 Good morning. I welcome you to this session of fluid machines. Now, we will be discussing in this session mainly the governing of reaction turbines and we will solve some examples on reaction turbines. Before that, I like to show you a typical plot of the efficiency of reaction and impulse turbines as a function of load applied to the turbine. Well, you see this figure, it is clear that this figure in the abscissa it is the percent of full load and this efficiency is the ordinate. As you see this curve refers to the pelton wheel, I can I think you can see. Now, this is the pelton wheel curve, this is the propeller turbine, this is the Francis turbine, this is the Kaplan turbine. Now, here of course you may have a question, it is a propeller turbine and Kaplan turbine are different. Usually in practice we refer to a turbine whose entry is mainly radial and tangential, but exit is purely axial to that of a Kaplan turbine. We refer to that as Kaplan turbine, this one for Kaplan turbine. And we refer to a propeller turbine when the entire flow through the turbine is axial and this type of turbine have got vanes or blades which are similar in nature to the propeller of a ship. And they are very less in numbers usually 4 to 6 the number of blades, these are propeller turbine, this is purely axial turbine. So, you see from this curve this give you an idea that how the efficiency varies, you see the pelton turbine gives almost a very high efficient in a wider range of load. So, therefore, pelton turbine in this respect we can tell is a flexible with respect to the variation in the load, that it operates at a fairly high efficiency during a wider range of load. This is not so in case of Francis turbine or propeller turbine, it attains almost a maximum efficiency at certain load. And then a change in load makes or decreases the efficiency, there is a drop in efficiency either side. However, the Kaplan turbine is a compromise between these two that it is almost like pelton turbine is efficiency does not fall in this side, the that means beyond a certain load you see that it fairly it operates at a fairly high efficiency. Now, after that I will show you another curve this is a propeller, this is propeller turbine P R O P L L E R this is pelton, this is the propeller one, this one is the pelton, this is the Kaplan turbine and this one is the Francis turbine. Now, after that I will show you another very interesting plot is this plot between the efficiency and the dimensionals dimension less specific speed. These are the efficiency values this is I think 0.82, 0.86, 0.90, 0.94, 0.98, 0.82 this is the this is a typical plot 0.86, this is 0.86, this is 0.98, this is 0.94, obviously this is 0.90. So, you see this curve is pelton, this curve is Francis, this curve is axial flow or Kaplan. Now, we have to recall that the specific speed dimension less for a turbine is N P to the power half divided by rho to the power half rho is there and G is there in dimension less specific speed G H to the power 5 by 4. So, we have seen that this specific speed earlier that it is a dimension less parameter and it indicates the similarity conditions that means we can say that it is a similarity parameter also. And a machine of a particular homologous series has different characteristics curve as far as this variation of specific speed with the efficiency. So, this corresponds to pelton all pelton turbines follow this curve all Francis turbine follow this curve all axial flow Kaplan turbines follow this curve. Here we see that the efficiency of the pelton turbine is high is a range of 0.94, 94 percent when the specific speed is relatively low which means that pelton turbines are efficient only at low specific speed that means at a high head this already we discussed earlier. So, this gives the typical ranges of the specific speed this is one dimension less this is two this is three this is four. The next is the Francis that means Francis turbine operates at a relatively lower head as compared to that of pelton turbine. So, if you operate Francis turbine at a high head what will happen in this range pelton turbine will give an efficiency below 90 percent. So, it is not wise it is not a wise decision to choose a Francis turbine in this specific speed ranges because pelton turbine will be giving a higher efficiency. So, axial flow turbine on the other hand this efficiency drops too much to lower specific speed that means higher head, but at a higher specific speed or at a lower head axial flow Kaplan turbine gives fairly high efficiency as compared to the Francis one or pelton one. So, this we have discussed earlier, but in this regard immediately we can see a very straight forward application of this through a problem straight forward application of this concept and this figure through a problem like this just you note this problem is a very straight forward and simple application of the problem which give you a clear concept in the design of the type of turbines hydraulic turbines for a particular operation which I told you earlier, but now this gives you a clear example a reservoir has a head of 40 meter and a channel leading from the reservoir permits a flow rate of 34 meter cube per second. Well that means the head is available that I have this much amount of head available that is the total energy of the fluid per unit weight and the permissible flow rate that means available flow rate is 34 meter cube per second. Now if the rotational speed of the rotor has to be 150 rpm that means rotational speed of the rotor is fixed 150 rpm what is the most suitable type of turbine to use. That means we have the operating conditions like this head 40 meter flow rate 34 meter cube per second and the rotational speed of the rotor is 150 rpm this three are the operating conditions which are usually specified for a turbine. Now it is a very simple practical example that we have here that means we have to just calculate the specific speed let us calculate the non dimensional or dimensional. Now whether you will calculate the dimensional or non dimensional depends upon the ready made figures or ready made what is called the table is available to you that whether you have the values of the dimensionless or dimensional specific speed. So this is the problem. So now what we do we have to calculate the specific speed that means n p to the power half rho to the power half g h to the power 5 by 4. So h is given 40 meter in the problem you see rotational speed is given. So you have to calculate the power what is power power is density flow g h. So power developed is rho q g h. So this can be equal to 1000 density kg per meter cube 34 meter cube per second is q g is 9.81 you take meter per second square and head is 40 so therefore we get the power developed from the available head as this one which is in what. So this comes to be 13 point if you calculate it 34 mega watt. Now if you calculate the specific speed dimensionless then what you do n is 150 rpm you better put it in r p s then 13.34 into 10 to the power 6 in terms of watt it is nothing only this algebraic calculations mathematical calculations g into h 40 whole to the power 5 by 4. Now if you calculate it it will come as 0.16 I am sorry 1000 to the power half I am sorry 1000 to the power half. If you calculate it I am giving you the answer it will come like this or in terms of revolution this is very simple school level thing just a straight forward application that means you find out the k s t with the values and then you see from our earlier figure 1.037 radian it is given in terms of radian k s t in radian. So which one we will now if we have this chart in our hand which one we will select France is very good. So answer is that France is turbine. So France is we will choose. So for this operation we will choose a France is turbine because it runs at the higher efficiency very good. Now I will come to the governing of turbine well governing of turbine. Now governing of reaction turbine as I have told you earlier the governing of turbine means to change the flow to the turbine accordingly with the change in the load to the turbine. When the load increases we have to increase the liquid flow to the turbine similarly when the load is decrease we have to decrease the flow of liquid to the turbine. So that the speed of the rotor remains constant to maintain the constant in the frequency of the electrical output. This this is done in a reaction turbine by changing the position gate position or changing the positions of the guide blades of the guide vanes of the wicket gates. So that the flow rate to the turbine is altered by changing the gate openings how it is done. Now I already told you earlier I have already told you earlier that the wicket gates or the fixed vanes of the turbine are pivoted so that this can be rotated this can be moved. Now this you see this is the volute casing this is volute this is the volute casing of the turbine. This is the inlet that means the liquid flows through this. Now this is the turbine wicket gates this is the rotor of the turbine. So this wicket gates or vanes which are not shown here they are connected through levers to a regulating ring. This circular portion this is written here is regulating ring. This regulating ring is connected to this regulating rod this two sides these are the regulating rod these are the connecting pins. So these regulating rods are connected to these regulating rings at one ends and the other ends are connected to a regulating lever. This is a regulating lever which is keyed to a shaft this is the typical this type of thing. This is I show you again that if you cannot make it this is a regulating this is the shaft here the regulating rods are going like this and this is keyed to a shaft which is turned by a servo motor. This is a servo motor this is the servo motor piston which is actuated by the pressurized oil the pressurized oil the pressure of the oil which comes to the servo motor and actuates the piston is actually controlled by a governor mechanism. That means that sense the pressure and in that proportion it controls the motion of the servo motor piston which ultimately actuates or turns this shaft to which this lever is keyed and ultimately through this regulating rod and through the lever mechanism which is ultimately connected to the wicked gates or the static vans of the rotor they are shifted. That means automatically when the load changes the oil pressure changes and ultimately through this mechanism the position of the gate is changed. So that the gate openings vary and accordingly flow liquid flow enters or the liquid flow rate entering to the turbine changes. This is the mechanism by which the governing of reaction turbine is made apart from that there is another valve another valve bypass valve or relief valve bypass or relief valve bypass valve that is being used in this circuit to this along with this phenomena along with this sorry along with this mechanism to discharge or bypass some amount of water to a different line. That means it is not allowed to go to the turbine this is known as double regulation this is known as double regulation. That means simultaneously the flow rate is controlled through this mechanism by altering the position of the stator stator vans or the wicked gates along with the bypassing or changing the direction of flow or converting or bypassing certain amount of flow in other directions not allowing them to enter to the turbine this is known as double regulation. That depends upon the amount to which the change in the flow rate has to be made that means when the load changes drastically then the double regulation is required. So this is all about the governing of reaction turbine. Now we will solve certain important problems because now we have almost completed the reaction turbines. So therefore at the end we like to go through certain important problems. So that yes please please yes yes yes just just just a moment let me say this diagram otherwise it will be difficult for you or for me to understand other efficiency versus KST diagram yes please tell this is actually the overall efficiency it does not matter it may be hydraulic efficiency or overall efficiency when the picture is there you will have to take care of that if the hydraulic efficiency is more means for that case overall efficiency will be also more it is usually coated in terms of overall efficiency. In the problem that we solved we took the hydraulic efficiency as 100 percent because P we wrote as Q yes very good in the problem we solved we do not know the efficiency we have not assumed that is a trend it is nothing coming quantitatively correct what we are finding out a trend that whether we will use which type of turbine here we have used the power developed we have considered the hydraulic or overall efficiency to be one that means this is the power developed so if we use this power developed we will be getting a KST this amount. Now any turbine if you choose this may be reduced because power developed may be reduced we may add we may make a multiplication of it out that we will not change this order of this KST that means we are interested to find out the ranges of the specific speed where we will tell that in this range Francis is very high that means we are not coming to a range where the specific speed is in the range of 0.1 or 0.2 so it hardly matters so quantitatively definitely this is not the power developed this is the power available so exact power developed will be multiplied by the overall efficiency so at the beginning a priori we are not knowing the power developed so therefore we take this as the power developed there may be some reduction with that you can add we can multiply with 0.9 some approximate value we can take it hardly matters we want the range that means the specific speed will be in the range of 1 we can write in the range of 1 where Francis turbine is the most suitable that is why you have assumed it and we have straight forward written P definitely from the theoretical point of view you should write here that P is this provided we assume the overall efficiency to be 100 percent very good correct. Now next problem next stage is that we should go through certain very important problems which will give a clear idea of the Francis turbines which we have already learnt Francis and Kaplan turbine look into this problem a Francis turbine has a diameter of 1.4 meter this is the diameter of the Francis turbine 1.4 meter 1.4 meter is the diameter and rotates at 430 rpm that means the revolutionary speed or rotational speed is given water enters the turbine runner without shock with a flow velocity of 9.5 meter per second this is the flow velocity and leaves the runner without whirl that means without any tangential component this is usually done in the design of all runner blades that they do not have any whirling velocity if in a problem for your examination purpose does not mention that without whirl that you can assume that the outlet is without whirl and you can draw the velocity triangle at the outlet according with an absolute velocity of 7 meter per second. So, this is the absolute velocity of water at the discharge that difference between the sum of the static and potential heads the difference between the sum of static and potential heads at entrance to the runner and at the exit from it is 62 meter that means it is difference of static and potential heads understand static and potential head that means p by rho g plus g that is the difference between the static and potential head this sum of this two quantities between the entrance and exit is 62 meter the turbine develops 12.25 megawatt this is the power developed by the turbine the flow rate to the turbine is 12 meter cube per second. So, flow rate is 12 meter cube per second for a net head of 115 meter net head is 115 meter find the following. So, problem is a long one usually a francis turbine problem gives so many data sometimes it is very long, but you will have to understand these things what are the data giving that rotational speed the flow velocity without wheel discharge 7 meter per second is the absolute velocity of discharge difference between the heads except the velocity head the difference between the static and potential head is 62 meter the net head 115 meter. So, this is the net head available find the following what are the following that you have to find out the absolute velocity of water these are the things that you will have to find out absolute velocity of water at entry to the runner and the angle of the inlet guide the absolute velocity of the water at entry to the runner and the angle of the inlet guide veins the entry angle of the runner blades the entry angle of the runner blades and the loss of head in the this you will have to find out you have written the problem let us solve the problem again I am telling if any difficulty you just please I repeat the thing have you completed yes now again I repeat a francis turbine has a diameter of 1.4 meter and rotates that 430 r p m water enters the turbine runner without shock with a flow velocity of 9.5 meter per second and leaves the runner without wheel with an absolute velocity of 7 meter per second the difference between the sum of the static and potential heads at entrance to the runner and at the exit from it is 62 meter the turbine develops 12.25 megawatt the flow rate through the turbine is 12 meter cube per second for a net head of 115 meter find the following what are the following the absolute velocity of water at entry to the runner and the angle of the inlet guidance the entry angle of the runner blades and the loss of head in the runner let us find out that how we can solve the problem let us think of the velocity triangle that means if you consider the runner blade like this if you remember this the runner blade so so please see that that this is your u 1 inlet this is v r 1 and this is v 1. So, this will be v f 1 that is the flow velocity at the inlet. So, therefore, this is u 1 u 1 is this one and this is v w 1. So, immediately you will have to think in terms of the velocity triangle the outlet velocity will be like this triangle though this will be now without wheel it is already mentioned in the problem. So, therefore, v 2 and v f 2 is same and this is the u 2 which will be lower than the u 1 in a francis turbine radial flow in what radial flow turbine as you know and this is your v r 2. So, therefore, this angle is the angle of the blade at the outlet this is the angle of the blade at the inlet beta 1 and this is the guide van angles guide van angles are always specified as the angle of the guide van at outlet that means the angle at which the fluid leaves the guide or this is the angle of the absolute velocity of the fluid approaching the runner and all the angles are referred whenever we tell angle we do not tell with what direction because it is always referred to the direction with the direction of the tangential velocity that means the tangent to the rotor at the point in the tangential direction. So, if this be the velocity triangle now we what are the quantities and we know that we know u u 1 we know what is u 1 u 1 is pi have you calculated n d 1 by 60. So, it is given in the problem is it given in the problem 4 30 r p m. So, pi into 4 30 divided by 60 here n is the r p m and what is the diameter of the runner has a diameter of 1.4 meter now here there is a problem that is why I like to point out the problem in particular when this diameter of the runner is given you will have to consider this diameter at the inlet diameter because it is not very explicit in its mathematical language is the diameter is the inlet diameter outlet diameter because the flow takes place with varying diameter is a radial flow machines for a pelton wheel or for an axial flow machines we mean diameter means that the mean diameter where both the inlet and outlet is taking place, but here it is very difficult to know that whether it is inlet diameter outlet diameter. So, it is a convention the diameter of the radial turbine means the outlet diameter in case of radial flow machines diameter is the outlet diameter. So, it is very important thing. So, you find out you what is the value of you 31.52 meter per second. So, then what you will do v w 2 is 0. So, what you will find out power given to the runner by the water this power given that means power available power available or power developed you can write by runner what is the power developed by runner it is rho into q into you know from the Euler turbine equation v w 1 u 1. So, u 1 we know we know v w 1 how do you know v w 1 v w 1 we do not know, but we know the power developed that is 12.25 10 to the power 6 and we get v w 1 as 32.39 v w 1 we get because we know that the turbine develops 12.25 megawatt. So, power developed by the turbine is rho q v w here also mechanical efficiency we take to be 100 percent, because this power developed is the final power developed and this v w 1 u 1 is the power developed by the runner. So, we consider the mechanical efficiency the mechanical efficiency is not given we can consider this to be 1 and we can find out v w 1. So, now it is easy to find out the alpha 1. So, when we know the v w 1. So, we can find out the alpha 1, because we know the flow velocity at the inlet is 9.5 meter per second. That means v f 1 is equal to 9.5 meter per second. So, we can calculate tan alpha 1 is 9.5 divided by 32.39 and this comes to be not tan alpha alpha 1 comes to be 16.35. Now, how to find out tan that beta 2 or beta 1 which we will have to find out please tell me which we will have to find out the at entry to the runner and the angle at entry to the runner. That means this beta 1 that means we will have to find out beta 1 how to find out beta 1 if I write tan beta 1 is v f 1 by very good v w 1 minus u 1. So, you know v f 1 you know v w 1 you know u 1 v w 1 you know since you know this alpha 1. So, v w 1 is v 1 cos alpha 1. So, this will give you a value of beta 1 is equal to 84.77 degree all right. So, next is the head loss to the runner we have to find out ok. Now, we see that total head now head loss to the runner how to find out head loss in the runner this is the last thing that you will have to find out what are the things we have found out absolute velocity of water at entry the angle of the inlet guide vends and runner vends the entry angle of the runner blades the loss of head in the runner. So, now loss of head in the runner how can you find out loss of head can be found out written like that head across the runner across the runner that means available head this terminology you have to understand very clearly available head head across the runner is equal to work head work head means that is the work produced per unit weight that means the head equivalent to work plus the head loss. Now, head across the runner available head is the difference of head between inlet and outlet that means if I give the suffix 1 as inlet and suffix 2 at the outlet from the runner that means the head available is this is the inlet head which comprises the static head that is the pressure head the kinetic head the velocity head and the potential head minus well p 2 by rho g plus v 2 sorry sorry sorry very good and what is this v w 1 in this case there is no whirl tangential component of velocity this is h l rather head loss. So, this things are clear we have been given that this head available what is the value head available to the runner. So, we can find out this thing from a different way that we know this p 1 by rho g minus p 2 by rho g plus z 1 minus z 2 rather we can take p 1 by rho g plus v 1 square minus v 2 square is equal to the right hand part. Now, this is given as 62 meter 62 meter well this is given as 62 meter. So, this part we can calculate from our velocity triangle we can calculate we have already calculated v 1 have we not we have not calculated v 1 know so far. So, v 1 we have calculated so we know the value of v 1 we can calculate the v 2 also how we can calculate the v 2 v f 1 is equal to v 2 very good v f 1 is equal to v f 2 is equal to v 2. So, if we know both v 1 and v 2 from the velocity triangle with the trigonometric relations then I can find out the net head across the runner well this thing is already calculated this thing is already known to us. So, we can find out h l so ultimately if you calculate the h l will come to 13.49 meter all right this we have already calculated earlier 32.39 this is 32.39 what is that into 31.52 divided by 9.80 this is v w 1 this is v 1 so plus h a. So, this way we can solve this problem next we come to another problem this is another interesting problem this you can take it francis turbine quick you write this problem time is of this problem the diameter of the runner the diameter of the runner the diameter of the runner the diameter of the runner the diameter of the runner of a vertical shaft turbine the diameter of the runner of a vertical shaft turbine you please write this problem is 450 millimeter is 450 millimeter at the inlet the diameter of the runner of a vertical shaft turbine is 450 millimeter at the inlet the width width of the runner at inlet the width of the runner at inlet is 50 millimeter is 50 millimeter. The diameter 50 millimeter the diameter and width at the outlet r the diameter and width at the outlet r 300 millimeter r 300 millimeter and 75 millimeter respectively 300 millimeter that means the diameter and width at the outlet are 300 millimeter and 75 millimeter respectively this is the diameter this is the width well the blades occupy next is that the blades occupy 8 percent of the circumference which will give you the idea about the flow area. So, flow area is not the entire circumference times the width the blade occupy 8 percent of the circumference stop the guide when angle is 24 degree which is alpha 1 that is the angle of the absolute velocity at the entrance to the turbine the guide when angle is 24 degree the inlet angle of the runner blade is 95 degree well the inlet angle of the runner blade is 95 degree and the outlet angle is 30 degree and the outlet angle is 30 degree the fluid leaves the runner the fluid leaves the runner without any whirl the fluid leaves the runner without any whirl the pressure head at inlet is 55 millimeter above that at exit the pressure head at inlet is 55 millimeter above that at exit from the runner the fluid friction the fluid friction losses the fluid friction losses account for sorry 18 percent of the pressure head at inlet the fluid friction losses account for 18 percent of the pressure head at inlet calculate the speed of the runner and the output power calculate the speed of the runner and the output power well calculate the speed of the runner and the output power you write another problem for this is the last one axial flow hydraulic machine quick and axial flow hydraulic turbine you have to be little first in writing because time is short and axial flow hydraulic turbine has a net head of 23 meter across it axial flow hydraulic turbine has a net head of 23 meter across it and when running at a speed of 150 rpm develops 23 megawatt that is the power develop that is the net head across the turbine and running at a speed of 150 rpm the blade tip and half diameters which is very important here the blade tip and half diameters are 4.75 and 2.0 meter respectively that means the tip and half diameter determines the flow area because here the flow is axial axial flow turbine the blade tip and half diameters are 4.75 2.0 meter respectively but inlet and outlet is through the entire height of the blades through the entire diameter that means this is not a radial flow machines if the hydraulic efficiency is 93 percent and the overall efficiency 85 percent calculate the inlet and outlet blade angles at the mean radius assuming axial flow at outlet assuming axial flow at outlet the entire flow through the turbine is axial assuming axial flow at outlet alright. So, you please try this problem next class if you have any difficulty in solving the problems we can discuss before starting the new topic pumps rotodynamic pumps. So, next class I will discuss the rotodynamic pumps but I have given you this two problems one problem I have solved other two problems which I have given you please try to your hall. So, if you have any difficulties next class we will discuss about this and then we will start the new topic rotodynamic pump. Thank you.