 Welcome back, so up till now we have discussed how the mass, the energy and the entropy of a control volume change with time and we had derived equation for the same. So though we will consider situations where the mass, energy and the entropy of the control volume do change with time, many of the problems that we will tackle at least for this course involved what is known as steady state which means that the states of the control volume do not change and the flow rates are constant. That means I can write that m dot i and m dot e are constant similarly we can also write that the heat transfer rate is not just a constant over delta t but over the whole length of the process. Similarly, W dot s is constant over the whole length of the process and we write this as following and we can write that the mass in the control volume does not change with time, the energy in the control volume does not change with time similarly the entropy does not change with time. So this what is known as steady state situation and most of our analysis regarding turbines, nozzles, compressors, heat exchangers, boilers and all other systems that we normally consider in mechanical engineering we will be considering a steady state analysis. There do occur situations where we will not encounter steady state which are known as transient situations of course those are going to be few as far as this course is concerned. So what happens when we have a steady state? You will realize that the left hand side of all the three equations have been put to 0 as far as the mass, energy and entropy were concerned. So what do we get once we get this? So we can write the mass conservation equation as follows because this is 0 it implies that the incoming mass flow rate is equal to the outgoing mass flow rate. So for the first law similarly we have DECV by DT is 0 what we will do is we will transpose Q dot and W dot S in the equation to the left side and write it as follows that is Q dot minus W dot S is equal to M dot notice that I am writing only M dot this is because the mass flow rate coming in is the same as mass flow rate going out and I could have just written this here as M dot. So I will not put a subscript i or e I will write it as M dot H e plus V e square by 2 plus G z e minus M dot H i plus V i square by 2 plus G z i and often one combines these terms and writes as follows. So we have just combined the terms and this you can see we could have as well written as delta H where delta H represents H at the exit minus H at the inlet. Similarly we could have written this as delta e k that is the change in the kinetic energy at the exit and inlet that is the difference between the kinetic energy at the exit and inlet and this one we can write as delta e p that is the change in the potential energy and we could have a very simple equation like this and depending on the situation some of these would be important some of these would not be important and finally for the second law we can write that the entropy change with time for the control volume is 0 and hence we can write the equation as follows. So we have just transposed M dot S e minus S i on the left side and we have used only one expression for M dot and this is the equation a very simple equation that we get once we have assumed that S of the control volume does not change with time. So these are the sets of equations we will be using often during our analysis now and please have a look at these equations again because we will use them reasonably for problem solving. Thank you.