 I welcome you to this session today we will discuss the force exerted and the power generation the force exerted on the runner of a francis turbine and the power generation last class we discussed the head across the runner or across a francis turbine what is meant by net head across a francis turbine or the head or the work developed by the francis turbine and the variation in the head during the flow in the francis turbine today we will discuss the force exerted by the water on the francis runner and the corresponding power generation let us see a typical francis runner first analyze let us see how a francis runner blade if you look a sectional view is like this this is the inlet if I draw the inlet velocity triangle it will be like this let me draw the inlet velocity triangle now let this francis runner is moving is rotating let this is the rotor this is the inlet this is the outlet this rotates is the r p m so this is the direction of velocity at the outlet also we can draw the diagram like this so this is the rotor velocity at the inlet u 1 this is the relative velocity of the liquid with respect to the rotor v r 1 and this is the absolute velocity v 1 the outlet velocity triangles look like this this is the velocity v 2 this is the outlet relative velocity at the outlet v r 2 and this is the rotor velocity u 2 now if we take a particular vane this is like this the section of the vane is like this through which the water flows between two vans so we can draw the velocity vector diagram of the velocity triangles at inlet and outlet like this now the function of the guide vane is to direct the velocity in such a way that the relative velocity always makes the same angle as that of the angle of the vane at the inlet that means this angle if we denote at beta 1 is the angle which the vane at its inlet makes with the tangential direction the u 1 is the rotor velocity in the tangential direction the linear rotor velocity which depends upon the r p m of the rotor and the radial distance at this inlet point from the axis of rotation that this is the axis of rotation when the fluid flows through the runner vane and comes out the runner vane is designed in such a way that fluid comes out without any tangential component it is perpendicular to the tangential direction so therefore this is v 2 and this is the rotor velocity or the tangential velocity of the rotor at the outlet which depends upon the rotational speed and the radius radius at the outlet from the axis of rotation and this is the corresponding relative velocity of the liquid with respect to the rotor and this also should match with the vane angle at the outlet that means if I denote this as beta 2 this is the vane outlet angle now the flow through the runner you have to understand the main direction of flow through the runner is radial and tangential at the inlet radial and tangential at the inlet but while it flows out of the runner the tangential velocity is almost diminished there is no tangential velocity so the flow becomes radial or little axial this is very important that means if you look a turbine for example the turbine runner in a horizontal plane with the shaft at the shaft being the vertical so the inlet is in a horizontal plane in a direction such that it has got both tangential component and the radial component while flowing through the rotor it comes out mostly in a different plane with a radial flow and an axial flow and ultimately it is turned completely in a axial direction which is vertical downward in case of a vertical shaft that is along the draft so you have to understand this way it enters in a direction which is combination of radial and tangential this is known as a mixed flow radial and tangential so the flow through the runner is usually termed as mixed flow which is which has got both radial component and the tangential component while it flows out of the runner the tangential component is reduced almost to zero it is discharged mostly in the radial direction and axial direction little axial component is there that means when it comes out of the runner it has got almost radial component without any tangential component which is immediately turned by the pipe into the axial direction at the inlet to the draft so this you will have to understand this is the nature of the flow now if you look into this diagram we see that therefore this is your let this is v r 1 so this is your v f 1 as we know that is the flow velocity that means this is the velocity in the direction perpendicular to the tangential direction that means it is radial direction similarly here the flow velocity itself is the absolute velocity because there is no tangential component of the flow at the outlet so v 2 is equal to v f 2 now one of the main design constraint of the flow is v f 1 is equal to v f 2 which is the flow velocity flow velocity this component is sometimes referred to as meridional velocity meridional that means the velocity component perpendicular to the tangential direction the design is made in such a way that the meridional component or the flow velocity component at inlet becomes equal to that at outlet and ultimately this becomes equal to the axial velocity v a at the inlet to the draft now let us see that the typical velocity from the typical velocity tangential what is the power that is being developed or that is being given by the fluid to the runner as we know that the energy per unit mass is given by the expression from the Euler's equation v w 1 u 1 minus v w 2 u 2 what is v w 1 in this case this is our v w 1 this is our v w 1 which we can write in our case v w 1 from this triangle v w 1 can be let this angle is alpha 1 where alpha 1 is the angle of the guide vanes at its exit because the incoming velocity the direction of the incoming velocity is in the direction of the angle of the guide blade guide vanes are fixed. So, therefore, when fluid coming out from the guide vanes its direction of velocity coming from the guide vanes that means there is no relative velocity relative velocity is the actual velocity or absolute velocity because guide vanes are fixed is equal to the angle of the guide vanes that is outlet, which is the angle of the inlet velocity absolute velocity with the tangential direction that is alpha 1. So, we can write v w 1 from this triangle inlet velocity triangle as v f 1 cot alpha 1. Now, our main aim will be to express the energy per unit mass developed in terms of the angles of the vanes at inlet and outlet v f 1 cot alpha. Then we can write similarly, what is the value of u 1 u 1 u 1 is this one this one is u 1 u 1 we can write v f 1 cot alpha 1 that means this one minus this part minus v f 1 in terms of v f 1 if I write cot 180 degree minus beta 1 cot of 180 degree if I define beta 1 this angle if I define this angle by beta 1. So, this becomes v f 1 cot of alpha 1 this is minus cot beta 1. So, this becomes plus now in this case we see the design is made in such a way that the fluid has 0 tangential motion tangential 0 component in the tangential direction 0 velocity component of velocity in the tangential direction which means the whirling component of the velocity at outlet is 0 that means this term is 0. Now, you can understand very well if this term is made 0 we get the maximum work or the maximum power developed by the turbine because this becomes 0. So, this expression becomes maximum. So, in this case we can write e by m is simply v w 1 u 1 and if we substitute this value of v w 1 and this value of u 1 this becomes equal to v f 1 square cot alpha 1 to cot alpha 1 plus cot beta 1 well. So, this is the expression for the power generation or the energy per unit mass rather we can tell this is the energy per unit mass. So, energy per unit mass that is being released by the fluid as it flows through the vanes or runner of the Francis turbine is given by v f 1 square cot alpha 1 cot alpha 1 plus cot beta. So, it becomes a function of the inlet angles of the absolute velocity or the exit angles of the guide vanes and the inlet angle of the runner along with the flow velocity which determines the rate of flow through the runner. Well, now we like to find out what is the efficiency eta h what is the efficiency hydraulic efficiency. Now, what is hydraulic efficiency as you know the hydraulic efficiency the numerator is the power developed power developed or rather I can write the head developed rather I can write the head that is the energy per unit weight head developed divided by the head available. That means, the energy available energy available at the nozzle entrance. Now, power developed we know that is e by m if I express this per unit mass. So, this is the energy developed per unit mass. So, what is the energy available per unit mass what is this energy available per unit mass energy available please tell me just I will ask you energy available per unit mass we have to find out. Now, if we consider that friction to be 0 then the energy available at the inlet to the runner becomes exactly equal to the energy available at the guide vanes because if we neglect any frictional loss the total energy remains same. So, what is that energy at the inlet to the guide vanes or at the inlet to the runner how do you find out please can we tell this v square v 1 square by 2 just like the pelton will pelton will we told that it is v 1 square by 2, but in this case if I have to find out the energy available at the runner inlet what is the value of the energy available at the runner inlet how can I find out yes very good it is p 1 by rho g plus v 1 square by 2 plus the data net if at all any that means, that depends upon our choice of the reference data that means, it comprises kinetic energy pressure energy and the potential energy very good, but how to find out it is true that this quantity corresponds to v 1 square by 2 plus p 1 by rho 1 per unit mass basis and g z 1. So, these two quantities were ascended in case of pelton this is because the entire energy was in the form of the kinetic energy, but here at the entrance to the runner the pressure energy and the potential energy is there. So, to find this we have to write in this fashion you can find this in this fashion this energy available per unit mass can be written as the energy develop plus the energy lost that means, v 2 square by 2 that means, e by m plus v f 2 because v 2 is v f 2 v f 2 square by 2 this is this is because the this is the energy which is going out of the runner. So, therefore, this is the energy which is extracted by the runner. So, if you discard the friction we can tell this plus this this is the total amount of energy which the runner received at its inlet that means, this is the energy which comprises the kinetic energy pressure energy and the potential energy understand. So, therefore, tactfully we find this without going for evaluating what is the pressure energy or what is this term p 1 by rho 1 we just simply add the amount of energy with which the turbine is going. So, therefore, v 2 square by 2 e by m. So, if I add this thing. So, we can get this is the amount of energy which was at its inlet because this amount of energy and the energy develop equals to the total amount of energy that the turbine received or the runner received. So, therefore, I can write tactfully eta h is equal to what is that e by m divided by e by m plus v f 2 square well by 2. Now, if I substitute again if I write this thing well if I write this eta h is equal to e by m e by m well divided by e by m plus v f 2 square by 2 then it is ok. So, here lies the concept now is algebraic steps. Now, if I substitute this expression for e by m and consider that v f 2 is equal to v f 1. So, what I will get v f 1 will cancel. So, I get 2 cot e by m e by m plus v f 2 square by 2 then it is ok. So, here lies the concept now is algebraic steps. Now, if I substitute this expression for e by m and consider that v f 2 is equal to v f 1. So, what I will get v f 1 will cancel. So, I get 2 cot alpha can you see this line 2 cot alpha 1 into cot alpha 1 plus cot beta 1 that means e by m is v f 1 square cot alpha 1 into cot alpha 1 plus cot beta 1 that I am substituting in place of e by m. Then therefore, what I will be getting 2 that means v f 2 square plus 2 of this quantity again cot alpha 1. Oh this has cancelled good good good because v f 2 is equal to v f 1 which is cancelled very good 1 plus very good cot alpha 1 plus cot beta 1. This is because v f 1 square and v f 1 square can say. So, this can be written in a manner 1 minus 1 by that means if you add 1 and subtract 1 in the numerator 1 plus 2 cot alpha 1 into cot alpha 1 plus cot alpha 1 into cot beta 1. So, this is the expression conventional this is the most useful and popular expression of the hydraulic efficiency or the runner efficiency in terms of the angles of the vane at inlet and the guide vane angle at the outlet. That means alpha 1 is the angle which the absolute velocity makes with the tangential direction and this is the angle which the relative velocity makes with the tangential direction which means alpha 1 is the angle of the guide vane at the inlet at the at this outlet. So, all the angles are referred with respect to tangential direction and beta 1 is the angle of the runner at the inlet. So, this is the useful expression for hydraulic efficiency or runner efficiency this is the hydraulic efficiency high hydraulic efficiency. Or runner efficiency provided or runner efficiency well or runner efficiency provided the friction is neglected provided the friction is neglected. That means the runner efficiency becomes the hydraulic efficiency friction neglected in both the guide vane and in the runner blades. Now, next we come to the expression of degree of reaction well. R degree of reaction degree of reaction how did you define degree of reaction in a turbo machines or hydraulic machines please tell me the degree of reaction is the change in energy due to please tell me change in energy change in energy exactly change in energy due to static head in the rotor that is most important static head in rotor divided by the total change in change in total energy. That means change in total energy is the power developed change in total energy. That means it is true that change in total energy is e by m that is the energy per unit mass which we get as power that is the change in total energy of the fluid. But what fraction of it is change due to static head that means due to the change in the relative velocity and change in the centrifugal head that means in fact it is the change in the pressure of the liquid pressure energy of the liquid. And the change in the pressure energy of the liquid takes place because of the two things we have already recognized earlier this very important that why the pressure pressure will change in a liquid and the liquid flows if there is no swirling or tangential velocities. So, pressure changes only because of the change in the velocity of the flow velocity of the liquid and this is with respect to a fixed duct. So, therefore when the duct is moving or the liquid flows through a passage where the solid has a motion that means there is a slip velocity between the liquid and the solid. So, therefore in that case what happens that it is the relative velocity whose change will cause a change in the pressure. Number two if along with that liquid possesses a swirling motion or the motion in the tangential direction which comes because of a rotation in the passage rotation of the duct I gave an example. Then what happens if the liquid flows in such a way that there is a change in its radial location from the axis of rotation in course of its flow. Then a centrifugal head is either released or imposed on the fluid for which there will be a change in its pressure energy or a change in the static pressure. So, therefore change in the pressure energy either released or gained depends upon two factors change in the relative velocity and change in the tangential centrifugal head that means change in the tangential velocity. So, when the relative velocity in decreases then there is an increase in the pressure in a turbine the relative velocity increases. So, there is a decrease in pressure or decrease in pressure energy. Similarly, in course of flow the centrifugal head is also decreased because the fluid comes from a higher radial location is a radial inward flow higher radial location from the axis of rotation to a lower radial location. So, these two quantity cause constitutes the change in its static head. Now, it is difficult to evaluate the change in the relative velocity that you know that v r 2 square minus v r 1 square and also the change in the centrifugal head that is u 1 square minus u 2 square of course, this part is simple, but without doing that what we do change in energy due to static head can be found out if we deduct from the total change in energy the change in the dynamic head which is simple. That means we can write this as e by m minus that is the change per unit mass minus this quantity is simple the change in the dynamic head. That means this part is the energy change due to static head that means this comprises the change in the pressure head due to both the change in relative velocity and change in the centrifugal head because of the change in the rotor velocity from inlet to outlet. So, now it becomes very simple that we can write this in a fashion 1 minus v 1 square minus v 2 square well by 2 into e by m. Now, we substitute the value of e by m again. So, let me write r is equal to 1 minus v 1 square minus v 2 square well 2 e by m well if we again see the value of e by m if we again see the value of e by m is v f 1 square cot alpha 1 into cot alpha 1 plus cot beta 1. Then what we can write 1 minus v 1 sorry v 1 square minus v 2 square by 2 v f 1 square cot alpha 1 into cot alpha 1 plus cot beta 1. Now, v 2 is equal to v f 2 and that is equal to v f 1 why if you see the velocity triangles that we see the design constant is that v f 1 is v f 2. So, v f 2 is equal to v f 1 now if it is so then this quantity v 1 square minus v 2 square. Therefore, I can write here v 1 square minus v 2 square can be written as v 1 square minus v f 1 square which can be written as v f 1 square cot square alpha 1 why because from this diagram v 1 square minus v f 1 square is v w 1 square this one and v w 1 is v f 1 cot alpha well. So, therefore, we can write that v w 1 square is v f 1 square cot square alpha 1. So, v f 1 square cot square alpha 1. So, simply we can write 1 minus therefore, v f 1 square will cancel out 1 by 2 cot alpha 1 well into this simple algebraic steps only concepts are there when we write this term numerical numerator of this term degree of reaction plus cot well very good cot alpha 1 by 2 cot very good because v f 1 square will cancel. So, cot alpha 1 will not be there in the denominator at all very good happy. So, if you write this v f 1 square will cancel and cot square alpha 1 will be cancel to give cot alpha 1. So, 2 I am sorry. So, 2 cot alpha 1 plus cot beta 1. So, again let me write the expression 1 minus cot alpha 1 divided by I think this is the degree of reaction r. Now, we come to the deduction of specific speed specific well specific specific speed n s t if you recall the definition of specific speed for a turbine what is its value n s t whenever you will talk about specific speed you will always refer to the dimensional specific speed until and unless it is told as non dimensionless non dimensional specific speed we will always refer to as dimensional specific speed. So, what is the value for a turbine n p to the power half very good h to the power 5 by 4. In the similar way as we did for pelton wheel we will find out first we will find out power in terms of the hydraulic efficiency well and the head available row q q is the volume flow rate g h here by definition h is the head available at the inlet to the turbine. That means this is the head available to the turbine. So, by definition h is like that. So, here I will write in terms of the definition itself row q g h eta h well. So, if I substitute this value then I will get n s t is equal to n p to the power half means eta h row q g to the power half h I will take out because h to the power 5 by 4 is there in the denominator. So, therefore, I will write h to the power minus 3 the manipulation that h to the power half will come out and h to the power 5 by 4 which will give h to the power minus 3 by 4. Then again with the same technique you can write we can write the n as the rotational speed u 1 divided by pi d 1. That means by equating the tangential velocity of the rotor at its inlet with the rotational speed I can write u 1 by pi d 1 it can be written at the outlet also u 2 by pi d 2, but I write at the inlet from which I get n is equal to u 1 by pi d 1 and u 1 if you recall the value I show you that we got the value of u 1 earlier u 1 is v f 1 cot alpha 1 plus cot beta 1 this is from the velocity triangle this is u 1. So, this u 1 is v 1 v f 1 cot alpha 1 minus v f 2 v f 1 sorry cot alpha 1 cot beta 1. So, therefore, v f 1 cot alpha 1 plus cot beta 1 this we already derived. So, you just substitute this value of u 1 as v f 1 cot alpha 1 plus cot beta 1 divided by pi d 1. Now, you substitute this value in this n s t that is the specific speed. So, this becomes equal to u 1 that is v f 1 v f 1 cot alpha 1 plus cot beta 1 divided by pi d 1. So, this is the value of u 1 that is v f 1 into eta h rho q g to the power half and h to the power minus 3 by 4. Now, this h we can write as what that is the head available if you remember we found this head available as now if you see this this is the head available I do not write again because this is the work develop. That means this is the change in the energy in the turbine runner and this is the energy with which it is going out. So, this is the denominator of this expression of the hydraulic efficiency that is the head develop that means e by m plus v f s square by 2. So, which is again if we write then we get that h is equal to by its definition h is the head develop which is e by m we recognized earlier plus v f 2 square by 2 what is e by m we derived earlier please head available that is the head available yes head available head this is head available this in this definition this is the available head n p to the power half h to the power 5 by 4 this is the available head this is these are in all the this is the this is the power develop this is the head available. These are all out input parameters or you can tell these are the operating conditions that are to be specified for a turbine this is the power develop this is the head available correct. This is the by definition head available what is e by m please tell me the value of e by m which we derived earlier e by m is v f 1 square cot alpha 1 cot alpha 1 plus cot beta 1 and v f 2 is v f 1. So, v f 1 square by well. So, that becomes 2 v f 1 square 2 v f 1 square 1 plus cot alpha 1 into cot alpha 1 plus cot beta 1 divided by 2 2 v f square plus v f 1 square. So, if I take. So, half will come here. So, 2 v f 1 square or 2 we have take v f 1 square as common. So, 2 will be well. So, 2 very good divided by 2 divided by 2. Yes sir that is all right that is all right that is all right v f square divided by 2 v f square divided by 2 just let me write this is equal to 2 v f 1 square. Let us write this one if you take v f 1 square as common then half 2 v f 1 square as common then what will happen. If we take 2 v f 1 square common then it is all right 1 plus why half plus because if I take v f 1 square common then what will happen that is half plus cot alpha 1 well cot alpha 1 plus cot beta 1 ok 1 plus half. So, that we can write this is ok v f 1 square into what we can write 2 plus same ok all right we keep it like this do not worry we keep it like this v f 1 square common 1 plus half plus half plus half plus this all ok. So, therefore v f 1 square by 2 into 1 plus 2 you are correct cot alpha 1 simple algebraic step nothing great in it cot beta 1 well. So, this is the thing that means if you take v f 1 square by 2 common 1 plus 2 yes cot alpha 1 cot alpha 1 plus cot beta 1 sorry. So, now if you put this value to this n s t now we got this value then n s t is equal to n s t carefully we will have to do otherwise we will again mistake in this simple algebraic steps at this stage you have to be very careful while doing this pi d 1 then we can write eta h rho q you please see rho q into g also g eta h rho q g to the power half then h to the power minus 3 by 4. So, 2 is there in the denominator. So, 2 to the power 3 by 4 and v f 1 to the power minus 3 by 2 very good minus 3 by 2 minus 3 by 2 3 by 4 minus 3 by 2 and there is another v f 1. So, it will be ultimately v f 1 to the power minus half very good into what will be there 1 plus 2 cot alpha 1 well into cot alpha 1 plus cot beta 1 all right this to the power minus 3 by 2. So, this is the final expression let me write again. So, this will be 2 to the power 3 by 4 and all this almost constant quantities eta h rho g q to the power half then v f 1 to the power minus half divided by pi d 1 this remains as it is then this big quantity in the bracket to cot alpha 1 into cot alpha 1 plus cot beta 1 to the power. Now, in this context I like to tell you that this expression of n s t this is a conventional expression which expresses the n s t in terms of the angles of the vends or the guide vends guide vends angles are outlet and the rotor vends angles are inlet and the flow velocity v f 1 which is the measure of the flow rate the rotor diameter at the inlet, but you do not have to memorize this formula, but you have to know the steps how it is being deduced. So, the expression for well the expression for hydraulic efficiency and the expression for degree of reactions all these things the hydraulic efficiency expressions as you have seen the efficiency of degree of reaction you do not have to memorize this equation to use in the examination. So, you have to deduce this equation and problems will tell you that how your concepts will be applied to deduce these equations under different situations not that you will have to memorize this typical equations for specific speed degree of reaction or hydraulic efficiency and all this. Any question please where please yes yes yes I am sorry one extra term one term is missing here you can write cot alpha 1 cot beta 1 this will be as it is without any index today I cannot tell you there is nothing that where I can check the final expression there is no point of memorizing this expression any more, but you see in this stage that whether it is ok or not. So, it is all right cot alpha 1 plus cot beta 1 thank you any question please any question today today any question whatever I have told degree of reaction in the deduction of degree of reaction hydraulic efficiency no question