 When I'm simplifying the Reynolds transport theorem for the conservation of mass, I'm plugging in a beta value of 1 and a b value of m. That means my conservation of mass form is dm dt of the system is equal to the time rate of change of the integral of density with respect to volume, plus the integral across the control surface of density times the velocity vector with respect to area. Then we can start to simplify this depending on our circumstance. So if we were talking about a control volume, like we are in chapter 3, then dm dt of our system is 0. So I can say 0 is equal to the time rate of change of the integral of density with respect to volume, plus the integral across the control surface of density times velocity vector with respect to area. Furthermore, if I have incompressible flow where the density doesn't change, then I can write density out front of all my integrals by pulling it out and by dividing all three terms by density, I have 0 is equal to the time rate of change of the integral of the volume plus the integral of the control surface of velocity with respect to area. If I simplify my control volume because the integral of dv is just going to be volume, then I would have 0 is equal to the partial derivative of volume with respect to time plus the integral across the control surface of the velocity vector with respect to area. I can simplify this a little bit further if I have say a fixed volume and incompressible flow, so 0 would equal the integral across the control surface of velocity with respect to area, or if I have steady flow but incompressible, I can write 0 is equal to the integral across the control surface of density times velocity with respect to area. Just like in thermo, I'm calling my volumetric flow rate the velocity times the area, but here we have to keep track of velocity as it might change depending on where in the area you're looking. I mean, for example, the flow rate in the center is going to be higher than the flow rate toward the edges, so keeping it as an integral will account for that. Therefore, volumetric flow rate is equal to the integral of velocity with respect to area. Then that allows me to write the average velocity across an orifice is going to be the volumetric flow rate divided by area, which is 1 over area, times the integral of velocity with respect to area. Note the standard notation for fluids, volumetric flow rate, is q. That could mean that we could confuse between heat transfer and volumetric flow rate down the line, but we will try to differentiate, especially by writing specific heat transfer as a lowercase q to avoid that issue. If I have uniform flow at each section, where my velocity is the same everywhere across the orifice, the same in the center as it is toward the edges, I have one same velocity at the entire inlet or exit, then I can simplify the integral by writing density times velocity times area. That would be either a positive term or negative term depending on if it's an inlet or an outlet, which means that I can write my conservation of mass as zero is equal to the time rate of change of the integral of density with respect to volume plus the summation of density times area times velocity minus the summation of density times area times velocity. Note over here that this equation down here is the same mass balance we applied in thermal. It is the change in mass is equal to the mass entering minus the mass exiting. We are just writing it in terms of quantities that we are more likely to know in a fluid mechanics problem. Furthermore, we are allowing for the density to change in the event of a compressible flow and by starting at the Reynolds transport theorem and simplifying, we are less likely to fall into a trap of hidden assumptions behind equations that we've already written down.