 Welcome back. So, in the last snippet we saw the normal diagram that is shown for an open system representing turbines and compressors. Now we will look at how the open system is simplified for such devices in steady state. So, let me just write the steady state equation again here. So, this is the steady state equation that was written earlier and in such devices normally there is good insulation especially when the temperature is high, even otherwise when there is no insulation required for example, in some pumps we assume that the system is adiabatic. So, it is not a perfect assumption, but it is a pretty good assumption and hence what we do is we say this is nearly equal to 0. What we also say is that the change, the net change in the kinetic energy is not really significant, it is really the work transfer that is important and the change in enthalpy that is important and though the inlet and exit velocities are not really negligible we say that they are small in this compared to the work transfer and whenever they are not negligible one should of course, appropriately check that this is true or not. So, we will put this term as 0 that is nearly equal to 0 that is delta ke and it is a reasonably good assumption again to assume that the potential energy change is also nearly 0. There is not too much variation in the inlet and exit height, it is normally in open system such as boilers where there is a reasonable change in height. Of course, whenever there is a change in height in such system one should not neglect it, but it is a reasonably good assumption to assume that this is nearly equal to 0 and if I now rearrange the equation it just turns out to be w dot s is m dot h i minus h e. So, this is the simplified form of the open system steady state equation for a work transfer device. Of course, the mass transfer equation is just m dot i is m dot e I should not say the mass transfer equation the mass conservation equation is just m dot i is m dot e is m dot. We also notice that because we have assumed that q dot is nearly 0 even the steady state equation for the second law gets simplified and we will see that if the q dot term is 0 the equation just simplifies to m dot s e minus s i is equal to s dot p that is because the term involving q dot has gone to 0. Similarly, the entropy change as a function of time has is also 0 it is a steady state problem. And of course, there is the rider here that s dot p is greater than equal to 0, but this is the simplified form of the second law for open system. This implies that if the system is adiabatic then because the right hand side is a positive quantity s e necessarily must be greater than equal to s i. Of course, this is assuming q dot equal to 0. If q dot is not equal to 0 and if it is actually negative that is it leaves the system then one of course cannot come to this conclusion and s e can actually be less than s i, but those are very rare cases for example, if the turbine is losing lot of heat such a thing is possible, but for the devices that we consider for the analysis that we have s e is greater than s i for the world transfer devices that we are considering. So, what are the implications of these that we will consider in the next snippet. So, right now we end here. Thank you.