Flow Characteristics; Uniform vs Non-uniform Flow, Incompressible vs Compressible

So now, let's look at the special cases of the conservation of mass, okay? So I usually call the CO mass, okay? Some people call COM, and sometimes do that as well, but be careful. Conservation of momentum, that also can be COM. I will actually have three special cases, okay? The very first one is, I'm gonna call this a steady flow, okay? And I covered steady in module number four. There was a segment unsteady, steady 1D, 2D, 3D. I'm gonna repeat over here. But basically, if I have something as a function of time, now that is not, okay? So it is not changing over time. So now let me go back and show you my equation over here. So what do you think is is gonna happen to in here? As I mentioned, this is the time rate of change of mass over the control volume. Can I have this changing over time? The answer is no. I get myself del delt of one two three control volume Rho d volume will be equal to zero. And hence, from here, first one drops out. Look at it, first one drops out. So I'm all let fit this, so this needs to be equal to zero. So, okay, let's write it: Double integral over the control surface Rho V dotted with n dA is zero, okay? So, and also just a few minutes ago, what I did was I called this m dot inlet and m dot exit. Look at it. So I can simply call this whole thing as this as well: m dot exit minus m dot inlet is equal to zero. Or I can call this m dot exit is equal to m dot inlet. The second one is, I'm gonna still have steady plus constant density. What I wanna do here though is, before I go ahead, you may see in some books or some professor notes, where they are saying that this is incompressible. That is technically not correct, okay? I'll show you a difference. So incompressible means that I cannot really compress it, right? So if I have a system like this, right, if I have a density one over here and density two over here, I cannot compress this anymore. I cannot compress this anymore. So this is gonna be a constant value, and this gonna be constant value. But this incompressible doesn't mean or say that this density must be equal to this density, okay? I want you to be aware of that. The reason why I bring this up is, if you see incompressible, they mean to say they are constant density. That's quite all right. Okay, with the steady, I start with this equation, okay? And with density being constant. Look at, look at this equation. I'm saying that density is constant. Can I take that out of the integral? Oh yeah, I sure can. Let's do it then. So then it becomes: Density double integral over the CS V dotted with n dA is equal to zero this time. So this is a case where we have this as basically this is A times B is equal to zero. What does this mean? Either A is equal to zero, or B is equal to zero, or both are zero. A is equal to B equal to zero. Now the question is, does it make sense to have my density zero? So what that means is, I have an arbitrary space, and inside of it, I don't have any mass. I don't have any molecule. I don't have any particles. That is a definition of vacuum, right? In real life, we don't have that. So for that reason, density cannot be zero. The only opportunity that I have is this B is zero. What it means is, I look at it. So: Double integral over the control surface V n dA is equal to zero. Okay, so similar to this over here, where I call this m dot, right over here, you have another name. This is called the volumetric flow rate, okay? And it's abbreviated as Q. So the Q dot exit minus Q dot inlet will be equal to zero, just following the same. Right, look at this. m dot exit, now it's Q dot exit, okay? And then if I basically equate them, you can see I get myself: Q dot exit is equal to Q dot inlet. So that will be my final formula for the steady and constant density case. Let's write over here: This is volumetric flow rate. Sometimes in the books and all, I see people just simply refer to this as flow rate. That's not perfect, because there is also a mass flow rate, right? This is called, let's actually go ahead and write this over here as well: This is called mass flow rate. When people call it flow rate, they mean to say this, okay? They mean to say this, not the other one. And the third special case is, I'm still gonna have steady. I'm still gonna have constant density. And I will also have uniform flow. So the question here is, what is uniform flow? We didn’t talk about it. So let me draw my system again. So I have a nozzle. I keep giving the same example, but whatever. So I have an inlet over here, and I have an exit over there. I have my control volume that I selected. Okay, so what's going on over here is I have an inlet, I have an exit. But what does uniform flow mean? And I sometimes see this as a mistake. I had this very question to students when I teach in class. They say velocity is constant, velocity here is equal to velocity there. Okay, that's not quite right. Actually, when you'll find out that that's not even physically possible. Here is what it means. The velocity is a vector. The vectors are defined at a point, not an area, okay? So how am I gonna translate when I say the velocity in here is equal to, let's say, 2 meter per second, okay? But then I'm multiplying this by area, right? Take a look here, up here, V times A. So I'm multiplying a something that you find at a point with an area. I'm generalizing this velocity is everywhere over the inlet is 2 meter per second. Take a look at over here. So this will be, let's say, 3 meter per second. You will see why it needs to be larger soon. So it's like this, okay? And it is uniform. So you can see this is uniform, this is uniform. That both doesn't have to be uniform. I can have one uniform, and I can have one non-uniform. Now take a look over here. Can I take this V outside of the integral? Answer is yes. This a number. They are not the same, but I'm evaluating one by one. At the exit I call it positive V, I call it at the inlet negative V, right? And zero otherwise. So this 2, can I take 2 out of the integral in this particular case? Oh yeah. Can I take 3 out of the integral? Oh yeah. Okay, let's do it then. V dotted with n goes out of the integral, and I have double integral over the control surface dA. That is equal to zero. What is this? In order to explain that to the students, I do a tactic. Say this, what is dx? Because sometimes I don't hear positive responses. So this is x. We kind of know that already, right? Okay, so over here then area is like dx dy, all right? So what is this? When I ask, or you can do this, if this is independent of that, you can call it like that, right? Right. So this is x, this is y. What is x times y? Well, that's area, okay? We just called it the answer, this area. I'm gonna go over here and call this is my area. But remember, there needs to be a summation sign because I may have multiple inlets and multiple exits. If that's not the case, I have one inlet, one exit just like I'm drawing over here, you don't have to differentiate or sum. Summation over the, let's call this inlets, and I need to put negative sign in front of it. That will be V n inlet A inlet, plus over the exits, V n exit A exit will be equal to zero. And remember, I'm not doing anything new over here. This is Q dot exit. This is Q dot inlet. And if I rewrite this, you can write two way: Q dot exit is equal to Q dot inlet, because I can move this term to this side of the equation, okay? So that's option one. So this is option number one. And option number two is writing it like this: Summation over the inlet, V n inlet A inlet will be equal to summation over the exit, V n exit A exit, okay? So this is the second option as well. This is much more direct as opposed to this. Okay, are there only three special cases in real life? I don't know, who said that. How can I classify things into three categories only, right? So now, but before closing it off, I want to also highlight something called Vn. And why do I call n over here? Well, the origination is V dot n, right? Just to remind myself that the origin was V dot n, okay? And this is a different, this distinction, okay? And one other thing that I want to note before I close off the segment is, sometimes I see students making mistakes. You know this is scalar equation, like 1 is equal to 1, or 5 minus 5 is equal to 0. It's not like a vector where it has components in the x, y, and z, okay? So do not get mistaken over that. You're not gonna have sine or cosine of the velocity in the x, y, etc.