 Welcome back. So, we have discussed enough about open systems. We saw the conservation of mass, we saw the first law and we saw the second law. Then we saw how the equations look like if we assume steady state. And then finally, we took a few examples of heat transfer systems, work transfer systems, nozzles and a duct. The last three were assumed to be adiabatic, but we will solve problems now, which may or may not assume these to be adiabatic. So, let us get on to a reasonably simple problem right now. It involves a steam turbine and we will make many of the assumptions that we did initially when we discussed a turbine. After that we will solve more generic problems where we will not neglect some of the terms, so that you realize that not every time we need to neglect such terms. So, let us just write down the problem now. So, the problem can be read out as a steam turbine receives steam at a rate of 5400 kilogram per hour and develops power of 600 kilowatt. Neglecting heat losses determine the change in specific enthalpy of the steam flowing through the turbine if the entrance and exit velocities are negligible. The difference in heights between the inlet and exit is also negligible. So, this is a reasonably straightforward problem, something that we discussed when we were discussing turbines. So, how do we go about it? We just write down the first law for steady state. So, it is not mentioned it is a steady state problem, but one can assume that the turbine is operating at steady state. So, we just write down the first law for steady state which is as follows. So, the first law was written as q dot minus w dot s is equal to m dot multiplied by the difference in the specific enthalpy, the difference in the kinetic energy and the difference in the potential energy. Now, notice that all the assumptions mentioned in the problem are the same as what we assumed while we were discussing an adiabatic turbine and these are as follows. They have mentioned that you should neglect the heat losses. So, we will put this q dot term as 0. It is also mentioned that the entrance and exit velocities are negligible. So, not just the difference these velocities themselves are negligible. So, overall we say this term is a reasonably small term we put it as 0 and it is mentioned that the difference in heights is also negligible that means z e minus z i is also a small quantity. So, we put that equal to 0 and we come down to our simple equation which is minus w dot s is m dot h e minus h i or we can write w dot s by m dot is equal to h i minus h e now. So, what is w dot s? It is given in kilowatt it is 600 kilowatt m dot is given in kilogram per hour it is 5400 kilogram per hour which can be written in kilogram per second because those are our preferred units. And we notice that w dot by m dot is 600 kilowatt which I will write down as kilo joule per second upon 1.5 kilogram per second and we end up with 400 kilo joule per kg and this is the difference in specific enthalpies that is being asked. Have a look at this number what we will do is in another snippet we will take a look at commonly found values of differences in enthalpies in steam turbines and gas turbines and figure out how the velocities and the change in heights compare with these values and why it is reasonably ok to assume that the difference in kinetic energies and potential energies are negligible. Thank you.