 The next hydraulic turbine that we are going to look at is the Pelton deal. You may recall that we have already discussed this in the introductory part of the lecture. So let us quickly review what we discussed and move on to learn more about this device and also do some calculations involving the machine. So the Pelton wheel has a runner on which are mounted the buckets or the buckets are actually shaped in the form of a cow and the water from the nozzle or the water jet from the nozzle hits the buckets and the buckets actually turn the water around almost through 180 degrees and the resulting change in momentum on the water imparts force on the buckets and hence the runner and turns it around. This is coupled to an electric generator which is used for generating electricity. Now the water from the dam reservoir is brought to a penstock and then the penstock terminates in a nozzle. So the gravity head that is available in the reservoir is converted entirely into kinetic energy of the water and the flow rate through the nozzle is regulated by means of a sphere which looks like this. So by moving the sphere in or out the flow rate may be decreased or increased. This is not the best way to regulate the flow rate because the losses can be quite high especially at low flow rates. So we will look at much more efficient forms of controlling the flow rate in the Pelton wheel in the next slide. The other important thing to be noticed in contrast to the Francis and the Kaplan turbine is that in the Pelton wheel the pressure remains constant. So when the water comes out of the nozzle it comes out of atmospheric pressure but at high speed and hence the entire casing actually need not be very thick because the pressure inside the casing is actually atmospheric. So it need not withstand hydrospatic stresses due to the gravity head in the dam which actually can be quite high in the case of a Pelton turbine installation because this is usually utilized in locations where the head available is very high. So it's fortunate that in such a situation we can operate at constant atmospheric pressure so the casing can be quite lightweight. Now here is a look at an actual Pelton wheel installation by Siemens hydropower and this design can be seen to be quite different from what we just saw. So in fact you can see that instead of a single jet that we saw before this layout has 1, 2, 3, 4, 5, 6 jets around the periphery of the wheel. So the advantages to having 6 jets around the periphery is that the water actually can hit the runner at more than one location and hence the runner actually more or less runs full in this case which is an advantage in contrast to this situation where the water hits only one bucket so the rest of the runner is actually not developing any work. So by distributing the water around the circumference of the runner the stressors, mechanical stressors on the runner is also reduced, asymmetry in the mechanical stressors is reduced to a large extent that is a big advantage. The second advantage is that in case the flow needs to be reduced or increased we can turn off or turn on one of the jets to accomplish this with the minimal flow losses instead of using the sphere that we saw in the previous illustration. So having multiple jets in a Pelton wheel has several advantages and this is invariably utilized in many of the Pelton wheel installations that are in operation today. The most, the modern installations all have multiple jets. In addition to having multiple jets another feature that is quite widely used these days is the double overhung design of the Pelton rotor as can be seen here in a normal situation you would have a Pelton wheel rotor like this that the water jet one or more of the water jet hits the runner and the runner shaft is connected to the generator. Now this actually is a single overhung design and better mechanical balancing and better mechanical stress distribution can be achieved if we add another rotor on the other side of the generator. So both these shafts can actually run the generator and this results in more uniform distribution of the mechanical stresses. Now the power that this runner has to develop is half of what it was doing before and that the stresses are also more uniformly balanced and the axial thrust on the shaft is also now zero because the axial thrust acts in opposite directions on these two rotors so it is mechanically well balanced. So for the power that generates this is extremely compact and a very, very efficient design. So modern installations use both these strategies to improve the efficiency of Pelton turbine and their multiple jets as well as the double overhung rotor design. So here we have a load at the bucket in a Pelton rotor so it's quite clear that these are massive buckets and they are capable of generating hundreds of megawatts of power and as can be seen the bucket is shaped like two cups with the ridge in the middle so the water jet impacts the ridge in the middle then flows through both the cups on both sides of the the cups on both sides of the ridge and then comes out like this. So the turning angle usually is about 150 to 160 degrees so it's quite a lot of turning and that generates the power that is expected and more importantly since there is only change in direction of the water and hence change in momentum the pressure remains constant. Okay so let us summarize what we have seen so far. Pelton wheel is usually used for high head and low discharge installations and the pressure remains constant because there is only a change in direction of the water and the axial velocity remains constant the relative velocity also remains constant so Pelton wheel is an impulse machine. The water jet as we already mentioned leaves the nozzle at atmospheric pressure so the entire turbine operates at atmospheric pressure consequently the rotor need not run full and the casing can be lightweight although making the rotor run full by using different by using multiple jets has its own advantages so that is practiced in real life applications yet another advantage with the Pelton wheel is that it is ideally suited for silty water as the buckets can be removed easily for servicing to fix the damage caused by the silt. Now here is a blade element of one of the buckets and as we mentioned earlier the water enters and impacts the central ridge and then flows on both sides and it is turned through an angle gamma 2 as shown here. So if we apply Euler's turbine equation to this blade element we have p equal to rho times q times v theta 1 u1 minus v theta 2 u2 u1 is equal to u2 so that can be taken outside and we can actually simplify this using the velocity triangle shown here and finally show that the power is actually equal to rho u times v1 minus u times 1 minus cosine gamma 2 okay notice that the relative velocity also remains constant here c2 equal to c1 equal to v1 minus u now with everything else held constant the power produces a maximum when gamma 2 is equal to 180 so in this case this becomes 1 minus out minus 1 plus 1 so that becomes equal to 2 so the power is a maximum when gamma 2 is 180 but this is practically not viable since the water exiting a bucket will hit the following bucket so this is counterproductive so typically gamma 2 is between 155 to 155 degrees and this results only in 2% reduction from the maximum power so it's quite acceptable okay so that is the value that is used now with everything else held constant the power is a maximum when the blade speed is half of v1 okay so we simply take this expression differentiate this with respect to u and set the derivative to zero so we can show that the power produces the maximum when the blade speed is half of v1 and it must be borne in mind that v1 is equal to vj which is the velocity of the jet as it leaves the nozzle so the power is a maximum when the blade speed is half of the jet speed okay and since the overall height of the available to the turbine rotor is h the speed with which the water leaves the nozzle is quite up to the edge at least theoretically without the absence of any losses so with u equal to vj over 2 we can rewrite this expression as follows p max equal to root 2 times vj square over 4 times 1 minus cosine gamma 2 as we mentioned earlier the output power may be controlled by controlling the discharge using the spear multiple jets actually is a much better option for a much better strategy for controlling the discharge now if we denote the length of the penstock as l sub p its diameter is d sub p and the velocity of fluid in the penstock as vp and apply Bernoulli's equation between the free surface of the dam reservoir and the penstock nozzle exit we get the following where we are accounting for frictional losses in the penstock pipe so this is the friction factor in the penstock pipe so we are accounting for frictional losses in the penstock pipe okay so q is the flow rate ap here in the cross sectional area of the penstock so from this we can get an expression for vj which looks like this vj square equal to 2gh minus uh f l over vp times q square over ap square so in the case of an ideal flow and there are no losses vj is equal to uh root 2gh as before now if we further set the blade speed to be half of the jet speed then we get the power as uh row times 2 over 4 times vj square times 1 minus cosine gamma 2 okay row times 2 times vj square over 4 times 1 minus cosine gamma 2 now notice that in this case if we increase the flow rate the power increases as a result of this term here but the power decreases as a result of the q square term here which appears after negative so this increases the power whereas the term in the square bracket as a as an opposing effect so it suggests that there must be an optimum value for q that gives a maximum power okay so everything else remaining the same the power increases as q increases but the term in the square bracket uh opposes this effect because the jet velocity decreases on account of the head loss in the principle okay so basically the power uh contains a product term 2 times vj square so when we increase the flow rate this increases but uh in the because of the friction losses uh which increase with q the jet velocity decreases so vj square decreases q increases it suggests that there must be an optimum value for q okay so we differentiate this expression with respect to q and set the derivative to zero and we can get the optimum flow rate to be this and the jet velocity in this case to be four thirds uh times g h so the the theoretical value is square root of 2 g h now we have four thirds that g h which suggests that the effective head that is available in this case is uh two thirds g h and the head loss due to friction is one-third g h so this value of flow rate gives the maximum power with everything else remaining constant let us now look at our example involving the pelton turbine the problem statement is uh given here effective head is given flow rate is given uh we also introduce a couple of new uh terms in this example nozzle coefficient which accounts for losses in the nozzle what is that in this derivation we accounted for nozzles in the penstock pipe but uh i'm sorry we accounted for losses in the penstock but not in the nozzle itself so losses in the nozzle uh maybe accounted for uh by using a nozzle coefficient uh the deflection flow deflection angle is given and the velocity coefficient in the bucket uh is also given here and we are asked to calculate efficiency of the turbine nozzle diameter and the number of buckets in the wheel okay so the jet velocity dj which is usually square root of 2 g times h e is now c n times square root of 2 g times h e which works out to 75.19 meter per second uh the blade speed in this case may be calculated uh using the given value the radius and rpm as a 2 pi uh or n divided by 60 which is 35 point v 4 meter per second so the relative velocity at the inlet c1 is equal to v1 minus u that's equal to 39.85 meter per second also v theta 1 equal to v1 in this case so the velocity triangle in this case is very simple as you can see here uh c1 is equal to v1 minus u and v theta 1 is equal to v1 entire component is tangential along the u direction and the relative velocity at exit which normally would have been equal to c1 is now cv times c1 that is 0.9 times c1 so that is 35.87 meter per second therefore v theta 2 is equal to from this velocity triangle uh v theta 2 equal to u minus c theta 2 so u minus c theta 2 and the c theta 2 is nothing but c2 times cosine 180 minus gamma 1 that is the blade angle 180 minus gamma 1 so if I substitute the numbers we get this to be 2.83 meter per second notice that uh both uh v theta 1 and uh v theta 2 are in the same direction so there is no change in the sign uh of v theta 2 in the Euler's turbine equation so if we substitute the numbers we get this to be 10740 kilowatts the efficiency is the is the actual power divided by the hydraulic power which is rho g q times h e and we get this to be 87 percent it's also given that in the problem statement that five jets are used in this installation so we take the total flow rate and divide that amount five jets so and then based on this we can evaluate the diameter of each jet using this expression which gives us 0.1193 meters for the diameter of the jet uh knowing the diameter of the runner and the diameter of the jet we can use a correlation to calculate the number of buckets that are used uh which may be written like this where d is the diameter of the runner or wheel diameter dj is a jet diameter so uh this tells us that the number of buckets needed in this case number runner is 21 that concludes our discussion on the pelton wheel what we will do next is to say a few things about turbine specific speed selection of turbines and the turbine specific speed okay in the case of turbines in contrast to centrifugal machines where we used the specific speed relating q and h here we use the power specific speed which relates the bhp and the head uh head h or effective head h e okay so this is a more uh meaningful metric for turbines and it is defined in this manner note that this is dimensionless and all the quantities here are used in their appropriate uh si units bhp here is of course the efficiency times the hydraulic power which is rho q g times h e where h e is the effective head as we mentioned before this is obtained by uh eliminating the rotor diameter between uh the dimensionless head coefficient c h and the power coefficient cp the the reason for uh preferring this form of specific speed or specific speed for turbines is as follows in any practical application the initial decision making uh would be based on known quantities such as available head the speed of the rotor because in this case the turbine runs a generator which is then used for generating electrical power the speed of the rotor actually is related to the supply frequency okay normally at the voltage output voltage is either 50 or 60 hertz and the rpm of the rotor is related to the supply frequency and the number of poles in the generator okay so one uh does not really have too much freedom in selecting the rpm cannot just be any arbitrary number so the known quantities during the decision making on the available head the speed of the rotor and the power that is expected to uh produce uh basically we know the flow rate so we probably know the power that it is expected to produce at the size of the rotor and the type of the rotor are both not known so that is what needs to be selected okay both the type and the size of the rotor okay so what is normally done is using the known quantities we evaluate the uh the uh power specific speed and uh then uh we go to a chart like this which lists the different uh types of turbines and associated uh power specific speed and the peak efficiency for that we can expect for each case okay so based on this chart or based on this table we select the type of turbine we want a number of jets for a pelton wheel or low speed francis turbine uh medium speed whatever we select the type of turbine type of turbine from this and then we go to the dimensionless characteristics of that particular turbine okay so once we have the dimensionless characteristics of the particular turbine we select the maximum efficiency point and the corresponding uh uh quantity from the dimensionless characteristic which could be for instance cp so uh once we know the value for cp we can actually evaluate all the other quantities for instance once you know the value for cp we go back to the definition of cp uh now in this particular case once we know the value for cp from the turbine characteristic corresponding to the maximum efficiency point uh we know bhp we know density we know omega only d remains to be calculated so we can calculate that using this okay so uh once we have the dimensionless characteristics uh the specific speed may be used to select the type of turbine and the dimensionless characteristics may be used to actually size the rotor so we select the type of turbine then we size the rotor using the dimensionless characteristic okay so the size of the rotor may be evaluated for different types of turbines from their characteristics so for a given application a turbine of appropriate shape and size may then be selected so using the known quantities we calculate nsp then we use a table like this to select the type of turbine we then go to the dimensionless characteristics of this turbine and select the value for say cp from the point of uh corresponding to the point of maximum efficiency and then evaluate the random diameter once the random diameter is known and the speed is known uh and the angular uh speed is known the rpm is known the blade speed may then be calculated the jet speed is then twice the blade speed for optimum power in the case of a belt and turbine and we can just go on like this okay although we have stated this in very simple terms here selection turbine selection is actually much more detailed and involved and is usually discussed in in greater depth in full-fledged course on turbo machines so we will not be doing that here another aid that we have for selecting turbines given for example the head and the q would be uh would be a chart like this uh notice that you know this start broadly summarizes some of the points that we mentioned earlier that pelton turbines are used for high heads and low flow rates and francis turbines are used for medium head and medium flow rates and capitol turbines are used for low heads and high flow rate okay so we see that in in this chart also that the specific speed of pelton wheel uh is small because it has a high head and the specific speed of francis turbine uh lies in the middle mid-range and the specific speed of the capitol turbine uh is high because the head is low and this chart also helps in determining or selecting the type of turbine number of jets and so on for a particular installation where h and q are generally known so this completes our discussion of hydraulic turbine