 Myself, Professor S.M. Mudd from Walsh Institute of Technology, Swalapur from Mechanical Indian Department. Today, I am going to engage regarding the next lecture of the study flow energy equation. In the previous lecture, I am going to explain the basic thing regarding mass entering in that system and mass living from that system. So, the basic concept already I am explaining. So, in that basic concept, whatever the flow of mass entering in the system and mass living from that system, all the energy transactions with the physical and other properties are remain the same. So, this concept will be used for the different purposes in our daily life. So, now from today's lecture, I am going to engage regarding the water turbine that is a hydraulic turbine and the centrifugal pump. So, now I am going to explain these things. So, in the first diagram here that water is moving from this is this basic concept is water turbine is basically used in the hydroelectric power plant. The water is moving from higher side to the lower side. When the water is to be trapped at the higher side in the dam, when it allows from higher side to the lower side through this penstock pipe and that water is to be injected on this turbine blades and the water is to be get collides with that blade it impart its own motion. So, due to that motion its energy blade rotates. So, when the blades are rotated the shaft get rotated and then shaft get rotated you get a rotational mechanical energy and this rotational mechanical energy is transmitted to the generator. So, now when I have to apply the steady flow energy equation. So, what are the basic formulas and how to calculate the different forms of energy related to the hydraulic turbine and water turbine. Now, basically the general equation of the steady flow energy are whatever the inputs and output values are here the input values and output values the input values and the output values are mass is equal to m 1 and output value is m 2. So, here the output also called as the exit and the input is at the inlet then the pressure it is P 1 and at the exit level it is P 2 then velocity that is C 1, C 2 specific volume at the inlet V 1, V 2 height of the inlet from the datum that for the inlet you can note it as Z 1 and at the exit it is Z 2 then enthalpy at the exit you can note it as H 1 and at the exit H 2 and internal energy that is a U 1 and U 2. So, these are the values for the inlet and exit mass m 1 is m 2 pressure P 1 exit P 2 velocity C 1 at the inlet at the exit is C 2 specific volume V 1 at the exit it is V 2 height is Z 1 it is Z 2 enthalpy H 1 or the inlet noted as a exit for H 2 and internal energy U 1 and U 2. So, these are the cross section for inlet and this is a cross section for a exit. So, when the mass is entering in that system and the same mass is leaving from that system. So, here is a m 1 and here is a m 2. So, now you say is that m mass is equal to m 1 is equal to m 2. So, when you know all these values the general equation you can write for this one as. So, when that constant flow mass of that system steady flow mass of that system here you get the work noted as a shaft work or you have to supply the heat or rejects the heat both are vice versa. So, Q and work this is the first law of thermodynamics by the first law of thermodynamics says that Q is equal to U plus W where U is noted as a internal energy of that system. Now, here the kinetic energy is equal to half m c square potential energy is equal to m z z internal energy noted as a U enthalpy H is equal to flow of work is equal to U plus P V. So, these are the values these are the formulas we can use for the steady flow energy equation. So, when you know this all these basic I recall all these things that is general equations mass entering that system mass leaving from that system. Now, I have to apply for this for the this hydroelectric turbine this hydroelectric turbine when water entering in the system and water leaving from that system it imparts its motion you get the work output from that system and when the work output from that system you assume it is the positive work. Now, for this hydroelectric turbine or water turbine the general steady flow energy equation S F E E general steady flow equation is Q plus mass into bracket kinetic energy at the inlet plus potential energy at the inlet plus flow of work one at the inlet is equal to work done at the exit to mass m 1 here it is a mass m 2. So, kinetic energy at the exit plus potential energy at the exit plus flow of work at the exit. So, it is a general equation general steady flow energy equation. So, when this no general steady flow equation here in this case the some values are to be 0 or closer to the 0. So, before that first we have to put the values of kinetic energy potential energy and flow of work. So, therefore, the formula you can write as a Q is equal to m mass 1 into bracket kinetic energy is a V C 1 square by 2 potential energy is a G Z 1 plus flow of work you are called as a this value is called as a enthalpy H 1 is equal to work plus mass m 2 into bracket kinetic energy C 2 square by 2 plus potential energy G Z 2 plus enthalpy H 2. So, this is a general equation where we put the values and all these things now rearrange these values. So, Q minus W is equal to mass here in this case m 1 minus m 2 into bracket as one more thing as m 1 is equal to m 2. So, write down here mass m 1 into bracket change in kinetic energy that is C 2 square minus C 1 square by 2 plus potential energy G Z 2 minus Z 1 plus H 2 minus H 1 is a general equation. So, from this value Q minus W is equal to mass m 1 is equal to m 2 is equal to write as a mass m because mass ending in that system and mass leaving from the system is the same one. So, it is a C 2 square minus C 1 square by 2 and all these things. So, for this condition here the change in internal energy is 0 and velocity is equal to 0. So, it is enthalpy is you write as H is equal to flow of work where H is equal to U plus P V. So, put these values C 2 square minus C 1 square by 2 plus G into bracket Z 2 minus Z 1 plus instead of H 2 you can write U 2 plus P V 2 minus U 1 plus P V 1 it rearrange that mass into bracket C 2 square minus C 1 square by 2 G into bracket Z 2 minus Z 1 plus U 2 minus U 1 plus P 2 V 2 minus P 1 V 1 where Q minus W is equal to. So, for this turbine here Q is equal to 0 internal energy is equal to 0 and velocity is equal to 0. So, therefore work done Q is equal to 0 work done minus W m into bracket you can write C 2 square minus C 1 square by 2 plus G into bracket Z 2 minus Z 1 plus U 2 minus U 1 is equal to 0 plus P 2 V 2 minus P 1 V 1. So, these are the values when I go for a change of work for turbine. So, this is a general equation when I can use steady flow energy equation we can applied for turbines. Thank you.