 So we're now going to start with the first basic law, taking our system formulation and applying it to a control volume and we're going to begin with the conservation of mass. So if you recall the extensive property we said was capital N equals big M and our intensive property it's then mass per unit mass which is just one. So we can write out our time rate of change equation of our extensive property defined for a system and we know from conservation of mass mass can either be created or destroyed so we say that that's equal to zero and the mass of the system is then defined as an integral either over dm or through our volume of the system that's where we had rho dv. Okay so what we're going to do we're going to rewrite this and this is where we combine the system formulation with the control volume formulation that we spent so long coming up with the equation for. So here on the left we have dm by dt for a system and we said that that is equal to zero and then on the right hand side it's going to be equal to the control volume formulation. So we can rewrite this. So what that is that is conservation of mass expressed for a control volume and so you can see it's relatively easy to come up with our equations now that we have the relationship between the system and the control volume through Reynolds transport theorem. So what we're going to do we're going to take a look at this equation and a few notes about it. We have a vector here d.da and so we're going to make a few comments about that. First one is that the velocity vector is measured relative to the control surface. Things will deal with whether mass is flowing into or out of the control surface and so we have to be a little careful with that. And what we can say is that v.da or rho v.da is positive when the flow is out and we have the dot product here and so it makes sense in terms of the the math of the dot product but let's say this is our surface and we have our area vector dA being normal to the surface and let's say we have some velocity vector v here that is not parallel it's at a different direction. The angle between these two we specified as being alpha and if we express what the dot product is it's the magnitude of v multiplied by the magnitude of dA cos alpha and this is a case where alpha is less than 90 degrees or less than pi over 2. Another example of that is where you have flow coming in so again drawing our surface we have one vector and then we have our area vector on the other side that would be the expression of the dot product and here alpha is greater than 90 degrees and consequently cos of alpha greater than 90 degrees between 90 and 180 is going to be a negative value and so that negative will go in there and that is how we are able to get a negative when flow is into the control volume. So the area for our control surface always points out as we see here so that's the control surface and that's the control surface. The area is always pointing outward but it depends which way the velocity vector is going and with this convention we can ensure that we don't get mixed up. So that's how you handle the dot product term. Now we are going to look at special cases of our conservation of mass equation and we are going to take a look at the first one being one where we have constant density so let's take a look at some special cases. So the first one where density is a constant. Now we call this a special form of flow and we refer to that as being incompressible so what that implies is density does not change. Looking at the conservation of mass equation I can pull density out of the integral for both the control volume as well as the control surface. The next thing I can do I can look at this here once we pull the density out that is just the volume because it's the integral and the control volume itself does not change with time consequently that can be rewritten. Now the time rate of change of density times the volume we said it's incompressible so row is not changing or control volume is not changing so this term actually disappears and that is because the size of control volume is fixed. So with that for the case of incompressible flow what happens is our continuity equation or mass conservation reduces to the following form so this applies if we have incompressible flow and it could be either steady or unsteady where the velocity is changing as a function of time or the conditions are changing and so that is a special case of the mass conservation equation for incompressible. The next one we're going to look at is the special case of a steady flow and if you have steady flow what happens is anything that has a time derivative goes to zero and consequently with that the first term in the continuity equation let me write it out so that is our continuity equation the first term oops let me undo that the first term here that goes to zero because the time rate of change no longer exists and what we're left with then is this expression here and this applies for compressible or incompressible flows. The last special case that we'll look at is one where we have uniform flow over an area so that would be where the velocity vector does not change so it's a fixed amount over an area and looking at our integral that we have in the mass conservation equation so this is the one over the control surface and I'm just going to say over some generic area a n if the velocity is fixed what we then get is something like that and n then would denote it could be a number of different inlets or exits into our control volume so we don't need to integrate the density or sorry not the density the velocity across that area now if you do have density variation then you have to be a little careful with that but assuming that your density is constant and the velocity is uniform over you just can do that calculation directly and sometimes in fluid mechanics we'll refer to this as being plus or minus m dot and that is the mass flow rate and be careful when we say m dot that is not dm by dt the time derivative of mass m dot is not the same as the time derivative what it reflects is the mass flow rate so this would be in units of kilograms per second going across a boundary and so that's what that is referring to so those are some special cases of the continuity equation what we're going to do in the next segment is we're going to solve an example problem with the continuity equation applied to a control volume