 Hello friends, I am Professor S.P. Mankani, Assistant Professor, Department of Mechanical Engineering, Wall Change Institute of Technology, SolarPore. Today we are going to discuss the solution to the Pelton wheel as a turbine. At the end of this session, students will be able to analyze velocity diagram and also students will be able to solve the numerical on Pelton wheel. So before going to the rest of the equation, we have to study the little bit about the layout of hydroelectric power plant. So here this diagram indicates the total layout of the hydraulic power plant. So this part is related with the water storage and this is the part related with the turbine part. And here the total, this is the gross head available in the dam. That is from the tail raise up to the head raise, we are going to call this as a gross head HG. The water flows through this particularly penstock. Penstock is made with the metal as well as this reinforcement. And here the water flowing through this particularly pipe has a frictional losses. And even there will be additionally the losses are also going to be occurring with this particularly the pipeline. So some of the losses are going to be neglected in this particular part. And the head loss due to the friction we are going to represent this as a HF. So that is H is equal to HG gross head minus HF. So this H is nothing but net head. So here this HF is nothing but FLV square upon twice GD. So we are going to take this as the FLV square upon twice GD as a head loss due to friction. So here the whatever the water is flowing through this penstock and exit is going to be taken place through this nozzle. And the whatever the energy available at this particular stage is not going to be transferred total energy to this buckets which is fixed on the runner. So here the energy, total energy available at this particularly nozzle is not going to be transferred to this particularly as the buckets. And here there will be loss of energy is going to be taken place and that loss we are going to be considering while the studying of efficiency of the turbine. And the energy transferred to this particular bucket it transferred with the runner and runner energy is going to be transferred to the shaft. And shaft is connected to the generator. So here the first one is hydraulic efficiency. So here the hydraulic efficiency are going to explain it as the whatever the energy available at the end of the nozzle and the whatever the energy is available at the exactly at the blades. So here this hydraulic efficiency so we are going to be considering it as energy at the outlet of the penstock at the outlet of the penstock. So this is the output sorry this is the input and the output you are going to be getting it as energy at bucket. So efficiency we are going to represent it as hydraulic efficiency. Hydraulic efficiency are going to write it as the output upon input. So this is equal to we are going to take it as the runner power divided by water power. Water power is nothing but the power available by the water at the outlet of the nozzle. So this is what we are going to write it as the hydraulic efficiency. So then we are going with the mechanical efficiency. This mechanical efficiency we are going to explain in the power available at the runner we are going to compare with the power available at the shaft. So output is coming at the shaft that is nothing but we are going to take it as a shaft work as output that is a shaft power divided by power given by the runner that we are going to be writing it as a runner power. So this ratio we are going to call it as a mechanical efficiency because the total energy available at the runner is not going to be transferred to the shaft there will be a mechanical loss that we are going to be representing it as a mechanical efficiency here then this is the volumetric efficiency. So volumetric efficiency we are going to explain in this way as the total whatever the water at outlet of the nozzle is not going to be utilized for the total water on the buckets of the runner. Some part of the water is going to be wasted that we are going to representing it as the volumetric efficiency. That is the total water hitting the bucket divided by the total water is coming out from the nozzle that is the volumetric efficiency. So then overall efficiency we are going to study it as the overall efficiency is nothing but the total energy are going to consider the in the turbine. So that is the whatever the energy available at the shaft that is the shaft power and whatever the energy given by the through the pen stock. So that we are going to be comparing it as the overall efficiency new O is equal to that is shaft power divided by water power. So this total the detail we are going to study it in the next slide. So here the hydraulic efficiency is runner power divided by water power. So this runner power is nothing but W into bracket VW1 U1 plus or minus VW2 U2 divided by G into 1000. This is a kilowatt we are going to write in terms of kilowatts. So this plus or minus sign we are going to be studying the detail in the numerical part of it and this work power. So work power we are going to take it sorry this is water power we are going to consider it as a WH divided by 1000 kilowatts. The comparison as RP upon WP that is a runner power divided by water power. Next is mechanical efficiency. Mechanical efficiency we are going to consider it as power at the shaft of the turbine divided by power delivered by the water to the runner. So this we are representing it as a SP upon RP. So then volumetric efficiency as volume of water actually striking the runner divided by volume of water supplied to the turbine. The next is overall efficiency new suffix O shaft power divided by water power shaft power is the actual output power from the shaft. So divided by water power this is the inlet this is the inlet power from water. So this is the equation we are going to take it as SP upon WP we are going to multiplying and dividing by RP that is a runner power. So this the adjustment of the equation we are going to make it as SP divided by RP into RP divided by WP. So this is related with the mechanical efficiency and this part is related with the hydraulic efficiency. The overall efficiency is equal to mechanical efficiency into hydraulic efficiency. So that we are going to write it as a overall efficiency. The water power is this W whatever you are taking it as so that nothing but a rho gq into h this h as it is divided by 1000 this is the weight of the water. So now we are going to discuss the velocity triangle. The water whatever hitting on the bucket it is going to split up into two parts because it is going to hit at the center of the bucket. The part of the water is going to be flowing through this way part of the water we are going to flowing through this way. We are representing this as a outlet as only one side this side also the same outlet is coming. So here the water is entering axially in this direction we are going to represent this as an inlet triangle and this as an outlet triangle. So here the important part as this is u1 as an inlet velocity vr1 is a relative velocity v1 is an inlet velocity and vw1 is a virle velocity. So this is at inlet side and the corresponding value is at the outlet side. The most of the important points to be remembered while you are going to solve the problems on the Pelton wheel. So v1 is equal to Cv under root of 2 gs this Cv indicating it as the value we are going to consider it as a 0.98 to 0.99. So here u is equal to phi under root of 2 gs this phi value we are going to consider it as around 0.43 to 0.48. So the important thing is this is purely trigonometrical part the comparison of this triangle as well as this triangle. So vr1 we are going to write it as vr1 is equal to v1 minus u1. So vr1 is nothing but this value. So this is total v1 minus this part is going to be getting the value as relative velocity at inlet this is vr1 is equal to v1 minus u1. So vw1 is equal to v1 this is the value we are going to consider it for the Pelton wheel that is the vw in the wheel velocity and v1 is the velocity at this v1. So this value we have to remember it as for the particular as a Pelton wheel. So the relation between the inlet triangle and outlet triangle is vr1 is equal to vr2 the relation for inlet triangle to outlet triangle while you are going to solve the unknown values from the outlet triangle we have to consider this as a vr1 is equal to vr2. So if you know this value then we can move to the outlet triangle. So here vw2 wheel velocity at outlet. So this value we are going to consider it as a vw2 is equal to vr2 cos phi this this line vr2 cos phi vr2 cos phi will give the line as vr2 cos of this particular minus this value. So that gives the value for vw2 which is the outlet wheel velocity. So then one more value we are going to consider it as a q is equal to pi by 4 d square v1. So this value as a this value as area and this is the velocity of a water at the inlet. So if you know these values we can calculate the value of discharge. So here we are considering it as a the discharge of single nozzle sometimes there will be a possibility of multiple nozzle we are going to use it. So that time we are having as a additional the calculation. So major information with a velocity triangle you are expected to remember these equations perfectly if you know these equations you can solve the value related with the outlets as well as the rest of the information. Thank you. So for a further information you are expected to refer a textbook of fluid mechanics and hydraulics machines by Dr. R. K. Bansal.