively the bucket can convert the incoming kinetic energy which he receives in the form of power developed. Now, you can ask me sir why is the discrepancy is it due to friction? No, even if you make k is equal to 1 you see this is the value of eta is not always 1. That means, it is not the friction it is not the friction, but it is it depends upon the deflection of the jet by the bucket. So, that the absolute kinetic energy at the outlet or the discharge should be minimum, because the numerator is equal to the change in the kinetic energy of the jet. That means, some energy in the form of kinetic energy is always rejected at the discharge. So, how effectively it can utilize this kinetic energy in the form of work. So, that the discharge kinetic energy which is loss is which is a loss is minimum. So, that is the physical concept of the efficiency and it becomes a function function of pertinent dimensionless parameter. One is the k that is the friction which takes care of the friction another is the bucket outlet angle and the ratio of u by v 1. That means, the speed of the bucket and the speed of the incoming water jet or the inlet velocity of the water jet. Now, for a fixed bucket bucket with a fixed geometry k and beta 2 are fixed. So, therefore, the efficiency of the bucket is a function of the ratio u by v 1, which becomes a pertinent dimensionless parameter representing the operating conditions. That means, it is a function of u by v 1. Now, if you plot this functions you will see you will get a curve like this. If you plot this function eta versus u by v 1 correct the parabola you get a and exactly the maximum efficiency is attained when this value is 0.5 correct 0.5 and this become equal to 0 when u by v 1 is equal to 1. It is very simple that if you make here sorry I can give a suffix or subscript w usually given to denote that it is the efficiency of the wheel d eta w d u by v 1. That means, if I differentiate this eta with respect to u by v 1 for a fixed blade fixed geometry of the blade we get that it is 1 minus 2 u by v 1. If I make sorry I should not write that if I make this is equal to 0 it becomes 1 minus 2 u by v 1 is equal to 0. Because the expression of this is not 1 minus 2 u by v 1 it is multiplied with a constant factor 2 into 1 plus k cos beta 2 if we make this is equal to 0. So, it is obvious that u by v 1 is equal to half. And we can check the maximum condition analytically by checking that or noticing this fact that d u by that second derivative is less than 0 or simply by plotting the curve for the expression we will see this expression has a maximum. So, therefore, we see when u by v 1 reaches half the blade efficiency is maximum. Again when u by v 1 is equal to 1 blade efficiency is 0 no power will be developed as you know that when the blade velocity is exactly equal to the incoming velocity of the water water cannot reach the blade. So, flow will be 0 and more than that the blade will recede away from the jet before it strikes the blade. So, therefore, it again falls at 1. So, this is the maximum value. Now, in actual cases we define a overall efficiency eta o which is equal to the power that is being delivered at the shaft divided by the input power input energy rather input energy to the wheel input energy to the turbine pelton turbine before coming to this I like to tell you that the actual power developed by the wheel depends upon friction not only the fluid to blade friction there are other frictions frictions in the bearing in the mechanical system and the windage losses that is the friction with the air and the parts of the fluid machines and these losses also increase with the increase in speeds. So, therefore, it has been found practically that the maximum efficiency becomes lower than the theoretical efficiency calculated and it attains a maximum value always lower than that and also becomes 0 even if the bucket velocity is less than v 1 and maximum is also attained at a value which is less than 0.5 the value is 0.4 c. So, this is the theoretical curve theoretical curve and this is the actual one. That means in actual case considering the frictional losses we see that the maximum efficiency occurs at a value of u by v 1 that is the ratio of the bucket speed to the incoming water speed equal to 0.46. Now, we come to the definition of eta o that is overall efficiency it is overall efficiency how you define now overall efficiency overall efficiency of a turbine as you know that is the p at shaft coupling that means this is the shaft power final power. So, this is less than this quantity obviously because this is the power developed by the rotor. So, this gets reduced by the mechanical losses that is losses in the mechanical system. So, that at finally, at the shaft end the shaft coupling we get this power which is the numerator of the overall efficiency expression. Now, input energy to pelton turbine now you tell me which should what should be the input energy to a pelton turbine what should be the input energy you see this figure what is the input energy to the turbine. If we consider the nozzles at the part of the turbine or the fluid machines which are the fixed parts that are known as stator and this is this wheel with the buckets at the moving parts known as rotor and both the nozzles and the moving wheels with the buckets from the fluid machines. So, what is the input energy to this machine input energy is what where is the input to the machine at entrance to the nozzle which means the energy at entrance to the nozzle. So, what should be the energy at entrance to the nozzle energy here this is the pressure energy of the fluid. So, this we can calculate like this if we consider that the what happens in practice that the liquid from a very high head it is stored in a high head the value of which is h from a fixed reference datum then it comes to the nozzle and if we consider that the friction loss is h f while flowing through this this line this pipeline which is known as pen stock in our terminology of fluid machine pen stock. If h f is the loss of energy per unit mass. So, therefore, energy per unit mass which comes at the inlet to the nozzle that means this is the input point at inlet to the machine is equal to h minus g h minus h f per energy per unit mass energy per unit mass g h minus h f or you can consider in terms of head that is h or rather you consider h f is the loss of it energy per unit weight. So, this will be g into h minus h f and h minus h f is the energy per unit weight that means the head that means this is the head at inlet to the machine and this is the head available at the reservoir at a high. So, these are known as gross and net head. So, the head this h which is the energy of the fluid at the reservoir at a great height is known as the gross head which is the gross energy per unit weight and this energy is purely in the form of potential energy these are the terminology these are extremely simple, but I tell it with a categorically because these are the terminology used and this h minus h f that means the energy per unit weight at entrance to the fluid machines that means at inlet to the nozzle is known as the net head. So, net head that means energy per unit weight available at the inlet to the nozzle which is in the form of both pressure and velocity energy because the flow velocity is there through the pipeline leading from the reservoir to the nozzle of the machine. So, therefore, this is the main energy main energy per unit weight coming to them. So, therefore, the input energy will be this one. So, therefore, if we write the overall energy eta o. So, eta o will be the power total power at the shaft coupling and the input energy that means the rho q g into h minus h f where h f is the friction loss in the pipeline leading from the reservoir to the nozzle inlet that means the inlet to the fluid machines. Now, nozzle is a part of the fluid machine which converts the energy into kinetic energy. So, therefore, if I want to find out the kinetic energy at the inlet to the bucket at the inlet to the bucket if I want to find out the if I have to find out the kinetic energy at inlet to the bucket then this kinetic energy can be found out by application of the Bernoulli's equation by the application of the Bernoulli's equation between the nozzle inlet and outlet. That means if I apply the Bernoulli's equation between the nozzle inlet and outlet simply per unit mass the total energy here or per unit weight total head here is h minus h f and that becomes is equal to v square by 2 g if I write the Bernoulli's equation for an inviscid fluid this total head is the head above the atmospheric pressure. So, therefore, here the pressure is 0 and we take the datum as the through the central line of the buckets that means the plane at where the jet is striking the buckets. So, therefore, it is very simple that v 2 becomes equal to 2 g that means the net energy at the inlet is exploited fully in terms of the velocity, but it is multiplied with a term known as coefficient of velocity which is same as that term k we discussed regarding the flow through flow of the liquid through the bucket which took care of the friction between the liquid and the bucket surface here also c v takes care of the friction in the nozzle which reduces the velocity from that of 2 g h minus h f which has been deduced by considering the fluid to be inviscid and by the application of Bernoulli's equation. So, therefore, you will have to know very carefully that v 1 is c v 2 g h minus h f. So, therefore, v 1 square by 2 is equal to c v into c v square rather into g into h minus h f. So, you see there is a relationship between the kinetic energy at the inlet to the bucket and the energy at the entrance to the nozzle. So, this is they are not equal because of this c v quantity c v square because of the friction. So, these are the stages. So, initially we get the energy at the entrance to the nozzle or entrance to the machine in the form of h minus h f per unit weight. This is being converted to kinetic energy v 1 square by 2 that means kinetic energy per unit weight that means kinetic head or the dynamic head through the expansion of the flow or the through the flow of the liquid through the nozzle and we get this head which is less than the head at the entrance by the factor c v square which takes care of the friction in the nozzle. And the shaft power is the power which is developed at the shaft coupling. So, you see this is the power that is developed by the fluid to the rotor. The rotor extract this power because of the deflection of the fluid flow through it and this power is getting reduced to p because of mechanical friction. So, these are the stages how the energy is transferred from the inlet to the fluid machines inlet to the pelton wheel down to the shaft coupling final power. So, I think today I will finish here stop here. So, next class I will discuss the specific speed. Thank you.