 So, now we will analyze a few very simple engineering systems and these systems will be operating mostly in steady state and hence I will be using only the steady state form of the three laws that we have considered up till now for open system. So, what we do is that we will make a few assumptions that either there is negligible heat transfer or there is negligible change in kinetic or potential energy depending on the situation, we can later on always check whether these assumptions were correct and verify that whatever we had done was right. So, the first set of devices that we will consider are heat transfer devices. The primary objective here is to transfer heat from one medium to the other and what we get in these class are boilers, heaters, heat exchangers, condensers used in power plants, refrigeration systems etc. And we will be analyzing the steady state form of these as I mentioned whenever there is a need for a transient form we can always get back to the transient set of equation. So, let us see what we would write for such a system. Let me first draw a very simplified diagram, I will just draw a control volume and show what is important. So, we have a control volume here, there is an inlet, there is an exit, there is W dot S and there is Q dot. So, in such devices as boilers, W dot S is normally 0 and hence we will put this. So, there is really no work that is of interest to us in such devices, we are not extracting any work out of such devices. So, only work that is involved is the flow work involved in pushing in the mass and pushing out the mass which anyway is taken care of in the enthalpy expression. And hence we can write down the following expression for steady state. The mass m dot i is the same as mass m dot e is m dot and this is conservation of mass. Let me write down the first law on another slide, Q dot minus W dot is equal to m dot h e minus h i plus v e squared by 2 minus v i squared by 2 plus g z e minus z i and we said that W dot S is 0. What we assume now is that there is negligible change in the kinetic energy. So, even if mass is coming in at a certain rate, the kinetic energy of what is coming in is nearly the same as kinetic energy of what is going out and there is not really any much difference. This is not a very bad assumption and hence for our analysis, this suits us perfectly. Hence we write this as follows, delta e k nearly equal to 0. Similarly, we will write delta e p nearly equal to 0. This depends on the situation. In some situation there could be a change in the potential energy depending on the height of the inlet and exit, but in many cases, this is a reasonably good assumption and whenever it is not, we will revisit it and not put it equal to 0. What remains now you see is just q dot is m dot h e minus h i. It is a very simple form of the first law for heat transfer devices at steady state. What about the second law? It turns out that you cannot simplify it any further apart from assuming steady state. The second law remains the same as far as heat transfer devices go and I would not like to repeat writing it, it was already presented. That will remain as it is and whenever we need it, we will use it. Often you will find that in our analysis, the second law is not so much needed whenever we are analyzing heat transfer devices, but definitely the first law is really needed whenever we are analyzing such systems and we will come to it when we are doing those problems. Thank you.