 Good morning. I welcome you to this session, where we will discuss the specific speed and governing and limitations of pelton wheel. First we come to the specific speed of a pelton wheel. Let us come here, the specific speed of a pelton turbine. As you know the specific speed for any turbine in the dimensional form is defined as n s t is equal to the rotational speed n, the power developed p to the power half divided by the available head h to the power five by four. Now to have an expression for the specific speed what we do the power developed by a pelton wheel can be expressed in terms of the hydraulic efficiency like this rho q g h into eta h, where eta h is the hydraulic efficiency and rho q g h is the available power h is the available head q is the volume flow rate through the pelton turbine. So, rho q g h represents the total available energy to the turbine times the hydraulic efficiency gives you the power developed well. Now if we substitute the h in terms of the incoming jet velocity then we can write that incoming jet velocity is equal to coefficient of velocity into root over two g h. Now this expression comes from the application of Bernoulli's equation between the inlet and outlet of the fixed nozzle, where c v is known as the velocity coefficient which takes care of the frictional loss in the nozzle and h is the available head. So, therefore, we see from this equation g h becomes v one square by two c v c v square correct two c v square this is the g h again keeps the q can be written in terms of the diameter of the jet and the inlet velocity of the water jet as pi d square by four into v one pi d square by four is the cross sectional area of the incoming jet where d is the diameter of the jet and v is the velocity of jet incoming velocity of jet v one v one and the rotational speed n can be written as pi d square by n is equal to u by pi d, where d is the rotor diameter and u is the rotor velocity that is the tangential velocity of the rotor at the mean bucket height as we have explained earlier. So, therefore, the rotational speed satisfies the relationship with n as n is equal to u by pi d. Now if you express this expression of g h in terms of v one v one v one q in terms of v one and the diameter of the jet in this expression of power and this you substitute here and the value of n as u by pi d here what we get let us see that we get n s t is equal to n n is u by pi d then p to the power half what is p p is rho pi d square by four q and g h is v one square by two c v square rho q g h into eta h to the power half and h to the power five by four again h can be written from this expression h is v one square by two c v square g that means v one square by two c v square by two c v square into g this comes from again the same expression h is equal to v one square by two c v square g whole to the power five by four here it will be v one q very good because we have this v one yes v one square into v one because q is equal to pi d square by four into v one so that will be v one q very good. So, if you make a simplification of this you get n s t is equal to g to the power five by four you do not have to remember this expression, but you should know how it can be derived what are the substitutions that you have to make that means you have to express everything all the terms in terms of the inlet velocity of the z and diameter of the z. So, this becomes two to the power one four c v to the power three by two into u by v one d by d eta h to the power half rho to the power half. Now we see that a turbine handling a particular fluid in our case if we consider the water turbine the pelton will the working fluid is water rho is constant coefficient of velocity usually varies within a very small range g is substantially constant. So, therefore, this two operating parameters decide the value of the specific speed of the turbine we also know that for the maximum efficiency to have the maximum efficiency of the turbine will efficiency u by v one is fixed we know that we know u by v one is equal to point four six to have maximum eta h or eta o. So, therefore, to have a max to have a maximum hydraulic efficiency or overall efficiency or even the blade efficiency the value of u by v one is fixed it is almost equal to point four six as obtained in practice. So, therefore, we see for the maximum for then optimum operation with the maximum efficiency of a pelton turbine handling water or a fixed liquid the most important parameter which decides the value of specific speed is the ratio of d by d. So, the ratio of d by d is very important in designing the turbine usually if we consider the value the typical values like this u by v one point four six this equation can be reduced to a form if we take a value of c v point nine seven the typical values of this parameters and hydraulic efficiency which lies between point eight two point nine for a calculation if I take a value of zero point eight five then this expression can be written as n s t the specific speed of turbine that means if we substitute those values we get hundred five is approximately hundred five into d by d here hundred five the constant is a dimensionless constant because d by d is non dimensional n s t is also dimensionless the revolutionary speed a rotational speed sorry it is the rotational speed same thing u by v the dimensional less value c v is the dimensionless value eta h is the dimensionless values which are substituted in this expression along with the value of g we get one zero five d by d and here the specific speed comes the dimension is k g to the power half s to the power minus this is the dimension of I am sorry and sorry I told you that this is dimensionless this is the dimensional constant because the specific speed is the dimensional term. So, here d by d is non dimensional so hundred five is the dimensional constant because it includes the value of g it includes the value of rho which I have not written the value of g will be use well nine point eight one meter per second square in the value of rho you can take thousand k g per meter cube. So, considering those values for a water turbine and with the typical values of these operating parameters u by v one c v eta h we can express n s t is equal to hundred five which is definitely a dimensional coefficient into d by d where n s t is expressed by usual dimension like this. So, this is the values for specific speeds and usually the specific speed of a pelton turbine varies from four to sixteen to n s t varies from four to sixteen the specific speed of a pelton turbine usually varies from four to sixteen depending upon its operational efficiency. Now, if we use this value from the equation this equation this equation n s t hundred five d by d we get the value of d by d between fourteen to sixty that means if we use the value of n s t as four to six we get the value of d by d as fourteen two sorry I am sorry the value of d by d is six to twenty six six to twenty six the value of d by d lies between six to twenty six. Now, how you will fix the value of d by d now fixation is like that if you make the d by d very large d by d very large that means a very large rotor diameter compared to the jet of the jet diameter of the nozzle. So, a large rotor diameter will make unnecessarily friction loss if the rotor diameter is large the friction loss will be more over for a given rotor speed the r p m of the rotor will be low. Similarly, if we make d by d small or d by d low then what will happen the for a fixed jet diameter the diameter of the rotor will be less which will give a very close spacing of the buckets for a given number of buckets the bucket spacing will be close which will again increase the frictional losses and will reduce the mechanical efficiency. So, therefore, we see if we increase the value of d by d there also the mechanical efficiency will be reduced by increasing the frictional losses more over the rotational speed of the turbine will be reduced and again if we use a lower value of d by d then also the frictional losses will be increased and the mechanical efficiency will increase because of the close spacing of the bucket. So, these two phenomena makes a compromisation for d by d which is optimum optimum d by d varies between 14 to 16 this is the optimum d by d this varies between 14 to 16 well now I will come to governing of pelton turbine governing of pelton turbine well before coming to the governing of pelton turbine I like to tell you what is meant by governing of turbines in general what is meant by governing of turbine. Now, when a turbine drives an electrical generator or alternator the primary requirement is that the rotational speed of the shaft in which both the turbine rotor and the generator or alternator are coupled should maintain a constant speed the rotational speed of the shaft will be fixed well. So, therefore, the speed of the rotor rotational speed of the rotor will also be fixed why it is so this is because the frequency of the electrical output has to be maintained same to maintain the constancy in the frequency of the electrical output the rotational speed of the shaft coupled with the electric generator or alternator has to be fixed. So, what happens when the electrical load changes according to the demand then the speed of the turbine also changes why this is because the shaft in which the turbine and generator alternator are fixed rotates at a constant speed because of a balance between the resisting torque given by the electrical load and the driving torque depending upon the change of the angular momentum of the fluid flowing through the turbine clear. Now, when the electrical load changes the resisting torque changes while the driving torque remains the same. So, therefore, the balance is distorted and the revolutionaries revolutionary speed or rotational speed changes for an example if the load increase load is increased then the resisting torque on the shaft increases while the driving torque remains the same which comes from the change in the angular momentum of the fluid. So, therefore, what happens the speed falls similarly the reverse happens when the electrical load is reduced then the resisting torque is reduced while the driving torque remains the same. So, speed is increased so therefore, to maintain a constant in the speed we have to change the driving torque accordingly with the change in the resisting torque because of a change in load which means that if the load is increased so driving torque has to be increased or load is decreased driving torque has to be decreased. Now, driving torque comes from the change in the angular momentum now you see the deflection of the fluid velocity is fixed as far as the rotor is fixed because the design of the blades and everything is fixed. So, therefore, the change in angular momentum can be accomplished only by a change in the flow rate of the fluid. So, therefore, it is the flow of the fluid coming to the turbine is changed to change the driving torque which means if the load is increased the flow of water to the turbine is increased so that the driving torque is increased accordingly to maintain the constant in the speed of the shaft. Similarly, load is decreased so the flow rate going to the turbine or flow of water to the turbine is decreased. So, this change in the flow to the turbine with the load is done automatically and that is known as the governing of the turbine. This response in the flow to the turbine with the change in the load is known in general as governing of turbine. So, hence forth whenever we will refer to governing of turbine this is not one hydraulic turbine for any turbine to run at a constant speed rotational speed to maintain the constant in cycle frequency if it is coupled with electric generator or alternator the main criteria is that the when the electrical load changes according to the demand the driving torque has to be changed by the change in the flow of the working fluid to the turbine it may not be a liquid it may be a gas that is known as the governing of turbines in general or governing of any turbo machines in general. Now, therefore, if we come to the governing of Pelton when one important requirement now you see that flow rate can be changed in two ways. If you write the flow rate in any hydraulic or fluid circuit is the product of the flow velocity times the cross sectional area area normal to the flow velocity. So, therefore, from the simple equation you see the flow rate can be changed either by a change in the flow velocity or by a change in the flow area. Now, in case of a Pelton wheel the inlet velocity of flow is fixed for a given rotor speed why because the value of u by v 1 is fixed because of the requirement of the maximum efficiency of the turbine. So, therefore, this is fixed that means the inlet velocity of the water jet that means in case of Pelton wheel if I write that q is the inlet velocity of the water jet times the cross sectional area of the jet we cannot touch this. So, long the rotor speed is fixed and we want its optimum performance that means for maximum efficiency the ratio u by v 1 is fixed. So, therefore, we cannot touch the value of v 1 that means the incoming water velocity has to be fixed for a Pelton turbine. Therefore, this change in q is brought by a change in the flow area how it is done this is done by a by the movement of a sphere valve in a nozzle this is the peculiar or the typical diagram this is a sphere valve which is moved axially in a nozzle. So, that the effective flow area changes according to the change in the load. So, you see this is a condition at high load where the sphere gives a relatively higher flow area this is the effective flow area of the nozzles where the sphere moves further to give a less effective flow area when the load is low that means this is for high load this is for high load condition and this is for low load condition. Now, the sphere is designed in such a way the sphere valve the shape that the flow of water after passing through it coalesces into the form of a jet again. So, therefore, we can see we can say that the in the effect of the movement of the sphere valve in this converging nozzle is to make a change in the diameter of the jet and hence in the cross sectional flow cross sectional area. Now, sometimes it happens that the change in the load is so drastic that it may not be possible to change the water flow rate by changing the flow area only in that those cases a jet deflectors are used some plates as the deflector. So, this is the jet deflected from bucket what happens is that a portion of the jet is deflected from going to the bucket here that portion of the jet is shown this is a deflector plate which is being deflected by the deflector plate that means this jet this portion of the jet is not allowed to go through the go to the bucket that means this is a short of bypassing then we can change the flow area to control further on the flow rate. So, these are the arrangements which are made in controlling the flow rate or to govern the pelton turbine accordingly as the load changes well. So, now we will discuss the certain geometrical constraints on the or geometrical specifications of the pelton wheel now geometry of pelton wheel well geometry of pelton wheel usually the bucket width the bucket width bucket width these are very simple there is no much physical thing this depends upon the practical criteria bucket width if you tell it as b the bucket width is usually optimized in a fashion that if the bucket width is too small jets are not smoothly deflected if the bucket width what is bucket width let us have a look this is the now let me tell you that this is the width of the bucket this is the bucket width this is the bucket width sorry I am sorry this is the length of the bucket now this is the bucket yes this is the bucket width this is the length of the bucket length of the bucket is this one bucket width is this one this is the plan view and this is the bucket depth and this is the bucket depth let us tell it as h let us call it as b and this is the bucket length. So, with this terminology now we become if the bucket width is very small as compared to the z diameter, then liquid z's are not deflected properly. Similarly, if the bucket width is unnecessarily large with respect to the z diameter, then the frictional losses are more. So, therefore, we make a compromisation to make this bucket width let is be b by d ratio of bucket width to z diameter which is within 4 to 5. Similarly, the length of the bucket to z diameter these are the typical values maintained in practice lies between 2 to 3 and the depth to z diameter is around 0.8, these are the typical values maintained in practice. Now, we will come to limitations of pelton wheel. What is the limitation of a pelton wheel? What is the limitation of a pelton wheel you must know. Now, the specific speed of the pelton wheel is determined as a function of the u by v 1 the ratio of the rotor speed to incoming z speed and d by d small d by capital D. That means, the ratio of the water z incoming water z to the diameter of the rotor. Now, specific speed of the pelton wheel is usually very low. What does it mean? That means the if we consider the ranges of running of the pelton wheel under efficient conditions under high efficiencies we see that the specific speed n s t is usually very low is usually very low. Now, you see the expression of n s t is n p to the power half h to the power 5 by 4 which means that the pelton wheel is more efficient at higher head because it gives a lower specific speed in the range of high efficiencies. So, therefore, the pelton wheel works at a very low head. If I want that a pelton wheel a low on a sorry high head if I want that pelton speed a pelton wheel should operate at a high head what will be the problem? If head is very high then to develop a given power this flow rate through the pelton wheel has to be very high, but the requirement of the flow rate and z velocity is already fixed by its maximum efficiency condition. So, what we will have to make the diameter of the pelton wheel has to be made very high. So, the diameter of the pelton wheel if you make high it will be unnecessarily bulky and slow running or slow moving machine. That is the reason for which it is not used at a lower head. So, pelton wheel is therefore, always referred to the operating conditions with a very high head where pelton wheel runs at is maximum efficiency clear. So, this is about the limitation of the pelton wheel. Now, next of course, we have finished this pelton wheel about the pelton wheel finish everything about pelton wheel. Now, this limitation that I have told that pelton wheel cannot be operated at a low head to generate a finite amount of power or high power and also running at maximum efficiency. Different types of machines have been thought of or have been developed known as reaction machines which will operate at a lower head and yielding to a relatively higher specific speed. So, that at a higher head this turbines develops finite power and runs at higher efficiency these are the reaction turbines, but before coming to the reaction turbines in this class I like to show you certain examples. So, you please see how you can solve certain example let us consider the example one. Well if you consider the example one here the mean bucket speed of a pelton turbine is 15 meter per second the rate of flow of water supplied by the jet under a head of 42 meter is 1 meter cube per second. If the jet is deflected by the buckets at an angle of 165 degree find the power and efficiency of the turbine take coefficient of velocity c d is equal to 0.985 this is a very straight forward application of the theory of pelton turbine again you will see that mean bucket speed is given as 15 meter per second I tell you the relevant data and the flow of water is 1 meter cube per second and the head available is 42 meter and the jet is deflected at an angle of 165 degree that is the angle at the outlet of the bucket and coefficient of velocity in the nozzle is 0.985. So, now if we want to find out the power and efficiency of the turbine the usual way of attacking a problem is that you should write the expression for the output quantities that you will have to find out can you write this thing from there I can wait well you please write please then only I will start please write please write please write little fast the mean bucket speed is 15 meter per second the head available is 42 meter the flow rate is 1 meter cube per second and the deflection of the jet by the bucket is 165 degree coefficient of velocity c v is 0.985 all of you have taken the data now you see the usual way is that if you write the power what is the power expression you know that it is the flow rate q dot into v w 1 u minus v sorry minus v w 2 u that means q dot into v w 1 minus v sorry sorry v w 2 into u and what is efficiency efficiency hydraulic efficiency is power developed divided by the total energy at the inlet that means rho q dot into g into h. Now our first job will be to find out the value of v w 1 and v w 2 the bucket speed is given u is 15 meter per second. So, how to find out v w 1 or v w 2 let us find out what is the value of v 1 the incoming jet speed how to know you know that h is equal to 42 meter given let me write what are the given quantities q is 1 meter cube per second what are the given quantities u is 15 meter per second and c v is 0 point. Now you know the formula by the application of Bernoulli's equation at the inlet and outlet of the nozzle we can write the v 1.985 into root over 2 into 9.81 into 42. So, this comes out to be if you calculate this I can tell you this value comes out to be well what is the value please tell me just a minute I will 28. So, I will check this with my calculations 28.26 very good the value of v 1 comes out to be I have calculated it as 28.27 let us say that meter per second will depends upon the approximation the last third place of decimal. Now what is v r 1 now you write the inlet diagram inlet velocity diagram. So, this is v 1 28.27 meter per second what is u 1 u 1 is 15 that means this is u or u 1 they are same 15 meter per second. So, what is v 1 v r 1 v r 1 which equals to v 1 minus u and that becomes equal to 13.27 meter per second 28.27 minus 15. Now if nothing is told about the friction loss within the bucket then we will consider that means friction in the bucket to be 0 and v r 2 is equal to v r 1 is 13.27 meter per second all right can you see now if you draw the outlet velocity triangles the outlet velocity triangles will look like this is 165 degree this is u is equal to 15 meter per second this is your v r 2 which is equal to 13.27 meter per second and this will be v 2. Now here I will tell you one tidbit which sometimes comes as a confusion to the students at your level that we know that the velocity triangle my may be either like this u v r 2 and v 2 which type of triangle will be there either like this or like this that means the velocity triangle will be an obtuse angle or an acute angle triangle how to judge it here it will not be of this type because what is the difference between these two you see the v r 2 cos phi cos beta rather if you tell this is as this as beta beta 2 rather this is 180 degree minus beta 2 in this case beta 2 is 15 degree. So, v r 2 if this case is valid when v r 2 cos beta 2 is greater than u and this is valid when v r 2 cos beta 2 is less than u. So, therefore, you first calculate this thing before drawing the velocity triangle otherwise what will happen you draw a velocity triangle and then calculate again you cut it and draw this velocity triangle. So, better to check because v r 2 you know cos beta 2 you know and you know u. So, therefore, you decide which will be your velocity triangle in this case the velocity triangle will be like this. So, this is your v w 2. So, what is the value of v w 2 then v w 2 is equal to. So, in this case v w 2 is equal to v r 2 that is v r 2 cos u minus v r 2 cos beta 2 that means is equal to 15 minus 13 degree 0.27 into cos of 15 degree. This gives a value of v w 2 as 2.18 meter per second now the problem is solved. Now, it becomes a simple arithmetic calculations that p is equal to rho into q into v w 1 minus v w 2 into u. So, v w 1 is equal to v w 2 minus v w 2 when you put what is the value of v w 1 we got please 28.27 this is 2.18 this is 15 q is straight forward given as 1 meter cube per second all are in consistent unit and rho in kg per meter cube is 1000. So, that the p comes in terms of what and if you calculate this with these values it will be coming a very huge value 391.35 kilo watt please calculate it and efficiency eta will be this power develop total power that means denominator will be the total energy available that means it will be also rho that means 1000 q that means what is the value of q 1 and 9.81 into 42 that means the total energy rho q g h and if you calculate this value this value will be coming as 0.95 that means 95 percent. So, this is a straight forward application of a problem in pelton turbine I give you another problem you can try your house that example 2 that I will not solve, but you see that you write this problem example 2 a single jet well a single jet pelton turbine this is an example 2 which you try to solve I hope that you will be able to solve it next class I will ask you a single jet pelton turbine is required to drive a generator to develop 10 mega watt that means the generator is developing 10 mega watt this is the power at the generator end try to understand is required to drive a generator to develop 10 mega watt the available head at the nozzle is 762 meter that means this is the total energy per unit weight at the nozzle that means at the entrance of the nozzle the available head at the nozzle is 762 meter well all right then assuming electric generator efficiency 95 percent the electric generator efficiency 95 percent that means this is the efficiency by which the mechanical power as the input to the generator is converted into electrical power pelton wheel efficiency 87 percent these are the input values coefficient of velocity 0.97 well mean bucket velocity 0.46 of jet velocity usually this value is almost fixed the mean bucket velocity means you the rotor speed at the mean height of the bucket that means mean bucket velocity is 0.46 of jet velocity. So, whenever the what jet velocity means it is the inlet velocity of the water because water strikes the pelton wheel in the form of jet that is inlet 0.46 of jet velocity outlet bucket angle 15 degree now whether it is the 15 or 165 we will have to decide from the triangles so either it is given the obtuse angle or the acute angle the obtuse angle may be given or the acute angle will be given. So, 15 degree and depending upon the relative magnitude of the rotor speed or the mean bucket speed and the tangential component of the relative speed at the outlet you should draw the velocity triangle this is a very important point this is a very important point in solving the problem and drawing the velocity triangle at the outlet. And the friction of the bucket reduces the relative velocity by 15 percent is a very important data friction of the bucket reduces the relative velocity by 15 percent that means relative velocity at the outlet is 0.85 times than that it is inlet. So, this is an important data find the flow rate of water through turbine find the flow rate of water through turbine the diameter of the jet the diameter of the jet and the force exerted by the jet on the bucket and the force exerted by the jet on the bucket all right. And I think today up to this well thank you all.