Solved Example Problem - Bernoulli's Equations

Okay, today I'm gonna solve a question illustrating how to use Bernoulli's equation along with some other conservation principles that we deem necessary. Okay, so here's a question reads: water flows through a horizontal tee as shown in the figure below. Okay, inlet velocity and pressure are five meter per second and 200 kilopascal, respectively. Okay, determine the pressure at the exit section of the tee marked as two if the pressure at exit section three is 100 kilopascals. Okay, and I also supplied you the diameter information in the question already in the figure. Okay, when I start to solve a question like this, I always look at what I know, what I don't know. Okay, let's look at the areas. Okay, area one is known. Area two is known. Area three is known as well, so I'm good to go from the area perspective. The second thing that I do assess is to look at whether I know every velocity. So I know my velocity one. Hmm. I do not know velocity two. I do not know velocity three. So I'm gonna actually take a note myself that V2 is question, V3 is question mark as well. Okay, fine, let's continue. The next thing I do is look at the pressures. Pressure one is known. Pressure two is unknown. I don't see it in here. Pressure three is known. So I'm gonna still go back over here and I'm gonna ask myself, what is P2? Actually, when you look at the question, this is what it is testing you on. Okay, this is what being asked to do. The first thing is to follow the steps. Okay, so basically I just showed you how to do the step one, which is to read the problem and comprehend what it is asking you to do. The second is to draw your control volume. So I'll actually proceed from that direction. Okay, so as you can see in here, I'm including all the liquid. It says water in the question statement, but I'm not including any solid or the pipes. Okay, so that's the second step. The third step is to look at the assumptions or special cases that we can apply to this particular question. So, well, let's take a look. Assumptions slash special cases. Okay, is this steady? Yes. Okay, I don't see any time dependence. Is this constant density? I would say so because it's water. And the third one is, is this uniform flow? Yes. I'm only given velocity in one section anyways, and I wasn't given any velocity distribution, so I will go ahead and say that this is uniform flow. Okay, good. So, so far so good. So with this, I will now move on to the next step, which is the most important step in my opinion, and that is to decide which basic principles am I gonna use. Okay, and then I'm gonna revisit the special cases and assumptions sections depending on what equations I need to do. Okay, now as the pressure is being asked me, the only equation that I can reasonably find the pressure is the Bernoulli's equation. Okay, so let me see whether I can do Bernoulli’s equation. I'll draw a streamline like this, one to two. Right, so let's find out what I know. The Bernoulli's equation will have velocities. Okay, velocity one is known. Here's the issue. Velocity two is not known, so I'm not gonna be able to simply use the Bernoulli's equation between one and two and find my P2 because V2 is unknown. Interesting. Okay, let's go back. Let's go back to the conservation of mass then. Okay, do I need to use conservation of mass? The answer is yes. I do not know two of the velocities. So let's start with that and let's see where it takes us. Okay, let's call this conservation of mass. And with the three assumptions that we has a we have established, what I will obtain is inlets, basically over the inlets, V inlet A inlet will be equal to over the exits, V exit A exit. Okay, I have one inlet. V1 A1 will be equal to two exits: V2 A2 plus V3 A3. And then I'm gonna plug in what I know from here. So what I will get is V1 is 5, that's given in the question statement. A1 is Pi over 4, diameter is 1 square will be equal to V2, which is actually an unknown, times D2. Let's take a look what it is: 0.8. Pi over 4, 0.8 square, right, plus V3, which I don't know as well. So this is not going well. Pi over 4. Area 3 is 0.3 square. These are diameters, so I'm dividing one Pi over 4. Obviously, pi over 4 is cancelled and they're multiplied by each. And from here I will get: 5 is equal to 0.64 V2 plus 0.09 V3. Okay, so this is nice and all, but I can't still find my V2 and V3 values. Right, I still have to do something about it. I'm not even being asked to find V2. My ultimate goal is to remember to find my P2 value. Okay, then what am I gonna do? I have V2 and V3 missing. Hmm. I can't use my Bernoulli from here to here because then I'm gonna get an issue of finding P2, which I don't know. So I cannot use the force equation. The reason being that if I use the force equation, that will be the reaction forces in the x-direction, assuming x is horizontal, y is vertical, and you will see that I will have some reaction forces over here. So then I cannot use the conservation of momentum either. So what am I left with over here? Well, let's take a look here. Why don't I take a streamline that goes from 1 to 3? Okay, 1 to 3, and then apply my Bernoulli's equation. If I do that, note that both my pressures are given, so that's fine. Diameters or area are not used in the Bernoulli's equation. V1 is known, and this says that this is a horizontal tee, so the z is same. So that will cancel out. So then, oh okay, I will be able to find myself V3. So that's nice. So that's what I need to do. Okay? But one thing we need to be careful about when using the Bernoulli's equation is that I have to be consistent in my assumptions. Right, one thing that we need to be careful is Bernoulli's equation is only applicable for inviscid, no loss, no work, and density constant we already have it. And I need to draw—that's not an assumption or special case—but I have to draw in my figure my streamline to illustrate what am I doing. Okay, so let's do that. I think that would be sufficient for my assumption section. If I need to, I can always revisit that. Right, okay. So let me go ahead and draw a streamline arbitrarily like this because I don't know any better. Right, so I'm gonna call this basically 1, 2, 3. Right, and I'm gonna use my Bernoulli's equation 1 to 3. So if I do that, I will get myself P1 over rho plus V1 square over 2 plus g z1 will be equal to P3 over rho plus V3 square over 2 plus g z3. Okay, now if I go back to the question statement itself, what it says is this is a horizontal tee. What it means by horizontal tee is that I don't have to worry about these gravitational changes or the z changes, right, the elevation changes. You can, if you want to, to be consistent, you can add it to the assumption section, but you don't have to because that's not an assumption—that's given to me in the question statement. Okay, so I simply get rid of that. So then I will go ahead and insert what I know to this equation. So P1 in the question statement was given to me as 200 kilopascal. P3 is 100 kilopascal. V1 is 5. Okay, let's insert those. Okay, here is what it says. 200,000. Please do not forget to convert kilopascal to pascal. I see this often, okay. And this is gonna be 1000 kilogram per meter cube, which is the density of water. That will be supplied to you. Plus V1 square, which is 5 square divided by 2, will be equal to P3 is 100,000. 100 kilopascals. Then divided by the density of water, 1000 kilogram per meter cube, plus V3 square over 2. Okay, then if I simply go ahead and cancel this, I cancel this. So you can see over here, I'm gonna get 200 plus 12.5, you know, 25 divided by 2, will be equal to 100 plus V3 square over 2. Okay, so then if I move 100 to the other side, I will get myself V3 square over 2 is equal to 112.5. Right, so from here I get V3 square is equal to 225. Actually, I do get myself a nice number. Right, V3 becomes 15 meter per second, because 15 times 15 is 225. Right, okay, good. So I got myself a nice V3 number. Now what I can do is I can simply take this and plug it into here, right, because that's the V3, and I will be able to obtain my V2 from here. So let’s do that. Here's what it says. 5 will be equal to 0.64 times V2 plus 0.09 times 15, right? And if I go ahead and do that, I will get my V2 as 5.7 meter per second. Okay, so I got myself. So now, I know my V2 and V3. So now I am actually feeling much better about myself. The reason is that if I look over here, I can write my Bernoulli's equation between 1 and 2 now. I will have my V1, I will have my V2, which I did find; 5.7. I know P1, and the only unknown will be P2. So from here I will be able to obtain my P2 value. So let's just do it. Let's not forget and include our streamline, and that would be from 1 to 2. Okay, so I will write my Bernoulli's equation for that particular streamline. Okay, and what you do notice is that this process will be a replication of what I've done, so I have been experienced. So I should go ahead and find an easy answer. P1 over rho plus V1 square over 2 plus g z1 will be equal to P2 over rho plus V2 square over 2 plus g z2. As we discussed in the previous Bernoulli's equation discussion, these zs, the elevation changes are negligible because the tee is located in a horizontal plane. Okay, P1 was given to me as 200,000 divided by density, which is the 1000, right, kilogram per meter cube, plus V1 square, 5 square divided by 2, will be equal to P2, which is what I'm being asked to find, 1000 plus V2, which is 5.7 square, divided by 2. Okay, so from here, you will see this will become 200 plus 12.5. So this becomes 212.5, and this is P2 over 1000 plus 16.25. Right, and if I do this calculation, I will get my P2 as 196250 approximately, okay, pascals, but I usually want to express this in terms of kilopascals. It's gonna be 196.25 kilopascals. Okay, I would like to answer the questions that I sometimes receive. Can I use Bernoulli’s between 2 and 3? Okay, now look at this; 2 and 3, they are not on the same streamline, right, because that both are exits. Right, but that's physical explanation. Then I look at the mathematics. I'll give you a hint. If I use my Bernoulli's 1 to 3, basically this parenthesis at 1 is equal to parenthesis at 3, and then I say that 1 is equal to 2. If 1 is equal to 3, then 1 is equal to 2, then mathematically speaking, 2 must be equal to 3. So from the mathematical standpoint, yes, you can do it. It doesn't help me in this particular case, okay, because I wasn't given enough information for section 3 and 2, but it's something that you may wanna rely on going forward in other questions.