 Welcome back. So, we will now discuss conservation of mass with the open system. So, I hope it is clear that we had a control volume and we considered an inlet plug. So, the mass that was in the inlet plug and the control volume is the same mass that occupies the control volume and the exit plug after time delta t. So, that part can be considered a closed system whose boundary has shifted at the inlet plug side and gone towards the exit on the exit plug side. So, let me just draw it again for you so that it becomes very clear what is the closed system that we are considering. So, let me draw another version of my open system control volume here. So, this was the control volume. At time t we had a certain mass occupying this inlet plug here as well as the control volume here. At time t plus delta t what has happened is that the situation has changed and I will just try to redraw this thing on the right here. Hopefully it looks similar. These are the boundaries of the control volume. What has happened is that the mass which was at the inlet plug has completely entered the CV. So, as far as the closed system is concerned the boundary has moved here and no mass has crossed this boundary. It is just that the boundary has moved here whereas this boundary has now moved out towards and it is occupying the entire exit plug here. So, the mass that was at time t here is at time t plus delta t occupying this exit plug plus the control volume and you will notice that there is no more inlet plug at this time. So, this is the same mass that occupied at the control volume and the inlet plug at time t. Now it is occupying the control volume and the exit plug at time t plus delta t. So, what we can do is have a few properties defined at time t and time t plus delta t and we will define the property separately for the control volume and separately for the closed system. So, for the control volume for example, we can have m c v which is the mass of the control volume and we can put in brackets t to tell that it is a function of time. Similarly, we could have other property t rho for the closed system we will say for the closed system we will say its mass is m system and we can say presently that it is a function of time t but of course, for the closed system the mass remains constant there is no mass flowing across its boundary. So, the mass at time t is the same as mass at time at time t plus delta t for the closed system. So, what I will do is now I will try to take this forward by considering that the mass of the closed system remains the same and figure out what is the mass of the control volume as we move from time t to time t plus delta t. So, let me just write down the equations now. So, I will write down m system which is the mass of the closed system at time t plus delta t is the same as mass of the closed system at time t. But what is the mass of the system at time t plus delta t? It is the mass in the control volume at time t plus delta t plus the mass in the exit plug that is what is there at time t plus delta t plus mass. So, let me write a small m here exit plug. This should be equal to m c v at time t plus mass in the inlet plug. So, we have now come down to a closed system we know mass does not cross its boundary it is a fixed mass system. But you realize that now we have got the mass of the control volume into the picture and then there is this mass of the two plugs. So, you will realize that the mass of the inlet plug and the exit plug need not be the same it depends on the velocities and densities at the inlet and exit. So, the mass in the control volume can actually change and it is reasonably obvious because these need not be the same. So, what do we do now? We have to now figure out what is an expression for finding out the mass at the exit plug. So, what we can do is we write m inlet plug will be just the density times the area times the velocity that is the flow rate coming in at the inlet side multiplied by delta t. So, whatever is at the inlet we had already said that we had a definite density a definite velocity that we could assign and we are going to use that and say over a small time delta t that remains constant and hence the mass that is in the inlet plug is nothing, but the mass that is contained in that plug which will just be rho A v at the inlet multiplied by delta t. So, one can consider it as follows rho i A i v i times delta t. Similarly, the mass at the exit plug can be written in terms of the quantities at the exit side. So, I will just write it as m exit plug rho e A e v e times delta t and hence we will have m C v t plus delta t plus rho e A e v e delta t is equal to m C v at time t plus rho i A i v i delta t. So, what we do now is that we divide the whole expression by delta t and bring the terms containing the control volume on one side and take the other terms on the other side. So, we write it as such m C v t plus delta t minus m C v t upon delta t is equal to rho i A i v i minus rho e A e v e. Now, what do we do? We say we will reduce delta t to an infinitesimal quantity which means the left hand side is nothing, but the time derivative of the mass in the control volume. So, I can just write the left hand side as such as delta t tends to 0. This is my expression for what is the mass changing in the control volume as a function of time. So, it depends on what is the inlet flow rate and what is the exit flow rate. So, you can realize that whatever is on the right hand side which is rho into A into V can be written as a mass inlet flow rate at the inlet and a mass exit flow rate at the exit and that is how we will write it now and we will make it very simple. So, you will realize that we have just considered a 1D situation at both inlet and exit. Assumed uniform velocities we know the area, we know the rho at both the inlet and exit and we have written it as such. If things varied and if there are multiple inlets and exits, you can just generalize it that you could just go ahead and integrate over all the inlets and exits. So, right now for the single inlet single exit system, we can write the following expression. We write m dot i is rho i A i V i and m dot e is rho e A e V e and thus we write D m C v by D t is nothing but m dot i minus m dot e and this is our basic form for conservation of mass. Notice we could have taken a differential approach as we have done in fluid mechanics and had an integral and a divergence term but this is very simple. We have just taken one inlet and one exit and assumed uniform flows at inlet and exit and just come up with an expression as to how the mass in the control volume will change by considering this inlet and exit plugs. Coming up with a closed system figuring out that the mass in the closed system cannot change with time and hence figuring out how the mass in the control volume will change as a function of time. So, it just depends on how much mass is coming in at the inlet and how much mass is going out at the exit. So, once we are sure about how to go about doing this, we can take up the first law as the next step. Thank you.