Flow Characteristics; Compressible

Okay, let's solve a question. Air; careful, it's not water. Air flows in a nozzle with dimensions indicated in the figure below. The pressures and temperatures at the nozzle inlet and outlet are also indicated in the figure below. If the nozzle inlet velocity is 10 meters per second, what is the nozzle exit velocity? Okay, so this is a nozzle. You can see over here, the diameter over here is what, 0.5 meters, and then it gets half of it, 0.25, right over there, okay? The inlet velocity is 10 meters per second. Absolute pressure is 50 kilopascal at entrance, and T1 is equal to 250 Kelvin. And at the exit, I have my pressure 2 is 40 kilopascal, while the temperature slightly increases to 260 Kelvin. Okay, first thing is to draw the control volume, as always, after reading the problem, and we did accomplish that together. Okay, so this is my control volume. Let's write here assumptions or special cases. Most of the time, I call them assumptions because it's not given, you assume something. Okay, so is this steady? Well, you see this information is not supplied to me, and you need to go ahead and assume that. Why? If I don't see any time dependence anywhere. Okay, so for that reason, I'm gonna go ahead and say that, yep, this is steady. Is this constant density? Well, now this is gas, air, and I gave myself pressure and temperature. I'll actually talk about that. I can assume that the pressure and temperature change are fairly, you know, similar, and I'll talk about how much error you are gonna incorporate into your analysis if you assume that. But I'm gonna do the proper way first, where I don't assume the densities are the same at the inlet and exit, okay? Okay. Next is, is this uniform flow? Well, I wasn't given enough information to understand that, and I only given velocities, ten at the inlet, and question mark at the outlet. So yeah, I'm gonna assume uniform. With these two assumptions, I don't have a particular special case number. Okay, so what I'm gonna do is, I would like to look at the special case number three, which I had, and I'm gonna multiply that with the density. The reason is that it is missing the constant density assumption. And if I do that, you will find out that here's what it's gonna look.At the inlet, Rho inlet, V inlet, A inlet will be equal to summation over the exit, Rho, V exit, A exit. You can call this V ne as well. It's not gonna make a difference in the conservation of mass, but it is going to do the difference in the conservation of momentum. So we'll see. Okay, let's proceed. Do you see the difference over here? If I go ahead and cancel these two, as they are the same for some applications, I will get the special case number three, okay? All right, so this actually the mathematics seems simple. So the inlet is 1. Rho1 V1 A1 will be equal to Rho2 V2 and A2. My question is asking you about V2. So let's write it this way. Maybe carefully here, V2 will be equal to V1 times. So I'm gonna write like this, ratios over here. So V1 times, you can see over here, I'm gonna get Rho1 divided by Rho2, and then over here, I'm gonna get A1 over A2. So take a look. I did V1 A1 Rho1. So this is this term. I left V2 in here, and the remaining Rho2 and A2 is down there. So that's good. So far, so good. Okay, the next question is, let's see what we know and what we don't know. The area relationship is given to me. Actually, the diameter is given, diameter is given. So I'm good with that. This is not a problem for me. Velocity 1 is 10, that's not a problem for me. What about density relation? Yeah, let's hold on for a second right over there. This is not given to me. So now what do I do? What do I do? So, there a few approaches, and the approach that I'm gonna take is the ideal gas law. You may remember this from your physics slash thermodynamics courses. It's fairly common. So let's go ahead and write my ideal gas law over here. Ideal gas law. So it's gonna be P is equal to Rho R T. If you don't know why this is, you can actually use the one version before this, and this will be this. P V is equal to m R T. You know, when you move this to over here, right, think about it. What is m divided by volume? That will be my density. So I'm talking about the same thing over here, okay? So then I know my pressures. You see over here. Okay, so let's call this 1 1 1 1. And then, let’s call this P2 is equal to Rho2 R2, well, R is a constant number. So I don't have to differentiate between 1 and 2. Times T2. So that’s what I'm gonna do is, I'm gonna divide this side by side, okay? And what you will find out then is: Rho1 over Rho2 will be equal to, careful, P1 over P2 times T2 over T1. Let's assess this together. I don't think it's very clear. So, Rho1 is over here. As Rho1 is over here, what I do is, I want to move T1 to down here. Right? That's exactly where it is. Same logic. I want to have Rho2 at the denominator, so now I have to move T2 up here. Right over there. And P1 over P2. P1 over P2. What happened to the Rs? Gone, right? They're the same. So now I'm gonna pick this up, and I'm gonna insert it right over there. Well, let’s do it. So then, V2 will be equal to V1 times Rho1 over Rho2. So that's gonna be P1 over P2 times T2 over T1 times A1 over A2. Okay, so let's now. I think it's time to plug the numbers in. V 1 is 10. I remember that. P1 over P2. P1 was, let's double check this one. I go up. 50, 40, 250, 260. Okay, so it's gonna be 50 kilopascal divided by 40 kilopascals. Do I need to convert them to pascals? Yeah, if you want to be consistent, but you know it's gonna cancel each other, right? 50,000 divided by 40,000 is equal to 50 divided by 40, right? So I don't really need it. So T 2 is 260, T 1 is 250, and A 1 is pi D square. D was 0.5 at the 1. Over 4, this is pi over 4, 0.25 square. Alright, let's see what's happening in here. So, and also you can see, this is meter per second, right? And this is meter square over here. This is meter square over here. Okay, so in terms of the units, you can see meter squares cancel, kelvins cancel, kilopascals cancel. You get meter per second. That sounds like a plan. So it's gonna be 10. What is 50 divided by 40? That is 1.25, right? I will not be able to find this one. So let's proceed. Pi over 4 cancels. So I get 0.5. So that is 1, 2 squared, which is 4, right? Because this is number is double of this, and then I'm squaring it. It's 4. Okay, so let's do this. So 10 times 1.25 times 4. That is what? 40, 50. So 50 times 260 divided by 250. You see? 5. Okay, this is easy. 260, 2. 26 times 2 is 52 meter per second. So that's what I obtain as my V 2 over here. Okay, does this make sense that it is going up? Oh yeah, I'm in a nozzle. So now what I'm gonna do is, I'm gonna do kind of like a bad student, you know. And I will assume that, hey, you know, what I'm gonna disregard the whole thing. Life is too short. I'm gonna, you know, not worry about these things. I see how much error I have for incorrect assumption of constant density. So actually it's easy to find because I have the... let me go here. I have the equation here. So basically, or actually let's go one more step right here. So you're assuming that density 1 is equal to density 2. So you can see over here, this becomes V 2 is equal to V 1 times A 1 over A 2, right? So that's, if I go back, so then I will have V 1 A 1 is equal to V 2 A 2. V 1 is 10. Pi over 4, 0.5 square will be equal to V 2 times A 2, which is pi over 4, 0.25 square. Alrighty then. Pi over 4 gone. This becomes what? So this becomes 1. This becomes 4, right? It's twice of it. So then I got myself V 2 is equal to 40 meter per second. Okay, is this close? Can I simply not worry about this difference and put an engineering safety factor? I don't like that, but what is error that I have here? So okay, the real answer is 52. So this is 40. Then what I need to do is to find the error. I need to do, you know, 52 minus 40 or 40 minus 52. It doesn't matter. Divided by 52, is real answer. So what, this is 12. 12 by 52 is, let's say, 22 percent. Something like that. I don't have the calculator with me now. This is gonna be over 20 percent. So are you okay with more than 20 percent error? I feel I am not. But you know, you may be. But also note this. I was being nice. I was being nice. I made this question, and these numbers were close, right? Future changes fairly, pretty minimal, right? And how about the pressure? So you can see that ratio will determine the error that I incorporate. But over here, I have over 20 percent. I don't like that. You have to stick to the proper approach that we have established.