 Consider steady incompressible flow through the arbitrary device shown below. Determine the velocity at the outlet, which is state 3. So I have cross-sectional areas at all three state points. I was told an average velocity at state 1 and 2. I know that those are average velocities because they have the horizontal bar above the V. And it's assumed that I'm looking for an average velocity at state 3 because I don't have enough information to actually determine a velocity profile. But I guess I will list that as an assumption. Assumption's problem is looking for average velocity at state 3. Which I'll write as V3 is average velocity at state 3. Okay. And I want to solve through this with two methods, just to bring you guys around to the way that we're thinking about problems in chapter 3. So first I will work through it like we would have in thermal 1 with a mass balance and then we will apply our Reynolds transport theorem for the conservation of mass and simplify until we end up with the same answer. Okay, so first of all, in our mass balance, we recognize that we are performing a mass balance on the control volume defined here with our dashed lines. And I can say the change in mass of our control volume is equal to the entering mass minus the exiting mass. And then because I have steady state analysis, because I was told it's steady, I am going to divide everything by dt, at which point I have dm dt is equal to m dot in minus m dot out. And then I can say dm dt is zero because for steady flow, nothing can change with respect to time. Therefore, m dot in is equal to m dot out. I have two state points with an entering mass that states 1 and 2 and I have one state point with an exiting mass. So I will write this as m dot 1 plus m dot 2 is equal to m dot 3. Cool. Now how do I write m dot in terms of stuff that I know? Well, m dot we would traditionally write as density times volumetric flow rate and volumetric flow rate would be written as cross-sectional area times average velocity for our purposes. So we would have written m dot 1 as density 1 times cross-sectional area at state 1 times average velocity 1. We would have written m dot 2 as density 2 times cross-sectional area 2 times velocity on average at state 2 and m dot 3 as density 3 times cross-sectional area 3 times average velocity 3. Then next I note that my horizontal line didn't come across. Then I note that I have incompressible flow. Incompressible flow means that my density is assumed to be constant. I don't have to list that as an assumption because it was part of the problem statement, but if the density is the same at all three state points then dividing my density would yield area times average velocity at state 1 plus area times average velocity at state 2 is equal to area at state 3 times average velocity at state 3. Now we jump to that equation a lot in thermal 1, but it's important to note that that comes out as a result of the conservation of mass simplified for incompressible flow. There is a such thing as the law of conservation of mass. There is not a such thing as the law of conservation of volumetric flow rate despite the fact that we're saying volumetric flow rate at 1 plus volumetric flow rate at 2 is equal to the volumetric flow rate at state 3. Anyway, I'm looking for the average velocity at state 3, so I'll write this as cross-sectional area at state 1 divided by the cross-sectional area at state 3 times the average velocity at state 1 plus cross-sectional area at state 2 divided by the cross-sectional area at state 3 times the average velocity at state 2. Now I know all of those quantities. I have 0.05, 0.01, and 0.06 as the cross-sectional areas for states 1, 2, and 3 respectively, and then I know the average velocities at state 1 and state 2. So if I pop up my calculator here, scroll up so that I can read it, I am going to say the area at state 1 which was 0.05 divided by the area at state 3, 0.06 times the velocity on average at state 1, which is 4 plus the cross-sectional area at state 2, which is 0.01 divided by the cross-sectional area at state 3, 0.06 times the velocity at state 2, which is 8. Therefore, my answer is 4.6 repeating meters per second. That's the average velocity at state 3. So that was the mass balance approach, which is what we would have done in thermal 1, which works for this specific problem because this specific problem is not unlike something that we would have considered in chapter 4 of thermal 1. However, for fluids, we have to have a little bit more of a well-rounded approach to solving the conservation of mass because we have to be able to account for more things changing. We can't just use a nice packaged, simplified form. We have to step back to what we know and simplify it for each problem in front of us. So for that, let's start a new analysis here and I'll scooch up my assumptions all the way to here. Actually, all the way to here just to give us some nice working space. And then I can jump over to our conservation of mass simplified for the Reynolds transport theorem. Remember that that means beta is 1 and b is m. So I am going to be taking this equation as simplified for a control volume form wherein dm dt is 0, but I will copy that over. So we have dm dt of our system is equal to d dt of the integral of our control volume and we are integrating density with respect to volume and then we are evaluating the integral across the control surface which remember is the surface of the entire control volume. So we're going to split that apart into the three orifices but that integral encompasses all of them for now and that's density times velocity vector times the area vector or rather with respect to the area vector. Okay, and then we are going to get rid of dm dt because this is a control volume. So dm dt of the system is 0. And then because I have steady state and nothing can change with respect to time including this entire quantity because remember that's d of that with respect to time. That is also 0. So this is because it's a control volume. This is because it's steady. Meaning I have 0 is equal to the integral across the entire control surface of density times velocity vector times area vector rather with respect to area vector. And then I recognize that I have incompressible flow so I'm going to bring out my density and then divide both sides by density where I have 0 on the left and now integral across the control surface of velocity vector with respect to area. And this control surface encompasses the entire thing so I will split it across all three orifices and I have the integral across area one that's the velocity vector at 1 da1 plus the integral across area 2 is the velocity vector at state 2 da2 plus the integral across area 3 the velocity vector at state 3 da3 and now I am going to simplify for uniform flow. Now that's not because I definitely have uniform flow it's just that by using the average velocity the average velocity is uniform across the entire area so it doesn't matter if v1 is actually higher in the center of a1 than it is toward the edges because average velocity is the same everywhere. So I am going to be splitting this by using the uniform flow and when I get rid of that vector when I'm using uniform flow I have to keep track of how the velocity vector and area vectors appear. So my rule of thumb is as follows when they are in the same direction the velocity and area vector then when I simplify my vector I'm left with a positive quantity when they are in opposite directions that vector simplifies to a negative quantity so it doesn't matter if it's an inlet or an outlet it really matters the direction of the vector is relative to one another furthermore remember that the area vector is always defined in the outward direction so when a1 and v1 are in opposite directions like at state 1 then I simplify my vector to a negative quantity and when they are in the same direction like over here at state 3 that simplifies to a positive quantity now I know what you are thinking well John why don't I just remember that the outward direction is always positive and inward direction is negative well I mean that's fine you can do that but the reason it is that way is because of the direction of the vectors it's a whole cross product thing but anyway so v1 and a1 are in opposite directions so that's a negative quantity and I'm taking that as the average velocity state 1 times the area state 1 plus v2 and a2 are in opposite directions so that's again a negative quantity and then v3 and a3 are in the same direction so that's finally a positive quantity so simplifying for uniform flow yields 0 is equal to negative v1, a1 plus negative v2, a2 plus v3, a3 so I could bring over v1, a1 and v2, a2 I have average velocity state 1 times area 1 plus average velocity state 2 times area 2 is equal to average velocity state 3 times area 3 solving for v3 yields average velocity at state 3 is equal to a1 over a3 times average velocity state 1 plus a2 over a3 times the average velocity at state 2 which is exactly the same as this equation neat, huh?