Flow Characteristics; Non-uniform Flow

Let's do an example for the conservation of mass. As I wrote, the question statement reads: two pipes merge into a single pipe. So I have this pipe over here like that, and I have another pipe over here, and they just connected like this, okay? Before the flow is fully developed, so we are gonna cover this in module 11, the flow needs some time or some distance to fully develop, okay? Before that, as you will see, there will be some mixed flow conditions over here, okay? And the velocity profile is given to me. Now the question is asking me, what is V is? So my D1 is given as 1 ft. Velocity 1 is 2 feet per second. The velocity 2 is 3 feet per second. Diameter 2 is 0.8 feet. And over here, this is the tricky part. Let's be careful over here. So this entire diameter is 1.2, okay? And the central 0.4-foot diameter section has a velocity of 2V, and the remaining of that will have the velocity of V. So I want to actually go out and plot this so that we are on the same page. This section is 2V. The outside that I have here is V. So I'm looking this direction over here, okay? Do you see what it is? So 2V is at the central, and the diameter to here, this is 0.4, the central section, and the whole thing from here to all the way to here is 1.2 feet. I will establish steps after the module 6 because things are gonna get more complicated, and I would like you to clearly follow the steps. But it's never too early to start. The first step is to read the question. You'll see, you look at the question. You read the question. It's a dummy step for now because I'm reading the question, solving the question for you. What I expect you to do in step two is draw your control volume. My expectation is, as the name of the course suggests, this is a fluid mechanics course, or fluid systems course, so I'm not going to include solid mechanics into my analysis. So this is my control volume. The next step is, I would like you to write it very clear, assumptions section to me. So the first question that I have is: this steady? The answer is: Yes, it is steady. Why? Well, do you see any time dependence in the question? No. So that’s the reason. Okay, I don't see anything related to time. If I had over here, let's say that in section number 2, my V2 is 3 feet per second plus 0.1t, then it would be unsteady flow, okay? Second is: Is this constant density? Yeah, I can say that because this is water. Liquids, it's a good assumption to have constant density. For gases, you need to be careful. And is this uniform flow? So if you follow what the numbers recommend you, the answer is 1 section 1 and section 2, yes. Section 3, no. Okay, so I have to deal with this integral formula, which I don't really like. So what I'm gonna do is, this is a trick that I would like you to know. I'm gonna separate the section 3 into two separate parts. At the center of this, let's call right over here, I'm gonna call this is section 4. And the remaining of this, I'm gonna call this section number 5. My goal is to ensure that over here, at exit 4, I have a uniform velocity. Velocity is not changing. At section number 5, the velocity is also not changing. It's different from that 4, but it's not changing. So now I can go out and say that uniform flow I’m gonna, you know like 4, 1, 2, 4, 5, just to be clear. The next question is, what happens to my conservation of mass in this particular assumptions special case 3 that we established? So if I go ahead and write it, it's going to be conservation mass. Now let's go write this. So it's gonna be: Summation over the inlets, V inlet, and I may have called this Vni Ai will be equal to exits, V exit A exit, okay? You can call this ni and ne, doesn't really matter. So let's look at inlets. So the two and one, those are the inlets. So then this will become like this: V1 A1 plus V2 A2. So far so good. How about exits? Remember that I separated it into two over here, so it's going to be 4 and 5. So let's write this: V4 A4 plus V5 A5. Simply, I go and write my numbers in. So let's go over here. V1 is 2, I can see barely over there, right? A1 is, I think it was one, right? Yeah. Pi 1 square divided by 4. V2 is right over there, is 3. A2 is Pi 0.8 square divided by 4. V4, 4 is the central region, so that's 2V. What about A4? That is gonna be Pi 0.4 square over 4, right? So that part is easy. What about V5? So V5 is V, that's given to me. Okay, so 2V is for the section 4. V is for section 5. What about the area 5? It will be the entire area of the circle minus this interior one. So think of this as like a doughnut, right? This is the region that you'd eat, right? I’m gonna take this entire area minus this area. That will leave me what I need over here. Pi 1.2 square divided by 4 minus 0.4 square divided by 4. And I close the bracket. Let's this write this nicely. So, 2 times. First thing is, I will cancel the Pis. Do you see it? Every term is multiplied by Pi, so I don't want to rewrite this. So it's going to be: 2 1 square over 4 plus 3 0.8 square divided by 4 will be equal to 2V 0.4 square divided by 4 plus V times, okay, this is a bit tricky, 1.2 square divided by 4 minus 0.4 square divided by 4. This becomes to me: this is 0.25, right? How about this? This becomes 0.8 times 0.8 is 0.64 divided by 4 is 0.16. 0.4 times 0.4 divided by 4 is 0.04. How about here? So 1.2 square is 1.44, minus 0.16, 1.28. 1.28 divided by 4 is 0.32. Okay. Now let me rewrite this nicely. So this becomes 0.5, and this becomes what 0.48 will be equal to, this will become 2 times 0.04 is 0.08V plus 0.32V, right? So the left-hand side is 0.98. The right-hand side becomes 0.4V. Then, basically, if I do it over here, I’m going to multiply 0.98 by 2.5. The final answer is 2.45 feet per second. So this will be the final answer for my question.